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  • Tensor product of Quantum States using Dirac's Bra-Ket Notation - 2018

     

    There has been increasing interest in the details of the Maple implementation of tensor products using Dirac's notation, developed during 2018. Tensor products of Hilbert spaces and related quantum states are relevant in a myriad of situations in quantum mechanics, and in particular regarding quantum information. Below is a presentation up-to-date of the design and implementation, with input/output and examples, organized in four sections:

     

    • 

    The basic ideas and design implemented

    • 

    Tensor product notation and the hideketlabel option

    • 

    Entangled States and the Bell basis

    • 

    Entangled States, Operators and Projectors

     

    Part of this development is present in Maple 2018.2. To reproduce what you see below, however, you need a more recent version, as the one distributed within the Maplesoft Physics Updates (version 272 or higher).

     

    The basic ideas and design implemented

     

     

    Suppose A and B are quantum operators and Ket(A, n), et(B, m) are, respectively, their eigenkets. The following works since the introduction of the Physics package in Maple

    with(Physics)

    Setup(op = {A, B})

    `* Partial match of  '`*op*`' against keyword '`*quantumoperators*`' `

     

    _______________________________________________________

     

    [quantumoperators = {A, B}]

    (1.1)

    A*Ket(A, alpha) = A.Ket(A, alpha)

    Physics:-`*`(A, Physics:-Ket(A, alpha)) = alpha*Physics:-Ket(A, alpha)

    (1.2)

    B*Ket(B, beta) = B.Ket(B, beta)

    Physics:-`*`(B, Physics:-Ket(B, beta)) = beta*Physics:-Ket(B, beta)

    (1.3)

    In previous Maple releases, all quantum operators are supposed to act on the same Hillbert space. New: suppose that A and B act on different, disjointed, Hilbert spaces.

     

    1) To represent that situation, a new keyword in Setup , hilbertspaces, is introduced. With it you can indicate the quantum operators that act on a Hilbert space, say as in hilbertdspaces = {{A}, {B}} with the meaning that the operator A acts on one Hilbert space while B acts on another one.

     

    The Hilbert space thus has no particular name (as in 1, 2, 3 ...) and is instead identified by the operators that act on it. There can be one or more, and operators acting on one space can act on other spaces too. The disjointedspaces keyword is a synonym for hilbertspaces and hereafter all Hilbert spaces are assumed to be disjointed.

     

    NOTE: noncommutative quantum operators acting on disjointed spaces commute between themselves, so after setting - for instance - hilbertdspaces = {{A}, {B}}, automatically, A, B become quantum operators satisfying (see comment (ii) on page 156 of ref.[1])

     

    "[A,B][-]=0"

     

    2) Product of Kets and Bras that belong to different Hilbert spaces, are understood as tensor products satisfying (see footnote on page 154 of ref. [1]):

     

    `⊗`(Ket(A, alpha), Ket(B, beta)) = `⊗`(Ket(B, beta), Ket(A, alpha)) 

     

    `⊗`(Bra(A, alpha), Ket(B, beta)) = `⊗`(Ket(B, beta), Bra(A, alpha)) 

     

    while

    Bra(A, alpha)*Ket(A, alpha) <> Bra(A, alpha)*Ket(A, alpha)

     

    3) All the operators of one Hilbert space act transparently over operators, Bras and Kets of other Hilbert spaces. For example

     

    A*Ket(B, n) = A*Ket(B, n)

      

    and the same for the Dagger of this equation, that is

    Bra(B, n)*Dagger(A) = Bra(B, n)*Dagger(A)

     

      

    Hence, when we write the left-hand sides of the two equations above and press enter, they are automatically rewritten (returned) as the right-hand sides.

     

    4) Every other quantum operator, set as such using Setup , and not indicated as acting on any particular Hilbert space, is assumed to act on all spaces.

     

    5) Notation:

     

    • 

    Tensor products formed with operators, or Bras and Kets belonging to different Hilbert spaces (set as such using Setup  and the keyword hilbertspaces), are now displayed with the symbol 5 in between, as in Ket(A, n)*Ket(B, n) instead of Ket(A, n)*Ket(B, n), and `&otimes;`(A, B) instead of A*B. The product of an operator A of one space and a KetNULL of another space Ket(B, n) however, is displayed AA, without 5.

    • 

    A new Setup option hideketlabel , makes all the labels in Kets and Bras to be hidden at the time of displaying Kets, Bras and Bracket, so when you set it entering Setup(hideketlabel = true),

     "Ket(A,m,n,l"  

      

    is displayed as

    Ket(A, m, n, l)

     

      

    This is the notation frequently used when working with angular momentum or in quantum information, where tensor products of Hilbert spaces are used.

    Design details

       

    Tensor product notation and the hideketlabel option

     

     

    According to the design section, set now two disjointed Hilbert spaces with operators A, C acting on one of them and B, C on the other one (you can think of  C = `&otimes;`(A, B))

     

    Setup(hilbertspaces = {{A, C}, {B, C}})

    [disjointedspaces = {{A, C}, {B, C}}]

    (2.1)

     

    Consider a tensor product of Kets, each of which belongs to one of these different spaces, note the new notation using"&otimes;"

    Ket(A, 1)*Ket(B, 0)

    Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0))

    (2.2)
    • 

    As explained in the Details of the design section, the ordering of the Hilbert spaces in tensor products is now preserved: Bras (Kets) of the first space always appear before Bras (Kets) of the second space. For example, construct a projector into the state (2.2)

    Physics[`*`](Physics[Ket](A, 1), Physics[Ket](B, 0))*Dagger(Physics[`*`](Physics[Ket](A, 1), Physics[Ket](B, 0)))

    Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))

    (2.3)

    You see that in the product of Bras, and also in the product of Kets, A comes first, then B.


    Remark: some textbooks prefer a diadic style for sorting the operands in products of Bras and Kets that belong to different spaces, for example, `&otimes;`(Ket(A, 1)*Bra(A, 1), `&otimes;`(Ket(B, 0), Bra(B, 0))) instead of the projector sorting style of  (2.3). Both reorderings of Kets and Bras are mathematically equal.

     

    • 

    Because that ordering is preserved, one can now hide the label of Bras and Kets without ambiguity, as it is usual in textbooks (e.g. in Quantum Information). For that purpose use the new keyword option hideketlabel

    Setup(hide = true)

    `* Partial match of  '`*hide*`' against keyword '`*hideketlabel*`' `

     

    _______________________________________________________

     

    [hideketlabel = true]

    (2.4)

    The display for (2.3) is now

    Physics[`*`](Physics[Ket](A, 1), Physics[Ket](B, 0), Physics[Bra](A, 1), Physics[Bra](B, 0))

    Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))

    (2.5)

    Important: this new option only hides the label while displaying the Bra or Ket. The label, however, is still there, both in the input and in the output. One can "see" what is behind this new display using show, that works the same way as it does in the context of   CompactDisplay . The actual contents being displayed in (2.5) is thus (2.3)

    show

    Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))

    (2.6)

    Operators of each of these spaces act on their eigenkets as usual. Here we distribute over both sides of an equation, using `*` on the left-hand side, to see the product uncomputed, and `.` on the right-hand side to see it computed:

    (`*` = `.`)(A, Ket(A, 1))

    Physics:-`*`(A, Physics:-Ket(A, 1)) = Physics:-Ket(A, 1)

    (2.7)

    (`*` = `.`)(A, Ket(A, 0))

    Physics:-`*`(A, Physics:-Ket(A, 0)) = 0

    (2.8)
    • 

    The tensor product of operators belonging to different Hilbert spaces is also displayed using 5

    A*B

    Physics:-`*`(A, B)

    (2.9)
    • 

     As mentioned in the preceding design section, using the commutativity between operators, Bras and Kets that belong to different Hilbert spaces, within a product, operators are placed contiguous to the Kets and Bras belonging to the space where the operator acts. For example, consider the delayed product represented using the start `*` operator

    'Physics[`*`](A, B)*Physics[`*`](Physics[Ket](A, 1), Physics[Ket](B, 0), Physics[Bra](A, 1), Physics[Bra](B, 0))'

    Physics:-`*`(A, B, Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))

    (2.10)

    Release the product

    %

    Physics:-`*`(A, Physics:-Ket(A, 1), B, Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))

    (2.11)

    The same operation but now using the dot product `.` operator. Start by delaying the operation

    'Physics[`*`](A, B).Physics[`*`](Physics[Ket](A, 1), Physics[Ket](B, 0), Physics[Bra](A, 1), Physics[Bra](B, 0))'

    Parse:-ConvertTo1D, "invalid input %1", Typesetting:-mprintslash([A*B.Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))], [A*B.Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0), Physics:-Bra(A, 1), Physics:-Bra(B, 0))])

    (2.12)

    Recalling that this product is mathematically the same as (2.11), and that

    B.Ket(B, 0)

    0

    (2.13)

    by releasing the delayed product (2.12) we have

    Typesetting[delayDotProduct](Physics[`*`](A, B), Physics[`*`](Ket(A, 1), Ket(B, 0), Bra(A, 1), Bra(B, 0)))

    0

    (2.14)

    Reset hideketlabel

    Setup(hideketlabel = false)

    [hideketlabel = false]

    (2.15)

    Implementation details

       

    Entangled States and the Bell basis

     

     

    With the introduction of disjointed Hilbert spaces in Maple it is possible to represent entangled quantum states in a simple way, basically as done with paper and pencil.

     

    Recalling the Hilbert spaces set at this point are,

    Setup(hilbert)

    `* Partial match of  '`*hilbert*`' against keyword '`*hilbertspaces*`' `

     

    _______________________________________________________

     

    [disjointedspaces = {{A, C}, {B, C}}]

    (3.1)

    where C acts on the tensor product of the spaces where A and B act. A state of C can then always be written as

    Ket(C, m, n) = Sum(Sum(M[j, p]*Ket(A, j)*Ket(B, p), j), p)

    Physics:-Ket(C, m, n) = Sum(Sum(M[j, p]*Physics:-`*`(Physics:-Ket(A, j), Physics:-Ket(B, p)), j), p)

    (3.2)

    where M[j, p] is a matrix of complex coefficients. Bra  states of C are formed as usual taking the Dagger

    Dagger(Ket(C, m, n) = Sum(Sum(M[j, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j), p))

    Physics:-Bra(C, m, n) = Sum(Sum(conjugate(M[j, p])*Physics:-`*`(Physics:-Bra(A, j), Physics:-Bra(B, p)), j), p)

    (3.3)

     

    • 

    By definition, all states Ket(C, alpha, beta) that can be written exactly as `&otimes;`(Ket(A, alpha), Ket(B, beta)), that is, the product of a arbitrary state of the subspace A and another of the subspace B, are product states, and all the other ones are entangled states. Entangelment is a property that is independent of the basis `&otimes;`(Ket(A, j), Ket(B, p))used in (3.2).

    The physical interpretation is the standard one: when the state of a system constituted by two subsystems A and B is represented by a product state, the properties of the subsystem A are well defined and all given by "Ket(A,alpha),"while those for the subsystem B by NULL. When the system is in an entangled state one typically cannot assign definite properties to the individual subsystems A or B, each subsystem has no independent reality.

    To determine whether a state Ket(C, alpha, beta) is or not entangled it then suffices to check the rank R of the matrix M[j, p] (see LinearAlgebra:-Rank ): when R = 1 the state is a product state, otherwise it is an entangled state. When the state being analized belongs to the tensor product of two subspaces, R = 1.is equivalent to having the determinant of M[j, p] equal to 0. The condition R = 1, however, is more general, and suffices to determine whether a state is a product state also on a Hilbert space that is the tensor product of three or more subspaces: "`&Hscr;`^()=`&Hscr;`^((1))&otimes;`&Hscr;`^((2))&otimes;`&Hscr;`^((3))... `&Hscr;`^((n))", in which case the matrix M will have more rows and columns and a determinant equal to 0 would only warrant the possibility of factorizing one Ket.

     

    Example: the Bell basis for a system of two qubits

     

    Consider a 2-dimensional space of states acted upon by the operator A, and let B act upon another, disjointed, Hilbert space that is a replica of the Hilbert space on which A acts. Set the dimensions of A, B and C respectively equal to 2, 2 and 2x2 (see Setup)

    Setup(quantumbasisdimension = {A = 2, B = 2, C[1] = 2, C[2] = 2})

    [quantumbasisdimension = {A = 2, B = 2, C[1] = 2, C[2] = 2}]

    (3.4)

    The system C with the two subsystems A and B represents the a two qubits system. The standard basis for C can be constructed in a natural way from the basis of  Kets of A and B, {Ket(A, 0), Ket(A, 1), Ket(B, 0), Ket(B, 1)}, by taking their tensor products:

    seq(seq(Ket(A, j)*Ket(B, k), k = 0 .. 1), j = 0 .. 1)

    Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0)), Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1)), Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)), Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1))

    (3.5)

    Set a more mathematical display for the imaginary unit

    interface(imaginaryunit = i)

     

    The four entangled Bell states also form a basis of C and are given by

    Setup(op = `&Bscr;`)

    `* Partial match of  '`*op*`' against keyword '`*quantumoperators*`' `

     

    _______________________________________________________

     

    [quantumoperators = {`&Bscr;`, A, B, C, E}]

    (3.6)

    Ket(`&Bscr;`, 0) = (Ket(A, 0)*Ket(B, 0)+Ket(A, 1)*Ket(B, 1))/('sqrt')(2)

    Physics:-Ket(`&Bscr;`, 0) = (Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))/sqrt(2)

    (3.7)

    Ket(`&Bscr;`, 1) = (Ket(A, 0)*Ket(B, 1)+Ket(A, 1)*Ket(B, 0))/('sqrt')(2)

    Physics:-Ket(`&Bscr;`, 1) = (Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))/sqrt(2)

    (3.8)

    Ket(`&Bscr;`, 2) = i*(Ket(A, 0)*Ket(B, 1)-Ket(A, 1)*Ket(B, 0))/('sqrt')(2)

    Physics:-Ket(`&Bscr;`, 2) = I*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))/sqrt(2)

    (3.9)

    Ket(`&Bscr;`, 3) = (Ket(A, 0)*Ket(B, 0)-Ket(A, 1)*Ket(B, 1))/('sqrt')(2)

    Physics:-Ket(`&Bscr;`, 3) = (Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))/sqrt(2)

    (3.10)

    There is no standard notation for denoting a Bell state (the linar combinations of the right-hand sides above). The convention used here relates to the definition of the Bell states related to the Pauli matrices shown below. Regardless fo the convention used, this basis is orthonormal. That can be verified by taking dot products, for example:

    Dagger(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2)).(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))

    1 = 1

    (3.11)

    In steps, perform the same operation but using the star (`*`) operator, so that the contraction is represented but not performed

    Dagger(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))*(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))

    Physics:-`*`(Physics:-Bra(`&Bscr;`, 0), Physics:-Ket(`&Bscr;`, 0)) = (1/2)*Physics:-`*`(Physics:-`*`(Physics:-Bra(A, 0), Physics:-Bra(B, 0))+Physics:-`*`(Physics:-Bra(A, 1), Physics:-Bra(B, 1)), Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

    (3.12)

    Evaluate now the result at `*` = `.`, that is transforming the star product into a dot product

    eval(Physics[`*`](Bra(`&Bscr;`, 0), Ket(`&Bscr;`, 0)) = (1/2)*Physics[`*`](Physics[`*`](Bra(A, 0), Bra(B, 0))+Physics[`*`](Bra(A, 1), Bra(B, 1)), Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1))), `*` = `.`)

    1 = 1

    (3.13)

    Dagger(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))*(Ket(`&Bscr;`, 1) = (Physics[`*`](Ket(A, 0), Ket(B, 1))+Physics[`*`](Ket(A, 1), Ket(B, 0)))/sqrt(2))

    Physics:-`*`(Physics:-Bra(`&Bscr;`, 0), Physics:-Ket(`&Bscr;`, 1)) = (1/2)*Physics:-`*`(Physics:-`*`(Physics:-Bra(A, 0), Physics:-Bra(B, 0))+Physics:-`*`(Physics:-Bra(A, 1), Physics:-Bra(B, 1)), Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))

    (3.14)

    eval(Physics[`*`](Bra(`&Bscr;`, 0), Ket(`&Bscr;`, 1)) = (1/2)*Physics[`*`](Physics[`*`](Bra(A, 0), Bra(B, 0))+Physics[`*`](Bra(A, 1), Bra(B, 1)), Physics[`*`](Ket(A, 0), Ket(B, 1))+Physics[`*`](Ket(A, 1), Ket(B, 0))), `*` = `.`)

    0 = 0

    (3.15)

    The Bell basis and its relation with the Pauli matrices

     

    The Bell basis can be constructed departing from Ket(`&Bscr;`, 0) using the Pauli matrices sigma[j]. For that purpose, using a Vector representation for Ket(A, j),

    Physics:-Ket(`&Bscr;`, 0)

    (3.16)

    Ket(B, 0) = Vector([1, 0]), Ket(B, 1) = Vector([0, 1])

    Physics:-Ket(B, 0) = Vector[column](%id = 18446744078301209294), Physics:-Ket(B, 1) = Vector[column](%id = 18446744078301209414)

    (3.17)

    Multiplying Ket(B, 0)by each of the sigma[j] Pauli Matrices and performing the matrix operations we have

    "[seq(Psigma[j] . ?[1], j=1..3)]"

    [Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0)) = Physics:-Psigma[1].Vector[column](%id = 18446744078301209294), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 0)) = Physics:-Psigma[2].Vector[column](%id = 18446744078301209294), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 0)) = Physics:-Psigma[3].Vector[column](%id = 18446744078301209294)]

    (3.18)

    "map(u -> lhs(u) =Library:-PerformMatrixOperations(rhs(u)),?)"

    [Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0)) = Vector[column](%id = 18446744078376366918), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 0)) = Vector[column](%id = 18446744078376368838), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 0)) = Vector[column](%id = 18446744078376358606)]

    (3.19)

    In this result we see that sigma[1] and sigma[2] flip the state, transforming Ket(B, 0) into Ket(B, 1), sigma[2] also multiplies the state by the imaginary unit I, while sigma[3] leaves the state Ket(B, 0) unchanged.

    We can rewrite all that by removeing from (3.19) the Vector representations of (3.17). For that purpose, create a list of substitution equations, replacing the Vectors by the Kets

    "map(rhs = lhs,[?, i *~ ?])"

    [Vector[column](%id = 18446744078301209294) = Physics:-Ket(B, 0), Vector[column](%id = 18446744078301209414) = Physics:-Ket(B, 1), Vector[column](%id = 18446744078376351494) = I*Physics:-Ket(B, 0), Vector[column](%id = 18446744078376351734) = I*Physics:-Ket(B, 1)]

    (3.20)

    So the action of sigma[j] in Ket(B, 0) is given by

    "Library:-SubstituteMatrix(?,?)"

    [Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0)) = Physics:-Ket(B, 1), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 0)) = I*Physics:-Ket(B, 1), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 0)) = Physics:-Ket(B, 0)]

    (3.21)

    For Ket(B, 1), the same operations result in

    "[seq(Psigma[j] . ?[2], j=1..3)]"

    [Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1)) = Physics:-Psigma[1].Vector[column](%id = 18446744078301209414), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 1)) = Physics:-Psigma[2].Vector[column](%id = 18446744078301209414), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 1)) = Physics:-Psigma[3].Vector[column](%id = 18446744078301209414)]

    (3.22)

    "map(u -> lhs(u) =Library:-PerformMatrixOperations(rhs(u)),?)"

    [Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1)) = Vector[column](%id = 18446744078464860518), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 1)) = Vector[column](%id = 18446744078464862438), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 1)) = Vector[column](%id = 18446744078464856182)]

    (3.23)

    "Library:-SubstituteMatrix(?,?)"

    [Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1)) = Physics:-Ket(B, 0), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 1)) = -I*Physics:-Ket(B, 0), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 1)) = -Physics:-Ket(B, 1)]

    (3.24)

    To obtain the other three Bell states using the results (3.21) and (3.24), indicate to the system that the Pauli matrices operate in the subspace where B operates

    Setup(hilbert = {{B, C, Psigma}})

    `* Partial match of  '`*hilbert*`' against keyword '`*hilbertspaces*`' `

     

    _______________________________________________________

     

    [disjointedspaces = {{A, C}, {B, C, Physics:-Psigma}}]

    (3.25)

     

    Multiplying Ket(`&Bscr;`, 0) given in (3.7) by each of the three sigma[j] we get the other three Bell states

    Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2)

    Physics:-Ket(`&Bscr;`, 0) = (1/2)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

    (3.26)

    Psigma[1]*(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))

    Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics:-`*`(Physics:-Psigma[1], Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

    (3.27)

    Substitute in this result the first equations of (3.21) and (3.24)

    [Physics[`*`](Physics[Psigma][1], Ket(B, 0)) = Ket(B, 1), Physics[`*`](Physics[Psigma][2], Ket(B, 0)) = I*Ket(B, 1), Physics[`*`](Physics[Psigma][3], Ket(B, 0)) = Ket(B, 0)][1], [Physics[`*`](Physics[Psigma][1], Ket(B, 1)) = Ket(B, 0), Physics[`*`](Physics[Psigma][2], Ket(B, 1)) = -I*Ket(B, 0), Physics[`*`](Physics[Psigma][3], Ket(B, 1)) = -Ket(B, 1)][1]

    Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0)) = Physics:-Ket(B, 1), Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1)) = Physics:-Ket(B, 0)

    (3.28)

    map(rhs = lhs, [Physics[`*`](Physics[Psigma][1], Ket(B, 0)) = Ket(B, 1), Physics[`*`](Physics[Psigma][1], Ket(B, 1)) = Ket(B, 0)])

    [Physics:-Ket(B, 1) = Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0)), Physics:-Ket(B, 0) = Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1))]

    (3.29)

    subs([Ket(B, 1) = Physics[`*`](Physics[Psigma][1], Ket(B, 0)), Ket(B, 0) = Physics[`*`](Physics[Psigma][1], Ket(B, 1))], Physics[`*`](Physics[Psigma][1], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics[`*`](Physics[Psigma][1], Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1))))

    Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics:-`*`(Physics:-Psigma[1], Physics:-`*`(Physics:-Ket(A, 0), Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 1)))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(B, 0))))

    (3.30)

    factor(Simplify(Physics[`*`](Physics[Psigma][1], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics[`*`](Physics[Psigma][1], Physics[`*`](Ket(A, 0), Physics[`*`](Physics[Psigma][1], Ket(B, 1)))+Physics[`*`](Ket(A, 1), Physics[`*`](Physics[Psigma][1], Ket(B, 0))))))

    Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))

    (3.31)

    This is Ket(`&Bscr;`, 1) defined in (3.8)

    Ket(`&Bscr;`, 1) = (Physics[`*`](Ket(A, 0), Ket(B, 1))+Physics[`*`](Ket(A, 1), Ket(B, 0)))/sqrt(2)

    Physics:-Ket(`&Bscr;`, 1) = (1/2)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))

    (3.32)

    (Physics[`*`](Physics[Psigma][1], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 1))+Physics[`*`](Ket(A, 1), Ket(B, 0))))-(Ket(`&Bscr;`, 1) = (1/2)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 1))+Physics[`*`](Ket(A, 1), Ket(B, 0))))

    Physics:-`*`(Physics:-Psigma[1], Physics:-Ket(`&Bscr;`, 0))-Physics:-Ket(`&Bscr;`, 1) = 0

    (3.33)

    Multiplying now by sigma[2] and substituting Ket(B, j) using the 2^nd equations of (3.21) and (3.24) we get Ket(`&Bscr;`, 1)

    Psigma[2]*(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))

    Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics:-`*`(Physics:-Psigma[2], Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

    (3.34)

    [Physics[`*`](Physics[Psigma][1], Ket(B, 0)) = Ket(B, 1), Physics[`*`](Physics[Psigma][2], Ket(B, 0)) = I*Ket(B, 1), Physics[`*`](Physics[Psigma][3], Ket(B, 0)) = Ket(B, 0)][2], [Physics[`*`](Physics[Psigma][1], Ket(B, 1)) = Ket(B, 0), Physics[`*`](Physics[Psigma][2], Ket(B, 1)) = -I*Ket(B, 0), Physics[`*`](Physics[Psigma][3], Ket(B, 1)) = -Ket(B, 1)][2]

    Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 0)) = I*Physics:-Ket(B, 1), Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 1)) = -I*Physics:-Ket(B, 0)

    (3.35)

    zip(isolate, [Physics[`*`](Physics[Psigma][2], Ket(B, 0)) = I*Ket(B, 1), Physics[`*`](Physics[Psigma][2], Ket(B, 1)) = -I*Ket(B, 0)], [Ket(B, 1), Ket(B, 0)])

    [Physics:-Ket(B, 1) = -I*Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 0)), Physics:-Ket(B, 0) = I*Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(B, 1))]

    (3.36)

    factor(Simplify(subs([Ket(B, 1) = -I*Physics[`*`](Physics[Psigma][2], Ket(B, 0)), Ket(B, 0) = I*Physics[`*`](Physics[Psigma][2], Ket(B, 1))], Physics[`*`](Physics[Psigma][2], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics[`*`](Physics[Psigma][2], Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1))))))

    Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(`&Bscr;`, 0)) = ((1/2)*I)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))

    (3.37)

    The above is Ket(`&Bscr;`, 2) defined in (3.9)

    Ket(`&Bscr;`, 2) = I*(Physics[`*`](Ket(A, 0), Ket(B, 1))-Physics[`*`](Ket(A, 1), Ket(B, 0)))/sqrt(2)

    Physics:-Ket(`&Bscr;`, 2) = ((1/2)*I)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 1))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 0)))

    (3.38)

    Expand((Physics[`*`](Physics[Psigma][2], Ket(`&Bscr;`, 0)) = ((1/2)*I)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 1))-Physics[`*`](Ket(A, 1), Ket(B, 0))))-(Ket(`&Bscr;`, 2) = ((1/2)*I)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 1))-Physics[`*`](Ket(A, 1), Ket(B, 0)))))

    Physics:-`*`(Physics:-Psigma[2], Physics:-Ket(`&Bscr;`, 0))-Physics:-Ket(`&Bscr;`, 2) = 0

    (3.39)

    Finally, multiplying Ket(`&Bscr;`, 2) by sigma[3]

    Psigma[3]*(Ket(`&Bscr;`, 0) = (Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2))

    Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics:-`*`(Physics:-Psigma[3], Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))+Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

    (3.40)

    Substituting

    [Physics[`*`](Physics[Psigma][1], Ket(B, 0)) = Ket(B, 1), Physics[`*`](Physics[Psigma][2], Ket(B, 0)) = I*Ket(B, 1), Physics[`*`](Physics[Psigma][3], Ket(B, 0)) = Ket(B, 0)][3], [Physics[`*`](Physics[Psigma][1], Ket(B, 1)) = Ket(B, 0), Physics[`*`](Physics[Psigma][2], Ket(B, 1)) = -I*Ket(B, 0), Physics[`*`](Physics[Psigma][3], Ket(B, 1)) = -Ket(B, 1)][3]

    Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 0)) = Physics:-Ket(B, 0), Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 1)) = -Physics:-Ket(B, 1)

    (3.41)

    (rhs = lhs)((Physics[`*`](Physics[Psigma][3], Ket(B, 0)) = Ket(B, 0), Physics[`*`](Physics[Psigma][3], Ket(B, 1)) = -Ket(B, 1))[1]), (rhs = lhs)(-(Physics[`*`](Physics[Psigma][3], Ket(B, 0)) = Ket(B, 0), Physics[`*`](Physics[Psigma][3], Ket(B, 1)) = -Ket(B, 1))[2])

    Physics:-Ket(B, 0) = Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 0)), Physics:-Ket(B, 1) = -Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(B, 1))

    (3.42)

    We get ``

    factor(Simplify(subs(Ket(B, 0) = Physics[`*`](Physics[Psigma][3], Ket(B, 0)), Ket(B, 1) = -Physics[`*`](Physics[Psigma][3], Ket(B, 1)), Physics[`*`](Physics[Psigma][3], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*Physics[`*`](Physics[Psigma][3], Physics[`*`](Ket(A, 0), Ket(B, 0))+Physics[`*`](Ket(A, 1), Ket(B, 1))))))

    Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

    (3.43)

    which is Ket(`&Bscr;`, 2)

    Ket(`&Bscr;`, 3) = (Physics[`*`](Ket(A, 0), Ket(B, 0))-Physics[`*`](Ket(A, 1), Ket(B, 1)))/sqrt(2)

    Physics:-Ket(`&Bscr;`, 3) = (1/2)*2^(1/2)*(Physics:-`*`(Physics:-Ket(A, 0), Physics:-Ket(B, 0))-Physics:-`*`(Physics:-Ket(A, 1), Physics:-Ket(B, 1)))

    (3.44)

    Expand((Physics[`*`](Physics[Psigma][3], Ket(`&Bscr;`, 0)) = (1/2)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 0))-Physics[`*`](Ket(A, 1), Ket(B, 1))))-(Ket(`&Bscr;`, 3) = (1/2)*2^(1/2)*(Physics[`*`](Ket(A, 0), Ket(B, 0))-Physics[`*`](Ket(A, 1), Ket(B, 1)))))

    Physics:-`*`(Physics:-Psigma[3], Physics:-Ket(`&Bscr;`, 0))-Physics:-Ket(`&Bscr;`, 3) = 0

    (3.45)

    Entangled States, Operators and Projectors

     

     

    Consider a fourth operator, H, that is Hermitian and acts on the same space of C, and then it has the same dimension,

    Setup(additionally, hermitian = H, basisdimension = {H[1] = 2, H[2] = 2}, hilbertspaces = {{A, C, H}, {B, C, H}})

    `* Partial match of  '`*hermitian*`' against keyword '`*hermitianoperators*`' `

     

    `* Partial match of  '`*basisdimension*`' against keyword '`*quantumbasisdimension*`' `

     

    _______________________________________________________

     

    [disjointedspaces = {{A, C, H}, {B, C, H}, {B, C, Physics:-Psigma}}, hermitianoperators = {H}, quantumbasisdimension = {A = 2, B = 2, C[1] = 2, C[2] = 2, H[1] = 2, H[2] = 2}]

    (4.1)

    To operate in a practical way with these operators, Bras and Kets, however, bracket rules reflecting their relationship are necessary. From the definition of C as acting on the tensor product of  spaces where A and B act (see (3.2)) and taking into account the dimensions specified for A, B and C we have

    Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Ket(A, j)*Ket(B, p), j = 0 .. 1), p = 0 .. 1)

    Physics:-Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics:-`*`(Physics:-Ket(A, j), Physics:-Ket(B, p)), j = 0 .. 1), p = 0 .. 1)

    (4.2)

    Bra(A, k).(Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j = 0 .. 1), p = 0 .. 1))

    Physics:-Bracket(Physics:-Bra(A, k), Physics:-Ket(C, a, b)) = Sum(M[a, k, b, p]*Physics:-Ket(B, p), p = 0 .. 1)

    (4.3)

    Bra(B, k).(Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j = 0 .. 1), p = 0 .. 1))

    Physics:-Bracket(Physics:-Bra(B, k), Physics:-Ket(C, a, b)) = Sum(M[a, j, b, k]*Physics:-Ket(A, j), j = 0 .. 1)

    (4.4)

    Bra(A, k).Bra(B, l).(Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j = 0 .. 1), p = 0 .. 1))

    Physics:-`*`(Physics:-Bra(A, k), Physics:-Bracket(Physics:-Bra(B, l), Physics:-Ket(C, a, b))) = M[a, k, b, l]

    (4.5)

    The bracket rules for A, B and C are the first two of these; Set these rules, so that the system can take them into account

    Setup(Bracket(Bra(A, k), Ket(C, a, b)) = Sum(M[a, k, b, p]*Ket(B, p), p = 0 .. 1), Bracket(Bra(B, k), Ket(C, a, b)) = Sum(M[a, j, b, k]*Ket(A, j), j = 0 .. 1))

    [bracketrules = {%Bracket(%Bra(A, k), %Ket(C, a, b)) = Sum(M[a, k, b, p]*Physics:-Ket(B, p), p = 0 .. 1), %Bracket(%Bra(B, k), %Ket(C, a, b)) = Sum(M[a, j, b, k]*Physics:-Ket(A, j), j = 0 .. 1)}]

    (4.6)

    If we now recompute (4.5), the left-hand side is also computed

    Bracket(C, i, j, H, C, k, l) = `&Hscr;`

    Physics:-Bracket(Physics:-Bra(C, I, j), H, Physics:-Ket(C, k, l)) = `&Hscr;`

    (4.7)

    Bra(A, k).Bra(B, l).(Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j = 0 .. 1), p = 0 .. 1))

    M[a, k, b, l] = M[a, k, b, l]

    (4.8)

    Suppose now that you want to compute with the Hermitian operator H, that operates on the same space as C, both using C using the operators A and B, as in

     

    Bracket(Bra(C, I, j), H, Ket(C, k, l)) = `&Hscr;`[i, j, k, l]

     

    `&otimes;`(Bra(A, I), Bra(B, j))*H*`&otimes;`(Ket(A, k), Ket(B, l)) = H[I, j, k, l]

     

    where `&Hscr;`[i, j, k, l] = H[I, j, k, l] when Ket(C, a, b) is a product (not entagled) state.

     

    For Bracket(Bra(C, I, j), H, Ket(C, k, l)) = `&Hscr;`[I, j, k, l] it suffices to set a bracket rule

    Setup(%Bracket(Bra(C, a, b), H, Ket(C, c, d)) = `&Hscr;`[a, b, c, d], real = `&Hscr;`)

    `* Partial match of  '`*real*`' against keyword '`*realobjects*`' `

     

    _______________________________________________________

     

    [bracketrules = {%Bracket(%Bra(A, k), %Ket(C, a, b)) = Sum(M[a, k, b, p]*Physics:-Ket(B, p), p = 0 .. 1), %Bracket(%Bra(B, k), %Ket(C, a, b)) = Sum(M[a, j, b, k]*Physics:-Ket(A, j), j = 0 .. 1), %Bracket(%Bra(C, a, b), H, %Ket(C, c, d)) = `&Hscr;`[a, b, c, d]}, realobjects = {`&Hscr;`, x, y, z}]

    (4.9)

    After that, the system operates taking the rule into account

    Bra(C, j, k).H.Ket(C, m, n)

    `&Hscr;`[j, k, m, n]

    (4.10)

    Regarding `&otimes;`(Bra(A, I), Bra(B, j))*H*`&otimes;`(Ket(A, k), Ket(B, l)) = H[I, j, k, l]NULL, since H belongs to the tensor product of spaces A and B, it can be an entangled operator, one that you cannot represent just as a product of one operator acting on A times another one acting on B. A computational representation for the operator Bra(B, j)*H*Ket(A, k) (that is not just itself or as abstract) is not possible in the general case. For that you can use a different feature: define the action of the operator H on Kets of A and B.

     

    Basically, we want:

     

    "H*Ket(A,k)-> H[k]"

    "H[k]*Ket(B,l)->H[k,l]"

    A program sketch for that would be:


    if H is applied to a Ket of A or B then

        if H itself is indexed then
            return H accumulating its indices, followed by the index of the Ket
        else

            return H indexed by the index of the Ket;
    otherwise
        return the dot product operation uncomputed, unevaluated

     

    In Maple language, that program-sketch becomes

     

    "H := K ->   if K::Ket and op(1, K)::'identical(A,B)' then      if procname::'indexed' then         if nops(procname) <4 then             H[op(procname), op(2, K)]    #` accumulate indices`         else             'H . K'         fi     else          H[op(2, K)]     fi  else      'procname . K'  fi:"

     

    Let's see it in action. Start erasing the Physics performance remember tables, that remember results like  computed before the definition of H

     

    Library:-Forget()

    H.Ket(A, k)

    H[k]

    (4.11)

    Recalling that H is Hermitian,

    Bra(B, j).H

    H[j]

    (4.12)

    Bra(B, j).H.Ket(A, k)

    H[j, k]

    (4.13)

    Bra(B, j).H.Ket(A, k).Ket(B, l)

    H[j, k, l]

    (4.14)

    Bra(A, i).Bra(B, j).H.Ket(A, k).Ket(B, l)

    H[I, j, k, l]

    (4.15)

    Note that the definition of H as a procedure does not interfer with the setting of an bracket rule for it with Ket(C, a, b), that is still working

    Bra(C, i, j).H.Ket(C, k, l)

    `&Hscr;`[I, j, k, l]

    (4.16)

    but the definition takes precedence, so if in it you indicate what to do with a C Ket, that will be taken into account before the bracket rule. Finally, In the typical case, the first four results, (4.11), (4.12), (4.13) and (4.14) are operators while (4.15) is a scalar; you can always represent the scalar aspect by substituing the noncommutative operator H by a related scalar, say H.

     

    • 

    You can set the projectors for all these operators / spaces. For example,

    `&Iopf;__A` := Projector(Ket(A, i)); `&Iopf;__B` := Projector(Ket(B, i)); `&Iopf;__C` := Projector(Ket(C, a, b))

    Sum(Physics:-`*`(Physics:-Ket(A, n), Physics:-Bra(A, n)), n = 0 .. 1)

     

    Sum(Physics:-`*`(Physics:-Ket(B, n), Physics:-Bra(B, n)), n = 0 .. 1)

     

    Sum(Sum(Physics:-`*`(Physics:-Ket(C, a, b), Physics:-Bra(C, a, b)), a = 0 .. 1), b = 0 .. 1)

    (4.17)

    Since the algebra rules for computing with eigenkets of A, B and C were already set in (4.6), from the projectors above you can construct any subspace projector, for example

    Bra(A, m).`&Iopf;__C`

    Sum(Sum(Sum(M[a, m, b, p]*Physics:-`*`(Physics:-Ket(B, p), Physics:-Bra(C, a, b)), p = 0 .. 1), a = 0 .. 1), b = 0 .. 1)

    (4.18)

    `&Iopf;__C`.Ket(A, m)

    Sum(Sum(Sum(conjugate(M[a, m, b, p])*Physics:-`*`(Physics:-Ket(C, a, b), Physics:-Bra(B, p)), p = 0 .. 1), a = 0 .. 1), b = 0 .. 1)

    (4.19)

    The conjugate of M[a, m, b, p] is due to the contraction or attachment from the right of (4.18), that is with

    Dagger(Ket(C, a, b) = Sum(Sum(M[a, j, b, p]*Physics[`*`](Ket(A, j), Ket(B, p)), j = 0 .. 1), p = 0 .. 1))

    Physics:-Bra(C, a, b) = Sum(Sum(conjugate(M[a, j, b, p])*Physics:-`*`(Physics:-Bra(A, j), Physics:-Bra(B, p)), j = 0 .. 1), p = 0 .. 1)

    (4.20)

     

    The coefficients M[a, m, b, p] satisfy constraints due to the normalization of  Kets of A and B. One can derive these contraints by inserting the unit operator `#msub(mi("&Iopf;"),mi("C"))` constructing this identity

    Sum(Sum(Physics:-`*`(Physics:-Ket(C, a, b), Physics:-Bra(C, a, b)), a = 0 .. 1), b = 0 .. 1)

    (4.21)

    Bra(A, m).Bra(B, n).`&Iopf;__C`.Ket(A, r).Ket(B, s) = Bra(A, m).Bra(B, n).Ket(A, r).Ket(B, s)

    Sum(Sum(conjugate(M[a, r, b, s])*M[a, m, b, n], a = 0 .. 1), b = 0 .. 1) = Physics:-KroneckerDelta[m, r]*Physics:-KroneckerDelta[n, s]

    (4.22)

    Transform this result into a function P  to explore the identity further

    P := unapply(subs(Sum = sum, Sum(Sum(conjugate(M[a, r, b, s])*M[a, m, b, n], a = 0 .. 1), b = 0 .. 1) = Physics[KroneckerDelta][m, r]*Physics[KroneckerDelta][n, s]), m, n, r, s)

    proc (m, n, r, s) options operator, arrow; sum(sum(conjugate(M[a, r, b, s])*M[a, m, b, n], a = 0 .. 1), b = 0 .. 1) = Physics:-KroneckerDelta[m, r]*Physics:-KroneckerDelta[n, s] end proc

    (4.23)

    The first and third indices refer to the quantum numbers of A, the second and fourth to B, so the the right-hand sides in the following are respectively 1 and 0

    P(1, 0, 1, 0)

    conjugate(M[0, 1, 0, 0])*M[0, 1, 0, 0]+conjugate(M[1, 1, 0, 0])*M[1, 1, 0, 0]+conjugate(M[0, 1, 1, 0])*M[0, 1, 1, 0]+conjugate(M[1, 1, 1, 0])*M[1, 1, 1, 0] = 1

    (4.24)

    P(1, 0, 0, 0)

    conjugate(M[0, 0, 0, 0])*M[0, 1, 0, 0]+conjugate(M[1, 0, 0, 0])*M[1, 1, 0, 0]+conjugate(M[0, 0, 1, 0])*M[0, 1, 1, 0]+conjugate(M[1, 0, 1, 0])*M[1, 1, 1, 0] = 0

    (4.25)

    To get the whole system of equations satisfied by the coefficients M[a, m, b, n], use P to construct an Array with four indices running from 0..1

    Array(`$`(0 .. 1, 4), P)

    _rtable[18446744078376377150]

    (4.26)

    Convert the whole Array into a set of equations

    "simplify(convert(Typesetting:-msub(Typesetting:-mi("_rtable",italic = "true",mathvariant = "italic"),Typesetting:-mrow(Typesetting:-mn("18446744078376377150",mathvariant = "normal")),subscriptshift = "0"),setofequations))"

    {abs(M[0, 0, 0, 0])^2+abs(M[1, 0, 0, 0])^2+abs(M[0, 0, 1, 0])^2+abs(M[1, 0, 1, 0])^2 = 1, abs(M[0, 0, 0, 1])^2+abs(M[1, 0, 0, 1])^2+abs(M[0, 0, 1, 1])^2+abs(M[1, 0, 1, 1])^2 = 1, abs(M[0, 1, 0, 0])^2+abs(M[1, 1, 0, 0])^2+abs(M[0, 1, 1, 0])^2+abs(M[1, 1, 1, 0])^2 = 1, abs(M[0, 1, 0, 1])^2+abs(M[1, 1, 0, 1])^2+abs(M[0, 1, 1, 1])^2+abs(M[1, 1, 1, 1])^2 = 1, conjugate(M[0, 0, 0, 0])*M[0, 0, 0, 1]+conjugate(M[1, 0, 0, 0])*M[1, 0, 0, 1]+conjugate(M[0, 0, 1, 0])*M[0, 0, 1, 1]+conjugate(M[1, 0, 1, 0])*M[1, 0, 1, 1] = 0, conjugate(M[0, 0, 0, 0])*M[0, 1, 0, 0]+conjugate(M[1, 0, 0, 0])*M[1, 1, 0, 0]+conjugate(M[0, 0, 1, 0])*M[0, 1, 1, 0]+conjugate(M[1, 0, 1, 0])*M[1, 1, 1, 0] = 0, conjugate(M[0, 0, 0, 0])*M[0, 1, 0, 1]+conjugate(M[1, 0, 0, 0])*M[1, 1, 0, 1]+conjugate(M[0, 0, 1, 0])*M[0, 1, 1, 1]+conjugate(M[1, 0, 1, 0])*M[1, 1, 1, 1] = 0, conjugate(M[0, 0, 0, 1])*M[0, 0, 0, 0]+conjugate(M[1, 0, 0, 1])*M[1, 0, 0, 0]+conjugate(M[0, 0, 1, 1])*M[0, 0, 1, 0]+conjugate(M[1, 0, 1, 1])*M[1, 0, 1, 0] = 0, conjugate(M[0, 0, 0, 1])*M[0, 1, 0, 0]+conjugate(M[1, 0, 0, 1])*M[1, 1, 0, 0]+conjugate(M[0, 0, 1, 1])*M[0, 1, 1, 0]+conjugate(M[1, 0, 1, 1])*M[1, 1, 1, 0] = 0, conjugate(M[0, 0, 0, 1])*M[0, 1, 0, 1]+conjugate(M[1, 0, 0, 1])*M[1, 1, 0, 1]+conjugate(M[0, 0, 1, 1])*M[0, 1, 1, 1]+conjugate(M[1, 0, 1, 1])*M[1, 1, 1, 1] = 0, conjugate(M[0, 1, 0, 0])*M[0, 0, 0, 0]+conjugate(M[1, 1, 0, 0])*M[1, 0, 0, 0]+conjugate(M[0, 1, 1, 0])*M[0, 0, 1, 0]+conjugate(M[1, 1, 1, 0])*M[1, 0, 1, 0] = 0, conjugate(M[0, 1, 0, 0])*M[0, 0, 0, 1]+conjugate(M[1, 1, 0, 0])*M[1, 0, 0, 1]+conjugate(M[0, 1, 1, 0])*M[0, 0, 1, 1]+conjugate(M[1, 1, 1, 0])*M[1, 0, 1, 1] = 0, conjugate(M[0, 1, 0, 0])*M[0, 1, 0, 1]+conjugate(M[1, 1, 0, 0])*M[1, 1, 0, 1]+conjugate(M[0, 1, 1, 0])*M[0, 1, 1, 1]+conjugate(M[1, 1, 1, 0])*M[1, 1, 1, 1] = 0, conjugate(M[0, 1, 0, 1])*M[0, 0, 0, 0]+conjugate(M[1, 1, 0, 1])*M[1, 0, 0, 0]+conjugate(M[0, 1, 1, 1])*M[0, 0, 1, 0]+conjugate(M[1, 1, 1, 1])*M[1, 0, 1, 0] = 0, conjugate(M[0, 1, 0, 1])*M[0, 0, 0, 1]+conjugate(M[1, 1, 0, 1])*M[1, 0, 0, 1]+conjugate(M[0, 1, 1, 1])*M[0, 0, 1, 1]+conjugate(M[1, 1, 1, 1])*M[1, 0, 1, 1] = 0, conjugate(M[0, 1, 0, 1])*M[0, 1, 0, 0]+conjugate(M[1, 1, 0, 1])*M[1, 1, 0, 0]+conjugate(M[0, 1, 1, 1])*M[0, 1, 1, 0]+conjugate(M[1, 1, 1, 1])*M[1, 1, 1, 0] = 0}

    (4.27)

    Reference

       

    NULL


     

    Download Tensor_Products_of_Quantum_States_-_2018.mw

    Edgardo S. Cheb-Terrab
    Physics, Differential Equations and Mathematical Functions, Maplesoft

    It can be considered as continuation of these topics: “Determination of the angles of the manipulator with the help of its mathematical model. Inverse problem.”  and  “The use of manipulators as multi-axis CNC machines”.
    There is a simple two-link manipulator with three degrees of freedom. We change its last link with a fixed length to a variable length link. By this action we added one degree of freedom to our manipulator. Now we have a two-link manipulator with 4 degrees of freedom.
    In a particular mathematical model the variable length of the link is related to the x7, and in the figure the total length of the last link is displayed as the result of subtracting the fixed part of the link (L2) from x7 (equation f2). At the same time, the value of the variable x7 depends on the inclination of the first link (equation f4).  x1, x2, x3 are the coordinates of the moving point of the first link, x4, x5, x6 are the coordinates of the operating point.  The  equation f2 defines the type of connection between the links, the equations f5 and f6  are the trajectory of the working point.

    MAN_2_4_Variable_length_MP.mw

    Here is a classic puzzle:
    You are camping, and have an 8-liter bucket which is full of fresh water. You need to share this water fairly into exactly two portions (4 + 4 liters). But you only have two empty buckets: a 5-liter and a 3-liter. Divide the 8 liters in half in as short a time as possible.

    This is not an easy task and requires at least 7 transfusions to solve it. 

    The procedure  Pouring  solves a similar problem for the general case. Given n buckets of known volume and the amount of water in each bucket is known. Buckets can be partially filled, be full or be empty (of course the case when all is empty or all is full is excluded). With each individual transfusion, the bucket from which it is poured must be completely free, or the bucket into which it is poured must be completely filled. It is forbidden to pour water anywhere other than the indicated buckets.

    Formal parameters of the procedure: BucketVolumes  is a list of bucket volumes,  WaterVolumes  is a list of water volumes in these buckets. The procedure returns all possible states that can occur during transfusions in the form of a tree (the initial state  WaterVolumes  is its root).

    restart;
    Pouring:=proc(BucketVolumes::list(And(positive,{integer,float,fraction})),WaterVolumes::list(And(nonnegative,{integer,float,fraction})), Output:=graph)
    local S, W, n, N, OneStep, j, v, H, G;
    uses ListTools, GraphTheory;
    
    n:=nops(BucketVolumes); 
    if nops(WaterVolumes)<>n then error "The lists should be the same length" fi;
    if n<2 then error "Must have at least 2 buckets" fi;
    if not `or`(op(WaterVolumes>~0)) then error "There must be at least one non-empty bucket" fi;
    if BucketVolumes=WaterVolumes then error "At least one bucket should not be full" fi;
    if not `and`(seq(WaterVolumes[i]<=BucketVolumes[i], i=1..n)) then error "The amount of water in each bucket cannot exceed its volume" fi;
    W:=[[WaterVolumes]];
    
    OneStep:=proc(W)
    local w, k, i, v, V, k1, v0;
    global L;
    L:=convert(Flatten(W,1), set);
    k1:=0; 
    for w in W do
    k:=0; v:='v';
    for i from 1 to n do
    for j from 1 to n do
    if i<>j and w[-1][i]<>0 and w[-1][j]<BucketVolumes[j] then k:=k+1; v[k]:=subsop(i=`if`(w[-1][i]<=BucketVolumes[j]-w[-1][j],0,w[-1][i]-(BucketVolumes[j]-w[-1][j])),j=`if`(w[-1][i]<=BucketVolumes[j]-w[-1][j],w[-1][j]+w[-1][i],BucketVolumes[j]),w[-1]); fi;
    od; 
    od; 
    v:=convert(v,list);
    if `and`(seq(v0 in L, v0=v)) then k1:=k1+1; V[k1]:=w else 
    for v0 in v do  
    if not (v0 in L) then k1:=k1+1; V[k1]:=[op(w),v0] fi;
    od;
    fi;
    L:=L union convert(v,set);
    od;
    convert(V,list);
    end proc:
    
    S[0]:={};
    for N from 1 do
    H[N]:=(OneStep@@N)(W);
    S[N]:=L;
    if S[N-1]=S[N] then break fi;
    od;
    if Output=set then return L else
    if Output=trails then interface(rtablesize=infinity);
    return <H[N-1]> else
    G:=Graph(seq(Trail(map(t->t[2..-2],convert~(h,string))),h=H[N-1]));
    DrawGraph(G, style=tree, root=convert(WaterVolumes,string)[2..-2], stylesheet = [vertexcolor = "Yellow", vertexfont=[TIMES,20]], size=[800,500])  fi; fi;
    
    end proc: 
    

    Examples of use:

    Here is the solution to the original puzzle above. We see that at least 7 transfusions are  
    required to get equal volumes (4 + 4) in two buckets

    Pouring([8,5,3], [8,0,0]);
               

     

     With an increase in the number of buckets, the number of solutions is extremely 
     increased. Here is the solution to the problem: is it possible to equalize the amount of water (7+7+7+7) in the following example? 

    Pouring([14,10,9,9],[14,10,4,0]);
    S:=Pouring([14,10,9,9],[14,10,4,0], set);
    is([7,7,7,7] in S);
    nops(S);
             

     

     

    Download Pouring.mw

     

     

    Am pondering how to best provide user-configurable options for a few packages I've written. The easiest method is to use global variables, preassign their default values during the package definition (but don't protect them) and save them with the mla used for the package. A user could then assign new values, say in their Maple initialization file or in a worksheet.  For example

      FooDefaultBar := true:
      FooDefaultBaz := false:
    

    That works for a few variables, but is unwieldy if there are many, as the names generally have to be long and verbose to avoid accidental collision. Better may be to use a single record

      FooDefaults := Record('Bar' = true, 'Baz' = false):
    

    To change one or more values, the user could do

       use FooDefaults in
          Bar := false;
       end use:
    

    A drawback of using a global variable or record is that the user can assign any type to the variable, so the using program will have to check it. While one could use a record with typed fields, for example,

      FooDefaults := Record('Bar' :: truefalse = true, 'Baz' :: truefalse := false):
    

    that only has an effect on assignments if kernelopts(assertlevel) is 2, which isn't the default.

    A different approach is to use a Maple object to handle configuration variables. The object should be defined separate from the package it is configuring, so that the target package doesn't have to be loaded to customize its configuration. I've created a small object for this, but am not satisfied with its usage. Here is how it is currently used

    # Create configuration object for package foo
    Configure('fooDefaults', 'Bar' :: truefalse = true, 'Baz' :: truefalse = false):
    

    The Assign method is used to reassign one or more fields

    Assign(fooDefaults, 'Bar' = false, 'Baz' = true):
    

    If a value does not match the declared type, an error is raised. Values from the object are available via the index operator:

       fooDefaults['Bar'];
    

    Am not wild about this approach, the assignment seems clunky and would require a user to consult a help page to learn about the existence of the Assign method, though that would probably be necessary, regardless, to learn about the defaults themselves. Any thoughts on improvements? Attached is the current code.

    Configure := module()
    
    option object;
    
    local Default # record of values
        , Type    # record of types
        , nomen   # string corresponding to name of assigned object
        , all :: static := {}
        ;
    
    export
        ModuleApply :: static := proc()
            Object(Configure, _passed);
        end proc;
    
    export
        ModuleCopy :: static := proc(self :: Configure
                                     , proto :: Configure
                                     , nm :: name
                                     , defaults :: seq(name :: type = anything)
                                     , $
                                    )
        local eq;
            self:-Default := Record(defaults);
            self:-Type    := Record(seq(op([1,1], eq) = op([1,2], eq), eq = [defaults]));
            self:-nomen   := convert(nm,'`local`');
            nm := self;
            protect(nm);
            self:-all := {op(self:-all), self:-nomen};
            nm;
        end proc;
    
    export
        ModulePrint :: static := proc(self :: Configure)
        local default;
            if self:-Default :: 'record' then
                self:-nomen(seq(default = self:-Default[default]
                                , default = exports(self:-Default)
                               ));
            else
                self:-nomen();
            end if;
        end proc;
    
    export
        Assign :: static := proc(self :: Configure
                                 , eqs :: seq(name = anything)
                                 , $
                                )
        local eq, nm, val;
            # Check eqs
            for eq in [eqs] do
                (nm, val) := op(eq);
                if not assigned(self:-Default[nm]) then
                    error "%1 is not a default of %2", nm, self:-nomen;
                elif not val :: self:-Type[nm] then
                    error ("%1 must be of type %2, received %3"
                           , nm, self:-Type[nm], val);
                end if;
            end do;
            # Assign defaults
            for eq in [eqs] do
                (nm, val) := op(eq);
                self:-Default[nm] := val;
            end do;
            self;
        end proc;
    
    export
        `?[]` :: static := proc(self :: Configure
                                , indx :: list
                                , val :: list
                               )
        local opt;
            opt := op(indx);
            if not assigned(self:-Default[opt]) then
                error "'%0' is not an assigned field of this Configure object", indx[];
            elif nargs = 2 then
                self:-Default[opt];
            elif not val :: [self:-Type[opt]] then
                error "value for %1 must be of type %2", opt, self:-Type[opt];
            else
                self:-Default[opt] := op(val);
            end if;
        end proc;
    
    export
        ListAll :: static := proc(self :: Configure)
            self:-all;
        end proc;
    
    end module:
    

    Later: Observing that this is just a glorified record with an assurance that the values match their declared types, but with less nice methods to set and get the values, I concluded that what I really want is a record that enforces types regardless the setting of . Maybe created with

       FooDefaults := Record[strict]('Bar' :: truefalse = true, 'Baz :: truefalse = false):
    

    In the meantime, I'll probably just use a record and not worry about whether a user has assigned an invalid value.

    Hoping to get your very own copy of Maple for Christmas? 

    No, I'm not about to announce a sale. No crass commercialism allowed on MaplePrimes!  But we have created this handy-dandy letter for Santa that students can use to try to convince him to put Maple under the tree this year*.

     

     

    Happy holidays from Maplesoft!

     

    *Results not guaranteed, but we hope you will agree we made a valiant attempt. :-)

     

    On the convex part of the surface we place a curve (not necessarily flat, as in this case). We divide this curve into segments of equal length (in the text Ls [i]) and divide the path that our surface will roll (in the text L [i]) into segments of the same length as segments of curve. Take the next segment of the trajectory L [i] and the corresponding segment on the curve Ls [i], calculate the angles between them. After that, we perform well-known transformations that place the curve in the space so that the segment Ls [i] coincides with the segment L [i]. At the same time, we perform exactly the same transformations with the equation of surface.

    For example, the ellipsoid rolls on the oX1 axis, and each position of the ellipsoid in space corresponds to the equation in the figure.
    Rolling_of_surface.mw

     

    And similar examples:

    Exact solutions for PDE and Boundary / Initial Conditions

     

    Significant developments happened during 2018 in Maple's ability for the exact solving of PDE with Boundary / Initial conditions. This is work in collaboration with Katherina von Bülow. Part of these developments were mentioned in previous posts.  The project is still active but it's December, time to summarize.

     

    First of all thanks to all of you who provided feedback. The new functionality is described below, in 11 brief Sections, with 30 selected examples and a few comments. A worksheet with this contents is linked at the end of this post. Some of these improvements appeared first in 2018.1, then in 2018.2, but other ones are posterior. To reproduce the input/output below in Maple 2018.2.1, the latest Maplesoft Physics Updates (version 269 or higher) needs to be installed.

     

    1. PDE and BC problems solved using linear change of variables

     

    PDE and BC problems often require that the boundary and initial conditions be given at certain evaluation points (usually in which one of the variables is equal to zero). Using linear changes of variables, however, it is possible to change the evaluation points of BC, obtaining the solution for the new variables, and then changing back to the original variables. This is now automatically done by the pdsolve command.

     

    Example 1: A heat PDE & BC problem in a semi-infinite domain:

    pde__1 := diff(u(x, t), t) = (1/4)*(diff(u(x, t), x, x))

    iv__1 := u(-A, t) = 0, u(x, B) = 10

     

    Note the evaluation points A and B. The method typically described in textbooks requires the evaluation points to be A = 0, B = 0. The change of variables automatically used in this case is:

    transformation := {t = tau+B, x = xi-A, u(x, t) = upsilon(xi, tau)}

    {t = tau+B, x = xi-A, u(x, t) = upsilon(xi, tau)}

    (1)

    so that pdsolve's task becomes solving this other problem, now with the appropriate evaluation points

    PDEtools:-dchange(transformation, [pde__1, iv__1], {tau, upsilon, xi})

    [diff(upsilon(xi, tau), tau) = (1/4)*(diff(diff(upsilon(xi, tau), xi), xi)), upsilon(0, tau) = 0, upsilon(xi, 0) = 10]

    (2)

    and then changing the variables back to the original {x, t, u} and giving the solution. The process all in one go:

    `assuming`([pdsolve([pde__1, iv__1])], [abs(A) < x, abs(B) < t])

    u(x, t) = 10*erf((x+A)/(t-B)^(1/2))

    (3)

     

    Example 2: A heat PDE with a source and a piecewise initial condition

    pde__2 := diff(u(x, t), t)+1 = mu*(diff(u(x, t), x, x))

    iv__2 := u(x, 1) = piecewise(0 <= x, 0, x < 0, 1)

    `assuming`([pdsolve([pde__2, iv__2])], [0 < mu, 0 < t])

    u(x, t) = 3/2-(1/2)*erf((1/2)*x/(mu^(1/2)*(t-1)^(1/2)))-t

    (4)

     

    Example 3: A wave PDE & BC problem in a semi-infinite domain:

    pde__3 := diff(u(x, t), t, t) = diff(u(x, t), x, x)

    iv__3 := u(x, 1) = exp(-(x-6)^2)+exp(-(x+6)^2), (D[2](u))(x, 1) = 1/2

    `assuming`([pdsolve([pde__3, iv__3])], [0 < t])

    u(x, t) = (1/2)*exp(-(-x+t+5)^2)+(1/2)*exp(-(-x+t-7)^2)+(1/2)*exp(-(x+t-7)^2)+(1/2)*exp(-(x+t+5)^2)+(1/2)*t-1/2

    (5)

     

    Example 4: A wave PDE & BC problem in a semi-infinite domain:

    pde__4 := diff(u(x, t), t, t)-(1/4)*(diff(u(x, t), x, x)) = 0

    iv__4 := (D[1](u))(1, t) = 0, u(x, 0) = exp(-x^2), (D[2](u))(x, 0) = 0

    `assuming`([pdsolve([pde__4, iv__4])], [1 < x, 0 < t])

    u(x, t) = piecewise((1/2)*t < x-1, (1/2)*exp(-(1/4)*(t+2*x)^2)+(1/2)*exp(-(1/4)*(t-2*x)^2), x-1 < (1/2)*t, (1/2)*exp(-(1/4)*(t+2*x)^2)+(1/2)*exp(-(1/4)*(t-2*x+4)^2))

    (6)

     

    Example 5: A wave PDE with a source:

    pde__5 := diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x)) = f(x, t)

    iv__5 := u(x, 1) = g(x), (D[2](u))(x, 1) = h(x)

    pdsolve([pde__5, iv__5], u(x, t))

    u(x, t) = (1/2)*(Int(Int((diff(diff(h(zeta), zeta), zeta))*c^2*tau+(diff(diff(g(zeta), zeta), zeta))*c^2+f(zeta, tau+1), zeta = (-t+tau+1)*c+x .. x+c*(t-1-tau)), tau = 0 .. t-1)+(2*t-2)*c*h(x)+2*g(x)*c)/c

    (7)

    pdetest(u(x, t) = (1/2)*(Int(Int((diff(diff(h(zeta), zeta), zeta))*c^2*tau+(diff(diff(g(zeta), zeta), zeta))*c^2+f(zeta, tau+1), zeta = (-t+tau+1)*c+x .. x+c*(t-1-tau)), tau = 0 .. t-1)+(2*t-2)*c*h(x)+2*g(x)*c)/c, [pde__5, iv__5])

    [0, 0, 0]

    (8)

    2. It is now possible to specify or exclude method(s) for solving

     

    The pdsolve/BC solving methods can now be indicated, either to be used for solving, as in methods = [method__1, method__2, () .. ()] to be tried in the indicated order, or to be excluded, as in exclude = [method__1, method__2, () .. ()]. The methods and sub-methods available are organized in a table,
    `pdsolve/BC/methods`

    indices(`pdsolve/BC/methods`)

    [1], [2], [3], [2, "Series"], [2, "Heat"], ["high_order"], ["system"], [2, "Wave"], [2, "SpecializeArbitraryFunctions"]

    (9)


    So, for example, the methods for PDEs of first order and second order are, respectively,

    `pdsolve/BC/methods`[1]

    ["SpecializeArbitraryFunctions", "Fourier", "Laplace", "Generic", "PolynomialSolutions", "LinearDifferentialOperator"]

    (10)

    `pdsolve/BC/methods`[2]

    ["SpecializeArbitraryFunctions", "SpecializeArbitraryConstants", "Wave", "Heat", "Series", "Laplace", "Fourier", "Generic", "PolynomialSolutions", "LinearDifferentialOperator", "Superposition"]

    (11)

     

    Some methods have sub-methods (their existence is visible in (9)):

    `pdsolve/BC/methods`[2, "Series"]

    ["ThreeBCsincos", "FourBC", "ThreeBC", "ThreeBCPeriodic", "WithSourceTerm", "ThreeVariables"]

    (12)

    `pdsolve/BC/methods`[2, "Heat"]

    ["SemiInfiniteDomain", "WithSourceTerm"]

    (13)

     

    Example 6:

    pde__6 := diff(u(r, theta), r, r)+diff(u(r, theta), theta, theta) = 0

    iv__6 := u(2, theta) = 3*sin(2*theta)+1

    pdsolve([pde__6, iv__6])

    u(r, theta) = -_F2(-I*r+2*I+theta)+1-3*sin((2*I)*r-4*I-2*theta)+_F2(I*r-2*I+theta)

    (14)

    pdsolve([pde__6, iv__6], method = Fourier)

    u(r, theta) = ((3/2)*I)*exp(2*r-4-(2*I)*theta)-((3/2)*I)*exp(-2*r+4+(2*I)*theta)+1

    (15)

    Example 7:

    pde__7 := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

    iv__7 := u(x, 0) = Dirac(x)

    pdsolve([pde__7, iv__7])

    u(x, y) = Dirac(x)-(1/2)*Dirac(2, x)*y^2+_C3*y

    (16)

    pdsolve([pde__7, iv__7], method = Fourier)

    u(x, y) = invfourier(exp(-s*y), s, x)

    (17)

    convert(u(x, y) = invfourier(exp(-s*y), s, x), Int)

    u(x, y) = (1/2)*(Int(exp(-s*y+I*s*x), s = -infinity .. infinity))/Pi

    (18)

    pdsolve([pde__7, iv__7], method = Generic)

    u(x, y) = -_F2(-y+I*x)+Dirac(x+I*y)+_F2(y+I*x)

    (19)

    3. Series solutions for linear PDE and BC problems solved via product separation with eigenvalues that are the roots of algebraic expressions which cannot be inverted

     

    Linear problems for which the PDE can be separated by product, giving rise to Sturm-Liouville problems for the separation constant (eigenvalue) and separated functions (eigenfunctions), do not always result in solvable equations for the eigenvalues. Below are examples where the eigenvalues are respectively roots of a sum of  BesselJ functions and of the non-inversible equation tan(lambda[n])+lambda[n] = 0.

     

    Example 8: This problem represents the temperature distribution in a thin circular plate whose lateral surfaces are insulated (Articolo example 6.9.2):

    pde__8 := diff(u(r, theta, t), t) = (diff(u(r, theta, t), r)+r*(diff(u(r, theta, t), r, r))+(diff(u(r, theta, t), theta, theta))/r)/(25*r)

    iv__8 := (D[1](u))(1, theta, t) = 0, u(r, 0, t) = 0, u(r, Pi, t) = 0, u(r, theta, 0) = (r-(1/3)*r^3)*sin(theta)

    pdsolve([pde__8, iv__8])

    u(r, theta, t) = `casesplit/ans`(Sum(-(4/3)*BesselJ(1, lambda[n]*r)*sin(theta)*exp(-(1/25)*lambda[n]^2*t)*(BesselJ(0, lambda[n])*lambda[n]^3-BesselJ(1, lambda[n])*lambda[n]^2+4*lambda[n]*BesselJ(0, lambda[n])-8*BesselJ(1, lambda[n]))/(lambda[n]^3*(BesselJ(0, lambda[n])^2*lambda[n]+BesselJ(1, lambda[n])^2*lambda[n]-2*BesselJ(0, lambda[n])*BesselJ(1, lambda[n]))), n = 0 .. infinity), {And(-BesselJ(1, lambda[n])+BesselJ(2, lambda[n])*lambda[n] = 0, 0 < lambda[n])})

    (20)

     

    In the above we see that the eigenvalue `&lambda;__n` satisfies -BesselJ(1, lambda[n])+BesselJ(2, lambda[n])*lambda[n] = 0. When `&lambda;__n` is the root of one single BesselJ or BesselY function of integer order, the Maple functions BesselJZeros and BesselYZeros are used instead. That is the case, for instance, if we slightly modify this problem changing the first boundary condition to be u(1, theta, t) = 0 instead of (D[1](u))(1, theta, t) = 0

    `iv__8.1` := u(1, theta, t) = 0, u(r, 0, t) = 0, u(r, Pi, t) = 0, u(r, theta, 0) = (r-(1/3)*r^3)*sin(theta)

    pdsolve([pde__8, `iv__8.1`])

    u(r, theta, t) = `casesplit/ans`(Sum(-(4/3)*BesselJ(1, lambda[n]*r)*sin(theta)*exp(-(1/25)*lambda[n]^2*t)*(lambda[n]^2+4)/(BesselJ(0, lambda[n])*lambda[n]^3), n = 1 .. infinity), {And(lambda[n] = BesselJZeros(1, n), 0 < lambda[n])})

    (21)

    Example 9: This problem represents the temperature distribution in a thin rod whose left end is held at a fixed temperature of 5 and whose right end loses heat by convection into a medium whose temperature is 10. There is also an internal heat source term in the PDE (Articolo's textbook, example 8.4.3):

    pde__9 := diff(u(x, t), t) = (1/20)*(diff(u(x, t), x, x))+t

    iv__9 := u(0, t) = 5, u(1, t)+(D[1](u))(1, t) = 10, u(x, 0) = -40*x^2*(1/3)+45*x*(1/2)+5

    pdsolve([pde__9, iv__9], u(x, t))

    u(x, t) = `casesplit/ans`(Sum(piecewise(lambda[n] = 0, 0, (80/3)*exp(-(1/20)*lambda[n]^2*t)*sin(lambda[n]*x)*(lambda[n]^2*cos(lambda[n])+lambda[n]*sin(lambda[n])+4*cos(lambda[n])-4)/(lambda[n]^2*(sin(2*lambda[n])-2*lambda[n]))), n = 0 .. infinity)+Int(Sum(piecewise(lambda[n] = 0, 0, 4*exp(-(1/20)*lambda[n]^2*(t-tau))*sin(lambda[n]*x)*tau*(cos(lambda[n])-1)/(sin(2*lambda[n])-2*lambda[n])), n = 0 .. infinity), tau = 0 .. t)+(5/2)*x+5, {And(tan(lambda[n])+lambda[n] = 0, 0 < lambda[n])})

    (22)

    For information on how to test or plot a solution like the one above, please see the end of the Mapleprimes post "Sturm-Liouville problem with eigenvalues that are the roots of algebraic expressions which cannot be inverted" 

     

    4. Superposition method for linear PDE with more than one non-homogeneous BC

     

    Previously, for linear homogeneous PDE problems with non-periodic initial and boundary conditions, pdsolve was only consistently able to solve the problem as long as at most one of those conditions was non-homogeneous. The superposition method works by taking advantage of the linearity of the problem and the fact that the solution to such a problem in which two or more of the BC are non-homogeneous can be given as

    u = u__1+u__2 + ...,  where each u__i is a solution of the PDE with all but one of the BC homogenized.

     

    Example 10: A Laplace PDE with one homogeneous and three non-homogeneous conditions:

    pde__10 := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

    iv__10 := u(0, y) = 0, u(Pi, y) = sinh(Pi)*cos(y), u(x, 0) = sin(x), u(x, Pi) = -sinh(x)

    pdsolve([pde__10, iv__10])

    u(x, y) = ((exp(2*Pi)-1)*(Sum((-1)^n*n*(exp(2*Pi)-1)*exp(n*(Pi-y)-Pi)*sin(n*x)*(exp(2*n*y)-1)/(Pi*(n^2+1)*(exp(2*Pi*n)-1)), n = 1 .. infinity))+(exp(2*Pi)-1)*(Sum(2*sin(n*y)*exp(n*(Pi-x))*n*sinh(Pi)*((-1)^n+1)*(exp(2*n*x)-1)/(Pi*(exp(2*Pi*n)-1)*(n^2-1)), n = 2 .. infinity))+sin(x)*(exp(-y+2*Pi)-exp(y)))/(exp(2*Pi)-1)

    (23)

     

    5. Polynomial solutions method:

     

    This new method gives pdsolve better performance when the PDE & BC problems admit polynomial solutions.

     

    Example 11:

    pde__11 := diff(u(x, y), x, x)+y*(diff(u(x, y), y, y)) = 0

    iv__11 := u(x, 0) = 0, (D[2](u))(x, 0) = x^2

    pdsolve([pde__11, iv__11], u(x, y))

    u(x, y) = y*(x^2-y)

    (24)

     

    6. Solving more problems using the Laplace transform or the Fourier transform

     

    These methods now solve more problems and are no longer restricted to PDE of first or second order.

     

    Example 12: A third order PDE & BC problem:

    pde__12 := diff(u(x, t), t) = -(diff(u(x, t), x, x, x))

    iv__12 := u(x, 0) = f(x)

    pdsolve([pde__12, iv__12])

    u(x, t) = (1/4)*(Int((4/3)*Pi*f(-zeta)*(-(x+zeta)/(-t)^(1/3))^(1/2)*BesselK(1/3, -(2/9)*3^(1/2)*(x+zeta)*(-(x+zeta)/(-t)^(1/3))^(1/2)/(-t)^(1/3))/(-t)^(1/3), zeta = -infinity .. infinity))/Pi^2

    (25)

     

    Example 13: A PDE & BC problem that is solved using Laplace transform:

    pde__13 := diff(u(x, y), y, x) = sin(x)*sin(y)

    iv__13 := u(x, 0) = 1+cos(x), (D[2](u))(0, y) = -2*sin(y)

    pdsolve([pde__13, iv__13])

    u(x, y) = (1/2)*cos(x-y)+(1/2)*cos(x+y)+cos(y)

    (26)

    To see the computational flow, the solving methods used and in which order they are tried use

    infolevel[pdsolve] := 2

    2

    (27)

    Example 14:

    pde__14 := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

    iv__14 := u(x, 0) = 0, u(x, b) = h(x)

    pdsolve([pde__14, iv__14])

    * trying method "SpecializeArbitraryFunctions" for 2nd order PDEs
       -> trying "LinearInXT"
       -> trying "HomogeneousBC"
    * trying method "SpecializeArbitraryConstants" for 2nd order PDEs
    * trying method "Wave" for 2nd order PDEs
       -> trying "Cauchy"
       -> trying "SemiInfiniteDomain"
       -> trying "WithSourceTerm"
    * trying method "Heat" for 2nd order PDEs
       -> trying "SemiInfiniteDomain"
       -> trying "WithSourceTerm"
    * trying method "Series" for 2nd order PDEs
       -> trying "ThreeBCsincos"
       -> trying "FourBC"
       -> trying "ThreeBC"
       -> trying "ThreeBCPeriodic"
       -> trying "WithSourceTerm"
          * trying method "SpecializeArbitraryFunctions" for 2nd order PDEs
             -> trying "LinearInXT"
             -> trying "HomogeneousBC"
                Trying travelling wave solutions as power series in tanh ...
                   Trying travelling wave solutions as power series in ln ...
          * trying method "SpecializeArbitraryConstants" for 2nd order PDEs
             Trying travelling wave solutions as power series in tanh ...
                Trying travelling wave solutions as power series in ln ...
          * trying method "Wave" for 2nd order PDEs
             -> trying "Cauchy"
             -> trying "SemiInfiniteDomain"
             -> trying "WithSourceTerm"
          * trying method "Heat" for 2nd order PDEs
             -> trying "SemiInfiniteDomain"
             -> trying "WithSourceTerm"
          * trying method "Series" for 2nd order PDEs
             -> trying "ThreeBCsincos"
             -> trying "FourBC"
             -> trying "ThreeBC"
             -> trying "ThreeBCPeriodic"
             -> trying "WithSourceTerm"
             -> trying "ThreeVariables"
          * trying method "Laplace" for 2nd order PDEs
             -> trying a Laplace transformation
          * trying method "Fourier" for 2nd order PDEs
             -> trying a fourier transformation

          * trying method "Generic" for 2nd order PDEs
             -> trying a solution in terms of arbitrary constants and functions to be adjusted to the given initial conditions
          * trying method "PolynomialSolutions" for 2nd order PDEs

          * trying method "LinearDifferentialOperator" for 2nd order PDEs
          * trying method "Superposition" for 2nd order PDEs
       -> trying "ThreeVariables"
    * trying method "Laplace" for 2nd order PDEs
       -> trying a Laplace transformation
    * trying method "Fourier" for 2nd order PDEs
       -> trying a fourier transformation

       <- fourier transformation successful
    <- method "Fourier" for 2nd order PDEs successful

     

    u(x, y) = invfourier(exp(s*(b+y))*fourier(h(x), x, s)/(exp(2*s*b)-1), s, x)-invfourier(exp(s*(b-y))*fourier(h(x), x, s)/(exp(2*s*b)-1), s, x)

    (28)

    convert(u(x, y) = invfourier(exp(s*(b+y))*fourier(h(x), x, s)/(exp(2*s*b)-1), s, x)-invfourier(exp(s*(b-y))*fourier(h(x), x, s)/(exp(2*s*b)-1), s, x), Int)

    u(x, y) = (1/2)*(Int((Int(h(x)*exp(-I*x*s), x = -infinity .. infinity))*exp(s*(b+y)+I*s*x)/(exp(2*s*b)-1), s = -infinity .. infinity))/Pi-(1/2)*(Int((Int(h(x)*exp(-I*x*s), x = -infinity .. infinity))*exp(s*(b-y)+I*s*x)/(exp(2*s*b)-1), s = -infinity .. infinity))/Pi

    (29)

    Reset the infolevel to avoid displaying the computational flow:

    infolevel[pdsolve] := 1

    7. Improvements to solving heat and wave PDE, with or without a source:

     

    Example 15: A heat PDE:

    pde__15 := diff(u(x, t), t) = 13*(diff(u(x, t), x, x))

    iv__15 := (D[1](u))(0, t) = 0, (D[1](u))(1, t) = 1, u(x, 0) = (1/2)*x^2+x

    pdsolve([pde__15, iv__15], u(x, t))

    u(x, t) = 1/2+Sum(2*cos(n*Pi*x)*exp(-13*Pi^2*n^2*t)*(-1+(-1)^n)/(Pi^2*n^2), n = 1 .. infinity)+13*t+(1/2)*x^2

    (30)

    To verify an infinite series solution such as this one you can first use pdetest

    pdetest(u(x, t) = 1/2+Sum(2*cos(n*Pi*x)*exp(-13*Pi^2*n^2*t)*(-1+(-1)^n)/(Pi^2*n^2), n = 1 .. infinity)+13*t+(1/2)*x^2, [pde__15, iv__15])

    [0, 0, 0, 1/2+Sum(2*cos(n*Pi*x)*(-1+(-1)^n)/(Pi^2*n^2), n = 1 .. infinity)-x]

    (31)

    To verify that the last condition, for u(x, 0) is satisfied, we plot the first 1000 terms of the series solution with t = 0 and make sure that it coincides with the plot of  the right-hand side of the initial condition u(x, 0) = (1/2)*x^2+x. Expected: the two plots superimpose each other

    plot([value(subs(t = 0, infinity = 1000, rhs(u(x, t) = 1/2+Sum(2*cos(n*Pi*x)*exp(-13*Pi^2*n^2*t)*(-1+(-1)^n)/(Pi^2*n^2), n = 1 .. infinity)+13*t+(1/2)*x^2))), (1/2)*x^2+x], x = 0 .. 1)

     

    Example 16: A heat PDE in a semi-bounded domain:

    pde__16 := diff(u(x, t), t) = (1/4)*(diff(u(x, t), x, x))

    iv__16 := (D[1](u))(alpha, t) = 0, u(x, beta) = 10*exp(-x^2)

    `assuming`([pdsolve([pde__16, iv__16], u(x, t))], [0 < x, 0 < t])

    u(x, t) = -5*exp(x^2/(-t+beta-1))*((erf(((t-beta-1)*alpha+x)/((t-beta+1)^(1/2)*(t-beta)^(1/2)))-1)*exp(4*alpha*(-x+alpha)/(-t+beta-1))+erf(((t-beta+1)*alpha-x)/((t-beta+1)^(1/2)*(t-beta)^(1/2)))-1)/(t-beta+1)^(1/2)

    (32)

     

    Example 17: A wave PDE in a semi-bounded domain:

    pde__17 := diff(u(x, t), t, t)-9*(diff(u(x, t), x, x)) = 0

    iv__17 := (D[1](u))(0, t) = 0, u(x, 0) = 0, (D[2](u))(x, 0) = x^3

    `assuming`([pdsolve([pde__17, iv__17])], [0 < x, 0 < t])

    u(x, t) = piecewise(3*t < x, 9*t^3*x+t*x^3, x < 3*t, (27/4)*t^4+(9/2)*t^2*x^2+(1/12)*x^4)

    (33)

     

    Example 18: A wave PDE with a source

    pde__18 := diff(u(x, t), t, t) = diff(u(x, t), x, x)+x*exp(-t)

    iv__18 := u(0, t) = 0, u(1, t) = 0, u(x, 0) = 0, (D[2](u))(x, 0) = 1

    pdsolve([pde__18, iv__18])

    u(x, t) = Sum(((-Pi^2*(-1)^n*n^2+Pi^2*n^2+2*(-1)^(n+1)+1)*cos(n*Pi*(t-x))-Pi*(-1)^n*n*sin(n*Pi*(t-x))+(Pi^2*(-1)^n*n^2-Pi^2*n^2+2*(-1)^n-1)*cos(n*Pi*(t+x))+Pi*n*(2*exp(-t)*(-1)^(n+1)*sin(n*Pi*x)+sin(n*Pi*(t+x))*(-1)^n))/(Pi^2*n^2*(Pi^2*n^2+1)), n = 1 .. infinity)

    (34)

     

    Example 19: Another wave PDE with a source

    pde__19 := diff(u(x, t), t, t) = 4*(diff(u(x, t), x, x))+(1+t)*x

    iv__19 := u(0, t) = 0, u(Pi, t) = sin(t), u(x, 0) = 0, (D[2](u))(x, 0) = 0

    pdsolve([pde__19, iv__19])

    u(x, t) = ((Sum(-2*((1/2)*cos(n*x-t)*n^3-(1/2)*cos(n*x+t)*n^3+((-2*n^4-(1/2)*Pi*n^2+(1/8)*Pi)*sin(2*n*t)+(t-cos(2*n*t)+1)*n*(n-1/2)*Pi*(n+1/2))*sin(n*x))*(-1)^n/(Pi*n^4*(4*n^2-1)), n = 1 .. infinity))*Pi+x*sin(t))/Pi

    (35)

    8. Improvements in series methods for Laplace PDE problems

     

    "  Example 20:A Laplace PDE with BC representing the inside of a quarter circle of radius 1. The solution we seek is bounded as r approaches 0:"

    pde__20 := diff(u(r, theta), r, r)+(diff(u(r, theta), r))/r+(diff(u(r, theta), theta, theta))/r^2 = 0

    iv__20 := u(r, 0) = 0, u(r, (1/2)*Pi) = 0, (D[1](u))(1, theta) = f(theta)

    `assuming`([pdsolve([pde__20, iv__20], u(r, theta), HINT = boundedseries(r = [0]))], [0 <= theta, theta <= (1/2)*Pi, 0 <= r, r <= 1])

    u(r, theta) = Sum(2*(Int(f(theta)*sin(2*n*theta), theta = 0 .. (1/2)*Pi))*r^(2*n)*sin(2*n*theta)/(Pi*n), n = 1 .. infinity)

    (36)

     

    Example 21: A Laplace PDE for which we seek a solution that remains bounded as y approaches infinity:

    pde__21 := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

    iv__21 := u(0, y) = A, u(a, y) = 0, u(x, 0) = 0

    `assuming`([pdsolve([pde__21, iv__21], HINT = boundedseries(y = infinity))], [a > 0])

    u(x, y) = ((Sum(-2*A*sin(n*Pi*x/a)*exp(-n*Pi*y/a)/(n*Pi), n = 1 .. infinity))*a-A*(x-a))/a

    (37)

     

    9. Better simplification of answers:

     

     

    Example 22: For this wave PDE with a source term, pdsolve used to return a solution with uncomputed integrals:

    pde__22 := diff(u(x, t), t, t) = A*x+diff(u(x, t), x, x)

    iv__22 := u(0, t) = 0, u(1, t) = 0, u(x, 0) = 0, (D[2](u))(x, 0) = 0

    pdsolve([pde__22, iv__22], u(x, t))

    u(x, t) = Sum(2*(-1)^n*A*sin(n*Pi*x)*cos(n*Pi*t)/(n^3*Pi^3), n = 1 .. infinity)+(1/6)*(-x^3+x)*A

    (38)

     

    Example 23: A BC at x = infinityis now handled by pdsolve:

    pde__23 := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0

    iv__23 := u(0, y) = sin(y), u(x, 0) = 0, u(x, a) = 0, u(infinity, y) = 0

    `assuming`([pdsolve([pde__23, iv__23], u(x, y))], [0 < a])

    u(x, y) = Sum(2*piecewise(a = Pi*n, (1/2)*Pi*n, -Pi*(-1)^n*sin(a)*n*a/(Pi^2*n^2-a^2))*exp(-n*Pi*x/a)*sin(n*Pi*y/a)/a, n = 1 .. infinity)

    (39)

     

    Example 24: A reduced Helmholtz PDE in a square of side "Pi. "Previously, pdsolve returned a series starting at n = 0, when the limit of the n = 0 term is 0.

    pde__24 := diff(u(x, y), x, x)+diff(u(x, y), y, y)-k*u(x, y) = 0

    iv__24 := u(0, y) = 1, u(Pi, y) = 0, u(x, 0) = 0, u(x, Pi) = 0

    `assuming`([pdsolve([pde__24, iv__24], u(x, y))], [0 < k])

    u(x, y) = Sum(-2*sin(n*y)*(-1+(-1)^n)*(exp(-(-2*Pi+x)*(n^2+k)^(1/2))-exp((n^2+k)^(1/2)*x))/((exp(2*(n^2+k)^(1/2)*Pi)-1)*Pi*n), n = 1 .. infinity)

    (40)

     

    10. Linear differential operator: more solutions are now successfully computed

     

     

    Example 25:

    pde__25 := diff(w(x1, x2, x3, t), t) = diff(w(x1, x2, x3, t), x1, x1)+diff(w(x1, x2, x3, t), x2, x2)+diff(w(x1, x2, x3, t), x3, x3)

    iv__25 := w(x1, x2, x3, 1) = exp(a)*x1^2+x2*x3

    pdsolve([pde__25, iv__25])

    w(x1, x2, x3, t) = (x1^2+2*t-2)*exp(a)+x2*x3

    (41)

     

    Example 26:

    pde__26 := diff(w(x1, x2, x3, t), t)-(D[1, 2](w))(x1, x2, x3, t)-(D[1, 3](w))(x1, x2, x3, t)-(D[3, 3](w))(x1, x2, x3, t)+(D[2, 3](w))(x1, x2, x3, t) = 0

    iv__26 := w(x1, x2, x3, a) = exp(x1)+x2-3*x3

    pdsolve([pde__26, iv__26])

    w(x1, x2, x3, t) = exp(x1)+x2-3*x3

    (42)

     

    Example 27:

    pde__27 := diff(w(x1, x2, x3, t), t, t) = (D[1, 2](w))(x1, x2, x3, t)+(D[1, 3](w))(x1, x2, x3, t)+(D[3, 3](w))(x1, x2, x3, t)-(D[2, 3](w))(x1, x2, x3, t)

    iv__27 := w(x1, x2, x3, a) = x1^3*x2^2+x3, (D[4](w))(x1, x2, x3, a) = -x2*x3+x1

    pdsolve([pde__27, iv__27], w(x1, x2, x3, t))

    w(x1, x2, x3, t) = x1^3*x2^2+3*x2*(-t+a)^2*x1^2+(1/2)*(-t+a)*(a^3-3*a^2*t+3*a*t^2-t^3-2)*x1-(1/6)*a^3+(1/2)*a^2*t+(1/6)*(-3*t^2+6*x2*x3)*a+(1/6)*t^3-t*x2*x3+x3

    (43)

     

     

    11. More problems in 3 variables are now solved

     

     

    Example 28: A Schrödinger type PDE in two space dimensions, where Z is Planck's constant.

    pde__28 := I*`&hbar;`*(diff(f(x, y, t), t)) = -`&hbar;`^2*(diff(f(x, y, t), x, x)+diff(f(x, y, t), y, y))/(2*m)

    iv__28 := f(x, y, 0) = sqrt(2)*(sin(2*Pi*x)*sin(Pi*y)+sin(Pi*x)*sin(3*Pi*y)), f(0, y, t) = 0, f(1, y, t) = 0, f(x, 1, t) = 0, f(x, 0, t) = 0

    pdsolve([pde__28, iv__28], f(x, y, t))

    f(x, y, t) = 2^(1/2)*sin(Pi*x)*(2*exp(-((5/2)*I)*`&hbar;`*t*Pi^2/m)*cos(Pi*x)*sin(Pi*y)+exp(-(5*I)*`&hbar;`*t*Pi^2/m)*sin(3*Pi*y))

    (44)

     

    Example 29: This problem represents the temperature distribution in a thin rectangular plate whose lateral surfaces are insulated yet is losing heat by convection along the boundary x = 1, into a surrounding medium at temperature 0 (Articolo example 6.6.3):

    pde__29 := diff(u(x, y, t), t) = 1/50*(diff(u(x, y, t), x, x)+diff(u(x, y, t), y, y))

    iv__29 := (D[1](u))(0, y, t) = 0, (D[1](u))(1, y, t)+u(1, y, t) = 0, u(x, 0, t) = 0, u(x, 1, t) = 0, u(x, y, 0) = (1-(1/3)*x^2)*y*(1-y)

    `assuming`([pdsolve([pde__29, iv__29], u(x, y, t))], [0 <= x, x <= 1, 0 <= y, y <= 1])

    u(x, y, t) = `casesplit/ans`(Sum(Sum((32/3)*exp(-(1/50)*t*(Pi^2*n^2+lambda[n1]^2))*(-1+(-1)^n)*cos(lambda[n1]*x)*sin(n*Pi*y)*(-lambda[n1]^2*sin(lambda[n1])+lambda[n1]*cos(lambda[n1])-sin(lambda[n1]))/(Pi^3*n^3*lambda[n1]^2*(sin(2*lambda[n1])+2*lambda[n1])), n1 = 0 .. infinity), n = 1 .. infinity), {And(tan(lambda[n1])*lambda[n1]-1 = 0, 0 < lambda[n1])})

    (45)

     

    Articolo's Exercise 7.15, with 6 boundary/initial conditions, two for each variable

    pde__30 := diff(u(x, y, t), t, t) = 1/4*(diff(u(x, y, t), x, x)+diff(u(x, y, t), y, y))-(1/10)*(diff(u(x, y, t), t))

    iv__30 := (D[1](u))(0, y, t) = 0, (D[1](u))(1, y, t)+u(1, y, t) = 0, (D[2](u))(x, 0, t)-u(x, 0, t) = 0, (D[2](u))(x, 1, t) = 0, u(x, y, 0) = (1-(1/3)*x^2)*(1-(1/3)*(y-1)^2), (D[3](u))(x, y, 0) = 0

     

    This problem is tricky ... There are three independent variables, therefore two eigenvalues (constants that appear separating variables by product) in the Sturm-Liouville problem. But after solving the separated system and also for the eigenvalues, the second eigenvalue is equal to the first one, and in addition cannot be expressed in terms of known functions, because the equation it solves cannot be inverted.

     

    pdsolve([pde__30, iv__30])

    u(x, y, t) = `casesplit/ans`(Sum((1/6)*(lambda[n]^2*sin(lambda[n])-lambda[n]*cos(lambda[n])+sin(lambda[n]))*cos(lambda[n]*x)*(exp((1/10)*t*(-200*lambda[n]^2+1)^(1/2))*(-200*lambda[n]^2+1)^(1/2)+exp((1/10)*t*(-200*lambda[n]^2+1)^(1/2))+(-200*lambda[n]^2+1)^(1/2)-1)*exp(-(1/20)*t*((-200*lambda[n]^2+1)^(1/2)+1))*(cos(lambda[n]*y)*lambda[n]+sin(lambda[n]*y))*(6*lambda[n]^2*cos(lambda[n])^2+cos(lambda[n])^2-5*lambda[n]^2-1)/((-200*lambda[n]^2+1)^(1/2)*(-cos(lambda[n])^4+(lambda[n]*sin(lambda[n])-1)*cos(lambda[n])^3+(lambda[n]^2+lambda[n]*sin(lambda[n])+1)*cos(lambda[n])^2+(lambda[n]^2+2*lambda[n]*sin(lambda[n])+1)*cos(lambda[n])+lambda[n]*(lambda[n]+sin(lambda[n])))*lambda[n]^4*(cos(lambda[n])-1)), n = 0 .. infinity), {And(tan(lambda[n])*lambda[n]-1 = 0, 0 < lambda[n])})

    (46)

    ``

     


     

    Download PDE_and_BC_during_2018.mw

    Edgardo S. Cheb-Terrab
    Physics, Differential Equations and Mathematical Functions, Maplesoft

    Error when running 'm_model.m' to generate simulink function block when using 'Group all parameters into nested structure'. No errors when using 'Do not group parameters'.

    Error using m_fullSwing (line 125)
    Invalid setting in 'MapleSim_fullSwing/MapleSim_fullSwing/MapleSimParameters' for parameter 'Value'.
    Caused by:
        Error using m_fullSwing (line 125)
        Expression '[Param.Address.ArmPlane, Param.Address.ClubRelHand,  ... ]' for
        parameter 'Value' in 'MapleSim_fullSwing/MapleSim_fullSwing/MapleSimParameters' cannot be evaluated.
            Error using m_fullSwing (line 125)
            Error: Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for
            mismatched delimiters.
     

    Yesterday, I accidentally discovered a nasty bug in a fairly simple example:

    restart;
    Expr:=a*sin(x)+b*cos(x);
    maximize(Expr, x=0..2*Pi);
    minimize(Expr, x=0..2*Pi);
                                        

    I am sure the correct answers are  sqrt(a^2+b^2)  and  -sqrt(a^2+b^2)  for any real values  a  and  b .  It is easy to prove in many ways. The simplest method does not require any calculations and can be done in the mind. We will consider  Expr  as the scalar product (or the dot product) of two vectors  <a, b>  and  <sin(x), cos(x)>, one of which is a unit vector. Then it is obvious that the maximum of this scalar product is reached if the vectors are codirectional and equals to the length of the first vector, that is, sqrt(a^2+b^2).

    Bugs in these commands were noted by users and earlier (see search by keywords bug, maximize, minimize) but unfortunately are still not fixed. 

    The attached worksheet develops a procedure for extrapolating boundary data from a square to its interior.  Specifically, let's consider the square [−1,1]×[−1,1] and the continuous functions u1, u2, u3, u4 defined on its edges. The procedure constructs a continuous function u(x,y) in the interior of the square which matches the boundary data.

    The function u(x,y) is necessarily discontinuous at a corner if the boundary data of the edges meeting at that corner are inconsistent.  However, if the boundary data are consistent at all corners, then u(x,y) is continuous everywhere including the boundary.

    Here is what the extrapolating function looks like:

    proc (x, y) options operator, arrow; (1/2)*(1-y)*(u__1(x)-(1/2)*(1-x)*u__1(-1))+(1/2)*(x+1)*(u__2(y)-(1/2)*(1-y)*u__2(-1))+(1/2)*(y+1)*(u__3(x)-(1/2)*(x+1)*u__3(1))+(1/2)*(1-x)*(u__4(y)-(1/2)*(y+1)*u__4(1))+(1/4)*(u__1(-1)-u__4(-1))*(-x^2+1)*(1-y)/((x+1)^2+(y+1)^2)^(1/2)+(1/4)*(u__2(-1)-u__1(1))*(-y^2+1)*(x+1)/((y+1)^2+(1-x)^2)^(1/2)+(1/4)*(u__3(1)-u__2(1))*(-x^2+1)*(y+1)/((1-x)^2+(1-y)^2)^(1/2)+(1/4)*(u__4(1)-u__3(-1))*(-y^2+1)*(1-x)/((1-y)^2+(x+1)^2)^(1/2) end proc

     

    The worksheet Extrapolant.mw contains the details of the derivation, and an example.

    Edit: The worksheet Extrapolant-ver2.mw presents a slightly different version of the previous result. Here the square's corners are treated symmetrically, leading to a more pleasing interpolating function.

    PS: A challenge.  Extend the result to 3D, that is, construct a function u(x,y,z) on the cube [−1,1]×[−1,1]×[−1,1] which matches prescribed functions on the cube's six faces.

     

    Maple can solve the easiest two problems of the Putnam Mathematical Competition 2018.  link

     

     

    Problem A1

     

    Find all ordered pairs (a,b) of positive integers for which  1/a + 1/b = 3/2018

     

    eq:= 1/a + 1/b = 3/2018;

    1/a+1/b = 3/2018

    (1)

    isolve(%);

    {a = 1358114, b = 673}

    (2)

    # Unfortunalely Maple fails to find all the solutions; eq must be simplified first!

    (lhs-rhs)(eq);

    1/a+1/b-3/2018

    (3)

    numer(%);

    -3*a*b+2018*a+2018*b

    (4)

    s:=isolve(%);

    {a = -678048, b = 672}, {a = 0, b = 0}, {a = 672, b = -678048}, {a = 673, b = 1358114}, {a = 674, b = 340033}, {a = 1009, b = 2018}, {a = 2018, b = 1009}, {a = 340033, b = 674}, {a = 1358114, b = 673}

    (5)

    remove(u ->(eval(a,u)<=0 or eval(b,u)<=0),[s]);

    [{a = 673, b = 1358114}, {a = 674, b = 340033}, {a = 1009, b = 2018}, {a = 2018, b = 1009}, {a = 340033, b = 674}, {a = 1358114, b = 673}]

    (6)

    # Now it's OK.

     

    Problem B1

     

    Consider the set of  vectors  P = { < a, b> :  0 ≤ a ≤ 2, 0 ≤ b ≤ 100, a, b in Z}.
    Find all v in P such that the set P \ {v} can be partitioned into two sets of equal size and equal sum.

     

    n:=100:
    P:= [seq(seq([a,b],a=0..2), b=0..n)]:

    k:=nops(P): s:=add(P):
    numsols:=0:

    for i to k do
      v:=P[i]; sv:=s-v;
      if irem(sv[1],2)=1 or irem(sv[2],2)=1 then next fi;
      cond:=simplify(add( x[j]*~P[j],j=1..k))-sv/2;
      try
        sol:=[];
        sol:=Optimization:-Minimize
             (0, {x[i]=0, (cond=~0)[], add(x[i],i=1..k)=(k-1)/2 }, assume=binary);
        catch:
      end try:
      if sol<>[] then numsols:=numsols+1;
         print(v='P'[i], select(j -> (eval(x[j],sol[2])=1), {seq(1..k)})) fi;
    od:
    'numsols'=numsols;

    [1, 0] = P[2], {5, 10, 13, 14, 16, 19, 21, 23, 24, 26, 28, 31, 32, 35, 36, 38, 40, 41, 43, 46, 47, 49, 52, 53, 55, 57, 59, 62, 64, 67, 69, 70, 73, 75, 76, 79, 81, 82, 85, 86, 88, 90, 92, 93, 94, 95, 97, 98, 99, 100, 102, 103, 106, 107, 109, 112, 113, 115, 118, 120, 121, 126, 128, 135, 138, 144, 150, 154, 159, 165, 168, 170, 171, 172, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 216, 219, 222, 225, 228, 229, 230, 231, 235, 236, 237, 238, 240, 241, 242, 243, 246, 248, 252, 253, 254, 258, 260, 261, 264, 266, 267, 270, 273, 274, 275, 276, 279, 280, 284, 287}

     

    [1, 2] = P[8], {3, 7, 10, 13, 14, 16, 19, 21, 22, 25, 27, 28, 31, 32, 34, 36, 37, 40, 42, 44, 46, 48, 50, 52, 55, 58, 59, 61, 62, 64, 66, 69, 70, 73, 74, 76, 77, 81, 83, 85, 87, 88, 91, 93, 94, 97, 98, 100, 102, 104, 106, 108, 110, 112, 115, 117, 118, 121, 122, 124, 126, 132, 135, 139, 140, 141, 144, 151, 153, 159, 171, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 215, 216, 217, 218, 219, 220, 222, 223, 225, 229, 230, 231, 234, 236, 237, 238, 240, 241, 242, 243, 246, 249, 250, 252, 255, 257, 258, 259, 261, 263, 266, 267, 269, 273, 276, 279, 281, 282, 284}

     

    [1, 4] = P[14], {7, 10, 13, 15, 19, 21, 22, 24, 26, 28, 31, 32, 33, 37, 38, 40, 42, 44, 47, 48, 50, 52, 55, 56, 59, 60, 63, 64, 66, 68, 71, 72, 75, 76, 78, 81, 82, 87, 88, 90, 91, 94, 95, 96, 97, 100, 102, 103, 105, 106, 109, 110, 113, 114, 115, 116, 117, 118, 121, 122, 124, 127, 132, 135, 138, 142, 144, 145, 154, 156, 159, 160, 163, 165, 170, 171, 172, 176, 177, 178, 179, 180, 181, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 216, 219, 220, 222, 225, 226, 228, 229, 231, 232, 234, 235, 237, 239, 240, 241, 244, 245, 247, 248, 250, 251, 253, 254, 255, 258, 261, 262, 264, 265, 267, 268, 270, 272, 273, 276, 284}

     

    [1, 6] = P[20], {10, 14, 15, 16, 19, 22, 24, 26, 28, 30, 31, 34, 35, 37, 40, 43, 44, 46, 47, 51, 52, 54, 56, 58, 60, 62, 65, 66, 68, 70, 73, 75, 76, 79, 82, 83, 84, 85, 86, 87, 88, 90, 93, 94, 96, 97, 99, 102, 103, 106, 107, 109, 110, 114, 116, 117, 120, 121, 123, 126, 127, 128, 129, 130, 135, 138, 142, 144, 147, 152, 153, 154, 156, 157, 162, 163, 172, 174, 177, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 223, 224, 225, 226, 228, 229, 233, 234, 235, 236, 237, 241, 242, 246, 248, 249, 250, 251, 252, 253, 255, 258, 259, 262, 263, 265, 267, 269, 270, 273, 275, 276}

     

    [1, 8] = P[26], {4, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 29, 30, 34, 36, 37, 38, 40, 43, 44, 46, 48, 51, 54, 55, 56, 58, 61, 63, 64, 66, 70, 71, 73, 74, 78, 79, 80, 82, 85, 86, 88, 90, 92, 94, 97, 100, 102, 103, 106, 108, 109, 112, 113, 115, 118, 120, 121, 123, 129, 133, 135, 141, 144, 150, 158, 159, 160, 163, 168, 169, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 216, 219, 220, 222, 225, 226, 228, 230, 231, 232, 233, 234, 237, 240, 241, 243, 245, 246, 249, 250, 255, 257, 258, 259, 260, 261, 262, 264, 265, 267, 269, 270, 273, 279, 281, 282}

     

    [1, 10] = P[32], {3, 4, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 37, 38, 40, 43, 44, 46, 49, 51, 52, 54, 58, 61, 63, 64, 65, 67, 68, 70, 73, 74, 78, 79, 81, 82, 85, 86, 88, 91, 94, 96, 97, 100, 103, 104, 106, 108, 109, 112, 113, 115, 117, 119, 121, 126, 131, 132, 136, 138, 144, 147, 148, 153, 156, 160, 161, 168, 174, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 216, 217, 219, 220, 222, 224, 225, 226, 227, 228, 231, 233, 234, 236, 240, 242, 243, 245, 246, 248, 249, 251, 252, 254, 255, 257, 258, 261, 262, 264, 266, 267, 269, 270, 273, 275, 276, 279, 282}

     

    [1, 12] = P[38], {7, 10, 14, 16, 19, 22, 25, 28, 29, 31, 34, 35, 36, 37, 40, 43, 46, 47, 48, 49, 52, 54, 57, 59, 61, 63, 64, 67, 68, 70, 72, 75, 76, 79, 81, 82, 85, 87, 88, 90, 93, 94, 96, 98, 99, 100, 101, 103, 106, 108, 109, 111, 112, 114, 117, 118, 121, 122, 124, 132, 135, 141, 150, 151, 158, 159, 162, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 219, 221, 222, 223, 224, 225, 228, 229, 230, 231, 232, 234, 235, 237, 239, 240, 243, 244, 246, 249, 250, 251, 252, 255, 258, 259, 260, 261, 264, 267, 270, 273}

     

    [1, 14] = P[44], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 49, 51, 52, 54, 57, 58, 61, 62, 64, 65, 69, 71, 72, 73, 76, 78, 80, 81, 84, 85, 88, 90, 93, 94, 95, 100, 102, 103, 106, 109, 111, 112, 113, 115, 117, 120, 121, 124, 127, 132, 133, 138, 141, 145, 150, 153, 159, 162, 165, 168, 171, 173, 174, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 224, 225, 226, 227, 228, 229, 230, 234, 235, 236, 238, 241, 242, 243, 244, 247, 248, 249, 252, 256, 258, 259, 260, 261, 264, 267, 268, 270, 272, 273, 275, 276}

     

    [1, 16] = P[50], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 53, 54, 56, 58, 61, 63, 64, 67, 68, 70, 72, 74, 76, 78, 82, 84, 85, 87, 88, 90, 92, 94, 96, 99, 103, 104, 106, 107, 109, 111, 115, 116, 118, 121, 123, 127, 132, 138, 141, 143, 147, 156, 159, 160, 168, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 215, 216, 217, 218, 219, 222, 225, 226, 227, 228, 231, 232, 233, 235, 238, 240, 242, 243, 244, 246, 247, 249, 250, 252, 253, 255, 256, 258, 259, 260, 261, 264, 267, 270, 272, 273, 275, 276, 281}

     

    [1, 18] = P[56], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 60, 62, 64, 66, 68, 70, 73, 75, 76, 78, 80, 82, 85, 86, 88, 89, 93, 94, 96, 98, 100, 102, 105, 106, 109, 110, 113, 114, 116, 118, 119, 121, 123, 129, 132, 136, 144, 148, 153, 168, 169, 170, 171, 172, 173, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 217, 218, 219, 221, 222, 223, 225, 229, 230, 234, 235, 237, 239, 240, 243, 244, 245, 246, 249, 250, 251, 252, 254, 255, 256, 257, 258, 259, 260, 261, 263, 264, 266, 267, 269, 270, 273, 276, 279}

     

    [1, 20] = P[62], {3, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 63, 64, 67, 68, 72, 73, 74, 76, 79, 80, 81, 82, 85, 87, 88, 90, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 114, 116, 117, 120, 121, 126, 127, 132, 135, 141, 147, 153, 155, 163, 169, 172, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 228, 229, 232, 234, 235, 237, 238, 240, 243, 244, 246, 248, 249, 253, 254, 255, 258, 261, 262, 264, 267, 268, 270, 273, 274, 275, 276, 279, 282}

     

    [1, 22] = P[68], {12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 73, 74, 75, 76, 78, 79, 82, 83, 84, 85, 88, 90, 91, 93, 94, 96, 98, 100, 103, 104, 109, 111, 112, 114, 115, 117, 119, 121, 124, 129, 132, 136, 138, 141, 142, 150, 153, 156, 159, 160, 168, 169, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 216, 217, 218, 219, 220, 224, 225, 226, 228, 230, 231, 232, 233, 237, 238, 240, 241, 243, 247, 249, 251, 252, 253, 254, 255, 258, 261, 262, 264, 267, 270, 273, 274, 275, 276, 279, 282}

     

    [1, 24] = P[74], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 78, 79, 81, 82, 85, 87, 88, 90, 92, 94, 97, 99, 102, 105, 106, 109, 111, 112, 113, 115, 117, 120, 122, 126, 127, 129, 132, 133, 138, 139, 144, 147, 153, 160, 165, 166, 168, 169, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 225, 227, 228, 229, 230, 231, 232, 233, 234, 237, 239, 240, 244, 245, 246, 247, 249, 251, 252, 253, 255, 258, 261, 262, 264, 267, 270, 273, 275, 276, 277}

     

    [1, 26] = P[80], {3, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 82, 83, 84, 86, 88, 91, 92, 95, 96, 99, 100, 103, 104, 106, 108, 110, 112, 115, 117, 120, 121, 124, 125, 126, 127, 128, 130, 132, 138, 139, 141, 145, 150, 153, 159, 162, 166, 174, 178, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 216, 217, 218, 219, 221, 222, 223, 225, 226, 228, 229, 231, 233, 234, 235, 236, 237, 240, 242, 243, 245, 246, 247, 251, 252, 253, 255, 258, 261, 262, 264, 267, 268, 270, 273, 274, 275, 276, 279, 282}

     

    [1, 28] = P[86], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 90, 91, 92, 94, 97, 98, 101, 102, 104, 106, 109, 111, 112, 115, 116, 119, 120, 123, 124, 129, 132, 134, 136, 141, 147, 153, 160, 162, 173, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 218, 219, 220, 221, 224, 226, 227, 228, 233, 234, 236, 237, 240, 241, 243, 244, 245, 246, 247, 249, 250, 251, 252, 254, 255, 257, 258, 259, 260, 261, 263, 264, 266, 267, 270, 273, 276, 279}

     

    [1, 30] = P[92], {7, 10, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 93, 94, 97, 98, 100, 103, 105, 107, 108, 109, 110, 111, 113, 115, 117, 119, 120, 121, 122, 123, 126, 127, 132, 135, 141, 147, 153, 157, 165, 171, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 225, 226, 228, 229, 231, 234, 235, 236, 237, 240, 241, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 262, 264, 267, 268, 270, 273, 275, 276, 279, 281}

     

    [1, 32] = P[98], {10, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 102, 105, 106, 110, 111, 112, 113, 114, 118, 120, 121, 123, 124, 126, 127, 128, 129, 130, 135, 138, 144, 150, 153, 156, 157, 160, 162, 166, 167, 171, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 214, 215, 216, 217, 219, 221, 222, 223, 225, 227, 228, 229, 232, 233, 234, 235, 237, 238, 240, 243, 244, 245, 246, 247, 249, 251, 252, 253, 255, 258, 261, 262, 264, 267, 270, 273, 275, 276, 281}

     

    [1, 34] = P[104], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 95, 97, 100, 101, 103, 106, 108, 110, 113, 115, 116, 118, 120, 122, 126, 132, 134, 138, 141, 147, 152, 160, 161, 171, 172, 173, 174, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 216, 217, 219, 220, 222, 223, 225, 226, 228, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 248, 249, 251, 252, 254, 255, 257, 258, 260, 261, 264, 267, 270, 273, 276}

     

    [1, 36] = P[110], {10, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 117, 118, 120, 121, 124, 125, 129, 134, 135, 136, 138, 139, 141, 142, 144, 145, 147, 150, 151, 154, 156, 162, 171, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 218, 219, 220, 221, 222, 223, 225, 226, 227, 228, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 264, 267, 270, 273, 275, 276, 279, 281}

     

    [1, 38] = P[116], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 120, 121, 124, 125, 129, 135, 136, 141, 147, 150, 156, 158, 165, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 218, 219, 220, 222, 223, 225, 226, 228, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 256, 258, 261, 262, 264, 267, 268, 270, 273, 274, 275, 276, 279, 282}

     

    [1, 40] = P[122], {7, 10, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 124, 126, 129, 135, 136, 141, 147, 153, 156, 159, 160, 163, 166, 168, 170, 171, 172, 173, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 218, 219, 220, 222, 223, 225, 226, 228, 231, 233, 234, 235, 237, 240, 242, 243, 244, 245, 246, 247, 249, 251, 252, 255, 258, 261, 264, 267, 270, 273, 275, 276, 279, 281}

     

    [1, 42] = P[128], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 130, 132, 138, 143, 144, 150, 153, 154, 157, 171, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 218, 219, 220, 222, 223, 225, 226, 228, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 256, 258, 261, 262, 264, 267, 270, 272, 273, 276, 278, 279, 281, 282}

     

    [1, 44] = P[134], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 138, 140, 144, 150, 159, 162, 168, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 228, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 249, 250, 251, 252, 253, 255, 256, 258, 261, 262, 264, 267, 268, 270, 273, 275, 276, 281}

     

    [1, 46] = P[140], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 150, 159, 162, 168, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 228, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 249, 250, 251, 252, 253, 255, 257, 258, 261, 262, 264, 267, 268, 270, 273, 275, 276, 277}

     

    [1, 48] = P[146], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 150, 155, 156, 159, 162, 165, 166, 168, 172, 176, 177, 178, 179, 180, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 228, 229, 231, 233, 234, 235, 237, 238, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 262, 264, 267, 269, 270, 273, 275, 276, 281}

     

    [1, 50] = P[152], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 156, 159, 160, 162, 171, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 228, 231, 232, 233, 234, 235, 237, 238, 240, 241, 243, 244, 245, 246, 248, 251, 252, 254, 255, 257, 258, 260, 261, 263, 264, 267, 270, 273, 276}

     

    [1, 52] = P[158], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 150, 156, 157, 160, 163, 165, 171, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 228, 231, 232, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 264, 267, 270, 273, 275, 276, 279, 281, 282}

     

    [1, 54] = P[164], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 150, 156, 157, 160, 162, 163, 172, 173, 174, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 228, 229, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 250, 251, 252, 254, 255, 258, 261, 264, 267, 270, 273, 275, 276, 279, 281, 282}

     

    [1, 56] = P[170], {7, 10, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 150, 156, 157, 160, 162, 163, 165, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 228, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 251, 252, 255, 258, 261, 264, 266, 267, 270, 273, 275, 276, 279, 281}

     

    [1, 58] = P[176], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 150, 156, 157, 160, 162, 163, 165, 171, 174, 175, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 228, 231, 233, 234, 235, 237, 238, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 264, 267, 270, 273, 275, 276, 279, 280, 282}

     

    [1, 60] = P[182], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 150, 156, 157, 160, 162, 163, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 228, 231, 233, 234, 235, 237, 238, 240, 243, 244, 245, 246, 247, 249, 251, 252, 255, 258, 261, 264, 267, 268, 270, 273, 276, 279, 282, 285}

     

    [1, 62] = P[188], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 150, 156, 168, 171, 172, 173, 174, 175, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 228, 229, 231, 233, 234, 235, 236, 237, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 262, 264, 265, 267, 270, 273, 275, 276, 279, 281}

     

    [1, 64] = P[194], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 153, 156, 157, 160, 162, 163, 171, 173, 174, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 225, 226, 228, 229, 231, 233, 234, 235, 237, 240, 242, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 264, 267, 270, 273, 275, 276, 279, 281}

     

    [1, 66] = P[200], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 132, 137, 141, 144, 150, 153, 157, 168, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 225, 226, 228, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 250, 251, 252, 253, 255, 258, 261, 262, 264, 267, 270, 273, 275, 276, 277, 279, 282}

     

    [1, 68] = P[206], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 128, 133, 138, 141, 144, 150, 156, 157, 159, 160, 162, 168, 172, 173, 174, 175, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 218, 219, 220, 221, 222, 223, 225, 226, 228, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 262, 264, 267, 268, 270, 273, 275, 276, 279, 282}

     

    [1, 70] = P[212], {7, 10, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 126, 132, 135, 138, 141, 145, 150, 153, 157, 159, 161, 165, 166, 168, 171, 174, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 249, 250, 251, 252, 255, 258, 261, 264, 267, 270, 275, 281}

     

    [1, 72] = P[218], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 112, 114, 118, 119, 121, 122, 124, 126, 132, 133, 138, 141, 144, 150, 159, 166, 169, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 215, 216, 219, 220, 221, 222, 223, 224, 225, 226, 228, 231, 233, 234, 235, 237, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 259, 260, 261, 263, 264, 266, 267, 269, 270, 272, 273, 276, 279}

     

    [1, 74] = P[224], {3, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 101, 103, 105, 107, 109, 113, 114, 116, 117, 121, 122, 124, 126, 129, 135, 141, 143, 147, 152, 153, 160, 165, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 217, 219, 220, 221, 222, 226, 228, 231, 233, 234, 235, 237, 240, 241, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 262, 264, 267, 268, 270, 271, 273, 275, 276, 279, 281, 282}

     

    [1, 76] = P[230], {7, 10, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 92, 93, 95, 97, 100, 102, 103, 105, 107, 109, 111, 116, 117, 120, 121, 123, 125, 127, 131, 135, 141, 143, 144, 148, 153, 156, 157, 163, 168, 169, 171, 172, 174, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 228, 229, 231, 233, 234, 235, 236, 237, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 264, 267, 270, 273, 275, 276, 290}

     

    [1, 78] = P[236], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 88, 89, 91, 93, 95, 97, 100, 103, 104, 107, 108, 110, 113, 114, 116, 118, 120, 122, 126, 128, 134, 135, 139, 141, 147, 150, 153, 156, 159, 162, 165, 167, 174, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 214, 215, 216, 217, 218, 219, 220, 223, 225, 226, 228, 229, 231, 232, 233, 234, 235, 237, 238, 240, 243, 244, 245, 246, 247, 251, 252, 253, 255, 258, 261, 262, 264, 267, 268, 270, 273, 275, 276, 281}

     

    [1, 80] = P[242], {7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 75, 79, 81, 82, 84, 85, 87, 90, 91, 96, 97, 99, 100, 102, 104, 107, 109, 110, 112, 114, 117, 120, 121, 125, 126, 127, 130, 131, 138, 141, 144, 150, 152, 153, 154, 159, 163, 166, 174, 175, 176, 177, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 216, 217, 218, 219, 220, 221, 222, 223, 228, 229, 231, 234, 235, 236, 237, 238, 241, 243, 244, 245, 246, 247, 249, 251, 252, 253, 255, 258, 261, 262, 264, 267, 270, 272, 273, 275, 276, 281}

     

    [1, 82] = P[248], {7, 10, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 62, 63, 65, 67, 70, 71, 73, 76, 77, 79, 81, 83, 85, 88, 90, 92, 93, 96, 97, 99, 102, 103, 105, 107, 108, 109, 110, 112, 113, 117, 118, 121, 122, 126, 127, 132, 133, 136, 138, 144, 147, 150, 156, 158, 162, 168, 171, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 215, 216, 217, 218, 219, 221, 222, 226, 227, 229, 230, 232, 234, 235, 237, 238, 240, 241, 243, 244, 246, 249, 251, 252, 253, 255, 258, 261, 262, 264, 267, 270, 271, 273, 275, 276, 280, 281}

     

    [1, 84] = P[254], {7, 10, 14, 16, 19, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 51, 53, 55, 58, 59, 61, 63, 65, 68, 69, 71, 72, 73, 74, 76, 79, 81, 82, 85, 86, 91, 93, 94, 96, 97, 99, 101, 103, 108, 109, 110, 111, 113, 115, 118, 120, 124, 127, 132, 133, 135, 141, 144, 147, 148, 150, 151, 153, 154, 156, 159, 162, 165, 168, 169, 171, 172, 174, 175, 176, 177, 178, 179, 180, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 216, 217, 219, 220, 221, 222, 223, 224, 226, 228, 230, 231, 232, 234, 235, 239, 240, 242, 246, 247, 249, 250, 252, 255, 258, 261, 264, 267, 270, 273, 275, 276, 279, 281}

     

    [1, 86] = P[260], {10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 52, 53, 55, 58, 60, 63, 64, 67, 68, 70, 71, 74, 75, 77, 79, 80, 81, 82, 85, 87, 89, 93, 94, 96, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 126, 127, 132, 133, 135, 141, 142, 144, 150, 151, 153, 159, 162, 163, 165, 166, 172, 174, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 219, 220, 221, 222, 223, 224, 228, 229, 231, 232, 234, 235, 236, 239, 240, 241, 243, 246, 247, 248, 249, 251, 253, 254, 255, 258, 261, 263, 264, 267, 270, 272, 273, 275, 276, 281}

     

    [1, 88] = P[266], {12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 36, 37, 38, 40, 43, 46, 47, 49, 52, 53, 55, 57, 59, 61, 63, 65, 67, 68, 71, 72, 74, 76, 79, 80, 82, 84, 87, 88, 91, 93, 94, 96, 100, 102, 103, 104, 108, 109, 111, 113, 117, 119, 121, 123, 127, 128, 135, 141, 145, 147, 151, 152, 153, 157, 161, 162, 163, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 203, 204, 205, 206, 207, 208, 209, 210, 213, 214, 215, 216, 217, 220, 222, 223, 225, 228, 229, 231, 233, 237, 238, 239, 240, 244, 246, 247, 252, 253, 254, 255, 258, 261, 263, 264, 267, 268, 269, 270, 273, 275, 276, 281}

     

    [1, 90] = P[272], {10, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 35, 39, 40, 41, 43, 46, 49, 50, 51, 52, 54, 55, 58, 59, 60, 62, 65, 66, 68, 71, 73, 76, 77, 79, 80, 84, 85, 87, 90, 91, 94, 97, 99, 100, 101, 103, 106, 109, 110, 111, 112, 113, 114, 115, 117, 120, 121, 123, 129, 130, 132, 138, 144, 147, 148, 150, 153, 155, 162, 163, 168, 171, 172, 173, 174, 175, 176, 177, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 214, 216, 217, 219, 220, 222, 223, 224, 225, 228, 229, 230, 231, 232, 233, 234, 235, 237, 239, 240, 243, 244, 246, 247, 250, 251, 252, 254, 259, 261, 263, 264, 267, 268, 269, 270, 273, 276}

     

    [1, 92] = P[278], {4, 7, 10, 12, 14, 15, 16, 19, 21, 22, 25, 28, 29, 31, 34, 36, 37, 39, 41, 43, 46, 47, 49, 51, 55, 56, 58, 59, 60, 64, 65, 67, 69, 71, 73, 74, 79, 81, 82, 83, 85, 87, 89, 91, 94, 96, 97, 100, 101, 104, 105, 107, 111, 114, 115, 116, 118, 120, 122, 124, 126, 130, 135, 138, 141, 144, 154, 157, 159, 168, 169, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 214, 216, 217, 219, 221, 223, 224, 225, 226, 228, 230, 231, 233, 237, 239, 240, 241, 243, 244, 246, 247, 249, 252, 253, 254, 255, 259, 261, 263, 264, 267, 270, 273, 275, 276, 279, 281, 282}

     

    [1, 94] = P[284], {3, 4, 7, 10, 12, 14, 15, 16, 19, 22, 24, 25, 28, 30, 31, 34, 35, 36, 39, 41, 43, 46, 47, 49, 52, 53, 55, 58, 59, 61, 63, 66, 67, 70, 71, 73, 76, 78, 79, 82, 84, 85, 88, 90, 93, 95, 97, 99, 101, 103, 106, 108, 110, 112, 115, 118, 120, 121, 123, 127, 129, 135, 138, 139, 144, 145, 150, 156, 157, 159, 166, 168, 171, 174, 175, 176, 177, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 213, 214, 216, 217, 219, 220, 222, 223, 225, 226, 227, 228, 229, 231, 232, 234, 235, 237, 238, 239, 240, 241, 242, 243, 246, 247, 251, 252, 254, 255, 257, 258, 260, 261, 264, 267, 269, 270, 273, 274, 275, 279}

     

    [1, 96] = P[290], {7, 10, 12, 13, 16, 17, 20, 21, 23, 25, 27, 29, 31, 32, 35, 37, 39, 43, 44, 46, 48, 49, 52, 53, 55, 58, 60, 61, 64, 65, 66, 67, 69, 70, 73, 75, 77, 80, 81, 84, 85, 88, 89, 91, 94, 96, 98, 101, 102, 104, 106, 109, 111, 112, 114, 116, 118, 121, 124, 126, 127, 129, 133, 135, 141, 142, 144, 150, 156, 161, 163, 166, 168, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 216, 217, 219, 220, 222, 224, 226, 229, 230, 231, 232, 234, 236, 237, 238, 239, 243, 244, 245, 246, 247, 249, 250, 251, 252, 255, 256, 258, 261, 264, 266, 267, 270, 273, 276, 279, 285}

     

    [1, 98] = P[296], {8, 10, 13, 17, 19, 20, 22, 25, 27, 29, 31, 32, 34, 37, 38, 39, 41, 42, 43, 45, 46, 49, 51, 52, 54, 57, 58, 61, 63, 64, 67, 68, 73, 74, 76, 77, 78, 82, 83, 85, 88, 89, 92, 94, 95, 97, 101, 102, 106, 108, 109, 110, 112, 115, 118, 119, 120, 121, 124, 126, 127, 132, 135, 137, 138, 144, 145, 150, 156, 162, 165, 166, 169, 171, 174, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 213, 214, 215, 216, 217, 218, 219, 220, 222, 223, 225, 226, 227, 228, 230, 231, 232, 233, 234, 236, 237, 240, 241, 245, 246, 247, 248, 249, 252, 253, 254, 258, 260, 264, 267, 269, 270, 273, 276, 279}

     

    [1, 100] = P[302], {6, 8, 10, 13, 15, 17, 19, 20, 22, 25, 27, 29, 31, 32, 34, 37, 38, 39, 41, 42, 43, 45, 46, 49, 51, 52, 54, 57, 58, 61, 63, 64, 67, 68, 71, 72, 74, 76, 78, 80, 82, 84, 87, 88, 91, 92, 94, 96, 98, 100, 102, 104, 106, 109, 112, 113, 115, 117, 119, 121, 126, 128, 132, 136, 144, 150, 157, 165, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 216, 217, 219, 220, 222, 225, 226, 227, 228, 229, 230, 231, 233, 234, 235, 237, 239, 241, 243, 244, 246, 248, 252, 253, 254, 258, 259, 260, 264, 265, 267, 268, 270, 273, 276, 278, 279, 282}

     

    numsols = 51

    (7)

     


    Download putnam2018.mw

     

    Edit.
    Maple can be also very useful in solving the difficult problem B6; it can be reduced to compute the (huge) coefficient of x^1842 in the polynomial

    g := (1 + x + x^2 + x^3 + x^4 + x^5 + x^9)^2018;

    The computation is very fast:

    coeff(g, x, 1842):   evalf(%);
            
    0.8048091229e1147

     

     

    While generating a 3D plot of the solution of an ODE with a parameter, I noticed that better performance could be obtained by calling the plot3d command with a procedure argument, done in a special manner.

    I don't recall this being discussed before, so I'll share it. (It it has been discussed, and this is widely known, then I apologize.)

    I tweaked the initial conditions of the original ODE system, to obtain a non-trivial solution. I don't think that the particular nature of the solution has a bearing on this note.

    restart;

    Digits := 15:

     

     

    The ODE system has two parameters. One, A, will get a fixed value. The

    other, U0, will be used like an independent variable for plotting.

     

     

    eq1:= diff(r[1](t), t, t)+(.3293064114+209.6419478*U0)*(diff(r[1](t), t))
          +569.4324330*r[1](t)-0.3123434112e-2*V(t) = -1.547206836*U0^2*q(t):
    eq2:= 2.03*10^(-8)*(diff(V(t), t))+4.065040650*10^(-11)*V(t)
          +0.3123434112e-2*(diff(r[1](t), t)) = 0:
    eq3 := diff(q(t), t, t)+1047.197552*U0*(q(t)^2-1)*(diff(q(t), t))
           +1.096622713*10^6*U0^2*q(t) = -2822.855019*A*(diff(r[1](t), t, t)):

     

    ics:=r[1](0)=0,D(r[1])(0)=1e-1,V(0)=0,q(0)=0,D(q)(0)=0:

     

    res := dsolve({eq1,eq2,eq3, ics},numeric,output=listprocedure,parameters=[A,U0]):

     

    I will call the procedure returned by dsolve, for evalutions of V(t), as the

    dsolve numeric solution-procedure in the discussion below.

     

    WV := eval(V(t), res):

     

    WV(parameters=[A=1e0]):

     

     

    The goal is to produce a 3D plot of V(t) as a function of t and the parameter U0.

     

     

    tlo,thi := 0.0, 2.0;
    U0lo,U0hi := 1e-3, 2e-1;

    0., 2.0

    0.1e-2, .2

     

    This is the grid size used for plot3d below. It is nothing special.

     

    (m,n) := 51,51;

    51, 51

     

    First, I'll demonstrate that a 3D plot can be produced quickly, by populating a
    Matrix for floating-point evaluations of V(t), depending on t in the first
    Matrix-dimension and on parameter U0 in the second Matrix-dimansion.

     

    The surfdata command is used. This is similar to how plot3d works.

     

    This  computes reasonably quickly.

     

    But generating the numeric values for U0 and t , based on the i,j positions

    in the Matrix, involves the kind of sequence generation formulas that are

    error prone for people.

     

    str := time[real]():
    M:=Matrix(m,n,datatype=float[8]):
    for j from 1 to n do
      u0 := U0lo+(j-1)*(U0hi-U0lo)/(n-1);
      WV(parameters=[U0=u0]);
      for i from 1 to m do
        T := tlo+(i-1)*(thi-tlo)/(m-1);
        try
          M[i,j] := WV(T);
        catch:
          # mostly maxfun complaint for t above some value.
          M[i,j] := Float(undefined);
        end try;
      end do:
    end do:
    plots:-surfdata(M, tlo..thi, U0lo..U0hi,
                    labels=["t","U0","V(t)"]);
    (time[real]()-str)*'seconds';

    1.686*seconds

     

    So let's try it using the plot3d command directly. A 2-parameter procedure
    is constructed, to pass to plot3d. It's not too complicated. This procedure
    will uses one of its numeric arguments to set the ODE's U0 parameter's

    value for the dsolve numeric solution-procedure, and then pass along

    the other numeric argument as a t value.


    It's much slower than the surfdata call above..

     

    VofU0 := proc(T,u0)
           WV(parameters=[U0=u0]);
           WV(T);
         end proc:

    str := time[real]():
    plot3d(VofU0, tlo..thi, U0lo..U0hi,
           grid=[m,n], labels=["t","U0","V(t)"]);
    (time[real]()-str)*'seconds';

    37.502*seconds

     

    One reason why the previous attempt is slow is that the plot3d command

    is changing values for U0 in its outer loop, and changing values of t in its

    inner loop. The consequence is that the value for U0 changes for every

    single evaluation of the plotted procedure. This makes the dsolve numeric

    solution-procedure work harder, by losing/discarding prior numeric
    solution details.

     

    The simple 3D plot below demonstrates that the plot3d command chooses
    x-y pairs by letting its second supplied independent variable be the one
    that changes in its outer loop. Each time the value for y changes the counter

    goes up by one.

     

    glob:=0:
    plot3d(proc(x,y) global glob; glob:=glob+1; end proc,
           0..3, 0..7, grid=[3,3],
           shading=zhue,  labels=["x","y","glob"]);

     

    So now let's try and be clever and call the plot3d command with the two
    independent variables reversed in position (in the call). That will make

    the outer loop change t instead of the ODE parameter U0.

     

    We can use the transform command to swap the two indepenent
    axes in the plot, if we prefer the axes roles switched. Or we could use the
    parametric calling sequence of plot3d for the same effect.

     

    The problem is that this is still much slower!

     

    VofU0rev := proc(u0,T)
           WV(parameters=[U0=u0]);
           WV(T);
         end proc:

    str := time[real]():
    Prev:=plot3d(VofU0rev, U0lo..U0hi, tlo..thi,
                 grid=[n,m], labels=["U0","t","V(t)"]):
    (time[real]()-str)*'seconds';

    plots:-display(
      plottools:-transform((x,y,z)->[y,x,z])(Prev),
      labels=["t","U0","V(t)"],
      orientation=[50,70,0]);

    34.306*seconds

     

    There is something else to adjust, to get the quick timing while using

    the plot3d command here.

     

    It turns out that setting the parameter's numeric value in the
    dsolve numeric solution-procedure causes the loss of previous details
    of the numeric solving, even if the parameter's value is the same.

     

    So calling the dsolve numeric solution-procedure to set the parameter

    value must be avoided, in the case that the new value is the same as

    the old value.

     

    One way to do that is to have the plotted procedure first call the

    dsolve numeric solution-procedure to query the current parameter

    value, so as to not reset the value if it is not changed. Another way

    is to use a local of an appliable module to store the running value

    of the parameter, and check against that. I choose the second way.

     

    And plot3d must still be called with the first independent variable-range

    as denoting the ODE's parameter (instead of the ODE's independent

    variable).

     

    And the resulting plot is fast once more.

     

    VofU0module := module()
           local ModuleApply, paramloc;
           ModuleApply := proc(par,var)
             if not (par::numeric and var::numeric) then
               return 'procname'(args);
             end if;
             if paramloc <> par then
               paramloc := par;
               WV(parameters=[U0=paramloc]);
             end if;
             WV(var);
           end proc:
    end module:

     

    For fun, this time I'll use the parameter calling sequence to flip the

    axes, instead of plots:-transform. That's just because I want t displayed

    on the first axis. But for the performance gain, what matters is that it

    is U0 which gets values from the first axis plotting-range.

     

    str := time[real]():
    plot3d([y,x,VofU0module(x,y)], x=U0lo..U0hi, y=tlo..thi,
           grid=[n,m], labels=["t","U0","V(t)"]);
    (time[real]()-str)*'seconds';

    1.625*seconds

     

    And, naturally, I could also use the parametric form to get a fast plot

    with the axes roles switched.

     

    str := time[real]():
    plot3d([x,y,VofU0module(x,y)], x=U0lo..U0hi, y=tlo..thi,
           grid=[n,m], labels=["U0","t","V(t)"]);
    (time[real]()-str)*'seconds';

    1.533*seconds

     

    Download ode_param_plot.mw

    We have just released updates to Maple and MapleSim, which provide corrections and improvements to product functionality.

    As usual, the Maple update is available through Tools>Check for Updates in Maple, and is also available from our website on the Maple 2018.2.1 download page, where you can also find more details.

    For MapleSim users, use Help>Check for Updates or visit the MapleSim 2018.2.1 download page.

    Just a simple use of DataFrames.  Looking for a new flashlight, I decided to compile a small selection of flashlights for comparison.
     

    interface(rtablesize = 20)

    Type := `<,>`(LED, LED, LED, Incandescent, Incandescent)``

    BType := `<,>`(AAA, AA, AAA, AA, AAA)

    Make := `<,>`(Maglite, Maglite, Maglite, Maglite, Maglite)

    Batteries := `<,>`(1, 2, 3, 2, 1)

    Lumens := `<,>`(47, 245, 200, 14, 2)

    Model := `<,>`(Solitaire, `Pro+`, XL50, mini, Solitaire)

    Minutes := `<,>`(105, 135, 405, 315, 225)

    Cost := `<,>`(17.49, 41, 46.99, 16.50, 10)

    NULL

    Note:The values for cost used is in Canadian dollars, and the stores for the prices taken were from the following - MEC, Lowes, Canadian Tire and Amazon.ca

    NULL

    NULL

    NULL

    Flashlight := DataFrame(`<|>`(Make, Model, Type, BType, Batteries, Lumens, Minutes, Cost), columns = `<,>`("Make", "Model", "Type", "BType", "Batteries", "Lumens", "Minutes", "Cost"))

    _m652460160

    (1)

    NULL

    numelems(Flashlight)

    40

    (2)

    sort(Flashlight, "Cost")

    _m653259008

    (3)

    sort(Flashlight, "Lumens")

    _m629071232

    (4)

    NULL

    No surprise that the cheapest and lowest light outputs were incandescent flashlights.  The list isn't very large so lets add a few more flashlights to our list

    NULL

    new1 := DataFrame(`<,>`(`<|>`(Thrunite, Ti3, LED, AAA, 1, 130, 30, 26.95), `<|>`(Police*Security, Stealth, LED, AA, 1, 80, 60, 7.99), `<|>`(Police*Security, Shield, LED, AA, 1, 120, 90, 15.99), `<|>`(Fenix, E12, LED, AA, 1, 130, 90, 37), `<|>`(Fenix, E20, LED, AA, 2, 265, 30, 50.75), `<|>`(Fenix, E05, LED, AAA, 1, 85, 45, 31.99), `<|>`(Maglite, mini, LED, AAA, 2, 84, 345, 22)), 'columns' = ["Make", "Model", "Type", "BType", "Batteries", "Lumens", "Minutes", "Cost"], 'rows' = [6, 7, 8, 9, 10, 11, 12])

    _m624778752

    (5)

    F1 := Append(Flashlight, new1)

    _m647087136

    (6)

    sort(F1, "Cost")

    _m627084608

    (7)

    F2 := convert(sort(F1, "Cost", `<`), DataFrame)

    _m650479744

    (8)

    Interesting to note that Police Security generally seems to be the cheapest.

     

     

    sort(F2, "BType")

    _m629097504

    (9)

     

     

    Adding yet more flashlights

     

    new2 := DataFrame(`<,>`(`<|>`(Pelican, 1910*Gen3, LED, AAA, 1, 106, 150, 38), `<|>`(Pelican, 1920*Gen3, LED, AAA, 2, 224, 135, 41)), 'columns' = ["Make", "Model", "Type", "BType", "Batteries", "Lumens", "Minutes", "Cost"], 'rows' = [13, 14])

    _m629911872

    (10)

    F3 := Append(F2, new2)

    _m625417184

    (11)

    sort(F3, "Cost")

    _m641386752

    (12)

    NULL

    Now lets select just the Maglite models

     

    F3[`~`[`=`](F3[() .. (), "Make"], Maglite)]

    _m650462944

    (13)

    ``

    Or select the flashlights that have a burn time longer than 100 minutes

     

    F3[`~`[`>`](F3[() .. (), "Minutes"], 100)]

    _m689998912

    (14)

    NULL


     

    Download flashlight3.mw

    From time to time, people ask me about visualizing knots in Maple. There's no formal "Knot Theory" package in Maple per se, but it is certainly possible to generate many different knots using a couple of simple commands. The following shows various examples of knots visualized using the plots:-tubeplot and algcurves:-plot_knot commands.

    The unknot can be defined by the following parametric equations:

     

    x=sin(t)

    y=cos(t)

    z=0

     

    plots:-tubeplot([cos(t),sin(t),0,t=0..2*Pi],
       radius=0.2, axes=none, color="Blue", orientation=[60,60], scaling=constrained, style=surfacecontour);

     

    plots:-tubeplot([cos(t),sin(t),0,t=0..2*Pi],    radius=0.2,axes=none,color=

     

    The trefoil knot can be defined by the following parametric equations:

     

    x = sin(t) + 2*sin(2*t)

    y = cos(t) + 2*sin(2*t)

    z = sin(3*t)

     

    plots:-tubeplot([sin(t)+2*sin(2*t),cos(t)-2*cos(2*t),-sin(3*t), t= 0..2*Pi],
       radius=0.2, axes=none, color="Green", orientation=[90,0], style=surface);

     

    plots:-tubeplot([sin(t)+2*sin(2*t),cos(t)-2*cos(2*t),-sin(3*t),t= 0..2*Pi],    radius=0.2,axes=none,color=

     

    The figure-eight can be defined by the following parametric equations:


    x = (2 + cos(2*t)) * cos(3*t)

    y = (2 + cos(2*t)) * sin(3*t)

    z = sin(4*t)

     

    plots:-tubeplot([(2+cos(2*t))*cos(3*t),(2+cos(2*t))*sin(3*t),sin(4*t),t=0..2*Pi],
       numpoints=100, radius=0.1, axes=none, color="Red", orientation=[75,30,0], style=surface);

     

    plots:-tubeplot([(2+cos(2*t))*cos(3*t),(2+cos(2*t))*sin(3*t),sin(4*t),t=0..2*Pi],    numpoints=100,radius=0.1,axes=none,color=

     

    The Lissajous knot can be defined by the following parametric equations:

     

    x = cos(t*n[x]+phi[x])

    y = cos(t*n[y]+phi[y])

    z = cos(t n[z] + phi[z])

    Where n[x], n[y], and n[z] are integers and the phase shifts phi[x], phi[y], and phi[z] are any real numbers.
    The 8 21 knot ( n[x] = 3, n[y] = 4, and n[z] = 7) appears as follows:
     

    plots:-tubeplot([cos(3*t+Pi/2),cos(4*t+Pi/2),cos(7*t),t=0..2*Pi],
       radius=0.05, axes=none, color="Brown", orientation=[90,0,0], style=surface);

     

    plots:-tubeplot([cos(3*t+Pi/2),cos(4*t+Pi/2),cos(7*t),t=0..2*Pi],    radius=0.05,axes=none,color=

     

    A star knot can be defined by using the following polynomial:
     

    f = -x^5+y^2

     

    f := -x^5+y^2
    algcurves:-plot_knot(f,x,y,epsilon=0.7,
       radius=0.25, tubepoints=10, axes=none, color="Orange", orientation=[60,0], style=surfacecontour);

     

     

    By switching x and y, different visualizations can be generated:

     

    g=(y^3-x^7)*(y^2-x^5)+y^3

     

    g:=(y^3-x^7)*(y^2-x^5)+y^3;
    plots:-display(<
    algcurves:-plot_knot(g,y,x, epsilon=0.8, radius=0.1, axes=none, color="CornflowerBlue", orientation=[75,30,0])|
    algcurves:-plot_knot(g,x,y, epsilon=0.8, radius=0.1, axes=none, color="OrangeRed", orientation=[75,0,0])>);

     

     

    f = (y^3-x^7)*(y^2-x^5)

     

    f:=(y^3-x^7)*(y^2-x^5);
    algcurves:-plot_knot(f,x,y,
      epsilon=0.8, radius=0.1, axes=none, orientation=[35,0,0]);

     

     

    h=(y^3-x^7)*(y^3-x^7+100*x^13)*(y^3-x^7-100*x^13)

     

    h:=(y^3-x^7)*(y^3-x^7+100*x^13)*(y^3-x^7-100*x^13);

    algcurves:-plot_knot(h,x,y,
       epsilon=0.8, numpoints=400, radius=0.03, axes=none, color=["Blue","Red","Green"], orientation=[60,0,0]);

     

    Please feel free to add more of your favourite knot visualizations in the comments below!

    You can interact with the examples or download a copy of these examples from the MapleCloud here: https://maple.cloud/app/5654426890010624/Examples+of+Knots

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