Maple Questions and Posts

The Zassenhaus formula and the Pauli matrices

by: ecterrab 14702 Product: Maple

January 11 2019

6 0

The Zassenhaus formula and the algebra of the Pauli matrices

Edgardo S. Cheb-Terrab¹ and Bryan C. Sanctuary²

(1) Maplesoft

(2) Department of Chemistry, McGill University, Montreal, Quebec, Canada

The implementation of the Pauli matrices and their algebra were reviewed during 2018, including the algebraic manipulation of nested commutators, resulting in faster computations using simpler and more flexible input. As it frequently happens, improvements of this type suddenly transform research problems presented in the literature as untractable in practice, into tractable.

As an illustration, we tackle below the derivation of the coefficients entering the Zassenhaus formula shown in section 4 of [1] for the Pauli matrices up to order 10 (results in the literature go up to order 5). The computation presented can be reused to compute these coefficients up to any desired higher-order (hardware limitations may apply). A number of examples which exploit this formula and its dual, the Baker-Campbell-Hausdorff formula, occur in connection with the Weyl prescription for converting a classical function to a quantum operator (see sec. 5 of [1]), as well as when solving the eigenvalue problem for classes of mathematical-physics partial differential equations [2].
To reproduce the results below - a worksheet with this contents is linked at the end - you need to have your Maple 2018.2.1 updated with the Maplesoft Physics Updates version 280 or higher.

References

[1] R.M. Wilcox, "Exponential Operators and Parameter Differentiation in Quantum Physics", Journal of Mathematical Physics, V.8, 4, (1967.

[2] S. Steinberg, "Applications of the lie algebraic formulas of Baker, Campbell, Hausdorff, and Zassenhaus to the calculation of explicit solutions of partial differential equations", Journal of Differential Equations, V.26, 3, 1977.

[3] K. Huang, "Statistical Mechanics", John Wiley & Sons, Inc. 1963, p217, Eq.(10.60).

Formulation of the problem

The Zassenhaus formula expresses as an infinite product of exponential operators involving nested commutators of increasing complexity

=

Given , and their commutator , if and commute with , for and the Zassenhaus formula reduces to the product of the first three exponentials above. The interest here is in the general case, when and , and the goal is to compute the Zassenhaus coefficients in terms of , for arbitrary finite n. Following [1], in that general case, differentiating the Zassenhaus formula with respect to and multiplying from the right by one obtains

This is an intricate formula, which however (see eq.(4.20) of [1]) can be represented in abstract form as

$"0=((∑)(lambda^n)/(n!) {A^n,B})+2 lambda ((∑) (∑)(lambda^(n+m))/(n! m!) {A^m,B^n,C[2]})+3 lambda^2 ((∑) (∑) (∑)(lambda^(n+m+k))/(n! m! k!) {A^k,B^m,(C[2])^n,C[3]})+ ..."$

from where an equation to be solved for each is obtained by equating to 0 the coefficient of . In this formula, the repeated commutator bracket is defined inductively in terms of the standard commutator by

${B, A^0} = B, {B, A^(n+1)} = %Commutator(A, {A^n, B^n})$

${C[j], B^n, A^0} = {C[j], B^n}, {C[j], A^m, B^n} = %Commutator(A, {A^`-`(m, 1), B^n, C[j]^k})$

and higher-order repeated-commutator brackets are similarly defined. For example, taking the coefficient of and and respectively solving each of them for and one obtains

This method is used in [3] to treat quantum deviations from the classical limit of the partition function for both a Bose-Einstein and Fermi-Dirac gas. The complexity of the computation of grows rapidly and in the literature only the coefficients up to have been published. Taking advantage of developments in the Physics package during 2018, below we show the computation up to and provide a compact approach to compute them up to arbitrary finite order.

Computing up to

Set the signature of spacetime such that its space part is equal to +++ and use lowercaselatin letters to represent space indices. Set also , and to represent quantum operators

>

(1)

To illustrate the computation up to , a convenient example, where the commutator algebra is closed, consists of taking and as Pauli Matrices which, multiplied by the imaginary unit, form a basis for the group, which in turn exponentiate to the relevant Special Unitary Group . The algebra for the Pauli matrices involves a commutator and an anticommutator

>

(2)

Assign now and to two Pauli matrices, for instance

>

(3)

>

(4)

Next, to extract the coefficient of from

$"0=((∑)(lambda^n)/(n!) {A^n,B})+2 lambda ((∑) (∑)(lambda^(n+m))/(n! m!) {A^m,B^n,C[2]})+3 lambda^2 ((∑) (∑) (∑)(lambda^(n+m+k))/(n! m! k!) {A^k,B^m,(C[2])^n,C[3]})+..."$

to solve it for we note that each term has a factor multiplying a sum, so we only need to take into account the first terms (sums) and in each sum replace by the corresponding . For example, given to compute we only need to compute these first three terms:

$0 = Sum(lambda^n*{B, A^n}/factorial(n), n = 1 .. 2)+2*lambda*(Sum(Sum(lambda^(n+m)*{C[2], A^m, B^n}/(factorial(n)*factorial(m)), n = 0 .. 1), m = 0 .. 1))+3*lambda^2*(Sum(Sum(Sum(lambda^(n+m+k)*{C[3], A^k, B^m, C[2]^n}/(factorial(n)*factorial(m)*factorial(k)), n = 0 .. 0), m = 0 .. 0), k = 0 .. 0))$

then solving for one gets .

Also, since to compute we only need the coefficient of , it is not necessary to compute all the terms of each multiple-sum. One way of restricting the multiple-sums to only one power of consists of using multi-index summation, available in the Physics package (see Physics:-Library:-Add ). For that purpose, redefine sum to extend its functionality with multi-index summation

>

(5)

Now we can represent the same computation of without multiple sums and without computing unnecessary terms as

$0 = Sum(lambda^n*{B, A^n}/factorial(n), n = 1)+2*lambda*(Sum(lambda^(n+m)*{C[2], A^m, B^n}/(factorial(n)*factorial(m)), n+m = 1))+3*lambda^2*(Sum(lambda^(n+m+k)*{C[3], A^k, B^m, C[2]^n}/(factorial(n)*factorial(m)*factorial(k)), n+m+k = 0))$

Finally, we need a computational representation for the repeated commutator bracket

${B, A^0} = B, {B, A^(n+1)} = %Commutator(A, {A^n, B^n})$

One way of representing this commutator bracket operation is defining a procedure, say F, with a cache to avoid recomputing lower order nested commutators, as follows

>

(6)

>

For example,

>

(7)

>

(8)

>

(9)

We can set now the value of

>

(10)

and enter the formula that involves only multi-index summation

>

H := sum(lambda^n*F(A, B, n)/factorial(n), n = 2)+2*lambda*(sum(lambda^(n+m)*F(A, F(B, C[2], n), m)/(factorial(n)*factorial(m)), n+m = 1))+3*lambda^2*(sum(lambda^(n+m+k)*F(A, F(B, F(C[2], C[3], n), m), k)/(factorial(n)*factorial(m)*factorial(k)), n+m+k = 0))

(1/2)*lambda^2*%Commutator(Physics:-Psigma[1], %Commutator(Physics:-Psigma[1], Physics:-Psigma[3]))+2*lambda*(lambda*%Commutator(Physics:-Psigma[1], I*Physics:-Psigma[2])+lambda*%Commutator(Physics:-Psigma[3], I*Physics:-Psigma[2]))+3*lambda^2*C[3]

(11)

from where we compute by solving for it the coefficient of , and since due to the mulit-index summation this expression already contains as a factor,

>

C[3] = (2/3)*Physics:-Psigma[3]-(4/3)*Physics:-Psigma[1]

(12)

In order to generalize the formula for H for higher powers of , the right-hand side of the multi-index summation limit can be expressed in terms of an abstract N, and H transformed into a mapping:

>

H := unapply(sum(lambda^n*F(A, B, n)/factorial(n), n = N)+2*lambda*(sum(lambda^(n+m)*F(A, F(B, C[2], n), m)/(factorial(n)*factorial(m)), n+m = N-1))+3*lambda^2*(sum(lambda^(n+m+k)*F(A, F(B, F(C[2], C[3], n), m), k)/(factorial(n)*factorial(m)*factorial(k)), n+m+k = N-2)), N)

(13)

Now we have

>

(14)

>

(15)

The following is already equal to (11)

>

(1/2)*lambda^2*%Commutator(Physics:-Psigma[1], %Commutator(Physics:-Psigma[1], Physics:-Psigma[3]))+2*lambda*(lambda*%Commutator(Physics:-Psigma[1], I*Physics:-Psigma[2])+lambda*%Commutator(Physics:-Psigma[3], I*Physics:-Psigma[2]))+3*lambda^2*C[3]

(16)

In this way, we can reproduce the results published in the literature for the coefficients of Zassenhaus formula up to by adding two more multi-index sums to (13). Unassign first

>

We compute now up to in one go

>

for j to 4 do C[j+1] := Simplify(solve(H(j), C[j+1])) end do

(2/3)*Physics:-Psigma[3]-(4/3)*Physics:-Psigma[1]

-(8/9)*Physics:-Psigma[1]-(158/45)*Physics:-Psigma[3]-((16/3)*I)*Physics:-Psigma[2]

(17)

The nested-commutator expression solved in the last step for is

>

(18)

With everything understood, we want now to extend these results generalizing them into an approach to compute an arbitrarily large coefficient , then use that generalization to compute all the Zassenhaus coefficients up to . To type the formula for H for higher powers of is however prone to typographical mistakes. The following is a program, using the Maple programming language , that produces these formulas for an arbitrary integer power of :

Formula := proc(A, B, C, Q)

This Formula program uses a sequence of summation indices with as much indices as the order of the coefficient we want to compute, in this case we need 10 of them

>

(19)

To avoid interference of the results computed in the loop (17), unassign again

>

Now the formulas typed by hand, used lines above to compute each of , and , are respectively constructed by the computer

>

sum(lambda^n*F(Physics:-Psigma[1], Physics:-Psigma[3], n)/factorial(n), n = N)+2*lambda*(sum(lambda^(n+m)*F(Physics:-Psigma[1], F(Physics:-Psigma[3], C[2], n), m)/(factorial(n)*factorial(m)), n+m = N-1))

(20)

>

sum(lambda^n*F(Physics:-Psigma[1], Physics:-Psigma[3], n)/factorial(n), n = N)+2*lambda*(sum(lambda^(n+m)*F(Physics:-Psigma[1], F(Physics:-Psigma[3], C[2], n), m)/(factorial(n)*factorial(m)), n+m = N-1))+3*lambda^2*(sum(lambda^(n+m+k)*F(Physics:-Psigma[1], F(Physics:-Psigma[3], F(C[2], C[3], n), m), k)/(factorial(n)*factorial(m)*factorial(k)), n+m+k = N-2))

(21)

>

(22)

Construct then the formula for and make it be a mapping with respect to N, as done for after (16)

>

(23)

Compute now the coefficients of the Zassenhaus formula up to all in one go

>

for j to 9 do C[j+1] := Simplify(solve(H(j), C[j+1])) end do

(2/3)*Physics:-Psigma[3]-(4/3)*Physics:-Psigma[1]

-(8/9)*Physics:-Psigma[1]-(158/45)*Physics:-Psigma[3]-((16/3)*I)*Physics:-Psigma[2]

(1030/81)*Physics:-Psigma[1]-(8/81)*Physics:-Psigma[3]+((1078/405)*I)*Physics:-Psigma[2]

((11792/243)*I)*Physics:-Psigma[2]+(358576/42525)*Physics:-Psigma[1]+(12952/135)*Physics:-Psigma[3]

(87277417/492075)*Physics:-Psigma[1]+(833718196/820125)*Physics:-Psigma[3]+((35837299048/17222625)*I)*Physics:-Psigma[2]

-((449018539801088/104627446875)*I)*Physics:-Psigma[2]-(263697596812424/996451875)*Physics:-Psigma[1]+(84178036928794306/2197176384375)*Physics:-Psigma[3]

(3226624781090887605597040906/21022858292748046875)*Physics:-Psigma[1]+(200495118165066770268119656/200217698026171875)*Physics:-Psigma[3]+((2185211616689851230363020476/4204571658549609375)*I)*Physics:-Psigma[2]

(24)

Notes: with the material above you can compute higher order values of . For that you need:

1.	Unassign as done above in two opportunities, to avoid interference of the results just computed.

2.	Indicate more summation indices in the sequence in (19), as many as the maximum value of n in .

3.	Have in mind that the growth in size and complexity is significant, with each taking significantly more time than the computation of all the previous ones.

4.	Re-execute the input line (23) and the loop (24).

>

Download The_Zassenhause_formula_and_the_Pauli_Matrices.mw

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

Problem with solving for C1 and C2...

Asked by: wlferguson19 90

January 11 2019

2 2

ode_problem.mw Can someone please look at this? I am having a problem with the ordinary differential equations.

'fsolve' statement is only rewritting the equation...

Asked by: danyheatley 20

January 11 2019

3 8

Hello there,

I'm pretty new to MAPLE so this may probably an "easy" mistake, but I don't know what's the problem anyway...

I'd like to find a solution to a system of four equations with four unknowns:

fsolve({(T[1]-T[0])/(10000-T[1]+T[0]) = -2.000000000, (T[2]-T[0])/(20000-T[2]+T[0]) = 0, (T[3]-T[0])/(50000-T[3]+T[0]) = 50, .1*T[0]+.3*T[1]+.55*T[2]+0.5e-1*T[3]-5000 = 0}, {T[0], T[1], T[2], T[3]})

Problem: Instead of providing the single solutions, MAPLE is simply just rewriting the fsolve-statement and does not solve. There is no error-message.

Does anybody know, what the problem is here? Is there no solution after all?

How to check that an n*n matrix is an MDS matrx wh...

Asked by: Magma 70

January 11 2019

1 10

An n*n matrix A is called an MDS matrix over an arbitrary field if all determinant of square sub-matrices of A are non-zero over the field. It is not difficult to prove that the number of all square sub-matrices of A is binomial(2*n, n)-1. The code that I use to check whether A is an MDS matrix is in the following form

u := 1;
for k to n do
P := choose(n, k);
for i to nops(P) do
for j to nops(P) do
F := A(P[i], P[j]);
r := Determinant(F);
if r = 0 then

u := 0; k:=n+1;

i := nops(P)+1; j := nops(P)+1;

end if;

end do;

end do;
if u = 1 then

print(A is an MDS Matrix)

end if;
end do:

When I run the mentioned code for n=16, it takes long time since we need to check binomial(32, 16)-1=601080389 cases to verify that A is an MDS matrix or not.

My Question: Is there a modified procedure which can be used to check that an n*n matrix is whether an MDS matrix for n>=16.

how to fix bug on DirectSearch package?...

Asked by: aradell27 5

January 10 2019

1 4

i'm doing an optimaztion for my final project. when i execute the coding, it says:

Warning, initial point [m1 = .900000000000000, m2 = .900000000000000, m3 = .900000000000000, sigma1 = .900000000000000, sigma2 = .900000000000000, sigma3 = .900000000000000] does not satisfy the inequality constraints; trying to find a feasible initial point
Error, (in DirectSearch:-Search) cannot find feasible initial point; last infeasible trial point is [m1 = HFloat(0.9), m2 = HFloat(1.9), m3 = HFloat(1.9), sigma1 = HFloat(1.9), sigma2 = HFloat(0.9), sigma3 = HFloat(0.9)].

here my full coding:

restart;
with(linalg);
with(Optimization);
with(DirectSearch);
[BoundedObjective, CompromiseProgramming, DataFit,

ExponentialWeightedSum, GlobalOptima, GlobalSearch, Minimax,

ModifiedTchebycheff, Search, SolveEquations, WeightedProduct,

WeightedSum]
q := Array(0 .. .111, [0.586e-2, 0.67475e-3, 0.52476e-3, 0.419785e-3, 0.354814e-3, 0.324859e-3, 0.319948e-3, 0.31e-3, 0.295013e-3, 0.28e-3, 0.259974e-3, 0.25496e-3, 0.26e-3, 0.270027e-3, 0.280026e-3, 0.284987e-3, 0.274934e-3, 0.279893e-3, 0.294824e-3, 0.324765e-3, 0.374672e-3, 0.439555e-3, 0.509439e-3, 0.56934e-3, 0.60923e-3, 0.634201e-3, 0.634233e-3, 0.624319e-3, 0.61442e-3, 0.624491e-3, 0.6495e-3, 0.684465e-3, 0.714452e-3, 0.729463e-3, 0.749451e-3, 0.789387e-3, 0.864347e-3, 0.964331e-3, 0.1064261e-2, 0.119413e-2, 0.1333876e-2, 0.150354e-2, 0.1683293e-2, 0.1883023e-2, 0.2102591e-2, 0.2356929e-2, 0.2656062e-2, 0.3004981e-2, 0.3403637e-2, 0.3841874e-2, 0.4349482e-2, 0.4931339e-2, 0.5587622e-2, 0.6293742e-2, 0.7044467e-2, 0.780967e-2, 0.8563865e-2, 0.9291539e-2, 0.9987678e-2, 0.10677379e-1, 0.1139506e-1, 0.12195833e-1, 0.13108819e-1, 0.14262384e-1, 0.15557439e-1, 0.17002799e-1, 0.18574592e-1, 0.20263971e-1, 0.22103523e-1, 0.24043337e-1, 0.26165465e-1, 0.28548531e-1, 0.31489568e-1, 0.3458584e-1, 0.37891201e-1, 0.41511365e-1, 0.45444649e-1, 0.49867283e-1, 0.54877269e-1, 0.60478547e-1, 0.66532745e-1, 0.7397857e-1, 0.81900977e-1, 0.90034141e-1, 0.98921462e-1, .108746024, .11839162, .130752255, .14368791, .157201822, .169770195, .177233464, .188823516, .204733363, .224639756, .249617627, .273409998, .298674064, .304378882, .319592743, .342821691, .367512872, .395257421, .426270115, .460795318, .498484563, .537417055, .577542345, .618093683, .660441764, .703856817, 1]);
variables = [m1, m2, m3, sigma1, sigma2, sigma3];
variables = [m1, m2, m3, sigma1, sigma2, sigma3]
psi1 := 0.11480601e-1; psi2 := 0.8890123e-2; psi3 := .979629276;
0.011480601
0.008890123
0.979629276
s1(x):=exp(-(x/(m1))^((m1/(sigma1)))): s1(x+1):=exp(-((x+1)/(m1))^((m1/(sigma1)))): s2(x):=1-exp(-(x/(m2))^((-m2/(sigma2)))): s2(x+1):=1-exp(-((x+1)/(m2))^((-m2/(sigma2)))): s3(x):=exp(exp(-m3/(sigma3))-exp((x-m3)/(sigma3))): s3(x+1):=exp(exp(-m3/(sigma3))-exp((x+1-m3)/(sigma3))) : s(x):=(psi1*s1(x)+psi2*s2(x)+psi3*s3(x)): s(x+1):=(psi1*s1(x+1)+psi2*s2(x+1)+psi3*s3(x+1)): qtopi(x):=1-((s(x+1))/(s(x))):
fungsikerugian := add((1-qtopi(x)/q[x])^2, x = 0 .. .110);
for x from 0 to 111 do c1[x] := 0 <= s1(x+1); c2[x] := 1 >= s1(x+1); c3[x] := 0 <= s2(x+1); c4[x] := 1 >= s2(x+1); c5[x] := 0 <= s3(x+1); c6[x] := 1 >= s3(x+1); c7[x] := 0 <= s1(x); c8[x] := 1 >= s1(x); c9[x] := 0 <= s2(x); c10[x] := 1 >= s2(x); c11[x] := 0 <= s3(x); c12[x] := 1 >= s3(x); c13[x] := 0 <= s(x); c14[x] := 1 >= s(x); c15[x] := 0 <= s(x+1); c16[x] := 1 >= s(x+1); c17[x] := 0 <= qtopi(x); c18[x] := 1 >= qtopi(x) end do;

constr := {m2 > sigma2, m3 > sigma3, seq(c1[x], x = 0 .. .110), seq(c10[x], x = 0 .. .111), seq(c11[x], x = 0 .. .111), seq(c12[x], x = 0 .. .111), seq(c13[x], x = 0 .. .111), seq(c14[x], x = 0 .. .111), seq(c15[x], x = 0 .. .110), seq(c16[x], x = 0 .. .110), seq(c17[x], x = 0 .. .110), seq(c18[x], x = 0 .. .110), seq(c2[x], x = 0 .. .110), seq(c3[x], x = 0 .. .110), seq(c4[x], x = 0 .. .110), seq(c5[x], x = 0 .. .110), seq(c6[x], x = 0 .. .110), seq(c7[x], x = 0 .. .111), seq(c8[x], x = 0 .. .111), seq(c9[x], x = 0 .. .111), m1 < sigma1, 0 <= m1 and m1 < 17, 17 <= m2 and m2 < 33, 33 <= m3 and m3 < 111};
solusi := DirectSearch:-Search(fungsikerugian, constr, assume = positive, evaluationlimit = 5555);
Warning, initial point [m1 = .900000000000000, m2 = .900000000000000, m3 = .900000000000000, sigma1 = .900000000000000, sigma2 = .900000000000000, sigma3 = .900000000000000] does not satisfy the inequality constraints; trying to find a feasible initial point
Error, (in DirectSearch:-Search) cannot find feasible initial point; last infeasible trial point is [m1 = HFloat(0.9), m2 = HFloat(1.9), m3 = HFloat(1.9), sigma1 = HFloat(1.9), sigma2 = HFloat(0.9), sigma3 = HFloat(0.9)]

thanks before

taylor( (2 * pi * n) ^ (1/(2n), n = infinity, 2)...

Asked by: Vitreg2 0

January 10 2019

0 7

The command

taylor( (2 * pi * n) ^ (1/(2n), n = 2)

return 1+((1/2)*ln(2*Pi)+(1/2)*ln(n))/n+O(1/n^2)

I think, this is incorrect, in my oppinion must be:

1+((1/2)*ln(2*Pi)+(1/2)*ln(n))/n+O(ln(n)^2/n^2)

Is this error in maple?

How to Complete the Square in Maple

by: TechnicalSupport 810 Product: Maple

January 10 2019

6 5

A common question to our tech support team is about completing the square for a univariate polynomial of even degree, and how to do that in Maple. We’ve put together a solution that we think you’ll find useful. If you have any alternative methods or improvements to our code, let us know!

restart;

# Procedure to complete the square for a univariate
# polynomial of even degree.

CompleteSquare := proc( f :: depends( 'And'( polynom, 'satisfies'( g -> ( type( degree(g,x), even ) ) ) ) ), x :: name )

       local a, g, k, n, phi, P, Q, r, S, T, u:

       # Degree and parameters of polynomial.
       n := degree( f, x ):
       P := indets( f, name ) minus { x }:

       # General polynomial of square plus constant.
       g := add( a[k] * x^k, k=0..n/2 )^2 + r:

       # Solve for unknowns in g.
       Q := indets( g, name ) minus P:

       S := map( expand, { solve( identity( expand( f - g ) = 0, x ), Q ) } ):

       if numelems( S ) = 0 then
              return NULL:
       end if:

       # Evaluate g at the solution, and re-write square term
       # so that the polynomial within the square is monic.

       phi := u -> lcoeff(op(1,u),x)^2 * (expand(op(1,u)/lcoeff(op(1,u),x)))^2:  
       T := map( evalindets, map( u -> eval(g,u), S ), `^`(anything,identical(2)), phi ):

       return `if`( numelems(T) = 1, T[], T ):

end proc:


# Examples.

CompleteSquare( x^2 + 3 * x + 2, x );
CompleteSquare( a * x^2 + b * x + c, x );
CompleteSquare( 4 * x^8 + 8 * x^6 + 4 * x^4 - 246, x );

m, n := 4, 10;
r := rand(-10..10):
for i from 1 to n do
       CompleteSquare( r() * ( x^(m/2) + randpoly( x, degree=m-1, coeffs=r ) )^2 + r(), x );
end do;

# Compare quadratic examples with Student:-Precalculus:-CompleteSquare()
# (which is restricted to quadratic expressions).

Student:-Precalculus:-CompleteSquare( x^2 + 3 * x + 2 );
Student:-Precalculus:-CompleteSquare( a * x^2 + b * x + c );

For a higher-order example:

f := 5*x^4 - 70*x^3 + 365*x^2 - 840*x + 721;
g := CompleteSquare( f, x ); # 5 * ( x^2 - 7 * x + 12 )^2 + 1
h := evalindets( f, `*`, factor ); 4 * (x-3)^2 * (x-4)^2 + 1
p1 := plot( f, x=0..5, y=-5..5, color=blue ):
p2 := plots:-pointplot( [ [3,1], [4,1] ], symbol=solidcircle, symbolsize=20, color=red ):
plots:-display( p1, p2 );

tells us that the minimum value of the expression is 1, and it occurs at x=3 and x=4.

A command to evaluate the non-negative integer sol...

Asked by: Magma 70

January 10 2019

1 4

Assume that a[1],a[2],..a[k] are positive integer numbers. Let n be a positive integer number.

Suppose that igcd(a[1],a[2],..a[k])=1.

My question: Is there a command in Maple such that the output of the this command be true provided that there are "non-negative integer numbers" x[1],x[2],..x[k] which satisfy the following condition:

a1*x1+a2*x2+...+ak*xk=n

Thanks in advance

How to let sqrt(5x+5+y) become sqrt(5(x+1)+y) au...

Asked by: linpc0128 10

January 10 2019

1 5

How to let sqrt(5*x+5+y) become sqrt(5*(x+1)+y) automatically?

derivative of an implicit function...

Asked by: gaurav_rs 65

January 09 2019

0 4

Dear all,

Following the comments I am editing this post:

I have a function F of variables (r1,r2,theta1,theta2,r,theta,a). r1, r2,theta1,theta2 are function of r, theta and a. I want to take derivative of F with respect to a. r and theta are independent of a . I expressed everything in terms of 'a' as a function of 'a' at first. Then I use diff(F, a). I see there is an error in the final expression G .There is a restriction that theta1 should lie between -Pi to Pi and theta2 between 0 to 2*Pi. I speculate this is the source of error. Work sheet is attached. Reason: value of G: integration in 0 to pi/4 gives some value but for 0 to pi it evaluates to zero and so is the case with 0 to 2*Pi. As "G "physically represents energy it must be a positive value.

>

restart;

>	theta1 := unapply(arctan(rsin(theta)/(rcos(theta)-a)), a);

(1)

>

>	## theta1 -->[-Pi,Pi] and theta2-->[0,2*Pi]

>

>	theta2 := unapply(arctan(rsin(theta)/(rcos(theta)+a)), a);

(2)

>

>	r1:=unapply(sqrt((rcos(theta)-a)^2+r^2(sin(theta))^2),a);r2:=unapply(sqrt((rcos(theta)+a)^2+r^2(sin(theta))^2),a);

(3)

>	sigma12:=0;sigma22:=sigma;

(4)

>

>	## I have to use a constraint that

>	assume(theta1(a) < Pi, theta1(a) > -Pi, theta2(a) > 0, theta2(a) < 2*Pi, a>0,r>0)

>

u1:=(1+nu)*sigma22*sqrt(r1(a)*r2(a))*(4*(1-2*nu)*cos((theta1(a)+(theta2(a)))/2)-4*r*(1-nu)*cos(theta)/sqrt(r1(a)*r2(a))-2*r^2/(r1(a)*r2(a))*(cos((theta1(a)+(theta2(a)))/2)-cos(2*theta-theta1(a)/2-(theta2(a))/2)))/(4*E)+(1+nu)*sigma12*sqrt(r1(a)*r2(a))*(2*(1-2*nu)*sin((theta1(a)+(theta2(a)))/2)-2*r*(1-nu)*sin(theta)/sqrt(r1(a)*r2(a))+1*r^2/(r1(a)*r2(a))*sin(theta)*cos(theta-theta1(a)/2-(theta2(a))/2))/(E);

>

(5)

>

u2:=(1+nu)*sigma*sqrt(r1(a)*r2(a))*(8*(1-nu)*sin((theta1(a)+(theta2(a)))/2)-4*r*(nu)*sin(theta)/sqrt(r1(a)*r2(a))-2*r^2/(r1(a)*r2(a))*(sin((theta1(a)+(theta2(a)))/2)+sin(2*theta-theta1(a)/2-(theta2(a))/2)))/(4*E)+(1+nu)*sigma12*sqrt(r1(a)*r2(a))*((1-2*nu)*cos((theta1(a)+theta2(a))/2)+2*r*(1-nu)*cos(theta)/sqrt(r1(a)*r2(a))-1*r^2/(r1(a)*r2(a))*sin(theta)*sin(theta-theta1(a)/2-theta2(a)/2))/(E);

>

(6)

>

>	## get u_r and u_theta as u[1] and u[2]

>	u[1] := u1cos(theta)+u2sin(theta);

(7)

>	u[2] := -sin(theta)u1+cos(theta)u2;

(8)

>	Diff_ur := simplify(diff(u[1], a));

(9)

>

>	Diff_ut := simplify(diff(u[2], a));

(10)

>

>	# find the limiting case

>	Att := limit(Diff_utrsin(2*theta), r = infinity);

(2*a*sigma*cos(theta)^3*sin(theta)^2*nu+2*a*sigma*cos(theta)*sin(theta)^4*nu+2*a*sigma*cos(theta)^3*sin(theta)^2+2*a*sigma*cos(theta)*sin(theta)^4-8*a*sigma*cos(theta)*sin(theta)^2*nu^2-6*a*sigma*cos(theta)*sin(theta)^2*nu+2*a*sigma*cos(theta)*sin(theta)^2)/(((cos(theta)^2+sin(theta)^2)/cos(theta)^2)^(1/2)*E)

(11)

>	Arr := limit(Diff_urr(1-cos(2*theta)), r = infinity);

(12)

>	G := (1/8)(int(Arr+Att, theta = 0 .. Pi/2))sigma*4;

-(1/8)*Pi*a*sigma^2*(4*nu^2-nu-5)/E

(13)

>	simplify(G)

-(1/8)*Pi*a*sigma^2*(4*nu^2-nu-5)/E

(14)

>

Download Derivative_implicit_maplePrime.mw

Thanks,

Adding Arrows to Parametrics to Show Direction...

Asked by: wlferguson19 90

January 09 2019

1 3

Hi,

Is there any way to add arrows to the parametrics plot to show the direction they are going? Thanks.

Finding Intersection of Parametrics;...

Asked by: wlferguson19 90

January 09 2019

0 3

Hi,

How would I tell maple to solve for the variable t? The answer should be 4.

[t-2,(t-2)^2]

[(3t/2)-4,(3t/2)-2]

I need to set them equal to eachother and then solve for t.

Thanks!

How to create recursive loops in procedure ...

Asked by: Magma 70

January 09 2019

1 4

Suppose that n is a positive integer number and [k1,k2,...,kn] is a list of non-negative integer numbers.

My Question: How to create the procedure proc([k1,k2,...,kn]) such that the output of this procedure is in the following form:

For i1 from 0 to k1 do

for i2 from 0 to k2 do

......................

for in from 0 to kn do

i1+i2+...+in

end do; end do;...end do;

How can this Maple application be generalized from...

Asked by: Earl 985

January 09 2019

1 2

This application describes the motion of a pendulum attached to a moving pivot, all in 2D.

https://www.maplesoft.com/applications/view.aspx?SID=4888

How can this situation be generalized to a pendulum attached to a pivot which moves along a 3D spacecurve?

Maplesoft at Joint Math 2019 in Baltimore

by: Daniel Skoog 1776 Product: Maple

January 09 2019

6 1

The Joint Mathematics Meetings are taking place next week (January 16 – 19) in Baltimore, Maryland, U.S.A. This will be the 102nd annual winter meeting of the Mathematical Association of America (MAA) and the 125th annual meeting of the American Mathematical Society (AMS).

Maplesoft will be exhibiting at booth #501 as well as in the networking area. Please stop by to chat with me and other members of the Maplesoft team, as well as to pick up some free Maplesoft swag or win some prizes.

This year we will be hosting a hands-on workshop on Maple: A Natural Way to Work with Math

This special event will take place on Thursday, January 17 at 6:00 -8:00 P.M. in the Holiday Ballroom 4 at the Hilton Baltimore.

There are also several other interesting Maple related talks:

MYMathApps Tutorials

MAA General Contributed Paper Session on Mathematics and Technology

Wednesday January 16, 2019, 1:00 p.m.-1:55 p.m.

Room 323, BCC
Matthew Weihing*, Texas A&M University
Philip B Yasskin, Texas A&M University

The Logic Behind the Turing Bombe's Role in Breaking Enigma.

MAA General Contributed Paper Session on Mathematics and Technology

Wednesday January 16, 2019, 1:00 p.m.-1:55 p.m.
Room 323, BCC
Neil Sigmon*, Radford University
Rick Klima, Appalachian State University

On a software accessible database of faithful representations of Lie algebras.

MAA General Contributed Paper Session on Algebra, I

Wednesday January 16, 2019, 2:15 p.m.-6:25 p.m.
Room 348, BCC
Cailin Foster*, Dixie State University

Discussion of Various Technical Strategies Used in College Math Teaching.

MAA Contributed Paper Session on Open Educational Resources: Combining Technological Tools and Innovative Practices to Improve Student Learning, IV

Friday January 18, 2019, 8:00 a.m.-10:55 a.m.
Room 303, BCC
Lina Wu*, Borough of Manhattan Community College-The City University of New York

An Enticing Simulation in Ordinary Differential Equations that predict tangible results.

MAA Contributed Paper Session on The Teaching and Learning of Undergraduate Ordinary Differential Equations

Friday January 18, 2019, 1:00 p.m.-4:55 p.m.
Room 324, BCC
Satyanand Singh*, New York City College of Technology of CUNY

An Effort to Assess the Impact a Modeling First Approach has in a Traditional Differential Equations Class.

AMS Special Session on Using Modeling to Motivate the Study of Differential Equations, I
Saturday January 19, 2019, 8:00 a.m.-11:50 a.m.

Room 336, BCC
Rosemary C Farley*, Manhattan College
Patrice G Tiffany, Manhattan College

If you are attending the Joint Math meetings this week and plan on presenting anything on Maple, please feel free to let me know and I'll update this list accordingly.

See you in Baltimore!

Daniel

Maple Product Manager

E-Mail Address:
Password:
Remember Me:	Automatically sign in on future visits

E-Mail Address:
Password:
Remember Me:	Automatically sign in on future visits

Ask a Question

Create a Post