Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Hello,

how can I tell Maple that "u" has the dimension of volts? 

Thanks for every good advice.

 

Dear MaplePrimes team,

 

For display a beautiful parametric surface with hole (or pierced surface), unfortunately unlike CAD (computer-aided design) tool, Maple requires a large grid number. Therefore the computing is too long and requires huge RAM only with grid of [500,500] on a modern computer.

      

Fig. 1: For a complex surface such an aircraft fuselage here, Needs high grid, therefore, the compilation is too long with modern computer (Intel Core i7, 2.4 GHz CPU, 16 GB RAM). If displays supplementary windows holes on fuselage, the software may bug when the RAM is full.

 

Question 1:

In matter of low cost computation, do you have better solution to create a fast surface with smooth hole from given a parametric surface equation S(x,y) and given any convex/concave-closed parametric curve C(t) that is projected on surface S(x,y)?

Fig. 2: These surfaces are been created in Maple 2016 from given initial parametric surface equation and curve projected on the surface. But requires huge grid, huge RAM, and long time to get smooth hole when free CAD tools are fast and low RAM.

 

Question 2:

Because I noted that the Maple’s view option renderers smooth edges of any surface. My question is: Is it possible to use the internal structure of display/view used by Maple to apply at free boundaries? I think it will be possible to create many class of view, for examples: cylindrical sector, spherical sector views or more rather to be limited with conventional cubic view. And thus, I think it will be possible to create own smooth hole or multiple-hole on surface and get smooth pierced surface.

 

I know that almost of technicians/engineers will recommend me to use a CAD tool to create surface with hole. But the objective here is to keep purely a mathematical mind and work with exact equations (analytic equations).

On a free CAD tool, even the complex holes are created very fast only with low RAM. Why not on Maple?

 

Best.

 

Guy.

Hi everyone,

as a very basis example I've tried to get the equipotential lines of an 1/r² potential in the x-y-plane using contourplot. However, I obtained the following very strange result. Do you get the same mistake or is there something wrong on my end?

Thanks in advance,
Sören


 

with(plots):

plots:-contourplot(1/(x^2+y^2), x = -2 .. 2, y = -2 .. 2)

 

plots:-contourplot(1/(x^2+y^2), x = -2 .. 2, y = -2 .. 2, contours = 5, numpoints = 100)

 

plots:-contourplot(1/(x^2+y^2), x = -2 .. 2, y = -2 .. 2, contours = 5, numpoints = 1000)

 

plots:-contourplot(1/(x^2+y^2), x = -2 .. 2, y = -2 .. 2, contours = 5, numpoints = 10000)

 

``

Download contourplot.mw

Hi everybody,

My worksheet begins with these two instructions (this is a notional example)
Digits := 25:
interface(displayprecision = 5):

Then the instruction  evalf(1/3+x/6)  returns  0.33333+0.16667x … which is what I had expected.

I now  want to do a formatted print of  evalf(1/3+x/6)  in some file.
Unfortunately the command  

 fprintf(fd, "%a\n", evalf(1/3+x/6)) 

generates a print with the full digits precision (0.333…..33 with 25 "3" for instance).

How could I force fprintf to print the floating point numbers with the number of digits set by interface(displayprecision = 5) ?
 
(PS : for now I have written some kind of “white elephant” stuff which works well only for very simple algebraic expressions)

Any suggestion or answer will be greatly appreciated

 


 

Hello everyone, why is maple not able to solve this easy system of equations?

eq1 := 1 = abs(1/(I*Pi*z*(1+I*Pi*a)));
eq2 := -(5/9)*Pi = argument(1/(I*Pi*z*(1+I*Pi*a)));

`assuming`([fsolve({eq1, eq2}, {a, z})], [a::real, z::real]);

 Thanks for helping me! Fabian

Given

Tour2:=[[[1, 2, 3, 4]], [[1, 2], [1, 3, 4]], [[1, 3], [1, 2, 4]], [[1, 4], [1, 2, 3]], [[1, 2], [1, 3], [1, 4]]]:

Kitonums code

seq(max(seq(add(d[s[i]]*x[s[i],s[i+1]], i=1..nops(s)-1) , s=t))<=K, t=Tour2); produces

18*x[1, 2]+26*x[2, 3]+11*x[3, 4] <= K, max(18*x[1, 2], 26*x[1, 3]+11*x[3, 4]) <= K, max(26*x[1, 3], 18*x[1, 2]+11*x[2, 4]) <= K, max(11*x[1, 4], 18*x[1, 2]+26*x[2, 3]) <= K, max(18*x[1, 2], 26*x[1, 3], 11*x[1, 4]) <= K

correctly. The problem is I want to pass this through LPSolve and it doesn't like it. So what I need is to convert it to a constraint it will accept. I found a method on stackexchange,

http://stackoverflow.com/questions/10792139/using-min-max-within-an-integer-linear-program

i'm not sure it even works.i would be great if a procedure could could convert the max's in the above expression, or change the code in bold.

max(6*x[1, 4],4*x[1, 2]+3*x[2, 3]), where x[i,j] are binary vars.
x4 >= 6*x[1, 4];
x4 >= 4*x[1, 2]+3*x[2, 3];
Objective:= x4;

 

how to count the number of decimal places if input a decimal number

Error, (in unapply) variables must be unique and of type name

The command

fd:=fopen("C:/temp/",'WRITE');

suddenly does not work. I keep getting a 'permission denied' error message. I haven't had to write anything to a file for a while so do you think the recent windows update screwed something up? In fact, I can't seem to write any file anywhere. It was working weeks ago. My apology if this ends up being a lame question. :)

 

Hi , I have question about poincare's theorem which needs to change the veriable and that transform the coordinates , but when I change the variable from x to y if doesn't defined any suggestion ? Please 

How do I assume variables as  Matrixs for calculation?
Please help me to solve this problem :

assume(M11::Matrix);
assume(M12::Matrix);
assume(M21::Matrix);
assume(M22::Matrix);
assume(A::Matrix);


A := M12*M21+M11;
                         M12 M21 + M11
solve(A = 0, M12);
Warning, solve may be ignoring assumptions on the input variables.
                               M11
                             - ---
                               M21
solve(A = 0, M12, useassumptions = true);
                               M11
                             - ---
                               M21


How can I get solution being a matrix?
Thank you very much !!!

Each day
a:=a+5% of a -(0.1% of a+b)-10

b:=b+5%of b -(0.1% of a+b)-10
In how many days i have a+b=0?

Sirs.

Probably a brain fade, but I cant seem to code what i want.

constraints.mw
 

Tour2:=[[[1, 2, 3, 4]], [[1, 2], [1, 3, 4]], [[1, 3], [1, 2, 4]], [[1, 4], [1, 2, 3]], [[1, 2], [1, 3], [1, 4]]];M:=nops(Tour2):

[[[1, 2, 3, 4]], [[1, 2], [1, 3, 4]], [[1, 3], [1, 2, 4]], [[1, 4], [1, 2, 3]], [[1, 2], [1, 3], [1, 4]]]

(1)

interface(rtablesize=M):
  maxEnt:=max([seq(nops(Tour2[i]),i=1..M)]):
  Tours_Distances := Matrix
                     ( maxEnt,
                       M,
                       [ seq
                         ( [ seq
                             ( `if`( numelems(Tour2[i])>=j,
                                     d[i]*x[op(Tour2[i,j])]<=K,
                                    0
                                   ),
                               i=1..M
                             )
                           ],
                           j=1..maxEnt
                         )
                       ]
                     );

Tours_Distances := Matrix(3, 5, {(1, 1) = d[1]*x[1, 2, 3, 4] <= K, (1, 2) = d[2]*x[1, 2] <= K, (1, 3) = d[3]*x[1, 3] <= K, (1, 4) = d[4]*x[1, 4] <= K, (1, 5) = d[5]*x[1, 2] <= K, (2, 1) = 0, (2, 2) = d[2]*x[1, 3, 4] <= K, (2, 3) = d[3]*x[1, 2, 4] <= K, (2, 4) = d[4]*x[1, 2, 3] <= K, (2, 5) = d[5]*x[1, 3] <= K, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (3, 5) = d[5]*x[1, 4] <= K})

(2)

convert( (2), 'list', 'nested' );

[[d[1]*x[1, 2, 3, 4] <= K, d[2]*x[1, 2] <= K, d[3]*x[1, 3] <= K, d[4]*x[1, 4] <= K, d[5]*x[1, 2] <= K], [0, d[2]*x[1, 3, 4] <= K, d[3]*x[1, 2, 4] <= K, d[4]*x[1, 2, 3] <= K, d[5]*x[1, 3] <= K], [0, 0, 0, 0, d[5]*x[1, 4] <= K]]

(3)

 

But what I want is:

d[1]*x[1,2]+d[2]*x[2,3]+d[3]*x[3,4]<=K,d[1]*x[1,2]<=K,d[1]*x[1,3]+d[3]*x[3,4]<=K;      #.....etc.

 

 

 

 

 

d[1]*x[1, 2]+d[2]*x[2, 3]+d[3]*x[3, 4] <= K, d[1]*x[1, 2] <= K, d[1]*x[1, 3]+d[3]*x[3, 4] <= K

(4)

NULL


 

Download constraints.mw

 

Hello guys,

I´ve a little bit problem by plotting an pyramid with an triangle in it. When I want to display both

parts, in the drawing the left and the right part and the top of the pyramid is missing. What is the problem ???
Abituraufgabe_2016_B2_Pyramide.mw

Thanks

I wish to express the Maxwell equations in potential fields deriving from a Lagrangian in cartesian coordinates, but expressed in vectorial form; But I am receiving an message error that I do not understand.

Field_A-MaplePrimes.mw
 

 

Initial Definitions:

 

``

I wish in this work to express the Maxwell equations deriving from the Lagrangian in the fields phiand "A,"

phi

(1.1)

but expressed in vectorial form;

 

restart; clear; with(Physics); with(Physics[Vectors]); with(Library)interface(imaginaryunit = I)

clear

 

I

(1.2)

Physics:-Coordinates(X = [t, x, y, z])

`Detected \`t\`, the time variable, in position 1. Changing the signature of the spacetime metric accordingly, to: + - - - `

 

`Default differentiation variables for d_, D_ and dAlembertian are: `*{X = (t, x, y, z)}

 

`Systems of spacetime Coordinates are: `*{X = (t, x, y, z)}

 

{X}

(1.3)

Physics:-Vectors:-Setup(math = true, Physics:-Vectors:-diff = X)

`* Partial match of  'math' against keyword 'mathematicalnotation'`

 

`* Partial match of  'Physics:-Vectors:-diff' against keyword 'differentiationvariables'`

 

`Default differentiation variables for d_, D_ and dAlembertian are: `*{X = (t, x, y, z)}

 

[differentiationvariables = [X], mathematicalnotation = true]

(1.4)

``

Some definitions

 

 

Defining the Maxwell tensor

 

Physics:-Define(F):``

`Defined objects with tensor properties`

(2.1)

 

Defining the field A with their components:

 

A[mu] = Vector(4, [phi(X), A__1(X), A__2(X), A__3(X)]); Define(%)

A[mu] = Vector[column](%id = 18446744074366759750)

 

`Defined objects with tensor properties`

 

{F, A[mu], Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], Physics:-g_[mu, nu], Physics:-KroneckerDelta[mu, nu], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}

(2.2)

Physics:-CompactDisplay(phi(X), A__1(X), A__2(X), A__3(X))

phi(t, x, y, z)*`will now be displayed as`*phi

 

A__1(t, x, y, z)*`will now be displayed as`*A__1

 

A__2(t, x, y, z)*`will now be displayed as`*A__2

 

A__3(t, x, y, z)*`will now be displayed as`*A__3

(2.3)

 

Applying:

 

 

``

F[alpha, beta] := Physics:-Vectors:-`+`(Physics:-d_[beta](A[alpha]), -Physics:-d_[alpha](A[beta]));

Physics:-d_[beta](A[alpha], [X])-Physics:-d_[alpha](A[beta], [X])

(3.1)

``

 

NULL

NULL

Term 1:

 

 

eq1 := Physics:-d_[alpha](F[alpha, beta])

Physics:-d_[alpha](Physics:-d_[beta](A[`~alpha`], [X]), [X])-Physics:-dAlembertian(A[beta], [X])

(4.1)

Physics:-SumOverRepeatedIndices(Physics:-d_[alpha](Physics:-d_[beta](A[`~alpha`], [X]), [X])-Physics:-dAlembertian(A[beta], [X]))

Physics:-d_[beta](diff(phi(X), t), [X])-Physics:-d_[beta](diff(A__1(X), x), [X])-Physics:-d_[beta](diff(A__2(X), y), [X])-Physics:-d_[beta](diff(A__3(X), z), [X])-Physics:-dAlembertian(A[beta], [X])

(4.2)

Physics:-SubstituteTensorIndices(beta = 4, Physics:-d_[beta](diff(phi(X), t), [X])-Physics:-d_[beta](diff(A__1(X), x), [X])-Physics:-d_[beta](diff(A__2(X), y), [X])-Physics:-d_[beta](diff(A__3(X), z), [X])-Physics:-dAlembertian(A[beta], [X]))

diff(diff(phi(X), t), z)-(diff(diff(A__1(X), x), z))-(diff(diff(A__2(X), y), z))-(diff(diff(A__3(X), z), z))-Physics:-dAlembertian(A__3(X), [X])

(4.3)

``

Physics:-SubstituteTensorIndices(beta = 1, Physics:-d_[beta](diff(phi(X), t), [X])-Physics:-d_[beta](diff(A__1(X), x), [X])-Physics:-d_[beta](diff(A__2(X), y), [X])-Physics:-d_[beta](diff(A__3(X), z), [X])-Physics:-dAlembertian(A[beta], [X]))

diff(diff(phi(X), t), t)-(diff(diff(A__1(X), t), x))-(diff(diff(A__2(X), t), y))-(diff(diff(A__3(X), t), z))-Physics:-dAlembertian(phi(X), [X])

(4.4)

SubstituteTensorIndices(beta = 2, Physics[d_][beta](diff(phi(X), t), [X])-Physics[d_][beta](diff(A__1(X), x), [X])-Physics[d_][beta](diff(A__2(X), y), [X])-Physics[d_][beta](diff(A__3(X), z), [X])-Physics[dAlembertian](A[beta], [X]))

Error, (in dchange/info) the number of new and old independent variables must be the same. Found {x, y} as new, while {} as old

 

SubstituteTensorIndices(beta = 3, Physics[d_][beta](diff(phi(X), t), [X])-Physics[d_][beta](diff(A__1(X), x), [X])-Physics[d_][beta](diff(A__2(X), y), [X])-Physics[d_][beta](diff(A__3(X), z), [X])-Physics[dAlembertian](A[beta], [X]))

Error, (in dchange/info) the number of new and old independent variables must be the same. Found {x, y} as new, while {} as old

 

Physics:-SubstituteTensorIndices(beta = 4, Physics:-d_[beta](diff(phi(X), t), [X])-Physics:-d_[beta](diff(A__1(X), x), [X])-Physics:-d_[beta](diff(A__2(X), y), [X])-Physics:-d_[beta](diff(A__3(X), z), [X])-Physics:-dAlembertian(A[beta], [X]))

diff(diff(phi(X), t), z)-(diff(diff(A__1(X), x), z))-(diff(diff(A__2(X), y), z))-(diff(diff(A__3(X), z), z))-Physics:-dAlembertian(A__3(X), [X])

(4.5)

``

What is the origin of the two error messages above? I did some wrong definition? Why this works in t and z but not in x and y?

 

NULL


 

Download Field_A-MaplePrimes.mw

 

 

 

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