MaplePrimes Questions

Hello everyone .

I asked Maple to compute a definite integral of a function, numerically. Unfortunately it does not give me any values . I briefly explain what I did.

As you can see in the picture first I define a real valued function BW_dilog(x) (with complex domain), where the function dilog  already exists in maple. Then I defined another function, V, (with real domain and codomain ). Now I need the approximate value of the bi integral of V, on a triangle. The Maple file is attached.

I would be grateful if any one could help me.

how do i Import part or all  PDF files into maple 

I believe this was easier in older versions of maple or worked partially.

I initialized a variable name using another variable name and := . I thought that Maple would assign the value, but I soon learned by debug that the address was assigned. This led to temporary problems.


Where can I learn the rules that Maple follows for assignment by value or address ?? Do I always have to initialize with a real number, like 0, to be safe ??


Mike Graber

Local a,

use XML Tools in 
end use;

end proc;


I have a problem, please help me to solve it.

I am trying to extract coefficients of some polynomial expressions but I receive Error.

For example:


>Setup(mathematicalnotation = true)

>S := a*(x^2)+b*x+c*(1/(x-h)):

>Coefficients(S, x, 1)

>Error, (in Physics:-Coefficients) unable to compute coeff.

**** Note that if instead of the term (1/(x-h)) we have (1/x), then we get the true result. The error occurs only when we have some extra things in the denominator namely "-h".


How I can pdsolve these equations in toroidal coordinates?

Initial conditions are arbitrary.

Please see attached pdf .




Can old maple version saved .m files in window be readable in maple 2015 Linux version?

Basically I need to include an if condition that assigns one set of values to a set of variables if the while loop completes a cycle without me interrupting it, and another set of values in the event I have pressed the interrupt button.

Is this able to be done?

Hello (again)

I thought I won't need help with that type of question but I came across an example that says otherwise.  Here it is


model7 := [x*(-RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)-5/4)+y*alpha[1, 2]-33/(32*RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)), x*z*alpha[2, 6]+y*RootOf(64*_Z^3+80*_Z^2+1104*_Z+561), x^2*(17*RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)+17)/(alpha[1, 2]*alpha[2, 6])-17*x*y/alpha[2, 6]+2*z*x-z-(163/32+RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)^2+5*RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)*(1/4))/(alpha[1, 2]*alpha[2, 6])]

then I issued the command 


to get 


[-33/(32*RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)), -RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)-5/4, alpha[1, 2], RootOf(64*_Z^3+80*_Z^2+1104*_Z+561), alpha[2, 6], -(163/32+RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)^2+5*RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)*(1/4))/(alpha[1, 2]*alpha[2, 6]), -1, (17*RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)+17)/(alpha[1, 2]*alpha[2, 6]), -17/alpha[2, 6], 2]

clearly the order does not follow model7's.  

I have also tried

[seq(coeffs(expand(model7[i]), indets(model7[i], suffixed({vars[]}))),i=1..nops(model7))];

Is there a solution to it?

Many thanks (again)





Dear all

I would like to compute a Laurent of a given function at an isolated point.

thanks for your help

I would like to analyse the graph I made and I cannot seem to find any command that gives me the average path length of an unweighted graph, is it perhaps referred as a different command? As I have tried searching Mean path length, distance and other derivatives with no useful result.


All too often I encounter an annoying issue, and when it happens it makes Maple slower than just scribbling on paper.  Unfortunately I don't understand why it happens or when it is going to happen so it creates a huge waste of time.

While most of the time expressions behave as expected, other times the exact same expression behaves as if it is something other than what is shown on screen.  I just ran into such an issue as an example:

What am I missing here? Why would phi not be interpreted as phi?

The only thing I could think of not being a Maple expert is assumptions.  I found I have an assumption earlier on phi (assume(phi>0,phi<45) so I removed it, but inexplicably(to me) that only fixes the first 2 lines, the last 2 lines still have the same issue, why I can't understand.  Why would having an assumption on phi, prevent me from evaluating the expression at some value of phi? The provided value is even within the assumed range.  And in either case, why can't I declare a value for phi and have that be understood?

Sorry for what I suspect is a dumb question, and thanks in advance.

EDITED to correct = -> :=

How to learn Maple Programming effectively, whether Python will help?

My objective is to write a procedure that will read data from xml at different mentioned condition, read ICBO value at different condition of VGS i.e ICBO value at VGS=5V

BJT_ICBO := proc(parsedXML, VGS at condition )


end proc;

can some one help me to write procedure 

I inadvertently discovered that it was possible to create variables in a kind of dynamic way.
I wonder if the way i did  is something which is documented somewhere?



vars   := [a, b__c, t[1]];

[a, b__c, t[1]]


k := 1:
for i in vars do
  ``||i := k;
  k := k+1:
end do:


[1, 2, 3]





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