Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

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 ??

Thanks.

Mike Graber

x:=Proc
Local a,
b,

C;
use XML Tools in 
end use;

end proc;

Hello,

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:

>with(Physics):

>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".

Regards

How I can pdsolve these equations in toroidal coordinates?

Initial conditions are arbitrary.

Please see attached pdf .

Thanks

Where.pdf

Weston1958

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

vars:=[x,y,z];

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 

map(w->coeffs(w,vars),model7);

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)

 

Ed

 

 

Dear all

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

laurent_series.mw

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.

Hello,

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 

Hi
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?

TIA
 

restart:

vars   := [a, b__c, t[1]];

[a, b__c, t[1]]

(1)

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

vars;

[1, 2, 3]

(2)

 


 

Download Assignement.mw

Hi, 

I have a 2nd order linear ODE that I solved firstly in a formal way.
The plot of the solution contains a "hole" which, in fact, corresponds to the range where the formal solution is not real.
This seemed strange to me as I expected a real solution.
So I solved this ODE numerically and this gave me a real solution.
Could you please explain that to me ?

Thanks in advance

PS : These results are obtained with Standard Worksheet Interface, Maple 2015.2, Mac OS X, December 21 2015 Build ID 1097895

restart:

with(plots):

perturbation := .9*(diff(epsilon(t), t, t)) = (-28.67085587*piecewise(0.2000000000e-1*t < 0.8e-1, 0., 0.2000000000e-1*t < .12, -0.8e-1+0.2000000000e-1*t, 0.2000000000e-1*t < .14, .16-0.2000000000e-1*t, 0.2e-1)-0.7645561571e-1*t-1.548363347)*(diff(epsilon(t), t))-0.7596363e-1*piecewise(0.2000000000e-1*t < 0.8e-1, 0., 0.2000000000e-1*t < .12, -0.8e-1+0.2000000000e-1*t, 0.2000000000e-1*t < .14, .16-0.2000000000e-1*t, 0.2e-1)-0.202569683e-3*t+0.26997403e-1-10.10*epsilon(t)

perturbation := .9*(diff(`&epsilon;`(t), t, t)) = (-28.67085587*piecewise(0.2000000000e-1*t < 0.8e-1, 0., 0.2000000000e-1*t < .12, -0.8e-1+0.2000000000e-1*t, 0.2000000000e-1*t < .14, .16-0.2000000000e-1*t, 0.2e-1)-0.7645561571e-1*t-1.548363347)*(diff(`&epsilon;`(t), t))-0.7596363e-1*piecewise(0.2000000000e-1*t < 0.8e-1, 0., 0.2000000000e-1*t < .12, -0.8e-1+0.2000000000e-1*t, 0.2000000000e-1*t < .14, .16-0.2000000000e-1*t, 0.2e-1)-0.202569683e-3*t+0.26997403e-1-10.10*`&epsilon;`(t)

(1)

Digits:=10:
EPS := rhs( dsolve({evalf(perturbation), epsilon(0)=0, D(epsilon)(0)=0}, epsilon(t)) ):

# a lengthy expression that involves Kummer's special functions

plot(EPS, t=0..10, gridlines=true);


Digits:=20:
evalf(eval(EPS, t=5));
Digits:=10:

 

-0.26529838913395427314e88+0.10340147081332027605e89*I

(2)

numsol := dsolve({perturbation, epsilon(0)=0, D(epsilon)(0)=0}, numeric):

odeplot(numsol, [t, epsilon(t)], t=0..10, gridlines=true, color=red)

 

 


 

Download dsolve_vs_numeric_dsolve.mw

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