MaplePrimes Questions

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

kindly send me OHAM source code in Maple for PDEs. I need it.

Hello

I have the following set of coefficients 

coef7 := [-1, 2, alpha[1, 2], alpha[2, 6], (17*RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)+17)/(alpha[1, 2]*alpha[2, 6]), -17/alpha[2, 6], -33/(32*RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)), -(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]), -RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)-5/4, RootOf(64*_Z^3+80*_Z^2+1104*_Z+561)]

 

Considering that alpha[1,2] and alpha[2,6] are always real, how can I extract only the real solution from coef7?  

Many thanks.

 

Ed

Hello everyone!

Is it possible to find index of first non-negative or first positive Vector's (or Array's) element with the help of one Maple build-in comand (for example: use something from ArrayTools or ListTools)?

 

In maple, how to change strings style? Thanks!  

In Mathematica , we can choose Style to change strings style.

Style["A", Blue, Italic, 24]

 

source of the  problem : I make an User Interface.

the color of   "mass "  may consider be changed to  red . 

 

 

"with(LinearAlgebra):  T := proc(n::integer)  local t0, tn, i, t, t1;  if (n>0) then t0 := Matrix(1, 1, 1); tn := Matrix(1,1,0);  for i from 1 to n do local z := 2^i;  t := Matrix([[Add(t0, tn), x .~ t0], [t0, ZeroMatrix(z/(2))]]);  t1 := Matrix([[y .~ t0, ZeroMatrix(z/(2))], [ZeroMatrix(z/(2)), ZeroMatrix(z/(2))]]);  t0 := t; tn := t1;  end do;  else 'T'(n) end if; t0; end proc;"

Error, unterminated loop

"with(LinearAlgebra):  T := proc(n::integer)  local t0, tn, i, t, t1;  if (n>0) then t0 := Matrix(1, 1, 1); tn := Matrix(1,1,0);  for i from 1 to n do local z := 2^i;  t := Matrix([[Add(t0, tn), x .~ t0], [t0, ZeroMatrix(z/2)]]);  t1 := Matrix([[y .~ t0, ZeroMatrix(z/2)], [ZeroMatrix(z/2), ZeroMatrix(z/2)]]);  t0 := t; tn := t1;  end do;  else 'T'(n) end if; t0; end proc;"

 

T(2)

T(2)

(1)

``

 

Download G2.mw

Hello everyone. I was trying to code the recurrence relation given in the image below (without taking the mth power):

 

But, as I run the program, I am getting an "unterminated loop" error. Can someone please point out the mistake(s)?

Is there any way to write the expression into the ratio of simple determinants? i..e,

 

Hi,

 

How do we remove the module name functions that massively pollute the output after using with(Units[Standard]):

So that I can have something presentable like this :

Thanks!

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