rsolve({c*(n+beta)*P(n+1)/(-1+c)+(c*(x+beta)/(1-c)-x)*P(n)+n*P(n-1)/(-1+c)=0,P(0)=1}, P(n));
return rsolve({c*(n+beta)*P(n+1)/(-1+c)+(c*(x+beta)/(1-c)-x)*P(n)+n*P(n-1)/(-1+c)=0,P(0)=1}, P(n));

Any other method besides rsolve to solve genfun?

i discover it can solve variable coefficient and n coefficient too.

but sometimes n*P(n+1) make it can not solve, i have to remove n to P(n+1)

any paper describing the mechanism of rsolve, where it store the maple code of rsolve in maple 15?

Please write for me another code for the following problem:
Write the equation of the plane (P) passing through the line (d): x = -t -1, y = t + 3, z = 2*t + 2, (P) cuts the axis Ox at A and the axis Oy at B (A, B different form origin O) so that OA = 2OB.

This is my code

restart:

f:=(x,y,z)->a*x+b*y+c*z+d:

A:=f(m,0,0):

B:=f(0,n,0):

C:=f(-1,3,2):

When one equation has no exact solution. I use rkf45 to solve it.

How can I plot the error figures. Can I use the order:s1 := seq([tt[p], Xrre[p] - Xrre[p - 1]], p = 2..N):

restart;
with(plots):
with(LinearAlgebra):
Digits:=15:
w := 500:
k1 := 1:
k2 := sqrt(2):
k3 := -1-sqrt(2):
u1(t):= t/(t + 1)*exp(I * k1 * w * t) + cos(t) * exp(I * k2 * w * t) + exp(I * k3 * w * t):
u2(t):= exp(I * k1 * w * t) + exp(I * k2 * w * t) + t * exp(I * k3 * w * t):

That's the question... 'cause I've tried to open it using

./maple14/bin/maple -cw

like a thousand times and nothing happens...

In status box between "Evaluating..." and "Memory, Time, Mode" there is enough space. Would be nice to have there smth like "user-defined status" of ~100 symbols (or so). Instead of long printed log you could see status of current stage of long-term process.

Hi, I have a question about LPSolve. I definately have a linear polynomial as objective function, but LPSolve seems to think otherwise. My best guess is that it has to do with that my variables are indexed like that? Here is the code, I attached it as well.

Hello,

I have a question. I need to rewrite the following function

Vo:= (1/((1/R2+1/(1.5*10^6))*(1/(1/R2+1/(1.5*10^6))+2.45*10^3-R2)));

To this form

R2:= ....Vo....

How can i do this in maple? 

Re random is it possible for man or machine to create any two objects the same i suggest not,we may be only talking about such small differences that are possible undetectable (number of atoms per object) however a difference and if there is then each object will have it's own characteristics ie two lottery balls if you could study trillions of thows i'm suggesting you would see a pattern it may be so slight but not totally truly radom

It is never a good error message but I have not been able to find out whatit means.

Thanks,

Is there a way to plot a function in a multithreaded way?

I have a computing-intensive function (eg. integral over Hermitian-Polynoms), which I want to plot. I couldn't find a way to do this such, that all my cpu-cores are used.

Any suggestions?

Thanks!

Dear All,

Thanks for the previous support and help.

I am trying to solve a poisson equation, but I am getting Bessel function in the solution. Derivation of this equation with respect to 'r' gives more complex form of Bessel function.

Actually I am not sure should I continue with the problem and Maple will take care of everything or I have to modify in the solution.  

Equation 9 and Equation 11 needs to...

Dear all,

Maple gives me the complete solution to the attached PDE. This solution is expressed in terms of KummerU and KummerM. What I want is to obtain the solution for real numbers only, i.e. I want to get rid of the possible imaginary numbers in the solution. This implies that the separation constant, _c1, must be less or equal to 2r. 

How can I do this?

Furthermore, the four arbitrary constants will need to be determined through boundary conditions....

(16y⁶)^1/2        - how can i simplfy this equation im sooo confused plz help xxx

I used the command minimize such

minimize(3*x^2+2*m*x+2*m^2-3*m-6,1 <= x)

But it can not. How can i find the least of the function g:=x>3*x^2+2*m*x+2*m^2-3*m-6,  here x >=1.

Please help me. Thank you.