Hi

I need to evalaute the Lauricella function of type F_A  for arbitrary value of   n , say around 2 to 10 or so. I think maple does not provide the ready function for the same and its is restricted to single variable series. Is there any method for computing the Lauricella functions in maple?

Thanks for any help

Vish

where can i get a german language pack? is one available?

I recently got my copy of maple in the university, but i cant choose german as my language.

To make everything a little easier (cause im new to maple), i was trying to find a language pack.  I hope there is a pack for download anywhere.

Thanks for your help.

Julian

Hello there, I have a problem solving the differential equation based on decomposition method. There are 3 operator which is N, R, G and u. N=4, g=-77, R=79. Below is the procedure, and yet there is an error at the last end statement. Please help asap.

Why does a^(3/2)*(1/a)^(3/2) not return 1 in maple? Sorry for the basic question, I'm just not sure what I'm missing.

I'm supposed to use a CAS to integrate (4*v*(m/(2*Pi*k*T))^(3/2)*Pi*v^2)*exp(-m*v^2/(2*k*T)) for v from -infinity to infinity. I know the answer should be sqrt(8*k*t/(Pi*m)), but I can't seem to get Maple to evaluate the limit once it integrates the expression. Any help is greatly appreciated. Let me know if I need to provide more information, I'm a bit new to the forums :)

Hi

I found the roots of a polynomial t^6+t^3+1 by hand, now i want to see if these values in the form exp(a*I/(b*Pi)) satisfy the polynomial, how can i do this?

Hey all. I'm looking at trying to generate a complete list of all maps from the integer plane to itself that have a maximal coordinate change (plus or minus) of n. Basically, I devised two wave functions that travel componentwise around the 8*n points d_inf=n from the origin, and am trying to use them to list every single equivalent map with such a metric-distance, from a point (x,y). Listed here is my current code, the problem I have on execution is that the i terms evaluate to that name as opposed to the assigned value.

Hi there

i'm trying to plot the phase portrait of a 2nd order ODE and i just cant get those damn arrows to come up. i've managed some solution curves but i'd like to see the phase portrait and maybe make an animation to see how it changes but i'll worry about that once i can get the arrows.

Dear all:

    I want to plot a function(say, z=x^2+y^2) on a nonsqure region(say, x^2+y^2<=1). What I did is to manually define a procedure so that it returns the function value if the sample point is inside the defined regioin, null otherwise.

region:=proc(x,y)
  if x^2 + y^2 <= 1 then
      return x^2+y^2;
   else
      return null;
   end if:
 
 end proc;

any info on the Jacobi iterative scheme

to get me started. please

consider the system

2x1 + x2     = 1

 x1 + 4x2 + x3 = -1

           x2 + 2x3=3

Find x using the Jacobi iterative scheme with x= (0,0,0)

My colleague at the university next week give a lecture where he tells about the Hardy-Ramanujan partition functions, the usual

notations are p(n) and q(n). He asked me, could I calculate by Maple a definite integral, because the answer was yes, then he

would say "by Maple".  I tried it in different ways, but no success. Could someone calculate it?

The exact result is  -1/2 ln (2 Pi) .

hey guys.. ive been trying to get the root of the following equation.. but watever i do i am getting only the positive value.. and i have another equation which has 5 answers.. but only 1 is being shown.. can u help me out??

>f0:=x->x/(sqrt(x^2 + cos(x-1))):
>f1:=D(f0):
>f2:=D(f1):
>a:=fsolve(f1(x)=0,x);
                                a:=3.639062685
 

hi, how do i perform the following operation and simplfy?

(4a- 3ab + 2b) + [ (2a- 4b) - ab ]

Hello all,

When I tried to plot with "semilogplot" command, I've found the ticks on the x-axis and the vertical grids do not line up. This is with Maple 10. Would somebody tell me if there is a way to line them up?

with(plots):

semilogplot(log[10](x), x=1..1000, gridlines=true);

Thank you.

Chin Li