## 499 Reputation

10 years, 364 days

## Fourier coefficients...

http://www.mapleprimes.com/questions/121551-Fourier-Serie-And-Discrete-Fourier-Transform

## Example...

restart:

s[1]:=x->diff(f[1](x),x,x)+f[1](x)=cos(x); s[2]:=x->diff(f[2](x),x,x)+f[2](x)=sin(x);

for ii from 1 to 2 do ;
ics[ii]:=f[ii](0)=0,D(f[ii])(0)=0;
ans := dsolve({s[ii](x), ics[ii]});
f[ii]:=unapply(rhs(ans),x);
od;

## for...

for ii from 1 to 10 do
ii*2:
f[ii]:=%:
end do;

OR

for ii from 1 to 10 do
ii*2:
f[ii]:=unapply(%,x):
end do;

## No error...

No problem for me with maple 15

restart;(a+3)^17;expand(%);

a^17+51*a^16+1224*a^15+18360*a^14+192780*a^13+1503684*a^12+

9022104*a^11+42532776*a^10+159497910*a^9+478493730*a^8+1148384952*a^7+

2192371272*a^6+3288556908*a^5+3794488740*a^4+3252418920*a^3+1951451352*a^2+

731794257*a+129140163

## Special question...

Thanks to send me directely by email your questions in relation with my post in the application center.

I will answer as soon as possible

Thank you

## Poisson equation and seprartion of varia...

Poisson_equation.mw

## n:=2*k+1...

In this type of problems, you can solve the Laplacian using k with:

M[r]:=Sum(M[2*k+1]*cos((2*k+1)*p*theta), k=0..infinity):

The potential u(r,theta) has a general form with Summation in k

I think M[r] is the magnetization of permanent magnet. Read also this application in application center:

http://www.maplesoft.com/applications/view.aspx?SID=120250

## Explore...

I don't know if command "Explore" exist in Maple 13, you can use

Explore(plot(cos(a*x^2)+x));   # for example with skip x

## One way...

restart:

ode:=diff(g(x),x)-g(x)=0;

sol:=dsolve(ode);

g:=unapply(rhs(sol),x);

g(0);

## Work in Maple 14 & 15...

Work and no error in Maple 14 and 15

## One way with piecewise...

f:=n->piecewise(And(n::integer, n::even),f3(n),0);

Statistics[ColumnGraph](<seq(f(n),n=0..20)>);

## Another way...

If I understand, what you want is:

f3 := (n) -> (2*n)!*(lambda)^n/(((n)!^2)*2^(2*n)*(lambda+1)^(n+0.5));
lambda:=10;

Statistics[ColumnGraph](<seq(f3(n),n=0..20)>);

## like this ?...

f:=j->proc(x,y,j)
x+y^(j)
end proc;

for j from 1 to 10 do
f(j)(x,y,j);
end do;

## or with procedure...

for j from 1 to 2 do
f[j]:=proc(x,y,j)
x+y^(j)
end proc;
od;

f[1](x,y,1);

x+y

f[2](x,y,2);

x+y^2

## function...

f:=(x,y)->x^2+y^2;indets(f(x,y));

{x, y}

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