Alger

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10 years, 174 days

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These are answers submitted by Alger

Look here in this link:

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

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 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 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

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

Start as in this worksheet below and add your boundary conditions.

Poisson_equation.mw

See also the link I gave you before where your work is I think the same

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

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

 

 

restart:

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

sol:=dsolve(ode);

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

g(0);

Work and no error in Maple 14 and 15

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

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

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)>);

@relativeway 

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;

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

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

                                     {x, y}

                           

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