Advanced Search

Question: "summand is singular"

Question:"summand is singular"

p.lindsay 4 Maple
April 02 2009
0
hi

I'm looking at Fourier coefficients:

Dn:= (2/L)* int(f*sin(n*Pi*x/L), x=0..L); where L = Pi

and the series

u:= Sum (Dn * sin(n*Pi*x/L)*cos(n*Pi*t/L), n=1..infinity);

When I try to sum the series:

u:= sum (Dn * sin(n*Pi*x/L)*cos(n*Pi*t/L), n=1..10);

I get the error

Error, (in NumericRange) summand is singular in the interval of summation

Now I can see the 0 / 0 thing, but the first term should be 1 and the other terms zero. The first term is 1 if you take limit as n->1

So I have to explicitly state that:

D1:= (2/L)* int(f*sin(Pi*x/L), x=0..L);

and by inspection see u = sin(x)*cos(t)

Can anyone help me past this block please ?

thanks

Peter

Please Wait...