Sum (sin(Pi*n)/(-1+n^2) , n=1infinity). n=1,2,3

[GerardEncina]

Hi everybody!

Solving a PDE for the damped wave equation I get with a solutions that part of it have the expression:

Sum (sin(Pi*n)/(-1+n^2) , n=1..infinity). n=1,2,3...

When calculating it with Maple I get this is equal to -1/2*Pi.

Can someone of you explain me why?

Thanks a lot!

 

[LCKurtz]

Hi everybody!

Solving a PDE for the damped wave equation I get with a solutions that part of it have the expression:

Sum (sin(Pi*n)/(-1+n^2) , n=1..infinity). n=1,2,3...

When calculating it with Maple I get this is equal to -1/2*Pi.

Can someone of you explain me why?

Thanks a lot!

Did you get that expression from a Fourier Series expansion? Sometimes these sums arise by simply evaluating the FS at some point.

 

[GerardEncina]

Yes, I'm trying to solve the PDE for the damped wave equations defined like:

u_tt+u_t-u_xx=0,

With Initial Conditions: u(x,0)=0 , u_t (x,0)=sin⁡(x)
And Boundary Conditions u(0,t)=u(Pi,t)=0

Finally I end with the solution u(x,t)=∑(n=1 to ∞) [exp^(-t/2)*(-4sin⁡(Pi*n))*sin⁡(sqrt(4n^2-1)*t/2)*sin(nx) / (Pi*sqrt(4n^2-1)*(-1+n^2))]

From here, there is the expression sin(Pi*n)/(-1+n^2), that for n=1 is 0. But solving it with Maple it gives u(x,t)=2/(3*sqrt(3))*exp^(-t/2)*sin(sqrt(3)*t/2)*sin(x).

I've also found that limit [when n-->1] of sin(Pi*n)/(-1+n^2) = -Pi/2; that plugged on the series give what Maple says.

But I'm still not sure if it is correct or not

Thanks!

 

[Mute]

Yes, I'm trying to solve the PDE for the damped wave equations defined like:

u_tt+u_t-u_xx=0,

With Initial Conditions: u(x,0)=0 , u_t (x,0)=sin⁡(x)
And Boundary Conditions u(0,t)=u(Pi,t)=0

Finally I end with the solution u(x,t)=∑(n=1 to ∞) [exp^(-t/2)*(-4sin⁡(Pi*n))*sin⁡(sqrt(4n^2-1)*t/2)*sin(nx) / (Pi*sqrt(4n^2-1)*(-1+n^2))]

From here, there is the expression sin(Pi*n)/(-1+n^2), that for n=1 is 0. But solving it with Maple it gives u(x,t)=2/(3*sqrt(3))*exp^(-t/2)*sin(sqrt(3)*t/2)*sin(x).

I've also found that limit [when n-->1] of sin(Pi*n)/(-1+n^2) = -Pi/2; that plugged on the series give what Maple says.

But I'm still not sure if it is correct or not

Thanks!

$\sin(n\pi)$ is zero for any integer n. So, all of your terms except for the first one is zero. The first is treated as non-zero because the denominator vanishes at n = 1. So, you use L'Hopital's rule to evaluate it instead.