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# limit of fourier series I'd like to compute the limit of $$\sum_{k=1}^{n}\frac{1}{k^2+1}\sin(kx)$$ I did the following already:

def b(k):
return(1/(k**2+1))
def a(k,x):
return(b(k)*sin(k*x))
def s(n,x):
s=0
for k in range(1,n+1):
s=s+a(k,x)
return(s)
var('x,n')
f=s(1000,x)
plot(f,0,2*pi)


I already tried computing it by hand, looked in Bronstein, searched the internet, but didn't find any solution. But I'm no specialist in Analysis, so perhaps somebody can help? Clearly the series converges for every x in [0,2*pi] ...

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## 1 Answer

Sort by » oldest newest most voted Here is what it looks like

sage: def f(x):
....:     return sum(sin(k*x)/(k**2+1) for k in range(1000))
sage: plot(f,0,2*pi)


Not clear if this is a known function. more

## Comments

Indeed, and the error is bounded by:

sage: var('k')
k
sage: sum(1/(k^2+1),k,1001,oo).n()
0.000999499833833166


Yes, but it does not answer my question ;-) I want to compute the limit function f(x), restricted to x from 0 to 2*\pi

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Asked: 2021-02-06 10:00:40 +0200

Seen: 67 times

Last updated: Feb 07