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limit of fourier series

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

dantetante gravatar image

updated 2021-02-07 23:10:42 +0200

Juanjo gravatar image

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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answered 2021-02-06 14:11:14 +0200

FrédéricC gravatar image

updated 2021-02-06 14:12:05 +0200

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.

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Comments

Indeed, and the error is bounded by:

sage: var('k')
k
sage: sum(1/(k^2+1),k,1001,oo).n()
0.000999499833833166
tmonteil gravatar imagetmonteil ( 2021-02-06 18:11:47 +0200 )edit

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

dantetante gravatar imagedantetante ( 2021-02-09 16:30:52 +0200 )edit

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

Seen: 74 times

Last updated: Feb 07