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limit of fourier series

asked 4 years ago

dantetante gravatar image

updated 4 years ago

Juanjo gravatar image

I'd like to compute the limit of nk=11k2+1sin(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 4 years ago

FrédéricC gravatar image

updated 4 years ago

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 ( 4 years ago )

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 ( 4 years ago )

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Asked: 4 years ago

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Last updated: Feb 07 '21