limit of fourier series

asked 4 years ago

dantetante
81 ●4 ●5 ●12

updated 4 years ago

Juanjo

3213 ●5 ●41 ●86

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] ...

answered 4 years ago

FrédéricC

5426 ●4 ●44 ●121

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.

image description

link

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 ( 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 ( 4 years ago )

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