Weilin Meng
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Homework Statement
Let f(x) = sin(x)/x for |x| <= pi with the obvious definition at x = 0
Extend it periodically. Will the Fourier series converge at x=0?
Homework Equations
Fourier coefficients:
ao = 1/\pi \int_{-\pi}^{\pi} (sin(x)/x)
an = 1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * cos(nx)
bn = 1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * sin(nx)
The Attempt at a Solution
ao = 1/\pi \int_{-\pi}^{\pi} (sin(x)/x) = 2Si(pi)/pi
an = 1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * cos(nx) = (-Si((n-1)pi) + Si((n+1)pi))/pi
bn = 1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * sin(nx) = 0
So the Fourier series will be:
f(x)=1/\pi*Si(\pi)+\sum_{1}^{\infty }1/\pi(-Si((n-1)\pi) + Si((n+1)\pi)
at x=0 because cos(n(0)) = 1
I figured that:
\sum_{1}^{\infty }1/\pi(-Si((n-1)\pi) + Si((n+1)\pi)
Will converge to zero and we get f(x) = si(pi)/pi...
I don't know what si(pi)/pi is but I don't think that converges to 1 as sin(x)/x does at x=0...did i mess up anything?