- #1
Weilin Meng
- 25
- 0
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 = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x)[/tex]
an = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * cos(nx)[/tex]
bn = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * sin(nx)[/tex]
The Attempt at a Solution
ao = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x)[/tex] = 2Si(pi)/pi
an = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * cos(nx)[/tex] = (-Si((n-1)pi) + Si((n+1)pi))/pi
bn = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * sin(nx)[/tex] = 0
So the Fourier series will be:
[tex]f(x)=1/\pi*Si(\pi)+\sum_{1}^{\infty }1/\pi(-Si((n-1)\pi) + Si((n+1)\pi)[/tex]
at x=0 because cos(n(0)) = 1
I figured that:
[tex]\sum_{1}^{\infty }1/\pi(-Si((n-1)\pi) + Si((n+1)\pi)[/tex]
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?