- #1
bolbteppa
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Whittaker (1st Edition, 1902) P.132, gives two proofs of Fourier's theorem, assuming Dirichlet's conditions. One proof is Dirichlet's proof, which involves directly summing the partial sums, is found in many books. The other proof is an absolutely stunning proof of Fourier's theorem in terms of residues, treating the partial sums as the residues of a meromorphic function and showing that, on taking the limit, we end up with Dirichlet's conditions.
(Note Later editions exclude the residue proof and prove the theorem under weaker conditions than Dirichlet's conditions in two ways, one way via the theory of summability, the other by modifying Dirichlet's proof.)
My question is about understanding the latter half of the residue proof, given here. The jist of the proof is to consider a trigonometric series with real coefficients, assume the coefficients are Fourier coefficients of a function ##f##, and then simplify the partial sum
\begin{align}
S_k(f) &= a_0 + \sum_{m=1}^k (a_m \cos(mz) + b_m \sin(mz)) \\
&= \frac{1}{2 \pi} \int_0^{2 \pi} f(t)dt + \frac{1}{\pi} \sum_{m=1}^k \int_0^{2 \pi} f(t)\cos[m(z-t)] dt \\
&= \sum_{m=-k}^k \frac{1}{2\pi} \int_0^{2 \pi} f(t)e^{im(z-t)} dt \\
&= \sum_{m=-k}^k \frac{1}{2\pi} \int_0^z f(t)e^{im(z-t)} dt + \sum_{m=-k}^k \frac{1}{2\pi} \int_z^{2 \pi} f(t)e^{im(z-t)} dt \\
&= U_k + V_k.
\end{align}
Next we try to turn ##U_k## into the sum of the residues of a meromorphic function derived from this, so try to modify it:
\begin{align}
U_k(z) &= \sum_{m=-k}^k \frac{1}{2\pi} \int_0^z f(t)e^{im(z-t)} dt \\
&= \sum_{m=-k}^k \frac{w}{2\pi w} \int_0^z f(t)e^{w(z-t)} dt |_{w = im, m \neq 0} \\
&= \sum_{m=-k}^k \frac{w}{1 + 2\pi w - 1} \int_0^z f(t)e^{w(z-t)} dt |_{w = im, m \neq 0} \\
&\to \frac{1}{1 + 2\pi w + \dots - 1} \int_0^z f(t)e^{w(z-t)} dt \\
&= \frac{1}{e^{2 \pi w} - 1} \int_0^z f(t)e^{w(z-t)} dt
\end{align}
to find
$$\phi(w) = \frac{1}{e^{2 \pi w} - 1} \int_0^z f(t)e^{w(z-t)} dt$$
so that, if ##C_k## is a circle in the ##w## plane containing ##0,i,-i,2i,-2i,\dots,ki,-ki## and no more poles, say of radius ##k+1/2##, we see
$$ \frac{1}{2 \pi i} \int_{C_k} \phi(w) dw = U_k.$$
From this we integrate over the boundary explicitly via ##w = (k + 1/2)e^{i\theta}## so that ##U_k## reduces to
$$U_k = \frac{1}{2 \pi} \int_0^{2 \pi} w \phi(w) d \theta$$
and from here on we are supposed to end up with Dirichlet's conditions.
Can anybody explain the rest of the proof? Since this aspect of the proof seems to be the crux of other flawed proofs, need to make sure I get the rest of it with no hand-waving.
Thanks!
(Note Later editions exclude the residue proof and prove the theorem under weaker conditions than Dirichlet's conditions in two ways, one way via the theory of summability, the other by modifying Dirichlet's proof.)
My question is about understanding the latter half of the residue proof, given here. The jist of the proof is to consider a trigonometric series with real coefficients, assume the coefficients are Fourier coefficients of a function ##f##, and then simplify the partial sum
\begin{align}
S_k(f) &= a_0 + \sum_{m=1}^k (a_m \cos(mz) + b_m \sin(mz)) \\
&= \frac{1}{2 \pi} \int_0^{2 \pi} f(t)dt + \frac{1}{\pi} \sum_{m=1}^k \int_0^{2 \pi} f(t)\cos[m(z-t)] dt \\
&= \sum_{m=-k}^k \frac{1}{2\pi} \int_0^{2 \pi} f(t)e^{im(z-t)} dt \\
&= \sum_{m=-k}^k \frac{1}{2\pi} \int_0^z f(t)e^{im(z-t)} dt + \sum_{m=-k}^k \frac{1}{2\pi} \int_z^{2 \pi} f(t)e^{im(z-t)} dt \\
&= U_k + V_k.
\end{align}
Next we try to turn ##U_k## into the sum of the residues of a meromorphic function derived from this, so try to modify it:
\begin{align}
U_k(z) &= \sum_{m=-k}^k \frac{1}{2\pi} \int_0^z f(t)e^{im(z-t)} dt \\
&= \sum_{m=-k}^k \frac{w}{2\pi w} \int_0^z f(t)e^{w(z-t)} dt |_{w = im, m \neq 0} \\
&= \sum_{m=-k}^k \frac{w}{1 + 2\pi w - 1} \int_0^z f(t)e^{w(z-t)} dt |_{w = im, m \neq 0} \\
&\to \frac{1}{1 + 2\pi w + \dots - 1} \int_0^z f(t)e^{w(z-t)} dt \\
&= \frac{1}{e^{2 \pi w} - 1} \int_0^z f(t)e^{w(z-t)} dt
\end{align}
to find
$$\phi(w) = \frac{1}{e^{2 \pi w} - 1} \int_0^z f(t)e^{w(z-t)} dt$$
so that, if ##C_k## is a circle in the ##w## plane containing ##0,i,-i,2i,-2i,\dots,ki,-ki## and no more poles, say of radius ##k+1/2##, we see
$$ \frac{1}{2 \pi i} \int_{C_k} \phi(w) dw = U_k.$$
From this we integrate over the boundary explicitly via ##w = (k + 1/2)e^{i\theta}## so that ##U_k## reduces to
$$U_k = \frac{1}{2 \pi} \int_0^{2 \pi} w \phi(w) d \theta$$
and from here on we are supposed to end up with Dirichlet's conditions.
Can anybody explain the rest of the proof? Since this aspect of the proof seems to be the crux of other flawed proofs, need to make sure I get the rest of it with no hand-waving.
Thanks!
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