Parseval–Gutzmer formula

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In mathematics, the Parseval–Gutzmer formula states that, if ƒ is an analytic function on a closed disk of radius r with Taylor series

f(z) = \sum^\infty_{k = 0} a_k z^k,

then for z = re on the boundary of the disk,

\int^{2\pi}_0 |f(re^{i\vartheta}) |^2 \, \mathrm{d}\vartheta = 2\pi \sum^\infty_{k = 0} |a_k|^2r^{2k}.


The Cauchy Integral Formula for coefficients states that for the above conditions:

a_n = \frac{1}{2\pi i} \int^{}_{\gamma} \frac{f(z)}{z^{n+1}} \, \mathrm{d}\ z

where γ is defined to be the circular path around 0 of radius r. We also have that, for x in the complex plane C,

\overline{x}{x} = |x|^2

We can apply both of these facts to the problem. Using the second fact,

\int^{2\pi}_0 |f(re^{i\vartheta}) |^2 \, \mathrm{d}\vartheta = \int^{2\pi}_0 {f(re^{i\vartheta})}\overline{f(re^{i\vartheta})}  \, \mathrm{d}\vartheta

Now, using our Taylor Expansion on the conjugate,

 = \int^{2\pi}_0 {f(re^{i\vartheta})}{\sum^\infty_{k = 0} \overline{a_k (re^{i\vartheta})^k}}  \, \mathrm{d}\vartheta

Using the uniform convergence of the Taylor Series and the properties of integrals, we can rearrange this to be

 = \sum^\infty_{k = 0} \int^{2\pi}_0 \frac{{f(re^{i\vartheta})}\overline{a_k} (r^k)}{(e^{i\vartheta})^k} , \mathrm{d}\vartheta

With further rearrangement, we can set it up ready to use the Cauchy Integral Formula statement

 = \sum^\infty_{k = 0} ({2\pi}{\overline{a_k} r^{2k}})(\frac{1}{2{\pi}i}\int^{2\pi}_0  \frac{{f(re^{i\vartheta})}}{(r e^{i\vartheta})^{k+1}} {rie^{i\vartheta}}) \mathrm{d}\vartheta

Now, applying the Cauchy Integral Formula, we get

 = \sum^\infty_{k = 0} ({2\pi}{\overline{a_k} r^{2k}}){a_k} = {2\pi} \sum^\infty_{k = 0} {|a_k|^2 r^{2k}}

Further Applications[edit]

Using this formula, it is possible to show that

\sum^\infty_{k = 0} |a_k|^2r^{2k} \le {M_r}^2 where M_r = \sup\{|f(z)| : |z| = r\}

This is done by using the integral

\int^{2\pi}_0 |f(re^{i\vartheta}) |^2 \, \mathrm{d}\vartheta \le 2\pi |max_{\vartheta \in [0,2\pi)}(f(re^{i\vartheta}))|^2 = 2\pi |max_{|z|=r}(f(z))|^2 = 2\pi(M_r)^2