Papyrus 15

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

$f(z)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{z}^k,$

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

$\underset{0}{\overset{2\pi }{\int }}|f(r{e}^{i\vartheta }){|}^2\mathrm{d}\vartheta =2\pi \sum _{k=0}^{\mathrm{\infty }}|a_k{|}^2{r}^{2k}.$

Proof

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

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

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,

$\bar{x}x=|x{|}^2$

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

$\underset{0}{\overset{2\pi }{\int }}|f(r{e}^{i\vartheta }){|}^2\mathrm{d}\vartheta =\underset{0}{\overset{2\pi }{\int }}f(r{e}^{i\vartheta })\overline{f(r{e}^{i\vartheta })}\mathrm{d}\vartheta$

Now, using our Taylor Expansion on the conjugate,

$=\underset{0}{\overset{2\pi }{\int }}f(r{e}^{i\vartheta })\sum _{k=0}^{\mathrm{\infty }}\overline{a_k(r{e}^{i\vartheta }{)}^k}\mathrm{d}\vartheta$

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

$=\sum _{k=0}^{\mathrm{\infty }}\underset{0}{\overset{2\pi }{\int }}\frac{f(r{e}^{i\vartheta })\overline{a_k}(r^k)}{(e^{i\vartheta }{)}^k},\mathrm{d}\vartheta$

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

$=\sum _{k=0}^{\mathrm{\infty }}(2\pi \overline{a_k}{r}^{2k})(\frac{1}{2\pi i}\underset{0}{\overset{2\pi }{\int }}\frac{f(r{e}^{i\vartheta })}{(r{e}^{i\vartheta }{)}^{k+1}}ri{e}^{i\vartheta })\mathrm{d}\vartheta$

Now, applying the Cauchy Integral Formula, we get

$=\sum _{k=0}^{\mathrm{\infty }}(2\pi \overline{a_k}{r}^{2k})a_k=2\pi \sum _{k=0}^{\mathrm{\infty }}|a_k{|}^2{r}^{2k}$

Further Applications

Using this formula, it is possible to show that

$\sum _{k=0}^{\mathrm{\infty }}|a_k{|}^2{r}^{2k}\le {M_r}^2$ where $M_r=\sup \{|f(z)\left|:\right|z|=r\}$

This is done by using the integral

$\underset{0}{\overset{2\pi }{\int }}|f(r{e}^{i\vartheta }){|}^2\mathrm{d}\vartheta \le 2\pi |ma{x}_{\vartheta \in [0,2\pi )}(f(r{e}^{i\vartheta })){|}^2=2\pi |ma{x}_{\left|z\right|=r}(f(z)){|}^2=2\pi (M_r{)}^2$

References

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