Remove adsIn mathematics, the Parseval–Gutzmer formula states that, if f {\displaystyle f} is an analytic function on a closed disk of radius r with Taylor series f ( z ) = ∑ k = 0 ∞ a k z k , {\displaystyle f(z)=\sum _{k=0}^{\infty }a_{k}z^{k},} then for z = reiθ on the boundary of the disk, ∫ 0 2 π | f ( r e i θ ) | 2 d θ = 2 π ∑ k = 0 ∞ | a k | 2 r 2 k , {\displaystyle \int _{0}^{2\pi }|f(re^{i\theta })|^{2}\,\mathrm {d} \theta =2\pi \sum _{k=0}^{\infty }|a_{k}|^{2}r^{2k},} which may also be written as 1 2 π ∫ 0 2 π | f ( r e i θ ) | 2 d θ = ∑ k = 0 ∞ | a k r k | 2 . {\displaystyle {\frac {1}{2\pi }}\int _{0}^{2\pi }|f(re^{i\theta })|^{2}\,\mathrm {d} \theta =\sum _{k=0}^{\infty }|a_{k}r^{k}|^{2}.} Remove adsProofSummarizePerspective The Cauchy Integral Formula for coefficients states that for the above conditions: a n = 1 2 π i ∫ γ f ( z ) z n + 1 d z {\displaystyle 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 origin of radius r. Also for x ∈ C , {\displaystyle x\in \mathbb {C} ,} we have: x ¯ x = | x | 2 . {\displaystyle {\overline {x}}{x}=|x|^{2}.} Applying both of these facts to the problem starting with the second fact: ∫ 0 2 π | f ( r e i θ ) | 2 d θ = ∫ 0 2 π f ( r e i θ ) f ( r e i θ ) ¯ d θ = ∫ 0 2 π f ( r e i θ ) ( ∑ k = 0 ∞ a k ( r e i θ ) k ¯ ) d θ Using Taylor expansion on the conjugate = ∫ 0 2 π f ( r e i θ ) ( ∑ k = 0 ∞ a k ¯ ( r e − i θ ) k ) d θ = ∑ k = 0 ∞ ∫ 0 2 π f ( r e i θ ) a k ¯ ( r e − i θ ) k d θ Uniform convergence of Taylor series = ∑ k = 0 ∞ ( 2 π a k ¯ r 2 k ) ( 1 2 π i ∫ 0 2 π f ( r e i θ ) ( r e i θ ) k + 1 r i e i θ ) d θ = ∑ k = 0 ∞ ( 2 π a k ¯ r 2 k ) a k Applying Cauchy Integral Formula = 2 π ∑ k = 0 ∞ | a k | 2 r 2 k {\displaystyle {\begin{aligned}\int _{0}^{2\pi }\left|f\left(re^{i\theta }\right)\right|^{2}\,\mathrm {d} \theta &=\int _{0}^{2\pi }f\left(re^{i\theta }\right){\overline {f\left(re^{i\theta }\right)}}\,\mathrm {d} \theta \\[6pt]&=\int _{0}^{2\pi }f\left(re^{i\theta }\right)\left(\sum _{k=0}^{\infty }{\overline {a_{k}\left(re^{i\theta }\right)^{k}}}\right)\,\mathrm {d} \theta &&{\text{Using Taylor expansion on the conjugate}}\\[6pt]&=\int _{0}^{2\pi }f\left(re^{i\theta }\right)\left(\sum _{k=0}^{\infty }{\overline {a_{k}}}\left(re^{-i\theta }\right)^{k}\right)\,\mathrm {d} \theta \\[6pt]&=\sum _{k=0}^{\infty }\int _{0}^{2\pi }f\left(re^{i\theta }\right){\overline {a_{k}}}\left(re^{-i\theta }\right)^{k}\,\mathrm {d} \theta &&{\text{Uniform convergence of Taylor series}}\\[6pt]&=\sum _{k=0}^{\infty }\left(2\pi {\overline {a_{k}}}r^{2k}\right)\left({\frac {1}{2{\pi }i}}\int _{0}^{2\pi }{\frac {f\left(re^{i\theta }\right)}{(re^{i\theta })^{k+1}}}{rie^{i\theta }}\right)\mathrm {d} \theta \\&=\sum _{k=0}^{\infty }\left(2\pi {\overline {a_{k}}}r^{2k}\right)a_{k}&&{\text{Applying Cauchy Integral Formula}}\\&={2\pi }\sum _{k=0}^{\infty }{|a_{k}|^{2}r^{2k}}\end{aligned}}} Remove adsFurther ApplicationsSummarizePerspective Using this formula, it is possible to show that ∑ k = 0 ∞ | a k | 2 r 2 k ⩽ M r 2 {\displaystyle \sum _{k=0}^{\infty }|a_{k}|^{2}r^{2k}\leqslant M_{r}^{2}} where M r = sup { | f ( z ) | : | z | = r } . {\displaystyle M_{r}=\sup\{|f(z)|:|z|=r\}.} This is done by using the integral ∫ 0 2 π | f ( r e i θ ) | 2 d θ ⩽ 2 π | max θ ∈ [ 0 , 2 π ) ( f ( r e i θ ) ) | 2 = 2 π | max | z | = r ( f ( z ) ) | 2 = 2 π M r 2 {\displaystyle \int _{0}^{2\pi }\left|f\left(re^{i\theta }\right)\right|^{2}\,\mathrm {d} \theta \leqslant 2\pi \left|\max _{\theta \in [0,2\pi )}\left(f\left(re^{i\theta }\right)\right)\right|^{2}=2\pi \left|\max _{|z|=r}(f(z))\right|^{2}=2\pi M_{r}^{2}} Remove adsReferences Ahlfors, Lars (1979). Complex Analysis. McGraw–Hill. ISBN 0-07-085008-9. This mathematical analysis–related article is a stub. You can help Wikipedia by expanding it.vteLoading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.Remove ads