Proof of Gauss's Mean Value Theorem
Theorem
Let’s say the function $f$ is analytic on a closed circle $| z - z_{0} | \le r$. Then, $$ f(z_{0}) = {{1} \over {2 \pi}} \int_{0}^{2 \pi} f(z_{0} + r e ^{i \theta } ) d \theta $$
Description
Just as the Mean Value Theorem for Derivatives evolved through generalizations and gave rise to various theorems named after mathematicians, the Mean Value Theorem for Integrals also has a version named after Gauss. Although it clearly follows the form of the Mean Value Theorem for Integrals, it’s a theorem whose concept is not so obviously intuitive.
Proof
Cauchy’s Integral Formula: $$f (z_{0}) = {{1} \over {2 \pi i }} \int_{\mathscr{C}} {{f(z)} \over { (z - z_{0}) }} dz$$
By the Cauchy Integral Formula, $$ f(z_{0}) = {{1} \over {2 \pi i }} \int_{ |z-z_{0} |= r } {{f(z)} \over { (z - z_{0}) }} dz $$ Substituting $z(\theta) = r e ^{ i \theta } + z_{0} , 0 \le \theta \le 2 \pi$ yields $$ \begin{align*} f(z_{0}) =& {{1} \over {2 \pi i }} \int_{0}^{2 \pi} {{f( z_{0} + r e^{i \theta} )} \over { r e ^{ i \theta} }} i r e^{i \theta } d \theta \\ =& {{1} \over {2 \pi }} \int_{0}^{2 \pi} f( z_{0} + r e^{i \theta} ) d \theta \end{align*} $$
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