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Analysis

Real & Complex Analysis, Calculus & Multivariate Calculus, Functional Analysis,
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Asis ghosh
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Analysis

#1

Post by Asis ghosh » Wed Feb 14, 2018 5:01 pm

solve the question

Let $f:[0, 2\pi] \rightarrow [0, 2\pi]$ be continuous such that $f(0)=f(2\pi)$. Show that there exists $x \in [0, 2\pi]$ such that

$$f(x)= f(x+\pi)$$
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Riemann
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Re: Analysis

#2

Post by Riemann » Thu Feb 15, 2018 1:59 pm

Asis ghosh wrote:solve the question

Let $f:[0, 2\pi] \rightarrow [0, 2\pi]$ be continuous such that $f(0)=f(2\pi)$. Show that there exists $x \in [0, 2\pi]$ such that

$$f(x)= f(x+\pi)$$
Hello my friend,

consider the function $g(t)=f(t) - f(t+\pi) \; , \; t \in [0, 2\pi]$. Clearly, $g$ is continuous as a sum of continuous functions. Note that

$$g(0)=f(0)-f(\pi) \quad , \quad g(\pi)=f(\pi)-f(2\pi)=f(\pi)-f(0)$$

Hence

$$g(0) g(\pi) = -\left ( f(0)- f(\pi) \right )^2 \leq 0$$

Hence by Bolzano's theorem we have that there exists an $x \in (0, \pi) \subset [0, 2\pi]$ such that $g(x)=0$. That is $f(x)=f(x+\pi)$. Can you now complete the proof of the existence of $x \in [0, 2\pi]$?
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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