A continuous function f [0, 1] → R satisfies f(0) = f(1). Show that for:each integer n ≥ 1 there exits x such that f(x+(1/n)) = f(x). Is the samestatement true for numbers other than 1/n?

Let fix δ=1n, n∈ℕ Consider the continuous function gnx:0,1-δ→ℝ; gnx:=fx+1n-fx and suppose that gnx≠0…

Answered: A continuous function ƒ : [0, 1] → R…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

A continuous function f [0, 1] → R satisfies f(0) = f(1). Show that for
:
each integer n ≥ 1 there exits x such that f(x+(1/n)) = f(x). Is the same
statement true for numbers other than 1/n?

Expert Solution

Step 1: Proof

Let fix $δ = \frac{1}{n}, n \in ℕ$

Consider the continuous function $g_{n} (x) : [0, 1 - δ] \to ℝ; g_{n} (x) : = f (x + \frac{1}{n}) - f (x)$ and suppose that $g_{n} (x) \neq 0$ on $I_{0} : = [0, 1]$

Then by intermediate value theorem, $g_{n} (x)$ does not changes sign in $I_{0}$ .

Without losing of generality if $g_{n} (x) > 0$ , then adding the $n$ inequalities

$g_{n} (1 - \frac{1}{n}) = f (1) - f (1 - \frac{1}{n}) > 0; g_{n} (1 - \frac{2}{n}) = f (1 - \frac{1}{n}) - f (1 - \frac{2}{n}) > 0; \dots g_{n} (0) = f (\frac{1}{n}) - f (0) > 0$