Interesting problem from my analysis class

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Discussion Overview

The discussion revolves around a problem from analysis concerning a continuous function defined on the interval [0,1] with the property that f(0) = f(1). The goal is to prove the existence of a horizontal chord of length 1/n in the graph of the function, specifically finding an x in the interval [0, (n - 1)/n] such that f(x + 1/n) = f(x).

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant suggests using a telescoping sum to express f(1) - f(0).
  • Another participant questions whether the function f is differentiable or merely continuous.
  • A different participant expresses that the problem seems trivial for values less than or equal to 1, but later rephrases this to "intuitively plausible" after considering the proof.
  • One participant notes that the problem appears to follow from the intermediate value theorem, although they find the details tedious and elusive.
  • Another participant proposes an assumption that f(x) is non-negative and asks what follows from that assumption.

Areas of Agreement / Disagreement

Participants express varying levels of confidence in the problem's difficulty, with some finding it trivial or intuitively plausible, while others note the complexity of the proof. There is no consensus on the approach or the assumptions regarding the function f.

Contextual Notes

Participants have not clarified whether the function f is differentiable, which may affect the proof's approach. The discussion also reflects uncertainty about the implications of assuming f(x) is non-negative.

StonedPanda
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Let n be a positive integer and suppose f is continuous on [0,1] and f(0) = f(1). Prove that the graph of f has a horizontal chord of length 1/n. In other words, prove there exists x \in [0,(n - 1)/n] such that f(x+1/n) = f(x)
 
Last edited:
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no one even wants to try?
 
hint: write f(1) - f(0) as a telescoping sum.
 
StonedPanda said:
no one even wants to try?

I would hazard a guess that people are suspicious that this is your homework.
 
Is f differentiable, or just continuous?
 
seems trivial. indeed trivial for all numbers less than or equal to 1, not just 1/n.

after thinking about an actual proof for a few minutes let me rephrase that as "intuitively plausible", rather than "trivial".

it apparently follows from the intermediate value theorem but the details seem tedious, even elusive. cute problem.
 
Last edited:
assume f(x) non negative, then what?
 

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