Interesting problem from my analysis class

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SUMMARY

The discussion centers on proving that for a continuous function f defined on the interval [0,1] with f(0) = f(1), there exists a horizontal chord of length 1/n. Specifically, it asserts that there exists an x in the interval [0, (n - 1)/n] such that f(x + 1/n) = f(x). The proof is linked to the Intermediate Value Theorem, although participants note that the details can be tedious. The problem is considered intuitively plausible, with some participants questioning the differentiability of f.

PREREQUISITES
  • Understanding of continuous functions on closed intervals
  • Familiarity with the Intermediate Value Theorem
  • Basic knowledge of telescoping sums
  • Concept of horizontal chords in graph theory
NEXT STEPS
  • Study the Intermediate Value Theorem in depth
  • Explore proofs involving telescoping sums
  • Investigate properties of continuous functions on closed intervals
  • Examine examples of horizontal chords in mathematical graphs
USEFUL FOR

Mathematics students, educators, and anyone interested in analysis, particularly those studying properties of continuous functions and their graphical representations.

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)
 
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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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