Solving for c: One Solution in (0,1)?

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Discussion Overview

The discussion revolves around the mathematical problem of demonstrating that the equation c = (n/pi) arccos [(n/pi)sin(pi/n)] + 2k(pi), where n is a positive integer, has precisely one solution in the interval (0,1). The focus is on exploring potential methods for proving this assertion, including limits and the behavior of the function involved.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses frustration in solving the equation and seeks to prove that c has one solution in (0,1).
  • Another participant suggests starting by testing positive integer values for n to gain insights into the behavior of the function.
  • A participant notes that testing specific values won't suffice for a complete proof and discusses examining limits of components within the function, such as the range of sin(pi/n).
  • There is a reiteration that proving the existence of a single solution is challenging and may require a theorem or its consequences rather than induction or contradiction.

Areas of Agreement / Disagreement

Participants generally agree that demonstrating the existence of a single solution in (0,1) is complex and may not be achievable through simple methods like induction. However, there is no consensus on the best approach to prove the claim.

Contextual Notes

Limitations in the discussion include the reliance on specific integer values for n and the challenges in addressing the behavior of the function as n varies. The discussion does not resolve the mathematical steps necessary for a complete proof.

martint
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Hello,

I've been trying to solve the following but with no luck its so frustrating!

if c= (n/pi) arccos [(n/pi)sin(pi/n)] + 2k(pi), where n is a positive integer,

how can I show that c has precisely one solution in (0,1)?

Thanks!
 
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What have you done to try and solve it?

Perhaps you can start by plugging in some positive integers for n...
 
that won't be a complete proof tho, need to show it holds for all values of n. Have tried breaking it down and looking at the limits of certain parts within the function, e.g sin (pi/n) lies between 0 and 1 for all n etc but this doesn't seem to get me anywhere as I get stuck when i reach the (n/pi) in front of the arccos!
 
martint said:
that won't be a complete proof tho, need to show it holds for all values of n. Have tried breaking it down and looking at the limits of certain parts within the function, e.g sin (pi/n) lies between 0 and 1 for all n etc but this doesn't seem to get me anywhere as I get stuck when i reach the (n/pi) in front of the arccos!

You are right, it won't be a complete proof. But by trying and working out the first few integer values, you will have a better understanding as to why it will always be between 0 and 1. Then perhaps you can produce a proof by induction.
 
i know that it lies between 0 and 1, but it is trying to show that there is only ONE solution between 0 and 1 that i haven't been able to achieve.I don't think it is possible by induction or contradiction but that a theorem or consequences of a theorem is required :confused:
 

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