Prove this equation with calculus knowledge

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Discussion Overview

The discussion revolves around proving a specific equation involving tangent and cotangent functions, presented in a summation format. The scope includes mathematical reasoning and exploration of finite calculus concepts.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant expresses difficulty in proving the equation involving a series of tangent functions and seeks assistance.
  • Another participant suggests a reformulation of the equation and provides a hint that relates the terms of the sum to differences of cotangent functions.
  • A question is raised about the relevance of calculus or derivatives to the problem, indicating a potential misunderstanding of the mathematical context.
  • A participant clarifies that the problem is more aligned with finite calculus rather than infinitesimal calculus, although they acknowledge the relationship between the two fields.
  • One participant claims to have proved the hint without needing finite calculus knowledge, suggesting an alternative approach to the problem.
  • A basic result in finite calculus is presented, emphasizing the importance of expressing terms as differences to facilitate the summation process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the necessity of calculus knowledge for the proof, with differing views on the relevance of finite versus infinitesimal calculus.

Contextual Notes

There are unresolved assumptions regarding the mathematical techniques required for the proof, and the discussion highlights the interplay between different branches of calculus without settling on a definitive approach.

aerograce
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I feel it quite difficult to prove this equation:
[itex]\frac{1}{2}[/itex]tan([itex]\frac{1}{2}[/itex]x)+[itex]\frac{1}{2^2}[/itex]tan([itex]\frac{1}{2^2}[/itex]x)+...+[itex]\frac{1}{2^n}[/itex]tan([itex]\frac{1}{2^n}[/itex]x)=[itex]\frac{1}{2^n}[/itex]cot([itex]\frac{1}{2^n}[/itex]x)-cotx

Can you help me with it?
 
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What have you tried?
$$\text{Let $0<x<\pi/2$, and n a positive integer} \\\text{prove that}\\
\sum_{k=1}^n \tan(x/2^k)/2^k=\cot(x/2^n)/2^n-cot(x)$$
hint
$$\tan(x/2^k)/2^k=\cot(x/2^k)/2^k-\cot(x/2^{k-1})/2^{k-1}$$
 
Last edited:
So it has nothing to do with calculus or derivative?
 
That is a usual problem in finite calculus which deals with sums and differences. It does not seem particularly related to infinitesimal calculus which deals with derivatives and integrals. Though the two are closely related.
 
Aww.. But I didn't use any finite calculus knowledge with your hint. I just proved your hint and then I found that I need nothing more.
 
A basic result in finite calculus is
$$\sum_{k=1}^n (a_k-a_{k-1})=a_n-a_0$$
Which my hint was an example of. The tricky part is figuring out how to write the terms of the sum as a difference.
There are some examples here.
https://en.wikipedia.org/wiki/Indefinite_sum
 

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