# Prove this equation with calculus knowledge

1. Aug 19, 2013

### aerograce

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

Can you help me with it?

2. Aug 19, 2013

### lurflurf

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: Aug 20, 2013
3. Aug 20, 2013

### aerograce

So it has nothing to do with calculus or derivative?

4. Aug 20, 2013

### lurflurf

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.

5. Aug 20, 2013

### aerograce

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.

6. Aug 20, 2013

### lurflurf

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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