Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In the rant I wrote at

http://ncatlab.org/nlab/show/trigonometric+identities+and+the+irrationality+of+pi

I asked: Are these four identities the first four terms in a sequence that continues?

This referred to the identities in the last bullet point above that question.

While we're at it, is there any intuitive geometric interpretation of the identity involving $f_2$?

OK, here are the functions involved:

$$ \begin{align} f_0(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 0} (-1)^{n/2} \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_1(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 1} (-1)^{(n-1)/2} \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_2(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 2} (-1)^{(n-2)/2} n \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_3(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 3} (-1)^{(n-3)/2} (n-1) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_4(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 4} (-1)^{(n-4)/2} n(n-2) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_5(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 5} (-1)^{(n-5)/2} (n-1)(n-3) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_6(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 6} (-1)^{(n-6)/2} n(n-2)(n-4) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_7(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 7} (-1)^{(n-7)/2} (n-1)(n-3)(n-5) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ & \vdots \end{align} $$ In each function the coefficient kills off the terms involving values of $n$ smaller than the index, so that for example we could have said "$\text{odd }n \ge 1$" instead of $\text{odd }n \ge 7$ and it would be the same thing.

Now some facts:

  • Each $f_k$ is a symmetric function of $\theta_1,\theta_2,\theta_3,\dots$.

  • 0 is an identity element for each of these functions, in the sense that $$f_k(0,\theta_2,\theta_3,\dots) = f_k(\theta_2,\theta_3,\dots).$$

  • $f_k(\theta_1,\theta_2,\theta_3,\dots) - f_k(\theta_1+\theta_2,\theta_3,\dots) = \left. \begin{cases} k & \text{if }k \ge 2\text{ is even} \\ k-1 & \text{if }k \ge 2\text{ is odd} \end{cases} \right\}\cdot \sin\theta_1\sin\theta_2 f_{k-2}(\theta_3,\theta_4,\dots)$ and $=0$ if $k = 0\text{ or }1$.

Now the sequence of identies: $$ \begin{align} f_0 & = \cos(\theta_1 + \theta_2 + \theta_3 + \cdots) \\ f_1 & = \sin(\theta_1 + \theta_2 + \theta_3 + \cdots) \\ \text{If } \sum_{i=1}^\infty \theta_i = \pi,\text{ then } f_2 & = \sum_{i=1}^\infty \sin^2\theta_i \\ \text{If } \sum_{i=1}^\infty \theta_i = \pi,\text{ then } f_3 & = \frac{1}{2} \sum_{i=1}^\infty \sin(2\theta_i) \end{align} $$ The QUESTION is whether these are the first four identities in a sequence that continues beyond this point.

share|improve this question
6  
Please try to make your question self-contained by not including a link that the reader must click on in order to parse it. This will greatly increase the readership of your question and thus the likelihood of getting a good answer. –  Pete L. Clark Jun 14 '10 at 14:46
    
Looking at the question and the answer below, I was intrigued how all these "trigonometric identities" could imply "the irrationality of pi". I am disappointed: there is nothing about irrationality nor about pi there... :-( –  Wadim Zudilin Jun 14 '10 at 15:17
    
Actually, if you're looking only at the question and not at the external link, you won't see any of that, and the external link has its own external link to a Wikipedia article that contains Mary Cartwright's proof of the irrationality of $\pi$, so I am somewhat guilty of non-self-containment as suggested above. –  Michael Hardy Jun 14 '10 at 16:11
2  
The nLab page is really hard to read. –  Qiaochu Yuan Jun 14 '10 at 20:30
1  
Michael, you still have a chance to edit your question by adding the necessary contents and, of course, the trig identity itself. –  Wadim Zudilin Jun 14 '10 at 23:27
show 5 more comments

1 Answer

With binomial theorem, the products on the right have closed form $$ \sum_{|A|=n} \prod_{i \in A} \sin \theta_i \prod_{i \notin A} \cos \theta_i = \prod_{i=1}^n ( \sin \theta_i + \cos \theta_i )^n = \prod_{i=1}^n \sqrt{2} \sin (\theta_i + \pi/4)^n$$ So we'll let $x = \sin \theta + \cos \theta$ so the first sum looks like $$ \sum_{n \geq 0, even} (-1)^{n/2} \prod_{i=1}^n x_i = 1 - x_1x_2 + x_1x_2x_3x_4 - \dots$$ This is not symmetric in the x's.

share|improve this answer
1  
John, I don't think this is right. What is true is that $$f_0+if_1=\prod_j(\cos\theta_j+i\sin\theta_j)$$ from which the $f_0$ and $f_1$ identities drop out immediately. –  Robin Chapman Jun 14 '10 at 15:04
    
The first two identities are of course universally known. The third one might be entirely novel for all I know. The simplest non-degenerate special case of the fourth one can at least be facetiously referred to as "well known", and possibly is actually well known within certain communities---I don't really know. (For the sake of at least a little bit of "self-containment", the simplest non-degenerate special case of the fourth one says that if $\alpha + \beta + \gamma = \pi$ then $4\sin\alpha\sin\beta\sin\gamma = \sin(2\alpha) + \sin(2\beta) + \sin(2\gamma)$.) –  Michael Hardy Jun 14 '10 at 16:19
    
Yeah I misread the A as running through subsets of {1, 2, \dots, n}. In fact A runs through all n-element subsets of the natural numbers. –  john mangual Jun 14 '10 at 17:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.