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This is similar to something else I posted, but this time we'll pretend we've never heard of infinite sets or infinite series. \begin{align} & \sin(\alpha+\beta+\gamma+\delta+\varepsilon+\zeta) \\[12pt] = {} & {}+\underbrace{\underbrace{{}\,\sin\alpha\,{}} \, \underbrace{\cos\beta\cos\gamma\cos\delta\cos\varepsilon\cos\zeta} + \cdots\cdots\cdots}_\text{6 terms} \\[15pt] & {} - \big(\ \underbrace{ \underbrace{\sin\alpha\sin\beta\sin\gamma}\,\underbrace{\cos\delta\cos\varepsilon\cos\zeta} +\cdots\cdots\cdots}_\text{20 terms} \ \big) \\[15pt] & {} + \big(\ \underbrace{ \underbrace{\sin\alpha\sin\beta\sin\gamma\sin\delta\sin\varepsilon}\,\underbrace{{}\,\cos\zeta\,{}} + \cdots\cdots\cdots}_\text{6 terms}\ \big) \end{align} Superficially, it looks as if sine and cosine play symmetrical roles in this trigonometric identity. And they do as long as the number of terms in $\alpha+\beta+\gamma+\delta+\varepsilon+\zeta$ remains fixed at six for all Eternity.

If we go to $\alpha+\beta+\gamma+\delta+\varepsilon+\zeta+\eta$, then the numbers of cosine factors in each term will be $6$, $4$, $2$, or $0$. If we go to $\alpha+\beta+\gamma+\delta+\varepsilon+\zeta+\eta+\theta$ then the number of cosine factors is again odd in each term, but this time the ones with five cosines have a minus sign rather than a plus sign. But the sine function sees no such changes as the number of terms changes: $$ \begin{array}{c|c} \text{number of sine factors} & \text{sign} \\ \hline 1 & + \\ 3 & - \\ 5 & + \\ 7 & - \\ 9 & + \\ \vdots & \vdots \end{array} $$

$$ \begin{array}{c|c} \text{number of cosine factors} & \text{sign} \\ \hline 0 & \text{depends on the number of terms in $\alpha+\beta+\gamma+\cdots$} \\ 1 & \text{depends on the number of terms in $\alpha+\beta+\gamma+\cdots$} \\ 2 & \text{depends on the number of terms in $\alpha+\beta+\gamma+\cdots$} \\ 3 & \text{depends on the number of terms in $\alpha+\beta+\gamma+\cdots$} \\ 4 & \text{depends on the number of terms in $\alpha+\beta+\gamma+\cdots$} \\ \vdots & \vdots \end{array} $$

So, my vague question is whether this is but one instance of some known phenomenon in combinatorics/logic/set theory/whatever, an account of which would not generally hint at trigonometry.

PS: If we had had a cosine of a sum rather than the sine of a sum, then it would still be the sine rather than the cosine for which the sign does not depend on the number of terms in $\alpha+\beta+\gamma+\cdots$. The difference would be that the number of sine factors would always be even (whereas the parities of the numbers of cosine factors would depend on the number of terms in $\alpha+\beta+\gamma+\cdots$).

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  • $\begingroup$ There don't exist well-formed formulas of infinite length (where by "length" I mean the number of symbols). So, I don't see how we could have such a phenomenon in logic. But also, what is the phenomenon precisely that you're describing? How would you describe it without referring to cosine or sine? $\endgroup$ – Doug Spoonwood Apr 30 '14 at 18:27
  • $\begingroup$ @DougSpoonwood : That is the question. $\endgroup$ – Michael Hardy Apr 30 '14 at 18:29
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Applying Euler's formula to $e^{i\sum_k \alpha_k}$, the expansion for sine of a finite sum is $$\sin\left(\sum_{k=1}^n \alpha_i\right) = \textrm{Im} \left[\prod_k^n \left( \cos \alpha_k + i \sin \alpha_k \right)\right].$$

Formally, this is very similar to the general case of expanding a product of binomials: $$\prod_{k=1}^n (a_k + b_k) = \sum_{A \subseteq \{1,\ldots, n\}} \prod_{i \in A} a_i \prod_{j \not\in A} b_j.$$

In each term, there is a choice between including $a_k$ or $b_k$ as a factor. Thus, the terms are in bijection with the set of all subsets of an $n$-element set.

Any product of a form similar to sine of a sum has the form $$\textrm{Im} \left[ \prod_{k=1}^n (a_k + i b_k)\right] = \textrm{Im} \left[\sum_{A \subseteq \{1,\ldots, n\}} i^{|A^c|} \prod_{i \in A} a_i \prod_{j \not\in A} b_j\right]$$

Now every term is in bijection with the subsets of odd cardinality of an $n$-element set. (The extraction of the real part gives an expansion whose terms are in bijection with the subsets of even cardinality, which yields the cosine expansion in the first product.) A term associated to a subset $A$ is positive if $|A| \equiv 1 \pmod 4$ and negative if $|A| \equiv 3 \pmod 4$.

The transformation of switching the role of sine and cosine in your example corresponds to transposing each $a_k$ with $b_k$ in the above expansion. In the correspondence with subsets of a set, this operation corresponds to mapping each subset $A$ to its complement. Thus, all examples of symmetry and anti-symmetry of these expansions can be understood in terms of the complementation operation.

In this post, you are interested in the disappearance of any hint of symmetry in the formula $$\sin\left(\sum_{i=1}^\infty \theta_i \right) = \sum_{\textrm{odd } k\geq 1} (-1)^{\frac{k-1}{2}} \sum_{A \subset \mathbb{N}; |A| = k} \left( \prod_{i \in A} \sin \theta_i \prod_{i \not\in A} \cos \theta_i \right)$$

As mentioned previously, each term has finitely many sine factors and infinitely many cosine factors, yet in each finite expansion, we don't see such a strong distinction between sine and cosine.

From the set point of view, we might imagine the terms in the above correspondence as being in bijection with the finite direct product $\prod_{k=1}^n \{0,1\}$. For an infinite product $\prod_{k \in \mathbb{N}} (a_k + b_k)$, a reasonable, purely formal version of the terms correspondence is with the set of all subsets of $\mathbb{N}$. However, this construction doesn't break any symmetry at the formal set level. If we took this idea seriously, we might write $$\sin\left(\sum_{i=1}^\infty \theta_i \right) = \textrm{Im} \left[ \sum_{A \subseteq \mathbb{N}} \left( \prod_{i \in A} i \sin \theta_i \prod_{i \not\in A} \cos \theta_i \right)\right]$$

However, that doesn't appear as asymmetric as the formula we began with. The symmetry-breaking must be due to other considerations.

In particular, since $\sum_{i=1}^\infty \theta_i$ converges, we have $\lim_{n \rightarrow \infty} \theta_i = 0$. Thus, any infinite product with an infinite number of sine factors must approach 0 and contribute nothing to the overall limiting sum. I don't think it quite makes sense to talk about it this way, since a sum over all subsets of $\mathbb{N}$ has uncountably many terms. More likely, we formalize the idea that for large $n$, the contributions of terms with lots of sine factors in the finite expansion become very small. In other words, the symmetry-breaking is due to the fact that $\sin(\theta_n)$ is near $0$ for large $n$ and $\cos(\theta_n)$ is near $1$ for large $n$. To divorce the phenomenon from trigonometry, we could impose similar hypotheses on the $a_k$ and $b_k$ of $\prod (a_k + b_k)$.

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  • $\begingroup$ If you type n \equiv 1 \; (\text{mod} 4) you see $n \equiv 1 \; (\text{mod} 4)$, but if you type n \equiv 1 \pmod 4 you see $n \equiv 1 \pmod 4$. The latter is simpler and provides parentheses automatically, and gives proper spacing. It is standard. (If more than one digit is involved after $\bmod$, you need braces to enclose the whole numeral in parentheses, e.g. \equiv5\pmod{43} yields $\equiv5\pmod{43}$. That is standard usage. I edited accordingly. (I'll digest the actual content here pretty soon.) $\endgroup$ – Michael Hardy Apr 30 '14 at 21:41
  • $\begingroup$ Just under the bottom display equation, it should be $i^n =i$ if $n \equiv 1 \pmod 4$ and $i^n=-i$ if $n \equiv 3 \pmod 4$ $\endgroup$ – Ross Millikan Apr 30 '14 at 23:09
  • $\begingroup$ This answer conspicuously falls short of addressing the question, but I haven't written a really good version of the question yet. $\endgroup$ – Michael Hardy May 7 '14 at 18:23

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