User:Michael Hardy/Euler's derivation of the sine series

From the identity


 * $$ \sin\left(\sum_{i=1}^\infty \theta_i\right)

=\sum_{\text{odd } k \ge 1} (-1)^{(k-1)/2} \sum_{\begin{smallmatrix} A \subseteq \{\,1,2,3,\dots\,\} \\ \left|A\right| = k\end{smallmatrix}} \left(\prod_{i \in A} \sin\theta_i \prod_{i \not \in A} \cos\theta_i\right) $$

one can infer that


 * $$ \sin(n\theta) = \sum_{\text{odd } k \in \{1,\dots,n\}} (-1)^{(k-1)/2} {n \choose k} \sin^k\theta \cos^{n-k}\theta. $$

Letting x = n&theta;, we have



\begin{align} \sin x & = \sum_{\text{odd } k \in \{1,\dots,n\}} (-1)^{(k-1)/2} {n \choose k} \sin^k\theta \cos^{n-k}\theta \\[12pt] & = \sum_{\text{odd } k \in \{1,\dots,n\}} \left[ (-1)^{(k-1)/2} \left({n(n-1)\cdots (n-k+1) \over k!}\right) \left({x \over n}\right)^k \right. \\ & {} \qquad\qquad\qquad\qquad\qquad\qquad\qquad \left. {} \cdot \left({\sin(x/n) \over x/n} \right)^k \cos^{n-k}(x/n) \right] \\[12pt] & = \sum_{\text{odd } k \in \{1,\dots,n\}} \left[ (-1)^{(k-1)/2}\ {x^k \over k!} \cdot {} \right.\\ & {} \qquad\qquad \left. {n(n-1)\cdots (n-k+1) \over n^k} \cdot \left({\sin(x/n) \over x/n} \right)^k \cos^{n-k}(x/n) \right] \end{align} $$

Having reached this point, Euler said that if n is an infinitely large integer, then the three factors on the last line above are equal to 1. In modern language, we would say that as n &rarr; &infin;, the factors on the last line approach 1 and we get


 * $$ \sin x = \sum_{\text{odd }k \ge 1} (-1)^{(k-1)/2}\ {x^k \over k!}. $$

However, a problem arises concerning the interchange in the order of two limiting operations.


 * $$ \cos x = \sum_{\text{even }k \ge 0} (-1)^{k/2}\ {x^k \over k!}. $$


 * $$ \cos\left(\sum_{i=1}^\infty \theta_i\right)

=\sum_{\text{even } k \ge 0} ~ (-1)^{k/2} \sum_{\begin{smallmatrix} A \subseteq \{\,1,2,3,\dots\,\} \\ \left|A\right| = k\end{smallmatrix}} \left(\prod_{i \in A} \sin\theta_i \prod_{i \not \in A} \cos\theta_i\right) $$