To provide context, the continued fraction in the form $\frac{a_0}{1-\frac{a_1}{1+a_1-\frac{a_2}{1+a_2-...}}}$ evaluated to the $n$th denominator equals $\sum_{k=0}^{n}\prod_{j=0}^{k}a_j$. If one wants to write $e^x$ as a continued fraction, you can take the Maclaurin series of it which is $1+\sum_{n=1}^{\infty}x^n/n!$ and rewrite that as $1+\sum_{n=1}^{\infty}\prod_{i=1}^{n}(x/i)$. Evaluating each $a_i$ to be in the form $x/i$, one can arrive at the continued series representation of $e^x$.
When I was trying to replicate this (rewriting other functions in the summation of products form), I had trouble with functions like log of x and trigonometric functions.
Can anybody help with getting $\sin{x}$ and $\cos{x}$ into the form $c+\sum_{k=0}^{n}\prod_{j=0}^{k}a_j$ where $c$ is either $0$ or $1$? The inverses of the function as well?
Lets write this up clearly, you want $\sin(x)$ to be written in a similar way to how you wrote $e^x$. This special form is essentially something like: $$ 1 + \frac{x}{1} + \frac{x}{1}\frac{x}{2} + \frac{x}{1}\frac{x}{2}\frac{x}{3} + \frac{x}{1}\frac{x}{2}\frac{x}{3}\frac{x}{4} + \frac{x}{1}\frac{x}{2}\frac{x}{3}\frac{x}{4}\frac{x}{5} + \cdots$$ $\sin(x)$ can also be written in this special way as well: $$ 0 + \frac{x}{1} + 0\cdot 0 - \frac{x}{1}\frac{x}{2}\frac{x}{3} + 0\cdot 0\cdot 0 \cdot 0 + \frac{x}{1}\frac{x}{2}\frac{x}{3}\frac{x}{4}\frac{x}{5} + \cdots$$ half of the products you want to write are all zeros. Hope that helps.