Question: Using $$ \sinh x = x + \tfrac{x^3}{3!}+ \tfrac{x^5}{5!} + \tfrac{x^7}{7!}+ \cdots$$ as the Standard Power Series. This series takes a very long time to run. Can it be written without using the exponentials divided by a huge factorial. The example functions in [Is there a way to get trig functions without a calculator?](https://math.stackexchange.com/q/501660/409) using the "Tailored Taylor" series representation for sin and cosine are very fast and give the same answers. I want to use it within my calculator program.

Thank you very much.

Answer: 
Note that $$\sinh x=\frac{e^x-e^{-x}}2$$
So all you need is a fast way to calculate the exponential $e^x$. You can use the regular Taylor series, but that's slow. So you can use the definition $$e^x=\lim\_{n\to\infty}\left(1+\frac xn\right)^n$$
For calculation purposes, use $n$ as a power of $2$, $n=2^k$. You calculate first $y=1+\frac x{2^k}$, then you repeat the $y=y\cdot y$ operation $k$ times. I've got the idea about calculating the fast exponential from this [article](https://codingforspeed.com/using-faster-exponential-approximation/).
Let me consider the problem from a computing point of view assumin that you do not know how to compute $e^x$.

The infinite series is 
$$\sinh(x)=\sum\_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}$$ If you compute each term independently of the other, for sure, it is expensive since you have to compute each power of $x$ as well as each factorial.

But suppose that you write instead
$$\sinh(x)=\sum\_{n=0}^\infty T\_n \qquad \text{where} \qquad T\_n=\frac{x^{2n+1}}{(2n+1)!}\qquad \text{and} \qquad T\_0=x$$ then
$$T\_{n+1}= \frac {t\,\, T\_n}{m(m+1)}\qquad \text{where} \qquad t=x^2\qquad \text{and} \qquad m=2n+2$$ This would be much less expensive in terms of basic operations and number of them.

You could use the same trick for most functions expressed as infinite series.