Taylor's series formula for $\sin i$

Question

After learning that some common functions can be extended to complex using Taylor's theorem, out of curiosity I tried evaluate $\sin i$. I got $$\sin(i)=i-\frac{i^3}{3!}+\frac{i^5}{5!}-...=i+\frac{i}{3!}+\frac{i}{5!}+...=i\sum_{j=0}^\infty \frac{1}{(2j+1)!}$$

How do I evaluate this? (I think) I know this converges, but what does it converges to?

Answer 1

I am not very clear on what you want to do. However, if you want to evaluate the complex sine function for a complex variable, you could use this formula obtained from Euler's formula:

$$\sin(iy) = \frac{e^{-y}-e^y}{2i} = i \left(\frac{e^y-e^{-y}}{2}\right) = i \sinh(y).$$

In your case, $y=1$.

Answer 2

Hint The Taylor series for the hyperbolic sine function $\sinh x = \frac{1}{2}(e^x - e^{-x})$ is $$\sinh x \sim x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = \sum_{j = 0}^\infty \frac{x^{2 j + 1}}{(2 j + 1)!} .$$

Answer 3

Now, make the same things with $e^1$ and $e^{-1}$.

Actually, $$e^x=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+...$$

Answer 4

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$ $$e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots$$ $$e^{-1}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots$$ $$\implies \frac{e-e^{-1}}{2}=\frac{1}{1!}+\frac{1}{3!}+\cdots$$