Taylor polynomial of degree 2n-1

Question

So I am asked to find the Taylor polynomial of degree 2n-1 for $f(x)=\sinh(x)$ centered at 0.

I have the general formula for a Taylor series but I'm not really sure how to proceed. I can't really take the 2n-1'st derivative or anything.

Can someone suggest how I start this problem?

Answer 1

Hint

Why not to start from $$\sinh(x)=\frac{e^x-e^{-x}}2$$

Edit

I suppose that your professor would not enjoy the following lazy way : starting from the trigonometric identity $$\sin(ix)=i \sinh(x)\implies \sinh(x)=-i \sin(ix)$$ then, using Taylor expansion of the sine,

$$\sinh(x) =- i\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} (ix)^{2n+1}=- \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!}i^{2n+2} x^{2n+1}= \sum_{n=0}^\infty \frac{ x^{2n+1}}{(2n+1)!}$$

If he/she has a sense of humour, it could be interesting to ask him/her what would have been his/her reaction facing such an approach.