Find $f(0)$ if $f(x)= \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}x^{2n-2}}{(2n+1)!} $

Question

Find $f(0)$ if $f(x)= \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}x^{2n-2}}{(2n+1)!} $

This seems trivial but I am not able to figure it out. Do I directly put $x=0$ or should I reindex $n=0$ and use Taylor Theorem. If I use Taylor theorem, then is $f(0)$ simply the $n$ part of the numerator?

Answer 1

Hint

$$f(x)= \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n-2}}{(2n+1)!}=\frac{1}{3!}-\frac{x^2}{5!}+\cdots$$

Answer 2

For $n\ge 2$ the summand is equal to zero, so you need only to consider the term for $n=1$ $$ \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n-2}}{(2n+1)!} =\frac{1}{(2\cdot1+1)!}+ \sum_{n=2}^\infty \frac{(-1)^{n+1}x^{2n-2}}{(2n+1)!}=\frac1{3!}+0=\frac16$$

Answer 3

We have $x^3f(x)=x-\sin x=\frac{x^3}{3!}-\frac{x^5}{5!}+...$

Divide by $x^3$ !