The taylor series is given determine $a_n$ for which x converges to f(x)

Question

The taylor series of the function f(x) = $1-\mathrm{e}^{-x^2}$ around x = 0 is given by $\sum_{n=0}^{\infty} a_nx^n$

determine $a_n$ for all n $\geq$ 0 and give for which value of x the series converges to f(x)

I've come to this result $\sum_{n=1}^{\infty} \dfrac{\left(-1\right)^{n+1}x^{2n}}{n!}$ = $\sum_{n=0}^{\infty} a_nx^n$

Anyone got ideas how to solve this further on?

Answer 1

It seems that next you want to find the radius of convergence $R$. You can use the ratio test to see that $R = \infty$. Consider $$\bigg\lvert \frac{(-1)^{n+2}x^{2(n+1)}}{(n+1)!} \cdot \frac{n!}{(-1)^{n+1}x^{2n}}\bigg\rvert= \frac{\lvert x \rvert^2}{n+1} \to 0 \text{ as } n \to \infty.$$

Edit: Since $R=\infty$, it follows that the power series converges (to $f(x)$) for all real $x$.

Answer 2

The exponential function has the well-known Taylor series

$$\sum_{k=0}^\infty \frac{x^k}{k!}$$ which converges for all real values.

Substituting $-x^2$ for $x$, you will get an entire series, which is indeed the Taylor series of $e^{-x^2}$.

You can conclude.