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

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?

• You are trying to find radius of convergence right? – Nimish Mar 28 at 20:27
• Unfortunately, you $a_n$'s are wrong. – Yves Daoust Mar 28 at 20:38
• @J.Doe: you just edited, didn't you ? – Yves Daoust Mar 28 at 20:40
• Also fix your $a^n$. – Yves Daoust Mar 28 at 20:42
• No, pay attention. $a^n\leftrightarrow a_n$. – Yves Daoust Mar 28 at 20:45

## 2 Answers

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 $$x$$.

• I misstated my question i meant you have to find an for which it converges – Hello there Mar 28 at 20:37

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.