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?

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

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 $x$.

  • $\begingroup$ I misstated my question i meant you have to find an for which it converges $\endgroup$ – 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.


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