Rewriting e as an infinite series

I'm given the function $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ and I'm supposed to prove that $\displaystyle f(x)=e^x$.

I was thinking that I should use a different representation of e, namely:

$\displaystyle e^x=\lim_{n \rightarrow \infty} (1+\frac{x}{n})^n$

and use the binomial theorem to expand it as an infinite series, but I can't get it to work properly. If it's possible, then how should it be done?

Otherwise, any hints on how I should prove it instead?

Re: Rewriting e as an infinite series

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**scounged** I'm given the function $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ and I'm supposed to prove that $\displaystyle f(x)=e^x$.

The answer to this depends upon the level of rigor required in *proof*.

There is one well respected calculus textbook simply points out that $\displaystyle f(x)=f'(x)$ in this case.

Re: Rewriting e as an infinite series

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**Plato** There is one well respected calculus textbook simply points out that $\displaystyle f(x)=f'(x)$ in this case.

Does that mean that $\displaystyle f(x)=e^x$ is the only function that equals its own derivative?

If that's the case, then I think that's the way I'm intended to prove it (in a previous part of this assignment I'm supposed to prove that f(x)=f'(x).

Re: Rewriting e as an infinite series

Quote:

Originally Posted by

**scounged** Does that mean that $\displaystyle f(x)=e^x$ is the only function that equals its own derivative?

If that's the case, then I think that's the way I'm intended to prove it (in a previous part of this assignment I'm supposed to prove that f(x)=f'(x).

Well there is a trivial function, $\displaystyle f(x)=0$. But clearly that is not the case here.

Re: Rewriting e as an infinite series

Then I think I'll ask my teacher tomorrow if proving f(x)=f'(x) is enough. Thanks for the help.

Re: Rewriting e as an infinite series

Quote:

Originally Posted by

**scounged** I'm given the function $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ and I'm supposed to prove that $\displaystyle f(x)=e^x$.

I was thinking that I should use a different representation of e, namely:

$\displaystyle e^x=\lim_{n \rightarrow \infty} (1+\frac{x}{n})^n$

and use the binomial theorem to expand it as an infinite series, but I can't get it to work properly. If it's possible, then how should it be done?

Otherwise, any hints on how I should prove it instead?

Instead of using the Binomial Theorem, expand $\displaystyle \displaystyle \begin{align*} e^x \end{align*} $ as a MacLaurin series.

$\displaystyle \displaystyle \begin{align*} f(x) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(0) \cdot x^n}{n!} \end{align*} $

Re: Rewriting e as an infinite series

Quote:

Does that mean that is the only function that equals its own derivative?

Quote:

Originally Posted by

**Plato** Well there is a trivial function, $\displaystyle f(x)=0$. But clearly that is not the case here.

Not quite. $\displaystyle f(x)= Ae^x$, for A any constant (which, of course, includes 1 and 0), has that property.

Re: Rewriting e as an infinite series

Yeah, I found it in my formula collection. If I had known about MacLaurin series yesterday I don't think I would've created this thread. Anyway, thanks for the help.