# Rewriting e as an infinite series

• Jun 3rd 2012, 12:29 PM
scounged
Rewriting e as an infinite series
I'm given the function $f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ and I'm supposed to prove that $f(x)=e^x$.
I was thinking that I should use a different representation of e, namely:

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

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?
• Jun 3rd 2012, 01:51 PM
Plato
Re: Rewriting e as an infinite series
Quote:

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

The answer to this depends upon the level of rigor required in proof.
There is one well respected calculus textbook simply points out that $f(x)=f'(x)$ in this case.
• Jun 3rd 2012, 02:13 PM
scounged
Re: Rewriting e as an infinite series
Quote:

Originally Posted by Plato
There is one well respected calculus textbook simply points out that $f(x)=f'(x)$ in this case.

Does that mean that $f(x)=e^x$ 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).
• Jun 3rd 2012, 02:26 PM
Plato
Re: Rewriting e as an infinite series
Quote:

Originally Posted by scounged
Does that mean that $f(x)=e^x$ 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, $f(x)=0$. But clearly that is not the case here.
• Jun 3rd 2012, 02:37 PM
scounged
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.
• Jun 3rd 2012, 06:19 PM
Prove It
Re: Rewriting e as an infinite series
Quote:

Originally Posted by scounged
I'm given the function $f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ and I'm supposed to prove that $f(x)=e^x$.
I was thinking that I should use a different representation of e, namely:

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

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 \begin{align*} e^x \end{align*} as a MacLaurin series.

\displaystyle \begin{align*} f(x) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(0) \cdot x^n}{n!} \end{align*}
• Jun 4th 2012, 06:12 AM
HallsofIvy
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, $f(x)=0$. But clearly that is not the case here.

Not quite. $f(x)= Ae^x$, for A any constant (which, of course, includes 1 and 0), has that property.
• Jun 4th 2012, 12:13 PM
scounged
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.