Thread: Rewriting e as an infinite series

I'm given the function and I'm supposed to prove that .
I was thinking that I should use a different representation of e, namely:



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?

Originally Posted by scounged

I'm given the function and I'm supposed to prove that .

The answer to this depends upon the level of rigor required in proof.
There is one well respected calculus textbook simply points out that in this case.

Originally Posted by Plato

There is one well respected calculus textbook simply points out that in this case.

Does that mean that 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).

Originally Posted by scounged

Does that mean that 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, . But clearly that is not the case here.

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

Originally Posted by scounged

I'm given the function and I'm supposed to prove that .
I was thinking that I should use a different representation of e, namely:



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 as a MacLaurin series.

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

Originally Posted by Plato

Well there is a trivial function, . But clearly that is not the case here.

Not quite. , for A any constant (which, of course, includes 1 and 0), has that property.

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.