# Back to basics II

1. Apr 26, 2012

### dextercioby

I'm looking for proofs to the 2 following results.

Let

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

Show that:

1. Universality of e.

$$\sum_{k=0}^{\infty} \frac{1}{k!} = e$$

2. Derivative of $e^x$.

$$(e^x)' = e^x, ~ \forall x\in\mathbb{R}$$

Searching google didn't get me satisfactory results.

2. Apr 26, 2012

### brydustin

The classic definition is based on the fact that d(2^x)/dx ~ 0.693147 and d(3^x)/dx~1.098612 at x=0. Then you guess that there must be a number such that f'(0)=1 where f=a^x (it turns out a =e and is unique).

===
g(x)=b^x

g'(x) =lim h->0 b^(x+h)-b^x/h
=lim h->0 (a^x)(a^h -1)/h
=(a^x) lim h->0 (a^h -1)/h
and the derivative is where at x=0 we get
lim h->0 (a^h -1)/h = f'(0)

f ' (0) = (2^x)(a^h -1)/h ~ 0.693147
f ' (0) = (3^x)(a^h -1)/h ~1.098612
So it seems like there should be a number that converges to 1.
But I don't know the proof that you mentioned. Hope that helps somewhat.

3. Apr 26, 2012

### brydustin

Also, if you assume the geometric definition of natural log and its derivative then the derivative of e^x can be computed from the fact that

d(ln(exp(x))/dx = (d/dx (e^x))*(1/e^x) = 1 >>> d/dx (e^x) = e^x
probably not the most satisfactory answer since it pushes off the proof to something contingent on the natural log. But its perfectly accurate.

Of course you can also try the taylor expansion but that's not really "first principals"

4. Apr 26, 2012

### D H

Staff Emeritus
Lemma:
$$\lim_{n\to\infty}\frac{n!}{n^k (n-k)!} = 1$$
Use the above in conjunction with the binomial expansion of
$$e=\lim_{n\to\infty}\left(1+\frac 1 n\right)^n$$

You need a definition of exp(x) for this. One definition of exp(x) is that it is the function that is equal to its own derivative such that exp(0)=1. That makes the proof a bit too easy. Try using
$$\exp(x)=\lim_{n\to\infty}\left(1+\frac x n\right)^n$$

5. Apr 29, 2012

### deluks917

First one follows from second if you use taylor series.