# Pfoof of e^x

1. Nov 17, 2004

### Alem2000

I was looking at my math text and I saw that $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ and it sugests that you prove it to yourself. So this is what I dont get if $$f(x)=e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ then $$\frac{d}{dx}f(x)$$ should equal that..now im not qeustioning that its true im just having a hard time proving it to myself this is what I did
$$\frac{d}{dx}(e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!})=\sum_{n=0}^{\infty}\frac{nx^n-1}{n!}$$ and after that I just set them equal. Now I will warn you im horrible at factorials so I didnt simplify and this is what turns out $$\sum_{n=0}^{\infty}\frac{nx^n-1}{n!}=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ now can somone tell me what I need to correct or simplify?

2. Nov 17, 2004

### NateTG

For $n>1$,
$$n!=\prod_{i=1}^{n}i=n \prod_{i=1}^{n-1}i=n (n-1)!$$
($\prod$ is like $\sum$ except that the numbers are multiplied instead of added.)
PS. You should put the exponent in {}'s so that you get
$$\sum_{n=0}^{\infty}\frac{nx^{n-1}}{n!}$$

3. Nov 17, 2004

### Alem2000

thanks alot, im gonna give it a shot

4. Nov 17, 2004

### James R

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$

Differentiating, we get:

$$\frac{d}{dx}e^x=\sum_{n=1}^{\infty}n\frac{x^{n-1}}{n!} = \sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!}$$

Put m=n-1, and that sum reduces to the original sum.

Note that the first term in the original series is 1, which, when differentiated, gives zero. This is why the sum in the differentiated series starts at n=1 rather than n=0.

5. Nov 18, 2004

### Alem2000

quick question $$cosx=\sum_{n=0}^{\infty}\frac{(-1^n)(x^2n)}{2n!}$$ so if you want to know $$\frac{cosx-1}{x^2}$$ all you have to

do is $$\sum_{n=0}^{\infty}\frac{(-1^n)(x^2n)-(2n!x^2)}{2n!x^2}$$

is that the way to go about it if you dont want to write out the terms?

6. Nov 18, 2004

### Galileo

I can't really see what you were doing to obtain that expression.

Notice the first term (n=0) in the series for cos(x) is 1, so:

$$\cos(x)-1 = \sum_{n=1}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$$.

and

$$\frac{\cos(x)-1}{x^2} = \sum_{n=1}^{\infty}\frac{(-1)^nx^{2n-2}}{(2n)!}$$.

7. Nov 18, 2004

### Alem2000

what I did was divide the whole thig on the right side with $$x^2$$ and subtracted $$1$$ from it then I simplify...can you do that to ther series?

8. Nov 22, 2004

### mruncleramos

Problem with the reasoning for the original question: You try to prove that the infinite series is equal to the aforementioned function, but doing so, you merely differentiate the series. This is invalid because all you have shown is that the series is it's own derivative, without proving that e^x is the only such function with the aforementioned property.

9. Nov 23, 2004

### mathwonk

once you have shown that the series defiens a differentiable function f which equals its own derivative, and such that f(0) = 1, it is easy to see that it equals e^x.

(Apply MVT to the result of differentiating f/e^x.)

The harder part is to show that this series defines a differentiable function and that you can differentiate the series term by term.

The easiest way to see this is to use the ratio test for convergence of a series in the sup norm, to show that the series above does converge to a continuous function, uniformly on every closed bounded interval. (On the interval [-r,r], compare with the geometric series with ratio r.) Then deduce that the series of indefinite integrals (starting from 0), converges to the indefinite integral of f. Since in fact the series of indefinite integrals visibly equals f-1, we get that f' = f.

Last edited: Nov 23, 2004