Prove the series (x^n)/(n!) is e^x

**uberbandgeek6** · November 16th 2009, 02:57 AM

Problem: Prove the series (x^n)/(n!) is e^x

I'm not really sure where to start here. We have just learned power series in class, but not specifically Taylor series yet, so I can use any of those techniques yet. I know that the derivative of the series remains the same, but I don't know how to apply this to prove it is e^x.

**tonio** · November 16th 2009, 03:02 AM

Originally Posted by uberbandgeek6

Problem: Prove the series (x^n)/(n!) is e^x

I'm not really sure where to start here. We have just learned power series in class, but not specifically Taylor series yet, so I can use any of those techniques yet. I know that the derivative of the series remains the same, but I don't know how to apply this to prove it is e^x.

I think the solution is already in your message, but obviously you haven't yet studied this or perhaps you forgot it: there is one unique function which is infinitely differentiable everywhere, which equals 1 at zero and which equals its own derivative, and this is the exponential function $e^x$ ...
Perhaps there's another way to solve your problem, but I can't remember it right now.

Tonio

**chisigma** · November 16th 2009, 04:42 AM

Two expressions of exponential function are...

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

$e^{x} = \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{x^{k}}{k!}$ (2)

Now we prove that (1) and (2) are equivalent. We start from the binomial expansion...

$(1+\frac{x}{n})^{n} = \sum_{k=0}^{n} \binom{n}{k}\cdot (\frac{x}{n})^{k} = 1 + x + \sum_{k=2}^{n} a_{n,k}\cdot \frac{x^{k}}{k!}$ (3)

... where...

$a_{n,k} = \prod_{i=1}^{k-1} (1-\frac{i}{n})$ (4)

In order to well understand the next step we write the sequence of polynomials taking in evidence the powers of x from 0 to 3...

$(1+\frac{x}{1})^{1} = 1+x$

$(1+\frac{x}{2})^{2} = 1+x + (1-\frac{1}{2})\cdot \frac{x^{2}}{2!}$

$(1+\frac{x}{3})^{3} = 1+x + (1-\frac{1}{3})\cdot \frac{x^{2}}{2!} + (1-\frac{1}{3})\cdot (1-\frac{2}{3})\cdot \frac{x^{3}}{3!}$

$\dots$

$(1+\frac{x}{n})^{n} = 1+x + (1-\frac{1}{n})\cdot \frac{x^{2}}{2!} + (1-\frac{1}{n})\cdot (1-\frac{2}{n})\cdot \frac{x^{3}}{3!} + \dots$

$\dots$ (5)

We now observe the coefficients of the $x^{k}$ 'by colums'. For $k=0$ and $k=1$ they are $1$ and they don't change by increasing $n$ . For $k=2$ the coefficient is $\frac{1-\frac{1}{n}}{2!}$ and for $n \rightarrow \infty$ it tends to $\frac{1}{2!}$ . For $k=3$ the coefficient is $\frac{(1-\frac{1}{n})\cdot (1-\frac{2}{n})}{3!}$ and for $n \rightarrow \infty$ it tends to $\frac{1}{3!}$ . In general is $\lim_{n \rightarrow \infty} a_{n,k}=1$ so that for $n \rightarrow \infty$ the coefficient of $x^{k}$ tends to $\frac{1}{k!}$ . The conlusion is that...

$\lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n} = \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{x^{k}}{k!}$ (6)

Kind regards

$\chi$ $\sigma$

**uberbandgeek6** · November 16th 2009, 02:03 PM

Wow, what a mouthful. I cant say I really understand it well, but I'm really impressed. Thanks!

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