Thread: Deriving the series of e using binomial theorem

1. Deriving the series of e using binomial theorem

I need help understanding how the series of e derives into the exponential series using the binomial theorem.

Here is a link to a pic of a page in my book, regarding the exponential series:

http://i46.tinypic.com/qz0oat.jpg

A couple of questions:

Where does the [1 + (1/k)]^k come from and why is it used?

Could you clarify the expansion of [1+(1/k)]^k?
I don't understand how it gets to ... k(1/k) + k(k-1)/2! (1/k^2) + ...

How does it end up with a 1 + 1 + 1[1-(1/k)]/2! + ...

Why are you finding the limit of the series?

And finally how do you end up with exponential series

x^n /n! = 1 + x + x^2/2! + ... ?

I'm confused and just really don't understand why or how you end up with everything. Try and keep it simple, please. Help is VERY appreciated.

2. Originally Posted by lobbyistboy
I need help understanding how the series of e derives into the exponential series using the binomial theorem.

Here is a link to a pic of a page in my book, regarding the exponential series:

http://i46.tinypic.com/qz0oat.jpg

A couple of questions:

Where does the [1 + (1/k)]^k come from and why is it used?

Could you clarify the expansion of [1+(1/k)]^k?
I don't understand how it gets to ... k(1/k) + k(k-1)/2! (1/k^2) + ...

How does it end up with a 1 + 1 + 1[1-(1/k)]/2! + ...

Why are you finding the limit of the series?

And finally how do you end up with exponential series

x^n /n! = 1 + x + x^2/2! + ... ?

I'm confused and just really don't understand why or how you end up with everything. Try and keep it simple, please. Help is VERY appreciated.
Remember that the exponential function is such that

$\displaystyle f'(x) = f(x)$.

Suppose you wanted to find the derivative of $\displaystyle a^x$.

Then $\displaystyle f'(x) = \lim_{h \to 0}\frac{a^{x + h} - a^x}{h}$

$\displaystyle = \lim_{h \to 0}\frac{a^xa^h - a^x}{h}$

$\displaystyle = \lim_{h \to 0}\frac{a^x(a^h - 1)}{h}$

$\displaystyle = a^x\lim_{h \to 0}\frac{a^h - 1}{h}$.

Now, if you wanted to find the value of $\displaystyle a$ that makes $\displaystyle f'(x) = f(x)$, you need to let $\displaystyle \lim_{h \to 0}\frac{a^h - 1}{h} = 1$ and solve for $\displaystyle a$.

Then $\displaystyle \lim_{h \to 0}\frac{a^h - 1}{h} = 1$

$\displaystyle \lim_{h \to 0}(a^h - 1) = \lim_{h \to 0}h$

$\displaystyle \lim_{h \to 0}a^h = \lim_{h \to 0}(1 + h)$

$\displaystyle \lim_{h \to 0}(a^h)^{\frac{1}{h}} = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$

$\displaystyle a = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$.

Now suppose $\displaystyle h = \frac{1}{k}$. Then as $\displaystyle h \to 0, k \to \infty$, so

$\displaystyle a = \lim_{k \to \infty}\left(1 + \frac{1}{k}\right)^k$.

Now we just give $\displaystyle a$ the name Euler's Number $\displaystyle e$, since Euler was the one who discovered it, and we can finally say

$\displaystyle e = \lim_{k \to \infty}\left(1 + \frac{1}{k}\right)^k$.

The rest involves using the Binomial theorem to expand $\displaystyle \left(1 + \frac{1}{k}\right)^k$ and letting $\displaystyle k \to \infty$.

3. Thanks, that was extremely helpful. I kind of understand how it was derived.

But one part that still confuses me is how the expanded form of [1+(1/k)]^k simplifies to:

1 + 1 + 1[1-(1/k)]/2! + ...

4. Originally Posted by lobbyistboy
Thanks, that was extremely helpful. I kind of understand how it was derived.

But one part that still confuses me is how the expanded form of [1+(1/k)]^k simplifies to:

1 + 1 + 1[1-(1/k)]/2! + ...
That's just an application of the Binomial Theorem.

$\displaystyle (a + b)^n = \sum_{r = 0}^n{{n\choose{r}}a^{n-r}b^r}$.