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 June 5th, 2010, 06:05 PM #1 Newbie   Joined: Jun 2010 Posts: 2 Thanks: 0 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.
 June 6th, 2010, 06:07 AM #2 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Re: Deriving the series of e using binomial theorem First, let me talk briefly about the nature of math. There is much more to math than simple rote memorization and arithmetic, no matter how complicated. There are moments of insight and creativity in the creation of new math, just as in other fields of research. So the answer to your first question, why is (1 + 1/k)^k used, is unfortunately unsatisfying. It is used because it works and yields the correct answer. How did someone first come up with this? That's a real mystery - it was quite an insight. Let's put things in the right context. You have a definition for e, which is that power series 1 + 1 + 1/2 + ... The point of this page is that if you happen to look at (1 + 1/k)^k using the binomial theorem, as k goes to infinity, we happen to get that series 1 + 1 + 1/2 + ... and accordingly we get e. But let's clarify the arithmetic involved, too. It's actually not so bad. Here is the binomial theorem. Follow it exactly to get the expansion of (1 + 1/k)^k. Then, the book simplifies the terms. For example, in the second term $\frac{(k(k-1))}{2!} * (\frac{1}{k})^2= \frac{(1 - \frac{1}{k})}{2!}$ Only by distributing the 1/k^2 term. Does that solve everything?
 June 6th, 2010, 07:11 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,981 Thanks: 1166 Math Focus: Elementary mathematics and beyond Re: Deriving the series of e using binomial theorem
 June 6th, 2010, 04:51 PM #4 Newbie   Joined: Jun 2010 Posts: 2 Thanks: 0 Re: Deriving the series of e using binomial theorem WOW! Thanks Jason for the in-depth post. That clears up pretty much everything I was confused about. One question though, could you explain why in the exponential series you replace the 1's with an x^n? I'm not that much of a math genius
 June 7th, 2010, 12:25 AM #5 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Re: Deriving the series of e using binomial theorem When we think of the Taylor series expansion for e^x, we think of 1 + x + (1/2)x^2 + ... I believe you are asking why in this expansion, we have all the x-terms. Unfortunately, as far as I know, there is no way to derive the series expansion from the series 1 + 1 + 1/2 + ... Instead, using Taylor's Formula from calculus, it is possible to derive 1 + x + (1/2)x^2 + ... and then, by plugging in 1 for x, we get the stated expansion for the value of e.

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# binomial expansion of exponential functions

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