How are $\displaystyle \lim_{x \to \infty}(1+\frac{1}{x})^x$ and $\displaystyle 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} ...$ similar or related to each other, other than their results being $\displaystyle e$?

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- Apr 25th 2006, 10:52 PM #1
## e question

How are $\displaystyle \lim_{x \to \infty}(1+\frac{1}{x})^x$ and $\displaystyle 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} ...$ similar or related to each other, other than their results being $\displaystyle e$?

- Apr 26th 2006, 11:30 AM #2

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- Apr 26th 2006, 12:58 PM #3

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Originally Posted by**cinder**

$\displaystyle

e_1(x)=\lim_{k \to \infty}e_1(x,k)

$

where:

$\displaystyle

e_1(x,k)=\left(1+\frac{x}{k}\right)^k

$

and:

$\displaystyle

e_2(x)=1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} ...

$

expand $\displaystyle e_1(x,k)$ and look at the behaviour of the coefficients of

powers of $\displaystyle x$ that occur in both, and consider what happens when

$\displaystyle k\to \infty$.

Maybe?

RonL

- Apr 26th 2006, 01:14 PM #4

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Originally Posted by**cinder**

$\displaystyle \ln x=\int^x_1 \frac{dx}{x}$ for $\displaystyle x>0$.

And define, $\displaystyle e$ such as $\displaystyle \ln e=1$ (it does exists and well-defined.)

Then we can*prove*that,

$\displaystyle e^x=1+x+\frac{x^2}{2!}+...$

Where $\displaystyle e^x$ is the inverse function of $\displaystyle \ln x$ because this function is bijective.

Thus if, $\displaystyle x=1$ then,

$\displaystyle e=1+1+\frac{1}{2!}+\frac{1}{3!}+...$

- Apr 26th 2006, 01:47 PM #5

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You don't have to define exp(x) as the inverse of log(x) which you defined through an integral. That's certainly

*possible*(and is often done) but is not of any use here since we can just as well start by defining exp(x) (and after that define log(x) as its inverse, though that's not necessary here).

What the topic starter was asking, I think, is to prove the equivalence between these two possible definitions. Of course, this follows trivially from the equivalence of the analogous definition for exp(x), but proving that equivalence would be a bit 'overkill' to prove this one.

- Apr 26th 2006, 02:04 PM #6

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Originally Posted by**TD!**

- Apr 26th 2006, 02:08 PM #7

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That's a bit strange, there aren't too many ways on how log(x) and exp(x) are conventionally defined. I know at least 3 books (and thus authors, i.c. professors) who start by defining log(x) as that integral; then defining exp(x) as its inverse (which is why I remarked that it is often done ).

One of them being "my professor" (the one I had for analysis); on the other hand you have for example Rudin (from 'Principles of Mathematical Analysis') who defines exp(x) first, as the series.

There are indeed different possibilities, but fundamentally there aren't too many different definitions in use (log through the integral, the series, the limit, the initial value probem y' = y with y(0) = 1).

- Apr 26th 2006, 02:13 PM #8

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- Apr 26th 2006, 02:18 PM #9

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I see, that's true of course although I still think you'll find that many functions are often defined in a similar way - within the rather limited number of fundamentally different ways of defining them of course.

Apart from that, I just read the original question carefully. What are you exactly looking for? As said before, both expressions are completely equivalent which means, in this case, that they're the same - being e as you said already. You can prove this, but is that what you're looking for?

- Apr 26th 2006, 06:12 PM #10
I couldn't tell you exactly what I'm looking for. My calculus teacher mentioned it might be part of a bonus question on a test, but wanted us to do figure it out. After looking but not really finding anything, I figured I'd ask here.

Thanks for your replies!