# What i call the second Euler's Number

1. Jul 30, 2007

### x2thay

Everyone of us know about the famous euler's number, which is e, which is aproximately 2.7182818...........(as far as i cam remember)...which is used for many things in calculus...

Well, i was wondering where the heck does e come from and i realize after searching in the web, that e is the limit as x aproaches to the infinite, of the function f(x)=(1+1/x)^x or f(x)=(1+x)^(1/x). And this is awesome! Because i had no idea e was a limit! lol

Anyway, next thing i was wondering about was: what if the '+' sinal in the function f(x)=(1+1/x)^x gets switched to an '-'? The function would become: f(x)=(1-1/x)^x.

Here are the functions: http://www.geocities.com/just_dre/e3e.GIF

Well, i graphed both functions and as expected, f(x)=(1+1/x)^x aproachs to e. But the other function (the grey one) has a different limit which is aproximately 0,3678794409875026009331610590813.......

My question is: does this new irrational constant have any meaning? If f(x)=(1+1/x)^x has, why can't f(x)=(1-1/x)^x?

Hope you can help me...^^

iMiguel

2. Jul 30, 2007

### quasar987

it's fairly easy to see that for any real number y,

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

In particular,

$$\lim_{x\rightarrow \infty}\left(1-\frac{1}{x}\right)^x = e^{-1}$$

Note also that e^y can equivalently be defined in terms of its power series expansion as the solution to the differential equation f '(y)=f(y).

Last edited: Jul 30, 2007
3. Apr 24, 2008

### a.a

The number is defined as the limit of (1+1/n)^n as n approched infinity.
My question is why is that limit the number e. If you evaluate it useing the subsistutaion method, you get the limit is 1.

E is also defined as the limit of (1+n)^(1/n) as n approaches 0. Using the substitution method isn't this limit also 1?

4. Apr 24, 2008

### CompuChip

One could take that as a definition. Or one could prove that it satisfies the definition you take for e, for example, if you define e to be the number such that $de^y/dy = e^y$ you can differentiate the expression in quasars post and show that it equals its own derivative in the limit $x \to \infty$, or you can plug in the power expansion and show that it is the same number.

You made me curious, can you show us how you do that?

5. Apr 24, 2008

### Hurkyl

Staff Emeritus
No you don't; $1^{+\infty}$ is an indeterminate form.