# e^2x = (1+2x/n)^n for large n?

• Jul 17th 2007, 07:36 AM
CrazyAsian
e^2x = (1+2x/n)^n for large n?
The task is to show why e^2x = (1+2x/n)^n for very large n. I'm not sure how to tackle this problem. I thought about creating the generating functions and doing some algebraic manipulation but couldn't finish it.
• Jul 17th 2007, 07:58 AM
galactus
Try using the fact that $\displaystyle \lim_{n\rightarrow{\infty}}\left(1+\frac{1}{n}\rig ht)^{n}=e$

Your problem is a variation on a theme.
• Jul 17th 2007, 07:58 AM
red_dog
The sequence $\displaystyle \displaystyle a_n=\left(1+\frac{2x}{n}\right)^n$ converges to $\displaystyle e^{2x}$ when $\displaystyle n\to\infty$
$\displaystyle \displaystyle \lim_{n\to\infty}\left(1+\frac{2x}{n}\right)^n=\li m_{n\to\infty}\left[\left(1+\frac{2x}{n}\right)^{\frac{n}{2x}}\right]^{\frac{2x}{n}\cdot n}=e^{2x}$
• Jul 17th 2007, 08:53 AM
CrazyAsian
Thanks but...
I don't think that's the method they are looking for because the next question is proving:

E^(-x^2/2) = Cos[x/(n^0.5)]^n as n approaches infinity.
• Jul 17th 2007, 10:08 AM
red_dog
$\displaystyle \displaystyle\lim_{n\to\infty}\left(\cos\frac{x}{\ sqrt{n}}\right)^n=\lim_{n\to\infty}\left(1+\cos\fr ac{x}{\sqrt{n}}-1\right)^n=\lim_{n\to\infty}\left(1-2\sin ^2\frac{x}{2\sqrt{n}}\right)^n=$
$\displaystyle \displaystyle =e^{\displaystyle -\lim_{n\to\infty}2\sin ^2\frac{x}{2\sqrt{n}}\cdot n}=e^{\displaystyle -\lim_{n\to\infty}\left(\frac{\sin \frac{x}{2\sqrt{n}}}{\frac{x}{2\sqrt{n}}}\right)^2 \cdot\frac{x^2}{2}}=e^{-\displaystyle \frac{x^2}{2}}$
• Jul 17th 2007, 10:09 AM
CaptainBlack
Quote:

Originally Posted by CrazyAsian
The task is to show why e^2x = (1+2x/n)^n for very large n. I'm not sure how to tackle this problem. I thought about creating the generating functions and doing some algebraic manipulation but couldn't finish it.

Suppose $\displaystyle n \gg 2x$, then expand the RHS using the binomial
expansion and compare with the power series expansion of the LHS.

RonL
• Jul 17th 2007, 10:39 AM
CaptainBlack
Quote:

Originally Posted by CrazyAsian
The task is to show why e^2x = (1+2x/n)^n for very large n. I'm not sure how to tackle this problem. I thought about creating the generating functions and doing some algebraic manipulation but couldn't finish it.

Another method is to define:

$\displaystyle f_n(x)=(1+2x/n)^n$

then

$\displaystyle f'_n(x)=2 f_n(x) (1+O(x/n))$

which is of seperable type so:

$\displaystyle \int \frac{1}{f_n} df_n = \int (2+O(x/n)) dx$

or:

$\displaystyle f_n(x)= Ae^{2x+O(x^2/n)}$

and as $\displaystyle f_n(0)=1,\ A=1$ (though a bit of care may be needed to show this)

Hence for fixed $\displaystyle x$ we have:

$\displaystyle \lim_{n \to \infty} f_n(x) = e^{2x}$

RonL
• Jul 17th 2007, 10:54 AM
ThePerfectHacker
You can use this.
• Jul 17th 2007, 11:33 AM
galactus
Here's a fairly simple proof I just scratched down.

$\displaystyle \lim_{n\rightarrow{\infty}}\left(1+\frac{2x}{n}\ri ght)^{n}$

Let $\displaystyle a=\frac{2x}{n}, \;\ n=\frac{2x}{a}$

$\displaystyle \lim_{a\rightarrow{0}}\left(1+a\right)^{\frac{2x}{ a}}$

$\displaystyle \lim_{a\rightarrow{0}}[\underbrace{\left(1+a\right)^{\frac{1}{a}}}_{\text {limit=e}}]^{2x}=e^{2x}$