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.

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- Jul 17th 2007, 07:36 AMCrazyAsiane^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 AMgalactus
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 AMred_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 AMCrazyAsianThanks 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 AMred_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 AMCaptainBlack
- Jul 17th 2007, 10:39 AMCaptainBlack
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 AMThePerfectHacker
You can use this.

- Jul 17th 2007, 11:33 AMgalactus
Here's a fairly simple proof I just scratched down.

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

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

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

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