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

  1. #1
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    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.
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  2. #2
    Eater of Worlds
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    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.
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  3. #3
    MHF Contributor red_dog's Avatar
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    The sequence $\displaystyle \displaystyle a_n=\left(1+\frac{2x}{n}\right)^n$ converges to $\displaystyle e^{2x}$ when $\displaystyle n\to\infty$
    $\displaystyle \displaystyle \lim_{n\to\infty}\left(1+\frac{2x}{n}\right)^n=\li m_{n\to\infty}\left[\left(1+\frac{2x}{n}\right)^{\frac{n}{2x}}\right]^{\frac{2x}{n}\cdot n}=e^{2x}$
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  4. #4
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    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.
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  5. #5
    MHF Contributor red_dog's Avatar
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    $\displaystyle \displaystyle\lim_{n\to\infty}\left(\cos\frac{x}{\ sqrt{n}}\right)^n=\lim_{n\to\infty}\left(1+\cos\fr ac{x}{\sqrt{n}}-1\right)^n=\lim_{n\to\infty}\left(1-2\sin ^2\frac{x}{2\sqrt{n}}\right)^n=$
    $\displaystyle \displaystyle =e^{\displaystyle -\lim_{n\to\infty}2\sin ^2\frac{x}{2\sqrt{n}}\cdot n}=e^{\displaystyle -\lim_{n\to\infty}\left(\frac{\sin \frac{x}{2\sqrt{n}}}{\frac{x}{2\sqrt{n}}}\right)^2 \cdot\frac{x^2}{2}}=e^{-\displaystyle \frac{x^2}{2}}$
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  6. #6
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    Quote Originally Posted by CrazyAsian View Post
    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
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by CrazyAsian View Post
    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
    Last edited by CaptainBlack; Jul 17th 2007 at 12:15 PM.
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  8. #8
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    You can use this.
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  9. #9
    Eater of Worlds
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    Here's a fairly simple proof I just scratched down.

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

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

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

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