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Proof for exponentials

  1. Mar 14, 2006 #1
    hello, I need the proof to show that:
    [tex]
    e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
    [/tex]

    Here's what I was able to come up with so far:
    since the derivative of [tex]e^x[/tex] is also [tex]e^x[/tex],

    then let f(x) = [tex]e^x[/tex]
    thus:
    D(f(x)) = [tex]\lim_{n\rightarrow\infty} \frac{f(x+h) - f(x)}{h} = \lim_{n\rightarrow\infty} \frac{e^{x+h} - e^x}{h} = e^x\lim_{n\rightarrow\infty} \frac{e^h - 1}{h}[/tex]

    so for the derivative of [tex]e^x[/tex] to equal itself,
    [tex]\lim_{n\rightarrow\infty} \frac{e^h - 1}{h} = 1[/tex]

    so for small values of h, we can write:
    [tex]e^h - 1 = h[/tex]
    and so
    [tex] e = (1+h)^{1/h}[/tex]

    Replacing h by 1/n, we get:
    [tex] e = (1 + 1/n)^n[/tex]
    As n gets larger and approaches infinity, we get:
    [tex]e = \lim_{n\rightarrow\infty} (1+1/n)^n[/tex]

    so, how do I get:
    [tex]
    e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
    [/tex]

    Also, is it true that:
    [tex](1+x/n)^n \leq e^x[/tex] and
    [tex](1-x/n)^n \leq e^{-x}[/tex] for every natural n and every x element of X?
    How can I prove this?

    thanks!
     
    Last edited: Mar 14, 2006
  2. jcsd
  3. Mar 14, 2006 #2

    Fermat

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    Have you tried the Binomial theorem ?
     
  4. Mar 14, 2006 #3

    Hurkyl

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    I would've taken the expression for e and raised both sides to the x power.

    You really ought to do this more rigorously...

    P.S. many of the limits in your limits are wrong.
     
  5. Mar 14, 2006 #4

    arildno

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    As a hint, set 1/u=x/n, and re-express in terms of u and x.
     
  6. Mar 15, 2006 #5
    which one of the limits are wrong?

    I was able to prove that:
    [tex] (1 + x/n)^n \leq e^x [/tex] and [tex] (1 - x/n)^n \leq e^{-x} [/tex]

    Using the Taylor expansion for [tex]e^x[/tex], I got

    [tex]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...[/tex]
    which means
    [tex]e^x \geq 1 + x[/tex]
    subtituting x by x/n, I get
    [tex]e^{x/n} \geq 1 + x/n[/tex]
    thus
    [tex]e^{(x/n)n} \geq (1 + x/n)^n[/tex]
    simplifying, I get:
    [tex]e^x \geq (1 + x/n)^n [/tex]

    Similarly I used the Taylor Expnansion to prove that
    [tex]
    e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
    [/tex]

    since [tex]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +...[/tex]

    and using binomial theorem for [tex]\lim_{n\rightarrow\infty} (1+x/n)^n[/tex]
    I got:
    [tex]\lim_{n\rightarrow\infty}1^n + \lim_{n\rightarrow\infty} x + \lim_{n\rightarrow\infty} \frac{n(n-1)(x^2)}{(n^2)(2!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(x^3)}{(n^3)(3!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(n-3)(x^4)}{(n^4)(4!)} + ...

    = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +...

    =e^x
    [/tex]

    these correct?
     
    Last edited: Mar 15, 2006
  7. Mar 15, 2006 #6

    Hurkyl

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    You've used n-->infinity in expressions that don't even have an n in them! (You meant h-->0)
     
  8. Mar 15, 2006 #7

    benorin

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    See this thread posts # 3 and 12.

    I take it that your definition for [tex]e^x[/tex] is the Taylor series about x=0, viz.

    [tex]e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}[/tex]
     
    Last edited: Mar 15, 2006
  9. Mar 15, 2006 #8

    benorin

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    Essentially, this

    and this,

     
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