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Mean Value Theorem Problem

  1. Jul 29, 2012 #1
    Given f(n) = (1 - (1/n))n

    I calculate that the limit as n -> infinity is 1/e.

    Also given that x/(1-x) > -log(1-x) > x with 0<x<1 (I proved this in an earlier part of the question) I want to show that:

    1 > (f(60)/f(infinity)) > e-1/59 > 58/59

    I have tried using my value for f (infinity) and f(60) = e60*log(59/60) in the original inequality but cannot seem to rearrange it to the last one.

    I get e(x/1-x) > 1/1-x > ex but am not sure where to go from there.

    Thanks!
     
  2. jcsd
  3. Jul 29, 2012 #2

    chiro

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    Hey sr3056 and welcome to the forums.

    Try subtracting 1/59 from both sides (let x = 1)
     
  4. Jul 29, 2012 #3

    mfb

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    Do you know e^x >= 1+x (with "=" just for x=0)? You can use it to show ex/(1-x) > 1/(1-x)

    1/1-x > ex... maybe the derivatives can give you some way to prove this.

    @chiro: x=1 is not in the range where the equation is well-defined.
     
  5. Jul 29, 2012 #4

    haruspex

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    Try inverting each term in that line (adjusting the relationships as appropriate), then plugging in a value for x.
     
  6. Jul 29, 2012 #5
    Thank you for your responses.

    Inverting, I get e(x/x-1) < 1 - x < e-x

    Letting x = 1/60 I then get e-1/59 < 59/60 < e-1/60

    This is starting to look a bit more like it, but I'm not sure how my f(60) and f(infinity) values are going to come in...
     
  7. Jul 29, 2012 #6

    mfb

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    "f(60)/f(infinity)" = e*(1-1/60)^60
    Take the log, and your inequality reads:

    0 > 1+60 log(1-1/60) > -1/59
    You get the left ">" with "-log(1-x) > x" and the right ">" with "x/(1-x) > -log(1-x)", both with the obvious choice for x.
     
  8. Jul 29, 2012 #7
    Thank you very much
     
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