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Power series of e

  1. Mar 28, 2006 #1
    Use a four-term Taylor approximation for e^h , for h near 0 , to evaluate the following limit.

    lim (e^h-1-h-h^2/2)/h^3
    h->0

    i know that e^h = 1+h+h^2/2+h^3/3+h^4/4......

    therefore, I say that e^h-1-h-h^2/2 = h^3/3+h^4/4........

    (h^3/3+h^4/4.....)/h^3 is approximately = 1/3

    but its wrong

    please give me some hints THANX!!!

    sorry for posting it at precalculus... please help urgent!
     
  2. jcsd
  3. Mar 28, 2006 #2

    Tom Mattson

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    It's not wrong.
     
  4. Mar 29, 2006 #3
    But It Is....
     
  5. Mar 29, 2006 #4

    Tom Mattson

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    Ack, you're right it is wrong. Your mistake is in the terms of the series.

    Nope, those denominators should have factorials in them, like so:

    e^h = 1+h+h^2/2!+h^3/3!+h^4/4!
    e^h = 1+h+h^2/2+h^3/6+h^4/24

    That should fix it.
     
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