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Expansion of log(1+x)

  1. Sep 17, 2011 #1
    We know that log(1+x) = x+((x^2)/2)+((x^3)/3)+.............((x^n)/n)+...........
    Could any body please tell me the proof
     
  2. jcsd
  3. Sep 17, 2011 #2

    micromass

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  4. Sep 17, 2011 #3

    mathman

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    log(1+x)=x - x2/2 + x3/3 - x4/4 + ....
    (Alternate signs)

    The easiest way to see it is by using an integral representation.

    log(1+x) = ∫dx/(1+x)

    Since 1/(1+x) = 1 - x + x2 - x3 + .....,

    integrating term by term gives the series for log(1+x), where the integration limits are [0,x].
     
  5. Sep 17, 2011 #4
    thankyou for the answers i am grateful to you
     
  6. Sep 17, 2011 #5
    In taylor series if f(x)=log(1+x) then is f'(a)=0?
    and is f(a)=log(1+a)
     
    Last edited: Sep 17, 2011
  7. Sep 18, 2011 #6

    gb7nash

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    Assuming a > -1:

    What do you get when you take the derivative of log(1+x)? What do you get when you plug a in?
     
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