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Thread: Taylor Series of log(1-t)

  1. #1
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    Mar 2012
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    Taylor Series of log(1-t)

    Hello
    How do you take the geometric/taylor (unsure which it is) series of log(1-t)?
    I do already have a result given to me of 0 -t -(t^2)/2 -(t^3)/3 -(t^4)/4 -...
    But don't know where its come from?
    Please help
    Also, I need to find log(1-t)-log(1-2t) in that form but getting different answers with my friend so would like some reassurement!
    is it just (-t -(t^2)/2 -(t^3)/3 -(t^4)/4 -...) - (-2t -((2t)^2)/2 -((2t)^3)/3 -((2t)^4)/4 -... ?

  2. #2
    Elite Member
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    12,630
    Quote Originally Posted by vjj View Post
    Hello
    How do you take the geometric/taylor (unsure which it is) series of log(1-t)?
    I do already have a result given to me of 0 -t -(t^2)/2 -(t^3)/3 -(t^4)/4 -...
    But don't know where its come from?
    Please help
    Also, I need to find log(1-t)-log(1-2t) in that form but getting different answers with my friend so would like some reassurement!
    is it just (-t -(t^2)/2 -(t^3)/3 -(t^4)/4 -...) - (-2t -((2t)^2)/2 -((2t)^3)/3 -((2t)^4)/4 -... ?
    Do you know the expansion of [TEX]\frac{1}{1-t}[/TEX]?

    Can you see a way to get from that expansion to expansion of loge(1-t)?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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