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Question about log/ln

  1. Sep 22, 2009 #1
    If you had something like ln(x+1) is there are way of breaking that term up?

    I have a question where I have to find a two term approximation and i'm at a stage similar to this one and I don't know what I can do.
     
  2. jcsd
  3. Sep 22, 2009 #2

    mathman

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    What do you mean by breaking it up? There is a power series for ln(1+x)

    x-x2/2+x3/3+....
     
  4. Sep 22, 2009 #3
    I have y=ln(3u+4) and need a 2 term approximation (we are given a hint 4 = 3+1) so I assume that we must do: y=ln(3u+1+3) and the only way we know to approximate is y=ln(1+u) ~ u so in order to get that I need to somehow get rid of the 3 in y=ln(3u+1+3)
     
  5. Sep 22, 2009 #4

    lurflurf

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    What kind of approximation.
    We know log(x+1)~x
    so expansion about x+1=1->x=0 would be nice
    let f(x)=log(x+1)
    f(x)~f(0)+f'(0)x+f''(0)x^2/2+f''(0)x^3/6+...
     
  6. Sep 22, 2009 #5
    The entire question is:

    Find a two term approximation for the following function near t=1.

    h = 5ln(3t+1)
    and after substituting u+1 for t

    I get:

    h = 5ln(3u+4)
    Now it gives us the hint: 4 = 3+1

    But im stuck after this.
     
  7. Sep 23, 2009 #6

    mathman

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    Let v=3u+1, so h=5ln(1+v)~5[v - v2/2 + ...]
    Take it from there.
     
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