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Taylor series of ln(x+2)

  1. Oct 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Using power series, expand ln(x + 2) about a = 0 (Taylor series)

    2. Relevant equations


    3. The attempt at a solution

    Is this appropriate?

    [tex] ln(x+2) = ln((x+1)+1) [/tex]

    [tex] x' = x+1 [/tex]

    [tex] ln(x'+1) = \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x')^{n+1} [/tex]

    or

    [tex] ln(x+2) = ln(\frac{x}{2}+1) [/tex]

    [tex] x' = \frac{1}{2}x [/tex]

    [tex] ln(x'+1) = \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x')^{n+1} [/tex]
     
    Last edited: Oct 28, 2013
  2. jcsd
  3. Oct 28, 2013 #2

    vanhees71

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    Well, you rather have
    [tex]\ln(2+x)=\ln[2(1+x/2)].[/tex]
    You can, however still use your trick with this correction made!
     
  4. Oct 28, 2013 #3
    So would your formula then be:

    [tex] ln2 + ln(1+(x/2)) = ln2 + \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x/2)^{n+1} [/tex]

    or

    [tex] ln(2(x/2+1)) = 2*\sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x/2)^{n+1} [/tex]
     
    Last edited: Oct 28, 2013
  5. Oct 28, 2013 #4

    vanhees71

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    This you should be able to answer yourself by giving the reasoning behind each of the two claims you made. Of course only one of them can be right!
     
  6. Oct 28, 2013 #5
    I believe that the first one is right because

    [tex] ln(AB) = lnA + lnB [/tex]

    [tex] A = 2, B = (1+x/2) [/tex]
     
  7. Oct 28, 2013 #6
    I think you want to do the expansion around x = -1, so that a = 0.

    So, the first few terms are: [tex]ln(x+2) = 0 + (x-1)f'(-1)+...=(x-1)+...[/tex]
     
  8. Oct 28, 2013 #7

    vanhees71

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    It was asked to do the expansion around [itex]0[/itex], and of course the first equation is correct, i.e.,
    [tex]\ln(x+2)=\ln 2 + \ln(1+(x/2)) = \ln 2 + \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}(x/2)^{n+1}.[/tex]
    Finally you should think about the domain, where the series is convergent!
     
  9. Oct 28, 2013 #8

    Ray Vickson

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    Why not just use the Maclaurin series formula
    [tex] f(x) = f(0) + f'(0)\,x + f''(0)\, \frac{x^2}{2!} + \cdots \:\:?[/tex]
    It is really very simple to use in this case.
     
  10. Oct 28, 2013 #9
    Yes, that's what I had in mind. For the x terms in your expansion. Can I do:



    [tex] f(x) = f(0) + f'(0)(\frac{x}{2}) + f''(0)\frac{1}{2!}(\frac{x}{2})^2 + ... [/tex]
     
  11. Oct 28, 2013 #10

    Ray Vickson

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    Your could write that, but you would be WRONG. The Correct form of the expansion is exactly as I wrote it. Your series above is for f(x/2), not for f(x).
     
    Last edited: Oct 28, 2013
  12. Oct 28, 2013 #11
    ok I think I get it now.

    I use your mclaurin series to expand the formula..

    Then based on the series outcome, we build the sum. So the following sum is still correct:

    [tex] ln2 + \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x/2)^{n+1} [/tex]
     
  13. Oct 28, 2013 #12
    Is that what it means when it says "expand ln(x + 2) about a = 0?" I'm not familiar with this type of terminology. What the heck is "a"?

    Chet
     
  14. Oct 28, 2013 #13

    Ray Vickson

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    My guess would be that it was a typo, and should have said "...about x = 0".
     
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