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Logarithmic Series Expansion Using Taylor's/McLaurin's

  1. Aug 1, 2007 #1
    Hi there guys. My first post here. I heard this forum was really helpful so i've signed up lol.I'm trying to get to grips with using Taylor's/McLaurin's formula for series expansions...My main problem lies with expansions of Logarithmic functions..

    I want to work out how to expand Logs when numbers are used instead of "x".
    for example i want to expand ln 2 instead of ln x
    or ln 12 instead of ln (1 + x). i can expand the function when there is an x involved inside the ln function.. but if its just a number such as (ln 2) then for some reason my expansion doesnt seem to work...


    for example i can work out ln (1 + x) using mclaurins:

    f(x) = ln (1 + x)
    [tex]f'(x) = 1/(1+x) [/tex]
    [tex]f''(x) = -1/(1+x)^2 [/tex]
    [tex]f'''(x) = 2/[(1+x)^3[/tex]

    so now we plug that into mclaurins we have:
    f(0) = ln 1 = 0
    [tex] f'(0)x = \frac{1}{1+0}x = x[/tex]
    [tex] \frac{f''(0)}{2!}x^2 = \frac{-1}{2!(1+0)^2}x^2 = -\frac{x^2}{2}[/tex]
    [tex] \frac{f'''(0)}{3!}x^3 = \frac{2}{3!(1+0)^3}x^3 = \frac{x^3}{3}[/tex]

    so we deduce that
    [tex] ln (1 + x) = x -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}...[/tex]


    so im fine with expanding logs that have an x involved...

    NOW HERES THE REAL PROBLEM
    how would i prove something like: ln 2 (which has no x involved in the ln function)..

    ln 2 = 1 - 1/2 + 1/3 - ... = [tex] \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[/tex]

    so far i can work out ..

    f(x) = ln 2
    f'(x) = 1/2
    f''(x) = 0...

    so i plug it into mclaurins i get:

    i can neglect f(0) since the sum is from 1 to infinity..
    [tex] f'(0)x = \frac{1}{2}x = \frac{1}{2}x[/tex]
    [tex] \frac{f''(0)}{2!}x^2 = \frac{0}{2!}x^2 = 0[/tex]
    as you can see here im immediately going wrong since the first term should just have been 1, then -1/2 ... as shown above on the BLUE text.

    please guide me on here. and how do you acutally deduce that the terms of ln 2 is equivalent to
    [tex] \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[/tex]

    thnk you very much guys =)
     
    Last edited: Aug 1, 2007
  2. jcsd
  3. Aug 1, 2007 #2
    So the general form for a taylor series expanded at a point a is:
    [tex]\sum^{\infty}_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^n[/tex].
    Since log(0) does not exist, choose a=1 and then expand log to get:
    [tex]\sum^{\infty}_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^n=log(1)+(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\cdots +\cdots[/tex]
    and then you plug in 2 for x (since you want log(2)) to get:
    [tex]\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}[/tex]
     
    Last edited: Aug 1, 2007
  4. Aug 1, 2007 #3

    bel

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    Okay, you got [tex] ln (1 + x) = x -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}...[/tex] and you want to evaluate [tex] ln(2) [/tex]. Therefore, equate the arguments of the logarithms, i.e., [tex] (1+x)=2 [/tex] to get [tex] x=1 [/tex], substitute that in to the series.
     
  5. Aug 1, 2007 #4
    thnx guys.. i totally understand both your methods :)

    just a quick question about : [tex]\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}[/tex]

    is there a proof or a way to actually prove the expanded terms of ln 2 are the same as the above sum... or do you just visually deduce that?
     
  6. Aug 1, 2007 #5
    oh and by the way. would bel's method be considered as a proof for ln2 = [tex]\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}[/tex]
     
  7. Aug 1, 2007 #6
    Both methods show that log(2) equals your series by definition of the Taylor Series, I suppose.
     
  8. Aug 2, 2007 #7

    Gib Z

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    Yes there is a way to prove that the Series are identical.

    The Taylor Series for ln(1+x) = [tex]\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}, 0<x<1[/tex] as you have already pointed out. Now let x =1 on both sides. Its exactly what you want.
     
  9. Aug 9, 2007 #8

    disregardthat

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    Is this correct?

    [tex]\ln{x} = \sum_{n=1}^{\infty} \frac{(x-1)^n}{x^n \cdot n} \ , \ \forall \ x > 0 \ , \ x \in \mathbb{R}[/tex]
     
    Last edited: Aug 9, 2007
  10. Aug 9, 2007 #9

    HallsofIvy

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    ?? Are you sure about that xn in the numerator?
     
  11. Aug 9, 2007 #10

    disregardthat

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    The denominator you mean? I don't know, I just used some integration by parts. It does work for many x-values, it's just that when x < 0.5 it doesn't work. I was hoping some of you could give me a clue.
     
    Last edited: Aug 9, 2007
  12. Aug 9, 2007 #11

    HallsofIvy

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    The difficulty, Jarle, is that you did not post the original post in this thread and your post does not seem to have anything to do with it.

    If you have a separate question yourself, please post it as a new thread.
     
  13. Aug 10, 2007 #12

    disregardthat

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    Ok, sorry...
     
  14. Aug 11, 2007 #13
    you know [tex] ln (1 + x) = x -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}...[/tex]

    Ask your self what it means for two functions to be the equal?

    proving ln(2)=[tex] \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}[/tex]

    should follow
     
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