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A difficult series expansion (finding a limit)

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data

    Find [itex]\lim_{x \to 0}\frac{ln(1+x^2)}{1-cos(x)}[/itex] by using series representations. Check using L'Hospitals rule.

    2. Relevant equations

    Taylor polynomial at x=0: [itex]\sum_{k=0}^{\infty}\frac{f^{k}(0)}{k!}(x)^{k} = f(0) + f'(0)(x) + f''(0)x^{2} +...[/itex]

    3. The attempt at a solution

    Using L'Hospital's rule gives me a limit of 2. Wolfram Alpha gives the following series expansion for the quotient: http://www.wolframalpha.com/input/?i=series+(ln(1+x^2))/(1-cosx), which obviously evaluates to 2 when [itex]x \to 0[/itex]

    What I don't get is how the program calculates the series. Differentiating the quotient given in the problem statement (to find out values of the derivatives to plug into the Taylor-formula) gives very complicated results that either equal zero or are indeterminate when x = 0.

    For example, [itex]\frac{d}{dx} \frac{ln(1+x^2)}{1-cos(x)} = \frac{2x(1+x^{2})^{-1}}{1-cos(x)} - \frac{ln(1+x^{2})sin(x)}{(1-cos(x))^{2}}[/itex] where both terms are indeterminate when x = 0. The same thing happens with the second derivative: http://www.wolframalpha.com/input/?i=second+derivative+(ln(1+x^2))/(1-cosx)

    So there are indeterminates where things would otherwise seem to evaluate to something that resembles 2, and zeroes everywhere else. It also doesn't help that f(0), the first term in the series, is again indeterminate.

    I really have no idea how to approach this problem beyond what I've already tried (which in addition to this approach included plugging in the series for ln(1+x2) and 1-cos(x) in their respective places in the fraction). Could someone point me in the right direction?
     
    Last edited: Feb 13, 2016
  2. jcsd
  3. Feb 13, 2016 #2

    blue_leaf77

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    You don't need to evaluate the series of the whole quotient. First write down each of the Taylor expansion of the numerator and denominator separately.
     
  4. Feb 13, 2016 #3
    I tried doing that as well, but a similar issue arises:

    If

    around x = 1: [itex]ln(1+x^{2}) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}(k-1)!\cdot x^{2k}}{k!}

    = -\sum_{k=1}^{\infty} \frac{(-1)^{k}(k-1)!\cdot x^{2k}}{k!}

    =-\sum_{k=1}^{\infty} \frac{(-1)^{k}\cdot x^{2k}}{k}[/itex]

    and

    around x = 0: [itex]cos(x) = \sum_{k=0}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}[/itex]

    then

    [itex]
    \lim_{x \to 0} \frac{ln(1+x^{2})}{1-cos(x)}

    = \lim_{x \to 0} \frac{ \sum_{k=1}^{\infty} \frac{(-1)^{k+1}x^{2k}}{k}}{1 - \sum_{k=0}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}}

    = \lim_{x \to 0} \frac{ -\sum_{k=1}^{\infty} \frac{(-1)^{k}x^{2k}}{k}}{1 - \sum_{k=0}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}}

    = \lim_{x \to 0} \frac{ -\sum_{k=1}^{\infty} \frac{(-1)^{k}x^{2k}}{k}}{1 - ( 1 + \sum_{k=1}^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!})}

    [/itex]
    where the numerator and denominator still both evaluate to zero (since I can't just cancel the xs because of the series notation). Therefore I'm still stuck with an indeterminate form.
     
  5. Feb 13, 2016 #4

    blue_leaf77

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    It may be more transparent if I write the expansions explicitly for first few terms
    $$
    \frac{\ln(1+x^{2})}{1-\cos(x)} = \frac{ x^2-\frac{x^4}{2}+\frac{x^6}{3}-\cdots }{ \frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}\cdots}
    $$
    Now what happens if you factor out ##x^2## from the numerator and denominator?
     
  6. Feb 13, 2016 #5
    Well, gee-whiz.

    If you factor out x2 and take the limit as [itex]x \to 0[/itex], we are left with [itex]\frac{1}{\frac{1}{2!}} = \frac{2!}{1} = 2[/itex].

    Thanks a bunch. I'm not used to working with series notation and have been simply thrown in the middle of it all. I should probably start writing the first few terms of a series whenever I'm faced with a problem like this.
     
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