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1st order expansion problem!

  1. May 26, 2014 #1
    I'm a bit frustrated at the moment, as this minor problem should be fairly easy. But I seem to go wrong at some point...

    So I've got to do a 1st order expansion of the function
    \begin{equation}
    f=\frac{\cos(\theta)}{\sin(\theta)}\ln(\frac{L\sin(\theta)}{d\cos( \theta)}+1)
    \end{equation}

    and my steps are:

    \begin{equation}

    f(0)=0 \\
    f^{\prime}(\theta)=-\frac{1}{\sin^2(\theta)}\ln(\frac{L\sin(\theta)}{d\cos(\theta)}+1)+ \frac{\cos(\theta)}{\sin(\theta)}\frac{d\cos(\theta)}{L\sin(\theta)+d \cos(\theta)}(\frac{\cos^2(\theta)+\sin^2(\theta)}{\cos^2(\theta)}\frac{L}{d}).
    \end{equation}

    When I insert theta=0 I end up divding by 0...
    Furthermore, when I make my computer do the expansion, I get the correct result from the assignment I'm working on.

    If anyone could help me out, I'd be grateful!
     
  2. jcsd
  3. May 26, 2014 #2

    ehild

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    f(0) is not defined, you need the limits at theta=0. Better expand the logarithm. You know that ln(1+x) ≈ x-x/2 if |x|<<1.

    ehild
     
    Last edited: May 26, 2014
  4. May 26, 2014 #3

    LCKurtz

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    Or, if you let ##x = \tan\theta## and ##k=\frac L d## you have ##\frac{\ln(1+kx)}{x}##. You can find the limit as ##x\to 0## using L'Hospital's rule.
     
  5. May 27, 2014 #4

    benorin

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    If you're not sure about the substitution, you would have:

    [tex]\lim_{\theta\to 0}f(\theta )=\lim_{\theta\to 0}\frac{\log\left(1+\frac{L}{d}\tan\theta\right)}{\tan\theta} = \lim_{\theta\to 0}\frac{\frac{d}{d\theta}\log\left(1+\frac{L}{d}\tan\theta\right)}{ \frac{d}{d\theta} \tan\theta}= \cdots[/tex]
     
  6. May 28, 2014 #5

    ehild

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    The function can be written in simpler form as benorin has shown:

    [tex]f(\theta )=\frac{\log\left(1+\frac{L}{d}\tan\theta\right)}{\tan\theta} [/tex]

    Use the Taylor expansion of log(1+x) = x - x2/2. Let be x=tan(theta).

    [tex]f(\theta )≈\frac{\frac{L}{d}\tan\theta-\left(\frac{L}{d}\tan\theta\right)^2/2}{\tan\theta} [/tex]

    Simplify by tan(θ): You get an expression linear in tan(θ). You can expand tan(θ) with respect to θ...

    ehild
     
  7. May 30, 2014 #6
    Thanks for all of your answers, it really helped me out. I've reread the series chapter of my calculus book, and I've come to a much better understanding of that subject in general.

    I basically expand the logarithm in the function, and then I expand the function, which gave the correct answer.
     
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