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Simplify equation

  1. Feb 1, 2014 #1
    1. The problem statement, all variables and given/known data

    In a paper I have found there is a simplification of an equation which I do not understand.
    How did they do this? Can someone explain this step by step?

    2. Relevant equations
    The euqation is as follows:

    X = [itex]e \cos\theta + \sqrt{(R^2 + e^2 \sin^2 \theta)}[/itex]

    X = distance
    e = eccentricity
    theta = angle
    R = radius

    Since the eccentricity is much than 1000 times smaller than the radius the above equation is simplified to:

    X = [itex]e \cos\theta + R + \frac{e^2}{2R} \sin^2 \theta[/itex]

    3. The attempt at a solution
    In the paper they say that they used the Taylor series expansion in terms of [itex]\sin^2 \theta[/itex] for this simplification.
     
    Last edited: Feb 1, 2014
  2. jcsd
  3. Feb 1, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    The Taylor expansion being made is that of ##(1+x)^{1/2}## with ##|x| \ll 1##:
    $$
    \begin{align}
    (R^2 + e^2 \sin^2 \theta)^{1/2} &= R \left( 1 + \frac{e^2}{R^2} \sin^2 \theta \right)^{1/2} \\
    &\approx R \left( 1 + \frac{e^2}{2 R^2} \sin^2 \theta \right) \\
    &= R + \frac{e^2}{2 R} \sin^2 \theta
    \end{align}
    $$
     
  4. Feb 2, 2014 #3
    Thank you DrClaude. Can you also help me to simplify the following equation using the Taylor series:

    [itex](R^2-e^2 \sin^2 \theta)^{1/2}[/itex]
     
  5. Feb 2, 2014 #4

    NascentOxygen

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    Staff: Mentor

    Follow the example DrClaude has given, except now replace every x by (-x) in the Taylor expansion:

    (1+x)^n = 1 + (n)(x) + (n)(n-1)(x^2)/2! + ...
     
  6. Feb 2, 2014 #5
    f(x) = (1-x)^(1/2) -> f(0)= 1
    f'(x) = (1/2)(1-x)^(-1/2) -> f'(0) = 1/2

    So the Tyalor expansion will become:
    (1-x)^(1/2) = 1 + (1/2) (x)

    and not
    (1-x)^(1/2) = 1 - (1/2) (x) as when I replace x by (-x) in the Taylor expansion for (1+x)^(1/2)

    Or what do I wrong?
     
  7. Feb 2, 2014 #6

    DrClaude

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    Staff: Mentor

    There is a minus sign missing:
    $$
    \begin{align}
    \frac{d}{dx} (1-x)^{1/2} &= \frac{1}{2} (1-x)^{-1/2} \frac{d}{dx} (1-x) \\
    &= -\frac{1}{2} (1-x)^{-1/2}
    \end{align}
    $$
     
  8. Feb 3, 2014 #7
    Thanks! That's the error I have made.
     
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