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Math Help - Help for a Perturbation problem

  1. #1
    Junior Member ginafara's Avatar
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    Help for a Perturbation problem

    Can someone please help me. I am studying from what is possibly the worst written book... I do not even know how to start the problem...

    Problem:

    Let
    f(y, \epsilon) = \frac{1}{(1+\epsilon y)^{\frac{3}{2}}},  y=y_{0}+\epsilon y_{1} +\epsilon^{2} y_{2}+...

    Expand f in powers of \epsilon up to O(\epsilon^{2})

    I know I let f(y, \epsilon) =0

    I know I need to find
    O(1), O(\epsilon), O(\epsilon^{2})

    The problem is that the examples to do this are only for very simple algebraic problems. Any help would be greatly appreciated...
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  2. #2
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    You can use the general binomial theorem

    \frac{1}{(1-x)^s} = \sum_{k=0}^\infty \binom{s+k-1}{s-1}x^k

    Here s = 3/2 and x = -\epsilon y


    \frac{1}{(1+\epsilon y)^{3/2}} = \sum_{k=0}^\infty \binom{k+1/2}{1/2}(-\epsilon y)^k

    Now it can be shown that

    \binom{k+1/2}{1/2} = \frac{(2k+1)!!}{(2k)!!}

    where !! is the double factorial operator (product of just the odd or even numbers up to and including (2k+1) or (2k)). So

    \frac{1}{(1+\epsilon y)^{3/2}} = \sum_{k=0}^\infty (-1)^k\frac{(2k+1)!!}{(2k)!!}(\epsilon y)^k

    And therefore you can expand in order of epsilon as

    \frac{1}{(1+\epsilon y)^{3/2}} = 1 - \frac{3y}{2}\epsilon + \frac{15y^2}{8}\epsilon^2 + O(\epsilon^3).

    Hope this helps.
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  3. #3
    Junior Member ginafara's Avatar
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    Thanks so much. if I understand correctly, this expansion is equivalent to the perturbation expansion. Am I correct? This book is really bad. sorry.
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  4. #4
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    Quote Originally Posted by ginafara View Post
    Thanks so much. if I understand correctly, this expansion is equivalent to the perturbation expansion. Am I correct? This book is really bad. sorry.
    My last line is the expansion in terms of y you should still substitute your expansion for y. For example,

    f(y,\epsilon) = 1 - \frac{3y}{2}\epsilon + \frac{15y^2}{8}\epsilon^2 + O(\epsilon^3).

    y = y_0 + \epsilon y_1 + \epsilon^2 y_2 + O(\epsilon^3).

    Substituting

    f(y,\epsilon) = 1 - \frac{3(y_0 + \epsilon y_1 + \epsilon^2 y_2)}{2}\epsilon + \frac{15(y_0 + \epsilon y_1 + \epsilon^2 y_2)^2}{8}\epsilon^2 + O(\epsilon^3).<br />

    Simplifying and keeping just terms less than O(\epsilon^3)

    f(y,\epsilon) = 1 - \frac{3 y_0}{2}\epsilon + \left(\frac{15y_0^2}{8} - \frac{3 y_1}{2}\right)\epsilon^2 + O(\epsilon^3).<br />

    So this gives the expansion of f(y,\epsilon) in terms of the expansion for y.
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  5. #5
    Junior Member ginafara's Avatar
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    That is what I did... Thanks again... I was reading through an example and the author stated that the binomial expansion is equivalent to the perturbation expansion in his example problem... It didn't make sense so I continued on with the rest... Thanks again so much for your help... May I say never use logan's applied mathematics 3rd edition.. Horrible...
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  6. #6
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    Quote Originally Posted by ginafara View Post
    May I say never use logan's applied mathematics 3rd edition.. Horrible...
    I've never seen it, but I'll bear that in mind for future reference.
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