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Help with power series solution

  1. Nov 2, 2003 #1
    I've done a power series solution to a differential equation and got the recursive formula for the coefficients below. Now I am to evaluate it for large j and I don't get the answer in the book.

    I'm not sure what method they are using to get the answer although their answer makes sense physically.

    a j+1 = aj * 2 * {(j + L + 1) - k }/ ( {j+1}(j + 2L + 2) )

    where L and k are constants and j is just an integer index number.

    If I consider large j I would say
    aj+1 approx. aj * 2 (j) / ( j * j) = aj * 2/j

    If I say j => infinity and use l'Hopital's rule I get

    aj+1 = aj * 2 / (2j + 2L + 3) approx. aj * 1/j

    The book gets

    aj+1 approx. aj * 2j / ( j*(j+1))

    What rules are they applying to get this?

    (of course a j+1 means aj+1 and aj means aj )
  2. jcsd
  3. Dec 2, 2003 #2


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    Science Advisor

    Essentially, they are ignoring terms "small" compared with the very large j: in this case that's just about everything that doesn't involve j.

    I must admit that I don't understand why they include the "+1" in "j+1" in the denominator. If j is large, then 1 is certainly small compared with it. Also, there is no reason not to cancel the js in the numerator and denominator.

    If j= 10000, say, then 2j/(j(j+1))= 20000/(10000*10001)= 20000/100100= 0.00019998 while 2/(j+1)= 2/(10001)= 0.00019998- exactly the same thing and 2/j= 2/10000= 0.0002.

    Your use of L'Hopital's rule IS incorrect. Your first result of
    aj * 2 (j) / ( j * j) = aj * 2/j is good.
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