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Jackson eqs. 3.25 and 3.26

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    How do you get from (3.25) to (3.26) in Jackson?

    2. Relevant equations

    Equation 3.25:

    [tex]
    A_l=\left(2l+1\right)\int_0^1P_l\left(x\right)dx
    [/tex]

    Equation 3.26:

    [tex]
    A_l=\left(-\frac{1}{2}\right)^{\frac{l-1}{2}}\frac{\left(2l+1\right)\left(l-2\right)!!}{2\left(\frac{l+1}{2}\right)!}
    [/tex]

    Rodriques:

    [tex]
    P_l\left(x\right)=\frac{1}{2^l l!}\frac{d^l}{dx^l}\left(x^2-1\right)^l
    [/tex]

    3. The attempt at a solution

    Put Rodrigues into (3.25):

    [tex]
    A_l=\frac{\left(2l+1\right)}{2^l l!}\int_0^1\frac{d^l}{dx^l}\left(x^2-1\right)^ldx
    [/tex]

    Evaluate:

    [tex]
    \frac{d^l}{dx^l}\left(x^2-1\right)^l
    [/tex]

    [tex]
    =\frac{d^l}{dx^l}\sum_k{\left(\frac{l!}{k!\left(l-k\right)!}\right)\left(x^2\right)^{l-k}\left(-1\right)^k}
    [/tex]

    We worked this out in class, and somehow that sum goes away. But I forgot how.

    As I work out the derivatives, I end up with:

    [tex]
    =\sum_k{\left(\frac{l!}{k!\left(l-k\right)!}\right)2\left(l-k\right)\left(2\left(l-k\right)-1\right)\left(2\left(l-k\right)-2\right)\left(2\left(l-k\right)-3\right)...\left(2\left(l-k\right)-\left(l-1\right)\right)x^{l-2k}\left(-1\right)^k
    [/tex]

    Any value of [tex]k>\frac{l}{2}[/tex] and the expression is zero. So the sum only goes from [tex]k=0[/tex] to [tex]\frac{l-1}{2}[/tex].

    How do I get rid of that sum completely?
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 18, 2009 #2

    gabbagabbahey

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    Looks like Jackson contains a typo; this should should be:

    [tex]
    A_l=(-1)^{l-1}\left(\frac{1}{2}\right)^{\frac{l-1}{2}}\frac{\left(2l+1\right)\left(l-2\right)!!}{2\left(\frac{l+1}{2}\right)!}
    [/tex]


    You've messed up the binomial expansion; you should have:

    [tex]
    =\frac{d^l}{dx^l}\sum_{k=0}^{\infty}{\left(\frac{l!}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}
    [/tex]
     
  4. Oct 18, 2009 #3
    That still doesn't get rid of the sum. It only gets rid of the values [tex]k=0[/tex] to [tex]k=\frac{l-1}{2}[/tex].
     
  5. Oct 19, 2009 #4

    gabbagabbahey

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    Okay, so that leaves you with

    [tex]A_l=\frac{2l+1}{2^ll!}\sum_{k=\frac{l-1}{2}}^\infty\frac{l!}{k!(l-k)!}}(-1)^{l-k}\int_0^1\frac{d^l}{dx^l}(x^{2k})dx[/tex]

    Now, I think what you want to do is make the substitution [itex]j=k-\frac{l-1}{2}[/itex]...
     
    Last edited: Oct 20, 2009
  6. Oct 20, 2009 #5
    [tex]\frac{d^l}{dx^l}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}[/tex]

    Differentiate once (1st time):

    [tex]\frac{d^{l-1}}{dx^{l-1}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2kx^{2k-1}\left(-1\right)^{l-k}}[/tex]

    Differentiate again (2nd time):

    [tex]\frac{d^{l-2}}{dx^{l-2}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)x^{2k-2}\left(-1\right)^{l-k}}[/tex]

    Differentiate again (third time):

    [tex]\frac{d^{l-3}}{dx^{l-3}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)x^{2k-3}\left(-1\right)^{l-k}}[/tex]

    ...
    ...

    Differentiate lth time:

    [tex]\frac{d^{l-l}}{dx^{l-l}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}[/tex]

    That leaves me with:

    [tex]\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}[/tex]

    Now, if [tex]k=0[/tex], the whole expression is zero. So we can rewrite it as:

    [tex]\sum_{k=1}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}[/tex]

    But if [tex]k=\frac{1}{2}[/tex], then the expression is also zero. But k can only be an integer. However, due to the term [tex]2k-2[/tex], if [tex]k=1[/tex] the expression is zero. So rewrite as:

    [tex]\sum_{k=2}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}[/tex]

    Now let's just cut to the chase:

    [tex](2k-(l-1)[/tex]

    If [tex]k=\frac{l-1}{2}[/tex] or less, the expression is zero.

    So rewrite as:

    [tex]\sum_{k=\frac{l-1}{2}+1}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}[/tex]

    Which can be written as:

    [tex]\sum_{k=\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}[/tex]

    That only works if l is odd. If l is even, then it would be written as:

    [tex]\sum_{k=\frac{l}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}[/tex]

    So we have the following:

    [tex]A_l=\frac{\left(2l+1\right)}{2^l l!}\int_0^1\frac{d^l}{dx^l}\left(x^2-1\right)^ldx[/tex]
    [tex]=\frac{\left(2l+1\right)}{2^l l!}\int_0^1\sum_{k=\frac{l}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}dx[/tex]
    [tex]=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))\left(-1\right)^{l-k}\int_0^1x^{2k-l}}dx[/tex] for even l
    [tex]=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))\left(-1\right)^{l-k}\int_0^1x^{2k-l}}dx[/tex] for odd l

    Now: [tex]\int_0^1x^{2k-l}dx=\frac{1}{2k-l+1}x^{2k-l+1}[/tex] evaluated from 1 to 0:

    That would equal:

    [tex]\int_0^1x^{2k-l}dx=\frac{1}{2k-l+1}[/tex]

    So now we have:

    [tex]A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l}{2},\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))\left(-1\right)^{l-k}\frac{1}{2k-l+1}[/tex]

    Which can be written as:

    [tex]A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l}{2},\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-2))\left(-1\right)^{l-k}[/tex]

    Now I still have that sum.

    smiley-bangheadonwall.gif
     
    Last edited by a moderator: Apr 24, 2017
  7. Oct 20, 2009 #6

    gabbagabbahey

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    You might want to re-read my last post; summing over [itex]j[/itex] instead of [itex]k[/itex] will probably make things easier for you...
     
  8. Oct 21, 2009 #7
    I don't see how. It changes the limits on the sum:

    [tex]A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{j=1}^{\frac{l+1}{2}}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-2))\left(-1\right)^{l-k}[/tex]

    And all the k's become more complicated:

    [tex]A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{j=1}^{\frac{l+1}{2}}{\left(\frac{l !}{\left(j+\frac{l-1}{2}\right)!\left(l-\left(j+\frac{l-1}{2}\right)\right)!}\right)2\left(j+\frac{l-1}{2}\right)(2\left(j+\frac{l-1}{2}\right)-1)(2\left(j+\frac{l-1}{2}\right)-2)...(2\left(j+\frac{l-1}{2}\right)-(l-2))\left(-1\right)^{l-\left(j+\frac{l-1}{2}\right)}[/tex]
     
  9. Oct 21, 2009 #8

    gabbagabbahey

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    Why are you summing from [itex]j=1[/itex] to [itex]j=(l+1)/2[/itex]?:confused:

    You started out with something like

    [tex]A_l=\sum_{k=0}^{\infty}c_k\int_0^1\frac{d^l}{dx^l}x^{2k}dx=\sum_{k=0}^{\infty}c_k\left.\left[\frac{d^{l-1}}{dx^{l-1}}x^{2k}\right]\right|_0^1[/tex]

    The terms for which [itex]2k<l-1[/itex] are all zero, and [itex]l[/itex] is odd (Remember, Jackson showed the [itex]A_l=0[/itex] for even [itex]l[/itex], so this sum is just to determine the odd coefficients), So the first [itex]\frac{l-1}{2}-1[/itex] terms of the sum are zero and you are left with:

    [tex]A_l=\sum_{k=\frac{l-1}{2}}^{\infty}c_k\left.\left[\frac{d^{l-1}}{dx^{l-1}}x^{2k}\right]\right|_0^1[/tex]

    So the substitution [itex]j=k-\frac{l-1}{2}[/itex]should give you

    [tex]A_l=\sum_{j=0}^{\infty}c_{j+\frac{l-1}{2}}\left.\left[\frac{d^{l-1}}{dx^{l-1}}x^{2j+l-1}\right]\right|_0^1=\sum_{j=0}^{\infty}c_{j+\frac{l-1}{2}}\left.\left[(2j+l-1)(2j+l-2)\ldots(2j+1)x^{2j}\right]\right|_0^1=\sum_{j=0}^{\infty}c_{j+\frac{l-1}{2}}\frac{(2j+l-1)!}{(2j)!}[/tex]
     
    Last edited: Oct 21, 2009
  10. Oct 21, 2009 #9
    The sum doesn't go to infinity. Using k as the index variable, it goes from 0 to l:

    [tex]
    \frac{d^l}{dx^l}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}
    [/tex]
     
  11. Oct 21, 2009 #10

    gabbagabbahey

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    No, it goes to infinity. You need to review the binomial expansion.

    [tex](x^2-1)^l=\sum_{k=0}^{\infty}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}\neq\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}[/tex]
     
  12. Oct 21, 2009 #11
    http://en.wikipedia.org/wiki/Binomial_theorem
     
  13. Oct 21, 2009 #12
    Because if [tex]k=\frac{l-1}{2}[/tex], that term is zero. So the sum has to start at [tex]k=\frac{l+1}{2}[/tex].

    If [tex]j=k-\frac{l-1}{2}[/tex], then the first term for j is when [tex]k=\frac{l+1}{2}[/tex]:

    [tex]j=k-\frac{l-1}{2} = \frac{l+1}{2}-\frac{l-1}{2} = 1[/tex].

    The last term is [tex]k = l[/tex]

    So [tex]j=k-\frac{l-1}{2} = l-\frac{l-1}{2} = \frac{l+1}{2}[/tex]
     
  14. Oct 21, 2009 #13

    gabbagabbahey

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    Contrary to popular belief, wikipedia is not always the most accurate source for information.

    Try this link instead. Or, better yet, break out your old calculus textbook.

    Edit: After my morning cup of coffee, I realized that there is nothing wrong with only summing up to [itex]k=l[/itex]. However, the sum can be extended to infinity since [tex]\begin{pmatrix}l\\k\end{pmatrix}[/tex] is zero for [itex]k>l[/itex].
     
    Last edited: Oct 22, 2009
  15. Oct 22, 2009 #14
    I would like to direct your attention to equation 2 in that link, and the text that immediately precedes it.

    It would only go to infinity if l goes to infinity.
     
  16. Oct 22, 2009 #15

    gabbagabbahey

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    Anyways, can you agree that

    [tex]A_l=-\frac{2l+1}{2^l}\sum_{j=0}^{j=\frac{l+1}{2}}\frac{(-1)^j(2j+l-1)!}{(j+\frac{l-1}{2})!(\frac{l+1}{2}-j)!(2j)!}[/tex]

    ?
     
  17. Oct 23, 2009 #16
    It looks clearer written in terms of k.
     
  18. Oct 23, 2009 #17

    gabbagabbahey

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    Since [itex]l[/itex] is odd, make the substitution [itex]l=2n+1[/itex]....
     
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