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Power Series Problem

  1. Jan 9, 2009 #1
    1. The problem statement, all variables and given/known data

    [itex]
    \begin{equation}
    1 - x + \frac{x^2}{(2!)^2} - \frac{x^3}{(3!)^2} + \frac{x^4}{(4!)^2} +.... = 0 \nonumber
    \end{equation}
    [/itex]

    2. Relevant equations

    To find out the power series in the LHS of the given equation.

    3. The attempt at a solution

    I have tried to solve it by constructing a differential equation for the LHS expression (=g(x) say) as:

    [itex]
    \begin{equation}
    (xg(x)')' + g(x) =0 \nonumber
    \end{equation}
    [/itex]

    which gives the solution for g(x) as Bessel function of first kind and zero order.
    But, I am still not fully convinced regarding the idea of "recognising" the power series in the LHS. Is there any other algebraic approach towards this problem ?
     
  2. jcsd
  3. Jan 9, 2009 #2

    gabbagabbahey

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    Bessel function? Yikes :eek:

    It's way easier than that. Hint: what is [tex]\frac{(-1)^0 x^0}{0!}[/tex]? How about [tex]\frac{(-1)^1 x^1}{1!}[/tex]? How about [tex]\frac{(-1)^2 x^2}{2!}[/tex]? Do you see the resemblance to the LHS?:wink:
     
  4. Jan 9, 2009 #3
    If you are trying to point to [itex]e^{-x}[/itex] by any chance, then let me bring to your attention the powers that are present over the factorial terms, which unlike [itex]e^{-x}[/itex] are not equal to one. Or else, I am too ignorant to "see" anything substantial for now. :confused:

    Thanks for replying!
     
  5. Jan 9, 2009 #4

    gabbagabbahey

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    Oops, I didn't see the powers in the factorial before.

    Anyways, it should be clear that the (n-1)th term on the LHS is [tex]\frac{(-1)^n x^n}{(n!)^2}[/tex] just by inspection. And so your LHS can be written as [tex]\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{(n!)^2}[/tex]

    If you compare that to the zero order Bessel function : [tex]J_0(u)=\sum_{n=0}^{\infty}\frac{(-1)^n (u/2)^{2n}}{(n!)^2}[/tex] You should see that your LHS is [tex]J_0(2\sqrt{x})[/tex]
     
  6. Jan 9, 2009 #5
    Yes, that's right. But is there any way to show this algebraically, and not by jumping directly to the form of the Bessel function per se.
     
  7. Jan 9, 2009 #6

    gabbagabbahey

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    I'm not sure what you mean by "show this algebraically". It is easy to show that your LHS is equivalent to [itex]J_0(2\sqrt{x})[/itex] by simply writing out the terms in the Bessel function's power series. I don't see any non-algebraic steps in my method?

    Do you mean that you want to explicitly sum the series on the LHS the way you would for the series of [itex]e^{-x}[/itex]? If so, I don't see how.

    I think the only way is to simply recognize what the series is by inspection, or set up a differential equation for the LHS.
     
  8. Jan 9, 2009 #7
    Yes exactly. All I want to know is if it is remotely possible to deliver the same solution by explicitly summing up the series, or splitting the series into seemingly identifiable products. Anyways, I think the approach that we have with us should do it for now. Thank you so much !
     
  9. Jan 9, 2009 #8
    Pardone my ignorance, for i am not that much experienced with power series, so i might just be throwing usless stuff here.

    However, would the following be correct ?

    [tex]e^{-x}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n[/tex]

    So,from here would the following hold(i'm not sure though):

    [tex]\frac{e^{-x}}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n!)^2}x^n[/tex]
     
  10. Jan 9, 2009 #9

    gabbagabbahey

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    No. On the RHS of your expression, n is a dummy variable; an index that is being summed over. On the LHS, n is not a dummy variable (in fact, you haven't defined what n is!).
     
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