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Are the sum and the hypergeometric equal?

  1. Sep 2, 2012 #1

    uart

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    I was looking at finding a series solution to a 2nd order DE the other day and came up with the following (for one of the solutions, and there was a somewhat similar series for the other solution).

    [tex]\sum_{k=0}^{\infty} \frac{x^{3k}}{(3k)!} \prod_{m=1}^{k-1} (3m+1)[/tex]

    Wolfram said the solutions were the Airy functions [itex]Ai[/itex] and [itex]Bi[/itex], and since these can be defined in terms of the Hypergeometric function, [itex]~_0F_1[/itex], I suspected that the series I had might be equivalent to a hypergeometric. http://mathworld.wolfram.com/AiryFunctions.html

    Wolfram lists one of the hypergeometrics involved in [itex]Ai[/itex] and [itex]Bi[/itex] as, [itex]~_0F_1(\frac{2}{3}, \frac{x^3}{9})[/itex]. And when I expand this as per the definition here http://mathworld.wolfram.com/ConfluentHypergeometricLimitFunction.html I get the following.

    [tex]\sum_{k=0}^{\infty} \frac{x^{3k}}{9^k \, k!} \left[ \prod_{m=0}^{k-1} (\frac{2}{3}+m) \right]^{-1}[/tex]

    When I sum these numerically in matlab I seem to get the same answer for both. I can see a lot of similarity but can't quite make out the equality of the two. Can anyone see any easy way to show they're equal?
     
    Last edited: Sep 2, 2012
  2. jcsd
  3. Sep 2, 2012 #2

    uart

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    Ok I figured it out, and yes they are identical :)

    I'm busy now but will post it later in anyone is interested.
     
  4. Sep 3, 2012 #3

    uart

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    OK, since I asked it I guess I should show a solution. It turned out to not be that difficult. Start in the obvious way by stripping away the bits that are already the same and writing just the bits that we require equality on.

    [tex](3k)!=9^k \, k! \prod_{m=0}^{k-1} (\frac{2}{3}+m) \prod_{m=1}^{k-1} (3m+1)[/tex]

    Splitting the [itex]9^k[/itex] into [itex]3^k \, 3^k[/itex] we can take one of these terms inside the fractional product to get.

    [tex](3k)! = 3^k \, k! \prod_{m=0}^{k-1} (3m + 2) \prod_{m=1}^{k-1} (3m+1)[/tex]

    Then with some simple rearrangements of the products we get.

    [tex](3k)! = 3^k \, k! \prod_{m=1}^{k} (3m - 1) \prod_{m=1}^{k-1} (3m+1)[/tex]

    [tex](3k)! = 3^k \, k! \, \, (3k-1) \prod_{m=1}^{k-1} \left\{ (3m - 1) (3m+1) \right\}[/tex]

    From here it's pretty easy to see the equality. The product, [itex](3k-1) \prod_{m=1}^{k-1} \left\{ (3m - 1) (3m+1) \right\} = 2 \cdot 4 \cdot 5 \cdot 7 \cdot 8 \cdot 10 \cdot 11 \ldots (3k-1)[/itex], provides all the non factor 3 terms of [itex](3k)![/itex], while the "[itex]3^k \, k![/itex]" fills in all the factor three terms as [itex]3^k \, k! = 3 \cdot 6 \cdot 9 \cdot 12 \ldots 3k[/itex].
     
    Last edited: Sep 3, 2012
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