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Inverse of (Summation)^2 ?

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to expand 1/y(x)2 , where y(x)=x1/2Ʃ(-1)n/(n!)2 * (3x/4)n for n=0 to ∞


    2. Relevant equations
    How does one arrive at the correct solution (-coefficients seem to vanish, only + remain)?


    3. The attempt at a solution
    I know that x1/2Ʃ(-1)n/(n!)2 * (3x/4)n expands to x1/2(1-3/4x+9/64x2-3/256x3...)

    also, I can follow the expansion of y(x)2 in that it expands to:
    x(1-3/2x+27/32x2-15/64x3+153/4096x4-27/8192x5+9/65536x6...)

    or at least Wolfram Alpha can lead me to this expansion which matches the answer.

    However, I'm stuck at seeing how 1/y(x)2 can expand to:
    x-1(1+3/2x-27/32x2+15/64x3+...+9/4x2-81/32+...+27/8x3+...)

    =x-1(1+3/2x+45/32x2+69/64x3+...)

    It seems to me that there should be some semblance of 1/y(x)2 to y(x)2, and therefore have the +/- coefficients. Wolfram Alpha doesn't seem to be able (willing?) to expand this expression either. Anyone have any tricks for calculating the inverse squared summation?

    Thanks in advance!
     
  2. jcsd
  3. Nov 6, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Maple easily gets the first few terms of the expansion:
    [tex] \frac{1}{y(x)^2} = \frac{1}{x}\left( 1 + \sum_{n=1}^{19} c_n x^n + \cdots \right),[/tex]
    where c_1, c_2, ..., c_19 are (each c on a different line):
    3/2
    45/32
    69/64
    6093/8192
    197451/409600
    1966419/6553600
    23325219/128450560
    70801156917/657666867200
    82637411991/1315333734400
    3810467243817/105226698752000
    526187311691121/25464861097984000
    7622948627022699/651900444108390400
    1034255551214515023/157387392934739968000
    3170886841011096265941/863742012425852944384000
    440593486864035364870749/215935503106463236096000000
    1997150701358677777480094517/1768943641448146830098432000000
    1017985739206517442921313503/1635919079611246188475029913600
    5591384715189383214803844678009/16359190796112461884750299136000000
    170038949461340219021318935958829/908564288830245960060747382784000000

    It would probably be better to convert to floating point:
    1.50000000000000
    1.40625000000000
    1.07812500000000
    .743774414062500
    .482058105468750
    .300051727294922
    .181589079876335
    .107655046115420
    .628261937102189e-1
    .362119812653020e-1
    .206632704441800e-1
    .116934245035661e-1
    .657140023688787e-2
    .367110409751349e-2
    .204039391635755e-2
    .112900753566332e-2
    .622271450888896e-3
    .341788587520973e-3
    .187151257816066e-3

    RGV
     
  4. Nov 6, 2012 #3
    Hmm... I suppose I should just give that a shot. Thanks!
     
  5. Nov 6, 2012 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    I don't know if it helps, but y(x) can be expressed in terms of known functions:
    [tex] y(x) = \sqrt{x} \, \text{BesselI}(0,\sqrt{-3x}),[/tex]
    where BesselI is the modified Bessel function of the first kind. The only point here is that lots of information is available about Bessel functions, including alternative representations, etc., and some of that might be useful.

    Here, I am using Maple's definition of the function BesselI. Its series expansion is
    [tex] \text{BesselI}(0,t) = 1 + \frac{1}{4} t^2 + \frac{1}{64} t^4
    + \frac{1}{2304} t^6 + \cdots .[/tex]
    Knowing this helps to check which definition of the Bessel function is being used, in case there are variations between different sources.

    RGV
     
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