How Do You Calculate the Inverse of the Square of a Summation Series?

[Chingon]

Homework Statement

I need to expand 1/y(x)² , where y(x)=x^1/2Ʃ(-1)ⁿ/(n!)² * (3x/4)ⁿ for n=0 to ∞

Homework Equations

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

The Attempt at a Solution

I know that x^1/2Ʃ(-1)ⁿ/(n!)² * (3x/4)ⁿ expands to x^1/2(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!

 

[Ray Vickson]

Homework Statement

I need to expand 1/y(x)² , where y(x)=x^1/2Ʃ(-1)ⁿ/(n!)² * (3x/4)ⁿ for n=0 to ∞

Homework Equations

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

The Attempt at a Solution

I know that x^1/2Ʃ(-1)ⁿ/(n!)² * (3x/4)ⁿ expands to x^1/2(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!

Maple easily gets the first few terms of the expansion:
\frac{1}{y(x)^2} = \frac{1}{x}\left( 1 + \sum_{n=1}^{19} c_n x^n + \cdots \right),
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

 

[Chingon]

Hmm... I suppose I should just give that a shot. Thanks!

 

[Ray Vickson]

Hmm... I suppose I should just give that a shot. Thanks!

I don't know if it helps, but y(x) can be expressed in terms of known functions:
y(x) = \sqrt{x} \, \text{BesselI}(0,\sqrt{-3x}),
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
\text{BesselI}(0,t) = 1 + \frac{1}{4} t^2 + \frac{1}{64} t^4<br /> + \frac{1}{2304} t^6 + \cdots .
Knowing this helps to check which definition of the Bessel function is being used, in case there are variations between different sources.

RGV