Expansion of generating function $\frac{1}{ \sqrt{1-12x+4x^2 } }$

Question

I came across this generating function

$$\frac{1}{ \sqrt{1-12x+4x^2 } }$$

How exactly does one expand this series? I have read through some notes, it seems like we need to factorize the denominator, but it doesn't look like this one can be factorized?

Answer 1

By the generalized Binomial theorem,

$$ \frac{1}{\sqrt{1+y}} =1-\frac{1}{2}y+\frac{3}{8}y^2-\frac{5}{16}y^3+\frac{35}{128}y^4-\frac{63}{256}y^5+\dots $$

Substitute $y=4x^2-12x$, expand and gather terms to get

$$ \frac{1}{ \sqrt{1-12x+4x^2 } }=1+6 x+52 x^2+504 x^3+5136 x^4+53856 x^5+\dots $$

Aside: a commenter pointed out that the coefficients of this series form a sequence with various interpretations: https://oeis.org/A084773.

Answer 2

Let $$f(x)=\frac{1}{\sqrt{1-12x+4x^2}}=\frac{1}{\sqrt{1-2tz+t^2}} \implies t=2x, z=3$$ Recall the generating function for the Legendre Polynomials: $$(1-2zt+t^2)^{-1/2}=\sum_{n=0}^{\infty} P_n(z) t^n, ~if~ |t|<min [z \pm \sqrt{z^2-1}]$$ And $$(1-2zt+t^2)^{-1/2}=\sum_{n=0}^{\infty} P_n(z) t^{-(n+1)}, ~if~ |t|> max [z \pm \sqrt{z^2-1}]$$ So $$f(x)=\sum P_n(3)~ 2^n ~x^n, ~if~ |x| <3-2\sqrt{2}$$ And $$f(x)=\sum P_n(3)~ 2^{-(n+1)} ~x^{-(n+1)}, ~if~ |x| >3+2\sqrt{2}$$

Answer 3

Make the long division

$$\frac{1}{ 1-12x+4x^2 }=1+12 x+140 x^2+1632 x^3+19024 x^4+221760 x^5+2585024 x^6+O\left(x^7\right)$$

Now (being patient), the binomial expansion $$\frac{1}{ \sqrt{1-12x+4x^2 } }=1+6 x+52 x^2+504 x^3+5136 x^4+53856 x^5+575296 x^6+O\left(x^7\right)$$