What is the expansion of $\sqrt{1+ux+vx^2}$ in powers of $x$?

Question

What is the expansion of $\sqrt{1+ux+vx^2}$ in powers of $x$?

This came up in my answer here: How to solve this recurrence$f(n)=A\cdot f(n-1)+B\sum{f(i)f(n-i)},\;1\leq i\leq n-1,$ and $f(1)=K$?

I just did a straightforward expansion and wondered if there is a better way to do it.

Here is my work:

Since $\sqrt{1+y} =\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}y^k $,

Then

$\begin{array}\\ \sqrt{1+ux+vx^2} &=\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}(ux+vx^2)^k\\ &=\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}x^k(u+vx)^k\\ &=\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}x^k\sum_{j=0}^k \binom{k}{j}v^jx^ju^{k-j}\\ &=\sum_{k=0}^{\infty} \sum_{j=0}^k\dfrac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}x^{k+j} \binom{k}{j}v^ju^{k-j}\\ &=\sum_{n=0}^{\infty} x^n\sum_{j=0}^n\dfrac{(-1)^{n-j+1}}{4^{n-j}(2(n-j)-1)}\binom{2(n-j)}{n-j} \binom{n-j}{j}v^ju^{n-2j} \qquad(k = n-j)\\ \end{array} $

We can play with $\binom{2(n-j)}{n-j} \binom{n-j}{j} =\dfrac{(2n-2j)!(n-j)!}{(n-j)!^2j!(n-2j)!} =\dfrac{(2n-2j)!}{(n-j)!j!(n-2j)!} $ but I don't see anything beyond this.

Also, any corrections appreciated.

Answer 1

We obtain a slightly simplified representation by using the binomial series expansion \begin{align*} \sqrt{1+ux+vx^2}&=\sum_{k=0}^\infty \binom{\frac{1}{2}}{k}(ux+vx^2)^k\\ &=\sum_{k=0}^\infty \binom{\frac{1}{2}}{k}(ux)^k\left(1+\frac{v}{u}x\right)^k \end{align*}

and the coefficient $[x^n]$ of the series expansion is \begin{align*} \color{blue}{[x^n]\sqrt{1+ux+vx^2}}&=\sum_{k=0}^n\binom{\frac{1}{2}}{k}u^k[x^{n-k}]\sum_{j=0}^k\binom{k}{j}\left(\frac{v}{u}x\right)^j\\ &=\sum_{k=0}^n\binom{\frac{1}{2}}{k}u^k\binom{k}{n-k}\left(\frac{v}{u}\right)^{n-k}\\ &\,\,\color{blue}{=\sum_{k=0}^n\binom{\frac{1}{2}}{k}\binom{k}{n-k}u^{2k-n}v^{n-k}} \end{align*}