How find this sum $\sum\limits_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$

Question

Find the sum close form $$f(x)=\sum_{i=0}^{2n}\dfrac{\binom{2n}{2i}\binom{2i}{i}x^{2i}}{2^{2i}}$$

if we let $$\dfrac{x}{2}=y$$ then $$f(y)=\sum_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$$

this PDF have this page 5 $$\sum_{k=j}^{n}\binom{n}{k}\binom{k}{j}=2^{n-j}\binom{n}{j}$$ the solution can see page 5

Maybe my problem can use this mathods?Thank you

Answer 1

Given any formal Laurent series $\;(???) = \sum \alpha_{k_1 k_2 \ldots k_n} t_1^{k_1} t_2^{k_2} \cdots t_n^{k_n}$, we will use the notation $[ t_1^{k_1} t_2^{k_2} \cdots t_n^{k_n} ](???)$ to denote the coefficient $\alpha_{k_1 k_2 \cdots k_n}$ in front of corresponding monomial.

Instead of $f(y)$, let us denote the polynomial we wish to find a closed form as $p_{2n}(y)$. We have

$$\begin{align} p_{2n}(y) &= \sum_{i=0}^{n}\binom{2n}{2i}\binom{2i}{i} y^{2i} = \sum_{i=0}^{n} \binom{2n}{2i} y^{2i}\bigg( [t^0](t + t^{-1})^{2i}\bigg)\\ &= \sum_{i=0}^{2n} \binom{2n}{i} \bigg( [t^0](y(t + t^{-1}))^i\bigg) = [\;t^0\;] \bigg( 1 + y(t+t^{-1})\bigg)^{2n} \end{align} $$

Substitute $t$ by $e^{i\theta}$ in above formal expression and notice for any $k \in \mathbb{Z}$, we have

$$\frac{1}{2\pi}\int_0^{2\pi} e^{ik\theta} d\theta = \begin{cases}1,&k = 0\\0,&\text{ otherwise }\end{cases}$$ We obtain an integral representation for $p_{2n}(y)$,

$$p_m(y) = \frac{1}{2\pi}\int_0^{2\pi} (1+2y\cos\theta)^{m} d\theta\quad\text{ for }\quad m = 2n$$

Treat this as a definition for $p_m(y)$ for general $m \in \mathbb{N}$ and consider following generating function:

$$p(y,\rho) = \sum_{m=0}^\infty p_m(y)\rho^m$$

It is easy to see $$ p(y,\rho) = \frac{1}{2\pi}\int_0^{2\pi}\frac{d\theta}{1-\rho(1+2y\cos\theta)} = \frac{1}{4\pi y\rho}\int_0^{2\pi}\frac{d\theta}{\frac{1-\rho}{2y\rho}-\cos\theta} $$ Using the identity $$\frac{1}{2\pi}\int_0^{2\pi}\frac{d\theta}{a - \cos\theta} = \frac{1}{\sqrt{a^2-1}}\quad\text{ for } a > 1$$ We get

$$\begin{align} p(y,\rho) &= \frac{1}{2y\rho}\frac{1}{\sqrt{\left(\frac{1-\rho}{2y\rho}\right)^2 - 1}} = \frac{1}{\sqrt{1-2\rho + (1- 4y^2)\rho^2}}\\ &= \frac{1}{\sqrt{1-2\frac{1}{\sqrt{1-4y^2}}(\rho\sqrt{1-4y^2}) + (\rho\sqrt{1-4y^2})^2}}\ \end{align} $$ Compare this with the generating function for Legendre polynomials,

$$\frac{1}{\sqrt{1-2zt+t^2}} = \sum_{k=0}^\infty P_k(z) t^k$$

We find

$$p(y,\rho) = \sum_{k=0}^\infty P_k\left(\frac{1}{\sqrt{1-4y^2}}\right) \left(\rho\sqrt{1-4y^2)}\right)^k$$ This leads to the expression we claimed in comment: $$p_{2n}(y) = (1-4y^2)^n P_{2n}\left(\frac{1}{\sqrt{1-4y^2}}\right)$$

For example, when $y = \frac12$, this leads to an interesting identity:

$$\begin{align} \sum_{i=0}^{n} \frac{\binom{2n}{2i}\binom{2i}{i}}{2^{2i}} &= p_{2n}\left(\frac12\right) = \lim_{y\to\frac12^{-}} (1-4y^2)^n P_{2n}\left(\frac{1}{\sqrt{1-4y^2}}\right)\\ &= [ t^{2n} ] P_{2n}(t) = \frac{(4n-1)!!}{(2n)!} \end{align} $$

Answer 2

Hint: A classical method is to specialize the general identity $$(1+a+b)^{2n}=\sum_{i,j}{2n\choose i,j}a^ib^j$$ to $a=ty$ and $b=y/t$. One gets $$(1+ty+y/t)^{2n}=\sum_{i,j}{2n\choose i,j}t^{i-j}y^{i+j},$$ and in particular $$[t^0](1+ty+y/t)^{2n}=\sum_{i}{2n\choose i,i}y^{2i}=\sum_{i}{2n\choose 2i}{2i\choose i}y^{2i}=f(y),$$ or, equivalently, $$f(y)=y^{2n}\cdot[t^{2n}](t^2+t/y+1)^{2n}.$$ Can you continue?