Chebyshev polynomials to hypergeometric function?

Question

I am trying to derive this hypergeometric version of the Chebyshev polynomials

https://en.wikipedia.org/wiki/Chebyshev_polynomials#Explicit_expressions

$$U_n(x) = \frac{\left (x+\sqrt{x^2-1} \right )^{n+1} - \left (x-\sqrt{x^2-1} \right )^{n+1}}{2\sqrt{x^2-1}} $$

$$ = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \binom{n+1}{2k+1} \left (x^2-1 \right )^k x^{n-2k} $$

$$ = x^n \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \binom{n+1}{2k+1} \left (1 - x^{-2} \right )^k $$

$$ = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \binom{2k-(n+1)}{k}~(2x)^{n-2k} \text{ for }~ n > 0$$

$$ = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} (-1)^k \binom{n-k}{k}~(2x)^{n-2k} \text{ for }~ n > 0 $$

$$ = \sum_{k=0}^{n}(-2)^{k} \frac{(n+k+1)!} {(n-k)!(2k+1)!}(1 - x)^k \text{ for }~ n > 0$$

$$ = (n+1) \ {}_2F_1\left(-n,n+2; \tfrac{3}{2}; \tfrac{1}{2}(1-x) \right) $$

How do you go from the floor function summation part to the hypergeometric part? Which binomial identities are used? How do you go from $(2x)^{n-2k}$ to $(1-x)^{k}$?

Answer 1

We show the following identity is valid for $n\geq 0$: \begin{align*} \color{blue}{\sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} (-1)^k \binom{n-k}{k}~(2x)^{n-2k} =\sum_{k=0}^{n}(-2)^{k}\binom{n+k+1}{2k+1}(1 - x)^k }\tag{1} \end{align*}

The bivariate generating function of the Chebyshev polynomials of the second kind is \begin{align*} \color{blue}{u(x,y)=\frac{1}{1-2xy+y^2}=\sum_{n=0}^{\infty}U_n(x)y^n} \end{align*} In the following we consider the generating function $\pi(x,y)$ \begin{align*} \pi(x,y)=\frac{1}{1-(2+x)y+y^2}=\frac{1}{(1-y)^2-xy}\tag{2} \end{align*} which is related with $u(x,y)$ by a transformation $x\to 2x-2$: \begin{align*} \color{blue}{\pi(2x-2,y)=u(x,y)} \end{align*}

Expanding the left-hand expression of $\pi(x,y)$ we obtain \begin{align*} \color{blue}{\pi(x,y)}&\color{blue}{=\frac{1}{1-(2+x)y+y^2}}\\ &=\sum_{n=0}^{\infty}\left((2+x)y-y^2\right)^n\tag{3.1}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\binom{n}{k}(2+x)^k(-1)^{n-k}y^{n-k}\right)y^n\tag{3.2}\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^n\binom{n}{k}(2+x)^{n-k}(-1)^ky^{n+k}\tag{3.3}\\ &=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\binom{n}{k}(2+x)^{n-k}(-1)^ky^{n+k}\tag{3.4}\\ &=\sum_{k=0}^{\infty}\sum_{n=2k}^{\infty}\binom{n-k}{k}(2+x)^{n-2k}(-1)^ky^n\tag{3.5}\\ &\,\,\color{blue}{=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} \binom{n-k}{k}(2+x)^{n-2k}(-1)^k\right)y^n}\tag{3.6} \end{align*}

Comment:

• In (3.1) we use a geometric series expansion.

• In (3.2) we apply the binomial theorem.

• In (3.3) we change the order of the inner sum $k\to n-k$.

• In (3.4) we exchange the order of the sums.

• In (3.5) we shift the index $n$ by $k$ to change $y^{n-k}\to y^n$.

• In (3.6) we shift again the order of the sums.

Expanding the right-hand expression of $\pi(x,y)$ we obtain \begin{align*} \color{blue}{\pi(x,y)}&\color{blue}{=\frac{1}{(1-y)^2-xy}} =\frac{1}{(1-y)^2}\,\frac{1}{1-\frac{xy}{(1-y)^2}}\\ &=\frac{1}{(1-y)^2}\sum_{n=0}^{\infty}\frac{(xy)^n}{(1-y)^{2n}}\tag{4.1}\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\binom{-2n-2}{k}(-y)^k(xy)^n\tag{4.2}\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\binom{2n+k+1}{k}x^ny^{n+k}\tag{4.3}\\ &=\sum_{n=0}^{\infty}\sum_{n=k}^{\infty}\binom{2n-k+1}{k}x^{n-k}y^n\tag{4.4}\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^n\binom{2n-k+1}{k}x^{n-k}y^n\tag{4.5}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\binom{n+k+1}{n-k}x^k\right)y^n\tag{4.6}\\ &\,\,\color{blue}{=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\binom{n+k+1}{2k+1}x^k\right)y^n}\tag{4.7}\\ \end{align*} It follows from (3.6) and (4.6) by comparing the coefficients of $y^n$ \begin{align*} \color{blue}{\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} \binom{n-k}{k}(2+x)^{n-2k}(-1)^k=\sum_{k=0}^{n}\binom{n+k+1}{2k+1}x^k} \end{align*} Replacing $x$ by $2x-2$ and the claim (1) follows.

Comment:

• In (4.1) we use a geometric series expansion.

• In (4.2) we use a binomial series expansion.

• In (4.3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (4.4) we exchange the order of the sums and we shift the index $n$ by $k$ to change $y^{n-k}\to y^n$.

• In (4.5) we exchange the order of the sums.

• In (4.6) we change the order of the inner sum $k\to n-k$.

• In (4.7) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

Note: This answer is a solution to chapter 2, problem 18.c in Combinatorial Identities by John Riordan.

Answer 2

Here is an alternate proof of the identity by @epi163sqrt. We seek

$$\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n-k\choose k} (2x)^{n-2k} = \sum_{k=0}^n (-2)^k {n+k+1\choose 2k+1} (1-x)^k.$$

We get for the coefficient on $[x^q]$ of the RHS where $0\le q\le n:$

$$\sum_{k=0}^n (-2)^k {n+k+1\choose 2k+1} (-1)^q {k \choose q} \\ = \sum_{k=0}^n (-2)^k {n+k+1\choose n-k} (-1)^q {k\choose q} \\ = [z^n] (1+z)^{n+1} \sum_{k=0}^n (-2)^k z^k (1+z)^k (-1)^q {k\choose q}.$$

Here may extend to infinity because of the coefficient extractor:

$$(-1)^q [z^n] (1+z)^{n+1} \sum_{k\ge 0} (-2)^k z^k (1+z)^k {k\choose q} \\ = (-1)^q [z^n] (1+z)^{n+1} \sum_{k\ge q} (-2)^k z^k (1+z)^k {k\choose q} \\ = (-1)^q [z^n] (1+z)^{n+1} (-2)^q z^q (1+z)^q \sum_{k\ge 0} (-2)^k z^k (1+z)^k {k+q\choose q} \\ = (-1)^q [z^n] (1+z)^{n+1} (-2)^q z^q (1+z)^q \frac{1}{(1+2z(1+z))^{q+1}} \\ = 2^q [z^{n-q}] (1+z)^{n+q+1} \frac{1}{(1+2z+2z^2)^{q+1}}.$$

This is

$$2^q \;\underset{z}{\mathrm{res}}\; \frac{1}{z^{n-q+1}} (1+z)^{n+q+1} \frac{1}{(1+2z+2z^2)^{q+1}}.$$

Now put $z/(1+z)=w$ so that $z=w/(1-w)$ and $dz = 1/(1-w)^2 \; dw$ to get

$$2^q \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n-q+1}} \frac{1}{(1-w)^{2q}} \frac{1}{(1+2w/(1-w)+2w^2/(1-w)^2)^{q+1}} \frac{1}{(1-w)^2} \\ = 2^q \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n-q+1}} \frac{1}{(1-w)^{2q}} \frac{(1-w)^{2q+2}}{((1-w)^2+2w(1-w)+2w^2)^{q+1}} \frac{1}{(1-w)^2} \\ = 2^q \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n-q+1}} \frac{1}{(1+w^2)^{q+1}} = 2^q [w^{n-q}] \frac{1}{(1+w^2)^{q+1}}.$$

Here we see that $n$ and $q$ must have the same parity, i.e. $n-q=2k$ where $0\le k\le \lfloor n/2 \rfloor.$ Letting $q=n-2k$ we find

$$2^{n-2k} [w^{2k}] \frac{1}{(1+w^2)^{n-2k+1}} = 2^{n-2k} [w^k] \frac{1}{(1+w)^{n-2k+1}} \\ = 2^{n-2k} (-1)^k {n-2k+k\choose n-2k} = 2^{n-2k} (-1)^k {n-k\choose n-2k} = 2^{n-2k} (-1)^k {n-k\choose k}.$$

This is the claim.