# Closed Form of the Chebyshev Polynomials of the First Kind [Proof Request]

$$T_n(x) = \frac{n}{2} \sum_{r=0}^{\lfloor \frac{n}{2} \rfloor} \frac{(-1)^r}{n-r} \binom{n-r}{r} (2x)^{n-2r}$$

Searching on the web yielded no results, and the result is given without proof on OEIS and Wolfram MathWorld

• I have in my pre-thesis. Jan 25 at 9:22

It is defined by $$T_n(x)=\frac{u^{n}+u^{-n}}{2}$$ where $$x=\frac{u+u^{-1}}{2}.$$ This implies that $$T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$$. So, it is essentially enough to check this recursive relation at $$n+1-2r$$-the power term: $$\frac{n+1}{2}\frac{(-1)^r}{n+1-r}{n+1-r\choose r}2^{n+1-2r}=2\frac{n}{2}\frac{(-1)^{r}}{n-r}{n-r\choose r}2^{n-2r}-\frac{n-1}{2}\frac{(-1)^{r-1}}{n-r}{n-r\choose r-1}2^{n+1-2r}$$ which is equivalent to $$\frac{n+1}{n+1-r}{n+1-r\choose r}=\frac{n}{n-r}{n-r\choose r}+\frac{n-1}{n-r}{n-r\choose r-1}$$ and it is true.

Starting from the OGF

$$\sum_{n\ge 0} T_n(x) t^n = \frac{1-tx}{1-2tx+t^2}$$

We extract coefficients $$[x^q] [t^n]$$ where $$0\le q\le n$$. First do the coefficient on $$q=0$$ which is $$[t^n] \frac{1}{1+t^2}.$$ Continue with $$q\ge 1$$

$$[x^q] [t^n] \frac{1-tx}{1-2tx+t^2} = [x^q] [t^n] \frac{1-tx}{1+t^2} \frac{1}{1-2tx/(1+t^2)} \\ = [t^n] \frac{1}{1+t^2} \frac{2^q t^q}{(1+t^2)^q} - [t^n] \frac{1}{1+t^2} \frac{2^{q-1} t^{q}}{(1+t^2)^{q-1}} \\ = [t^n] \frac{2^q t^q}{(1+t^2)^{q+1}} - [t^n] \frac{2^{q-1} t^{q}}{(1+t^2)^{q}}.$$

Now we see here that $$q$$ and $$n$$ must have the same parity. Supposing that is true we obtain

$$2^q [t^{n-q}] \frac{1}{(1+t^2)^{q+1}} - 2^{q-1} [t^{n-q}] \frac{1}{(1+t^2)^q} \\ = 2^q (-1)^{(n-q)/2} {(n-q)/2+q\choose q} - 2^{q-1} (-1)^{(n-q)/2} {(n-q)/2+q-1\choose q-1} \\ = 2^q (-1)^{(n-q)/2} {(n+q)/2\choose q} - 2^{q-1} (-1)^{(n-q)/2} \frac{q}{(n+q)/2} {(n+q)/2\choose q} \\ = \frac{n}{n+q} 2^q (-1)^{(n-q)/2} {(n+q)/2\choose q}.$$

Note that this also gives the correct value for $$q=0$$ so we can merge that case into our formula. Observe also that we get zero from the binomial coefficient when $$(n+q)/2\lt q$$ or $$n\lt q.$$

Now when $$q$$ and $$n$$ have the same parity with $$n$$ and $$q$$ non-negative then $$n-q=2r$$ where $$0\le r\le \lfloor n/2 \rfloor.$$ We have $$q=n-2r$$ and obtain summing over $$r$$

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

This is the claim.