Chebyshev Polynomials and the Hypergeometric Function

Question

A problem related to Chebyshev Polynomials and the Hypergeometric Function involves transformation from one function to another. The task is to transform the Chebyshev polynomial into its correct scaled Hypergeometric Function. $$ T_{2n} (x) = ( - 1)^n {}_2F_1 \left( { - n,n;\tfrac{1}{2};x^2 } \right). $$ Is there any way for me to make use of the relationship below? $$ T_n (x) = {}_2F_1 \left( { - n,n;\tfrac{1}{2};\tfrac{{1 - x}}{2}} \right). $$

Answer 1

The Chebyshev polynomials are defined by $$ T_n(\cos\theta)=\cos(n\theta). $$ Hence $$ T_{2n}(\cos(\theta))=\cos(2n\theta)=T_n(\cos(2\theta)). $$ We now use the relation $\cos(2\theta)=2\cos(\theta)^2-1$ and write $\cos\theta=x$ to get $$ T_{2n}(x)=T_n(2x^2-1). $$ This should prove the claimed relation.

Answer 2

Let me carry out my suggestion.
Start HERE with the differential equation for ${}_2F_1$:
The solution of $$ z(1-z)F''(z)+\left(\frac12 - z\right)F'(z)+n^2 F(z) = 0,\quad F(0)=1, F'(0)=-2n^2 \tag1$$ is $$ F(z) = {}_2F_1\left(-n,n;\frac12;z\right) . $$ Change variables $z=x^2$. Then we have: The solution of $$ (1-x^2)F''(x)-xF'(x)+4n^2F(x) = 0,\quad F(0)=1,F'(0)=0 \tag2$$ is $$ F(x) = {}_2F_1\left(-n,n;\frac12;x^2\right) . $$

Next, HERE we find that the Chebyshev polynomial $T_m(x)$ satisfies $$ (1-x^2)T_m''(x)-xT_m'(x)+m^2T_m(x)=0 \tag3$$ Put $m=2n$, then $(3)$ becomes $(2)$. Also $T_{2n}(0)=(-1)^{n}, T_{2n}'(0)=0$. Therefore $$ T_{2n}(x) = (-1)^n\;{}_2F_1\left(-n,n;\frac12;x^2\right) . $$