Chebyshev Polynomials and the Hypergeometric Function 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?
$$
T_n (x) = {}_2F_1 \left( { - n,n;\tfrac{1}{2};\tfrac{{1 - x}}{2}} \right).
$$
 A: 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.
A: 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
$$
T_{2n}(x) = (-1)^n\;{}_2F_1\left(-n,n;\frac12;x^2\right) .
$$
