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Discrete Chebyshev polynomials
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The classical Chebyshev polynomials (of the first kind), $\mathcal{T}_n$, solve the differential equation $(1-t^2)y''-ty'+n^2y=0$ The Chebyshev polynomials of the second kind solve $(1-t^2)y''-3ty'+n(n+2)y=0.$ Question 1: Define and solve the discrete analogue of these differential equations.
Answer: The differential equation for Chebyshev polynomials of the first kind has the following discrete analogue:
$t(t-1)\Delta^2 y(t-2) + 2t \Delta y(t-1) + t \Delta y(t-1) + \Delta y(t) - n^2 y(t) =0,$ and we call the polynomial solutions the discrete Chebyshev polynomials of the first kind, $T_n$.

The differential equation for Chebyshev polynomials of the second kind has the following discrete analogue:
$t(t-1)\Delta^2 y(t-2) + 2t\Delta^2 y(t-1) + 3t \Delta y(t-1) + 3\Delta y(t) - n(n+2)y(t)=0,$ and we call the polynomial solutions the discrete Chebyshev polynomials of the second kind, $U_n$.

Question 2: Prove various formulas that the classical Chebyshev polynomials obey have discrete analogues.
Answer: The discrete Chebyshev polynomials obey the following formulas:
$T_{n+1}(t)-2tT_n(t-1)-2T_n(t)+T_{n-1}(t)=0,$ $U_{n+1}(t)-2tU_n(t-1)-2U_n(t)+U_{n-1}(t)=0,$ $\Delta T_n(t) = nU_{n-1}(t),$ and $U_n(t)-U_{n-2}(t)=2T_n(t).$

Question 3: What conclusions about classic generalized hypergeometric functions follow from the investigation?
Answer: Both discrete Chebyshev polynomials are examples of ${}_3\mathcal{F}_1$ hypergeometric functions and we obtain
${}_3\mathcal{F}_1\left(-n-1,n+1,-t;\dfrac{1}{2};\dfrac{1}{2}\right)-2t{}_3\mathcal{F}_1\left(-n,n,-t+1;\dfrac{1}{2};\dfrac{1}{2}\right) -2{}_3\mathcal{F}_1\left(-n,n,-t;\dfrac{1}{2};\dfrac{1}{2}\right) + {}_3\mathcal{F}_1\left(-n+1,n-1,-t;\dfrac{1}{2};\dfrac{1}{2}\right)=0,$ $(n+2){}_3\mathcal{F}_1\left(-n-1,n+3,-t;\dfrac{3}{2};\dfrac{1}{2}\right)-2t(n+1){}_3\mathcal{F}_1\left(-n,n+2,-t+1;\dfrac{3}{2};\dfrac{1}{2}\right) - 2(n+1){}_3\mathcal{F}_1\left(-n,n+2,-t;\dfrac{3}{2};\dfrac{1}{2}\right) + n{}_3\mathcal{F}_1\left(-n+1,n+1,-t;\dfrac{3}{2};\dfrac{1}{2}\right)=0,$ and $(n+1){}_3\mathcal{F}_1\left(-n,n+2,-t;\dfrac{3}{2};\dfrac{1}{2}\right)-(n-1){}_3\mathcal{F}_1\left(-n+2,n,-t;\dfrac{3}{2};\dfrac{1}{2}\right)=2{}_3\mathcal{F}_1\left(-n,n,-t;\dfrac{1}{2};\dfrac{1}{2}\right).$