Chebyshev Polynomial of the First Kind in terms of Gaussian Hypergeometric Function

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Theorem

Let $\map F {a, b; c; x}$ denote the Gaussian hypergeometric function of $x$:

$\ds \sum_{k \mathop = 0}^\infty \dfrac {a^{\overline k} b^{\overline k} } {c^{\overline k} } \dfrac {x^k} {k!}$


Then:

$\map {T_n} x = \map F {n, -n; \dfrac 1 2; \dfrac {1 - x} 2}$

where:

$x$ is a real number such that $\size x < 1$
$n$ is a natural number
$\map {T_n} x$ is the Chebyshev polynomial of the first kind of order $n$.


Proof 1

From Solution to Hypergeometric Differential Equation, we have:

The hypergeometric series:

\(\ds \map F {a, b, c; x}\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dfrac {a^{\overline n} b^{\overline n} } {c^{\overline n} \, n!} x^n\)

defines a solution to the hypergeometric differential equation:

$x \paren {1 - x} \dfrac {\d^2 y} {\d x^2} + \paren {c - \paren {a + b + 1} x} \dfrac {\d y} {\d x} - a b y = 0$


Inputting $\map F {n, -n; \dfrac 1 2; \dfrac {1 - z} 2}$ into the hypergeometric differential equation, we obtain:

\(\ds 0\) \(=\) \(\ds x \paren {1 - x} \dfrac {\d^2 y} {\d x^2} + \paren {c - \paren {a + b + 1} x} \dfrac {\d y} {\d x} - a b y\) Solution to Hypergeometric Differential Equation
\(\ds \) \(=\) \(\ds x \paren {1 - x} \dfrac {\d^2 y} {\d x^2} + \paren {\frac 1 2 - \paren {n + \paren {-n} + 1} x} \dfrac {\d y} {\d x} - n \paren {-n} y\) Setting $a := n$, $b := -n$ and $c := \dfrac 1 2$
\(\ds \) \(=\) \(\ds x \paren {1 - x} \dfrac {\d^2 y} {\d x^2} + \paren {\frac 1 2 - x} \dfrac {\d y} {\d x} + n^2 y\)
\(\ds \) \(=\) \(\ds \frac {\paren {1 - z} } 2 \paren {1 - \frac {\paren {1 - z} } 2 } \dfrac {\d^2 y} {\paren {-\dfrac {\d z} 2 }^2} + \paren {\frac 1 2 - \dfrac {\paren {1 - z} } 2 } \dfrac {\d y} {\paren {-\dfrac {\d z} 2 } } + n^2 y\) Setting $x := \dfrac {\paren {1 - z} } 2$ and $\d x = -\dfrac {\d z} 2$
\(\ds \) \(=\) \(\ds \frac {\paren {1 - z} } 2 \frac {\paren {1 + z} } 2 \dfrac {\d^2 y} {\d z^2} \paren 4 + \frac z 2 \dfrac {\d y} {\d z} \paren {-2} + n^2 y\)
\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds \paren {1 - z^2} \dfrac {\d^2 y} {\d z^2} - z \dfrac {\d y} {\d z} + n^2 y\) Chebyshev's differential equation

We now see that $\map F {n, -n; \dfrac 1 2; \dfrac {1 - z} 2}$ simultaneously solves both the hypergeometric differential equation AND Chebyshev's differential equation!


We also have that the general solution to Chebyshev's differential equation is given by:

$y = \begin {cases} A \map {T_n} z + B \sqrt {1 - z^2} \, \map {U_{n - 1} } z & : n = 1, 2, 3, \ldots \\ \\ A + B \arcsin z & : n = 0 \end {cases}$

where:

$\map {T_n} z$ denotes the Chebyshev polynomial of the first kind of order $n$
$\map {U_n} z$ denotes the Chebyshev polynomial of the second kind of order $n$

Therefore, for $n > 0$, $\map F {n, -n; \dfrac 1 2; \dfrac {1 - z} 2}$ must be either $A \map {T_n} z$ or $B \sqrt {1 - z^2} \, \map {U_{n - 1} } z$.

Setting $z := \map \cos 0 = 1$, then

$\map F {n, -n; \dfrac 1 2; \dfrac {1 - 1} 2} = 1$ and
$A \map {T_n} 1 = A$ and
$B \sqrt {1 - 1^2} \, \map {U_{n - 1} } 1 = 0 $

Therefore, $A = 1$ and we obtain:

$\map F {n, -n; \dfrac 1 2; \dfrac {1 - z} 2} = \map {T_n} z$


$\blacksquare$


Proof 2

From De Moivre's Formula, we have:

$\cos n \theta + i \sin n \theta = \paren {\cos \theta + i \sin \theta}^n$

As $n \in \Z_{>0}$, we use the Binomial Theorem on the right hand side, resulting in:

$\ds \cos n \theta + i \sin n \theta = \sum_{k \mathop \ge 0} \binom n k \paren {\cos^{n - k} \theta} \paren {i \sin \theta}^k$

When $k$ is even, the expression being summed is real.

Equating the real parts of both sides of the equation, replacing $k$ with $2 k$ to make $k$ even, gives:

\(\ds \cos n \theta\) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k } \paren {\cos^{n - \paren {2 k } } \theta} \paren {\sin^{2 k } \theta}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {-1}^k \frac {n!} {\paren {n - 2 k}! } \frac 1 {\paren {2 k}!} \paren {\cos^{n - \paren {2 k } } \theta} \paren {\sin^{2 k } \theta}\) Definition of Binomial Coefficient
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {-1}^k \frac {n!} {\paren {n - k}! } \frac {\paren {n - k}! } {\paren {n - 2 k}! } \frac 1 {\paren {2 k}!} \paren {\cos^{n - \paren {2 k } } \theta} \paren {\sin^{2 k } \theta}\) $\dfrac {n!} {\paren {n - 2 k}! } = \dfrac {n!} {\paren {n - k}! } \dfrac {\paren {n - k}! } {\paren {n - 2 k}! }$
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {-1}^k \frac n {n - k} \frac {\paren {n - 1}!} {\paren {n - 1 - k}! } \frac {\paren {n - k}! } {\paren {n - 2 k}! } \frac 1 {\paren {2 k}!} \paren {\cos^{n - \paren {2 k } } \theta} \paren {\sin^{2 k } \theta}\) Definition of Factorial
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {-1}^k \frac n {n - k} \paren {n - k}^{\overline k } \paren {n - k}^{\underline k } \frac 1 {\paren {2 k}!} \paren {\cos^{n - \paren {2 k } } \theta} \paren {\sin^{2 k } \theta}\) Rising Factorial as Quotient of Factorials, Falling Factorial as Quotient of Factorials
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \frac n {n - k} \paren {n - k}^{\overline k } \paren {-\paren {n - k} }^{\overline k } \frac 1 {\paren {2 k}!} \paren {\cos^{n - \paren {2 k } } \theta} \paren {\sin^{2 k } \theta}\) Rising Factorial in terms of Falling Factorial of Negative
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \frac n {n - k} \paren {n - k}^{\overline k } \paren {-\paren {n - k} }^{\overline k } \frac 1 {k! \paren {\dfrac 1 2}^{\overline k} 2^{2 k } } \paren {\cos \theta}^{n - \paren {2 k } } \paren {\sin^{2 } \theta}^k\) Legendre's Duplication Formula: Corollary $2$
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \frac n {n - k} \paren {n - k}^{\overline k } \paren {-\paren {n - k} }^{\overline k } \frac 1 {k! \paren {\dfrac 1 2}^{\overline k} 2^{2 k } } \paren {\cos \theta}^{n - \paren {2 k } } \paren {1 - \cos^{2 } \theta}^k\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \frac n {n - k} \paren {n - k}^{\overline k } \paren {-\paren {n - k} }^{\overline k } \frac 1 {k! \paren {\dfrac 1 2}^{\overline k} } \paren {\cos \theta}^{\paren {n - k } - k } \paren {\frac {\paren {1 + \cos \theta} } 2}^k \paren {\frac {\paren {1 - \cos \theta} } 2}^k\) Square of Difference
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \frac {n + k} n \paren {n }^{\overline k } \paren {-n }^{\overline k } \frac 1 {k! \paren {\dfrac 1 2}^{\overline k} } \paren {\cos \theta}^{n - k } \paren {\frac {\paren {1 + \cos \theta} } 2}^k \paren {\frac {\paren {1 - \cos \theta} } 2}^k\) $\paren {n - k} \to n$
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {n }^{\overline k } \paren {-n }^{\overline k } \frac 1 {k! \paren {\dfrac 1 2}^{\overline k} } \paren {\frac {\paren {1 - \cos \theta} } 2}^k \paren {1 + \frac k n} \paren {\cos \theta}^{n - k } \paren {\frac {\paren {1 + \cos \theta} } 2}^k\)
\(\ds \) \(=\) \(\ds
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\)
\(\ds \) \(=\) \(\ds ?\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac { \paren {- n}^{\overline k} n^{\overline k} \paren {\dfrac {1 - x} 2}^k } { \paren {\dfrac 1 2}^{\overline k} k!}\)
\(\ds \) \(=\) \(\ds \map F {n, -n; \dfrac 1 2; \dfrac {1 - x} 2}\) Definition of Gaussian Hypergeometric Function

$\blacksquare$



Sources

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