Particular Values of Chebyshev Polynomials of the Second Kind/0

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Theorem

Let $\map {U_n} x$ denote the Chebyshev polynomial of the second kind of order $n$.

Then:

$\map {U_n} 0 = \begin {cases} \paren {-1}^{n / 2} & : \text {$n$ even} \\ 0 & : \text {$n$ odd} \end {cases}$


Proof

Even Order

Let $n = 2 m$ for some $m \in \N$.

The proof proceeds by induction.

For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:

$\map {U_{2 m} } 0 = \paren {-1}^m$


Basis for the Induction

$\map P 0$ is the case:

\(\ds \map {U_0} 0\) \(=\) \(\ds \paren {-1}\) Chebyshev Polynomial of the Second Kind: $U_0$
\(\ds \) \(=\) \(\ds \paren {-1}^0\)

Thus $\map P 0$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\map {U_{2 k} } 0 = \paren {-1}^k$


from which it is to be shown that:

$\map {U_{2 \paren {k + 1} } } 0 = \paren {-1}^{k + 1}$


Induction Step

This is the induction step:

\(\ds \map {U_{2 \paren {k + 1} } } 0\) \(=\) \(\ds 2 \times 0 \times \map {U_{2 k + 1} } 0 - \map {U_{2 k} } 0\) Recurrence Formula for Chebyshev Polynomials of the First Kind, setting $n = 2 \paren {k + 1} = 2 k + 2$
\(\ds \) \(=\) \(\ds \paren {-1} \times \map {U_{2 k} } 0\) simplifying
\(\ds \) \(=\) \(\ds \paren {-1} \times \paren {-1}^k\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {-1}^{k + 1}\) simplifyng

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall m \in \Z_{\ge 0}: \map {U_{2 m} } 0 = \paren {-1}^m$

$\blacksquare$


Odd Order

Let $n = 2 k + 1$ for some $k \in \N$.

\(\ds \forall x \in \Dom {U_n}: \, \) \(\ds \map {U_n} {-x}\) \(=\) \(\ds \paren {-1}^n \map {U_n} x\) Particular Values of Chebyshev Polynomials of the Second Kind: $-x$
\(\ds \leadsto \ \ \) \(\ds \map {U_n} 0\) \(=\) \(\ds \paren {-1}^n \map {U_n} 0\)
\(\ds \) \(=\) \(\ds -\map {U_n} 0\) as $n$ is odd: $\paren {-1}^n = -1$

The only number $x$ such that $x = -x$ is $x = 0$.

Hence the result.

$\blacksquare$


Sources

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