Particular Values of Chebyshev Polynomials of the Second Kind/-x

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Theorem

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

Then:

$\map {U_n} {-x} = \paren {-1}^n \map {U_n} x$


Proof

The proof proceeds by strong induction.

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

$\map {U_n} {-x} = \paren {-1}^n \map {U_n} x$


First we note that:

\(\ds \map {U_n} {-x}\) \(=\) \(\ds \paren {-1}^n \map {U_n} x\)
\(\ds \leadstoandfrom \ \ \) \(\ds \dfrac 1 {\paren {-1}^n} \map {U_n} {-x}\) \(=\) \(\ds \map {U_n} x\) dividing both sides by $\paren {-1}^n$
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {-1}^n \map {U_n} {-x}\) \(=\) \(\ds \map {U_n} x\) as $\paren {-1}^n = \dfrac 1 {\paren {-1}^n}$

That is:

$(1): \quad \map {U_n} {-x} = \paren {-1}^n \map {U_n} x \iff \map {U_n} x = \paren {-1}^n \map {U_n} {-x}$


Basis for the Induction

From Chebyshev Polynomial of the Second Kind: $U_0$:

\(\ds \map {U_0} x\) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds \map {U_0} {-x}\)
\(\ds \) \(=\) \(\ds \paren {-1}^0 \map {U_0} {-x}\)


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

$\Box$


From Chebyshev Polynomial of the Second Kind: $U_1$ we have:

\(\ds \map {U_1} x\) \(=\) \(\ds 2 x\)
\(\ds \map {U_1} {-x}\) \(=\) \(\ds -2 x\)
\(\ds \) \(=\) \(\ds -\map {U_1} x\)
\(\ds \) \(=\) \(\ds \paren {-1}^1 \map {U_1} x\)


Thus $\map P 1$ is seen to hold.

$\Box$


$\map P 0$ and $\map P 1$ form the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.


This is the induction hypothesis:

$\map {U_k} x = \paren {-1}^k \map {U_k} {-x}$

which from $(1)$ can be expressed as:

$\map {U_k} {-x} = \paren {-1}^k \map {U_k} x$


from which it is to be shown that:

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

which from $(1)$ can be expressed as:

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


Induction Step

This is the induction step:

\(\ds \map {U_{k + 1} } {-x}\) \(=\) \(\ds 2 \paren {-x} \, \map {U_k} {-x} - \map {U_{k - 1} } {-x}\) Recurrence Formula for Chebyshev Polynomials of the First Kind setting $x = -x$ and $n = k + 1$
\(\ds \) \(=\) \(\ds \paren {-2 x} \times \paren {-1}^k \map {U_k} x - \paren {-1}^{k - 1} \map {U_{k - 1} } x\) Induction Hypothesis: $\map {U_k} {-x} = \paren {-1}^k \map {U_k} x$ and $\map {U_{k - 1} } {-x} = \paren {-1}^{k - 1} \map {U_{k - 1} } x$
\(\ds \) \(=\) \(\ds \paren {-2 x} \times \paren {-1}^k \map {U_k} x - \paren {-1}^{k + 1} \map {U_{k - 1} } x\) as $\paren {-1}^{k - 1} = \paren {-1}^{k + 1}$
\(\ds \) \(=\) \(\ds \paren {-1}^{k + 1} 2 x \, \map {U_k} x - \paren {-1}^{k + 1} \map {U_{k - 1} } x\) simplifying
\(\ds \) \(=\) \(\ds \paren {-1}^{k + 1} \paren {2 x \, \map {U_k} x - \map {U_{k - 1} } x}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \map {U_{k + 1} } {-x}\) \(=\) \(\ds \paren {-1}^{k + 1} \map {U_{k + 1} } x\) Recurrence Formula for Chebyshev Polynomials of the Second Kind setting $n = k + 1$


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


Therefore:

$\forall n \in \Z_{\ge 0}: \map {U_n} x = \paren {-1}^n \map {U_n} {-x}$

$\blacksquare$


Sources

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