Particular Values of Chebyshev Polynomials of the Second Kind/1

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Theorem

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

Then:

$\map {U_n} 1 = n + 1$


Proof

The proof proceeds by strong induction.

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

$\map {U_n} 1 = n + 1$


Basis for the Induction

$\map P 0$ is the case:

$\map {U_0} 1 = 1$

which is demonstrated in Chebyshev Polynomial of the Second Kind: $U_0$.

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

$\Box$


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

$\map {U_1} x = 2 x$

Substituting $x = 1$ yields:

$\map {U_1} 1 = 2 = 1 + 1$


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:

$\map {U_k} 1 = k + 1$


from which it is to be shown that:

$\map {U_{k + 1} } 1 = k + 2$


Induction Step

This is the induction step:

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


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


Therefore:

$\forall n \in \Z_{\ge 0}: \map {U_n} 1 = n + 1$

$\blacksquare$


Sources

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