This looks awfully familiar

Calculus Level 5

Define a sequence of polynomials $\lbrace T_{n} \rbrace_{n=1}$ in $x$ by $T_{1} = x$ , $T_{2} = 2x^{2}-1$ and $2x \ T_{n} = T_{n+1}+T_{n-1}$ where $n \geq 2$ .

Let $S$ be the set of all $n$ for which there exists a polynomial function $f$ such that $T_{n} = f \circ f$ , and $T_{n}(\frac{\sqrt{3}}{2})$ is locally maximized.

Find the value of $\displaystyle \left\lfloor 1000 \cdot \sum_{n \in S} \frac{1}{n} \right \rfloor$ .

Note: $(f \circ g)(x) = f(g(x))$ .

The answer is 45.

1 solution

Jake Lai
May 19, 2015

Recognise that $\lbrace T_{n} \rbrace$ are the Chebyshev polynomials of the first kind.

$T_{n} = f \circ f$ iff $n$ is a perfect square; this is not hard to prove.
We have $T_{n}(\frac{\sqrt{3}}{2}) = \cos(n\cos^{-1}(\frac{\sqrt{3}}{2})) = \cos(\frac{n\pi}{6})$ . The maxima of $T_{n}(\frac{\sqrt{3}}{2})$ are, clearly, $n = 12k$ where $k \in \mathbb{Z}^{+}$ .

As such, we now know $S = \lbrace 6^{2}, 12^{2}, 18^{2}, \ldots \rbrace$ and so

$\sum_{n \in S} \frac{1}{n} = \sum_{k=1}^{\infty} \frac{1}{(6k)^{2}} = \boxed{\frac{\pi^{2}}{216}}$

This looks awfully familiar

The answer is 45.

1 solution

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