2023 A1

A problem with gratuitous use of the chain and product rules.

Problem Setup

For a positive integer \(n\), let

\[f_n(x) = \cos(x) \cos(2x) \cos(3x) \cdots \cos(nx).\]

Find the smallest \(n\) such that \(|f_n''(0)| > 2023\).

Solution

We can first attempt to generalize the first derivative of:

\[f(x) = \cos(x) \cos(2x) \cos(3x) \ldots \cos(nx)\]

using the product rule:

\[ f_{n}'(x) = (\cos x)' \left[\cos(2x) \cos(3x) \ldots \cos(nx)\right] + \cos(x) \left[\cos(2x) \cos(3x) \ldots \cos(nx)\right]' \]

This simplifies to:

\[ -\sin(x) \cos(2x) \ldots + \cos(x) \left[\cos(2x)' \left(\cos(3x) \ldots \cos(nx)\right) + \cos(2x) \left(\cos(3x) \ldots \cos(nx)\right)'\right] \]

Repeating this process, we achieve a series of \(f_{n}(x)\) similar terms with one \(- k \sin(kx)\) term replacing the respective cosine term:

\[ \begin{aligned} f_{n}'(x) &= \sum_{k=1}^{n} -k \sin(kx) \prod_{\substack{j=1 \\ j \neq k}}^{n} \cos(jx) \\ &= -\sin(x)\cos(2x)\ldots\cos(nx) \\ &\quad - 2\cos(x)\sin(2x)\cos(3x)\ldots\cos(nx) \\ &\quad - \ldots - n\cos(x)\cos(2x)\ldots\cos((n-1)x)\sin(nx) \end{aligned} \]

As the first derivative of \(f_{n}(x)\) is a collection of terms similar in form to \(f_{n}(x)\) itself, finding the second derivative is like finding the first derivative repeatedly:

First term of \(f_{n}'(x)\):

\[-\sin(x) \cos(2x) \cos(3x) \ldots \cos(nx)\]

Taking the derivative:

\[ \begin{aligned} \left(-\sin(x) \cos(2x) \cos(3x) \ldots \cos(nx)\right)' &= -\cos(x)\cos(2x)\cos(3x)\ldots\cos(nx) \\ &\quad + 2\sin(x)\sin(2x)\cos(3x)\ldots\cos(nx) \\ &\quad + 3\sin(x)\cos(2x)\sin(3x)\ldots\cos(nx) \\ &\quad + \ldots + n\sin(x)\cos(2x)\ldots\sin(nx) \end{aligned} \]

By a similar analysis, the second derivative of the second term is:

\[ \left(-2\cos(x)\sin(2x)\cos(3x)\ldots\cos(nx)\right)' \]

\[ = -2 \left[-\sin(x)\sin(2x)\ldots\cos(nx) + 2\cos(x)\cos(2x)\ldots\cos(nx) - \ldots - n\cos(x)\sin(2x)\ldots\sin(nx)\right]. \]

Thus, the general form of \(f_{n}''(x)\) looks like:

\[-f_{n}(x) - 2^2 f_{n}(x) - 3^2 f_{n}(x) - \ldots - n^2 f_{n}(x) + \text{forms of } f_{n}(x) \text{ containing sine terms.}\]

Evaluating \(f_{n}''(0)\) becomes straightforward since \(f_{n}(0) = (\cos 0)^n = 1\) and every sine term vanishes. Therefore:

\[f_{n}''(0) = -(1^2 + 2^2 + 3^2 + \ldots + n^2).\]

Finishing the Problem

There are two approaches:

1.) The Clever Way:

According to Kedlaya’s solution, the sum \(\left| f_n''(0) \right|\) simplifies to:

\[\frac{n(n + 1)(2n + 1)}{6}\]

Using this formula, finding the smallest \(n\) such that \(\left| f_n''(0) \right| > 2023\) is straightforward.

2.) The Way I Found \(n\):

I expanded the series term by term until reaching the required value. I won't bore you with the simple multiplication and addition. We find:

\[\left| f_{17}''(0) \right| = 1785, \quad \left| f_{18}''(0) \right| = 2109\]

Therefore, the solution to the problem is \(n = 18\). \(\square\)