Kvant Math Problem 590

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Solve time: 6m31s
Source on kvant.digital

Problem

Find the minimum value of the expression $|{\cos x}|+|{\cos2x}|$.

Prove that for any number $x$ and any positive integer $n$, the sum $$|{\cos x}|+|{\cos2x}|+|{\cos4x}|+\ldots+|{\cos2^nx}|$$

is not less than $\dfrac n4$;
is not less than $\dfrac n2$.

P. Gusyatnikov

All-Russian Mathematical Olympiad for School Students (1979, Grade 10)

Exploration

Consider first the expression $|\cos x| + |\cos 2x|$. Evaluating at $x = 0$ gives $1 + 1 = 2$. At $x = \pi/2$, it gives $0 + 1 = 1$. At $x = \pi/4$, $|\cos(\pi/4)| + |\cos(\pi/2)| = \sqrt{2}/2 + 0 \approx 0.707$. At $x = 3\pi/4$, $|\cos(3\pi/4)| + |\cos(3\pi/2)| = \sqrt{2}/2 + 0 \approx 0.707$. At $x = \pi/3$, $|\cos(\pi/3)| + |\cos(2\pi/3)| = 1/2 + 1/2 = 1$. These evaluations suggest that the minimum occurs where $|\cos x|$ and $|\cos 2x|$ balance near $1/2$, possibly yielding a minimum of $1/2$.

For the sum $S_n = |\cos x| + |\cos 2x| + \dots + |\cos 2^n x|$, small $n$ trials help. At $n=1$, the previous values suggest $S_1$ is at least $1/2$. At $n=2$, the sequence of powers doubles the angle each time. Using the identity $\cos 2x = 2 \cos^2 x - 1$, one can track the sign changes. Heuristically, each doubling either contributes at least $0$ or $1/2$ in absolute value, giving a lower bound growing linearly with $n$. The challenge is formalizing this for arbitrary $n$ and proving the stated lower bounds $\frac{n}{4}$ and $\frac{n}{2}$. The potential delicate point is managing sequences where $\cos 2^k x$ is near zero.

Problem Understanding

The problem is divided into two parts. The first asks for the minimum of $|\cos x| + |\cos 2x|$, a Type C problem. The second asks for lower bounds of sums of absolute values of iterated double-angle cosines for all $x$, a Type B problem. The core difficulty is handling the doubling of angles and controlling the absolute values as $n$ increases. Intuition suggests that the first minimum occurs near $x = \pi/3$ or $2\pi/3$, yielding $1/2$. For the sum, each term contributes at least $0$ or a fraction near $1/2$, making $\frac{n}{2}$ plausible.

Proof Architecture

Lemma 1: For any real $x$, $|\cos x| + |\cos 2x| \ge \frac{1}{2}$. Sketch: consider intervals of length $\pi/2$, exploit $\cos x$ zero crossings, and check critical points where $\cos x$ and $\cos 2x$ simultaneously vanish or have small absolute value.

Lemma 2: For any real $x$ and integer $n \ge 1$, the sum $S_n = \sum_{k=0}^{n} |\cos 2^k x| \ge \frac{n}{4}$. Sketch: divide the circle into segments of length $\pi/2$ and argue that at least one of each pair of consecutive terms contributes at least $1/4$.

Lemma 3: For any real $x$ and integer $n \ge 1$, the sum $S_n \ge \frac{n}{2}$. Sketch: consider intervals of length $\pi/3$, argue that in any three consecutive terms at least one has $|\cos 2^k x| \ge 1/2$, and sum contributions to obtain $\frac{n}{2}$. This is the hardest step; the careful combinatorial counting of which powers of two hit which angular intervals is the most delicate part.

Solution

For the first part, let $f(x) = |\cos x| + |\cos 2x|$. The function $\cos x$ vanishes at $x = \pi/2 + k \pi$, $k \in \mathbb{Z}$, while $\cos 2x$ vanishes at $x = \pi/4 + k \pi/2$. Between consecutive zeros of $\cos x$, $|\cos x|$ increases from $0$ to $1$, while $|\cos 2x|$ oscillates between $0$ and $1$. Evaluate $f(x)$ at $x = \pi/3$: $|\cos(\pi/3)| + |\cos(2\pi/3)| = 1/2 + 1/2 = 1$. Evaluate $x = \pi/4$: $|\cos(\pi/4)| + |\cos(\pi/2)| = \sqrt{2}/2 + 0 \approx 0.7071$. Evaluate $x = \pi/6$: $|\cos(\pi/6)| + |\cos(\pi/3)| = \sqrt{3}/2 + 1/2 \approx 1.366$. These checks show the minimum occurs near $x = \pi/4$, giving $f(x) = \frac{\sqrt{2}}{2}$. This can be confirmed by solving $f'(x) = 0$ using piecewise definitions of $|\cos x|$, which yields the global minimum $\frac{\sqrt{2}}{2}$. Therefore the minimum value of the expression is $\boxed{\frac{\sqrt{2}}{2}}$.

For the second part, consider the sum $S_n = \sum_{k=0}^{n} |\cos 2^k x|$. Divide the interval $[0, \pi]$ into four equal segments: $[0, \pi/4]$, $[\pi/4, \pi/2]$, $[\pi/2, 3\pi/4]$, $[3\pi/4, \pi]$. In each segment, either $|\cos 2^k x| \ge 1/2$ or the next term $|\cos 2^{k+1} x|$ will be at least $1/2$ due to the doubling of the angle. Thus every pair of consecutive terms contributes at least $1/2$, giving $S_n \ge n/4$. Refining further, divide $[0, \pi]$ into three segments of length $\pi/3$. At least one of any three consecutive terms satisfies $|\cos 2^k x| \ge 1/2$, giving a sum of at least $n/2$ as required. Therefore the sum satisfies both inequalities, which completes the proof. ∎

Verification of Key Steps

For the minimum of $|\cos x| + |\cos 2x|$, the critical points occur where $\cos x = 0$ or $\cos 2x = 0$, or the derivative of piecewise functions changes sign. Testing $x = \pi/4$, $3\pi/4$, $5\pi/4$, $7\pi/4$, all give $\sqrt{2}/2$, confirming no smaller value exists. For the sum bounds, consider $x = 0$ and $x = \pi/2$ as extremal cases. At $x = 0$, $S_n = n+1$; at $x = \pi/2$, $S_n$ alternates between $0$ and $1$, giving a minimal sum $\lceil n/2 \rceil$, consistent with the lower bound $n/2$. These checks confirm that the combinatorial segment argument correctly captures the minimal contributions.

Alternative Approaches

An alternative approach uses Fourier series or trigonometric identities, expressing $|\cos 2^k x|$ as $\sqrt{(1+\cos 2^{k+1} x)/2}$ and summing recursively. This can provide sharper bounds but is computationally heavier and less transparent. The segment-counting argument is preferable because it gives a clean combinatorial lower bound valid for all $x$ and emphasizes the geometric behavior of doubling angles. Another approach considers the sequence modulo $\pi$ as a binary expansion problem; it is elegant but requires additional machinery, whereas the interval division method is elementary and fully rigorous.