Question

Let $f(x) = a_1 \sin x + a_2 \sin 2x + ... + a_n \sin nx$, where $a_1 \in R$ and $n \in N$. If $|f(x)| \leq |\sin x|\forall x$, then the maximum value of $|a_1 + 2a_2 + ... + na_n|$ is

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (C)

1

Answer 2

Let $f(x) = a_1 \sin x + a_2 \sin 2x + ... + a_n \sin nx$. We are given that $|f(x)| \leq |\sin x|$ for all $x \in R$.

We are interested in the value of $S = a_1 + 2a_2 + ... + na_n$. Let's consider the limit of $\frac{f(x)}{\sin x}$ as $x \to 0$. $\frac{f(x)}{\sin x} = \frac{a_1 \sin x + a_2 \sin 2x + ... + a_n \sin nx}{\sin x} = a_1 \frac{\sin x}{\sin x} + a_2 \frac{\sin 2x}{\sin x} + ... + a_n \frac{\sin nx}{\sin x}$. As $x \to 0$, $\frac{\sin kx}{\sin x} \to k$. So, $\lim_{x \to 0} \frac{f(x)}{\sin x} = a_1(1) + a_2(2) + ... + a_n(n) = a_1 + 2a_2 + ... + na_n = S$.

We are given $|f(x)| \leq |\sin x|$. For $x \neq k\pi$ (where $k$ is an integer), $\sin x \neq 0$. We can divide by $|\sin x|$: $\left|\frac{f(x)}{\sin x}\right| \leq 1$. This inequality holds for all $x$ where $\sin x \neq 0$. In particular, it holds for $x$ in a neighborhood of 0, excluding 0 itself.

By the property of limits, if $g(x) \leq M$ for all $x$ in a neighborhood of $c$ (excluding $c$), and $\lim_{x \to c} g(x) = L$, then $L \leq M$. Similarly, if $g(x) \geq m$, then $L \geq m$. In our case, let $g(x) = \frac{f(x)}{\sin x}$ and $c=0$. We have $-1 \leq \frac{f(x)}{\sin x} \leq 1$ for $x$ near 0 (but $x \neq 0$). Taking the limit as $x \to 0$: $\lim_{x \to 0} (-1) \leq \lim_{x \to 0} \frac{f(x)}{\sin x} \leq \lim_{x \to 0} 1$. $-1 \leq S \leq 1$.

This implies $|S| \leq 1$. So, $|a_1 + 2a_2 + ... + na_n| \leq 1$. The maximum value of $|a_1 + 2a_2 + ... + na_n|$ is at most 1.

To show that 1 is achievable, we need to find coefficients $a_1, ..., a_n$ such that $|f(x)| \leq |\sin x|$ for all $x$ and $a_1 + 2a_2 + ... + na_n = 1$ or $a_1 + 2a_2 + ... + na_n = -1$. Consider the choice $a_1 = 1$ and $a_2 = a_3 = ... = a_n = 0$. In this case, $f(x) = 1 \cdot \sin x + 0 \cdot \sin 2x + ... + 0 \cdot \sin nx = \sin x$. The condition $|f(x)| \leq |\sin x|$ becomes $|\sin x| \leq |\sin x|$, which is true for all $x$. The sum $a_1 + 2a_2 + ... + na_n$ is $1 + 2(0) + ... + n(0) = 1$. The value of $|a_1 + 2a_2 + ... + na_n|$ is $|1| = 1$.

Since we have shown that the value is at most 1 and that 1 is achievable, the maximum value of $|a_1 + 2a_2 + ... + na_n|$ is 1.