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I've asked this question before a long time ago, but I didn't get a complete answer. This is the link to the incomplete answer: Prove the following property of $f(x)$?

Let $$f(x)=|a_1\sin(x)+a_2\sin(2x)+a_3\sin(3x)+\cdots+a_n\sin(nx)|.$$ Given that $f(x)$ is less than or equal to $|\sin(x)|$ for all $x$, prove that $|a_1+a_2+\cdots+a_n|$ is less than or equal to $1$. Please keep this at a calculus AB level because that's where I got the problem from. Thanks!

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up vote 3 down vote accepted

I believe that under the same assumptions the following result can be proved, that is, $|a_1+2a_2+\ldots+na_n|\leq 1$.

We will need the following result (which I believe can be proved using Calculus AB level): For every $k\in\mathbb{N}$, $\dfrac{\sin(kx)}{\sin(x)}$ can be extended to a continuous function. The proof goes like this:

Observe that $\dfrac{\sin(k(x+2\pi))}{\sin(x+2\pi)}=\dfrac{\sin(kx)}{\sin(x)}$. Observe that this function is continuous except possibly where $\sin(x)=0$. Because the above observation, we only need to worry about what happens at $x=0$. But in this case the following limit exists: $$ \lim_{x\to 0}\dfrac{\sin(kx)}{\sin(x)}=k $$ this limit follows using the well known result: $\lim_{x\to 0}\dfrac{\sin(x)}{x}=1$.

Since we have that $|a_1\sin(x)+a_2\sin(2x)+\ldots+a_n\sin(nx)|\leq |\sin(x)|$ for all $x$ we have that $$ |\sin(x)||a_1+a_2\dfrac{\sin(2x)}{\sin(x)}+\ldots+a_n\dfrac{\sin(nx)}{\sin(x)}|\leq|\sin(x)| $$ for all $x$. Subtracting $|\sin(x)|$ from both side we obtain the following inequality for all $x$: $$ |\sin(x)|(|a_1+a_2\dfrac{\sin(2x)}{\sin(x)}+\ldots+a_n\dfrac{\sin(nx)}{\sin(x)}|-1)\leq 0 $$ which implies that: $$ |a_1+a_2\dfrac{\sin(2x)}{\sin(x)}+\ldots+a_n\dfrac{\sin(nx)}{\sin(x)}|\leq 1 $$ for all $x\neq 2\pi m$. Taking $\lim_{x\to 0}$ in both sides we obtain: $$ |a_1+2a_2+\ldots+na_n|\leq 1 $$ Now if we take the inequality:

$$ |\sin(x)||a_1+a_2\dfrac{\sin(2x)}{\sin(x)}+\ldots+a_n\dfrac{\sin(nx)}{\sin(x)}|\leq|\sin(x)| $$ and divide by $x\neq 0$ both sides and take the $\lim_{x\to 0}$ we obtain: $$ |a_1+2a_2+\ldots+na_n|\leq 1 $$

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VERY nice, thank you very much. And actually I just remembered, the original problem was actually |a1+2a2+3a3+...| but I forgot about that since I was just writing this from memory. Thanks again, this is a nice solution that I can understand. – Ovi May 5 '13 at 12:04
You can even copy and paste this answer on the previous question and I'll accept it and vote it up. – Ovi May 5 '13 at 12:06
I did copied and pasted the answer on the previous question as you suggested, for completeness only. – Heberto del Rio May 8 '13 at 2:40
Thanks well I accepted it as the best answer anyway because it was – Ovi May 8 '13 at 20:01

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