How to find global Maximum of this function.

Question

This question which I am asking is to be used in another question and I am unable to do this. So, I am asking for help.

Consider the function$s_n(x)=\frac{2}{\pi} \int_{0}^{x} \frac{\sin (2nt)} {\sin (t)} dt $ . I proved that this function has Local Maxima at $x_1 ,x_3,x_5,...,x_{2n-1}$ where $x_m=\frac{m \pi } {2n}$.

But I am not able to prove that $s_n(\frac{\pi}{2n}) $is global maximium.

The method I thought was to compare all the local maximum but I am not able to compute the integral involving $s_n(x)$ .

I think that it has to be done in some other way( using some clever way and indirectly) and I am not able to figure out that .

So, Can you please help with that?

Answer 1

Above Figure illustrates the graph of $f(t):=\frac{\sin(2nt)}{\sin(t)}$ on $[0, \pi]$.

Idea. The idea to achieve the goal is very simple thank to the following geometric observations:

• Obs1. $f_n$ is symmetric with center $x=\pi/2$.
• Obs2. Obs1 implies that $s_n$ is symmetric with axis $x=\pi/2$. Thus, it is sufficient to prove that $s_n( \frac{\pi}{2n} )$ is the maximum of $s_n$ on $[0, \frac{\pi}{2}]$, i.e. prove that it is the maximum of the set $\{ s_n( \frac{i\pi}{2n}) : i=1, ... n\}$.
• Obs3. The area of $f_n$ over $[\frac{i\pi}{2n}, \frac{(i+1)\pi}{2n}]$, denoted by $s_n^i$, is decreasing for $i=0, ..., n-1$.
• Obs4. \begin{align} s_n( \frac{\pi}{2n} ) & = s_n^1\\ s_n( \frac{2\pi}{2n} ) & = s_n^1-s_n^2\leq s_n^1\\ s_n( \frac{3\pi}{2n} ) & = s_n^1-(s_n^2-s_n^3) \leq s_n^1\\ s_n( \frac{4\pi}{2n} ) & = s_n^1-(s_n^2-s_n^3) -s_n^4 \leq s_n^1\\ ..... \end{align}

Hence we conclude that $s_n^1$ is the global maximum and $\frac{\pi}{2n}$ is global maxima on $[0,\pi]$.

Based on geometric properties of $f_n$, similar arguments can be applied to prove that $\frac{\pi}{2n}$ is global maxima on $[0,2\pi]$. Since$f_n$ is periodic $2\pi$, thus $\frac{\pi}{2n}$ is global maxima on $\mathbb R$.

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We now move into the details.

Proof of Obs1. We need to show $f_n(x)=-f_n(\pi-x)$. Indeed, \begin{align} f_n(\pi-t) & = \frac{\sin(2n(\pi-t))}{\sin(\pi-t)}\\ & = -\frac{\sin(2nt)}{\sin(t)}\\ & = -f_n(t). \end{align}

Proof of Obs2. We need to show $s_n(x)=s_n(\pi-x)$. Indeed, \begin{align} s_n(x) & = \int_0^{x} f_n(t) dt\\ & = \int_0^{x} -f_n(\pi-t) dt\\ & = \int_{\pi-x}^\pi -f_n(t) dt\\ & = \int_0^{\pi-x} f_n(t) dt\\ & = s_n(\pi-x). \end{align} Here the fourth equation follows from the fact that $\int_0^\pi f_n(t)dt=0$ as $f_n$ is symmetric with center $x=\frac{\pi}{2}$.

Proof of Obs3. Let $x_i=\frac{i\pi}{2n}$ for $i=1, n$. \begin{align} s^i_n & = \int_{x_i} ^{x_{i+1}} |f_n(t)| dt\\ & = \int_{x_i} ^{x_{i+1}} \frac{|\sin(2nt)|}{\sin(t)} dt\\ & = \int_{x_i+x_1} ^{x_{i+1}+x_1} \frac{|\sin(2n(t+x_1))|}{\sin(t+x_1)} dt\\ & = \int_{x_{i+1}} ^{x_{i+2}} \frac{|\sin(2nt)|}{\sin(t+x_1)} dt\\ & < \int_{x_{i+1}} ^{x_{i+2}} \frac{|\sin(2nt)|}{\sin(t)} dt\\ & = s^{i+1}_n \end{align} Here we have used $\sin(t+x_1)>\sin(t)$ on $[0, \frac{\pi}{2}]$.

Proof of Obs4. This is easy because we just need to investigate the sign of $f_n$. Note that we are working on $[0, \frac{\pi}{2}]$ and $\sin(t)$ is positive now.