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Question:

let $\theta_{1},\theta_{2},\cdots,\theta_{n}\ge 0$,and such $$\theta_{1}+\theta_{2}+\theta_{3}+\cdots+\theta_{n}=\pi$$

find the $P$ the maximum of value $P(n)$ $$P=\sin^2{\theta_{1}}+\sin^2{\theta_{2}}+\cdots+\sin^2{\theta_{n}}$$

Find the closed $P(n)$

I found this

when $n=2$ then $$P=\sin^2{\theta_{1}}+\sin^2{\theta_{2}}=1-\dfrac{1}{2}(\cos{2\theta_{1}}+\cos{2\theta_{2}})=1-\cos{(\theta_{1}+\theta_{2})}\cos{(\theta_{1}-\theta_{2})}=1+\cos{(\theta_{1}-\theta_{2})}\le 2$$ when $\theta_{1}=\theta_{2}=\dfrac{\pi}{2}$. so $$P(2)=2$$

and for general $n$,maybe have use other methods,Thank you

I guess we can prove $$P(3)=P(4)=P(5)=\cdots=P(n)?$$

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Try using the Cauchy-Schwarz inequality. –  user41281 Jun 25 at 16:26
    
@user41281,I think this can't use Cauchy-Schwarz inequality –  math110 Jun 25 at 16:32
1  
    
You cannot conclude that... $$P(2) = 2$$ $$P(3) = \frac{9}{4}$$ $$P(4) = 2$$ $$P(5) = \frac{5}{8} \Big (5 - \sqrt{5} \Big)$$ $$P(n) = n \sin^2\Big(\pi/n\Big)$$ –  johannesvalks Jun 25 at 18:48
    
How can this be interpreted geometrically? Maximizing sum of squares of lengths of sides of an $n$-gon inscribed in a circle? –  Martin Sleziak Jun 28 at 13:56

5 Answers 5

We are told to find the maximum $\sigma_n$ of $$f(\theta):=\sum_{k=1}^n\sin^2\theta_k$$ on the $(n-1)$-dimensional simplex $$S:=\{\theta=(\theta_1,\ldots,\theta_n)\ |\ \theta_k\geq0, \quad \theta_1+\ldots+\theta_n=\pi\}\ .$$ Claim: One has (trivially) $\sigma_2=2$, and $\sigma_n={9\over 4}$ $(n\geq3)$. The latter value is realized by $\theta_k={\pi\over3}$ for $1\leq k\leq 3$ and $\theta_k=0$ for $k>3$.

Proof. Since the function $t\mapsto\sin^2 t$ has inflection points at $t={\pi\over4}$ and $t={3\pi\over4}$ we have to envisage a great number of conditionally stationary points of $f$, most of them on $\partial S$. In order to avoid these complications we only look at two variables at a time and make use of the inherent symmetry of the problem.

From $$h(\theta_1,\theta_2):=\sin^2\theta_1+\sin^2\theta_2=1-\cos(\theta_1+\theta_2)\cos(\theta_2-\theta_1)$$ we can draw the following conclusions:

(I) When $\theta\in S$ and $0<\theta_1\leq\theta_2<{\pi\over4}$ then $\cos(\theta_1+\theta_2)>0$. Therefore the value of $h$ can be increased by replacing $(\theta_1,\theta_2)$ by $\theta_1':=0$, $\theta_2':=\theta_1+\theta_2$, and the corresponding point $\theta':=(\theta_1',\theta_2',\theta_3,\ldots,\theta_n)$ is still feasible.

(II) When $\theta\in S$ and ${\pi\over4}\leq\theta_1<\theta_2$ then $\cos(\theta_1+\theta_2)<0$. Therefore the value of $h$ can be increased by replacing $(\theta_1,\theta_2)$ by $\theta_1':=\theta_2':={\theta_1+\theta_2\over2}$, and the corresponding point $\theta':=(\theta_1',\theta_2',\theta_3,\ldots,\theta_n)$ is still feasible.

Consider now a point $\theta=(\theta_1,\ldots,\theta_n)\in{\rm argmax}_S(f)$. Then the fact (I) implies that one has $0<\theta_k<{\pi\over 4}$ for (a) no $k$ or (b) exactly one $k$, because otherwise there would be a point $\theta'\in S$ with $f(\theta')>f(\theta)$. In a similar way the fact (II) implies that all $\theta_k\geq{\pi\over4}$ have the same value. Let there be $r\in\{2,3,4\}$ of them.

Case (a): When $r=2$ we have $\theta_1=\theta_2={\pi\over2}$, which leads to $f(\theta)=2<{9\over4}$. When $r=3$ we arrive at the case described in the claim. When $r=4$ we arrive at $f(\theta)=2$ again.

Case (b): Let $\theta_1:=\alpha\in\ \bigl]0,{\pi\over4}\bigr[\ $. Then we have to study the auxiliary functions $$g_2(\alpha):=\sin^2\alpha+2\sin^2{\pi-\alpha\over2},\qquad g_3(\alpha):=\sin^2\alpha+3\sin^2{\pi-\alpha\over3}$$ in the interval $0<\alpha<{\pi\over4}$. It turns out that $g_2$ is monotonically increasing and $g_3$ monotonically decreasing in this interval (as expected). This implies that case (b) contains no points $\theta\in{\rm argmax}_S(f)$.

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Hello,sorry,because my bad English.I can't understand your methods. –  math110 Jun 27 at 13:31

If you restrict in the interior of $ \tilde {\theta} = \{ ( \theta _1 , \dots , \theta _n)\mid \theta _1 + \dots + \theta _n=\pi\}$ the critical point is $ \theta _1= \dots =\theta _n$ where you get $P= n \sin^2 ( \pi/n)$.

For case $n=2$ you have found the maximum. For $n=3$, you compare the value of $P$ at the critical point and the maximum at the boundary. The maximum at the boundary is the same as the previous case namely for $n=2$. So you compare $3 \sin^2 \pi /3=9/4 >2$ and you conclude the the maximum for $n=3$ is $ 9/4$.

Now you can proceed inductively comparing the critical value you get from Lagrange multipliers with the max at the boundary, which is known.

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$n\sin^2(\pi/n)$ goes to 0 as n tends to infinity. It is clear we would be better off taking two $\pi/2$'s and rest of the values 0 to get at least 2 in the sum.

It is tempting to think that this value of 2 is best, but at least for n=3, $n\sin^2(\pi/n)$ does give a better answer namely 2.25.

I think the important point is to look at the function $f(x) = \sin^2(x)/x$. Note that if f(x) is the greatest at some $x = x_0$, then $\sin^2(x_0)$ is greater than any sum of the form $\sum \sin^2(x_i)$ where the $x_i$s sum to $x_0$.

With some help from Wolfram Alpha, this seems to be maximized around 1.16556, call this a. $\pi/a$ is around 2.7, which indicates that 2 and 3 are the really interesting cases. If the total sum was a multiple of a, then we would be almost done. Unfortunately, it is not. By some trial and error, we can see that ($\pi/3$, $\pi/3$, $\pi/3$) works better than other alternatives like (a, a, $\pi-a$) etc. So for n=3 we have to go with ($\pi/3$, $\pi/3$, $\pi/3$). As $\sin^2(x)/x$ is increasing in the interval $(0, \pi/3)$ we can now start to use the logic that motivated our search for this function. For n>3, instead of looking for further subdivisions, we are better off just taking the first three angles as $\pi/3$ and the rest as 0.

EDIT:

The kind of thing you had in mind can actually be proven for n = 6 onwards, i.e. P(n) = P(n-1) for $n \ge 6$ Here is why: for $n \ge 6$, there must be two thetas whose sum is less than $\pi/3$ and therefore less than a. For any pair of such thetas say x and y, $\sin^2(x+y) = (x+y)\frac{\sin^2(x+y)}{x+y} = x\frac{\sin^2(x+y)}{x+y} + y\frac{\sin^2(x+y)}{x+y} \ge x\frac{\sin^2(x)}{x} + y\frac{\sin^2(y)}{y} = \sin^2(x) + \sin^2(y)$.

($\frac{\sin^2(x)}{x}$ is increasing in the interval [0,a])

So we can replace (x, y) by (x+y, 0) and go to the n-1 case.

But the fact than P(5) = P(4) = P(3) needs to be seen on a case by case basis.

EDIT 2: actually I can prove even that.

If n is at least 4, there are two angles with sum no more than pi/2. Call these x and y. We can replace (x,y ) by (x+y, 0). Reason:

$\sin^2(x+y) - \sin^2(x) - \sin^2(y) = 2 \sin(x)\sin(y)\cos(x+y)$, which is non negative for x, y in [0, pi/2]

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Use a Lagrange multiplier to fix the constraint. It then falls out that all $\theta_i$ are the same and the rest is easy.

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Given

$$ \Theta = \sum_{k=1}^n \theta_k, $$

and $\Theta = \pi$, and

$$ P = \sum_{k=1}^n \sin^2(\theta_k). $$

Note that

$$ d\Theta = 0$, $$

so

$$ d\theta_n = - \sum_{k=1}^{n-1} d\theta_k $$

and

$$ dP = \sum_{k=1}^n \sin(2 \theta_k ) d\theta_k = \sum_{k=1}^{n-1} \Big( \sin(2\theta_k) - \sin(2\theta_n) \Big) d\theta_k, $$

so

$$ \theta_k = \theta_n + m \pi $$

Condition $\theta_k \ge 0$ and $\Theta = \ pi$ implies

$$ \theta_k = \theta_n, $$

so

$$ \theta_k = \frac{\pi}{n}, $$

whence

$$ P(n) = n \sin^2\Big( \frac{\pi}{n} \Big). $$

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