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I want the minimum value of A. (closed form)

$$A = \int_{0}^{p} \sin(x) dx + \int_{p}^{\frac{\pi}{2}} (\sin(x)-\sin(p))dx$$ $$0 \leq p \leq \frac{\pi}{2}$$

So... $$A = p \cdot \sin(p) - \frac{\pi}{2}\sin(p) + 1$$

$$\frac{d}{dp} A = \sin(p) + p \cdot \cos(p) - \frac{\pi}{2}\cos(p) = 0$$

$$\tan(p) + p - \frac{\pi}{2} = 0$$

I stopped here. I know that:

$$A = 1 - \tan(p) \cdot \sin(p)$$

$$\sin(p) = \cos(\tan(p))$$ $$\cos(p) = \sin(\tan(p))$$ $$\tan(p) = \cot(\tan(p))$$ $$\csc(p) = \sec(\tan(p))$$ $$\sec(p) = \csc(\tan(p))$$ $$\cot(p) = \tan(\tan(p))$$

$$p \approx 0.71046$$ $$A \approx 0.4389$$

EDIT:

$$\tan(p) + p - \frac{\pi}{2} = 0$$

$$\tan(p) = \frac{\pi}{2} - p$$

$$-cot\left(p-\frac{\pi}{2}\right) = \frac{\pi}{2} - p$$

$$cot\left(p-\frac{\pi}{2}\right) = p - \frac{\pi}{2}$$

$$B:=p-\frac{\pi}{2}$$

$$cot(B) = B$$

"This kind of equations which mix polynomial and trigonometric terms do not, in general, show solutions which have a closed form expression..." as said by Claude Leibovici here.

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  • $\begingroup$ Hint: $$\int_a^bf(x)\,\mathrm dx + \int_b^cf(x)\,\mathrm dx = \int_a^cf(x)\,\mathrm dx$$ since $[a,c] = [a,b]\cup[b,c]$ for $a< b< c$. In your case: $a=0$, $b=p$, $c=\dfrac{\pi}2$ and $f(x)=\sin (x)$. $\endgroup$
    – Mr Pie
    Mar 2 '21 at 23:19
  • $\begingroup$ Ok. I solved the integrals. I know an approximation of $p$ and an approximation of $A$. I'm looking for a closed form of A. $\endgroup$
    – Jorge Rael
    Mar 3 '21 at 0:38
  • $\begingroup$ I know that. My hint, upon application, allows you to find a closed form for $A$, irrespective of your approximation for $p$. It is basic integration beyond that. $\endgroup$
    – Mr Pie
    Mar 4 '21 at 23:26
  • $\begingroup$ You should derive @vitamin d's answer. If we have a function $f(x)$ with local vertices (maxima and minima), the $x$-coordinates of those vertices are the set of $x$ values such that $f'(x)=0$ (because the gradient at these vertices would be flat, or $0$, and the gradient is measured by a first derivative). The function $A$ has a local minimum. $\endgroup$
    – Mr Pie
    Mar 4 '21 at 23:32
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    $\begingroup$ I already did it. $\endgroup$
    – Jorge Rael
    Mar 4 '21 at 23:33
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$$A = \int\limits_{0}^{p} \sin(x)\,\mathrm{d}x + \int\limits_{p}^{\pi/2} (\sin(x)-\sin(p))\,\mathrm{d}x=\int\limits_{0}^{\pi/2} \sin(x)\,\mathrm{d}x-\sin p\int\limits_{p}^{\pi/2} \,\mathrm{d}x.$$ Here the first integral is equal to $1$. The second term simpliefies to $\sin p(\pi/2-p)$. So $$A=1-\sin p\left(\frac{\pi}{2}-p\right).$$ Now take the derivative of this function. We will get $$A'=-\frac{\pi}{2}\cos p+\sin p+p\cos p.$$ Set $A'=0$ and solve for $p$ and pick out the $p$-value that was in your original domain. I believe the solution has not a trivial closed form.

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  • 1
    $\begingroup$ I'm curious. Why the down-vote? $\endgroup$
    – vitamin d
    Mar 2 '21 at 23:16
  • $\begingroup$ I'm curious too. $+1$ $\endgroup$
    – Mr Pie
    Mar 2 '21 at 23:18
  • $\begingroup$ $sin(p) = p???$ $\endgroup$
    – Jorge Rael
    Mar 2 '21 at 23:19
  • $\begingroup$ @JorgeRael Sorry for the typo. $\endgroup$
    – vitamin d
    Mar 2 '21 at 23:20
  • $\begingroup$ @JorgeRael Edited everything. Are you now able to evaluate the minimum of $A$ or should I elaborate? $\endgroup$
    – vitamin d
    Mar 2 '21 at 23:23
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Starting from @vitamin d's answer, we need to solve for $p$ the equation $$\sin (p)+\left(p-\frac{\pi }{2}\right) \cos (p)=0$$ for which we cannot expect a closed form (this is already the case for $x=\cos(x)$).

First, for the fun of it, I shall use two approximations, $1,400$ years old, which are valid for the range under consideration; they are $$\sin(p)\simeq \frac{16 (\pi -p) p}{5 \pi ^2-4 (\pi -p) p} \qquad \text{and} \qquad \cos(p) \simeq\frac{\pi ^2-4p^2}{\pi ^2+p^2}$$

Unfortunately, this will lead to a quintic polynomial $$32p^5-16(3 \pi-2)p^4+16 \pi (3 \pi -2)p^3+8 (4-\pi ) \pi ^2 p^2-2 \pi ^3 (16+7 \pi )p+5 \pi ^5=0$$ the numerical solution of which being $0.710393$.

On the other side, we can make better and better explicit approximations of the solution performing one single iteration of high order method.

As a function of the order [$n=2$ means Newton, $n=3$ Halley, $n=4$ Householder, $n=5$ has no name], I give the expression obtained for $p$, its numerical representation and the value of the minimum value of $A$.

$$\left( \begin{array}{cccc} n & p_{(n)} & p_{(n)} \sim & A_{\text{min}} \\ 2 & \frac{-16+12 \pi +\pi ^2}{4 (8+\pi )} & 0.70835286 & 0.43890829 \\ 3 & \frac{-256+208 \pi +24 \pi ^2+3 \pi ^3}{704+64 \pi +12 \pi ^2} & 0.71064251 & 0.43890370 \\ 4 & \frac{-8448+6464 \pi +832 \pi ^2+124 \pi ^3+11 \pi ^4}{20480+3136 \pi +352 \pi ^2+44 \pi ^3} & 0.71045640 & 0.43890366 \\ 5 & \frac{-327680+236800 \pi +43776 \pi ^2+6944 \pi ^3+656 \pi ^4+57 \pi ^5}{820224+147456 \pi +24960 \pi ^2+1920 \pi ^3+228 \pi ^4} & 0.71046310 & 0.43890366 \\ \cdots & \cdots& \cdots & \cdots \\ \infty & &0.71046274 & 0.43890366 \end{array} \right)$$

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  • $\begingroup$ I know the Bhaskara I's sine approximation. And I have WolframAlpha, Python, Casio High Precision Calculator... $\endgroup$
    – Jorge Rael
    Mar 3 '21 at 17:27
  • $\begingroup$ I don't want another approximation. Sorry. $\endgroup$
    – Jorge Rael
    Mar 3 '21 at 17:34
  • $\begingroup$ @JorgeRael. Then, you will not get anything from any one. $\endgroup$ Mar 4 '21 at 2:38
  • $\begingroup$ In your opinion (and experience), should I give up? $\endgroup$
    – Jorge Rael
    Mar 4 '21 at 13:46
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    $\begingroup$ @JorgeRael. I think you must (remember that there is no explicit solution for $x=\cos(x)$. For the fun $\zeta (3)-\frac{2 \log (\pi )}{3} =0.438903646$ to be compared to the "exact" $0.438903662$ $\endgroup$ Mar 4 '21 at 14:08

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