How to Minimize the Integral of \(\int_0^{\pi/2} |\cos(x)-ax^2|\,dx\)?

[Saitama]

Problem:
Find the value of $a$ such that
$$\int_0^{\pi/2} |\cos(x)-ax^2|\,dx$$
is minimum.

Attempt:
Honestly, I don't know how to start. I tried the following:
$$\int_0^{\pi/2} |\cos(x)-ax^2|\,dx \geq \int_0^{\pi/2}|\cos(x)|\,dx-\int_0^{\pi/2}|a|x^2\,dx=1-\frac{|a|\pi^3}{24}$$
$$\Rightarrow \int_0^{\pi/2} |\cos(x)-ax^2|\,dx \geq 1-\frac{|a|\pi^3}{24}$$
but I am not sure if the above is a valid step. Even if it is, I don't know how to proceed from here. :(

Any help is appreciated. Thanks!

 

[ThePerfectHacker]

Let $P_0,P_1,P_2,...$ be orthonormal polynomials on $[0,\pi/2]$. Then to minimize your quantity it follows by general result about inner products that minimal would be projecting $\cos x$ onto $P_2$, i.e.
$$ \int_0^{\pi/2} \cos x \cdot P_2(x) ~ dx $$
So now it remains to compute the $P_n(x)$.

 

[I like Serena]

Problem:
Find the value of $a$ such that
$$\int_0^{\pi/2} |\cos(x)-ax^2|\,dx$$
is minimum.

Attempt:
Honestly, I don't know how to start. I tried the following:
$$\int_0^{\pi/2} |\cos(x)-ax^2|\,dx \geq \int_0^{\pi/2}|\cos(x)|\,dx-\int_0^{\pi/2}|a|x^2\,dx=1-\frac{|a|\pi^3}{24}$$
$$\Rightarrow \int_0^{\pi/2} |\cos(x)-ax^2|\,dx \geq 1-\frac{|a|\pi^3}{24}$$
but I am not sure if the above is a valid step. Even if it is, I don't know how to proceed from here. :(

Any help is appreciated. Thanks!

Hey Pranav! ;)

How about calculating the integral, taking the derivative with respect to $a$, and solve for being equal to zero?

You could almost use Leibniz integral rule or Differentiation under the integral sign, but the condition of Leibniz is not satisfied and otherwise it becomes too complicated.

 

[Saitama]

Hey Pranav! ;)

How about calculating the integral, taking the derivative with respect to $a$, and solve for being equal to zero?

I tried that. After differentiating wrt $a$, I got:
$$\int_0^{\pi/2} x^2\text{sgn}(\cos(x)-ax^2)\,dx=0$$
But I am not sure how to proceed after this.

 

[I like Serena]

I tried that. After differentiating wrt $a$, I got:
$$\int_0^{\pi/2} x^2\text{sgn}(\cos(x)-ax^2)\,dx=0$$
But I am not sure how to proceed after this.

Huh? How did you get that?

Suppose $b$ is the x coordinate where the expression changes its sign, or $\pi/2$ if it doesn't.
Then:
\begin{aligned}\int_0^{\pi/2} |\cos(x)-ax^2|\,dx
&=\int_0^b (\cos(x)-ax^2)\,dx +\int_b^{\pi/2} (ax^2 - \cos(x))\,dx \\
&= (-\sin x - \frac 1 3 ax^3)\Big|_0^b + (\frac 1 3 ax^3 +\sin x)\Big|_b^{\pi/2}
\end{aligned}

 

[Saitama]

Huh? How did you get that?

d/da(|cos(x)-ax^2| - Wolfram|Alpha

I used differentiation under the integral sign.

Suppose $b$ is the x coordinate where the expression changes its sign, or $\pi/2$ if it doesn't.
Then:
\begin{aligned}\int_0^{\pi/2} |\cos(x)-ax^2|\,dx
&=\int_0^b (\cos(x)-ax^2)\,dx +\int_b^{\pi/2} (ax^2 - \cos(x))\,dx \\
&= (-\sin x - \frac 1 3 ax^3)\Big|_0^b + (\frac 1 3 ax^3 +\sin x)\Big|_b^{\pi/2}
\end{aligned}

The definite integral comes out to be:
$$\frac{a\pi^3}{24}+1-2\sin b-\frac{2}{3}ab^3$$
How do I minimise this?

 

[Saitama]

Looks like I should have posted the original problem.

The original problem is as follows:
$$I=\int_0^{4\lim_{n\rightarrow \infty} \sum_{r=0}^{n} \frac{1}{16r^2+16r+3}} |\cos(x)-ax^2|\,dx$$
If the minimum value of I exists at $a=\cos \beta/\beta$. Then find $\beta$.

I computed the limit and that gave me the following integral:
$$I=\int_0^{\pi/2} |\cos(x)-ax^2|\,dx$$
Now I have to minimise this integral. I am not sure if posting the original problem statement helps.

 

[chisigma]

d/da(|cos(x)-ax^2| - Wolfram|Alpha

I used differentiation under the integral sign.The definite integral comes out to be:
$$\frac{a\pi^3}{24}+1-2\sin b-\frac{2}{3}ab^3$$
How do I minimise this?

What You call b is the solution of the equation $\displaystyle \cos x - a\ x^{2} = 0$, so that You can obtain b as function of a inverting the function $\displaystyle a = \frac{\cos b}{b^{2}}$...

Kind regards

$\chi$ $\sigma$

 

[Saitama]

What You call b is the solution of the equation $\displaystyle \cos x - a\ x^{2} = 0$, so that You can obtain b as function of a inverting the function $\displaystyle a = \frac{\cos b}{b^{2}}$...

Kind regards

$\chi$ $\sigma$

Thanks chisigma! :)

But I have stumbled upon a doubt, why did ILS assume that $\cos(x)>ax^2$ for x=0 to b? Does that make no difference?

 

[chisigma]

Thanks chisigma! :)

But I have stumbled upon a doubt, why did ILS assume that $\cos(x)>ax^2$ for x=0 to b? Does that make no difference?

The function $\displaystyle f(x) = \cos x - a\ x^{2}$ for a>0 is monotonically decreasing and it vanishes for x=b, being $\displaystyle 0 < b < \frac{\pi}{2}$. Once You find b as function of a solving the equation $\displaystyle a = \frac{\cos b}{b^{2}}$ You solve the problem computing the minimum of the integral with the formula You have written...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

The function $\displaystyle f(x) = \cos x - a\ x^{2}$ for a>0 is monotonically decreasing and it vanishes for x=b, being $\displaystyle 0 < b < \frac{\pi}{2}$. Once You find b as function of a solving the equation $\displaystyle a = \frac{\cos b}{b^{2}}$ You solve the problem computing the minimum of the integral with the formula You have written...

Kind regards

$\chi$ $\sigma$

More precisely if the integral as function of a and b is...

$\displaystyle I= \frac{a\ \pi^{3}}{24} + 1 -2\ \sin b - \frac{2}{3}\ a\ b^{3}\ (1)$

... and is also...

$\displaystyle a = \frac{\ cos b}{b^{2}}\ (2)$

... then combining (1) and (2) You obtain...

$\displaystyle I = \frac{\pi^{3}\ \cos b}{24\ b^{2}} - 1 - 2\ \sin b - \frac{2}{3}\ b\ \cos b\ (3)$

... and now You have to minimize the (3) respect to b...

Kind regards

$\chi$ $\sigma$