Maxima and Minima of Sin(x)/x

Question

I am trying to calculate the maximum and minimum points (between $-3\pi$ and $3\pi$) of $$f(x)=\frac{\sin(x)}{x}$$
I have found the derivative of the function and let it equal to zero.
$$f'(x)=\frac{x\cos(x) - \sin(x)}{x^2}$$
$$f'(x)=0$$ $$\frac{x\cos(x) - \sin(x)}{x^2}=0$$ $$x\cos(x) - \sin(x)=0$$ $$x\cos(x)=\sin(x)$$ $$x=\tan(x)$$

I am unaware as to how to find $x$. I assume that once I find $x$, I can use a sign diagram or second derivative test to determine the minimum and maximum values. Any help would be highly appreciated.

Answer 1

Equations which contains polynomial and trigonometric functions do not show explicit solutions and numerical methods should be used to find the roots.

The simplest root finding method is Newton; starting from a reasonable guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)}$$ So, in your case, $$F(x)=x \cos(x)-\sin(x)$$ I think it is better to let it under this form because of the discontinuities of $\tan(x)$.

You can notice that if $x=a$ is a root, $x=-a$ will be another root. So, let us just focus on $0\leq x \leq 3\pi$. If you plot the function, you notice that, beside the trivial $x=0$, there are two roots located close to $5$ and $8$. These would be the guesses.

Using $F'(x)=-x \sin (x)$, the iterative scheme then write $$x_{n+1}=x_n-\frac{1}{x_n}+\cot (x_n)$$ Let us start with $x_0=5$; the method then produces the following iterates : $4.50419$, $4.49343$, $4.49341$ which is the solution for six significant figures.

I let you doing the work for the other solution.

Answer 2

Without proof, the $m^{th}$ extremum of $y(x) =\dfrac{\sin(x)}{x}$ is given by the root of the following function: $$ \sqrt{1-y_m^2} +y_m \sin^{-1}(y_m)-(-1)^{m}\left(m+\dfrac{1}{2}\right)\pi y_m=0$$

It can be simplied to the following:

$$ \sin^{-1}(y_m)+\cot(\sin^{-1}(y_m))=(-1)^m \left(m+\dfrac{1}{2}\right)\pi$$

It is an exact expression and you can find all the extrema of $y(x) =\dfrac{\sin(x)}{x}$ using it. It can be easily verified that $y_0=1$ (which is the $0^{th}$ extremum) is a root of this equation. However, for $m\ge1$ you cannot solve this equation analytically. But no worries, if you would look at the graph of $y(x) =\dfrac{\sin(x)}{x}$ you would notice that it decays pretty fast owing to the $\dfrac{1}{x}$ factor and $y_m\ll1$ should not be a bad approximation. Under this approximatiom, we will have the following $O(y_m^3)$ approximations:

$$\sin^{-1}(y_m)\approx y_m+O(y_m^3)$$ $$\sqrt{1-y_m^2}\approx 1-\dfrac{1}{2}y_m^2+O(y_m^3)$$

Plugging these into the exact equation, we have:

$$ \dfrac{1}{2}y_m^2-(-1)^{m}\left(m+\dfrac{1}{2}\right)\pi y_m+1=0$$

We get a quadratic equation which we can exactly solve. So, on solving it we have found an expression that gives approximate extrema values for $y(x) =\dfrac{\sin(x)}{x}$.

$$\boxed{y_m\approx\pi\left(m+\dfrac{1}{2}\right)(-1)^m-(-1)^m\sqrt{\pi^2\left(m+\dfrac{1}{2}\right)^2-2}}$$

You can also convert this single equation into two equations, one for the maxima and one for the minima just by setting $m$ as even or odd integer.

Now, checking for how accurate our formula is. The exact and the approximate formulas are solved for their roots. The extrema $y_0$, as expected under our assumption of $y_m\ll1$, is the least accurate.

$$\begin{array}{|c|c|c|c|} \hline &exact &approx &error\% \\ \hline y_0 &+1.000000 &+0.887129 &11.2871\% \\ \hline y_1 &-0.217234 &-0.217213 &0.00963963\% \\ \hline y_2 &0.128375 &0.128373 &0.0011467\% \\ \hline y_3 &-0.0913252 &-0.0913249 &0.000291779\% \\ \hline y_4 &+0.0709135 &0.0709134 &0.000105792\% \\ \hline y_5 &-0.0579718 &-0.0579718 &0.0000471872\% \\ \hline y_6 &+0.0490296 &0.0490296 &0.0000241245\% \\ \hline y_7 &-0.0424796 &-0.0424796 &0.0000135875\% \\ \hline \end{array}$$

Answer 3

You don't need to solve the equation for $x$ to find the extreme point. You know that extremes happen when $x = \tan(x)$ and, therefore, taking the inverse $\tan$ on both sides, when $\mathrm{atan}(x) = x$. I will write $x_e$ to note the points where $f(x)$ has an extreme. Then

$\displaystyle f(x_e) = \frac{\sin(x_e)}{x_e} = \frac{\sin[\mathrm{atan}(x_e)]}{x_e}$.

But

$\displaystyle \sin[\mathrm{atan}(x_e)] = \frac{x_e}{\sqrt{1 + x_e^2}}$,

so, for each $x_e$ that is the point of an extreme, the value of the function at a point is

$\displaystyle f(x_e) = \frac{1}{\sqrt{1 + x_e^2}}$.

An alternative would be

$\displaystyle f(x_e) = \frac{\sin(x_e)}{x_e} = \frac{\sin(x_e)}{\tan(x_e)} = \cos(x_e)$, which is actually the same result, since, using again that $x_e = \mathrm{atan}(x_e)$, we have

$\displaystyle \cos[\mathrm{atan}(x_e)] = \frac{1}{\sqrt{1 + x_e^2}}$.

In any case, you see that $\displaystyle f(x_e) = \frac{1}{\sqrt{1 + x_e^2}}$ is maximum at $x_e = 0$, as it decreases everywhere else. At this point, $f(0) = 1$.