Given $f(x) = x \sin\frac1x$, find roots of $f'(x)$ in the interval $0\le x \le \frac 1{\pi}$.

rolles-theoremtrigonometry

This question is taken from book: Advanced Calculus: An Introduction to Classical Analysis, by Louis Brand. The book is concerned with introductory analysis.

If $f(x) = x \sin\frac1x\;(x\ne 0), f(0) = 0$, does Rolle's theorem guarantee a root of $f'(x)$ in the interval $0\le x \le \frac 1{\pi}$? Show that $f'(x)$ has an infinite number of roots $x_l \gt x_2 \gt x_3\gt \cdots$ in the given interval which may be put in one-to-one correspondence with the roots of $\tan y = y\,$ in the interval $\pi \le y \lt \infty$. Calculate $x_1$ to three decimal places.

Given $f(x) = x \sin\frac1x(x\ne 0), f(0) = 0$.

At $x=0, f(0) = 0 \sin(\infty)$, but $\sin(\infty)\in[-1,1]$, which means the range corresponding to $x=0$ is undefined.

But, the value of $f(0)$ is stated to be $0$. This is a point of confusion as how this range point is specified.

Also at $x =\frac 1{\pi}$, the fn. yields $f(x) = \frac 1{\pi} \sin(\pi) =0.$

So, $f(0)= f\left(\frac 1{\pi}\right) = 0$.

Rolle's theorem needs three conditions:

  1. Let $f(x)$ be continuous on a closed interval $[a, b]$,
  2. and, $f(x)$ be differentiable on the open interval $(a, b)$.
  3. If $f(a) = f(b)$, then there is at least one point $c$ in $(a, b)$ where $f'(c) = 0$.

By being a product of polynomial & a trigonometric function, both of which are differentiable & continuous, the product too is.

Hence, all three conditions are satisfied. So, root of $f'(x)$ is guaranteed in the given interval $\big[0,\frac{1}{\pi}\big]$.

First need calculate $x_1$, so find $f'(x)$.
It is given by $\sin\left(\frac 1x\right)-\frac 1x \cos \left(\frac 1x\right)$.
$f'(x)=0\implies x\sin\left(\frac 1x\right)=\cos \left(\frac 1x\right)\implies x=\cot\left(\frac 1x\right)$.

Unable to solve further.

I hope that the solution of the above equation can help with the rest two questions, although have doubts for each as stated below:

  1. $f'(x)$ has an infinite number of roots $x_l \gt x_2 \gt x_3\gt \cdots$ in the interval $0 \le x \le \frac 1{\pi}$.
    Unable to understand how it is possible to have the given scenario of infinite roots in a given order.

  2. These roots may be put in one-to-one correspondence with the roots of $\tan y = y\,$ in the interval $\pi \le y \lt \infty$.
    Here, the two equations whose roots are to be paired are:
    $x = \cot\left(\frac 1x\right)$ and $ y = \tan(y)$ with connection not visible.

Edit The book states the answer for $x_1=0.2225$. Still have no clue about attaining it.

Best Answer

For $x>0$,$$f'(x)=\sin\left(\frac1x\right)-\frac1x\cos\left(\frac1x\right)=\sin(y)-y\cos(y).$$

Hence the roots of $f'$ are the inverse solutions of $y=\tan(y)$ and $x_1$ corresponds to the smallest $y$ above $\pi$.

As $\dfrac{df'(y)}{dy}=\cos(y)-\cos(y)+y\sin(y)$ is negative between $\pi$ and $2\pi$, and $f'(\pi)$ and $f'(2\pi)$ differ in sign, we can approach the isolated root by the secant method.

The successive approximations are

$$4.1887902\cdots (f>0)\\ 4.5312881\cdots (f<0)\\ 4.4901885\cdots (f>0)\\ 4.4933831\cdots (f>0)\\ 4.4934095\cdots (f<0)\\ $$

and

$$\frac1{4.4934095\cdots}=0.2225\cdots.$$

Related Question