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

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.

• – Robert Z Jun 5 '19 at 8:41
• @RobertZ The book states the answer for $x_1=0.2225$. Still have no clue about attaining it. – jiten Jun 5 '19 at 10:00
• See Yves Daoust's answer below. – Robert Z Jun 5 '19 at 10:25

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

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

• Thanks. But, would still request one value's (in successive approximations) calculation in detail. This would help remove any doubts. – jiten Jun 5 '19 at 10:33
• Apart from my last request for one value's calculation details, also request showing how is the (sign) value calculated for $f$ at that point. – jiten Jun 5 '19 at 11:05
• @jiten: such a request is somewhat displaced. From my post, you have a proof of a single root in $[\pi,2\pi]$, and you can check for yourself the last two function signs, which give full guarantee. – Yves Daoust Jun 5 '19 at 11:49

Partial answer. $$f'(x)=\sin(\frac 1 x)-\frac 1 x \cos(\frac 1 x)$$ so $$\tan (\frac 1 x)=\frac 1 x$$. So the correspondence between roots of $$\tan \, y=y$$ and solutions of the given equation is obvious. Also $$tan\, y -y$$ changes sign in every interval of the type $$(2n\pi-\pi/2,2n\pi+\pi /2)$$ so it has a root in that interval. It follows that the given equation has infintely many roots.

• But, the interval for consideration (if change variable from $x$ to $1/x$, hence the equation changes to $y =\tan y$), for $y$ is $0\le y \le \pi$. Then, I hope you also mean asymptotic series expansion as by the comment of @RobertZ. – jiten Jun 5 '19 at 8:54
• @jiten $y=\frac 1 x$ lies in $[\pi, \infty)$ not in $[0,\pi]$. – Kavi Rama Murthy Jun 5 '19 at 9:01
• The book gives answer for $x_1 = 0.2225$. I still see no approach to get that. – jiten Jun 5 '19 at 9:03
• $x_1=\frac 1 {y_1}$ where $y_1$ is the smallest root of $tan\, y=y$. So look at the interval $(2\pi-\pi/2,2\pi+\pi/2)$. @jiten – Kavi Rama Murthy Jun 5 '19 at 9:14
• Only seeing a graph it is at $x=0$. But, how will it lead further to answer. – jiten Jun 5 '19 at 9:27

Guide:

• The part that you mention at $$x=0$$, $$f(0)=0\cdot \sin (\infty)$$ makes no sense. It has been stated that the rule $$f(x)=x\sin \frac1x$$ only holds if $$x\ne 0$$. Also, $$\frac10$$ is undefined.

• You haven't argued that it is continuous at $$0$$. You have to show that $$\lim_{x \to 0^+}f(x)=f(0)$$.

• To show that there are countably infinite solutions, after you partition the domain to countably many partitions, you want to check that each partition has at least $$1$$ and also finitely many solutions.

• Check that $$\tan y - y$$ increases on $$(\pi, \frac{3\pi}2)$$ and there is a unique root in that interval.

• Hence we know that $$x_1 \in \left(\frac{2}{3\pi} ,\frac1{\pi}\right)$$. Now, you can perform a binary search on that interval to narrow down $$x_1$$ up to $$3$$ decimal places.

• Thanks, please join chat for the stated issues. – jiten Jun 6 '19 at 23:33
• If free, please join chat room for this answer's raised issues. – jiten Jun 7 '19 at 0:29
• Please see my comments in chat. – jiten Jun 7 '19 at 2:59
• Please be in chat for bisection method related question for equation $\frac 32x -6 -\frac 12\sin(2x)=0$. The question states that it can be shown that it has a real root. It then asks to find an interval on which this unique real root is guaranteed to exist. First of all, I am not sure how the question is so sure about a real root of the equation. Second, I tried to find derivative ($f'(x)=\frac 32 -\cos(2x)=\frac 52-2\cos^2x$) of the equation as possibly the root will be in the interval having $f'(x)=0$. It leads to $\frac{\sqrt{5}}2=1.118=\cos x$ which is impossible, as $-1\le \cos(x)\le1$. – jiten Jun 11 '19 at 4:59
• Please tell here or in chat the reason for: Prove if $a^n\lt x^n\lt b^n, x,a,b \in \mathbb{R}, n \in \mathbb{N}$; then $-a^n\ge -x^n\ge -b^n$. I mean why the equality sign appears on multiplying by $-1$. Also, I want to state that this occurs in a book titled: Introduction_to_analysis, at page #27, by Traynor, in a proof for showing by completeness that irrational roots exist. Also, is there a name for this property. The chatroom is at: chat.stackexchange.com/rooms/94404/…. – jiten Jun 13 '19 at 19:36