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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.

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    $\begingroup$ See math.stackexchange.com/questions/110256/… $\endgroup$ – Robert Z Jun 5 '19 at 8:41
  • $\begingroup$ @RobertZ The book states the answer for $x_1=0.2225$. Still have no clue about attaining it. $\endgroup$ – jiten Jun 5 '19 at 10:00
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    $\begingroup$ See Yves Daoust's answer below. $\endgroup$ – Robert Z Jun 5 '19 at 10:25
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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

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

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  • $\begingroup$ Thanks. But, would still request one value's (in successive approximations) calculation in detail. This would help remove any doubts. $\endgroup$ – jiten Jun 5 '19 at 10:33
  • $\begingroup$ 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. $\endgroup$ – jiten Jun 5 '19 at 11:05
  • $\begingroup$ @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. $\endgroup$ – Yves Daoust Jun 5 '19 at 11:49
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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.

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  • $\begingroup$ 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. $\endgroup$ – jiten Jun 5 '19 at 8:54
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    $\begingroup$ @jiten $y=\frac 1 x$ lies in $[\pi, \infty)$ not in $[0,\pi]$. $\endgroup$ – Kavi Rama Murthy Jun 5 '19 at 9:01
  • $\begingroup$ The book gives answer for $x_1 = 0.2225$. I still see no approach to get that. $\endgroup$ – jiten Jun 5 '19 at 9:03
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    $\begingroup$ $x_1=\frac 1 {y_1}$ 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 $\endgroup$ – Kavi Rama Murthy Jun 5 '19 at 9:14
  • $\begingroup$ Only seeing a graph it is at $x=0$. But, how will it lead further to answer. $\endgroup$ – jiten Jun 5 '19 at 9:27
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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.

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  • $\begingroup$ Thanks, please join chat for the stated issues. $\endgroup$ – jiten Jun 6 '19 at 23:33
  • $\begingroup$ If free, please join chat room for this answer's raised issues. $\endgroup$ – jiten Jun 7 '19 at 0:29
  • $\begingroup$ Please see my comments in chat. $\endgroup$ – jiten Jun 7 '19 at 2:59
  • $\begingroup$ 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 $-1\le \cos(x)\le1$. $\endgroup$ – jiten Jun 11 '19 at 4:59
  • $\begingroup$ 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/…. $\endgroup$ – jiten Jun 13 '19 at 19:36

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