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Use the Newton Method to find the smallest and the second smallest positive roots of the equation $\tan x=4 x$, correct to 4 decimal places.
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Solution: Draw the curves $y=\tan x$ and $y=4 x$. The roots of our equation are the $x$-coordinates of the places where these two curves meet.

A glance at the picture shows that (for $x \geq 0$ ) the curves meet at $x=0$, then at a point with $x$ just shy of $\pi / 2$, and then again at a point with $x$ just shy of $3 \pi / 2$ (the pattern continues).

We first find the root that is near $\pi / 2$. Let $f(x)=\tan x-4 x$. The $f^{\prime}(x)=\sec ^{2} x-4$, and the Newton Method recurrence is
$x_{n+1}=x_{n}-\frac{\tan x_{n}-4 x_{n}}{\sec ^{2} x_{n}-4} .$
Some simplification is possible. For example, we can use the identity $\sec ^{2} x=1+\tan ^{2} x$ to rewrite the recurrence as
$x_{n+1}=x_{n}-\frac{\tan x_{n}-4 x_{n}}{\tan ^{2} x_{n}-3} .$
This trick cuts down on the computational work. This was a particularly important consideration in the old days when computations were done by hand, with the aid of tables and slide rules.

For the first root, a bit of fooling around suggests taking $x_{0}=1.4$. Then $x_{1}=1.393536477, x_{2}=1.393249609$, and $x_{3}=1.393249075$. This suggests that to 4 decimal places the root is $1.3932$. We can verify this by the sign change criterion in the usual way.

For the second root, after some work we can for example arrive at the initial estimate $x_{0}=4.66$. The computation is quite sensitive to the right choice of initial value. And then we get $x_{1}=4.658806388$ and $x_{2}=4.658778278$. To 4 decimal places the root is $4.6588$. We can verify that we are close enough by the sign change criterion.

Note. We may be nervous about using a casual sketch to locate the first two positive roots. If we are, we can analyze the behaviour of $f(x)$ by looking at its derivative $f^{\prime}(x)$. Recall that $f^{\prime}(x)=\sec ^{2} x-4$. The sec function increases steadily in the interval $(0, \pi / 2)$. It follows that $f^{\prime}(x)$ is negative in this interval up to the point where $f^{\prime}(x)=0$, which happens when $\sec x=2$, that is, when $\cos x=1 / 2$, at $\pi / 3$. So $f(x)$ decreases from $x=0$ to $x=\pi / 3$, then increases. Since $f(0)=0$, we conclude that $f$ is negative in the interval $(0, \pi / 3)$, then increases. It becomes very large positive near $x=\pi / 2$. So $f(x)=0$ in exactly one place in the interval $(0, \pi / 2)$. In a similar way, we can show that $f(x)=0$ at exactly one place in the interval $(\pi / 2,3 \pi / 2)$.

Note. We attacked the problem in the 'natural' way, making the most obvious choice for $f(x)$. It turns out that, particularly for the second positive root, and even more so say for the fifth positive root, we have to be very careful in our choice of $x_{0}$. The problem is that near the roots the tan function is growing at a violent rate. A quite small change in $x$ can have a dramatic effect on $\tan x$.

We can rewrite the equation $\tan x=4 x$ in ways that avoid most of the problems. For instance, rewrite it as $g(x)=0$ where $g(x)=$ $\sin x-4 x \cos x$. The Newton-Raphson recurrence becomes
$x_{n+1}=x_{n}-\frac{\sin x_{n}-4 x_{n} \cos x_{n}}{4 x_{n} \sin x_{n}-3 \cos x_{n}} .$
Calculations with this recurrence are quite a bit more numerically stable than calculations with the 'natural' recurrence that we used earlier.
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