10 Newton's Method

The root is approximately $3+\frac16-\frac{\left(3+\frac16\right)^2-10}{6+\frac13}$.

The root is approximately $\frac{28}{9}.$

A calculator tells us $\left(\frac{28}{9}\right)\approx 30.11$. This is pretty close to 30, which is the cube of the real root, so our approximation seems reasonable.

If we start with $x=1$, we find the critical point is approximately $\frac25=0.4$.

Before we start, we may want to get a feel for the problem with a quick sketch of $y=g(x)$. A full sketch, like we did in Chapter 7, isn't necessary (or really possible, since we don't know where the critical points are – although we can guess from the question text that there is only one critical point). But we can do a quick sketch using the ideas from Chapter 1. Namely: close to the origin, we expect $g(x) \approx x$; far from the origin, we expect $g(x) \approx -x^4$.

This helps us guess that our critical point should be at a positive value of $x$. The critical points of the function are the roots of its derivative. So, we set

and use Newton's Method to approximate the values of $x$ that make $f(x)$ close to 0. The formula for refining our guess will be

This question doesn't tell us where to start, so we have to figure that out on our own. From our sketch, it seems small nonnegative values of $x$ are good starting places. We'll try a few to see which are close to a root. (This is something we could also do without having first made the sketch.)

So it seems that $x=0$ is a good starting point. (We also see that $f$ changes from positive to negative somewhere between $x=0$ and $x=1$, so we expect our root to be somewhere between those two numbers.)

So, two iterations of Newton's method gives us an approximate critical point of $g(x)$ at $x=\frac25$. Since we know the root is between 0 and 1, we also could have started at $x=1/2$, which would have lead to the following:

Other starting points are possible, but don't offer much benefit. Starting at $x_0=1$ gives us an integer starting point, which we usually go for, but it's not as good at starting at $x_0=0$, since $f(0)$ is closer to 0 than $f(1)$ is.

Two reasonable answers are below.

Newton's Method finds roots, so the first thing to do is to rephrase the question in terms of root finding. So, we set

and find where $f(x)=0$.

Now, we need to find a starting place for Newton's method. That is, we should find a (preferable integer) value of $x$ such that $f(x)$ is reasonably close to 0. Rather than use a calculator to find exact values of $\arctan x$, recall $\lim\limits_{x \to \infty}\arctan x = \frac\pi2$. So, for large positive values of $x$, $\arctan x \approx\frac\pi2$. With these ideas in mind, let's start evaluating $f(x)$ at different places:

Two good candidates are $x_0=11$ (since $\frac\pi 2 - 1\approx 0.5$, which is relatively close to 0) and $x_0=10$ (since $f(10)$ is still relatively close to 0, but 10 is a `rounder' number than 11.)

In both cases, the formula we'll use to get an updated approximation is

The formula we'll use is

We know we want a root close to $x=2$, so ordinarily, $x_0=2$ would be our choice. However, $x=2$ is not in the domain of the function above. Geometrically, the function $f(x)$ has a horizontal tangent line at $x=2$. Newton's method finds roots of tangent lines, but horizontal tangent lines either have no roots (as is the case here) or infinitely many roots. So, we'll need a different starting point.

The two obvious choices are $x=1$ and $x=3$. Note $f(1) = 4$ and $f(3) = -25$, so $x=1$ seems like a better choice. But actually, if you start with $x=3$, you get close to a different root. So, we'll show both below.

Using the most obvious choices (that is: $f(x)=x^2-8$ and $x_0=3$), we approximate $\sqrt 8 \approx 3-\frac16$.

$\approx \frac{1}{3}-\frac{1}{90}$

$\approx \frac56$

Note $0.4^3=0.064$ and $6^3=216$.

$\sqrt[3]{0.065}\approx0.4+\frac{1}{480}$

$\sqrt[3]{215}\approx 6-\frac1{108}$

Make sure you're using Newton's Method on $g'(x)$, not $g(x)$, since you want roots of $g'(x)$.

A spreadsheet might also help you find a starting value for Newton's Method. Look for two positive numbers $x$ and $y$ that are close together with $g'(x)>0$ and $g'(y)<0$. There will be a root of $g'(x)$ between them.

The function whose roots we want to find is $g'(x)$, so we'll call that function $f(x)$. Specifically,

The most obvious function whose root is $\arcsin(0.1)$ is $\sin(x)-0.1$.