# Numerical solutions with Newton's method

## An optimization problem

Suppose we want to find the maximum value of $f(x)=x\sin(x)$ over $[0,\pi]$. A glance at graph shows there's exactly one.

So, how would we find that? I guess we just have to solve $$ f'(x) = \sin(x) + x\cos(x) = 0. $$ Unfortunately, that's hard!

## Resources for the numerical solutions of equations

There are plenty of tools for solving equations numerically. One of the simplest to use is WolframAlpha. To solve $\sin(x)+x\cos(x)=0$, for example, just type it in!

If you move on in a technical discipline, though, you'll eventually want to use a more programmatic tool. One broadly applicable tool for mathematical exploration and programming is SageMath.

## Newton's method

Newton's method is a technique to find numerical approximations to roots of functions; it is theoretical foundation on which numerical tools like Sage's `find_root`

works.
Given an initial guesss \(x_1\), Newton's method improves this guess by applying the function
\[N(x) = x - \frac{f(x)}{f'(x)}.\]
This produces \(x_2 = N(x_1)\). We then plug that back in to get \(x_3\) and continue. More
generally, we produce a sequence \((x_k)\) via \(x_k=N(x_{k-1})\).

## Example

Let \(f(x) = x^3-x-1\). It's evident from a graph that there's one root.

Newton's method works by riding the tangent line from an initial guess. If we note that \(x_1=2\) is pretty close to the root, we compute \(x_2 = N(x_2)\), where \[N(x) = x - \frac{f(x)}{f'(x)} = x - \frac{x^3-x-1}{3x^2-1}.\] Thus, \(x_2 = N(2) = 2-5/11 \approx 1.54545\). Geometrically, this point is obtained by riding the tangent line to the \(x\)-axis:

If we do that again, we end up even closer to the root:

That's why we iterate!

## Performing Newton's method

Here's how we can apply Newton's method to the previous example using Sage.

## Exercises

- Use a numerical tool to solve the following equations. Be sure to find all solutions
- \(x^5-x-1 = 0\)
- \(x^5-2x-1 = 0\)
- \(\sin(3x) = x/2\)

- For each of the following functons, take three Newton steps from the given initial point
- \(f(x) = x^2 - 2\) from \(x_1 = 2.0\)
- \(f(x) = \sin(x)\) from \(x_1 = 3.0\)
- \(f(x) = x^5-x-1\) from \(x_1 = 1.0\)

- The equation \(\sin(x)=x/9\) has 7 solutions. We want to find an approximation to the largest solution. Use a graph to find a good initial approximation for your \(x_1\) and apply Newton's method from that point obtain the approximation.