# Newton's Method

Newton's method provides a useful way to estimate the roots and solutions to equations. Suppose for instance we have a function $f(x) = x^10+2x^4+x+5$, and we want to find the roots for the equation.

When the equation is a quadratic or cubic, it is easy to find the roots. However, for higher power polynomials, and other similar functions, it becomes quite hard to do. Using Newton's method, we can more easily solve problems like these.

The idea of Newton's method is the following. We have a tangent line L, that is tangent to the curve f(x) at the point $(x_1,f(x_1))$. Because the line is tangent to the curve, it has an x intercept at some point, $x_2$. We know that f(x) also has an x intercept at some point r, which is what we are trying to find. If we pick some number, x_{1}, we can get an x intercept that exists somewhere close to the root, as shown below.

You can see that this line is tangent and
is close to the root of f(x). Using this idea, we can continue to change the
point x_{1} that we are tangent to, and our approximation can get
closer to the actual root of f(x).

So, to find the formula for x_{2}, the x-intercept of the tangent line, with respect to x_{1}, the point we are tangent to, we can use the following formula:

$y - f(x_1) = f'(x_1)(x_2-x_1)$

We know that x_{2} is the x intercept, therefore we know that y is 0.

$0 - f(x_1) = f'(x_1)(x_2-x_1)$

From here, we can rearrange for x_{2} to get:

$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}$

To apply Newton's method, we can use this formula as follows:

- Start with an initial estimate at the root, $x_1$ and find $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}$
- Using $x_2$, calculate $x_3 = x_2 - \frac{f(x_2)}{f'(x_2)}$
- Repeat this until you find a root that is sufficently accurate

The steps to Newton's method are simple and effective for finding roots. Often, root finding programs will use Newton's method or similar estimations to provide roots for complicated equations.

**Example: **Starting with $x_1 = 2$, find the third approximation to the root of the equation $f(x) = x^3 - 2x -5$

The first step is to find the derivative. $f'(x) = 3x^2-2$

Once this is done, we simply apply Newton's method 3 times.

$x_2 = 2 - \frac{f(2)}{f'(2)} = 2 + \frac{1}{10} = 2 + 0.1 = 2.1$

$x_3 = 3.1 - \frac{f(2.1)}{f'(2.1)} = 2.1 - \frac{0.061}{11.23} = 2.094$

Let's put our estimate into f(x) and see how it looks.

$f(2.094) = (2.094)^3 - 2(2.094) - 5 = -0.00615$

As you can see, the estimate is very close to a root, so Newton's method proves to be a good estimate for the functions root.