Newton Raphson Method

The Newton-Raphson method (also known as Newton's method) is a way to quickly find a good approximation for the root of a real-valued function \(f(x) = 0\). It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.

Suppose you need to find the root of a continuous, differentiable function \(f(x)\), and you know the root you are looking for is near the point \(x = x_0\). Then Newton's method tells us that a better approximation for the root is \[x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}.\] This process may be repeated as many times as necessary to get the desired accuracy. In general, for any \(x\)-value \(x_n\), the next value is given by \[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.\]

Note: the term "near" is used loosely because it does not need a precise definition in this context. However, \(x_0\) should be closer to the root you need than to any other root (if the function has multiple roots).

Here is a picture to demonstrate what Newton's method actually does:

We draw a tangent line to the graph of \(f(x)\) at the point \(x = x_n\). This line has slope \(f'(x_n)\) and goes through the point \(\big(x_n, f(x_n)\big)\). Therefore it has the equation \(y = f'(x_n)(x - x_n) + f(x_n)\). Now, we find the root of this tangent line by setting \(y = 0\) and \(x=x_{n+1}\) for our new approximation. Solving this equation gives us our new approximation, which is \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\).

Find the root of the equation \(x^2 - 4x - 7 = 0\) near \(x = 5\) to the nearest thousandth.

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We have our \(x_0 = 5\). In order to use Newton's method, we also need to know the derivative of \(f\). In this case, \(f(x) = x^2 - 4x - 7\), and \(f'(x) = 2x - 4\).

Using Newton's method, we get the following sequence of approximations:

\[\begin{align} x_1 &= 5 - \frac{5^2 - 4\times 5 - 7}{2\times 5 - 4} = 5 - \left(\frac{-2}{6}\right) = \frac{16}{3} \approx 5.33333\\ x_2 &= \frac{16}{3} - \frac{\left(\frac{16}{3}\right)^2 - 4\left(\frac{16}{3}\right) - 7}{2\left(\frac{16}{3}\right)-4} = \frac{16}{3} - \frac{\frac{1}{9}}{\frac{20}{3}} = \frac{16}{3} - \frac{1}{60} = \frac{319}{60} \approx 5.31667 \\ x_3 &= \frac{319}{60} - \frac{\left(\frac{319}{60}\right)^2 - 4\left(\frac{319}{60}\right) - 7}{2\left(\frac{319}{60}\right)-4} = \frac{319}{60} - \frac{\frac{1}{3600}}{\frac{398}{60}} \approx 5.31662. \end{align}\]

We can stop now, because the thousandth and ten-thousandth digits of \(x_2\) and \(x_3\) are the same. If we were to continue, they would remain the same because we have gotten sufficiently close to the root:

\[x_4 = 5.31662 - \frac{(5.3362)^2-4(5.3362)-7}{2(5.3362)-4} = 5.31662.\]

Our final answer is therefore 5.317. \( _\square \)

Newton's method may not work if there are points of inflection, local maxima or minima around \(x_0\) or the root.

For example, suppose you need to find the root of \(27x^3 - 3x + 1 = 0\) which is near \(x = 0\).
The correct answer is \(-0.44157265\ldots\) However, Newton's method will give you the following:

\[x_1 = \frac{1}{3}, x_2 = \frac{1}{6}, x_3 = 1, x_4 = 0.679, x_5 = 0.463, x_6 = 0.3035, x_7 = 0.114, x_8 = 0.473, \ldots.\]

This is very clearly not helpful. That's because the graph of the function around \(x = 0\) looks like this:

As you can see, this graph has a local maximum, a local minimum and a point of inflection around \(x = 0\). To see why Newton's method isn't helpful here, imagine choosing a point at random between \(x = -0.19\) and \(x = 0.19\) and drawing a tangent line to the function at that point. That tangent line will have a negative slope, and therefore will intersect the \(y\)-axis at a point that is farther away from the root.

In a situation like this, it will help to get an even closer starting point, where these critical points will not interfere.