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Root-hunting algorithm: Newton's method

Problem: Given a real-valued function \(f(x)\) in one real variable, what are the values \(x_0 \in \mathbb{R}\) such that \(f(x_0) = 0\)?

If the function \(f(x)\) is linear, then the problem is trivial. Explicitly, if \(f(x) = ax + b\) for some \(a, b \in \mathbb{R}\), then \(x_0 = -b/a\) gives a solution as long as \(a \neq 0\). However, when the function is nonlinear, the problem can get complicated very fast. For example, try solving when the function is \(f(x) = \sin(e^{x}) + \cos(\log x)\).

Newton's idea (an overview)

One way of solving this problem is to linearize the function \(f(x)\) around a certain point \(x_0\) of our choice so that we can easily solve the resulting linear equation. Say we get \(x_1\) as a solution, then we repeat linearizing \(f(x)\) around \(x_1\); so on and so forth. The initial point \(x_0\) is chosen such that it is close to our hoped solution, say, \(x^*\). The idea is that if \(x_0\) is suitably chosen, then the solutions \(x_1, x_2, x_3, \ldots\) to each linear approximation of \(f(x)\) approximates \(x^*\) better and better, and in the limit converges to \(x^*\). This whole process is known as the Newton's method. Here is a nice picture of the method applied to \(f(x) = x^2\) over \(n = 10\) iterations, starting at \(x_0 = 1\).

Newton's method illustration

We see from the illustration that the Newton's method converges towards \(x^* = 0\), as expected. For the code to generate this plot, see end of post.

Newton's idea (the algebra)

Let us make our discussion above more precise. Linearizing \(f(x)\) around \(x_0\) simply means Taylor expanding \(f\) around \(x_0\) and neglecting \(\mathcal{O}(h^2)\) terms. Of course, this is assuming that we can actually do Taylor expansion in the first place! Explicitly, the Taylor expansion of \(f\) around \(x_0\) yields

$$f(x) = f(x_0) + f'(x_0) (x - x_0) + \mathcal{O}(h^2). $$

So if we neglect \(\mathcal{O}(h^2)\) terms, we get (in approximation) the linear equation

$$f(x) = f(x_0) + f'(x_0) (x - x_0).$$

The solution to this linear equation is thus simply

$$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}.$$

We then repeat the process by linearizing \(f\) around \(x_1\). In this case we have

$$f(x) = f(x_1) + f'(x_1) (x - x_1) \implies x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}, $$

with \(x_2\) being a solution. Doing this iteratively yields a general formula

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)},$$

known as Newton's formula. Here is the Newton's method in one statement.

Theorem (Newton's method). Let \(x^* \in \mathbb{R}\) be a solution to \(f(x) = 0\). If \(x_n\) is an approximation of \(x^*\) and \(f'(x_n) \neq 0\), then the next approximation to \(x^*\) is given by

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)},$$

with initial condition, a suitably chosen \(x_0 \in \mathbb{R}\).

Code implementation

An iterative implementation of the Newton's method in Python is given below:

def iterative_newton(f, df, x0, n):
    """Solves f(x) = 0 using the Newton's method.

        f: A callable, the function f(x) of interest.
        df: A callable, the derivative of f(x).
        x0: Initial good point to start linearizing.
        n (Optional): Number of recursion steps to make.
    xs = [x0] # Sequence of xn.

    # Get latest x value in sequence and
    # apply the Newton recurrence formula.
    for _ in range(n):
        last = xs[-1]
        res = last - f(last)/df(last)

    return xs

Using the same parameters as above, we can also implement a one-liner recursive implementation:

def recursive_newton(f, df, x0, n):
    return x0 if n <= 0 else recursive_newton(f, df, x0 - f(x0)/df(x0), n-1)

Observe that both algorithms have \(\mathcal{O}(n)\) space complexity where \(n\) is the number of iterations or depth of the recursion. The time complexity for the iterative implementation is also \(\mathcal{O}(n)\), but for the recursive implementation, it is a bit tricky to compute (so we leave it as an exercise!).

Note that there is a small caveat to the Newton's method which we have implicitly highlight in this post, can you spot it?

Code to generate Newton's method plot

We will utilize the numpy and matplotlib packages.

import numpy as np
import matplotlib.pyplot as plt

We first need two helper functions.

def finite_diff(f):
    """ Returns the derivative of f(x) based on the
    finite difference approximation.
    h = 10**(-8)
    def df(x):
        return (f(x+h)-f(x-h)) / (2*h)
    return df

def tangent_line(f, x0, a, b):
    """ Generates the tangent line to f(x) at x0 over
    the interval [a, b].
    df = finite_diff(f)
    x = np.linspace(a, b, 300)
    ytan = (x - x0)*df(x0) + f(x0)

    return x, ytan

We then define the function \(f(x) = x^2\), compute its derivative and apply Newton's method over \(n = 10\) iterations, starting at \(x_0 = 1\).

f = lambda x: x**2
df = finite_diff(f)
res = iterative_newton(f, df, 1, 10)
res = np.array(res)  # To utilize numpy broadcasting later.

We finally plot the function.'bmh')

# Bounds on the x-axis.
lo = -0.1
hi = 1.1
mesh = abs(hi) + abs(lo)

fig, ax = plt.subplots(figsize=(10, 6))

# Points of the function f(x).
xs = np.arange(start=lo, stop=hi, step=0.01)
ys = f(xs)

# Tangent lines to f(x) at the approximations.
xtan0, ytan0 = tangent_line(f, res[0], 0.35*mesh, hi)
xtan1, ytan1 = tangent_line(f, res[1], 0.1*mesh, hi)
xtan2, ytan2 = tangent_line(f, res[2], lo, 0.7*mesh)

ax.plot(xs, ys, label="$f(x) = x^2$", linewidth=3)
ax.plot(xtan0, ytan0, label="Linearization 1", alpha=0.8)
ax.plot(xtan1, ytan1, label="Linearization 2", alpha=0.8)
ax.plot(xtan2, ytan2, label="Linearization 3", alpha=0.8)
ax.plot(res, f(res), color='darkmagenta',
        label="Newton's method\napproximations",
        marker='o', linestyle='None', markersize=6)

# Labels for occurring approximations.
for i in range(0, 4):
    ax.plot(res[i], 0, marker='+', color='k')
    ax.vlines(res[i], ymin=0, ymax=f(res[i]),
              linestyles='dotted', color='k', alpha=0.3)
                 (res[i], 0),
                 textcoords="offset points",
                 xytext=(0, -20),

# Grid and xy-axis.
ax.grid(True, which='both')
ax.axvline(x = 0, color='k')
ax.axhline(y = 0, color='k')

# Labels and legend.
ax.set_title("Newton's method applied to $f(x) = x^2$")
plt.legend(fontsize=12, loc=9)

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