from numpy import *
from scipy import linalg
from scipy.linalg import norm
from matplotlib.pyplot import *
from prettytable import PrettyTable
from timeit import default_timer as timer
import sys
sys.path.append('fncbook/')
import fncbook as FNC

# This (optional) block is for improving the display of plots.
# from IPython.display import set_matplotlib_formats
# set_matplotlib_formats("svg","pdf")
# %config InlineBackend.figure_format = 'svg'
rcParams["figure.figsize"] = [7, 4]
rcParams["lines.linewidth"] = 2
rcParams["lines.markersize"] = 4
rcParams['animation.html'] = "jshtml"  # or try "html5"

4.3.Newton’s method¶

Newton’s method is the cornerstone of rootfinding. We introduce the key idea with an example in Example 4.3.1.

Example 4.3.1 (Graphical interpretation of Newton’s method)

Suppose we want to find a root of this function:

f = lambda x: x * exp(x) - 2
xx = linspace(0, 1.5, 400)

fig, ax = subplots()
ax.plot(xx, f(xx), label="function")
ax.grid()
ax.set_xlabel("$x$")
ax.set_ylabel("$y$");

From the graph, it is clear that there is a root near $x=1$ . So we call that our initial guess, $x_1$ .

x1 = 1
y1 = f(x1)
ax.plot(x1, y1, "ko", label="initial point")
ax.legend()
fig

Next, we can compute the tangent line at the point $\bigl(x_1,f(x_1)\bigr)$ , using the derivative.

df_dx = lambda x: exp(x) * (x + 1)
slope1 = df_dx(x1)
tangent1 = lambda x: y1 + slope1 * (x - x1)

ax.plot(xx, tangent1(xx), "--", label="tangent line")
ax.set_ylim(-2, 4)
ax.legend()
fig

In lieu of finding the root of $f$ itself, we settle for finding the root of the tangent line approximation, which is trivial. Call this $x_2$ , our next approximation to the root.

x2 = x1 - y1 / slope1
ax.plot(x2, 0, "ko", label="tangent root")
ax.legend()
fig

y2 = f(x2)
print(y2)

0.06716266657572145

The residual (i.e., value of $f$ ) is smaller than before, but not zero. So we repeat the process with a new tangent line based on the latest point on the curve.

xx = linspace(0.83, 0.88, 200)

plot(xx, f(xx))
plot(x2, y2, "ko")
grid(), xlabel("$x$"), ylabel("$y$")

slope2 = df_dx(x2)
tangent2 = lambda x: y2 + slope2 * (x - x2)
plot(xx, tangent2(xx), "--")
x3 = x2 - y2 / slope2
plot(x3, 0, "ko")
title("Second iteration");

y3 = f(x3)
print(y3)

0.0007730906446230534

Judging by the residual, we appear to be getting closer to the true root each time.

Using general notation, if we have a root approximation $x_k$ , we can construct a linear model of $f(x)$ using the classic formula for the tangent line of a differentiable function,

q(x) = f(x_k) + f'(x_k)(x-x_k).

(4.3.1)

Finding the root of $q(x)=0$ is trivial. We define the next approximation by the condition $q(x_{k+1})=0$ , which leads to the following.

4.3.1Convergence¶

The graphs of Example 4.3.1 suggest why the Newton iteration may converge to a root: any differentiable function looks more and more like its tangent line as we zoom in to the point of tangency. Yet it is far from clear that it must converge, or at what rate it will do so. The matter of the convergence rate is fairly straightforward to resolve. Define the error sequence

\epsilon_k = x_k - r , \quad k=1,2,\ldots,

(4.3.3)

where $r$ is the limit of the sequence and $f(r)=0$ . Exchanging $x$ -values for $\epsilon$ -values in (4.3.2) gives

\epsilon_{k+1}+r = \epsilon_k + r - \frac{f(r+\epsilon_k)}{f'(r+\epsilon_k)}.

(4.3.4)

We assume that $|\epsilon_k|\to 0$ ; eventually, the errors remain as small as we please forever. Then a Taylor expansion of $f$ about $x=r$ gives

\epsilon_{k+1} = \epsilon_k - \frac{ f(r) + \epsilon_kf'(r) + \frac{1}{2}\epsilon_k^2f''(r) + O(\epsilon_k^3)}{ f'(r) + \epsilon_kf''(r) + O(\epsilon_k^2)}.

(4.3.5)

We use the fact that $f(r)=0$ and additionally assume now that $r$ is a simple root, i.e., $f'(r)\neq 0$ . Then

\epsilon_{k+1} = \epsilon_k - \epsilon_k \left[ 1 + \dfrac{1}{2}\dfrac{f''(r)}{f'(r)} \epsilon_k + O(\epsilon_k^2)\right] \, \left[ 1 + \dfrac{f''(r)}{f'(r)}\epsilon_k + O(\epsilon_k^2)\right]^{-1}.

(4.3.6)

The series in the denominator is of the form $1/(1+z)$ . Provided $|z|<1$ , this is the limit of the geometric series $1-z+z^2-z^3 + \cdots$ . Keeping only the lowest-order terms, we derive

\begin{align*} \epsilon_{k+1} &= \epsilon_k - \epsilon_k \left[ 1 + \dfrac{1}{2}\dfrac{f''(r)}{f'(r)} \epsilon_k + O(\epsilon_k^2) \right] \, \left[ 1 - \dfrac{f''(r)}{f'(r)} \epsilon_k + O(\epsilon_k^2) \right]\\ &= \frac{1}{2}\, \frac{f''(r)}{f'(r)} \epsilon_k^2 + O(\epsilon_k^3). \end{align*}

(4.3.7)

Recall that linear convergence is identifiable by trending toward a straight line on a log-linear plot of the error. When the convergence is quadratic, no such straight line exists—the convergence keeps getting steeper. As a numerical test, note that $|\epsilon_{k+1}|\approx K |\epsilon_{k}|^2$ implies that as $k\to\infty$ ,

\begin{split} \log |\epsilon_{k+1}| & \approx 2 \log |\epsilon_{k}| + \log K,\\ \frac{\log |\epsilon_{k+1}|}{\log |\epsilon_{k}|} &\approx 2 + \frac{\log K}{\log |\epsilon_{k}|} \to 2. \end{split}

(4.3.9)

Example 4.3.2 (Convergence of Newton’s method)

We again look at finding a solution of $x e^x=2$ near $x=1$ . To apply Newton’s method, we need to calculate values of both the residual function $f$ and its derivative.

f = lambda x: x * exp(x) - 2
df_dx = lambda x: exp(x) * (x + 1)

We don’t know the exact root, so we use nlsolve to determine a proxy for it.

from scipy.optimize import root_scalar
r = root_scalar(f, bracket=[0.8, 1.0]).root
print(r)

0.8526055020137255

We use $x_1=1$ as a starting guess and apply the iteration in a loop, storing the sequence of iterates in a vector.

x = ones(5)
for k in range(4):
    x[k + 1] = x[k] - f(x[k]) / df_dx(x[k])

print(x)

[1.         0.86787944 0.85278337 0.85260553 0.8526055 ]

Here is the sequence of errors.

err = x - r
print(err)

[1.47394498e-01 1.52739392e-02 1.77871403e-04 2.43551965e-08
 4.44089210e-16]

The exponents in the scientific notation definitely suggest a squaring sequence. We can check the evolution of the ratio in (4.3.9).

logerr = log(abs(err))
for i in range(len(err) - 1):
    print(logerr[i+1] / logerr[i])

2.1840144823399648
2.064863881067786
2.030299689916648
2.01651205997716

The clear convergence to 2 above constitutes good evidence of quadratic convergence.

Let’s summarize the assumptions made to derive quadratic convergence as given by (4.3.7):

The residual function $f$ has to have enough continuous derivatives to make the Taylor series expansion valid. Often this is stated as $f$ having sufficient smoothness. This is usually not a problem, but see Exercise 4.3.6.
We required $f'(r)\neq 0$ , meaning that $r$ must be a simple root. See Exercise 4.3.7 to investigate what happens at a multiple root.
We assumed that the sequence converged, which is not easy to guarantee in any particular case. In fact, finding a starting value from which the Newton iteration converges is often the most challenging part of a rootfinding problem. We will try to deal with this issue in Quasi-Newton methods.

4.3.2Implementation¶

Our implementation of Newton’s iteration is given in Function 4.3.1. It accepts functions that evaluate $f$ and $f'$ and the starting value $x_1$ as input arguments. Beginning programmers are tempted to embed $f$ and $f'$ directly into the code, but there are two good reasons not to do so. First, each new rootfinding problem would require its own copy of the code, creating a lot of duplication. Second, you may want to try more than one rootfinding algorithm for a particular problem, and keeping the definition of the problem separate from the algorithm for its solution makes this task much easier.

Algorithm 4.3.1 (newton)

newton.py

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def newton(f, dfdx, x1):
    """
    newton(f, dfdx, x1)

    Use Newton's method to find a root of f starting from x1, where dfdx is the
    derivative of f. Returns a vector of root estimates.
    """
    # Operating parameters.
    eps = np.finfo(float).eps
    funtol = 100 * eps
    xtol = 100 * eps
    maxiter = 40

    x = np.zeros(maxiter)
    x[0] = x1
    y = f(x1)
    dx = np.inf  # for initial pass below
    k = 0

    while (abs(dx) > xtol) and (abs(y) > funtol) and (k < maxiter):
        dydx = dfdx(x[k])
        dx = -y / dydx  # Newton step
        x[k + 1] = x[k] + dx  # new estimate

        k = k + 1
        y = f(x[k])

    if k == maxiter:
        warnings.warn("Maximum number of iterations reached.")

    return x[:k+1]

About the code

Function 4.3.1 accepts keyword arguments. In the function declaration, these follow the semicolon, and when the function is called, they may be supplied as keyword=value in the argument list. Here, these arguments are also given default values by the assignments within the declaration. This arrangement is useful when there are multiple optional arguments, because the ordering of them doesn’t matter.

The break statement, seen here in line 25, causes an immediate exit from the innermost loop in which it is called. It is often used as a safety valve to escape an iteration that may not be able to terminate otherwise.

Function 4.3.1 also deals with a thorny practical issue: how to stop the iteration. It adopts a three-part criterion. First, it monitors the difference between successive root estimates, $|x_k-x_{k-1}|$ , which is used as a stand-in for the unknown error $|x_k-r|$ . In addition, it monitors the residual $|f(x_k)|$ , which is equivalent to the backward error and more realistic to control in badly conditioned problems (see The rootfinding problem). If either of these quantities is considered to be sufficiently small, the iteration ends. Finally, we need to protect against the possibility of a nonconvergent iteration, so the procedure terminates with a warning if a maximum number of iterations is exceeded.

Example 4.3.3 (Using Newton’s method)

Suppose we want to evaluate the inverse of the function $h(x)=e^x-x$ . This means solving $y=h(x)$ , or $h(x)-y=0$ , for $x$ when $y$ is given. That equation has no solution in terms of elementary functions. If a value of $y$ is given numerically, though, we simply have a rootfinding problem for $f(x)=e^x-x-y$ .

Tip

The enumerate function produces a pair of values for each iteration: a positional index and the corresponding contents.

h = lambda x: exp(x) - x
dh_dx = lambda x: exp(x) - 1
y_ = linspace(h(0), h(2), 200)
x_ = zeros(y_.shape)
for (i, y) in enumerate(y_):
    f = lambda x: h(x) - y
    df_dx = lambda x: dh_dx(x)
    x = FNC.newton(f, df_dx, y)
    x_[i] = x[-1]

plot(x_, y_, label="$y=h(x)$")
plot(y_, x_, label="$y=h^{-1}(x)$")
plot([0, max(y_)], [0, max(y_)], 'k--', label="")
title("Function and its inverse")
xlabel("x"), ylabel("y"), axis("equal")
ax.grid()
legend();

4.3.3Exercises¶

For each of Exercises 1–3, do the following steps.

(a) ✍ Rewrite the equation into the standard form for rootfinding, $f(x) = 0$ , and compute $f'(x)$ .

(b) ⌨ Make a plot of $f$ over the given interval and determine how many roots lie in the interval.

(c) ⌨ Like in Example 4.3.2, find a reference value for each root using a native solver.

(d) ⌨ Use Function 4.3.1 to find each root.

(e) ⌨ For one of the roots, use the errors in the Newton sequence to determine numerically whether the convergence is roughly quadratic.