# The rootfinding problem¶

For the time being we will focus on single equations in one variable.

Definition 31 (Rootfinding problem)

Given a continuous scalar function $$f$$ of a scalar variable, find a real number $$r$$ such that $$f(r)=0$$.

We call $$r$$ a root of the function $$f$$. The formulation $$f(x)=0$$ is general enough to solve any equation, for if we are given an equation $$g(x)=h(x)$$, we can define $$f=g-h$$ and find a root of $$f$$.

Unlike the linear problems of the earlier chapters, the usual situation here is that the root cannot be produced in a finite number of operations, even in exact arithmetic. Instead, we seek a sequence of approximations that formally converge to the root, stopping when some member of the sequence seems to be “good enough” (more on that later). The NLsolve package for Julia has a function nlsolve for general-purpose rootfinding.

## Conditioning, error, and residual¶

In the rootfinding problem the data is a continuous function $$f$$ whose root we seek. Let’s assume $$f$$ has at least one continuous derivative near a particular root $$r$$. Say that $$f$$ is perturbed to $$\tilde{f}(x) = f(x) + \epsilon$$. As a result, the root will be perturbed to $$\tilde{r} = r + \delta$$ satisfying, by definition, $$\tilde{f}(\tilde{r})=0$$. We now compute an absolute condition number $$\kappa_r$$, which is the ratio $$\left | \frac{\delta}{\epsilon} \right|$$ as $$\epsilon\to 0$$.

Using Taylor series expansions,

$0 = f(r+\delta) + \epsilon = f(r) + f'(r) \delta + \epsilon + O(\delta^2).$

Given that $$f(r)=0$$, this implies

(75)$\kappa_r = \bigl| f'(r) \bigr|^{-1}.$

Recall that the absolute condition number for the evaluation of $$f$$ at $$x$$ is simply $$|f'(x)|$$. The condition number of the rootfinding problem is equivalent to that of evaluating the inverse function. The influence of $$|f'(r)|$$ on rootfinding is easily illustrated.

We must accept that when $$|f'|$$ is small at the root, it may not be possible to get a small error in a computed root estimate. As always, the error is not a quantity we can compute without knowing the exact answer. But we can compute the residual of a root estimate, which is the value of $$f$$ there. Since $$f(r)=0$$, it stands to reason that a small residual might be associated with a small error.

Suppose that we find an approximation $$\tilde{r}$$ to the actual root $$r$$. Define the new function $$g(x)=f(x)-f(\tilde{r})$$. Trivially, $$g(\tilde{r})=0$$, meaning that the root estimate is a true root of $$g$$. Since the difference between $$g$$ and the original $$f$$ is the residual value $$f(\tilde{r})$$, the residual is the distance to a rootfinding problem that our root estimate solves exactly. That is, the residual is the backward error of the estimate.

To summarize: In general, it is not always realistic to expect a small error in a root approximation. However, the backward error is the same as the residual of the estimate.

## Multiple roots¶

The condition number (75) naturally leads to the question of what happens if $$f'(r)=0$$ at a root $$r$$. Suppose first that $$f$$ is a polynomial of degree $$n>0$$, so that

$f(x) = (x-r)q(x),$

for a polynomial $$q$$ of degree $$n-1$$. If $$r$$ is a simple root of $$f$$—that is, it appears just once in the list of the $$n$$ roots—then it follows that $$q(r)\neq 0$$. Conversely, if $$q(r)=0$$, then $$r$$ appears among the roots of $$q$$ and is a multiple root of $$f$$. However, we don’t need to know the quotient polynomial $$q$$ explicitly in order to make the determination. Consider that

$f'(x) = (x-r)q'(x) + q(x),$

so that $$f'(r) = q(r)$$. Hence $$r$$ is simple if and only if $$f'(r)\neq 0$$.

This conclusion extends to non-polynomial differentiable functions $$f$$. If $$r$$ is a root of $$f$$, define $$q(x)=f(x)/(x-r)$$. By L’Hôpital’s rule, $$g$$ is well defined at $$x=r$$ as long as $$f'$$ is. Now we can again write $$f(x)=(x-r)q(x)$$ for $$x\neq r$$, and by continuity it works at $$x=r$$ as well. So the reasoning we applied to polynomials can be repeated: $$r$$ is a simple root of $$f$$ if and only if $$f'(r)\neq 0$$.

Now suppose that $$f'(r)=q(r)=0$$, so that $$r$$ is not simple. If $$q$$ is differentiable, we may apply the same logic to it that we did to $$f$$. Hence $$r$$ is not simple for $$q$$ if and only if $$q'(r)=0$$. Now observe that

$f''(x) = (x-r)q''(x) + 2q'(x),$

and thus $$f''(r)=2q'(r)$$, so $$f''(r)=0$$ if and only if $$r$$ is a multiple root of $$q$$. In general we define $$r$$ as a root of multiplicity $$m$$ if $$f(r)=f'(r)=\cdots=f^{(m-1)}(r)=0$$, but $$f^{(m)}(r)\neq 0$$. If $$m=1$$, we say $$r$$ is a simple root.

It’s useful to think through the consequences of these definitions for the Taylor series at the point $$r$$,

$f(x) = a_0 + a_1(x-r) + a_2(x-r)^2 + \cdots,$

where $$a_n=f^{(n)}(r)/n!$$. The fact that $$r$$ is a root implies $$f(r)=a_0=0$$. If $$r$$ is a simple root, then $$a_1\neq 0$$, and conversely. If $$r$$ is a double root, then $$a_2\neq 0$$, and so on. Simply put, if $$r$$ is a root of order $$m$$, then the series expansion begins with $$(x-r)^m$$.

When $$r$$ is a multiple root, the condition number (75) is apparently infinite.1 However, even if $$r$$ is technically simple, we should expect difficulty if the condition number is very large. This occurs when $$|f'(r)|$$ is very small, which means that quotient $$q$$ satisfies $$q(r)\approx 0$$ and another root of $$f$$ is very close to $$r$$. The situation is reminiscent of the linear system problem: the degenerate case (singular matrix/multiple root) is mathematically isolated when considered exactly, but the effect on fixed precision computation is just as drastic in a neighborhood of the singularity.

## Exercises¶

1. ✍ For each function, find the multiplicity of the given root. If it is a simple root, find its absolute condition number.

(a) $$f(x) = x^3-2x^2+x-2$$, root $$r=2$$

(b) $$f(x) = (\cos x + 1)^2$$, root $$r=\pi$$

(c) $$f(x) = \frac{\sin^2 x}{x}$$, root $$r=0$$ (define $$f(0) =0$$)

(d) $$f(x) =(x-1)\log(x)$$, root $$r=1$$

2. For any $$\epsilon>0$$, let $$f_\epsilon(x) = \sin[(x-1+\epsilon)(x-1)]$$. This function has roots at $$1-\epsilon$$ and $$1$$.

(a) ✍ Find $$|f_\epsilon'(1)|$$. According to (75), the condition number of the root $$r=1$$ is inversely proportional to this quantity.

(b) ⌨ Define a perturbation function by $$g(x) = \cos[10x+\sin(20x)]$$. Verify that $$f_\epsilon(x)+10^{-10}g(x)$$ has a root in the interval $$[1-\epsilon,1+10\epsilon]$$ for $$\epsilon=10^{-3}$$.

(c) ⌨ For $$\epsilon=10^{-3},10^{-4},10^{-5},10^{-6}$$, use nlsolve to find the root in the interval given in part (b). Make a table of $$\epsilon$$, $$1/|f_\epsilon'(1)|$$, $$|r-1|$$, and $$|(r-1)f_\epsilon'(1)|$$. The last value should be approximately constant.

3. (continuation) The condition number theory (75) suggests that if $$f$$ has simple roots at $$r_1$$ and $$r_2$$ that are close to each other, then the condition number is large. But then the function $$\tilde{f}(x) = f(x)/(x-r_2)$$ no longer has a root at $$r_2$$ and should have a much better condition number if there are no other nearby roots. This trick (called deflation) can work even if the division factor does not use the exact root $$r_2$$. For the steps below, use the same definitions as in the preceding problem.

(a) ✍ Define $$\tilde{f}_\epsilon(x) = f_\epsilon(x)/(x+1.01)$$. Find $$|\tilde{f}_\epsilon'(1)|$$. (You may want to use computer algebra for this.)

(b) ⌨ Repeat part (c) of the preceding problem, but using the interval $$[1-\epsilon/200,1+\epsilon/100]$$ each time. By what factor are the errors improved over using $$f_\epsilon$$? Do you still find that $$|(x-1)\tilde{f}_\epsilon'(1)|$$ is roughly constant?

1

Based on our definitions, this means that the relative change to the root when $$f$$ is changed by a perturbation of size $$\epsilon$$ is not $$O(\epsilon)$$ as $$\epsilon\to 0$$.