A Function $g(x)$, Written in Terms of $f(x)$, That Flips Sign when a Root of $f(x)$ is Encountered

Question

Let $f(x)$ be any continuous, differentiable function that has extrema and that $f(x) \geq 0$ for all $x$ and has a finite amount of roots (it is not always a polynomial). Let $g(x)$ be a function (written in terms of $f(x)$) that flips sign only when a root of $f(x)$ is encountered. For example, let's say that $f(x)$ has roots at $2,3,7,11$. That means if from $0$ to $2$, $g(x)$ is positive, from $2$ to $3$, $g(x)$ should be negative. From $3$ to $7$ it should be positive and from $7$ to $11$ it should be negative. $g(x)$ should be written in terms of $f(x)$; therefore $g(x)$ should be a general equation. You should not need to know the roots of $f(x)$ in order to construct $f(x)$. $g(x)$ does not have to be continuous nor differentiable. Hopefully, $g(x)$ is closed form (ie. no $\sum$ or $\prod$). What is one definition of $g(x)$? Or is such a function possible to construct?

In any answer, please use $f(x)=\sin^2(\frac{33}{x}\pi)+sin^2(x \pi)$ as the example function.

Background: I was thinking about this problem because I wanted to know if it was possible to find roots numerically for functions that are all positive and have other extrema other than the roots (so something like f(x)/f'(x) wouldn't work). Newton's, fixed point, and so on are only valuable if you start near the root, and even then, with a function like the example, it's not guaranteed to work. Bisection is the only one guaranteed to work, but I need to get the equations in a form so that can work.

EDIT: After giving it some thought, I'm thinking that such $g(x)$, if possible, will probably include sine, cosine, and/or tangent in some way, but I'm not sure how. I was hoping someone could help me.

Answer 1

$\require{mathtools}$Let $x_1 < x_2 < \dots < x_k$ be the roots of $f$. Write $\sigma$ for the following function:

$$\sigma(x) = \begin{cases} -1 & x \le x_1 \\ (-1)^i & x_{i-1} < x \le x_i,\; 2 \le i \le k \\ (-1)^{k+1} & x_k < x \end{cases}$$

Then $g(x) = \sigma(x)f(x)$. One way to think of $\sigma$ is as a sum of indicator functions and their negatives. I'm not sure if this is what you were looking for?

Answer 2

From context clues it seems like you are looking for an algorithm that can compute $g(x)$ from $f(x)$ without first having to compute the roots of $f(x)$, since that would defeat the purpose (as you would like to use $g(x)$ as a step in determining the roots of $f(x)$ numerically). I am not optimistic that you will find such an algorithm, simply for the reason that it seems no easier to find such a $g(x)$ than it is to find the roots using standard numerical techniques.

However, there are still interesting things to say here. First let me point out that in the case of a polynomial, we would necessarily have $$ f(x)=\prod_{i} (x-r_i)^{2n_i} $$ for some integers $n_i$ (up to multiplication by an overall constant), and then the most natural choice of polynomial satisfying your conditions would be $$ g(x)=\prod_{i} (x-r_i). $$ The two polynomials $f(x)$ and $g(x)$ bear the same relation as do the characteristic polynomial and the minimal polynomial of a matrix in linear algebra. There are computational approaches for computing the minimal polynomial of a matrix, which could be used in this problem: starting from the original polynomial $f(x)$, you can put the coefficients (no knowledge of the roots required!) into a so-called companion matrix (whose characteristic polynomial is automatically equal to $f(x)$) and then use the mentioned computational approach to calculate the minimal polynomial of the companion matrix, which will equal $g(x)$. (No claims for efficiency here...) This type of method could be made to apply to non-polynomials by using an approximation scheme.

To address your specific example $$ f(x)=\sin^2\left(\frac{33}{x}\pi\right)+\sin^2(x \pi), $$ let me start by pointing out the obvious, which is that $f(x)=0$ if and only if both of the two terms are equal to zero. The second term is zero precisely when $x\in\mathbb Z$, and the first term is zero precisely when $33/x\in\mathbb Z$. Thus the roots occur when $$ x=-33,-11,-3,-1,1,3,11,33. $$ If you heuristically apply a method similar to what I suggested above for polynomials, you would end up with $$ g(x)=(x^2-33^2)(x^2-11^2)(x^2-3^2)(x^2-1^2), $$ which can be compared with the expressions you get from heuristically manipulating Euler's infinite product

Answer 3

If your $f$ is a polynomial, here is a way to do this. Let $g(x) = \tfrac{f(x)}{\mathrm{GCD}(f(x), f'(x))}$. Note that the GCD of two polynomials can be computed by the Euclidean algorithm without factoring the polynomials.

Why this works: Since $\mathrm{GCD}(f(x), f'(x))$ divides $f$, the only zeroes of the GCD are zeroes of $f$. If $r$ is a zero of $f$ with multiplicity $m$, then $r$ is a zero of $f'(x)$ with multiplicity $m-1$, so $r$ is a zero of $\mathrm{GCD}(f(x), f'(x))$ with multiplicity $m-1$, so $r$ is a zero of $g(x)$ with multiplicity $1$. In other words, this is a way to turn $\prod (x-r_i)^{m_i}$ into $\prod (x-r_i)$, as in pre-kidney's answer.

Answer 4

It is impossible to write such a function $g(x)$ just in terms of $f(x)$. Suppose that the functions $f_1(x)$ and $f_2(x)$ are defined as follows:

$$f_1(x) = x^2,$$

$$f_2(x) = \begin{cases} 2x^2, & x \le -1, \\ 1 + x^4 & -1 < x < 1, \\ 2x^2 & x \ge 1. \end{cases}$$

Both functions are continuous and differentiable. On the other hand, let $g_1(x)$ and $g_2(x)$ denote the corresponding functions. Clearly $g_2(x)$ has the same sign for all $x$, but $g_1(x)$ has to change signs at $x = 0$. But since $f_1(x) = f_2(x)$ for $x \ge 1$, any possible construction of the $g_1(x)$ and $g_2(x)$ has to "know" what is happening on the interval $[-1,1]$ as an input, since it is not determined just from $f_i(x)$ in the range $x \ge 1$ given that you are only assuming it is differentiable. So it is certainly impossible to give $g(x)$ "as a function of $f(x)$" because it depends on more than $f$ at $x$ or even in a neighborhood of $x$.