7
$\begingroup$

I realise this is a general question. I am self-teaching mathematics and I have observed that the zeros of real and complex functions are of much interest.

Question: Why are the zeros of real or complex so important?

I was going to say that the zeros completely determine a function because the function can be factorised into factors, with each factor corresponding to one of the zeros. For example, $f(x) = (x-1)(x-3)$ is completely determined by its zeros at $x=1$ and $x=3$.

However this doesn't work for functions like $g(x) = (x-1)(x-3) + 2$.

I appreciate your patience with a question that might be naive.

$\endgroup$
9
  • 4
    $\begingroup$ $f(x)=2(x-1)(x-3)$ also has zeros at $x=1$ and $x=3$ $\endgroup$ Aug 4, 2021 at 23:29
  • 4
    $\begingroup$ So what you say about being completely determined is basically true for polynomials, not general functions. I say "basically" because a polynomial is determined by its zeros, their multiplicities, and one additional number, the leading coefficient. More generally though, we care about equations like $f(x)=0$ because we can reduce the problem of solving $f(x)=a$ to solving $g(x)=0$ where $g(x)=f(x)-a$. As for $f(x)=a$, that has plenty of applications, things like finding when an object hits the ground, how long a population takes to reach some threshold, optimization problems, etc. $\endgroup$
    – Ian
    Aug 4, 2021 at 23:30
  • 5
    $\begingroup$ Just a general remark: Solving any equation $$A=B$$ is equivalent to solving $$A-B=0$$ as soon as $A, B$ are anything which even vaguely resemble mathematical objects which can be subtracted from another. In particular, if $A-B$ happens to be a function, your task is now to $$\text{find the zeros of } A-B.$$ Hence, more often than you think, "solving an equation" immediately reduces to "finding zeros". Now you could ask why solving equations is important ... $\endgroup$ Aug 5, 2021 at 0:05
  • $\begingroup$ I might suggest that one reason to focus on zero, rather than other values, is that you might need to restrict your domain to exclude the zeros if you divide by the function. But it is more that other values can easily be rephrased in terms of zeros. $\endgroup$ Aug 5, 2021 at 0:07
  • 2
    $\begingroup$ @TorstenSchoeneberg - can we also say that if we can write f(x)=0 as AB=0 then this tells us that either A=0, B=0, or both. Whereas we can't say anything about A or B if AB=1. Thus "AB=0" is more informative than "AB=c not zero" ? $\endgroup$
    – Penelope
    Aug 5, 2021 at 0:24

4 Answers 4

4
$\begingroup$

Often the zeros carry special significance.

For example the location of non-trivial zeros of the Riemann-Zeta function has far reaching consequences in mathematics, especially for theories of prime numbers.

In ordinary linear differential equations, the zeros of these “functions of functions” are called the “homogenous solutions” and form a vector space of solutions (since every solution evaluates to zero you can take linear combinations and still end up with zero).

The study of the roots of polynomials led to Galois Theory which connected fields and groups and helped resolve longstanding issues of which polynomials are solvable and also issues about provability in geometry.

Finally, a trivial interest in zeros is that $f(x)=c \iff g(x):=f(x)-c=0$.

Just a smattering of examples to illustrate.

$\endgroup$
4
$\begingroup$

Yes, Weierstrass and Hadamard showed that "entire" functions (=holomorphic on the entire complex plane) do have product expansions in terms of their zeros, with some extra factors for convergence of that infinite product... and a leading exponential term.

$\endgroup$
2
  • $\begingroup$ does this not work for real (not-complex) functions? f(x)=2+sin(x) I guess doesn't work. $\endgroup$
    – Penelope
    Aug 13, 2021 at 22:23
  • $\begingroup$ @Tariq, a real-valued function of a real variable, if it extends reasonably to an "entire" function, does also have that property! $\endgroup$ Aug 13, 2021 at 23:19
2
$\begingroup$

A general remark: Solving any equation

$$A=B$$

is equivalent to solving

$$A-B =0$$

as soon as $A$ and $B$ are anything which at least vaguely resembles mathematical objects that can be subtracted from each other. In particular, if $A-B$ is something which you can express as a function, your task is now to

$$\text{ find the zeros of } A-B.$$

Hence, more often than one thinks, "solving an equation" immediately reduces to "finding zeros".

Now one could ask why solving equations is important, but I trust you don't ask that if you have ever worked in any STEM field. Or you and I trust people who claim that equations like $$\mathbf F + \mathbf{u} \frac{\mathrm{d} m}{\mathrm{d}t} = m {\mathrm{d} \mathbf v \over \mathrm{d}t}$$ or $$\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{l} = \iint_S \left( \mu_0 \mathbf{J}+ \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\right) \cdot \mathrm{d} \mathbf{S}$$ or $$G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu}$$ have importance, and it's useful to solve them occasionally, in order to hit a target or get a motor running or make a GPS useful or estimate how old the universe is.

$\endgroup$
1
  • $\begingroup$ Torsten - thanks for this answer. I hope you don't mind that for me the line of interest is the issue of "can a function be fully defined in terms of its zeros" which some comments/answers are exploring. $\endgroup$
    – Penelope
    Aug 13, 2021 at 22:26
1
$\begingroup$

Actually it does work perfectly well for $g(x)=(x-1)(x-3)+2$. This has zeros at $2\pm i$, and it is indeed true that $$g(x)=(x-2-i)(x-2+i).$$

The statement that the zeroes determine the function only holds for polynomials (and only up to multiplication by a constant). For polynomials this always works over the complex numbers, since any polynomial can be fully factorised over the complex numbers (this is the fundamental theorem of algebra), but not necessarily over the reals.

However, note that you may need to know the multiplicities of the roots, not just the roots themselves. $f_1(x)=x^3-x$ and $f_2(x)=x^3-x^2$ are not the same, even though they have the same roots $0$ and $1$, but really the roots of $f_1$ are $0,1,1$ and the roots of $f_2$ are $0,0,1$.

$\endgroup$
2
  • $\begingroup$ So if we restrict ourselves to the real domain, this doesn't work. So does the statement that zeros of a function completely define a function require the complex domain? $\endgroup$
    – Penelope
    Aug 13, 2021 at 22:25
  • 1
    $\begingroup$ @Tariq I've edited the answer to address this. $\endgroup$ Aug 14, 2021 at 8:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .