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Please excuse the naivity of this question, but it is a concept that I just have not been able to grasp entirely.

My question is, why are the roots of a function, or a system of polynomials so important? Why do they tell us so much about the function itself? Why, for example, do the zeros of the zeta function hold within them information about so many different number theoretical problems? Why would, say, x raised to the power of certain roots, encode seemingly unrelated information?

I can understand it on a problem to problem basis, but can't really grasp the concept on a general level.

Answers in English would be most appreciated!

If this question is too general or off topic, I will delete as requested.

martin
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    For polynomials, anyway, the roots tell us how to construct a factorization, which is a great help in understanding the structure and behaviour of a function. In fact, in this case, the roots essentially determine the function. – bubba Feb 25 '14 at 13:46

3 Answers3

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It is often very important for us to know when two functions are equal; i.e. to solve an equation $f(x) = g(x)$.

It turns out to greatly simplify the problem if you rewrite the question as $f(x) - g(x) = 0$; so if we learn how to solve equations $h(x) = 0$, that also lets us solve equations $f(x) = g(x)$ too.

As an example of simplification, imagine if you had to learn the quadratic formula as

To solve $a x^2 + b x + c = d x^2 + e x + f$ when $a \neq d$, the roots are $$ x = \frac{(e-b) \pm \sqrt{ (e-b)^2 + 4 (a-d)(f-c) }}{2(a-d)} $$

Gah, much worse!

There's a strong analogy between factoring integers into a product of primes, and factoring polynomials into a product of irreducible polynomials. Since the irreducible factors correspond to roots of the polynomial, this shows that finding roots of a polynomial is like finding the prime factors of an integer. Similarly, finding the roots and poles of a rational function is like finding the prime factorization (allowing negative exponents) of a rational number.

(the roots may be in an extension; e.g. the roots of the irreducible real polynomial $x^2 + 1$ are complex numbers)

The general properties of analytic functions -- roughly, functions equal to their Taylor series -- share a lot of properties with polynomials. Techniques for working with polynomials usually get generalized to work with analytic functions... such as finding the zeroes and poles.

The analysis zeta function is a whole subject in of itself.

Caleb Stanford
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  • This is great. Could you expand a little on functions' Taylor series sharing properties with polynomials? I understand now the importance of roots in understanding polynomials, but I am still at sea a little as regards roots of functions and functional analysis (as a general concept). And why is the analysis of zeta functions any different to the analysis of other functions (aside from its complexity)? – martin Feb 25 '14 at 14:27
  • Are all functions theoretically expandable to a corresponding Taylor series? – martin Feb 25 '14 at 14:29
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    @martin: Some functions are not infinitely differentiable at some points, and thus the Taylor series isn't defined at those points. Other functions have a Taylor series at all points, but for some or all of those points, its radius of convergence is zero. The analytic functions are actually somewhat rare amongst all functions, although they are disproportionately common amongst functions people actually study. –  Feb 25 '14 at 14:33
  • Thank you - this explanation is very clear on this point :) – martin Feb 25 '14 at 14:43
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    @martin: In a useful coincidence, I stumbled across a relevant theorem on the factorization of analytic functions whose name I always forget, and thus I couldn't reference. –  Feb 25 '14 at 14:46
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    @martin, Euler famously fooled around with series as "infinite polynomials" with given zeroes, and thus derived many wonderful results. It took a while to put that on solid foundations, though... – vonbrand Feb 25 '14 at 15:05
  • @ Hurkyl, Thank you very much for the link :) @ Vonbrand, - yes, I looked at some of the Euler product formulas, but because of the unpredictable (and for me, often impenetrable!) nature of primes and prime reciprocals, I can only get so far conceptually as many lesser mortals than Euler can! – martin Feb 25 '14 at 15:16
  • @ Hurkyl, having looked at the link you provided, this really hits at the heart of my question. Not that I fully understand it, I might add!! But it certainly provides future food for thought... – martin Feb 25 '14 at 15:24
  • @ Hurkyl, I have not come across the Hadamard Product before, so in light of your answer, and your introduction to me of this and the Weierstrass Product Theorem, I shall accept your answer. Again, many thanks :) – martin Feb 25 '14 at 15:43
  • I would not say "irreducible factors correspond to roots of the polynomial". You can extend the field to find roots, but then your irreducible factors change. Why not "degree one factors correspond to roots of the polynomial"? – Marc van Leeuwen Feb 25 '14 at 17:02
  • @Marc: The precise statement is that there is a one-to-one correspondence between irreducible factors of a polynomial over a field and Galois orbits of roots of the polynomial in its splitting field -- e.g. complex conjugate pairs for the non-real roots of real polynomials. –  Feb 25 '14 at 17:04
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For a polynomial $p(x)$ of degree $n$, if we know all the $n$ (complex) roots $r_1, r_2, r_3, ..., r_n$, where some may be equal, we know exactly what the polynomial is:

$$p(x) = C(x-r_1)\cdot(x-r_2)\cdot(x-r_3)\cdot...\cdot(x-r_n)$$

So, in a sense, all the information about the polynomial is contained "within" the roots of the polynomial.

naslundx
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The importance of zeros of functions comes into play when you start getting into complex analysis, where instead of dealing with functions of your everyday numbers, you deal with functions of complex numbers. In complex analysis, both zeros and their opposites (called poles, for which $1/f(z)) = 0$) allow you to know almost everything about the function. Pretty much everything you do is either talking about poles or talking about zeros, and from that you can solve much harder problems like finding infinite sums and integrals.

The main idea is that many times you can write a function as a product of other, simpler functions. This is the same as with factorization of polynomials, where you can factor a polynomial in terms of very simple functions like $x^2-1 = (x+1)(x-1)$. Because the right hand side is a product of two functions, it can only be zero when either $x+1=0$ or $x-1=0$.

Going all the way, if you have a nice function $f(z)$ (no poles and differentiable), you can factorize it as $g(z)=f(z) (z-a_1)(z-a_2)...$, where $\{a_i\}$ is a set of complex numbers (possibly repeated), and $f(z)$ is a function with no zeros. In particular, $f(z)$ will be either a constant or a product of $e^{(z^k)}$'s. This is the Weierstrass factorization theorem. A lot of the time however, the $f(z)$ will just be a constant, meaning that most of information needed to define the function is located with where the zeros are, and their multiplicity (the number of repeats of a zero in the above factorization). There are technicalities to worry about, but I'll skip them for clarity.

In the case of the Riemann-Zeta function $\zeta(z)$, the point at which primes and the importance of zeros comes into play is when you write the function as a product. $\zeta(z)$ is defined at the start as:

$$\zeta(z) = \sum_{n=1}^{\infty} \frac{1}{n^z}$$

So $\zeta(2) = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...=\frac{\pi^2}{6}$. Making a product out of it gives (after some clever tricks):

$$\zeta(z) = \prod_{\mbox{$p$ prime}} \frac{1}{1-p^{-z}}=\frac{1}{1-2^{-z}}\frac{1}{1-3^{-z}}\frac{1}{1-5^{-z}}...$$

So zeros of the zeta function become related to the prime numbers (although not in a straight forward manner since the factors above are not simple like $(z-c)$). There are more details left out, for instance the definition of $\zeta(z)$ has to be extended here, but this is where it starts.

If you get into complex analysis, you'll be able to solve integrals and sums that you couldn't do otherwise. Like getting that answer of $\frac{\pi^2}{6}$, I found out how that was calculated in a complex analysis class (it involved factorizing $\sin(z)/z$).

Barry Rowe
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