## 31 January 2009

How many solutions does a typical quadratic equation have? Discuss.

(I'm deliberately leaving the problem vague, because it's more interesting that way.)

Daniel Lecoanet said...

2 (counting multiplicity and allowing complex solutions)

thecooper said...

If we take 'typical' to mean average: one.

Considering only upward-opening parabolas (since the argument I'm about to make is symmetric with respect to concavity), we just need to see whether the vertex lies on, above, or below the x-axis.

There is a bijective correspondence between upward-opening quadratics with vertex below the x-axis and those with vertex above the x-axis. The former have two solutions; the latter have no solutions. Taking the mean pairwise, we have that on average the quadratics with vertex off the x-axis have one solution.

Clearly the quadratics with vertex on the x-axis have one solution, so this doesn't affect the mean number of solutions.

jd2718 said...

This average stuff is clever, but I think the question asked "typical" and I don't know that they really are that close.

The very idea of 'typical' is a tough one (I know you were vague on purpose, but still...)

Typical quadratics have 2 complex roots. Half the time, those roots are a ±0i. Typical quadratics have roots that are not rational.

I don't think I can say more.

Jonathan

Todd Trimble said...

The mean is infinite if we average over all finite commutative rings. I'll just point that the only constraint is that the number, for a fixed finite commutative ring R and counting multiplicity, be even; otherwise any number is possible depending on R. For example, in Z mod 15, there are four square roots of 4.

unapologetic said...

What about infinite commutative rings? I have no reason to believe that "most" rings allow "most" quadratics to have any solutions at all!

Anonymous said...

A "typical" quadratic is in how many variables? Surely more than two, maybe infinitely many if we average over all possible quadratics... but an equation in infinitely many variables isn't really properly called a quadratic, so I dunno.

Veky said...

n.

(I'm deliberately leaving the answer vague, because it's more interesting that way.;)

((I hope you get my point.))

Todd Trimble said...

I'll take this opportunity to admit that I goofed up big time: in a typical finite commutative ring, the elements which actually have square roots at all tend to be sparse. For example, in Z mod (p_1^k_1 ... p_n^k_n), the chance that an element has a square root at all is (about) 1 in 2^n. This is because the element must produce a square when you reduce modulo each p_i^k_i, which for each i is (about) 1 out of 2.

To get further with this, I'm gonna have to actually apply pen to paper!

nathan said...

Suppose we restrict to only considering quadratic equations in one variable with real coefficients, and only allow real zeroes.

If we let X_n={(A,B,C)|B^2-4AC>0 and A,B,C each in [-n,n]} as a set in R^3, then it might make sense to ask what the limit as n->infinity of m(X_n)/(2n)^3 is. (Note that forcing all three coefficients to be simultaneously bounded has some nasty drawbacks.)

If we restrict X_n to A,B,C each in [0,n], then I think the limit (now over only n^3) turns out to be 1/18+ln(2)/6.

Anonymous said...

I wonder if it's worth mentioning that the equation x² + 1 = 0 has an uncountable family of solutions over the quaternions. A quaternion ai + bj + ck is a solution whenever a² + b² + c² = 1.

Zygmund said...

How does one average over all commutative rings?

The answer can be anything you want. Consider the ring R[X_a]/I, where X_a is a series of variables with a in an index set A (R the real numbers) and I the ideal generated by X_a^2 + 1. Then the equation x^2 + 1 has card A solutions in the ring.

Aaron said...

thecooper - That's really clever! I'm going to remember that. :)

Kevin C. said...

If we were to approach the problem using random variables as Nathan suggests, it seems like a more natural setup might be to have all the coefficients be iid Gaussians with 0 mean and unit variance.

Anonymous said...

If we were to approach the problem using random variables as Nathan suggests, it seems like a more natural setup might be to have all the coefficients be iid Gaussians with 0 mean and unit variance.

The "right" normalization has variance (n choose i) for the i-th degree term in an n-th degree polynomial. Then the expected number of roots is exactly sqrt(n). See for example the Edelman-Kostlan article in Bulletin of the AMS (1995). Shub and Smale have a beautiful extension to real algebraic varieties: the expected number of real points in the intersection of random hypersurfaces of degrees d_1,...,d_k is sqrt(d_1*...*d_k).

So a typical quadratic has sqrt(2) real roots, in this model.