Archive for the ‘Math’ Category

How many quadratic equations with real roots are there? (Part II)

May 4, 2009

In the last post, we started looking at the question: of all possible quadratic equations with real coefficients, what fraction of them, f_{real}, have real roots?

To review, the roots of a quadratic equation, ax^2+bx+c are given by the quadratic formula: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. The roots are real when the thing inside the square-root, the discriminant, D=b^2-4ac is positive.

Last time, we tried using some probability arguments to see if we could make sense of f_{real}, but we didn’t get anywhere useful.  This time, let’s explore what we learn by noting that the condition that the disciminant is positive for real roots defines a surface in (a,b,c)-space that separates sets that give real roots from those that don’t, and so the ratio of the volume of the real-root space to the whole space ought to give us the fraction of possible quadratics with real roots.

How big is the volume of (a,b,c)-space? Infinite.  Let’s avoid that by assuming the space is finite and then we’ll take the limit to infinity later.  Specifically, let’s treat the space as a cube bounded in each direction by \pm r.  Then, the volume of (a,b,c)-space is:

V_{abc}=8r^3.

Now, to the space of real roots.  The condition that the discriminant is positive corresponds to the region defined by:

a\le \frac{b^2}{4c}.

The volume of this region is given by an integral:

\begin{array}{ll} V_{real} &=\displaystyle \int_{-r}^r dc \int_{-r}^ db \int_{-r}^{\frac{b^2}{4c}} da \\&\\&=\displaystyle \int_{-r}^r dc \int_{-r}^r db \left(\frac{b^2}{4c}+r\right) \\&\\ &=\displaystyle \int_{-r}^r dc \left(\frac{r^3}{6c}+2r^2\right) \\&\\& = 4r^3+\frac{r^3}{6}\left(\text{ln}(r)-\text{ln}(-r)\right) \\&\\& =4r^3+\frac{r^3}{6}\text{ln}(-1).\end{array}

Hmm, that’s funny.  What’s that \text{ln}(-1) doing there?  Doesn’t \text{ln}(-1)=i\pi?  But volume had better be real.  Did we make a silly mistake?

If we take a look at the calculation above, the only place shenanigans might show up is from the integration over c.  Let’s take a closer look.  One thing we ought to be able to do is split the integral like this:

\displaystyle \int _{-r}^r \frac{dc}{c}=\int_{-r}^0\frac{dc}{c}+\int_0^r\frac{dc}{c}.

Furthermore, in the integral over the negative range, we can make a change of variable:

\begin{array}{l} u=-c \\ dc=-du, \end{array}

and so we can rewrite the integral as:

\displaystyle\int_{-r}^r\frac{dc}{c}=\int_{r}^0\frac{du}{u}+\int_0^r\frac{dc}{c},

but there’s no reason we can’t also call the symbol u the symbol c instead!  So, let’s make that change and do a couple extra things to find:

\begin{array}{ll} \displaystyle\int_{-r}^r\frac{dc}{c} &=\displaystyle\int_{r}^0\frac{dc}{c}+\int_0^r\frac{dc}{c} \\ &\\&= \displaystyle-\int_{0}^r\frac{dc}{c}+\int_0^r\frac{dc}{c} \\&\\ & =0 .\end{array}

What the … happened?  Where’s the \text{ln}(-1)?  On the bright side, if this integral is zero, then V_{real}=4r^3, which is real.  And, on top of that, we find for the fraction of quadratics with real coefficients with real roots:

f_{real}=\frac{V_{real}}{V_{abc}}=1/2,

which is to say that there are exactly as many quadratics with real roots as without!

Since we’re asking a question about volumes, the answer had to be real, and so the second version of the integral over c as got to be right.  It’s also the answer you’d get if you picture the volume defined by a\le\frac{b^2}{4c} and think hard for a couple minutes.

But wait, so the integral of \frac{1}{c} over a symmetric domain should be zero you say, but we still have that result from before that \text{ln}(r)-\text{ln}(-r)=\text{ln}(-1)\not= 0.  What do we do about that?

Let’s take a look at what we mean by \text{ln}.  The logarithm is defined by:

\text{ln}\left(e^x\right)=x.

So, for what value of x do we get e^x=-1? Euler tells us that e^{i\theta} = \text{cos}(\theta)+i\text{sin}(\theta) (and it’s not too hard to prove it, but the proof is reasonably rigorous, and thus has no place in this post!).  We get -1 for \theta = \pi, and so \text{ln}(-1)=i\pi.

But wait! We also get e^x=-1 for x=-i\pi or, most generally, for x= (2n+1)i\pi for all integer n.

So, \text{ln}(-1)=(2n+1)i\pi!  So many answers, and none are zero!

I bet Euler would say, if we’ve got an infinite number of perfectly good answers, the average is really what we’re looking for.  What’s the average answer? For every positive answer, there’s a corresponding negative answer, and so the average is zero!

So there you have it, another reason to think that \displaystyle \int_{-r}^r\frac{dc}{c}=0. (In general this answer for that integral is known as it’s Cauchy Principle Value.)

If it’s not clear by now, I don’t understand how modern mathematicians’ clear this confusion up.  For the problem of interest, we get zero for that integral three ways (change of variables, picturing the volume, and averaging the multivalued function), but not all ways.  Anyone care to explain?

To to summarize, if you shenanigan the integral correctly, one finds that the fraction of quadratics with real coefficients that have real roots is a half.

And if you’re still unhappy, there’s (at least) one more reason to believe that f_{real}=\frac{1}{2}.  For every choice of (b,c) a choice of a is either above or below the boundary set by a=\frac{b^2}{4c}.  Since the space of possible a is infinite, then partioning that space at any one point yields two semi-infinite spaces, and so half of the total space leads to real roots.

How fun is it that such a simple question leads to such sneaky mathematics?

Next time, let’s look at something a little less messy.

Update (5/4/2009):  The internet has a little chatter about this question and it seems f_{real}=1/2 is not the most popular answer.  Take a look at this thread or this one to see why

f_{real}=\frac{1}{2} + \frac{1}{12} \left(\frac{5}{6}+ \text{ln}(2)\right)\approx 0.627

is a good answer too.

Question for the masses: how can there be multiple good answers?

How many quadratic equations with real roots are there? (Part I)

May 3, 2009

The answer: infinity.

Okay, that wasn’t a useful question, so let’s ask a better one.  Of all possible quadratic equations with real coefficients, what fraction of them, f_{real}, have real roots?

As all children know, the quadratic equation is a x^2 +bx +c and its roots are the values of x for which the equation equals zero.  Everyone, adults included, also knows that the roots can be found with the quadratic formula:

x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

For real coefficients, the roots are real only when the quantity inside the square-root is positive.  This quantity is called the discriminant, and so we’ll give it the symbol D and talk about it for the rest of the post.

D=b^2-4ac

The question about real roots is a question about the discriminant: of all possible choices of a, \, b,\, and c, how often will the discriminant be positive?

To get a feel for what the answer might be and to show an example of how to introduce probability into interesting places where it ought to have no business, as a first step, we might ask, what’s the average value of the discriminant?

More specifically, let’s assume a,\,b and c are each drawn from a uniform distribution, and, since infinite domains are a pain, let’s assume the allowed choices of each coefficient are bounded by \pm r.  We’ll take the limit as r\rightarrow \infty later.  This is a model for a process like: “I throw a dart at a number line three times, and each point of impact gives me a coefficient.”  We could choose other distributions, but this one is in some sense the most “symmetrical” and thus best.

Under the uniform selection assumption, the average value of the discriminant is easy to find:

\bar{D}=\frac{1}{8r^3}\displaystyle \int_{-r}^r dc\int_{-r}^r db\int_{-r}^r da \left(b^2-4ac\right)= \frac{r^2}{3}

Well, that’s interesting.  On the one hand, if we take r to infinity, we get \bar{D}\rightarrow \infty, which is bad—screwy stuff happens when we try to think about things with infinite mean.  That the mean is infinite probably isn’t suprising.  Unless it’s exactly zero, then either b^2 or 4ac is bigger on average, and since both can go to infinity, the only plausible answers for \bar{D} are 0,\pm\infty

On the other hand, for any finite r, \bar{D} is positive, and so we might expect that quadratics with real roots are more common than quadratics with imaginary roots, and so maybe f_{real}>\frac{1}{2}.

Further ignoring 200 years of mathematics, let’s look at the standard deviation, \sigma_D, of the discriminant, since, if it’s small relative to the mean, maybe we can trust the mean anyway.  The idea is that sometimes things that give mathematicians strokes are still informative.  As they say in regard to somewhat related issues:

There are in this world optimists who feel that any symbol that starts off with an integral sign must necessarily denote something that will have every property that they should like an integral to possess. This of course is quite annoying to us rigorous mathematicians; what is even more annoying is that by doing so they often come up with the right answer.

McShane, E. J.
Bulletin of the American Mathematical Society, v. 69, p. 611, 1963.

Calculated in the usual way, we find:

\sigma_D= \frac{2}{15}\sqrt{105}r^2

Well, that’s infinite too in the limit, but how big of an infinite is it?  If we look at the standard deviation over the mean, we see:

\frac{\sigma_D}{\bar{D}}=\frac{2}{5}\sqrt{105}\approx 4.1

and so, infinite or not, the standard deviation is bigger than the mean, which tells us that I probably shouldn’t bet on f_{real}>\frac{1}{2} no matter how much I adore a nice, sloppy argument.

Some of you may be screaming by now: this can’t be the best way to do this! And you’d be right.  The condition that the discriminant is positive defines a surface in (a,b,c)-space that separates coefficient sets that lead to real roots from those that don’t.  It seems plausible that the volume of real-root space relative to the volume of the whole space will tell us what fraction of quadratics have real roots.  We’ll pick this up in the next post, wherein we’ll also learn more about that quote by McShane.