Spinor Info

Enough of politics and cats. Time to blog about math and physics again.

Back in my high school days, when I was becoming familiar with calculus and differential equations (yes, I was a math geek) something troubled me. Why were certain expressions called “linear” when they obviously weren’t?

I mean, an expression like \(Ax+B\) is obviously linear. But who in his right mind would call something like \(x^3y + 3e^xy+5\) “linear”? Yet when it comes to differential equations, they’d tell you that \(x^3y+3e^xy+5-y^{\prime\prime}=0\) is “obviously” a second-order, linear ordinary differential equation (ODE). What gives? And why is, say, \(xy^3+3e^xy-y^{\prime\prime}=0\) not considered linear?

The answer is quite simple, actually, but for some reason when I was 14 or so, it took a very long time for me to understand.

Here is the recipe. Take an equation like \(x^3y+3e^xy+5-y^{\prime\prime}=0\). Throw away the inhomogeneous bit, leaving the \(x^3y+3e^xy-y^{\prime\prime}=0\) part. Apart from the fact that it is solved (obviously) by \(y=0\), there is another thing that you can discern immediately. If \(y_1\) and \(y_2\) are both solutions, then so is their linear combination \(\alpha y_1+\beta y_2\) (with \(\alpha\) and \(\beta\) constants), which you can see by simple substitution, as it yields \(\alpha(x^3y_1+3e^xy_1-y_1^{\prime\prime}) + \beta(x^3y_2+3e^xy_2-y_2^{\prime\prime})\) for the left-hand side, with both terms obviously zero if \(y_1\) and \(y_2\) are indeed solutions.

So never mind that it contains higher derivatives. Never mind that it contains powers, even transcendental functions of the independent variable \(x\). What matters is that the expression is linear in the dependent variable. As such, the linear combination of any two solutions of the homogeneous equation is also a solution.

Better yet, when it comes to the solutions of inhomogeneous equations, adding a solution of the homogeneous equation to any one of them yields another solution of the inhomogeneous equation.

Notably in physics, the Schrödinger equation of quantum mechanics is an example of a homogeneous and linear differential equation. This becomes a fundamental aspect of quantum physics: given two solutions (representing two distinct physical states) their linear combination is also a solution, representing another possible physical state.

Having just finished work on a major project milestone, I took it easy for a few days, allowing myself to spend time thinking about other things. That’s when I encountered an absolutely neat problem on Quora.

Someone asked a seemingly innocuous number theory question: are there two positive integers such that one is exactly the π-th power of the other?

Now wait a minute, you ask… We know that π is a transcendental number. How can an integer raised to a transcendental power be another integer?

But then you think about \(\alpha=\log_2 3\) and realize that although \(\alpha\) is a transcendental number, \(2^\alpha=3\). So why can’t we have \(n^\pi=m\), then?

As it turns out, we (probably) cannot, but the reason is subtle and it relies on a very important, but unproven conjecture from transcendental number theory.

But first, let us rewrite the equation by taking its logarithm:

$$\pi\log n = \log m.$$

We can also divide both sides by \(\log n\), which leads to

$$\pi = \frac{\log m}{\log n}=\log_n m,$$

but it turns out to be not very helpful. However, squaring the equation will help, as we shall shortly see:

$$\pi^2\log^2 n=\log^2 m.$$

Can this equation ever by true for positive integers \(n\) and \(m\), other than the trivial solution \(n=m=1\), that is?

To see why it cannot be the case, let us consider the following triplet of numbers:

$$(i\pi,\log n,\log m),$$

and their exponents,

$$(e^{i\pi}=-1, e^{\log n}=n, e^{\log m}=m).$$

The three numbers \((i\pi,\log n,\log m)\) are linearly independent over \({\mathbb Q}\) (that is, the rational numbers). What this means is that there are no rational numbers \(A, B, C, D\) such that \(Ai\pi+B\log n+C\log m + D=0\). This is easy to see as the ratio of \(\log n\) and \(\log m\) is supposed to be transcendental but both numbers are real, whereas \(i\pi\) is imaginary.

On the other hand, their exponents are all rational numbers (\(-1, n, m\)). And this is where the unproven conjecture, Schanuel’s conjecture, comes into the picture. Schanuel’s conjecture says that given \(n\) complex numbers \((\alpha_1,\alpha_2,…,\alpha_n)\) that are linearly independent over the rationals, out of the \(2n\) numbers \((\alpha_1,…,\alpha_n,e^{\alpha_1},…,e^{\alpha_n})\), at least \(n\) will be transcendental numbers that are algebraically independent over \({\mathbb Q}\). That is, there is no algebraic expression involving roots and powers of the \(\alpha_i\), \(e^{\alpha_i}\), and rational numbers that will yield 0.

The equation \(\pi^2\log^2 n=\log^2 m\), which we can rewrite as

$$(i\pi)^2\log^2 n + \log^2 m=0,$$

is just such an equation, and it can never be true.

I wish I could say that I came up with this solution but I didn’t. I was this close: I was trying to apply Schanuel’s conjecture, and I was of course using the fact that \(\pi=-i\log -1\). But I did not fully appreciate the implications and meaning of Schanuel’s conjecture, so I was applying it improperly. Fortunately, another Quora user saved the day.

Still I haven’t had this much fun with pure math (and I haven’t learned this much pure math all at once) in years.

The other day, I ran across a cute geometry puzzle on John Baez’s Google+ page. I was able to solve it in a few minutes, before I read the full post that suggested that this was, after all, a harder-than-usual area puzzle. Glad to see that, even though the last high school mathematics competition in which I participated was something like 35 years ago, I have not yet lost the skill.

Anyhow, the puzzle is this: prove that the area of the two semicircles below is exactly half the area of the full circle.

I am going to insert a few blank lines here before providing my solution.

I start with labeling some vertices on the diagram and also drawing a few radii and other lines to help.

Next, let’s call the radii of the two semicircles as \(a\) and \(b\). Then, we have
\begin{align}
(AC)&= a,\\
(BD)&= b.
\end{align}Now observe that
\begin{align}
(OA) = (OB) = r,
\end{align}and also
\begin{align}
(CD)&= a + b,\\
(OD)&= a + b~- (OC).
\end{align}The rest is just repeated application of the theorem of Pythagoras:
\begin{align}
(OC)^2&= r^2 – a^2,\\
(OD)^2&= r^2 – b^2,
\end{align}followed by a bit of trivial algebra:
\begin{align}
(OC)^2 + a^2&= [a + b – (OC)]^2 + b^2,\\
0&= 2(a + b)[b – (OC)],\\
(OC)&= b.
\end{align}Therefore,
\begin{align}
a^2+b^2=r^2,
\end{align}which means that the area of the full circle is twice the sum of the areas of the two semicircles, which is what we set out to prove.

I guess I have not yet lost my passion for pointless, self-serving mathematics.