Jun 082020



 Posted by at 2:11 am
May 222020






 Posted by at 12:16 am
May 052020









How about that.



And this one for the Bates plot:



 Posted by at 11:19 am
May 022020


Intriguing, isn’t it.















 Posted by at 8:26 pm
Jun 052019

And before I forget: Last week, wearing my release manager hat I successfully created a new version of Maxima, the open-source computer algebra system. As a result, Maxima is again named one of SourceForge’s projects of the week, for the week of June 10.

The release turned out to be more of an uphill battle than I anticipated, but in the end, I think everything went glitch-free.

Others have since created installers for different platforms, including Windows.

And I keep promising myself that when I grow up, I will one day understand exactly what git does and how it works, instead of just blindly following arcane scripts…

 Posted by at 1:50 pm
Sep 252018







A slide from Atiyah’s talk on September 24, 2018.










(%i1) assume(j>1);
(%o1)                               [j > 1]
                                  log(j) + 1
                   (- j) - 1          j
(%o3)                         0.02944508691740671
(%o4)                         0.02944508691730876
(%o5)                         0.02944508691730876
(%o6)                         0.1602598029967022




 Posted by at 12:27 pm
Feb 272018

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.

 Posted by at 11:22 am
May 312017

I am watching the morning news and it’s all about numbers. Some good, some not so good, some really bad. Here are a few, in descending order:

  • 2018: The year when Ottawa plans to introduce a new low-income transit fare.
  • 417: The provincial highway number of the Queensway, which has been reopened after yesterday’s huge crash.
  • 175.6: The amount of rain, in mm, that Ottawa received in the month of May.
  • 80: The estimated number killed by a massive ISIS terrorist bomb in Kabul.
  • 21: The highest expected temperature of the day and, incidentally, the entire week, in Centigrade.
  • 15: The new minimum wage, in Canadian dollars, as proposed by the Ontario provincial government.
  • 7: The age of a baby, in months, who died allegedly due to her mother’s negligence in Gatineau.

I thought of turning these bullet points into a numbered list, but that would have been too confusing.

 Posted by at 8:25 am
Dec 182016



 Posted by at 1:23 am
Apr 022016








 Posted by at 9:31 pm
Jun 182015

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.

 Posted by at 8:25 pm
May 242015




 Posted by at 1:38 pm
Oct 132014







 Posted by at 4:00 pm
Dec 312013

So the other day, I solved this curious mathematics puzzle using repeated applications of Pythagoras’s theorem and a little bit of algebra.

Now I realize that there is a much simpler form of the proof.

The exercise was to prove that, given two semicircles drawn into a bigger circle as shown below, the sum of the areas of the semicircles is exactly half that of the larger circle.

Again, I’m inserting a few blank lines before presenting my proof.















Once again I am labeling some vertices in the diagram for easy reference.

Our goal is to prove that the area of a circle with radius AO is twice the sum of the areas of two semicircles, with radii AC and BD. But that is the same as proving that the area of a circle with radius AO is equal to the sum of the areas of two circles, with radii AC and BD.

The ACO< angle is a right angle. Therefore, the area of a circle with radius AO is the sum of the areas of circles with radii AC and CO. (To see this, just multiply the theorem of Pythagoras by π.) So if only we could prove that CO = BD, our proof would be complete.

Since AO = BO, they are the sides of the isosceles triangle ABO. Now if we were to pick a point O on the line CD such that CO‘ = BD, the ACO and ODB triangles will be identical (CD being the sum of AC and BD by construction). Therefore, AO‘ = BO, and the ABO triangle would be another isosceles triangle with its third vertex on the CD line. Clearly that is not possible, so O = O, and therefore, CO = BD. This concludes the proof.

 Posted by at 8:16 am
Dec 292013

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
(AC)&= a,\\
(BD)&= b.
\end{align}Now observe that
(OA) = (OB) = r,
\end{align}and also
(CD)&= a + b,\\
(OD)&= a + b~- (OC).
\end{align}The rest is just repeated application of the theorem of Pythagoras:
(OC)^2&= r^2 – a^2,\\
(OD)^2&= r^2 – b^2,
\end{align}followed by a bit of trivial algebra:
(OC)^2 + a^2&= [a + b – (OC)]^2 + b^2,\\
0&= 2(a + b)[b – (OC)],\\
(OC)&= b.
\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.

 Posted by at 8:45 pm
Oct 112013






 Posted by at 9:40 pm
Sep 272013

It is now formally official: global surface temperatures did not increase significantly in the past 15 years or so.

But if skeptics conclude that this is it, the smoking gun that proves that all climate science is hogwash, they better think again. When we look closely, the plots reveal something a lot more interesting.

For starters… this is not the first time global temperatures stagnated or even decreased somewhat since the start of recordkeeping. There is a roughly 20-year period centered around 1950 or so, and another, even longer period centered roughly around 1890. This looks in fact like evidence that there may be something to the idea of a 60-year climate cycle. However, the alarming bit is this: every time the cycle peaks, temperatures are higher than in the previous cycle.

The just released IPCC Summary for Policymakers makes no mention of this cycle but it does offer an explanation for the observed stagnating temperatures. These are probably a result of volcanic activity, they tell us, the solar cycle, and perhaps mismodeling the effects of greenhouse gases and aerosols, but they are not exactly sure.

And certainty is characterized with words like “high confidence,” “medium confidence” and such, with no definitions given. These will be supplied, supposedly, in the technical report that will be released on Monday. Nonetheless, the statement that “Probabilistic estimates […] are based on statistical analysis of observations or model results, or both, and expert judgment” [emphasis mine] does not fill me with confidence, if you will pardon the pun.

In fact, I feel compelled to compare this to the various reports and releases issued by the LHC in recent years about the Higgs boson. There was no “expert judgment”. There were objective statistical analysis methods and procedures that were thoroughly documented (even though they were often difficult to comprehend, due to their sheer complexity.) There were objective standards for claiming a discovery.

Given the extreme political sensitivity of the topic, I think the IPCC should adopt similar or even more stringent standards of analysis as the LHC. Do away with “expert judgment” and use instead proper statistical tools to establish the likelihood of specific climate models in the light of the gathered data. And if the models do not work, e.g., if they failed to predict stagnating temperatures, the right thing to do is say that this is so; there is no need for “expert judgment”. Just state the facts.

 Posted by at 10:45 pm
Sep 272013

I’ve been hesitant to write about this, as skeptics will already have plenty to gripe about, I don’t need to pile on. And I swear I am not looking for excuses to bash the IPCC, not to mention that I have little sympathy or patience for skeptics who believe that an entire body of science is just one huge scam to make Al Gore and his buddies rich.

But… I was very disappointed to see plots in the latest IPCC “Summary for Policymakers” report that appear unnecessarily manipulative.

Wikipedia describes these as truncated or “gee-whiz” graphs: graphs in which the vertical axis does not start at zero. This can dramatically change the appearance of a plot, making small variations appear much larger than they really are.

To be clear, the use of truncated plots is often legitimate. Perhaps the plot compares two quantities that are of a similar magnitude. Perhaps the plot shows a quantity the absolute magnitude of which is irrelevant. Perhaps the quantity is such that “0” has no special meaning or it is not a natural start of the range (e.g., pH, temperature in Centigrade).

But in other cases, this practice can be viewed as misleading, intellectually dishonest (for instance, it is common for financial companies to manipulate plots this way to make their market performance appear more impressive than it really is) or outright fraudulent.

So here we are, the 2013 IPCC report’s summary for policymakers has been released in draft form, and what do I see in it? Several key plots that have been presented in truncated “gee-whiz” form, despite the fact that the quantities they represent are such that their absolute magnitudes are relevant, that their variability must be measured against their absolute magnitudes, and where zero is a natural start of the range.

I am presenting the original plots on the left and my crudely “untruncated” versions on the right:

This is not kosher, especially in a document that is intended for consumption by a lay audience who may not have the scientific education to spot such subtleties.

The document is still labeled a draft, with copy editing in particular yet to take place. Here’s to hoping that these plots (and any similar plots that may appear in the main report) are corrected before publication, to avoid the impression of trying to exaggerate the case for climate change. Scientists should be presenting the science objectively and leave the manipulation, even inadvertent manipulation, to politicians.

 Posted by at 10:22 pm
Apr 242012




Multiplying by 4 and switching the two sides gives


Adding these two equations together, we get


Subtracting \(9a\) from both sides, we obtain







(%i3) eq3:eq1+eq2$
(%i4) eq4:eq3-9*a$
(%i5) eq5:factor(eq4)$
(%o6)                                4 = 5


 Posted by at 9:22 am