Danny Ayalon can go fuck himself

Danny Ayalon is the Deputy Foreign Minister of Israel.  Both he and his boss, a racist from the Yisrael Beitanu [Israel Our Home] party named Avigdor Lieberman, represent an unfortunate trend of Israel being seen as less and less likable to its allies.  But it is Ayalon who is in my line of fire this week, for several reasons.

Ayalon made the news last month because of an incident which a skilled diplomat could have handled discretely and in a dignified manner.  Turkey, with whom relations with Israel have been going down the tubes, had one of its stations broadcast a program with anti-Semitic themes.  A quiet complaint and message of understanding the balance between freedom of expression and placing Jews in danger would have been appropriate.  But Ayalon, representing an increasingly tone-deaf Israel, made a mess of things:

Footage of Mr Ayalon urging journalists to make clear the ambassador was seated on a low sofa, while the Israeli officials were in much higher chairs, has been widely broadcast by the Israeli media.

He is also heard pointing out in Hebrew that “there is only one flag” and “we are not smiling”.

In an interview with Israel’s Army Radio on Tuesday, Mr Ayalon was unapologetic.

“In terms of the diplomatic tactics available, this was the minimum that was warranted given the repeated provocation by political and other players in Turkey,” he said, according to Reuters.

One Israeli newspaper marked the height difference on the photo, and captioned it “the height of humiliation”.

That, however, is not even my main problem with him.  Massachusetts Congressman Bill Delahunt is part of a contingent of Democratic lawmakers in Israel, traveling as part of a mission sponsored by liberal lobby group J-Street.  I will say that I have been warming to J-Street, although I do not like the arrogance of its identification as “pro-peace, pro-Israel”.  [So is the Zionist Organization of America…it just has vastly different definitions of these terms than does J-Street.]  Standard protocol is that such a delegation, if desired, may seek out an audience with Ministry officials.  However, Alayon, because he feels that J-Street is not in fact “pro-Israeli“, has locked the delegation out from any such meetings:

“We were puzzled that the Deputy Foreign Minister has apparently attempted to block our meetings with senior officials in the Prime Minister’s office and Foreign Ministry – questioning either our own support of Israel or that we would even consider traveling to the region with groups that the Deputy Foreign Minister has so inaccurately described as ‘anti-Israel,'” Delahunt continued.

“In our opinion this is an inappropriate way to treat elected representatives of Israel’s closest ally who are visiting the country – and who through the years have been staunch supporters of the U.S.-Israeli special relationship.”

Ayalon has clearly associated himself with the rabid right.  The combination of this tone-deaf outlook and utter incompetence as a diplomat is making for an explosive combination.  How dare he insult a delegation from the United States because their politics do not dovetail with his!  It is men like Ayalon that will be the downfall of Israel.  For that, he can go fuck himself.

Meanwhile, Michael Oren, in contrast to his dumbass administration colleague, is wisely making peace with J-Street.  Good for him.

An improper double integral

Problem: Evaluate the following double integral.

\(I=\displaystyle\int\limits_{-\infty}^{\infty}dx\int\limits_{-\infty}^{\infty}dy\; e^{-x^2-y^2-(x-y)^2}\)

Solution: This problem was taken from the collection “Berkeley Problems in Mathematics“, Problem 2.3.3.  Two solutions are given, neither of which are close to [stylistically] mine, which I give below.

First, change to polar coordinates; that is, \(x=r \cos \theta\), \(y=r \sin\theta\).   Using the Jacobian \(dx\,dy=r\,dr\,d\theta\), we get

\(I=\displaystyle\int\limits_{0}^{2\,\pi}d\theta\int\limits_{0}^{\infty}dr\,r\;e^{-2\,(1-\sin \theta\, \cos \theta)\,r^{2}}\)

Or…

\(I=\displaystyle\frac{1}{4}\,\int\limits_{0}^{2\,\pi}\frac{d\theta}{1-\sin \theta\,\cos \theta}=\frac{1}{4}\,\int\limits_{0}^{2\,\pi}\frac{d\theta}{1-\frac{1}{2}\sin \theta}\).

Now, there are two ways I can think of to go about evaluating this latter integral.  First, we can Taylor expand and hope for the best; it turns out that the resulting series converges to something well-known, but you have to be an expert at recognizing such things.  [I so hate solutions that require a deus ex machina like that.]  The other way is to convert to complex variables and use the Residue Theorem. [I’m afraid, however, if you are not familiar with the Residue Theorem, then we are back to a deus ex machina. But one has to draw the line somewhere, I guess…]

So, consider the following integral:

\(J(a)=\displaystyle\int\limits_{0}^{2\,\pi}\frac{d\theta}{1-a\,\sin \theta}\;:|a|<1\).

Observe that

\(\displaystyle \sin \theta=\frac{1}{2 i}\left (e^{i \theta}-e^{-i \theta} \right )\).

The trick is to recognize that we are integrating over the unit circle \(C\).  if we let \(z=e^{i \theta}\), and transform to an integral over \(z\), then the result is the following:

\(J(a)=-\displaystyle\frac{2}{a}\,\displaystyle\oint\limits_C \frac{dz}{z^2-i \frac{2}{a}\,z-1}\).

Recall that a residue of a function \(f\) at \(z=z_0\) is equal to

\(\mbox{Res}(f;z_0)=\displaystyle\lim_{z \to z_0} (z-z_0)\,f(z)\),

with the Residue Theorem stating that, for a function \(f\) having simple poles \(\displaystyle\{z_n\}_{n=1}^{N}\) within the simple closed curve \(C\), then

\(\displaystyle\oint\limits_C dz\,f(z)=i\,2\,\pi\,\sum\limits_{n=1}^{N} \mbox{Res}(f;z_n)\).

To compute \(J\) using the Residue Theorem, we must compute the roots of the quadratic in the denominator of the integrand.  These roots are at

\(z=z_{\pm}=\displaystyle\frac{i}{a}\left (1 \pm \sqrt{1-a^2} \right )\).

Note that \(\displaystyle |z_{+}|>1\), so that we need only consider the root \(z_{-}\).  Hence,

\(J(a)=-\displaystyle\frac{2}{a}\,i\,2\,\pi\,\frac{1}{z_{-}-z_{+}}=\frac{2 \pi}{\sqrt{1-a^2}}\).

Finally, the result is

\(I=\displaystyle\frac{1}{4}\,J \left (\frac{1}{2} \right )=\frac{\pi}{\sqrt{3}}\).

Is it cheating to use a symbolic math computer to do your homework?

Fascinating demonstration given by Conrad Wolfram of Wolfram Research at TEDx, concerning the question of whether or not one cheats by using Wolfram Alpha to do your integrals for you.

The short answer to the question is that there is cheating going on, but not in the way someone who asks this question would think. The gist is that, as Wolfram claims, about 80% of math education consists of hand computations: computing integrals, derivatives, limits, roots, matrix inverses, etc. But not only is this all incredibly boring, but it also ill-prepares students for the real mathematical challenges out there. Really, the challenge is to teach students how to translate real-world problems in business, engineering, etc., into a mathematical language. Once the pure computation problem is set up, then a machine like Wolfram Alpha can turn the crack and generate data. The remaining challenge is to figure out how to interpret the data, and such an interpretation does not lend itself to a black/white solution.

Another point that Wolfram makes is that calculus should be taught a lot earlier than it is now. When, he does not say, but he makes the case that there are concept in calculus, namely the limit, that a “3 or 4 year-old” could grasp. He points to a terrific visual example of using inscribed polygons to approximate \(\pi\).  The greater point is that math education in the US needs a radical reshaping, and that computers are crucial in this reshaping.  The cheating done, in the meantime, is not by the students, but to the students, because they are being told that the computational tools they will use int he real world to solve problems are viewed as verboten in school.

In my opinion, Wolfram has a number of terrific points and his demonstration is valuable and should be viewed by anyone with an interest in math education.  But ultimately, Wolfram’s proposals would create a generation of students with too much trust in the computer, and by extension the people who program the computer.  One must remember that Wolfram is in the business of providing computational engines, and the stock of his company rises if the people behind his company are seen as the gatekeepers to a mysterious technology.  It is not unlike the trend of making automobile engines more computerized and less able to be worked on by average people.  By saying that the messiness of computation is boring and turns off students, we increase the reliance of math professionals on the computer and leave out the crucial skill of checking the computer for errors.

I have had quite a bit of experience with this issue in my work at IBM.  In semiconductor lithography, one of the main challenges is to simulate the physical processes involved in imaging circuit patterns on a wafer.  The calculations involved in this simulation are extremely complex and very heavy-duty.  We did rely on software packages to do a lot of this, but most of the time, the models we built with these software packages led us astray.  Was it a problem with the data, or was the computer lying to us, or, even more subtly, was the computer telling us the truth but we were making false assumptions about that truth?  The problems in trying to answer these questions were severe: taking data on the few running machines we had was expensive and getting time was difficult.  The software vendors were always too busy to answer our difficult questions about the integrity of their computational models.  The only practical way to deal with this was for IBM to have someone who could devise simple tests that would reverse-engineer the engine’s algorithm and assess from where mistakes were coming.

That someone was invariably myself, as I had all the necessary background, both from my schooling and my work experience.  I knew how to look under the hood.  More importantly, I knew how to derive the equations that went under the hood.  And many of these equations weren’t simple expressions that could be typed into Wolfram Alpha.  Rather, such equations required careful geometrical reasoning and pattern matching that was difficult, if not impossible, with which to trust such a tool as Wolfram Alpha.  In fact, I found it best to be completely distrustful of the computer as I was building my test cases.  These test cases would be designed so as to be hand computable, yet nontrivial.  Once these test cases were designed and computed, then the diagnosing of problems could commence.

Furthermore, without someone to understand how to plumb the depths of how computations are done, we would not get users who can diagnose incorrect results at the chip level.  That’s right, recall Intel’s Pentium FDIV error.  Finding this error took a forensic approach to computation – an approach that none of us would have in Wolfram’s world, as none of us would deign to even think about so lowly an operation as division.  And, irony of all ironies, Wolfram’s flagship product, Mathematica, has not been without its own problems over the years – not just standard-issue software bugs, but incorrect algorithms.

As to the point Wolfram makes that calculus can be taught a lot earlier – making allusions to 3 or 4 year-olds.  I’m not so sure.  Yes, the basic calculus concept of the limit is easy to grasp, but beyond the most superficial level it is essentially a deus ex machina.  Further, applying those limits to sequences and series involves the culmination of everything a typical calculus student has learned.  Sloppy analytical techniques leads to an inability to solve problems, even if the calculus concepts are well understood.  I have a terrific example of this from my days as an undergraduate tutor in the Math Dept at UMass.  I used to sit in the calculus drop-in centers for students taking the business calc [Math 127/128 for those of you who know of which I speak].  Now, I admit, this was not the calculus that one with serious mathematical curiosity took, but still.  Anyway, at some point in time, the students were required to perform double integrations of polynomials over 2 variables, and come up with a number as an answer.  The drop-in center got real busy with folks who were simply perplexed.  A typical conversation would go like this:

  • Me: So, tell me, what’s troubling you?
  • Student: I can’t do these integrals!
  • Me: Well, why don’t you do this one in front of me, and let’s see what’s wrong.
  • Student: OK.  So first I do the integral over y…is that right?
  • Me: Yes.
  • Student: Now I do it over x.  is that right?
  • Me: Looks good.
  • Student: Now I plug in the limits and…it gives me a different answer than what the answer key tells me.
  • Me: that’s because you added wrong.  1/2 – 1/3 = 1/6, not what you wrote.
  • Student: huh?  I don’t understand?
  • Me: Do you know how I got 1/6?
  • Student: No.

So, what we learn here is that the student understood the mechanics of integration, but couldn’t add fractions.  How is such a student supposed to comprehend a result from Wolfram Alpha?

So, I disagree that the mechanics of computation are best left to the experts.  I do think that there is a place for learning the mechanics of a root solve, or an integration – in fact, many, many such operations – as a part of math education.  I do agree that computers should play a greater role in math education, and perhaps elements of calculus could be taught earlier.  But hand computation is essential if we are going to educate a class of people ready to question authority.

Which is bigger?

Problem: Which is bigger…\(3^{\pi}\) or \({\pi}^{3}\)?

Solution: Write …\(\pi=3+\delta\), where \(\delta>0\) is the fractional part of …\(\pi\).  Then, in terms of \(\delta\):

\({\pi}^{3}=(3+\delta)^3=27+27 \delta+9 \delta^2+\delta^3\).

On the other hand…

\(3^{\pi}=3^{3+\delta}=27 \times 3^{\delta}=27 e^{\delta \log 3}\).

Now we make the observation that \(\log 3>1\), and use the expansion for the exponential function

\(e^x=1+x+\frac{1}{2} x^2+\frac{1}{6} x^3+\ldots\)

Given these facts, we can then state the following:

\(3^{\pi}>27+27 \delta+\frac{27}{2} \delta^2+\frac{9}{2} \delta^3>27+27 \delta+9 \delta^2+\delta^3\)

And therefore \(3^{\pi}>{\pi}^3\).