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Archive for the ‘puzzle’ Category

What’s in a picture

Photograph

Today’s guessing game is:

What is this a picture of?

 

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I have written many papers in the last 20 years. Recently, I was taking a trip down memory lane and was reading some of my older stuff. I happened on a pretty identity on one of my papers. Something I had forgotten. That particular identity probably has someone elses name attached to it. I don’t believe that I was the first person who discovered this algebraic identity, but I wouldn’t know where to start looking for the correct precedence.  If the identity has a name attached to it, I wouldn’t be surprised if the culprit for finding it first is pushing daisies.

Mathematics keeps on getting rediscovered after all.

 

The identity is as follows.

Consider a set of of s numbers (they can be real, rational, algebraic, or even more messy commutative algebra objects so long as for the most part their multiplicative inverses are well defined). Let us call this set

S= \{ \alpha_1, \dots , \alpha_s\}

And consider the set of permutations of the first s integers, where \sigma is one such permutation

\sigma\in Perm\{1, \dots, s\}

We are then instructed to take the sum

{\sum_{\sigma\in Perm\{1, \dots, s\}} } {\frac 1{(\alpha_{\sigma(1)}+\alpha_{\sigma(2)}+\dots +\alpha_{\sigma(s)}) }}{\frac 1{(\alpha_{\sigma(2)}+\dots+\alpha_{\sigma(s)})}}\dots {\frac 1{\alpha_{\sigma(s)}}}

The stipulation is that the sum has no infinities (the numbers are generic).

This sum is equal to

{\frac 1{\alpha_1 \cdot \alpha_2 \dots \alpha_s}}

As everyone can see, it’s an  obvious identity so the proof is left to the reader 😉

For some reason, looking at it I can imagine it appearing miraculously in the twistor formulation of  Yang Mills scattering amplitudes… well, this is just a random speculation.

 

 

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Many times in physics one wants to solve systems of ordinary second order differential equations (equations of motion for example). If the dynamics comes from a Lagrangian,  it is standard to try to put them into first order formalism by going to the Hamiltonian formalism and working in “phase space”. Once you get to this stage, hyou can try putting the system on a computer by evolving the equations of motion discretely. Many times this destroys certain aspects of the dynamics. However, if you do things right, you can get some things to work better then expected.

For example, in the Hamiltonian formalism of the Kepler problem, one would have a Hamiltonian of the form

H= \frac {p_1^2 + p_2^2}{2m} - \frac{K}{r}

where

r= \sqrt{x_1^2+x_2^2}

The sign indicates that one has smaller energy where $r$ gets smaller (the potential energy is attractive).

A naive implementation of the evolution of the system is given by evolving

p_i [t+\delta t ] = p_i[t] - \partial_{x_i} V[r[t]] \delta t

and

r_i[t+\delta t]= r_i[t]+ \frac{p_i}{m} \delta t

However, after staring at this for a while, one notices that the dynamics is not reversible: both x,p have changed, so going back by changing \delta t\to -\delta t does not get you exactly back to where you started.

There is a very nice fix to this problem: you think of momenta as being evaluated at half times, and positions at full times. This is, we get

p_i[t+\delta t/2] = p_i[t-\delta t/2]+ \partial_{x_i} V(r[t]) \delta t

and

x_i[t+\delta t]= x_i[t]+ p_i[t+\delta t/2] \delta t

and even though this looks almost identical to what we had before, it is now time reversible (just send $\delta t\to -\delta t$ and do appropriate shifts to check that you really get back to where you started).

This is called the leap-frog algorithm. For problems like the one above, it has rather nice properties. The most important one is that it preserves Liouville’s theorem (it keeps the volume element of phase space constant).

In examples like the one above, it does something else that is quite amazing. If you remember Kepler’s second law (sweeping equal areas in equal time intervals), it is the law of angular momentum conservation. I’ll leave it to you to find a proof that the above system sweeps equal areas in equal times around the origin x_{1,2}= 0. I learned this fact recently in a conversation in my office and I was quite pleased with it, so I thought it would be nice to share it.

This algorithm does quite well on a lot of other systems (like the one I’m studying now for my research). If you have a system with a lot of symmetries, sometimes the leapfrog algorithm will preserve a lot of these symmetries and also the conserved quantities, so that you can evolve the system for much larger values of \delta t without loss of information.

 

 

 

 

 

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As I contemplate a set of bubble sheets and a midterm coming up for my class,   I have to think that all those multiple choice tests that I took in my life ended producing a valuable skill. They taught me how to take the tests so that when I design them I can solve them and fix the errors that crop up. I can now make them with great confidence. After all that training, I am a professional.

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Today’s puzzle is really simple. It is a single number, and don’t worry about the formating: it is not essential.

Your job, if you decide to take it, is to figure out how this number was chosen.


3983166922118810678205990336564718434224120605664143701183608

7681419052507877828777197836786790614849623650815370930359235

0129704516432578314074470098769495753238915840085811644293357

6455898753759925348456032608650278150327110180834816760596303

09728652360528042842943610644525298135991037793081877485511245

2474332249873412259136132368731917415053983296042291396691439150

43327960030646869030670768248956738775708735485960504256334453

736488021205358726905569574806264667486581193984051009097136483

506525271749484080035733863870881763567000457650973784539927240

2043442156827044195038802613229625783142492845706197046207004041

55134209534719408986385965928333597344225594042950352896

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As I told you previously, my project for the summer was to learn some Python language. I’ve been toying with it for a bit and with a few lines of code I was happily getting output. It’s great that I don’t have to declare types before using variables! And it took me a while before I realized that indentation substitutes for all the curly brackets. So far, I find it rather intuitive.

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In case you ever wondered what the most prestigious mathematical competition for high school students looks like, you get to solve 6 problems, in two sets of three problems. For each set, you get 4.5 hours of uninterrupted work, with all the scratch paper you might need, you bring your standard geometry drawing tools (ruler and compass) and you go.

For the list of problems, the IMO organization has a list of the official problems. Some are quite entertaining to do. This year, after being piqued a bit, I worked problem one in about half an hour to 45 minutes (I still got it). I used to compete in these types of competitions when I was in high school.

In any case, the problems are elementary: they just use algebra, trigonometry, basic plane geometry and elementary number theory. Here is problem one:

Let n be a positive integer and let

a_1 , . . . , a_k (k \geq 2)

be distinct integers in the set {1, . . . , n} such that n divides  a_i(a_{i+1}-1) for i = 1, . . . , k − 1. Prove that n does not divide a_k(a_1-1).

If you ever have tried to design these problems, you will notice that the conditions are somewhat contrived. n appears in two places: both in the size of the original set, and in the division condition. Could one do without the first?

The answer is no. If all of the a_k are multiples of n, then the problem does not follow. But having all of them small must be important. The next thing one needs to do is understand the relation of divisibility better so that one can solve the problem. I’ll let you figure the rest out…

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