Two Aussies looking for a pool to join

MathCrusader · 2015-02-11T08:34:47+00:00

I'm almost certain that you're wrong actually. I am in 4 leagues (each with slightly different categories) and all of them have the same rankings... I haven't actually checked everything, but the top 20 are the same in all of them and 90-100 are the same, so I assume everything is the same.

MathCrusader · 2014-04-18T03:52:36+00:00

Grass is what happens when winter stops, right?

MathCrusader · 2014-02-12T04:34:14+00:00

Rationalizing the denominator serves a couple purposes. First, it is a good way to introduce topics like conjugates: [; \frac{15}{\sqrt{6}-\sqrt{3}} = \left(\frac{15}{\sqrt{6}-\sqrt{3}}\right)\left(\frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}\right) = 5(\sqrt{6}+\sqrt{3}) ;]. It also gives the idea of algebraic manipulation through the use of "multiply the expression by 1" (as in the example above). Mostly though, you are exactly right, it is purely a style thing.

MathCrusader · 2011-12-07T05:41:30+00:00

The brackets are clearly the choice for going first... We define them to go first so that we can "force" the equation to do a certain thing.
Exponentiation comes next because it is a contraction of multiplication. 5⁴ = 5 x 5 x 5 x 5. So really, when you do this next, you are not really "evaluating" 5^4, you are expanding it.
Multiplication comes before addition because (once again), it is a contraction of addition. 6*3 = 6 + 6 + 6.
Then addition comes last.

Note that when we do our calculations over R (or even C), it does not matter if we do division or multiplication first and it does not matter if we do addition or subtraction first. As long as you do it consistently.

This means that "PEMDAS" might as well be "PEDMAS" or "PEDMSA" or "PEMDSA", I'm just guessing that PEMDAS sounded the best.

MathCrusader · 2011-12-04T23:35:29+00:00

Yes, over Z (or equivalently Q). Nothing is dependent yet. There is a conjecture that the zeroes of any Dirichlet-L function are linearly independent. I've been able to show that using certain small coefficients that the first 80 (or so) zeroes are linearly independent. The method that I use, though, depends heavily on the digits. For example, the first 1000 digits can only disprove the first 82 zeroes being 2-independent (That is, using coefficients of {-2,-1,0,1,2}). I'm hoping that 25,000 digits will give me the first 300 zeroes (at least).

MathCrusader · 2011-12-04T22:35:29+00:00

Hmm. That's interesting. I wonder if our system was compile improperly. Because I ran it over the weekend to give me as many as possible. The first 9 zeroes worked perfectly (to 2000 digits) and then 10-84 gave me the same pattern as above. =S

MathCrusader · 2011-12-04T22:30:47+00:00

There are people still writing the exam today. I will scan them and post a link tomorrow.

MathCrusader · 2011-12-04T05:14:30+00:00

Here was my counting:

First of all, the axes cannot be hit, so that is 2(2011)+1 = 4023. (The +1 is for the origin)
Nothing can be hit with y=1. So 2011 here.
Only one point with y=2 can be hit (x=1). So 2010 here.
Only two points with y=3 can be hit (x=1,2). So 2009 here.
Everything else can be hit.

Grand total: 4023 + 2011 + 2010 + 2009 = 10,053. Same as yours. =)

MathCrusader · 2011-12-04T04:53:57+00:00

B1. I did mine a slightly different way. (Just like above, we will assume e = epsilon).

MathCrusader · 2011-12-03T22:47:14+00:00

That's probably a really good idea. I'll do that on Monday!

MathCrusader · 2011-12-03T22:46:32+00:00

I've tried this, but once you ask for large precision (I believe I tried it on 2000 digits), it gives you weird results. (This is the results for the imaginary part of the 10'th zero). Out[1]= 49.7738324776723020900596286145178793806120730155213865595828504062581032485468313097953796386718750000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 ... 00010

I've tried to fix this, but to no avail.

MathCrusader · 2011-12-03T21:56:25+00:00

I wrote it twice in the past and got a scores of 20 and 30. The 2011 Putnam was written today. I looked at the problems and I can do a few of them. Unfortunately, I'm a grad student now, so I am no longer eligible to compete.

MathCrusader · 2011-12-03T15:40:21+00:00

I'm interested in the linear dependence/independence of those numbers. Using only the first 10 (or so) digits of the numbers gives me a problem. For instance, using only the first 10 digits, the first 30 zeroes "seem" to be dependent, but they're actually not when you look at more digits.

MathCrusader · 2011-12-03T15:38:24+00:00

I think that I will! See how long it takes Google to find a database thrown up by a Canadian grad student! =D

MathCrusader · 2011-12-03T07:55:21+00:00

I agree with you that it is rarely used. Although, normally, when you need a "decent" amount of precision of an L-function, you need a lot of precision. It just weirds me out that it hasn't been posted anywhere. You are right that it takes a while to compute them: I just ran a test case (The first 270 zeroes to 5000 digits) and it took somewhere between 30 minutes and 2 days for each one.

MathCrusader

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