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Ramo
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Austin, Texas, USA
Oct 1999
time: 23:16
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If it's all a and all b, you could simply insert any a and any b into the assumption, and you have your proof.
quote:
You have a circle of radius 1 in which is enclosed a parabola. Is it possible for the parabola's arc to be longer than 4? |
Where c: [0, 2pi] -> R^2 and c(t) = (cost, sint)
Arc Length of circle =
[integral]{over circle}||c'(t)||dt
= [integral]{0, 2pi}dt = 2pi
p[-i, i] -> R^2 and p(t) = (t, ct^2)
Maximize arc length
d([integral]{i, f}(1 + 2ct^2)^.5 dt)/dc = 0
(i^2 + ci^4) = 1
i^2(c + i^2) = 1
ermmm..
*solve for i* 
d([integral]{-i(c), i(c)}(1 + 2ct^2)^.5 dt)/dc = 0
*do math*
c = some constant c1
So if [integral]{-i(c), i(c)}(1 + 2(c1)t^2)^.5 dt <= 4, you've got your answer.
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loinburger
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Sweet Sauce Jones
Jul 1999
time: 00:16
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Here's a problem from my Comp Sci class. I've asked most of the math and comp sci professors here if they knew how to solve it, but no dice.
The @ function is the floor function: chop a positive number's decimal value to produce an integer.
Given the recursive relation:
A[0] = 0; A[1] = 0; A[2] = 1;
A[N] = A[@(N/2)] + R[@(N/2), @(N/2)+1];
R[0,U] = A[U]; R[S,0] = 0;
R[S,U] = U + (R[@(N/2),@(U/2)] + [sum from i = 0 to @(U/2) : 2 * R[@(S/2), U - i]]) / U;
For N approaches infinity, what does A[N] equal in terms of N? What does A[N] equal in terms only of N (not of A[N] or R[S,U])?
I've written a computer program to determine the answer, and running it for up to N = 64 it appears that A[N] is converging to N*log2(N), where log2 is log base 2. However, I can't test the problem any higher than around N = 64 because the recursive algorithm runs in exponential time; A[75] would take hours to compute, A[100] days or weeks, A[1000] might run until the end of time.
The background behind the problem is in relation to a sorting algorithm that I've come up with, where A[N] gives the algorithm's average case. I had to present the algorithm the other day, but couldn't say what the average case was other than "it looks like it's approaching N*log2(N), but I can't really say for certain since I've only tested it up to N=64."
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SnowFire
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New York City, NY
Jan 1970
time: 00:16
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I got like a 35 (out of possible 120) when I took it, and I was 2nd highest at McGill.
Darn. You get at least my congratulations... I took the stupid thing Freshman year when I was a lot more arrogant then I am now, and unsurprisingly got the most popular score on the Putnam, the good old zero.
Anyway, I've got a few questions for everybody, except that my Geometry teacher probably wouldn't approve of me posting questions from the take-home test before I hand it in. Anyway, the problem that should have been the easiest is probably the most annoying right now... a stupid little proof in projective geometry about some points being collinear, and it's such a problem that you don't even have to be inventive with placing points you know the coordinates of and using variables. It's just straight calculation with numbers, and it ISN'T COMING OUT! In fact, the solution keeps getting worse and worse. It started out with of the 3 points, two of the lines between them were the same. I made an adjustment after discovering one mistake, and there were still 2 lines the same. And after finding yet another one (or possibly 2), I now have 3 completely different lines going between these points.
Anyway, here's an enjoyable little problem: Does there exist such a function that f(x) is rational on every irrational number and irrational on every rational number? If there is, give an example, if there isn't, explain why.
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02-12-2001 07:47
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