Math Game

[KrazyHorse]

Oooppps....

you know what I meant, even though I put the brackets in the wrong place. Are the a and b generalised? All a and all b?

[loinburger]

Originally posted by KrazyHorse
Oooppps....

you know what I meant, even though I put the brackets in the wrong place. Are the a and b generalised? All a and all b?

Yes, all a and b within the domain of *

[Juggernaut]

Answer to mine:
n^2 - (n - 1)^3

[KrazyHorse]

For techno:

((b*a)*b)*(b*a) = a*(b*a) by performing reduction on the leftmost brackets

but, substituting c = b*a into the expression ((b*a)*b)*(b*a)
we get (c*b)*c = b by reduction

Therefore, by transitivity of =, we get b = a*(b*a)

[Ramo]

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.

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.

[loinburger]

Originally posted by Ramo
So if [integral]{-i(c), i(c)}(1 + 2(c1)t^2)^.5 dt <= 4, you've got your answer.

Woot! Got that one right.

[KrazyHorse]

For the {N} thing:

We see that for all n&N (natural numbers), (n^2-n)lessthanNlessthan(n^2+n) gives {N} = n (simple enough to prove...just compare to (n-.5)^2 and (n+.5)^2 and convince yourself that the boundaries don't ever cross form one natural number to the next)

So, each n has 2n N's associated with it (if you catch my meaning), and if you stare hard enough at the first 15 or so terms, you can see that we can rewrite the summation as a double summation, with the inside summations each having 2^(n+1) terms...

Anyhoo, this starts to get a bit nasty, and since I'm typing this, I'll just state what happens when you collapse the first summation:

sum from n=1 to infinity of (2^{-n^2+2n}+2^{-n^2})*(1-2^{-2^n})

This is sort of disgusting in itself. I'll come back when I'm able to evaluate it.

[Juggernaut]

Let lg A = x and lg B = y.
Define the expression AB + lg(A^3 * B^2) in terms of x and y.

[Ramo]

Log base what? e?

A = e^x
B = e^y

e^(x + y) + ln(e^(3x) * e^(2y))
e^(x + y) + 3x + 2y

[KrazyHorse]

Okay. Rearranged a slightly different way (you'll have to trust me; I can prove it, but it would take a while):

1 + 2*sum_{n=1 to infinity}2^(-n)

As this is much easier to evaluate, I'll do so:

3

[Juggernaut]

10th-logaritm, custom.

[Juggernaut]

but you're of course correct.

[loinburger]

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."

[KrazyHorse]

That's just nasty...

Did you get 3 for the {N} problem too?

[loinburger]

Originally posted by KrazyHorse
1 + 2*sum_{n=1 to infinity}2^(-n)

As this is much easier to evaluate, I'll do so:

3

Dammit. I knew the answer had to be three because it was converging there, but I spent an hour trying to turn the sum of the series into something possible to solve. (It wasn't really wasted time, though, since I finished the second part of the exam an hour early by virtue of the fact that I only turned in three of the six problems.)