Math Game

[Buck Birdseed]

Not what I meant. We're not talking a closed shape here, the top of the shape is just a straight, flat surface of any length. Think bowl/glass.

[KrazyHorse]

I think it's a half-sphere. I haven't proved it yet, but I'm getting there.

[KrazyHorse]

Okay. It is a half-sphere, and this is obvious from the closed-container question. Why? Because we can see that by doubling any open container we can get a closed container, with double the volume and double the surface area. If there were a more efficient form than the half-sphere, then it's doubled, closed form would be more efficient than the sphere, which we can't allow.

[loinburger]

Originally posted by shade
maybe you like this one:
create 24 by using +,-,x,:
and the numbers 1,3,4,6
You may only use the numbers 1 time each and you have to use them 1 time each.

This isn't a trick question, right?

[Dauphin]

What is the sum of the infinite series:

1 + ¹/₄ + ¹/₉ + ....+ ¹/_n²

[Roman]

Originally posted by Big Crunch
What is the sum of the infinite series:

1 + ¹/₄ + ¹/₉ + ....+ ¹/_n²

Hmm, isn't this just an ordinary geometric progression? I may be mistaken.

[Ramo]

No, a geometic series has a common ratio between each term.

What is the sum of the infinite series:
1 + 1/4 + 1/9 + ....+ 1/n2

As a Riemann sum, this can be approximated (I think ) by 1 + ([integral]{2, infinity}dn 1/n^2)/1

Which is 1 + (-1/infinity + 1/2) = 3/2

[Dauphin]

It passes 3/2 at n=7; try again.

[shade]

Originally posted by technophile

Originally posted by shade
maybe you like this one:
create 24 by using +,-,x,:
and the numbers 1,3,4,6
You may only use the numbers 1 time each and you have to use them 1 time each.

This isn't a trick question, right?

No it isn't,although it looks like it can not be that hard to solve,it took me about an hour to find the correct solution(yes really there is one)
(got it from a mathematesian in my class )

What is the sum of the infinite series:
1 + 1/4 + 1/9 + ....+ 1/n2

isn't that e ???(2,7182818284)


Shade

[SnowFire]

Well, I happen to know the sum of (sigma) 1/n^2 as n->infinity, but proving it is the problem. Apperantly it was originally proved back in the 19th century by some unknown mathematician called... Leonhard Euler. It was his big break in math and ever after he was pretty famous.

Anyway, it is pi^2/6. Always fun, you get to factor sin x, multiply it out, factor out a pi^2 over 6 and a 1/3! and you get them on the other side equalling the series 1/n^2. Really odd, I never would have thought to find the sum of that series through sin x of all functions.

Here's a REAL problem: What is (sigma) 1/n^3 for n->infinity? If you can figure this out, inform your nearest math department, since you might have won a prize in mathematics since I think that problem is still unsolved.

[Chowlett]

Originally posted by KrazyHorse
A sphere has the most efficient volume/surface area ratio.

You can prove this locally (show it must have constant Gaussian curvature), but it requires a pretty decent knowledge of differential geometry. There might be an easier way to prove it, but this is the method I've seen.

Is this calculus of variations? We've just been doing this and I reckon that would be a decent way to solve the problem. Couldn't flesh it out though.

I should be able to do most of the rest, and probably could when fully awake, and with course notes beside me - curvature is a case in point, we did it last year but I couldn't do it now without notes.

[Goingonit]

Originally posted by SnowFire
Argh, I should have mentioned that. It's continuous. Otherwise, the answer is obvious (let x be 0 for all irrational numbers, e for all rational numbers).

If f(x) is continuous, the answer is no. This is because the number of irrational numbers is an Aleeph-one infinity, and the number of rationals is aleph-0. If the function is continuous, then you'd need to have an infinity of rationals for each irrational. Wouldn't this be imopssible?

[KrazyHorse]

Originally posted by Chowlett


Is this calculus of variations? We've just been doing this and I reckon that would be a decent way to solve the problem. Couldn't flesh it out though

I guess so. Theoretically, though, you'll have to extend to functions of two variables. Shouldn't be that hard, but I've never actually done it myself.

[SnowFire]

Goingonit: You're basically correct. Look at the number of values f(x) takes on. f(x) takes on aleph-null values at most on the rationals because there are only aleph-null rationals. On the irrationals, f(x) takes on aleph-null values because we're restricted to rationals and there are only that many rationals. So our range has a size of aleph-null, but in order to be continuous, you have to have an interval, and any interval, no matter how small, has c values. Hence this function doesn't exist.

I have to admit I've never seen that notation before, aleph-one? I always knew a countable infinity as aleph-null, and the uncountable infinity as c (and of course you can make as many infinities bigger than c as you like, but we don't care about them. And there may or may not be an infinity bigger than A_0 but smaller than c, but it's impossible to find.).

[Goingonit]

Cantor defined three groups of infinities. Aleph-null is countable, Aleph-one is the infinity of points in a line,and other things. Aleph-two is the number of geometric curves, which is larger than the number of points in the line.