02-12-2001 22:19									#61
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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.

	02-12-2001 22:21									#62
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I think it's a half-sphere. I haven't proved it yet, but I'm getting there.

	02-12-2001 22:26									#63
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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.

	03-12-2001 00:07									#64
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quote: 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?

	03-12-2001 00:39									#65
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What is the sum of the infinite series:

1 + ¹/₄ + ¹/₉ + ....+ ¹/_n²

	03-12-2001 00:55									#66
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quote: 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.

	03-12-2001 01:25									#67
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No, a geometic series has a common ratio between each term.

quote: 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

	03-12-2001 02:19									#68
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It passes 3/2 at n=7; try again.

	03-12-2001 02:53									#69
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quote: Originally posted by technophile quote: 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 )

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

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


Shade

	03-12-2001 03:44									#70
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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.

	03-12-2001 04:57									#71
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quote: 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.

	03-12-2001 06:22									#72
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quote: 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?

	03-12-2001 07:22									#73
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quote: 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.

	03-12-2001 13:06									#74
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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.).

	03-12-2001 16:59									#75
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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.

	03-12-2001 18:41									#76
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I have one for you:

let k and p be n-dimensional vectors, a and c are real numbers, and eps is a real number which is very small.

What is the n-dimensional integral of (k^2 + 2*k.p +c -i*eps)^(-a) over the infinite volume. You may take eps->0 (it only acts as a regulator).

	09-12-2001 16:11									#77
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differential equations don't seem a problem but simple sums are

lets try an eaier one
create 20 by using four 4's.
(eg to create 8=> 4+4-4+4)
(this really is easy)

Shade

	09-12-2001 20:26									#78
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quote: Originally posted by shade differential equations don't seem a problem but simple sums are lets try an eaier one create 20 by using four 4's. (eg to create 8=> 4+4-4+4) (this really is easy) Shade

Ha!

(4+ 4/4)*4

That's too simple

	10-12-2001 02:03									#79
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Try this one:
A goat is tied to a pole on the circumference of a circular lawn with a 15 meter long leash. The radius of the lawn is 20 meter. How much of the lawn can the goat eat?

	10-12-2001 02:30									#80
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quote: Originally posted by uncle_funk Try this one: A goat is tied to a pole on the circumference of a circular lawn with a 15 meter long leash. The radius of the lawn is 20 meter. How much of the lawn can the goat eat?

All of it - given that goats will eat anything, it would eat through the leash that was preventing it from eating all the grass

	10-12-2001 02:46									#81
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a=Bgsin(7,5/20)
b=Pi/2 - a
==>[(4*a)/(2*Pi)]*20²*Pi
+[(2*b)/(2*Pi)]*15²*Pi - (20*(20*sin(2*a)))

voila solved

quote: Ha! (4+ 4/4)*4 That's too simple

well try the first one no-one even came up with a solution(only the question if it was a trickquestion and it isn't)

quote: 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.

Shade

	10-12-2001 04:36									#82
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quote: Originally posted by KrazyHorse 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.

I think we did that, and it wasn't too hard, as you suspected.

Ok, I have some problems from the Archimedeans (Cambridge mathmo society) problems drive, syndicated in the internal journal "Eureka". I'll post them one at a time. They are very hard, and I couldn't solve any of them. I do however have the answers.

1. Complete the following phrases, where each word has been replaced with its initial letter - eg 1=FP of a CM is 1 = Fixed Points in a Contraction Mapping.
a) 1729 = N of HT
b) 4.6692...=R of PD in some DS
c) 0.00000007151=P of W the NL
d) 2^3021376(2^3021377-1)=LKPN (possibly not true any more)
e) 3.14159...=LN
f) 13=AP
g) (1+æ5)/4=C of PBF
h) +æ2/12=V of RT of SO
i) 23=HP
j) 13=B of EE
k) 6=RS in FD
l) 25=PBOH

edit: arghh! The weird symbols are square roots. For some odd reason, charmap in XP no longer shows the keystrokes for the Symbol font...

	11-12-2001 00:26									#83
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quote: d) 23021376(23021377-1)=LKPN (possibly not true any more)

Largest Known Prime Number.

That -1 gave it away.

The others appear more obsure at first glance. For example

quote: e) 3.14159...=LN

I know its pi, but what does LN stand for? *Something* Number?

	11-12-2001 03:20									#84
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quote: Originally posted by Big Crunch Largest Known Prime Number. That -1 gave it away.

Hmm, a prime number which is a product of 2^3021376 and (2^3021377-1)...

	11-12-2001 06:36									#85
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(sorry BC)

All in good fun.

	11-12-2001 18:34									#86
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quote: Originally posted by Chowlett b) 4.6692...=R of PD in some DS

this is(translated from Dutch) 'the number of Feigenbaum'
best guess I can make for the 'english shortcut' is
Rate of periodic Dubbling in Some Dynamic Systems
those some are the unimodal images and the number is defined as
q=lim(i->infinity)[(r(i+1)-r(i))/(r(i+2)-r(i+1))]
( those (i+??) are indices)

(Noone for the simple +-*/?? )

Shade

	11-12-2001 18:37									#87
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quote: Originally posted by Chowlett Hmm, a prime number which is a product of 23021376 and (23021377-1)...

I made atypo

We can't all be "perfect".

	12-12-2001 01:58									#88
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Hmmm...that's an interesting way to construct perfect numbers. Good catch, BC.

	12-12-2001 03:14									#89
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Prove that Sqrt(11) does not exist as a rational number using only axiomatic methods.

	12-12-2001 03:46									#90
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quote: Originally posted by KrazyHorse Hmmm...that's an interesting way to construct perfect numbers. Good catch, BC.

The -1 and the 1 diffence in the powers of two reminded me of the method for making perfect numbers. I knew that the 2ⁿ-1 number had to be prime, unfortunately I got an idée fix on prime, and said prime when I meant perfect.

Oh well.

Edit. Damned idée fix.