ln(e^(what page this is))

[Edited on July 19, 2007 at 10:34 PM. Reason : .]

7/19/2007 10:31:58 PM

4.02

7/19/2007 10:33:38 PM

Aficionado: I love you, you made my night. Hahahahahaha.

7/19/2007 10:35:42 PM

Quote :

"when i was in 3rd grade, the teacher used to call ~10 students to the front of the class and do as many of these type problems as they could in 60 seconds. whoever won got a candy bar. obviously, i won most of the time "

ever play "Around the World"? I don't think I ever succeeded in going all the way around


[Edited on July 19, 2007 at 10:52 PM. Reason : /]

7/19/2007 10:39:27 PM

^grad school question plz

7/19/2007 10:40:55 PM

@jeff

oh yeah. i remember doing that in 4th/5th grade i think. AV Baucom ftw

7/19/2007 10:47:47 PM

V I gotcha

What is the minimum number of edges in simplicial complex structures K and L on S1 such that there is a simplicial map K -> L of degree n?

or how about

Show that a linear functional f on a normed linear space is bounded iff its kernel is closed.

There are very few numbers involved in graduate school math


[Edited on July 19, 2007 at 11:00 PM. Reason : /]

7/19/2007 10:52:39 PM

you are the grad student, not me

7/19/2007 10:53:27 PM

we played around the world in 3rd grade a lot

i won that shit pretty much every time we played

7/19/2007 10:55:14 PM

Which set has more elements: the open interval (0,.0000000000001), or the rationals?

7/19/2007 11:04:31 PM

both are infinite, no?

7/19/2007 11:04:58 PM

yes, but the rationals are countable, whereas the set of numbers in the open interval (0,0.0000001) is not. Therefore there are more numbers in (0,0.000000001) than there are rationals (or technically, the cardinality of (0,0.00000001) is greater)

7/19/2007 11:08:40 PM

touche

7/19/2007 11:11:38 PM

^^ yeah that stuff about infinite numbers gets very interesting and plays with your mind!

countably infinite
uncountably infinite
transfinite numbers

etc

7/20/2007 12:03:29 AM

let f: X ---> R be a non-zero functional from X, a normed
space (any topological vector space will do).

The following are equivalent:

1) f is continuous.
2) the null space (or kernel) N(f) is closed.
3) N(f) is not dense in X.
4) f is bounded on some neighborhood of 0.

1) --> 2): this is easy, as N(f) = f^-1[{0}] and {0}
is a closed set in R.

2) --> 3): as f is non-zero, N(f) != X, and N(f) is
closed by assumption, so N(f) is not dense.

3) --> 4): if x is some point not in cl(N(f)), then
for some e>0 : B(x,e) /\ N(f) is empty, where
B(x,e) = {y: ||x-y|| < e }.
Look at B(0,e). The image f[B(0,e)] is either bounded
(in which case we are done), or not. Let's assume the latter:
then f[B(0,e)] = R, as the image is convex and symmetric around 0
(f(x) = -f(-x)) (because B(0,e) is both and this is preserved by linearity)
and the only symmetric convex unbounded set in R is R itself.
But then there is a y in B(0,e) with f(y) = -f(x), and then
x+y is in N(f) and is in B(x,e), contradicting the way we chose e.
So f is bounded on B(0,e).

4) --> 1): Fix e>0 and M > 0 such that |f(x)| < M for all
x in B(0,e). But then, pick r>0: if ||x|| < e*r/M then
(M/r)*x has norm < e and so ||f((M/r) * x)|| = (M/r)||f(x)|| < M
or ||f(x)|| < r,
and this says that f is continuous at 0.
But then f is continuous at any p in X:
if ||x-p|| < e*r/M then apply this argument to (M/r)(x-p)
to conclude that ||f(x) - f(p)|| < r.

7/20/2007 5:00:28 AM

I won Around the World constantly until once in fourth grade when I got beat on the last question by a girl name Becky.

That was the day I swore off math for life.

7/20/2007 5:27:19 AM

hahah damn, becky left her mark

7/20/2007 7:55:30 AM

Quote :

"yeah that stuff about infinite numbers gets very interesting and plays with your mind! "

Yeah, it's interesting shit. I think I'll deliver a little lecture about countability here, since I'm bored.

We generally think of all infinities as being the same, but there are different kinds of infinities. The most fundamental division is between countable and uncountable infinities. An infinite set X is "countable" if we can map the positive integers onto X without missing any of the elements of X.

For example, the integers, despite being infinite, are countable; the function

f(n)= (1-n)/2 for odd n
n/2 for even n

will map from the postive integers to all integers. f(1)=0, f(2)=1, f(3)=-1, f(4)=2, f(5)=-2, etc.

It is surprising that the rational numbers (all numbers of form p/q, where p and q are integers) are also countable. The way one proves this is to consider that there are only n^2 possible fractions whose numerator and denominator are both <=n. For example, 1/1,1/2,2/1 and 2/2 are the only possible fractions whose numerators and denominators are all <=2; there are four such fractions, and 2^2=4. So, if we want to count the rationals, we are guaranteed to hit (p/q) by our r^2 count, where r is the max of p and q.

There are, however, uncountable sets. The classic example is the open interval (0,1), or any open interval. The standard method for proving (0,1) to be uncountable is:

Consider all real numbers on the interval (0,1). Every such number can be represented as

x1=0.x11 x12 x13 x14 x15 x16 etc.

where x1i is the ith digit of x1. So suppose for contradiction that we have counted all the numbers on (0,1). Then we can label them x1, x2, x3, etc. and we have

x1=0.x11 x12 x13 x14 x15 x16 etc.
x2=0.x21 x22 x23 x24 x25 x26 etc.
x3=0.x31 x32 x33 x34 x35 x36 etc.

etc.

Now consider the number y on (0,1) whose nth digit is xnn+1. (If xnn=9, let yn=0). Then y disagrees with every xn by at least one decimal place, and therefore y is not xn for any n, i.e. y was not counted. Therefore we cannot possibly count all real numbers on (0,1).

This is very weird shit, by the way, and very counterintuitive; by a basic argument one can show that there are exactly as many perfect squares (1,4,9,16...) as there are positive integers, even though the set of positive integers contains many numbers which the set of perfect squares does not.

7/20/2007 4:21:44 PM

Quote :

"hahah damn, becky left her mark"

Eh, yes and no. She also beat me in the Discovery Place scavenger hunt that year, but the next year I beat her and everyone else by seven points.

Plus, I don't remember her last name, so it's not like "Laces out, Dan!"

7/20/2007 4:25:35 PM

her last name is Warner.

I know, because...I AM THAT BECKY

7/20/2007 4:27:11 PM

COMEDY!

7/20/2007 4:48:08 PM

more like tragedy

7/20/2007 4:49:21 PM

Quote :

""Laces out, Dan!""

7/20/2007 8:14:59 PM

if you posted once every 12.2 seconds, (disregarding flood control), how many posts would you have in 2/3 of a year

[Edited on September 9, 2007 at 11:26 PM. Reason : /]

9/9/2007 11:20:23 PM

if you never slept, 1,723,278

9/10/2007 1:49:29 AM

Quote :

"If you posted every 30 seconds for a year, how many posts would you have? and GO"

you have to account for flood control

9/10/2007 1:53:21 AM

beer 30?????

9/10/2007 1:57:06 AM

Just finished this problem, its a good one

Assume that there are nine parking spaces next to one another in a parking lot. Nine cars need to be parked by an attendant. Three of them are fancy sports cars, the other 6 are not. If the attendant parks each car randomly, what is the probability that the three sports cars are parked adjacent to one another?

9/11/2007 10:06:10 PM

1 in 12

I've seen this one before in one of my classes...lots of factorials

9/11/2007 10:11:03 PM

close, but it's 1/24


9! possible ways of parking the cars

7 ways the 3 cars can be next to each other

3! ways for them to be next to each other

6! ways the other 6 can be parked when they are next to each other

so, [7(3!*6!)]/9!=1/24

9/11/2007 10:22:06 PM

hahaha...yea....if we were doing binomials

9/11/2007 10:22:50 PM

9/11/2007 10:34:45 PM

damn thanks, good catch

(7 * 3! * 6!) / 9! = (3! * 7!) / 9!

(3 * 7!) / 9! = 3 / (9 * 8)

i left off my factorial with the 3! when writing it down

9/11/2007 10:50:31 PM

^^ha. is that a cellphone

[Edited on September 11, 2007 at 10:51 PM. Reason : pssh]

9/11/2007 10:51:26 PM

um that'd be a ti-89 titanium foo

9/11/2007 10:52:03 PM

In a group of 11 people, no two people are the same age.

Suppose we select at random two teams each with 4 members, to carry out two different tasks. What is the probability that the youngest 4 people are all on one team or all on the other?

9/25/2007 8:27:04 PM

^ Quick Math

9/25/2007 8:30:10 PM

it's not hard

9/25/2007 8:30:28 PM

this isnt quick math but i need help


ok if given the following (disregard line M)



If i know AB=20, CD=7 and the diameter(in my problem) BD=25, how do I find BC and AD?

[Edited on October 4, 2007 at 9:27 PM. Reason : .]

10/4/2007 9:26:50 PM

^is AC the length of the radius?











ok

starting at exactly 3:00:00.0pm

if you counted 300 seconds

it would be 3:05:00.0pm

then counted 305 seconds

it would be 3:10:05.0pm

then counted 310.05 seconds

it would be 3:15:15.05pm

then counted 315.1505 seconds

it would be 3:20:30.2005pm

and so on

how long until it's exactly 3:00:00.0pm again?

10/5/2007 1:23:44 AM

^AC looks like the diameter to me

and 24 hours.

10/5/2007 1:46:18 AM

can't remember my basic geometry, so a little googling:
http://mathforum.org/dr.math/faq/faq.circle.segment.html#10

first solving for AD

d=25, r=25/2
c1=20
c2=??=chord AD
call arc AB length S1
call arc AD length S2
circumference = pi*d = 25pi
We know that S1+S2 = half the circumference = 25pi/2

use the following:
theta = 2 arcsin (c1/d) where theta is the angle from the center of the circle across the specified chord
s = r*theta

S1 = r*2*arcsin(c1/d)
S2 = 25pi/2 - S1 = 25pi/2 - r*2*arcsin(c1/d)
also S2 = r*2*arcsin(c2/d)
thus
r*2*arcsin(c2/d) = 25pi/2 - r*2*arcsin(c1/d) ------ only unknown is c2

AD = c2 = 15


Solving for BC we do the same process except with a known CD of 7 instead of known AB of 20

d=25, r=25/2
c1=7
c2=??=chord BC
call arc CD length S1
call arc BC length S2
circumference = pi*d = 25pi
We know that S1+S2 = half the circumference = 25pi/2

r*2*arcsin(c2/d) = 25pi/2 - r*2*arcsin(c1/d) ------ only unknown is c2

BC = c2 = 24




i think

10/5/2007 2:39:41 AM

I just started reading this thread, but Aficionado's argument of it being impossible to fold a piece of paper 50 times wrong.

You can fold a piece of paper, given a big enough piece of paper, as many times as you like.

You cannot fold a standard 8.5x11 piece of paper in HALF more than 6 or 7 times. Bigger sheets of paper CAN be folded in half many more times.

10/5/2007 3:07:57 AM

I'm amazed that it took well into page 2 for someone to point out that a year is actually more than exactly 365 days long.

10/5/2007 6:27:40 AM

^^^ Don't know why I took such a circuitous route to the answer there. No need to bring arc length and circumference into it.
Instead, it's a little simpler just to say theta1+theta2=pi and use the same theta equations. There will still only be one unknown in each equation. (The equations all simplify down to the same thing using the circumference route and this route, but there just a couple fewer steps in this route and less possibility for error.)

10/5/2007 10:27:57 AM

^^^ no it cant

watch the mythbusters episode about it

10/5/2007 10:29:43 AM

^ I saw the same episode...the started with a paper that took up the floor of an airplane hanger...i don't remember the number that they got stuck on though...

10/5/2007 10:31:36 AM

i think it was 8 or 9 but they needed a forklift to compress the paper to get the last one, so thats a bullshit fold

10/5/2007 10:35:19 AM

anti-derivative of 20/(x^2)

[Edited on October 17, 2007 at 1:40 PM. Reason : i actually messed this up in the middle of a problem is why im posting, so retarded]

10/17/2007 1:38:09 PM

sry, the ti-89 is in the other room...too hard

10/17/2007 1:43:15 PM