Problem 40 - Page 2

ThePerfectHacker · #11 $Old$ 11-11-2007, 10:07 AM

Quote:

Originally Posted by CaptainBlank $View Post$

These proofs are thinly disguised use of the Stone-Weierstrass theorem.

You got the main part right! Weierstrass theorem is the secret trick here.

Quote:

Originally Posted by Jhevon

Now i can see where the problem here would be, the "obviously" part.

It is really not that obvious.

Quote:

Originally Posted by janvdl $View Post$

1.) $(\frac{1}{2007} + \frac{1}{2007}) \times (\frac{1}{2007} + \frac{1}{2007})$

= $\frac{4}{4028049}$

= $\frac{1}{1007012,25}$

How did you get that? My solution is much longer.

janvdl · #12 $Old$ 11-11-2007, 11:32 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

How did you get that? My solution is much longer.

Haha, i guess what matters is if we got the same answer? And if mine is right?

ThePerfectHacker · #13 $Old$ 11-11-2007, 01:34 PM

Quote:

Originally Posted by janvdl $View Post$

Haha, i guess what matters is if we got the same answer? And if mine is right?

I have no idea what the answer is, I only created the problem and thus know how to solve it. I did not fully solve it.

Jhevon · #14 $Old$ 11-11-2007, 01:50 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

You got the main part right! Weierstrass theorem is the secret trick here.

are you saying we should approximate f(x) - g(x) with some other polynomial, say h(x), and show that we must have h(x) = 0 ?

ThePerfectHacker · #15 $Old$ 11-11-2007, 01:56 PM

Quote:

Originally Posted by Jhevon $View Post$

are you saying we should approximate f(x) - g(x) with some other polynomial, say h(x), and show that we must have h(x) = 0 ?

Hint: If $f(x)$ is continous on $[0,1]$ then there exists a sequence of polynomial $p_n(x)$ that converge uniformly to $f(x)$ . (That is the Stone-Weierstrass theorem).

kalagota · #16 $Old$ 11-13-2007, 07:08 AM

okay, i don't know yet what those theorems are.. Ü
but can you check this out.. (i think, this disproves the statement)

$\int_0^1 x^n \cdot (n+1) dx = 1$

and also

$\int_0^1 x^n \cdot (nx+2x) dx = 1$

and there, $f(x) = n+1 \neq (nx+2x) = x(n+2) = g(x)$

or, i just misunderstood the statement.. Ü

ThePerfectHacker · #17 $Old$ 11-13-2007, 09:04 AM

Quote:

Originally Posted by kalagota $View Post$

okay, i don't know yet what those theorems are.. Ü
but can you check this out.. (i think, this disproves the statement)

$\int_0^1 x^n \cdot (n+1) dx = 1$

and also

$\int_0^1 x^n \cdot (nx+2x) dx = 1$

and there, $f(x) = n+1 \neq (nx+2x) = x(n+2) = g(x)$

or, i just misunderstood the statement.. Ü

How can $f(x) = n+1$ ? It is a function, how is that a function?

kalagota · #18 $Old$ 11-13-2007, 06:57 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

How can $f(x) = n+1$ ? It is a function, how is that a function?

if $n \in Z^+$ , then it can be thought as a constant function, right?
even if $x^n$ is a sequence of fucntion, then that f(x) can be thought as a sequence of constant function..

then if convergence is at stake, then can we take $f(x) = x^2n + 3x^2 = (n+3)x^2$ .. this is just a hunch..

qspeechc · #19 $Old$ 11-14-2007, 02:11 PM

None of the solutions proposed have been that which kids like myself would have been abe to put forward. Are you suggesting that everyone look for simple proofs?
For 1), isn't the probability of picking a real number 0? Do you mean picking it from the natural numbers less than 2008?
Anyway I had a proof, but then my father came in, telling me to get off the computerwhile munching on something, consequently spitting his meal all over me, and allowing me the pleasure of watching the food in his mouth get reduced into a pulp. I am really not in a mood to go through the charades of Latex right now.

angel.white · #20 $Old$ 11-14-2007, 05:29 PM

As an example of why I don't think this is as easy as everyone seems to think:

shown for 1-20

Probability of a= any given number, 1/20.
Probablility of b = any given sum of a specific set of factors, 1/20

So each number 1-20 = 1/20 of being picked, then (1/20)*# of potential b's for the a.

(colours and underline are for aesthetic purposes only, intending to improve readibility)

# of potential b's can be found through prime factorization:

# -> Factors -> potential b's -> probability for b's
1 -> 1,1 -> 2 -> 1/20
2 -> 1,2 -> 3 -> 1/20
3 -> 1,3 -> 4 -> 1/20
4 -> 1,2,2 -> 5,4 -> 2/20
5 -> 1,5 -> 6 -> 1/20
6 -> 1,2,3 -> 7, 5 -> 2/20
7 -> 1,7 -> 8 -> 1/20
8 -> 1,2,2,2 -> 9, 6 -> 2/20
9 -> 1,3,3 -> 10, 6 -> 2/20
10 -> 1,2,5 -> 11, 7 -> 2/20
11 -> 1,11 -> 12 -> 1/20
12 -> 1,2,2,3 -> 13, 8, 7 -> 3/20
13 -> 1,13 -> 14 -> 1/20
14 -> 1,2,7 -> 15,9 -> 2/20
15 -> 1,3,5 -> 16,8 -> 2/20
16 -> 1,2,2,2,2 -> 17, 10, 8 -> 3/20
17 -> 1,17 -> 18 -> 1/20
18 -> 1,2,3,3 -> 19, 11, 9 -> 3/20
19 -> 1,19 -> 20 -> 1/20
20 -> 1,2,2,5 -> 12, 9 -> 2/20

so the answer is 1/20*(34/20) = .085

So for 1-20, there is an 8.5% chance of choosing an A, B pair where the product equals A and the sum equals B.

(unless I made a mistake, but still this shows the complexity, I think)

So I don't see how you can come up with a simple equation that will be able to take into account that every number will have a different number of potential B's. It seems to me that it would need to be able to find how many factors each number had, including repeats, and how many numbers they could sum to, excluding repeats. And this for every number through 2007.

(unless there is a pattern, but I doubt there is, because if there was, we would be able to use it to calculate prime numbers, to my knowledge, despite many efforts by many great mathematicians, this has not happened yet.)

If you can solve the problem for the first 20, like I have, without using tedious brute techniques, like I did, please post, I would be very interested to see it done. (If you get close, you might want to check my work, make sure I didn't miss something, and am giving you a false answer to shoot for)