Problem 40 - Page 2

kalagota · #16 $Old$ November 13th, 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 $Old$ November 13th, 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 $Old$ November 13th, 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 $Old$ November 14th, 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 $Old$ November 14th, 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)

ThePerfectHacker · #21 $Old$ November 14th, 2007, 07:57 PM

By the way the numbers to do not have to be natural numbers. I specifically said that.

angel.white · #22 $Old$ November 14th, 2007, 08:28 PM

Okay, in that case I have a start, maybe someone else can pick up what I don't know.

I would say we can find every instance by the following method:
A, B = random from the set {1,2,3,...,2007}
c,d = real numbers (this includes irrational and negative)

so:
$cd=A \Longrightarrow c=\frac{A}{d}$

$c+d=B$

Substitute:
$\frac{A}{d}+d=B$

Multiply by d:
$A+d^{2}=dB$

Subtract dB
$d^{2}-dB+A=0$

By Quadratic Formula:
$d=\frac{B\pm\sqrt{B^{2}-4A}}{2}$

-----

Therefore there will be a real solution so long as $B^{2} -4A \geq 0$

So I don't know how to find the probability that this will be the case, but I expect if you can find the probability of this, happening, then you can subtract that from100%, and that will be your answer.

angel.white · #23 $Old$ November 14th, 2007, 09:59 PM

Okay, I think I have found a way to calculate it, but I would need computer software to make it work

so we need to know the probability that $B^{2}-4A>0$ is true. Lets modify the equation to make it more workable.

$B^{2}-4A>0$

Add 4A
$B^{2}>4A$

Square root (B must be positive because it is taken from a set of positive integers)
$B>2\sqrt{A}$

So now we need to find the probability that this is true for A and B are elements of {1,2,3,...,2007}

So if we assume, A we can figure out the probability for this to be true for B, based on some specific A. In this case, $2007-floor(2\sqrt{A})$ will give us the number of possibilities that B can be which will satisfy the equations. Then divide that by 2007 (the total number of possibilities that B can be, including those that do and those that do not satisfy the equations), and that will give you the percentage that the equations can be satisfied. Meaning cd=A and c+d=B is true (if A is true).

So for any A, the probability that it has real factors whose sum is B is
$\frac{2007-floor(2\sqrt{A})}{2007}$

And so because any given A has a 1 in 2007 chance of being true, then we can multiply that probability against the probability that B will be true for that A, and then count the total probabilities for each A.
$\frac{1}{2007}\cdot\frac{2007-floor(2\sqrt{A})}{2007}$

$\frac{2007-floor(2\sqrt{A})}{4028049}$

Now we just need to count them all. I'm not very familiar with summation notation, but I think this is correct

$\sum^{2007}_{i=1}\frac{2007-floor(2\sqrt{i})}{4028049}$

As i cycles through every value, it will calculate the odds for an A of that value having a B that will satisfy cd=A and c+d=B, then add it to the running total, and when it is done there should be a percentage, between .96 and 1

Okay, so I don't actually know how to do that, (well, I know what it means, but I don't care to sit here for 2007 iterations and calculate the odds) but I think that it is the answer.

Does anyone know an online site which can do this type of equation? If not, perhaps someone can create a quick program to do it (I used to know how to, but it's been too long

)

Edit: Looks like if you have a graphing calculator, you can do it Summation (Sigma) Notation Using the Graphing Calculator I, unfortunately, do not have a graphing calculator, but if any of you would plug it in for me, I'd +rep you ^_^ (that's worth like an entire point)

ThePerfectHacker · #24 $Old$ November 14th, 2007, 10:02 PM

You are doing good. But I have a way without software. There is a formula.

angel.white · #25 $Old$ November 14th, 2007, 10:07 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

You are doing good. But I have a way without software. There is a formula.

I don't know how to do it without software, I'm still in Discrete Math 1, and this is the first summation notation equation I have written on my own, I had to look up how they work just to make sure I did that one right.

Perhaps you would be willing to educate me about the formula?

angel.white · #26 $Old$ November 15th, 2007, 12:23 AM

Okay, I found a site that will calculate these:
Sigma
$\sum^{2007}_{i=1}\frac{2007-floor(2\sqrt{i})}{4028049}=.970475036426$
So about 97.05%
^_^

Is that the correct answer?

ThePerfectHacker · #27 $Old$ November 15th, 2007, 08:53 AM

Quote:

Originally Posted by angel.white $View Post$

Is that the correct answer?

I have no idea I did not solve the problem yet numerically.

CaptainBlack · #28 $Old$ November 16th, 2007, 12:25 AM

Quote:

Originally Posted by angel.white $View Post$

Okay, I found a site that will calculate these:
Sigma
$\sum^{2007}_{i=1}\frac{2007-floor(2\sqrt{i})}{4028049}=.970475036426$
So about 97.05%
^_^

Is that the correct answer?

That is the correct value for the sum you give.

RonL

angel.white · #29 $Old$ November 16th, 2007, 11:26 AM

Quote:

Originally Posted by CaptainBlack $View Post$

That is the correct value for the sum you give.

RonL

Do you know whether the sum I've given the correct answer to the question?

ThePerfectHacker · #30 $Old$ November 18th, 2007, 07:32 PM

Here is solution to #1. We want $x+y=B, xy=A$ that means $x^2+y^2+2xy=B^2\implies x^2+y^2=B^2-2A$ . But then $(x-y)^2 = x^2 + y^2 - 2xy = B^2 - 2A - 2A = B^2 - 4A \implies x-y = \sqrt\pm {B^2-4A}$ . And the system of equations $x+y = B \mbox{ and }x-y = \pm \sqrt{B^2-4A}$ always has a solution provided $B^2-4A \geq 0 \implies B^2 \geq 4A$ .

So we need to count the number of ways we can get $B^2\geq 4A$ for all pairs $(A,B)$ . Note if $B\geq 90$ . Then the maximum value $4A$ is $4(2007) = 8028$ which is always true for $B\geq 90$ . We we just need to count all pairs $(A,B)$ where $B\leq 79$ .

If $B=1$ then there is no $A$ in the pair $(A,B)$ so that $B^2\geq 4A$ . So the count is 0.

If $B=2$ then there is just $A=1$ in the pair $(A,B)$ so that $B^2\geq 4A$ . So the count is 1.

If $B=3$ then there is just $A=1,2$ in the pair $(A,B)$ so that $B^2 \geq 4A$ . So the count is 2.

If $B=4$ then we can pick $A=1,2,3,4$ . So the count is 4.

If $B=5$ then we can pick $A=1,2,3,4,5,6$ . So the count is 6.

The question is whether we can find a pattern. Yes! It is based on looking at even and odd cases. Say $B$ is even so $B=2c$ then $B^2 \geq 4A\implies 4c^2\geq 4A\implies c^2 \geq A$ in that case $A=1,2,3,...,c^2$ . So the count is $c^2$ .

If $B$ is odd, so, $B=2c+1$ then $(2c+1)^2 \geq 4A \implies 4c^2+4c+1\geq 4A \implies c(c+1)+.25\geq A$ . Which means $A=1,2,3,...,c(c+1)$ . So the count is $c(c+1)$ .

Now if we write out the numbers as we did for $B=1,2,3,4,5$ out further we will get:
$0,1,2,4,6,9,12,16,...$
Where (alternatively) $0=0\cdot 1,2=1\cdot 1, 6=2\cdot 3,12=3\cdot 4,...$
And (again alternatively) $1=1^2,4=2^2,9=3^2,16=4^2,...$ .
This list continous until $B=78,79$ on $78$ the value of $c=78/2 = 39$ . And for $79$ is will be $39(40)$ .

If we split this sum into even terms in the sequence and odd terms in the sequence we get:
$\sum_{n=1}^{39}n^2 + \sum_{n=1}^{39}n(n+1) = \sum_{n=1}^{39}2n^2+n = \frac{39(40)(79)}{3}+\frac{39(40)}{2} = 41080+780=41860$ .
But this is for $B\leq 79$ (I just realized it should have been $B\leq 89$ but I am too lazy to change it now. You get the idea).

Now just find the number of pairs $(A,B)$ for $B\geq 80$ which is just $80(2007)$ and add to the answer.