Math Challenge!

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chuckywang

Lifer
Jan 12, 2004
20,133
1
0
Originally posted by: TuxDave
If the solution requires integrals, then I'm on the right track... if not, please stop me.

It can be done with integrals or without integrals. If you're doing it with integrals, be careful with the limits. Remember that the maximum that both x and y can be is 1.
 

mugs

Lifer
Apr 29, 2003
48,920
46
91
Originally posted by: chuckywang
Originally posted by: aux
about 45.467%

How did you come up with this answer? It's close, but not quite.

I bet he wrote a program that generated 100,000 pairs of random numbers, divided, rounded, modded, and counted the evens and odds. (I assumed 100,000 pairs because of his level of precision)

I just did the same and got similar results. Now my question is - why?
 

Gibson486

Lifer
Aug 9, 2000
18,378
2
0
Originally posted by: chuckywang
Originally posted by: TuxDave
If the solution requires integrals, then I'm on the right track... if not, please stop me.

It can be done with integrals or without integrals. If you're doing it with integrals, be careful with the limits. Remember that the maximum that both x and y can be is 1.

Makes my mathmatical ability seem teh suckay:(
 

Argo

Lifer
Apr 8, 2000
10,045
0
0
Originally posted by: chuckywang
Original post now edited with hint. Enjoy. :)

You need to draw lines of the following functions:

y = 2x
y = 2/3x
y = 2/5x
y = 2/(2n + 1)x

and shade the arear between each alternative line. The area of triangle equals 1/2 * height * base. The heights of all these triangles are 1. To find base we need to plug x=1 into each equation and solve it for y, then find the difference between each y. The first equation is an exception since the line will intersect y plane instead of x plane. So you need to find the sum of the following series as n goes from 1 to infinity:

1 / (4n - 1) - 1 / (4n + 1).

Simplifying this a little we get:

8n / (16n^2 -1) with n=1,2,3...

I don't remember the formula for finding sums of inifinite series, but I believe there is one. Once you find the sum, you'll need to add 0.25 to it, since that's the first element in the series. The result should be your answer.
 

chuckywang

Lifer
Jan 12, 2004
20,133
1
0
Originally posted by: Argo
Originally posted by: chuckywang
Original post now edited with hint. Enjoy. :)

You need to draw lines of the following functions:

y = 2x
y = 2/3x
y = 2/5x
y = 2/(2n + 1)x

and shade the arear between each alternative line. The area of triangle equals 1/2 * height * base. The heights of all these triangles are 1. To find base we need to plug x=1 into each equation and solve it for y, then find the difference between each y. The first equation is an exception since the line will intersect y plane instead of x plane. So you need to find the sum of the following series as n goes from 1 to infinity:

1 / (4n - 1) - 1 / (4n + 1).

Simplifying this a little we get:

8n / (16n^2 -1) with n=1,2,3...

I don't remember the formula for finding sums of inifinite series, but I believe there is one. Once you find the sum, you'll need to add 0.25 to it, since that's the first element in the series. The result should be your answer.

You are correct. It is a known summation that:

pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ....

Your summation is 1/3 - 1/5 + 1/7 - 1/9 + 1/11 - 1/13 + ....

which is therefore equal to 1-pi/4.

Therefore, the final probability is 5/4-pi/4.
 

HonkeyDonk

Diamond Member
Oct 14, 2001
4,020
0
0
Originally posted by: FoBoT
Originally posted by: chuckywang
Originally posted by: FoBoT
Originally posted by: chuckywang

Therefore, the final probability is 5/4-pi/4.

that isn't a number :confused:

How so?

you know, a number? like 3 or 6 or 2324 ? do you mean 544 ?

or is that some fancy "new" math thing?

Wow FoBot, you're pretty retarded.

5/4 - pi/4 is the exact format

if you want a estimate, then the number is 1.25-(3.14/4) or ~.465ish.
 

aux

Senior member
Mar 16, 2002
533
0
0
Originally posted by: mugs
Originally posted by: chuckywang
Originally posted by: aux
about 45.467%

How did you come up with this answer? It's close, but not quite.

I bet he wrote a program that generated 100,000 pairs of random numbers, divided, rounded, modded, and counted the evens and odds. (I assumed 100,000 pairs because of his level of precision)

I just did the same and got similar results. Now my question is - why?

Actually they were 10 million (if I remember correctly what I did last night), but I was too lazy to copy more digits.

 

chuckywang

Lifer
Jan 12, 2004
20,133
1
0
Originally posted by: aux
Originally posted by: mugs
Originally posted by: chuckywang
Originally posted by: aux
about 45.467%

How did you come up with this answer? It's close, but not quite.

I bet he wrote a program that generated 100,000 pairs of random numbers, divided, rounded, modded, and counted the evens and odds. (I assumed 100,000 pairs because of his level of precision)

I just did the same and got similar results. Now my question is - why?

Actually they were 10 million (if I remember correctly what I did last night), but I was too lazy to copy more digits.

Yeah, probably for the better. Those extra digits wouldn't have given you more accuracy.