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Math Challenge!

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Originally posted by: TuxDave
If the solution requires integrals, then I'm on the right track... if not, please stop me.
Click to expand...

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.
 
Originally posted by: chuckywang
Originally posted by: aux
about 45.467%
Click to expand...

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

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?
 
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.
Click to expand...

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.
Click to expand...

Makes my mathmatical ability seem teh suckay🙁
 
Originally posted by: chuckywang
Original post now edited with hint. Enjoy. 🙂
Click to expand...

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.
 
Originally posted by: Argo
Originally posted by: chuckywang
Original post now edited with hint. Enjoy. 🙂
Click to expand...

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.
Click to expand...

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.
 
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.
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that isn't a number 😕
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How so?
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you know, a number? like 3 or 6 or 2324 ? do you mean 544 ?

or is that some fancy "new" math thing?
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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.
 
Originally posted by: mugs
Originally posted by: chuckywang
Originally posted by: aux
about 45.467%
Click to expand...

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

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?
Click to expand...

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

 
Originally posted by: aux
Originally posted by: mugs
Originally posted by: chuckywang
Originally posted by: aux
about 45.467%
Click to expand...

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

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?
Click to expand...

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

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