Math Quiz 9

[z_beeblebrox]

My simple Python Monte Carlo script gave me an average of 0.521408 (or 0.52141 rounded to 5 digits) in ~5e9 iterations (10 miles of hiking worth).

Code is below:

import numpy as np

iterations = 10000000

avg = 0

for i in range(100000):
    for i in range(iterations):
        pos = np.random.rand(4)
        length = np.sqrt((pos[0]-pos[1])**2+(pos[2]-pos[3])**2)
        avg = (avg*i + length)/(i+1)
    with open("test.txt", "a") as f:
        f.write(str(avg)+'\n')

import numpy as np

f = np.loadtxt('test.txt')
print(np.average(f))
print(len(f))

I split it into two scripts just so I could stop the computation whenever without altering the result.

Edit: Reps to whoever finds the bug in my code and explains what it does.

[Ether]

Originally Posted by z_beeblebrox

0.52141 rounded to 5 digits

That is the correct answer. Reps to you

[Ether]

Originally Posted by Hitchhiker 42

I ran the Monte Carlo thing in a python script and the answers, each run with 1,000,000 trials, were all around 0.333

Your script has an error.

Compare it to the one-line AWK script I posted.

[GeeTwo]

OK, another follow-on challenge (cleared with Ether):

What is the average length of all the line segments which can be drawn within the unit circle (1 unit in radius, 2 units in diameter)?

Reps for both the first numeric (good to 1 part per million) and first closed form solution.

Edit: Of course, you must show your work in either case!

[GeeTwo]

Originally Posted by GeeTwo

What is the average length of all the line segments which can be drawn within the unit circle (1 unit in radius, 2 units in diameter)?

This can actually be solved in closed form with much less intricate calculus than the case of the square, with proper selection of coordinate system.

Some warmup questions to this if you can't figure out where to start:

• What is the average of the squares of the lengths of all the line segments which can be drawn within the unit circle? [taking out the square root simplifies the problem greatly, and helps provide an upper bound to the original question]
• What is the average of the lengths of all the chords of the unit circle? (That is, line segments with both endpoints on the border of the circle)
• What is the average of the lengths of all the line segments between a point on the [border of] a circle of radius R, and a point within the same circle?

OBTW, with proper scaling, Greg's answer for the 100000-gon of area 1 provides an answer to the original problem good to within 1 part per 10,000:

Originally Posted by Greg Woelki

Here is a generalized Monte Carlo simulation ....

And, OBTW2: if you can find the closed form, each of the four questions can be calculated in fewer than 10 keystrokes on a calculator with the following buttons:

0 1 2 3 4 5 6 7 8 9 + - / * π √ =

[GeeTwo]

I have posted this figure which I created to describe variables in my calculation of the length of the average line segment from a point on the edge of a circle to a point on the interior. I used this, or a very similar coordinate system, to solve all the "circle questions" except for the one about the mean square of the segments. I am also confident that the same questions can be answered for the sphere using a similar point of view.

[GeeTwo]

Originally Posted by GeeTwo

I am also confident that the same questions can be answered for the sphere using a similar point of view.

I did the integral for the average length of a line segment within a sphere, and found that the answer is rational! This problem was actually simpler to integrate (though just a bit trickier to set up) than any of the two-dimensional problems. If no one posts anything (yes, even a "hold on, I'm working it!") on the circle or sphere questions by Wednesday at 1800 US-CDT, I'll post my solutions so we can close out these quiz questions.

[Ether]

Originally Posted by GeeTwo

I did the integral for the average length of a line segment within a sphere, and found that the answer is rational!

The average length of sphere chords (line segments with both endpoints on sphere surface) is also rational.

[GeeTwo]

Originally Posted by Ether

The average length of sphere chords (line segments with both endpoints on sphere surface) is also rational.

Yes, as well as the intermediate case where one endpoint is on the sphere and the other is internal. I don't recall whether the mean square length was rational or not; I'll check this evening.

[AMendenhall]

0.9525383819936485
Sample size: 60 million

[Ether]

Originally Posted by AMendenhall

0.9525383819936485
Sample size: 60 million

Post your code and we'll show you where you went wrong.

[Aren Siekmeier]

Precisely ln(1+sqrt(2))/3 + sqrt(2)/15 + 2/15

Or approximately 0.52140543316
Rounded to 8 digits: 0.52140543
Rounded to 5 digits: 0.52141

A fun exercise in every integration tool I've ever learned, plus a cool rationalization trick I wasn't familiar with.

The computation took a bit to hash through some arithmetic errors and one silly differential error (chain rule!). In practice you probably wouldn't bother to do this (or use a symbolic manipulation tool like Mathematica - this is how I checked my work), and sometimes there is no closed form for the integral. To get 8 digits simply run a larger Monte Carlo simulation (a couple lines of code instead of several pages of derivation). In the end we are always limited to some finite number of digits, so numerical methods win out.

[GeeTwo]

Originally Posted by Aren Siekmeier

Precisely ln(1+sqrt(2))/3 + sqrt(2)/15 + 2/15

Or approximately 0.52140543316
Rounded to 8 digits: 0.52140543
Rounded to 5 digits: 0.52141

A fun exercise in every integration tool I've ever learned, plus a cool rationalization trick I wasn't familiar with.

The computation took a bit to hash through some arithmetic errors and one silly differential error (chain rule!). In practice you probably wouldn't bother to do this (or use a symbolic manipulation tool like Mathematica - this is how I checked my work), and sometimes there is no closed form for the integral. To get 8 digits simply run a larger Monte Carlo simulation (a couple lines of code instead of several pages of derivation). In the end we are always limited to some finite number of digits, so numerical methods win out.

Wonderful! I've been stuck on the ln(1+√(1+y2)) terms. Didn't think to keep the ln(y) terms with them. (I did the x integrals first, so was using y where you have x-bar).

Edit - Now that I've had a chance to spend more than a few minutes on it, I see that wasn't the trick at all. I just need to go back and re-learn integration by parts.

Edit2: Gathering up my paper notes, this is what I had so far (haven't double-checked everything yet):

Originally Posted by GeeTwo

4ʃʃ(1-x)(1-y)√(x²+y²)dxdy, both integrals being over 0..1

Checking the CRC Handbook integrals table (2000 edition):

Originally Posted by CRC 2000 Handbook integral #156, ‘+’ case, letting a=y

ʃ√(x²+y²)dx = (x√(x²+y²) + y² ln(x+√(x²+y²)))/2| = (√(1+y²) + y² ln(1+√(1+y²)) - y² lny)/2

Where the | is an implied evaluation over 0..1.

Originally Posted by CRC 2000 Handbook integral #163, ‘+’ case, letting a=y

ʃx√(x²+y²)dx = (x²+y²)3/2/3| = ((1+y²)3/2 - y³)/3

• ʃ(1-y)( (√(1+y²)/2+y²ln(1+√(1+y²))/2-y²lny/2+(1+y²)3/2/3-y3/3)dy

evaluated over y=0..1. This expands to ten terms:

1. ʃ√(1+y²)dy/2
2. -ʃy√(1+y²)dy/2
3. +ʃy²ln(1+√(1+y²))dy/2
4. -ʃy³ln(1+√(1+y²))dy/2
5. -ʃy²lnydy/2
6. +ʃy³lnydy/2
7. +ʃ(1+y²)^3/2dy/3
8. -ʃy(1+y²)^3/2dy/3
9. -ʃy³dy/3
10. +ʃy⁴dy/3

I know or quickly found all but the forms with ln(1+√(1+y²)) in an online table of integrals or the CRC table: http://2000clicks.com/mathhelp/Calcu...rals.aspx#CatL

Originally Posted by Term1

ʃ√(y²+1)dy = (√(y²+1) + ln(y+√(y²+1)))/2| = √2 – 1 + ln(1+√2)

Originally Posted by Term2

-ʃ√y(y²+1)dy = -(y²+1)^3/2/3| = 1/3 - 2√2/3

Originally Posted by Term5

+ʃy²lnydy/2 = +y³(lny/3 – 1/9)| = -1/9

Originally Posted by Term6

-ʃy³lnydy/2 = +y⁴(lny/4 – 1/16)| = 1/16

Originally Posted by Term7, CRC 164

+ʃ(1+y²)^3/2dy/3 = (y(1+y²) ^3/2 + 3y(√(y²+1))/2 + 3ln(y+√(y²+1)/2)/12| = √2/6 + √2/8 + ln(1+√2)/8

Originally Posted by Term8

-ʃy(1+y²)^3/2dy/3 = -(y²+1)^5/2/15| = 1/15 - 4√2/15

Originally Posted by Term9

-ʃy³dy/3 = -y⁴/12| = -1/12

Originally Posted by Term10

+ʃy⁴dy/3 = y⁵/15| = 1/15

I have also worked through term 3, using your same u for integration by parts, and term 4 looks trivially different. Now I'm going to look for a solution that exploits the symmetry between x and y.

[Ether]

Originally Posted by Aren Siekmeier

.

Originally Posted by GeeTwo

.

Really nice work Aren and Gus.

Here's how I did it.

[MechEng83]

Now that this has been satisfactorily solved, I'll post a youtube video of this problem that I literally saw the week before Ether posted. I didn't feel right in just posting it, or claiming it as my own solution.

The problem gets to the 4ʃʃ(1-x)(1-y)√(x^2+y^2)dxdy GeeTwo derived, but then it does a polar coordinate substitution to make the integration "easier"
It's another approach which gives a closed form solution, demonstrating that there can be multiple ways to validly solve a problem.