Chief Delphi - Math Quiz 9

Chief Delphi (http://www.chiefdelphi.com/forums/index.php)

- Math and Science (http://www.chiefdelphi.com/forums/forumdisplay.php?f=70)

- - Math Quiz 9 (http://www.chiefdelphi.com/forums/showthread.php?t=149436)

Quote:

Originally Posted by Aren Siekmeier (Post 1597068)

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):

Quote:

Originally Posted by GeeTwo (Post 1596789)

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

Checking the CRC Handbook integrals table (2000 edition):

Quote:

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.

Quote:

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:

ʃ√(1+y²)dy/2
-ʃy√(1+y²)dy/2
+ʃy²ln(1+√(1+y²))dy/2
-ʃy³ln(1+√(1+y²))dy/2
-ʃy²lnydy/2
+ʃy³lnydy/2
+ʃ(1+y²)^3/2dy/3
-ʃy(1+y²)^3/2dy/3
-ʃy³dy/3
+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

Quote:

Originally Posted by Term1

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

Quote:

Originally Posted by Term2

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

Quote:

Originally Posted by Term5

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

Quote:

Originally Posted by Term6

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

Quote:

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

Quote:

Originally Posted by Term8

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

Quote:

Originally Posted by Term9

-ʃy³dy/3 = -y⁴/12| = -1/12

Quote:

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.

Quote:

Originally Posted by Aren Siekmeier (Post 1597068)

Quote:

Originally Posted by GeeTwo (Post 1597092)

Really nice work Aren and Gus.

Here's how I did it.

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.

Quote:

Originally Posted by FiMFanatic (Post 1597049)

Makes no logical sense unless the question is improperly worded or bounded. You can fit an infinite number of lines in the box, and fact is, more lines closer to 0 in length fit in the box than lines averaging 0.52xxxx

There are uncountably infinite lines of length 0.01, just as there are an uncountably infinite number of lines of length 0.52.

It makes no logical sense to say that there are "more" or "less" of one uncountably infinite thing than another uncountably infinite thing.

Quote:

Originally Posted by Caleb Sykes (Post 1597166)

Excellent point. Hence the question was not worded well enough to allow for a proper solution. :D

Quote:

Originally Posted by FiMFanatic (Post 1597167)

Excellent point. Hence the question was not worded well enough to allow for a proper solution. :D

The question is worded just fine. It asks for an average length of all lines, not the length of the line of most frequent length.

For example, would you be able to answer the question: "What is the average of all numbers between 0 and 1?"

Also, we got several proper solutions...

Quote:

Originally Posted by Ether (Post 1597148)

Really nice work Aren and Gus.

Here's how I did it.

Cool, I much prefer how you dropped to the double integral, I don't think about stats enough when doing these things.

Quote:

Originally Posted by MechEng83 (Post 1597158)

While you can certainly say that there are multiple ways to solve a problem, I'll venture that the one presented in the video is considerably more elegant than mine :p The immediate trig sub works wonders on the rest of it.

Thanks for the link!

Quote:

Originally Posted by MechEng83 (Post 1597158)

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.

This is the route I was following. (I have not looked at the video, so perhaps this is exactly what is there.) My original thought was to substitute r for √(x²+y²), rcosθ for x, and rsinθ for y. dxdy then becomes rdrdθ. While driving to mom's house for Sunday dinner, I realized that I could take advantage of the symmetry of x and y by rotating θ by π/4, so that y is √2(cosθ + sinθ)r/2 and x is √2(cosθ - sinθ)r/2. Then the integration is over an interval symmetric over θ=0, and any odd terms in θ can be tossed (I already figured out while driving that there aren't any, however), and then the integration can be done from 0 to π/4. The fun part is the limit of integration - for r<=1, there's no issue. For 1<r<√2, the limits of one integration need to be set so that the maximum of r is √2/(cosθ + sinθ) (ignoring the required absolute values because I've limited θ to the first quadrant). Alternately, the limits of θ can be set based on r.

Quote:

Originally Posted by Caleb Sykes (Post 1597166)

It makes just as much sense as to ask "What fraction of integers are even?" There are uncountably many even integers and uncountably many integers, and it is possible to identify a different even integer for every integer (e=2i), so that the sets have the same number of elements. This does not change the equally meaningful statement that exactly half of the integers are even.

Quote:

Originally Posted by GeeTwo (Post 1597204)

There are uncountably many even integers and uncountably many integers...

You are using the word "uncountably" incorrectly here.

The integers are countably infinite.

The reals are uncountably infinite.

Quote:

This does not change the equally meaningful statement that exactly half of the integers are even.

It's meaningful only if you define what you mean :)

If you mean "the set of all even integers has an asymptotic density of ½", then yes, it's meaningful. Otherwise, not.

These solutions are quite above my mathematics level, but nonetheless I can somewhat follow along, thanks for the cool thread.

Quote:

Originally Posted by Caleb Sykes (Post 1597170)

For example, would you be able to answer the question: "What is the average of all numbers between 0 and 1?"

That has a clear answer, despite the infinite number of numbers between 0 and 1, as each number only occurs once.

The original question has no proper solution as worded.

However, I do applaud the high level of solutions that were provided - well done.

Quote:

Originally Posted by FiMFanatic (Post 1597226)

The original question has no proper solution as worded.

Quote:

Originally Posted by Ether (Post 1596692)

What's the average length of all the line segments which can be drawn inside a 1 inch square?

:confused:

I'm not sure what's missing... What extra information is being used in solving this problem?

Quote:

Originally Posted by Ether (Post 1597219)

You are using the word "uncountably" incorrectly here.

The integers are countably infinite.

The reals are uncountably infinite.

"Countably infinite" is still uncountable for us mortals. When you finish, please get back to me.

Quote:

Originally Posted by Ether (Post 1597219)

... the concept of half of an infinite set is not well-defined. It’s possible to pair up the even integers with the odd integers with none left over in either set, and if we were talking about finite sets, that would be a demonstration that each was half of the whole set of integers. However, it’s also possible to pair up the multiples of 100, say, with all the rest of the integers with none left over in either set, and the multiples of 100 are obviously only part of the set of even numbers. Clearly, then, this kind of pairing argument cannot lead to any very useful notion of half of the set of integers.

There is a notion of asymptotic density of a set of positive integers that does a pretty good job of capturing many people’s intuitive sense of what half (or any other fraction) of the set of positive integers should mean.

excerpted from http://math.stackexchange.com/questi...ll-numbers-odd

I fully admit to having (yes, consciously and deliberately) blown through the difference between aleph null and aleph one in my earlier reply, to make a statement that might make sense to shose who found the original problem "meaningless". (That is, I was more worried about enlightenment than mathematical rigor.) I specifically addressed your concern that the even integers and all integers are of the same cardinality.

A far as I am aware, any argument that decides the statement "Half of the integers are even." as meaningless can also be used to decide the concept of "average length of a line segment located within the unit square" as meaningless -- or worse.

Edit:

Quote:

Originally Posted by Ether, as edited after I prepared the response above (Post 1597219)

It's meaningful only if you define what you mean,

If you mean "the set of all even integers has an asymptotic density of ½", then yes, it's meaningful. Otherwise, not.

OK, I'll go with that -- and it's a remarkably fast asymptotic function, as of every pair of consecutive integers, exactly one is even.

Here's a bit more precise statement: For every set of consecutive integers with a non-zero, even number of members, exactly half are even.