To see cancellation issues really go off the deep end, review the 1999 paper by Cuyt, Verdonk, Becuwe, and Kuterna.
However, don't review that paper if you aren't willing to go off the deep end yourself. Gahhhh.
I thought I blogged this long ago, but then again I thought I learned this long ago, and rediscovered this morning that I needed to recreate it: how to solve A x^2 + B x + C = 0.
[more ]Given x1,x2,...xn are independent, and uniformly distributed in the range [a,b], what is the expectation of the maximum of (x1..xn)?
No, this doesn't particularly have anything to do with the Super Bowl, except that it is what I am doing instead of watching.
[more ]A discussion on Baseball Primer prompted me to examine the question in more detail. A contributor suggested that home field advantage was a necessary component of a first order approximation, which lead me to wonder if the order of the games mattered.
That is, does it change the overall chances of winning a series if you get to play all the home games first? More generally, if the odds of winning each game differ, does the order that you compete at those odds matter?
[more ]via Lambda the Ultimate, Gibson, Lester, and Bird on enumerating the rational numbers, without duplicates.
Now I need to find a problem to use this on....
My father asked a fairly straight forward question last week; "How do you calculate the probability of winning a seven game series, given that the probability of winning any individual game is p?"
There are two different ways to calculate it. The process that seems most like the real problem is to work out the odds of winning 3 of 3 and game 4, 3 of 4 and game 5, 3 of 5 and game 6, 3 of 6 and game 7 - the sum of these values is the likelyhood of winning a best of 7.
On the other hand, you can simply sum the probability of winning 4 of 7, 5 of 7, 6 of 7, 7 of 7, and get the same result.
But looking at the formulas, it wasn't obvious to me that had to be so.
[more ]Once again tripping over a misunderstanding of this puzzle, I've worked out another possible illustration which may confern enlightment on the reader.
[more ]I tried to toss Euler's solution to the Basel Problem onto the whiteboard this week, and didn't quite have it from memory.
[more ]A friend recommended Linear Algebra Done Right, which has within it an exercise in approximating sin(x) using orthogonal projections onto a P(5) basis generated using the Gram-Schmidt procedure.
I was curious about this approximation, and how closely the derivative of the approximation would approach cos(x). 21 pages of notes and an excel spreadsheet later, I have some observations (and shortcuts) to share.
[more ]While attempting to work out the probability of Joe DiMaggio's hitting streak, I tumbled across this handy looking formula, in Ask Dr. Math.
The probability of no run of length r in n trials is approximately
"So there's a 1/3 chance of Monty opening a door at random, and there's a 2/3 chance of Monty revealing information by opening the door he does." -- Aalbert Torsius @ Wiki
Update: Spelling corrected in title. My introduction to Monty Hall was a description of AD&D campaigns; I didn't really get the joke until Let's Make a Deal returned to afternoon television sometime in the 80s. When I'm not paying attention, the AD&D spelling pops back in.
More demonstrations of the Pythagorean Theorem than you are likely to need in any given lifetime.
Once you know something to be true, there are many ways to demonstrate it.
Via Off the Kuff.
What is the square root of 10?
1024 = 32^2
π is the ratio of the circumference of a circle to its diameter. The Greeks noticed that this ratio was the same for all circles, and named it. They also made efforts to calculate it, but the tools available in those times were too slow - you don't get enough digits for the calculation effort.
From Games Magazine: given a dart board with two scores (5 & 7), and as many darts as you like, what is the highest total score that cannot be achieved?
What's the general solution with available scores A & B?
[more ]A different perspective on the usual puzzle.
Our host M, offers contestant A a choice of three doors - only one of which reveals a winning outcome. A chooses door #1, and M opens door #3, revealing a non winning outcome.
At this point, contestant B, who has been watching from the audience, is invited on stage, and given a choice between doors #1 and #2.
Following this selection, we invite contestant C to the stage. Contestant C knows the rules of the game, but was absent from the room during the first two picks. Thus, he does not know which doors have been selected by A or B. Door #3 is ajar, and C does understand that M opened this door, in accordance with the rules, after A's selection was made.
Remove door #3, and invite contestant D to the stage, like C, D was out of the room during the earlier proceedings, but unlike C, he doesn't not know the game. We simply give D a choice between the two remaining doors.
At this point, the doors are opened. To keep the math simple, assume that the prize is not shared - each player gets full value if they selected the winning door.
The fun part of all this, of course, is that D chooses the same door as A 50% of the time, and D chooses the winning door 50% of the time (assuming D's choice is uniformly distributed).
Benford's Law - random samples of data do not have a random progression of first digits, but rather a logarithmic one, with 1 being the most likely.
Bizarre realization #1: this does not depend on the base used to represent the data.
If you are in the mood for a good math Soap opera, take a gander at the story of Tartaglia
Having tumbled into this (my notes are unclear. The immediately previous entry refers to Gaussians, and before that are several pages of Nash problems. I think I got to Tartaglia via Eratosthenes, but who can be sure.
11 pages of trash on cubics, mostly because I can't seem to keep track of negative signs. Once I got that working, and did a bit more reading, I started trying to work out which root would fall out of this mess (you are taking the cube root of a quadatic solution. The +- from the quadratic formula washes out, as it happens. So you just have to pick which cube root you want - oh, did we not mention that most of the time you are looking at complex solutions to the quadratic?
Getting some godawful mess, and converting it to cosine. 2 root 7 cosine of 1/3 of Acos ( root something else ). Bleah. Excel claims the solution is exact, and is in fact one of the roots I used to construct the original equation. Which suggests to me that I can find some identity to solve it exactly.
A brief burst of energy unveils the cos(3x) trig identity that I've never had a use for, but I soon discover that converting that to a cos(x/3) identity is solving a cubic [yes, to those of you snickering, I'm well on my way to duplicating de moivre the hard way]. And the minus signs continue to jump away, and weird things happen - that don't work when I plug them back in - until it finally occurs to me that this would all go a lot smoother if the quadratic had imaginary roots... Oh it does (another minus sign bites the dust).
Great, now I'm at de Moivre; realizing I've deadended myself, I step back where I should have to begin with: e^ix (I skipped a few physics courses I ought to have taken in college; this comes around to haunt me from time to time). Rotate the angle by 2pi/3, and the other loaded roots fall out.
OK, so with one real root, that's what the cubic gives us. When we have three real roots, the quadratic has imaginary solutions, so we're taking the cube root of a complex. Fine, but I still don't know which is which.
I think this means answering "which root will have the smallest theta", but I'm not sure.
If you happen to see a missing minus sign lying about, send it this way.
I've been reading a lot of Feyman lately, trying to acquire a better grasp of particle physics (purely for the entertainment value - yes, I'm a geek). This has revealed, among other things, that I know very little classical, newtonian physics.
For instance, how far away is the moon? Not the number, but how do you figure it out, given that you are an ancient greek? Eratosthenes put together a decent approximation.
Mark Pilgrim introduced Matchmaker, a fun bit of image generation - the puzzle is to find the unique match from a number of very similar tiles.
Given that there are so few color combinations, it seems reasonable to sample the pixels, use squash ordering to find an index for the image, then check an array to determine if that image has already appeared. On average N/(2^0.5) where N is the total number of images.
Thomas Andrews does an excellent job explaining how squashed ordering works.
I used this technique when working on the Fairy Chess exercise for Test Driven Development.
Mike coded a version of this puzzle as well, which is a lot easier, because his implementation allows you to specify the colors
My father was present in the lecture halls of Cal Tech when Richard Feynman was teaching the freshman physics course. I recently obtained CD versions of several lectures (the famous Lost Lecture on planetary bodies, Six Easy Pieces), that I've been listening to in the car. Wow - geometric constructions from Big Jule - let's shoot crap.
I need to work through the ellipse constructions, and decided to begin by reviewing the famous construction by Pythagorus. This got me to thinking about right triangles with sides of integer length. (3,4,5) is most well known, followed perhaps by (5,12,13), but I have never really explored how to find other triplets. It can't be all that hard...