Wikipedia:Reference desk/Archives/Mathematics/2007 September 2

= September 2 =

Fractals and Apophysis and Math...Oh My!
Hello, I wasn't sure where to put this question (so if I need to move it just say so!)

I recently downloaded the newest version of Apophysis (2.02) and I was wondering about this question: How Exactly do you make fractals?I already read the article on Fractals but it seems slightly ridiculous that I have Apophysis and not know how to make a fractal! I already can create thing from altering the code (I think it's a code) from fractals generated randomly by the program but is there any way to do it manually? Do you just type in random numbers like "01001001..." or is there more to it? And, will the fractal appear for a given parameter exactly the same on any other program or Apophysis? For example: if 01001001...appears as an "eye", will it look the same on another Apophysis or fractal program or will it be a completely different image for each one? P.S.- If anyone knows a good place to get started on fractals let me know!

Many thanks in advance for my confusing question- ♥   ECH3LON   ♥  00:30, 2 September 2007 (UTC)


 * I'm not familiar with Apophysis, but the screenshots I've seen make it look like an Iterated function system type of fractal whereit is generated by continually making copies of the whole that are scaled/rotated/distorted. We have other articles on Fractals, and you might want to read those. - Rainwarrior 05:09, 2 September 2007 (UTC)

Russian roulette
I've been cleaning up the Russian roulette article, and have removed the mathematics bit as it didn't agree with the answers I got. Can someone check out my logic (and the other person's) to check I've got this right? Whole shebang at Talk:Russian roulette. --h2g2bob (talk) 00:57, 2 September 2007 (UTC)


 * I'm not entirely certain why there should be an odds section in such an article at all. The idea that the odds change depending whether you start first or second is misguided. - Rainwarrior 04:42, 2 September 2007 (UTC)
 * Read the analysis more carefully.
 * If the chamber is not randomized after each trigger squeeze, what happens? Suppose, for simplicity, that we have two participants, one bullet, three chambers. If Alice goes first, she gets the bullet 1&frasl;3 of the time on the first squeeze. The remaining 2&frasl;3 of the time Bob gets to squeeze exactly once, and 1&frasl;2 of those times gets the bullet. Finally, 2&frasl;3×1&frasl;2 = 1&frasl;3 of the time Alice gets a second squeeze and is guaranteed to get the bullet. Said another way, Alice can get a bullet from chambers 1 and 3, while Bob can only get a bullet from chamber 2.
 * What if we have randomization, with two participants, one bullet, and two chambers? Half the time Alice gets the bullet on the first squeeze; Bob's cumulative risk, being a fraction of the remaining half, is always less. (And Alice gets more squeezes, so her risk is more than half.)
 * Either way, it's better to be Bob. --KSmrqT 09:06, 2 September 2007 (UTC)
 * With a standard 1 bullet, 6 chamber gun, Alice gets 1, 3 & 5 for 3/6 or 50% and Bob gets 2, 4 and 6 for 3/6 or 50%. So unrandomised it doesn't matter. For respinning, the odds are 1/6; 5/6*1/6; 5/6*5/6*1/6; 5/6*5/6*5/6*1/6 etc each term tending to 0. Alice gets 1/6; 5/6*5/6*1/6; ... Bob gets 5/6*1/6; 5/6*5/6*5/6*1/6; ... Note that Bobs terms = Alice terms *5/6, and as Alice+Bob=1 we have Alice dying at 6/11 and Bob at 5/11. -- SGBailey 10:08, 2 September 2007 (UTC)


 * In the non randomized version, the idea that the game would be played with three chambers is not realistic (does such a gun even exist?); a 6-chambered gun works for 2,3 or 6 players in this way and with 4 or 5 it should be immediately obvious that if one or two players have to take an extra pull they are more likely to die; we don't need to use symbols and terminology from probability to explain something that is just common sense. I was actually not aware that there was a "randomized" style of play, but yes, that would disadvantage the first player slightly since he will always have taken either the same number of pulls or one more than the others at any given time (this could be balanced: if player 1 loses, the gun is reloaded and player 2 has one more pull. either that or player 1 gets more money for winning, etc.). Maybe for this second unbalanced play style it would be worth showing how to calculate the odds for each round (or as it trends as it approaches infinity). - 17:52, 2 September 2007 (UTC) —Preceding unsigned comment added by Rainwarrior (talk • contribs)
 * That's a backwards way to agree you were wrong. "Not realistic"? For a "game" risking suicide? Notice I refused to dignify the game idea, instead referring to "participants" rather than "players". And my brief analysis did its job, reducing the question to such simple examples there could be no doubt about the results. This is a strategy I highly recommend; while an example cannot substitute for a proof (except a proof of existence!), one good counterexample can beat pages of reasoning. --KSmrqT 06:24, 3 September 2007 (UTC)


 * (After edit conflict> Let's analyse the one bullet game, with or without randomisation (spinning the chambers) after each turn. Suppose there are n chambers; let a be the probability that Alice (first player) wins and let b be the probability that Bob (second player) wins.
 * With randomisation after each turn: Bob wins immediately if Alice loses on her first turn - probability 1/n. If Alice survives her first turn - probability (n-1)/n - then Bob must take a turn, and by symmetry his probability of winning is now a. So


 * $$b=\frac{1}{n}+\frac{n-1}{n}a$$


 * But a+b=1. so


 * $$b=\frac{1}{n}+\frac{n-1}{n}\left(1-b\right)$$


 * $$\Rightarrow b=1-\frac{n-1}{n}b$$


 * $$\Rightarrow b=\frac{n}{2n-1}$$


 * $$\Rightarrow a=\frac{n-1}{2n-1}$$


 * Without randomisation after each turn: Call the chamber on which Alice starts chamber 1. Then Alice wins if the bullet is in an even numbered chamber and Bob wins if the bullet is in an odd numbered chamber. So if n is even, a and b are both 1/2, and if n is odd then


 * $$a=\frac{n-1}{2n}$$


 * $$b=\frac{n+1}{2n}$$


 * So Bob's probability of winning is at least as good as Alice's, and sometimes better. Interestingly, if Alice knows she has to go first but she can choose the style of play, then she should choose ramdomisation if there is an odd number of chambers, and no randomisation if there is an even number of chambers. Gandalf61 10:26, 2 September 2007 (UTC)


 * Doi! I've been an idiot - the original version switches back and forth between with and without spinning - no wonder I got confused :) Thanks for the great explanations, they're much easier to follow. Rainwarrior: you're right - I was thinking of a putting it in wikibooks, with just the answer in the article. --h2g2bob (talk) 17:13, 2 September 2007 (UTC)

Infinite Product
Consider the infinite product (1-x)(1-x2)(1-x3)(1-x4)... Is it convergent or divergent (and for what x is it so)?

I've attempted to expand the product but that seems fairly useless (and possibly not very rigorous in dealing with it) so my other possible approach was to take the log of it and then you get an infinite series (which are usually easier to evaluate). The ratio test is useless on the series but I thought the root test might work. Though I'm not very good at analysis so I was wondering whether this was the correct approach to the problem? (The other problem is that i would be taking the roots of negative numbers since 1-xn is always less than 1 so I'm not sure if that procedure is valid. Is there any other fairly elementary way of evaluating this product? --AMorris  (talk)  &#x25CF;  (contribs)  14:37, 2 September 2007 (UTC)
 * There is no reason to assume that $$x > 0$$, which you seem to have done in stating that $$1-x^n<1$$. There are, of course, some negatives, but this is not a problem since in the root test you take the absolute value. However, I recommend a different approach to show convergence. Hint - for sufficiently small y we have $$|\log(1-y)| < 2|y|$$.
 * Finding a closed form expression for the value of the product is a different matter entirely, and I doubt much can be said there. -- Meni Rosenfeld (talk) 15:41, 2 September 2007 (UTC)
 * As for expanding the product, it requires some more work but it can be shown to be possible; it also seems that the power series expansion of this function is $$\sum_{n=0}^{\infty}x^{6n^2+n}(1-x^{4n+1}-x^{6n+2}+x^{10n+5})$$, but proving this could be difficult. -- Meni Rosenfeld (talk) 16:03, 2 September 2007 (UTC)


 * See pentagonal number theorem and Euler function. -- Fredrik Johansson 17:58, 2 September 2007 (UTC)