Wikipedia:Reference desk/Archives/Mathematics/2011 February 20

= February 20 =

"Reducible" differential equation
I need to solve y '' =(x+y')^2. The instructions say it is reducible, so that p=y', p'=y ''. Substituting these, I get p'=(x+p)^2. I don't know where to go from here- isn't this a still a nonlinear equation? 149.169.134.102 (talk) 01:27, 20 February 2011 (UTC)


 * Yes, it's still nonlinear. But try setting q=x+p and see if you don't get something with a well-known solution anyway. –Henning Makholm (talk) 02:01, 20 February 2011 (UTC)

Geometric representation of square numbers
Please see this Google Books link (R. Fenn, Geometry, p. 6). Exercise 1.4 says "Show 1111111112 = 12345678987654321. A calculator will not help you here. A computer might but a few moments thought will convince you the equation is true." I'm not sure what the author is getting at here. I asked on IRC, but their primary suggestions – to use the formula for the sum of an arithmetic series or to try longhand multiplication – were not very geometric or compatible with the author's sentiment of "a few moments thought". Any suggestions would be appreciated. Thanks. — Anonymous Dissident  Talk 05:39, 20 February 2011 (UTC)
 * It's not square numbers in general, but that particular square number, which has that pattern because the multiplication can be seen as a convolution that doesn't involve any carries. Does that help? 71.141.88.54 (talk) 06:34, 20 February 2011 (UTC)
 * ... and, of course, looking at 112, 1112 and 11112 might help. This should also illustrate why 11111111112 breaks the pattern.    D b f i r s   08:30, 20 February 2011 (UTC)
 * I think what the author has in mind is to imagine a square with this side, partitioned into segments of 1, 10, 100 etc. This gives you a 9X9 grid with 1 rectangle on the first diagonal, 2 on the second, ..., 9 on the 9th, 8 on the 10th and so on. The rectangles on the nth diagonal have area $$10^{n-1}$$, hence the result. -- Meni Rosenfeld (talk) 09:19, 20 February 2011 (UTC)
 * I've drawn a picture of what you've described, and it fits together. Even if that's not what the author intended, it's a great geometric explanation for the result. Thanks. — Anonymous Dissident  Talk 10:56, 20 February 2011 (UTC)


 * That's a really beautiful construction Meni. Tinfoilcat (talk) 15:16, 20 February 2011 (UTC)

Just thinking about the usual multiplication algorithm you learned in third grade, applied to this particular case, seems to answer the question in an instant. Michael Hardy (talk) 20:49, 20 February 2011 (UTC)
 * That's of course right, but note the book mentioned is  'Geometry'  and Anonymous Dissident asks (implicitly) for more geometric suggestions. --CiaPan (talk) 06:58, 21 February 2011 (UTC)

complex vs chaotic
Is there a formal definition of "complex" dynamics that is distinct from "chaotic" dynamics, or do they mean the same thing? —Preceding unsigned comment added by 130.102.158.12 (talk) 09:47, 20 February 2011 (UTC)


 * I think it will depend on the context, but usually "chaotic" means more than just "complex". Chaotic dynamics has a fairly precise mathematical definition - see our article on chaos theory. A dynamical system with a large number of degrees of freedom could be described as complex, but is not nexessarily chaotic. Gandalf61 (talk) 11:16, 20 February 2011 (UTC)


 * Often complex dynamics refers to dynamical systems that arise by iterating holomorphic functions in a complex domain. These are typically chaotic discrete dynamical systems.   Sławomir Biały  (talk) 14:16, 20 February 2011 (UTC)

Is it optimal for the evaluation method used to change with frequency?
I must either take a bet or not: I have a 20% chance of winning £100000, otherwise I lose £10000. I don't have any money or much income, so if I lose I will have years of struggling to pay off the debt plus interest.

I know that if I have say one-hundred or more opportunities of taking the bet, then I should bet every time, as it has a positive expected value.

But what about if I will only ever have one opportunity to bet? The bet would still have a positive expected value, but I have an 80% chance of getting into heavy debt for years. Would it be optimal and correct to use maximin (?) instead of expected value when there is only one shot? Or should I stick with expected value in all circumstances? Thanks 92.28.245.90 (talk) 18:10, 20 February 2011 (UTC)


 * Your problem is somewhat similar to a St. Petersburg lottery, which has an infinite expected value in the long run, but meager payouts short term. As you indicate, £1 here isn't always worth the same as £1 there. £10 to someone who is pennyless is worth more (it has more utility) than £10 to a millionaire. Likewise, loosing £1000 is much more of a blow to someone who is just making ends meet than it is to a millionaire, or even someone who has £20000 in their savings account. Following the expected utility hypothesis, you first have to compute the raw monetary amounts in to the equivalent utility gain/loss for the person and situation, and then calculate the utility expected value, rather than the monetary expected value. If the bet has a positive expected utility, take it, if not, don't. The trick is what function to use for the money->utility conversion. When done properly, the conversion function will be different for each person, as a risk-adverse person will have more psychological trauma (and hence more negative utility) from going £10000 in debt than a more risk-loving person. -- 174.21.250.120 (talk) 18:53, 20 February 2011 (UTC)
 * You should not take risks you are not prepared to handle. A real world example is choosing to buy insurance. Since the insurance companies makes money the expected value of the insurance is less than you pay for it.  But because buying an insurance reduces the risks you have to take it will often make sense to buy one.  Taemyr (talk) 19:24, 20 February 2011 (UTC)
 * As 174 said, you need to maximize expected utility, not expected money. A reasonable function (far from perfect, though) to model the utility of money is the logarithm. So if your net worth, including your capacity to generate income, is £20000, then your utility on an arbitrary scale is $$\ln20000\approx9.9$$. If you take the bet your expected utility is $$0.2\ln120000+0.8\ln10000\approx9.7$$, so you shouldn't take it.
 * Using the maximin is equivalent to a utility of $$u(x)=-t^{-x}\;\!$$ as $$t\to\infty$$, which is not realistic.
 * A few points to consider:
 * To a first order approximation, log and any other smooth function is locally linear at any point; thus, when small amounts of money are concerned, maximizing utility is roughly equivalent to maximizing money.
 * To a second order approximation, log is a concave quadratic. Thus the utility of a Lottery (probability) (bet) depends on its expectation first, and on a negative term proportional to its variance second (and then on higher-order terms). For lotteries of similar expectation, the one with lower variance is preferred.
 * Your observation that taking multiple instances of the bet is good is not necessarily true. It is based on the intuition that averaging over many bets decreases the variance. But here you are not averaging, you're summing, which increases the total variance. Whether the outcome is good depends on the utility function, and I don't think it is for the log. If you have the option to take the bets sequentially and quit at any point, it is probably good for some values of the parameters. -- Meni Rosenfeld (talk) 19:42, 20 February 2011 (UTC)
 * Even if you take utility bets will make more sense if you spread them over several bets. For the simple reason that the probability that you will end up losing money decreases.  Taemyr (talk) 21:22, 21 February 2011 (UTC)
 * Your phrase "spread them over several bets" suggests that each of the several bets is smaller than the one bet. In this case yes, generally the multiple bets will be better, but that's not what the OP was suggesting.
 * If you take multiple large bets, your probability of losing money will indeed be small, but there is a positive probability of losing a large amount of money which will take you to a very deep place on your utility function. For example, if you currently have £1000000 and you take 100 of the OP's bets, there is a positive probability to end up with 0 which has $$-\infty$$ utility with the logarithmic function, thus the expected utility is $$-\infty$$.
 * You could argue that this is an artifact of using the logarithmic function. To replace it you will have to clarify if you are at all allowed to make bets you might not be able to pay, and if so, what happens in this case. Whatever the real-world situation is, you can model it as a utility function. And as I said, which lotteries are good simply depends on the details of the function, there's no sweeping generalization for it.
 * Lastly, "The probability that you will end up losing money" is a bad proxy for the utility of a lottery. It is equivalent to a utility function of $$M\cdot I(x\ge c)$$ where c is your current amount of money and $$M\to\infty$$.
 * -- Meni Rosenfeld (talk) 07:13, 22 February 2011 (UTC)

Loxodromic transformation: product of 4 inversions
Hi everyone,

I'm looking at ways of composing (circle) inversions to obtain Mobius maps: I've been able to show that hyperbolic, parabolic and elliptic transformations can all be written as the product of 2 inversions, but I'm having trouble showing that a loxodromic map can be written as the product of 4 inversions (it's not possible as 2) - could anyone explain how to produce a loxodromic map via 4 inversions?

Thankyou! Spalton232 (talk) 19:25, 20 February 2011 (UTC)


 * Loxodromic transformations are conjugate to a scaling-rotation $$z\mapsto \lambda z$$. You can write this as a reflection in two lines (to fix the argument of lambda) followed by a reflection in two circles centered at the origin ( fixing its modulus).  For the general case, you just need to conjugate with any möbius transformation that sends the fixed points of the loxodromic transformation to 0 and infinity.  Sławomir Biały  (talk) 19:53, 20 February 2011 (UTC)
 * Perfect, that's exactly what I was looking for; thank you very much! :) Spalton232 (talk) 20:56, 20 February 2011 (UTC)