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July 22

Showing inverse of a bijective holomorphic function on open set is holomorphic

I am pretty sure I proved this in detail, following a proof in my book. But, I am trying to get a quick proof in case I need to prove it on a qualifying exam just to use the result. By quick, I mean I want to explain it a bit but not show much of the detail, basically name a couple theorems. Is the following enough?

First, f(z) is never 0 since it is one-to-one. By the inverse function theorem, $(f^{-1})'(z)={\frac {1}{f'(f^{-1}(z))}}$ for each point. Since f(z) is holomorphic, f'(z) is also, so 1/f'(z) is also as the quotient of holomorphic functions is holomorphic as long as the denominator is not 0.

Thanks. StatisticsMan (talk) 01:04, 22 July 2009 (UTC)

Which inverse function theorem are you using, exactly? And how are you showing that injective holomorphic functions have nonzero derivative? Algebraist 01:08, 22 July 2009 (UTC)

First, I had the derivative formula incorrect and have changed it. As to which inverse function theorem I'm using, I do not know I guess. Are you asking because the formula was wrong? The detailed proof I got from the book mostly starts by showing the inverse is continuous at each point. Then, it shows the derivative is equal to 1 / f'(f^{-1}(z)) at each point. This is the result I want really, though I don't see the exact thing in the article on Inverse function theorem here. If I consider the total derivative, that is the complex derivative, and consider $\mathbb {C} =\mathbb {R} ^{2}$, then it says that the derivative of the inverse exists and is continuously differentiable. But, it does not give a formula for it. Is it always the same formula? As far as showing that an injective holomorphic function have nonzero derivatives, this is something that is tricky to me. My book gives a theorem that says if a nonconstant holomorphic function on a connected open set which has f(P) = Q with order k, then there exists a small disk around Q such that every point in the disk (other than Q) has exactly k distinct preimages in a small disk around P. But, I don't see how this implies a nonzero derivative implies that the function is locally one-to-one, as they say. StatisticsMan (talk) 01:45, 22 July 2009 (UTC)

I asked about your inverse function theorem because there is a complex inverse function theorem which contains the result you're trying to prove, and presumably would not be accepted as part of a proof of it in an exam. Algebraist 01:48, 22 July 2009 (UTC)

Well, I would like to know how to prove this directly and in a nice way. But, for now I'm working on a problem asking if there is a bijective holomorphic function which maps the unit disk onto the complex plane. The easy solution is, no, because its inverse is entire and bounded and thus constant. If I take some time to explain that, including using Liouville's theorem and the inverse function theorem and such, then perhaps it would not be trivial and would be good enough to get credit. If I have time later, I can come back to try to prove more. But, the point is, I want a quick proof of the fact to use in cases like this. StatisticsMan (talk) 01:56, 22 July 2009 (UTC)

Also, is the complex version just the same basic thing? I looked up "inverse" in my 3 main complex analysis books and found nothing basically. StatisticsMan (talk) 01:57, 22 July 2009 (UTC)

I would call the following result an inverse function theorem: let f be a holomorphic function from an open set U to the complex plane, let x be in U, and suppose f'(x)≠0. Then there is a neighbourhood V of x such that f is injective of V, the inverse map is holomorphic, and the derivative of the inverse is given by [whatever the formula is]. Algebraist 02:17, 22 July 2009 (UTC)

Here's how the book I have proves it:

First they prove an injective holomorphic function f on an open set U has non-zero derivative. Suppose f'(z₀) = 0. Then $f(z)-f(z_{0})=a(z-z_{0})^{k}+G(z)$ for some a ≠ 0, k ≥ 2, and G(z) a function that vanishes to order k+1 at z₀, so then we can choose a radius δ > 0 around z₀ small enough so that |G(z)| < (a/2)δ^k = c for all z within δ of z₀ and so that f'(z) ≠ 0 for z ≠ z₀. Let $F(z)=a(z-z_{0})^{k}-c$, so that f(z) - f(z₀) - c = F(z) + G(z). |F(z)| = c > |G(z)| along the circle of radius δ, so by Rouché's theorem, f(z) - f(z₀) - c has k (not necessarily distinct) zeros in the circle. z₀ is not a root, so f' is non-zero at each root, so the roots are distinct, so f is not injective. So I guess the basic idea is that if a holomorphic function has a multiple root at some point, then by adding a small constant we get a function that has distinct roots near the point, which shows that neither function is injective in a neighborhood of the point.

Then they prove the inverse function part just using the definition of the derivative. Let g = f^-1 on f(U) and w = f(z). $\lim _{w\to w_{0}}{\frac {g(w)-g(w_{0})}{w-w_{0}}}=\lim _{f(z)\to f(z_{0})}{\frac {z-z_{0}}{f(z)-f(z_{0})}}$ which converges to 1/f'(z). You don't need to show that 1/f' is holomorphic, just that the complex derivative of g exists everywhere in f(U).

(I hope I didn't screw any of this up.) Rckrone (talk) 03:54, 22 July 2009 (UTC)

What book is this from? StatisticsMan (talk) 00:58, 30 July 2009 (UTC)

July 23

A variation of the knapsack problem

I will describe the problem that I need to solve:

Given a budget I need to choose parts that will be used to build a car. To build a car I need a-

1. Engine

2. Body

3. Set of wheels

I need to choose one engine, once body, and one set of wheels from a set of 1000 engines, 1000 bodies, and 1000 sets of wheels. (I also know the price of each engine, each body, ... ).

In addition every part has a power ranking that is between 1-100.

I need to choose 3 parts so that the car that I get will cost less than (or equal to) my budget, and that its "power ranking" will be maximized. ("power ranking" := the sum of the power rankings of the engine, body, wheels.)

Can you please tell me if this can be reduced to the knapsack problem, and how I would solve this problem.

Thanks!

Quilby (talk) 03:11, 23 July 2009 (UTC)

We do have an article about the knapsack problem that describes some approaches. The problem is formally NP-hard, but instances that occur in practice appear to usually be tractable. 70.90.174.101 (talk) 05:04, 23 July 2009 (UTC)

Right, but my problem is different. Notice how in the regular knapsack problem, all weights are 'equal'. Here I have 3 different categories of weights and I need to choose 1 from each. If someone can give me some tips on how to solve this problem it would be helpful. 94.159.156.225 (talk) 05:23, 23 July 2009 (UTC)

The naive approach of trying all combinations is only n³ so it's clearly polynomial. An n² algorithm is not too hard. Sort each set of parts by cost and throw out the strictly worse choices, then for each engine, traverse the bodies list from most expensive to cheapest while traversing the wheels list in the opposite direction looking for the most expensive wheels that still put everything under the budget. Hopefully someone can think up something more clever though. Rckrone (talk) 06:40, 23 July 2009 (UTC)

By the way you are looking for an exact solution, right? Rckrone (talk) 06:44, 23 July 2009 (UTC)

I dont know what you mean by exact, but as I said - the price of the final car needs to be smaller than or equal to my budget. So I think this is not an exact solution. Quilby (talk) 21:38, 23 July 2009 (UTC)

For each possible power ranking (1..100) select the cheapest engine, the cheapest body, and the cheapest set of wheels, and discard the rest. This reduces the size of the problem from 10^9 to 10^6 combinations to consider. Also discard any engine (body, wheels) that costs more than some higher ranked engine (body, wheels). Next, subtract the price of the cheapest engine (body, wheels) from the prices of all the engines (bodies, wheels), and from your budget. Now your budget reflects how much extra money you can spend to buy something more expensive than the very cheapest car. Then discard all parts that individually exceed you budget. The rest is brute force. First consider the set of pairs, (the cartesian product), of engines and bodies. Discard all pairs that cost more than your budget. Sort on ascending ranking. Retain only the cheapest pair of each rank. Also discard more expensive pairs of lower ranks. Finally combine the pairs with the wheels. Bo Jacoby (talk) 10:52, 23 July 2009 (UTC).

PS. ("power ranking" := the sum of the power rankings of the engine, body, wheels.) Is that really what you want? Is it a good idea to use fancy wheels together with a lousy engine? I suppose you would be happier maximizing the minimum of the power rankings of the engine, body, wheels. Bo Jacoby (talk) 13:18, 23 July 2009 (UTC).

That algorithm is still worst case n³. For a hard data set it doesn't reduce the number of combinations tried. Rckrone (talk) 17:05, 23 July 2009 (UTC)

Oh I see what you're saying. Since there are only 100 power ratings but 1000 parts, there are guaranteed to be some stupid ones (assuming only integer power ratings which is not specified). Still once you've weeded out the bad ones, you can do better than brute force. Rckrone (talk) 17:43, 23 July 2009 (UTC)

Let's try to solve a more general problem while combining all of the above. We have k types of parts, for each we have n models, each having an integer power rating between 1 and m (in the OP $k=3,n=1000,m=100$).

We construct an $m\times k$ array of the identity and cost of the cheapest model of each type and power rating, by going over all $n\cdot k$ models. We then proceed in $k-1$ steps; in the first step we consruct an $2m\times 1$ array of the identity and cost of the cheapest pair of type 1 and type 2 models for each combined power rating between 1 and $2m$, by going over all $m^{2}$ pairs. In the ith step we combine the list of 1-i combinations with the list of $i+1$ type objects to a single $(i+1)m\times 1$ array by going over all $im^{2}$ pairs. In the end we have a list of the cheapest combination of parts for every possible power rating, in which we can binary-search whatever our budget is. All this takes $O(nk+m^{2}k^{2})$ operations.

Of course, in practice, if we know the budget in advance we can skip the last step and do only upwards\downwards traversal.

If the power rating need not be an integer, we might have to use a more sophisticated data structure to achieve better than $O(n^{k-1})$. -- Meni Rosenfeld (talk) 18:57, 23 July 2009 (UTC)

If power isn't an integer, $n=2$ and k is large, then this is the knapsack problem. If, as in the OP, n is large and k is small, then the problems seem unrelated. -- Meni Rosenfeld (talk) 19:23, 23 July 2009 (UTC)

Yes you are correct. Quilby (talk) 21:38, 23 July 2009 (UTC)

I would like to thank everyone who answered my question. Quilby (talk) 21:38, 23 July 2009 (UTC)

Units

Consider the differential equation
${\frac {dr}{dt}}=(1-r)^{3}+3p(1-r)r^{2}-3(1-p)(1-r)r^{2}.$
In this differential equation, r is a density (of anything, I don't think it matters). And p is a probability. My questions is how can one explain that the units in this differential are correct. I came across this while reading a paper and the author derives this equation. On the LHS, it looks like the units should be 1/time but the right hand side appears unitless so the equal sign doesn't make sense. Am I missing something? Even if non-dimensionalization was used, what was done? I mean if all the terms on RHS had p or something, one can say something about p being proportional to a constant (which will correct the units after division and redefining the variable t) but that isn't true anyway. Is it r? There are powers of r on the RHS. What could be the units of r? Anyone can shed some light on this? Thanks!-69.106.207.49 (talk) 07:19, 23 July 2009 (UTC)

I don't think there's anything here you're missing; maybe with some more context from the paper (or a link) we might be able to help better. Because the expressions "1 - r" and "1 - p" appear, r and p have to be unitless, making the right-hand side as a whole unitless; for the units to match, we have to make t unitless as well, which quite possibly is what is intended by the author (or possibly not). There also could be an implicit conversion factor in the right side, say, dividing all of it by one second or some such.

It looks like the right hand can be simplified? The last two terms can be combined. Eric. 76.21.115.76 (talk) 07:44, 23 July 2009 (UTC)

Yes, it can be simplified to ${\frac {dr}{dt}}=(1-r)^{2}(6p-r-2)$, unless I've made a mistake. --COVIZAPIBETEFOKY (talk) 12:25, 23 July 2009 (UTC)

Haha, I shouldn't have jinxed it like that! ${\frac {dr}{dt}}=(1-r)[(1-r)^{2}+r^{2}(6p-3)]$. This next one is not really a simplification, but expanding the part in square brackets gives ${\frac {dr}{dt}}=(1-r)[(6p-2)r^{2}-2r+1]$, which can't be factored nicely. --COVIZAPIBETEFOKY (talk) 12:31, 23 July 2009 (UTC)

Max mod principle problem

Hi, I'm working on a problem and there are 2 cases. One I have and the other I have no clue. This is a qual problem so I have old solutions from two different people. But, one person didn't do this case and the other one did something that makes no sense. So, I guess I'm wondering if the problem is even correct. If it is, can you help me figure out how to do it?

Assume that f, g are holomorphic in the disk D = D(0, 1), continuous on the closure of D and have no zeros in D. If $|f(z)|\equiv |g(z)|$ for |z| = 1, prove that f(z) = k g(z) in D, for some constant k of modulus 1.

Alright, so case 1 is f(z) is never 0 on the boundary. The condition given implies g is never 0 either. So, we can get $|f(z)/g(z)|\leq 1$ and $|g(z)/f(z)|\leq 1$ on D by the max modulus theorem since f = g on the boundary. Putting those together gives |f(z) / g(z)| = 1 on D. But, then the max occurs inside the disk so f / g is constant on D. And, clearly the constant is unit modulus.

Case 2: f(z) = 0 for at least one z in the boundary. I have no idea what to do here other than f(z) = 0 if and only if g(z) = 0. Any ideas? Thanks StatisticsMan (talk) 15:31, 23 July 2009 (UTC)

If f(z₀)=0, consider f(z)/(z-z₀) and g(z)/(z-z₀). Bo Jacoby (talk) 18:22, 23 July 2009 (UTC).

Off the top of my head, I think you're saying to get rid of the zeros by dividing by (z - z_0), one for each zero of f (or g), and then use the previous case. Is this right? But, what if f is 0 an infinite number of times around the boundary? Or, if f and g are both zero but when getting rid of the zero by this method, maybe the new f and g satisfy f(1) = 100 and g(1) = 50... so they no longer have equal magnitude around the boundary and case 1 doesn't apply? This is part of my problem. Since f and g are not holomorphic on the boundary, do I know anything about the zeros? StatisticsMan (talk) 19:28, 23 July 2009 (UTC)

${\frac {f(z)}{z-z_{0}}}$ is not even necessarily continuous - for example, $f(z)={\sqrt {z+1}}$ for an appropriate branch. -- Meni Rosenfeld (talk) 20:21, 23 July 2009 (UTC)

(edit conflict). Meni, you have got a point. $f(z)=\sum _{k=0}^{\infty }{\binom {1/2}{k}}z^{k}$ is holomorphic for |z| < 1 and zero for z = −1, and ${\frac {f(z)}{z+1}}=\sum _{k=0}^{\infty }{\binom {-1/2}{k}}z^{k}$ is not continuous on the closure of D. Statman, yes, that was what I was trying to say. Can a nonzero holomorphic function have an infinite number of zeroes around the boundary? The new f and g satisfy |f(z)| = |g(z)| for |z| = 1, z ≠ z₀; by continuity |f(z₀)| = |g(z₀)|. Bo Jacoby (talk) 08:14, 24 July 2009 (UTC).

Yes: by the Riemann mapping theorem. There exists a holomorphic function on the disk, continuous up to the closure, whose zero set at the boundary is any given closed subset you fixed. --pma (talk) 14:16, 24 July 2009 (UTC)

You may adapt your case 1 to circles of radius r<1, where you know that f and g do not vanish. By the uniform continuity of f and g on the unit closed disk, you have, uniformly for |z|=r, |f(z)| ≤ ||g||_∞,D + o(1) as r→1. Apply the max mod to f to extend the inequality on the disk of radius r as you were doing, and conclude. PS: Does not work --pma (talk) 07:44, 24 July 2009 (UTC)

Really? The article on the Riemann mapping theorem merely says that the disk D is mapped one-to-one to any simply connected open subset. Bo Jacoby (talk) 20:51, 24 July 2009 (UTC).

Yes, but it says more if the assumptions are strenghtened. As a matter of fact, any closed curve with finite length has a re-parametrization that extends to a holomorphic map on the disk, continuous up to the boundary. So it seems to me that this allows to construct a holomorphic function on the disk, continuous on the closure, with infinitely many zeros on the boundary (in fact now I'm not sure to get any given closed subset as a zero set, even if I think I remebmer that you can do that; I will think about this... --pma (talk) 06:57, 25 July 2009 (UTC)

How can a one-to-one mapping have more than one zero? Bo Jacoby (talk) 15:49, 25 July 2009 (UTC).

The holomorphic mapping of the open disk is one-to-one, its continuous extension to the boundary is not. — Emil J. 12:18, 27 July 2009 (UTC)

Multiple regression v. neural nets

Were neural nets just a fad? I would like to try predicting or forecasting the real-estate market using several time series of economic data. The dependant variable would be a time series of real estate average prices. The independant variables would be time series of things like unemployment, inflation, stock market index, and so on, and a lagged real-estate average price time series. Which would be the better methodology to use? And does anyone have an opinion on which of the various specific types of the two methods would be best for what I want to do? Would using independant variables which were correlated make any difference to the choice - I would also like to try using lagged real-estate average price time series from other geographic areas. Thanks. 89.240.217.9 (talk) 21:35, 23 July 2009 (UTC)

Software packages will happily perform multiple regression, but they don't correct for data anomalies. If your data exhibits (as most time series data does) heteroskedasticity, multicollinearity, non-stationarity, etc. (the list of anomalies is rather large), then the results that the software reports could be meaningless. Wikiant (talk) 22:52, 23 July 2009 (UTC)

But if you know what you're doing, you can usually use the software to diagnose those problems, and in some cases then transform the data in ways that make it amenable to multiple regression. Michael Hardy (talk) 23:49, 23 July 2009 (UTC)

Absolutely, but you'll probably need a one or two stats courses plus at least two econometrics courses to get there. My implied warning is that a little knowledge of regression is more dangerous than no knowledge of regression -- it enables one to use the software to produce results without knowing that the results can be meaningless. Wikiant (talk) 11:15, 24 July 2009 (UTC)

I dimly recall studying about heteroskedasticity, multicollinearity, and non-stationarity years ago - I need a refresher. But my question was, would neural nets be better than multiple regression in these circumstances? I'm not aware of any no-cost software that lets you use neural nets, (R may do it, but seems difficult to use), so I'd have to code it myself. 78.151.148.77 (talk) 14:38, 24 July 2009 (UTC)

I don't do neural nets, but my guess is that they would be as susceptible to these problems as is regression because the anomalies reside in the data, not in the analytic techniques. Wikiant (talk) 15:56, 24 July 2009 (UTC)

Both have their points, but talking about forecasting sales of a drink for instance, they would try doing multiple regression with some heuristic functions to try and get straight lines after correcting for seasonal variations - but there's all sorts of other things too. For instance they will also try predicting the amount sold for various types of holidays, they have to get a forecast for the next couple of weeks weather, if there is a big match there is also the variability depending on who wins - and that would affect the area with the match more. I'm sure the problem is the same with real estate plus of course there will be a much bigger dependency on the stock market. Sometimes it is fairly stable and other times it goes haywire. See Super Crunchers for a good read and an introduction if you want to get into this sort of thing. Regression might work better for drink but neural networks would I guess be more useful with real estate. It is very worthwhile trying to figure out why a neural network gets good results if it works well and that can sometimes be difficult. Dmcq (talk) 19:27, 24 July 2009 (UTC)

July 24

population proportion

When we make inferences about one population proportion, what assumptions do we need to make? Mark all that apply.

• a. Random samples.
• b. Normal distribution of the response variable.
• c. The sample size is 30 or greater.
• d. Counts of successes and failures at least 15 each.
• e. Counts of successes and failures at least 5 each.

Well, I do assume simple random sample (A). And since data is categorical (yes/no), it's not normally distributed (so not B). The sample size (n) of 30 is a population mean/sample mean assumption (so not C). But what about D or E? —Preceding unsigned comment added by 70.169.186.78 (talk • contribs) 05:34, 24 July 2009

Consider a population of $\scriptstyle N$ items of which $\scriptstyle I$ are special. Take a sample of $\scriptstyle n$ items of which $\scriptstyle i$ are special. This can be done in $\scriptstyle {\binom {I}{i}}{\binom {N-I}{n-i}}$ ways. This is the well known hypergeometric distribution formula.

Deduction is estimating sample information from population data. Knowing $\scriptstyle N,n,I$ the mean value of $\scriptstyle i$ is $\scriptstyle \mu ={\frac {nI}{N}}$, and the variance-to-mean ratio is $\scriptstyle \varepsilon ={\frac {(N-n)(N-I)}{N(N-1)}}$, so the estimate is $\scriptstyle i\approx f(N,n,I)=\mu \pm {\sqrt {\mu \varepsilon }}$.

Example: $\scriptstyle f(2,1,1)=\mu \pm {\sqrt {\mu \varepsilon }}$ where $\scriptstyle \mu ={\frac {(1)(1)}{2}}={\frac {1}{2}}$ and $\scriptstyle \varepsilon ={\frac {(2-1)(2-1)}{(2)(2-1)}}={\frac {1}{2}}$. So $\scriptstyle f(2,1,1)={\frac {1}{2}}\pm {\sqrt {({\frac {1}{2}})({\frac {1}{2}})}}={\frac {1}{2}}\pm {\frac {1}{2}}$.

This result is exactly what should be expected: if the population contains two items, ($\scriptstyle N=2$), one of which is special, ($\scriptstyle I=1$), you take a sample containing one item, ($\scriptstyle n=1$), then you don't know whether this selected item is special or not, so the estimate of the number of special items in the sample is $\scriptstyle i\approx {\frac {1}{2}}\pm {\frac {1}{2}}$.

Induction (or inference) is estimating population information from sample data. Knowing $\scriptstyle N,n,i$ you estimate $\scriptstyle I\approx F(N,n,i)=-1-f(-2-n,-2-N,-1-i)$. This formula is exact, for small or big samples, and for small or big populations. The only assumption is that the sample is random.

Example: $\scriptstyle F(1,0,0)=-1-f(-2-0,-2-1,-1-0)=-1-f(-2,-3,-1)=-1-(\mu \pm {\sqrt {\mu \varepsilon }})=(-1-\mu )\mp {\sqrt {\mu \varepsilon }}$ where $\scriptstyle \mu ={\frac {(-3)(-1)}{-2}}=-{\frac {3}{2}}$ and $\scriptstyle \varepsilon ={\frac {((-2)-(-3))((-2)-(-1))}{(-2)((-2)-1)}}=-{\frac {1}{6}}$. So $\scriptstyle F(1,0,0)=(-1-(-{\frac {3}{2}}))\pm {\sqrt {(-{\frac {3}{2}})(-{\frac {1}{6}})}}={\frac {1}{2}}\pm {\frac {1}{2}}$.

This result is exactly what should be expected: if you take no sample, ($\scriptstyle n=i=0$), and the population contains one item, ($\scriptstyle N=1$), then you don't know whether this item is special or not, so the estimate of the number of special items in the population is $\scriptstyle I\approx {\frac {1}{2}}\pm {\frac {1}{2}}$. Bo Jacoby (talk) 13:30, 24 July 2009 (UTC).

You need (a) or something like it. Let's say you want to estimate the proportion of voters who will vote Republican next week. If you take your sample from the group of Republicans who are meeting in the building next door, you're making a mistake.

It doesn't make sense to assume a normal distribution. Each person will either vote Republican or not, so you get either a 0 or a 1, and that's not normally distributed. But you might conclude that the total number who will vote Republican is approximated normally distributed—that depends in part on sample size and in part on how the sample was taken. But it's a logical inference, not an assumption.

For a binary response variable, the question of whether that sum is approximately normally distributed depends not only on sample size, but also on how close the proportion is to either of the two extremes–0 and 1. And there are ways of drawing inferences when it's not approximately normally distributed and the sample size is small.

One sometimes sees a rough rule of thumb that you shouldn't conclude approximate normality unless you've got at least five outcomes in each of the two categories. I would add that you should use a continuity correction unless the numbers in both categories are pretty big. Michael Hardy (talk) 23:52, 24 July 2009 (UTC)

July 27

Atoms in measure theory

So, there is a qualifying exam problem which defines an atom to be

$A_{x}=\{\cap B:x\in B\in {\mathcal {M}}\}$

But, in the article in Wikipedia, Atom_(measure_theory), it says that Lebesgue measure has no atoms. Is this a situation where both are right, some people define it one way and some another? StatisticsMan (talk) 00:12, 27 July 2009 (UTC)

It's defining atoms differently, and individual points in space are the "atoms" of the sigma-algebra of Lebesgue-measurable sets. Notice that A_x is not required to have positive measure here.Ray^Talk 00:27, 27 July 2009 (UTC)

If I hear the word "atom" in measure theory, the first thing that comes to mind is a set of strictly positive measure that contains only one point. Or it may be a set of strictly positive measure that has no measurable subsets. "Atom" traditionally means something that cannot be divided into smaller pieces, so it should mean something that cannot be divided into smaller measurable parts.

In particular, in the theory of Boolean algebras, one sometimes considers measures on the algebra of all clopen subsets of a Stone space (not generally a sigma-algebra), and in that context, the standard usage is a one-element set whose measure is strictly positive. Michael Hardy (talk) 02:47, 27 July 2009 (UTC)

Statistical terms.

Dear Wikipedians:

Within the context of statistics, what do the terms "degree of performance" and "degree of consistency" mean?

Thanks.

174.88.241.78 (talk) 00:01, 27 July 2009 (UTC)

I haven't encountered either term, but consistency (statistics) could have some relevant material. On the other hand, there are various terms having to do with application of statistics to particular fields, in which the term might have other meanings. Michael Hardy (talk) 02:41, 27 July 2009 (UTC)

Thanks Dr. Hardy for your response. 174.88.241.78 (talk) 21:02, 27 July 2009 (UTC)

Resolved

Chaos

I'm looking for an example of a 2D continuous chaotic map, but the only continuous maps I've found are in three dimensions. Does anybody know of a 2D example? If so, what are its defining equations? —Preceding unsigned comment added by 72.197.202.36 (talk) 05:21, 27 July 2009 (UTC)

Do you want a map from $\mathbb {R} ^{2}\rightarrow \mathbb {R} ^{2}$? I can do $\mathbb {R} \rightarrow \mathbb {R} ^{2}$. --Tango (talk) 14:18, 27 July 2009 (UTC)

For the R2->R2 case, try the Henon map. risk (talk) 15:11, 27 July 2009 (UTC)

A 2d phase space for a continuous dynamical system can't be chaotic, see Poincaré–Bendixson theorem. Dmcq (talk) 17:40, 27 July 2009 (UTC)

Ah, but he was asking for a continuous map, which would imply a discrete dynamical system, where the function defining it is continuous. At least that's how I interpreted it. risk (talk) 18:34, 27 July 2009 (UTC)

I must admit that sounds more likely, I think I've got in the habit of wondering if people mean something else and just taking key words out of what they say :) Dmcq (talk) 19:17, 27 July 2009 (UTC)

Is the Lorenz attractor (oops, 3D) Smale horseshoe not the most famous one? 70.90.174.101 (talk) 09:46, 28 July 2009 (UTC)

Easter eggs in baskets

Hey all,

My math teacher gave me a problem today: given twenty baskets, put a number of easter eggs in each basket such that no two baskets have the same number of eggs and the total number of eggs is minimized.

After confirming that the number of eggs in each basket had to be a positive integer, I said $\sum _{i=1}^{20}i=210$. He says that's wrong and that the answer is nontrivial, but refuses to tell me what he thinks it is.

Confused, I proved that the above sum is the optimal solution for a set of positive integers - he said my proof was valid, but my answer was not.

No idea what he's getting at, and beginning to think that it's semantics - any ideas? Thanks, Aseld ^talk 10:47, 27 July 2009 (UTC)

It's possible he's one of the people who considers zero to be positive. Algebraist 11:10, 27 July 2009 (UTC)

Can you put a basket in a basket? In which case you could get away with twenty eggs even without a basker of zero eggs. Dmcq (talk) 12:00, 27 July 2009 (UTC)

What was the question? You suppose that you are requested to provide the total number of eggs in the baskets, but why suppose so? Perhaps the question is in how may ways it can be done, and the answer is 2432902008176640000. Bo Jacoby (talk) 12:16, 27 July 2009 (UTC).

I think putting baskets inside each other is probably the answer wanted, however for a mathematician I think the correct answer is to put ever larger and larger IOU's for eggs in each basket, they can be paid back in coffee and biscuits and odd jobs ;-) Dmcq (talk) 12:36, 27 July 2009 (UTC)

The problem description requests a positive number of eggs to be put in each basket. — Emil J. 13:12, 27 July 2009 (UTC)

Oh I see, I thought that was just an assumption but it was confirmed. Very good you never know what kind of devilish tricky thing they might be up to. Twenty eggs in total it is then. Negative numbers can be used to solve problems as at Math Jokes and Archimedes in the one where the mathematician, biologist and physicist are sitting in a street cafe. 13:56, 27 July 2009 (UTC)

"Given twenty baskets" is the key, as the solution depends on the form of the baskets. If they are undeformable and the same size they can't be put inside each other, unless they are tapering and will fit together with an egg-sized space between. "Twenty shopping bags" would be better, as they may be presumed to be flexible.86.148.186.45 (talk) 14:12, 27 July 2009 (UTC)

Thanks for the responses everyone. I think the stacking baskets is the answer that's wanted - will find out tomorrow. Little disappointed in the problem, I was expecting some pretty mathematical solution. I would have never thought of that myself - thank you. ^_^ Aseld ^talk 14:28, 27 July 2009 (UTC)

Yes I'd like some quantum mechanical sort of solution where you had fewer than twenty eggs but whenever you looked at any two baskets they always had a different number of eggs. Can't see how to so something like that though. And they'd have to be very small eggs. Dmcq (talk) 14:37, 27 July 2009 (UTC)

Representing an affine transformation

I'm trying to find a good intuitive representation for an n-dimensional affine map, as a vector of real numbers. The simplest representation would be to represent the map as a transformation matrix and a translation vector (n^2 + n real values), but I'd like each value to have a single clearly defined function.

I'm thinking about splitting the transformation into a translation, a scaling, a rotation and various shears. My first question is, does this cover all affine operations? I've found a paper that says so, but I'm not sure if they're just referring to the 2D case.

The translation and scaling are each defined by n values with a straightforward interpretation. The rotation can be represented by (n^2 -n)/2 angles, but I can't find any good references for the shear. From shear matrix, I gather that there are at least (n^2 -n) ways to create an elementary skew matrix (one for each non-diagonal element). But then I started thinking, I can do a QR decomposition on a matrix, and end up with a rotation matrix, and an upper diagonal matrix, which means that the skew component can be encoded in just the upper non-diagonal elements of the matrix. I've done a lot of googling, but everything I've found so far is specific to the 2D case. It seems like this would be something that someone worked out a long time ago, but I can't find the right things to search for. 146.50.144.17 (talk) 15:01, 27 July 2009 (UTC)

The singular value decomposition expresses a matrix as a composition of a rotation, a scaling, and a rotation. JackSchmidt (talk) 15:10, 27 July 2009 (UTC)

Lebesgue Density

Qual prob as usual. I have a solution but I'm stuck on the last little bit. They define the Lebesgue density of E at a point $x\in \mathbb {R}$ by

 $\delta (x)=\lim _{h\to 0+}{\frac {m[(x-h,x+h)\cap E]}{2h}},$

provided this limit exists. Actually, in the problem they have $\lim _{h\to 0}$ which doesn't make sense. The question is: Let $E=\bigcup _{n=1}^{\infty }(1/(2n+1),1/(2n))$. Calculate the density of E at 0.

Proof is straightforwrd. The basic idea is for a given natural number i, assume 1 / (2i) < h < 1 / (2i-1) and define

 $S_{i}=m(E\cap (0,h))=\sum _{n=i}^{\infty }({\frac {1}{2n}}-{\frac {1}{2n+1}})={\frac {1}{4}}\sum _{n=i}^{\infty }{\frac {1}{n(n+1/2)}}$.

Then, this is clearly bounded by

 ${\frac {1}{4}}\sum _{n=i}^{\infty }{\frac {1}{(n+1)^{2}}}\leq S_{i}\leq {\frac {1}{4}}\sum _{n=i}^{\infty }{\frac {1}{n^{2}}}$.

Using the idea of upper and lower Riemann sums and the integral $\int _{i}^{\infty }1/s^{2}$, we get

 ${\frac {1}{4(n+1)}}\leq S_{i}\leq {\frac {1}{4(n-1)}}$.

Very last step says: "Hence $S_{i}/2h\to 1/4$ as $i\to \infty$." And, this is where I get confused. The thing is, in doing this limit, we are skipping infinitely many points as h gets closer and closer to 0. The way we defined h, h is always in some interval 1 / (2i) < h < 1 / (2i-1). But, not every h close to 0 is in such an interval as these intervals comprise 1/2 < h < 1, 1/4 < h < 1/3, 1/6 < h < 1/5. So, how does this show the limit as h goes to 0 is 1/4 if we don't use large chunks of h in any neighborhood about 0? StatisticsMan (talk) 16:29, 27 July 2009 (UTC)

You got all many of the order signs backwards, it should be 1/(2i) < h < 1/(2i − 1), 1/2 < h < 1, 1/4 < h < 1/3, etc. Now, back to the point: the exact same calculation works assuming just 1/(2i) ≤ h ≤ 1/(2i − 2), then you do not leave out any points. — Emil J. 16:54, 27 July 2009 (UTC)

By the way, notice that you can estimate

$\sum _{n=i}^{\infty }{\frac {1}{n(n+1/2)}}\geq \sum _{n=i}^{\infty }{\frac {1}{n(n+1)}}=\sum _{n=i}^{\infty }\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)={\frac {1}{i}}$

$\sum _{n=i}^{\infty }{\frac {1}{n(n+1/2)}}\leq \sum _{n=i}^{\infty }{\frac {1}{(n-1/2)(n+1/2)}}=\sum _{n=i}^{\infty }\left({\frac {1}{n-1/2}}-{\frac {1}{n+1/2}}\right)={\frac {1}{i-1/2}}$

without doing any integrals. — Emil J. 17:06, 27 July 2009 (UTC)

Okay, I think I fixed at least most of the mistakes. Thanks for noticing that. And, more importantly, thanks for the help on the problem. StatisticsMan (talk) 21:25, 27 July 2009 (UTC)

One small caveat to what Emil J said. For 1/(2i) ≤ h ≤ 1/(2i − 2), ${\frac {m[(-h,h)\cap E]}{2h}}$ isn't always equal to S_i but is between S_i and S_i-1, which still works just fine. Rckrone (talk) 00:59, 28 July 2009 (UTC)

Yea, I already noticed that, but thanks in case I didn't. I am considering 1/(2i) ≤ h ≤ 1/(2i − 2) and got rid of what I called S_i. I just used the summation stuff EmilJ mentioned to get the bound $1/4\cdot 1/(i+1)\leq m(E\cap (0,h))\leq 1/4\cdot 1/(i-1)$. Then, I divide through by 2h and use the first inequalities to get $i\leq 1/(2h)\leq i+1$ so I get

$\lim _{h\to 0+}{\frac {1}{4}}{\frac {i}{i+1}}\leq \delta (0)\leq \lim _{h\to 0}{\frac {1}{4}}{\frac {i+1}{i-1}}$.

Of course, the left side is slightly less than 1/4 and the right side is slightly more than 1/4 and both sides go to 1/4 as i goes to infinity, which happens as h goes to 0. So, limit exists and is 1/4. StatisticsMan (talk) 01:24, 28 July 2009 (UTC)

A beautiful solution?

Dear Wikipedians:

While trying to solve a question through the messy "cases" approach, I hit upon an alternative idea which led me to what I believe is a beautiful solution, but I need your expertise in validating my solution.

The question reads, "An elevator in a building starts with 6 people and stops at 7 floors. If each passenger is equally likely to get off at any floor and independent of other passengers, find the probability that exactly 3 people in total get off at the 3rd or 4th floor".

After trying, futilely, the messy aproach of P(3 on 3rd floor) + P(3 on 4th floor) + P(1 on 3rd & 2 on 4th) + P(2 on 3rd floor & 1 on 4th floor), I hit upon an elegant idea of combining floors 3 and 4 into a single floor with P(get off) = 2/7. Using the binomial distribution, I was able to solve the question using one short expression:

${6 \choose 3}\left({\frac {2}{7}}\right)^{3}\left({\frac {5}{7}}\right)^{3}=0.17$

Is it right?

Thank.

70.29.26.86 (talk) 21:16, 27 July 2009 (UTC)

Looks right to me. Rckrone (talk) 02:13, 28 July 2009 (UTC)

Thanks! 174.88.243.93 (talk) 15:22, 28 July 2009 (UTC)

Resolved

July 28

Right or left continuity, convergent sequence definition

So, we have an equivalent definition for continuity (which I copied and pasted from Continuous function):

This can also be formulated in terms of sequences and limits: the function f is continuous at the point c if for every sequence (x_n) in X with limit lim x_n = c, we have lim f(x_n) = f(c). Continuous functions transform limits into limits.

The question is, can we do a similar thing for right or left continuous only using sequences that are greater than or equal (for right) or less than or equal (for left) to c? StatisticsMan (talk) 14:16, 28 July 2009 (UTC)

Yes, with the same proof. Algebraist 14:18, 28 July 2009 (UTC)

Graph

How do I graph y<4x ? —Preceding unsigned comment added by 71.244.44.98 (talk) 14:51, 28 July 2009 (UTC)

Draw the line y=4x and then take any point not on the line, eg (1,0). If it satisfies the inequality (which (1,0) does as 0<4(1)) then shade the region containing the point (i.e. shade the whole side of the line where (1,0) lies.)--Shahab (talk) 14:56, 28 July 2009 (UTC)

this is a very vague response because of the word "region", which could be delimited by an axis or something. YOU SHOULD EITHER SHADE EVERYTHING ABOVE Y=4X OR EVERYTHING BELOW IT. I think you can figure out which :) —Preceding unsigned comment added by 82.234.207.120 (talk) 19:46, 28 July 2009 (UTC)

also any points you use to graph them, since it isn't less than or equals to, aren't part of it, and neither is the line. You could indicate this by drawing a dotted line instead of a solid one, and small empty circles instead of normal points for any points you graph. —Preceding unsigned comment added by 82.234.207.120 (talk) 19:48, 28 July 2009 (UTC)

Orthogonal Arrays and Codes

Where can I get more information on Orthogonal Arrays and their relation to error correcting codes. I want to find a proof of the following theorem: The minimum distance d(C) of a binary linear code C and the (maximal) strength t(C') of C' (where C' is the dual code of C) are related by d(C)=t(C')+1. If someone can find me the proof online I would be much obliged. Thanks--Shahab (talk) 14:52, 28 July 2009 (UTC)

Your link to "Orthogonal Arrays" didn't work because you incorrectly capitalized the initial "A" and used the plural instead of the singular. Wikipedia article naming conventions would require the article to be called orthogonal array, in the singular and with a lower-case "a" (the first letter of the article's name is case-insensitive). Orthogonal array redirects to orthogonal array testing. I've redirected your link. Michael Hardy (talk) 15:57, 28 July 2009 (UTC)

is it statistically possible to tell which is the world's best poker bot? whether it's better than the world's best Poker player? who the latter even is?

Is it statistically possible to objectively rank the poker bots of the world, or at least find the top of the rank? (I mean on standard hardware, in a standard configuration?)

Three reason I can think of for why it wouldn't be:

1) It might take a prohibitive number of hands to reach statistical significance. (Is this true for anything else we COULD statistically distinguish, if it didn't take so long or be so expensive because of the necessary sample size? I don't know statistics...)

2) poker bots could include hand-coded behavioral logic in such a way that

• program A is really good at beating program B, as the coder of A has studied and coded for B's behavior (but not vice versa).
• B's programmer hasn't studied A's code, but he has studied C's code and his bot beats C.
• While C's programmer hasn't studied B, he has studied A, and C beats A.

So A beats B beats C beats A. Which is the best? Also, the bots could consistently beat other bots for other reasons than their programmer having studied them. It could just look like a directed graph for whatever reason, where each edge you can determine with as much statistical confidence as you want (A really, we are very confident, beats B), but the overall graph is useless for ranking or coming up with a "best". And all this is even assuming that it is possible to statistically compare two bots!!! My other suspicion, in the absence of this, is:

3) Maybe poker bots going at it against each other look like random number generators going at it against each other. In this case one might still win on a particular run, but you can't statistically distinguish which is better.

This leads me to my question: when a poker player wins a tournament, are the tournament conditions such that the "best" player is statistically chosen with high confidence from among all the players (say 1300 of them!) Or, as I suspect, is it mostly luck, meaning that if you WENT BACK IN TIME and redid the tournament (so that the players don't have the knowledge gained from the tournament) you wouldn't have the same rankings. (maybe if you redid the whole universe 1000 times with different random hands, in the 1000 results one of the same 5 players would win almost all of the time, but which of the 5 players would be split... also maybe in 50 of the 1000 universes a non-"best" [the 5 players one of whom "should" win] ends up winning because of luck?)

Does anyone here know statistics??? Thank you for any help you might have in whether there is any statistical confidence in evaluating poker players and pokers. 82.234.207.120 (talk) 18:03, 28 July 2009 (UTC)

above question simplified: "Does a poker tournament reach statistical significance"

Does a poker tournament reach statistical significance in telling apart players' abilities? 82.234.207.120 (talk) 18:12, 28 July 2009 (UTC)

Statistical significance must be specified. For example, one can speak of "5% significance," "1% significance," or any other level of significance. The number of games players play in a tournament may be adequate for distinguishing significance at a higher level but not at a lower level. Wikiant (talk) 19:04, 28 July 2009 (UTC)

obviously I mean something reasonable. Skill may be such a small factor that in a typical tournament you have 52% confidence that #1 player is really better than #2 player. If they went at it heads up for 8 days, you would reach 75% confidence that it was in fact #2, another week or two for 95%/98%/99% -- the levels we reserve the term "statistical significance" for, and which in my example could really be #1 despite the initial impression at 75% confidence that #2 was better. So...is that the case?82.234.207.120 (talk) 19:32, 28 July 2009 (UTC)

Found seome evidence

some evidence from Greg Raymer:

“ At the 2004 World Series, Raymer defeated David Williams to win the $5,000,000 first prize in the $10,000 no limit Texas hold 'em WSOP main event. [...]

As defending champion, Raymer went on to finish in 25th place (out of 5,619 entrants) in the 2005 WSOP Main Event, earning him $304,680. No player since Johnny Chan has came so close to repeating as World Series champion, and his deep finish has been described by his peers as one of the most incredible achievements in poker. (empthasis added

”

What?? No "champion" of any year comes even close to being champion the next? Isn't this very strong evidence that a poker tournament like the World Series does not reach ANY statistical significance that the winner is better than the runners-up, let alone the best of them? 82.234.207.120 (talk) 19:41, 28 July 2009 (UTC)

There is an enormous amount of luck in no-limit tournament poker. The winner will, by the end, usually have gone "all in" several times. Any of those times it is possible that someone with a better hand will call them and they will be out. I would say that you need to be very good to do well a large poker tournament (smaller ones you can win by fluke) but being very good doesn't guarantee that you will do well. --Tango (talk) 19:49, 28 July 2009 (UTC)

but this clearly isn't true of a chess tournament, wouldn't you agree? Chess tournaments leave us with great statistical confidence that the winner would win if we went back in time and redid the tournament, isn't it so? In fact, it would be preposterous if the world champion of chess could NEVER successfully defend his title from year to year. On the contrary, the world's best chess player defends his title constantly. And often there is a period of about a decade or more with clearly one "best" chess player. So, poker tournaments fail to choose one where chess tournaments succeed. The question is "why"? Is it htat they aren't long enough (it would take 1000 days to find the true "best" [and not just luckiest] poker player?) or some other reason? Thanks for your input. 82.234.207.120 (talk) 20:51, 28 July 2009 (UTC)

There is no random element in chess (in the lingo [if I can remember it correctly], the players have "perfect and complete information"), so would expect more stability in the results. It's not a matter of the number of hands for tournament poker since the best player may go all-in on the first hand with a 99% chance of winning and lose and no matter how many more hands you play, the best player can't win the tournament. In fact, it's really the opposite of what you say. If you had a really long knockout chess tournament where you have to win, say, 1000 matches in a row, there is a very good chance that the best player will have one bad match and go out. You could be pretty sure of getting a very good winner, but not necessarily the best, just as in poker. --Tango (talk) 00:29, 29 July 2009 (UTC)

There is certainly a random element in chess, such as opening preparation. Maybe you are playing Black and you are very good at defending king pawn openings and not so good at queen pawn openings. OK, if you are lucky, your opponent opens with the king pawn. And he did that because he happened to see someone else analyzing that opening the previous night, etc, which is quite random. Or maybe you have a chance lapse of attention at just the wrong time in the game and miss a variation. There is a body of theory of chess ratings whose basic idea is that every player has a "strength" which is a normal probability distribution N(r,sd) associated with that player. For a strong amateur player ("Bob") you might have r=2000 ("expert" rating) and sd=50. "Alice" might have r=2050 and sd=75 (higher sd because she plays with a riskier style). You can model a game between Alice and Bob by drawing a number A from Alice's distribution and B from Bob's distribution, which are their performance levels for that game. Then if |A-B| >= 100 or so, the higher number wins. If |A-B| < 100 then the game is a draw.

For poker it should be possible to do the same thing. On the big name poker servers (I don't play myself so I'm going by what I've heard), bots are aggressively pursued and banned, but on some other servers they are welcomed. They can apparently beat some fairly good human players but not the very good ones. There are probably some WP articles with more info. 70.90.174.101 (talk) 02:57, 29 July 2009 (UTC)

Book recommendations?

I hope this is the right category.

I've always been a logical person, but usually intuitively logical. Lately I want to understand the philosophy of logic, how to think rationally, and how to apply logic and statistics to my own thinking and my life. Unfortunately, a lot of the web materials out there require a lot of background knowledge and understanding that I just don't have. For instance, the great blogs Less Wrong and Overcoming Bias have some extremely interesting discussion topics, but sometimes they just absolutely lose me.

Right now, I'm eighteen and in high school. My math skills are limited to algebra, very basic graphing (y=mx+b is about the limit of my understanding) and a very small amount of trig. I don't understand quadratics or polynomials or a lot of trig or geometry, and I still haven't taken any course that covers functions, or calculus. I know what set theory and formal logic are, and maybe a little bit about them, but not enough to say, explain it to my thirteen-year-old brother.

With that in mind, what would be a good, clearly-written, layman-oriented primer on some of these subjects, especially ones like the Less Wrong blog covers, but also basic statistics and stuff like that? Having some book to learn this from would make it so much easier than aimlessly clicking links and throwing half-assed, vaguely-written quries at Google only to get answers so academic I barely comprehend the page titles.

Again, many apologies if I didn't put this in the right place, but it didn't feel likely that I'd get good responses in any other category.

--✶♏‭ݣ 20:20, 28 July 2009 (UTC)

I think that it is nice that you are so interested in logic, so I am more than happy to help out. First of all, everyone uses logic all the time. For instance, "the cost of these bananas is too high" => "I will not buy them" is one possible logical implicaion at the shop. However, the philosophy of logic is something more "advanced". One need not know calculus to understand this, but I think that a satisfactory understanding of set theory is necessary (in fact, axiomatic set theory is something very philosophical in nature). In the context of mathematics, the logical implications that a mathematician makes often involve sets and order structures, and thus set theory and order theory are two "fundamentals" of mathematics.

With regards to your question, I do not think that you can apply logic to your real life, more than what you currently do, unless you learn philosophy. Philosophy is perhaps the best name for "logic in daily life" but of course, there are other ways to apply logic in daily life. Philosophy, however, is something a book cannot completely teach you. If you continue thinking about logic, and applying it in real life, that is the best preparation apart from reading a book on the subject. The article philosophy may be of some use. Hope this helps. --PST 01:15, 29 July 2009 (UTC)

On the other hand, the question regarding why people make particular logical implications, lies in the realm of psychology. --PST 01:17, 29 July 2009 (UTC)

I've read a bit about both subjects, but information on the Internet (especially about philosophy) tends to be either highly academic or highly polarized and messy. If you look up "epistemology" for instance you're certain to get a lot of postmodern crap about how there is no truth or logic. What I'm really looking for is a book, in particular, or maybe specific very good web resources. I already know "the sorts of things" I should read about but the recommendations I want are particular texts about them. --✶♏‭ݣ 02:03, 29 July 2009 (UTC)

If you don't mind a rather old book, start with How to Lie with Statistics by Darrell Huff. Of course it's really about how to tell when someone is lying to you with statistics. 70.90.174.101 (talk) 03:49, 29 July 2009 (UTC)

An alternate solution by radicals to the Quartic Equation

Roughly 2 years ago I inquired about whether it was well known that one could solve certain Quartic Equations by representing it as a composition of two Quadratic functions, the consensus I got from that was no, and that it wasn't particularly useful to have a non-general solution like that. Well, just in the last few days I've figured out how to solve the general case by using another 'relatively' simple technique that transforms any Quartic into one that can be represented as a composition of two Quadratics. I'm once again curious if this would be a significant result. A math-wiki (talk) 21:24, 28 July 2009 (UTC)

Try it on (x-1)(x³-2). Does it give the cube root of 2 as one possible solution? Dmcq (talk) 22:50, 28 July 2009 (UTC)

By "composition" do you mean "product"? If so, then over the complex numbers that is trivial, over anything else I don't believe you. Dmcq gives a good example of a quartic you almost certainly can't solve - the solutions to quadratics are at most square roots, you'll never get a cube root (assuming we're working over the rationals, which is the usual assumption). As for your first inquiry, you seem to have got the wrong answer - it is very well known and very useful, factorising a polynomial to solve it is usually the first thing people try (unless they have a computer handy), that applies to quartics as much as any other degree. --Tango (talk) 00:07, 29 July 2009 (UTC)

I'm not talking about factoring here, I'm talking about Function composition, in this case, both being Quadratics yielding a Quartic naturally, I'm quite certain it works. A couple years back I figured out how to solve an equation of the form f(g(x))=0 if I know how to solve f, and how to solve an equation of the form g(x)+C for arbitrary constant C. But for the Quartic, the middle terms had to have a certain relation, by doing a change of variables of the form $y-\lambda =x$, I can transform any Quartic into the correct form, this involves solving a Cubic, the Resolvent Cubic in fact. A math-wiki (talk) 00:34, 29 July 2009 (UTC)

Monty Hall problem

Since we know that Monty will disclose a goat why is the contestant's original chance of picking the car not one half?

Also -

What if there were two contestants? If the two thirds chance of winning by swapping was correct then since one of them winning is a certainty, we have two thirds plus two thirds is greater than one, which is not correct.

Bob Coote

Wentworth Falls NSW Australia —Preceding unsigned comment added by 203.164.18.191 (talk) 22:00, 28 July 2009 (UTC)

Have you read the explanation at Monty Hall problem? It's because if you stick to the strategy of not changing, then your odds can never change from being 1/3, hence the odds from the strategy of always changing must be 2/3. Confusing Manifestation(Say hi!) 22:56, 28 July 2009 (UTC)

First: Because the contestant's original chance of picking the car does not depend on if Monty will disclose a goat or not. Second: Both contestants cannot swap, so the swapping one wins with probability two thirds and the other one wins with probability one third. Bo Jacoby (talk) 23:04, 28 July 2009 (UTC).

If they picked different doors originally they could both swap, but it makes no difference. --Tango (talk) 00:19, 29 July 2009 (UTC)

The section Monty Hall problem#Increasing the number of doors is particularly worth a read. Recasting the problem with more doors can help change your perspective, and might overcome the situation where you know what the answer should be intellectually, but it still doesn't sit right intuitively. -- 76.201.158.47 (talk) 03:56, 29 July 2009 (UTC)

You can't just add the probabilities like that since the events are not independent. If the first player wins then you know whether the second player has won or lost based on whether they chose the same door, so the probability is either 0 or 1. If they choose the same door the probability of at least one of them winning is 2/3+1/3*0=2/3, if they choose different doors the probability of at least one of them winning is 2/3+1/3*1=1. That is exactly what you would expect. --Tango (talk) 00:19, 29 July 2009 (UTC)

Tango, Adding probabilities does not require independence, but mutually exclusive events. (It is the multiplication of probabilities that requires independence). Bo Jacoby (talk) 00:32, 29 July 2009 (UTC).

Here's another way of thinking of this, to start off you pick 1 of 3 doors, so the chances that door has the car are 1/3. And the other pair of doors has a 2/3 chance of containing the car collectively. Then the host, who knows where the car is opens one (possibly the only one, 1/3 of this) that doesn't have a car behind it, this is very important that the host knows which has the car and therefore which ones don't. The door you originally chose still only has a 1/3 of having the car behind, that's because the event of you picking said door had that chance of having the car behind it. So the pair has the 2/3 chance still as well, but one of the two that make up that pair is now open and it's the one w/ the goat, so the remaining door has the 2/3 chance of the car. A math-wiki (talk) 02:03, 29 July 2009 (UTC)