Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

June 24

Constrained solutions of a linear equation

Hello...My today's question is as follows. Consider the 4 variable equation v + x + y - z = 4. If I desire to find solutions to it, I can easily do it by assigning arbitary values to x, y and z and solving for v. But what I wish for, is to find solutions where the variables all belong to the set {1,2,3}. How to do that systematically? Also if I generalize the problem, keep my equation as v + x + y - z = b, b any positive even integer and search for solutions within {1,2,...,n} is there a systematic way to do that. Thanks. - Shahab (talk) 06:26, 24 June 2010 (UTC)

There's a fairly obvious simple systematic way to do that: three nested loops for each of x, y and z in 1…n, solve for v = b - x - y + z, output x, y, z, v if v is in [1,n]. That scales as n³, so should be perfectly feasible for values of n up to several hundreds or thousands on modern CPUs when implemented in any decent programming language or numerical computing environment, and you could do it with pencil and a single sheet of paper for n = 3 (<small(>19 solutions i reckon when b=4. Ok i admit i used a computer even for that). I'm sure that simple algorithm can be made at least a bit more efficient, but with modern computing power it's probably not worth the extra mental and programming effort unless you're interested in larger values of n. Qwfp (talk) 09:50, 24 June 2010 (UTC)

(edit conflict) If your variables are constrained to be members of a finite set, then a brute force solution is to create all the tuples of the appropriate size that can be formed from that set, and test each one of them in turn to see if it is a solution. So if v, x, y and z have to come from the set {1,2,...,n} then you construct each of the n⁴ 4-tuples (p, q, r, s) with members drawn from this set, and then check to see if p + q + r - s = b. If you want a more efficient algorithm, then you can reduce the size of the search space by exploiting the symmetries of your equation - the result for (p, q, r, s) is the same as the result for (q, p, r, s), (r, p, q, s) etc. - or the properties of your candidate set - if solutions have to come from {1,2,...,n} then p + q + r must lie between b + 1 and b + n, so max(p, q, r) ≥ (b + 1)/3 and min(p, q, r) ≤ (b + n)/3. Gandalf61 (talk) 10:04, 24 June 2010 (UTC)

Also 3 ≤ p+q+r ≤ 3n. Bo Jacoby (talk) 14:28, 24 June 2010 (UTC).

Simple formula

Not a homework question. If 0.82(s1-p)=0.72(s2-p) then s2/s1= what? I have not been able to work out a solution. I'd be interested in how a solution is obtained as well as the answer. S1, s2 and p must all be greater than zero, and s1>p and s2>p. Thanks. 92.24.186.235 (talk) 09:33, 24 June 2010 (UTC)

1. Simplify the equation to 41s₁ − 36s₂ − 5p = 0.
2. The inequality s₁>p implies that 0 = 41s₁ − 36s₂ − 5p > 41s₁ − 36s₂ − 5s₁ = 36s₁ − 36s₂.
3. The inequality p>0 implies that 41s₁ − 36s₂ > 0.
4. So 1 < s₂/s₁ < 41/36.

I hope it is correct now. Bo Jacoby (talk) 13:18, 24 June 2010 (UTC).

Probability Distributions

Hello. If the probability of a computer chip failing quality-control testing is 0.015, then what is the probability that one of the first three chips off the line will fail? Do I use a geometric or binomial distribution? Thanks in advance. --Albert (talk) 17:27, 24 June 2010 (UTC)

You can use either if you use it correctly. But it's best not to consider distributions at all, but just basic probability: The probability of a chip to pass is 0.985. If they're independent, the probability they all pass is $0.985^{3}\approx 0.95567$, so the probability that at least one fails is $0.04433...$.

Unless you meant the probability that exactly one fails, in which case use binomial. -- Meni Rosenfeld (talk) 18:24, 24 June 2010 (UTC)

June 25

Probability question

n identical shapes (think of them as cut out of paper), each of area S, are placed independently and uniformly at random over a region of area A, overlapping as necessary. What is the probability, p, that every point in the region will be covered by at least one shape? I am only interested in cases where n will be very large (millions or billions, say), before p becomes non-zero, and a decent approximation would be fine. Edge effects can be ignored. I assume p does not depend on the (non-pathological) shape of the region being covered, but it's not obvious to me whether it depends on the (non-pathological) shape of the covering shapes. If it does, assume circles. —Preceding unsigned comment added by 81.151.34.16 (talk) 03:49, 25 June 2010 (UTC)

would you clarify better "...uniformly over a region A". A possible interpretation is: $n$ points of $A$ are choosen indep. and uniformly on $A$, and the $n$ congruent shapes are $S_{i}=x_{i}+S$, for $i=1,..n$ (having assumed $0\in S$). This allows some of them to partially get out of $A$. Or are you imposing $S_{i}\subset A$ (this should make things more complicated). And are you also considering rotations or just translates (simpler option)?

Thanks for your reply. I'm afraid I do not understand your notation "$n$ congruent shapes are $S_{i}=x_{i}+S$". As I mentioned, I am not concerned about what happens at the edges, including whether shapes can partially overlap the boundary. The shapes are so small compared to the containing region that it's irrelevant for my purposes. I assume that there is a limiting distribution that holds as the covering shapes get indefinitely small, and this limiting distribution is really what I'm after. If the shape of the covering pieces matters, and it matters whether we consider rotations, then assume circles. 86.135.29.110 (talk) 14:12, 25 June 2010 (UTC).

When S out of A square centimeters are covered, each cm² is covered on average S/A times. When n shapes are placed, the average cm² is covered nS/A times. The actual number of times has a poisson distribution. The probability that some particular cm² is uncovered is e^−nS/A. The average uncovered area is L = Ae^{− nS/A}. Bo Jacoby (talk) 07:27, 25 June 2010 (UTC).

Does this lead to an answer to the original question? 86.135.29.110 (talk) 14:12, 25 June 2010 (UTC).

Yes, I think so. The dependence on shape enters here. Consider the case where each shape is a square and the shapes are placed on a square grid like a chess board. Then the average number of uncovered squares is L / S and the answer to the original question is

$p=e^{-{\frac {L}{S}}}=e^{-{\frac {Ae^{-{\frac {nS}{A}}}}{S}}}$

Bo Jacoby (talk) 19:18, 25 June 2010 (UTC).

Are you assuming that the squares are always placed exactly on the grid lines? If so, this won't work (I mean, it isn't in the spirit of the original question). The squares can be placed anywhere. 86.185.77.226 (talk) 19:29, 25 June 2010 (UTC).

The shape does matter unfortunately. This can be seen by comparing two circles of area S/2 joined by a line as the basic shape area S. When rotations are allowed the distance between is large this can be consiered almost as two independent circles of area S/2. If we split up the original area A in two then if n circles of area S have probability p of fully covering area S then the same can be said of n circles of area S/2 covering area A/2. So the probability of the two circle combinations of area S covering area A is p².

I think we should just consider the circles or squares form of the original question. I'll imagine the circle form as if I'm spray painting what is my chance of covering the object completely, I've protected the area around so I can go evenly up to the edges. Of course spray paint is rather more uiform and the paint won't form exact circles but were in maths land here. Dmcq (talk) 08:51, 25 June 2010 (UTC)

p.s. I had been wondering if this could be applied to how long till Jackson Pollock completely covered every spot of the canvas but since he dribbled bits around rather than using drops I suppose not :) Dmcq (talk) 08:56, 25 June 2010 (UTC)

I don't quite follow this. I'm not sure how, in the two-circles-joined-by-a-line case, you arrange that the two circles always fall in different halves of A. In any case, in my problem we can assume that the extent (as well as the area) of the covering shapes is negligibly small compared to the region A (otherwise there are numerous edge-related complexities which I specifically want to ignore), so I don't think this argument can apply. 86.185.77.226 (talk) 18:26, 25 June 2010 (UTC).

I wasn't saying they would fall in different halves of A, just that allowing rotations the two halves of the long dumbell would act as two independent circles to all intents and purposes when covering the area. And I believe Bo Jacoby's formula would apply generally when the area has a reasonable chance of being fully covered. For n shapes of area S in an area A you would have probability of something like exp(-K*A*exp(-nS/A)/S)) of completely covering the area where K is some constant depending only on the shape. This does depend on some big assumptions, the main one being that as the flecks of uncovered space get separated they are small and are wholly covered with probability S/A for each added shape. I wouldn't have thought the difference between a circle and a square would make a big difference to the outcome. Dmcq (talk) 19:49, 25 June 2010 (UTC)

I wasn't saying they would fall in different halves of A, just that allowing rotations the two halves of the long dumbell would act as two independent circles to all intents and purposes when covering the area. I understand that, but wouldn't this just tell us that p(A,S,n) = p(A,S/2,2n)?? Why would there be a problem with that? (By the way, I'm not particularly arguing that the formula shouldn't be shape-dependent, I just can't see how your argument demonstrates that it is.) 86.185.77.226 (talk) 21:02, 25 June 2010 (UTC).

It does tell you that,but you can then split up that 2n lot into two halves and (S/2)/(A/2)=S/A. Each half has p(A/2,S/2,n) of being completely filled which will be same as p(A,S,n) with two of those areas stuck together the probability of the two areas stuck together being filled is p(A/2,S/2,n)p(A/2,S/2,n) and this should be the same as p(A,S/2,2n). However if shape doesn't matter this last should be the same as p(A,S,n). So we have p(A,S,n)=p(A,S,n)^2 Dmcq (talk) 21:25, 25 June 2010 (UTC)

When you talk about splitting A into two halves, each of area A/2, how are those halves defined? I've tried the exp(-K*A*exp(-nS/A)/S)) formula against simulations with square-shaped covering shapes and, while there is always the possibility that I made a mistake, the results do not look good. If I haven't made a mistake then the formula looks definitely wrong. How confident do you feel about it? 86.174.161.139 (talk) 23:08, 25 June 2010 (UTC).

Nothing special, just chop in half. It shouldn't matter if A is a square or rectangle or circle if it is significantly larger than S. What did you simulate? Dmcq (talk) 23:18, 25 June 2010 (UTC)

I'm sorry if I'm being slow, but I just don't see it. If you chop A in half, then, in order for your logic to work, half of the dumbell ends need to fall into one half, and half into the other, don't they? How is that arranged? A typical simulation was placing 5 x 5 squares (anywhere) on a 100 x 100 grid. The results I get are nothing like a shape that could be produced by the proposed formula (it feel too wrong to be just due to the fact that this a way off insignificantly small S). However, placing 1 x 1 squares on a 20 x 20 grid, for example, gives fairly plausibly similar results to the formula with K = 1. This suggests something is going wonky with the logic once it becomes possible for the squares to overlap. I am not 100% certain about any of this, by the way! 86.174.161.139 (talk) 23:53, 25 June 2010 (UTC).

I'm pleased that your 1x1 simulation confirmed my analysis. How did you treat the edges in your 5x5 simulation? Bo Jacoby (talk) 06:49, 26 June 2010 (UTC).

I think Bo's reasoning may be correct for the 1×1 case but break down for larger shapes because the probabilities of two nearby squares being covered (or uncovered) is not independent. For example, if there are only two squares left to cover, if they are neighbours there's a good chance that both will be covered at once. If they are far apart this is clearly impossible. I also notice that Bo's answer above to the original question takes the form of the cumulative distribution function of a Gumbel distribution, implying the distribution of the number of shapes placed before every square is covered would follow this distribution. Qwfp (talk) 08:52, 26 June 2010 (UTC)

For the dumbell the two parts would be in the same half practically all the time, Try an A4 sheet with small circles and then two halves of an A4 sheet with dumbells of circles 1/sqrt(2) the radius and I was thinking the smaller circles on the A5 sheet should behave to all intents and purposes like the bigger circles on the A4 sheet. I wasn't trying to make the two bits go in the different halves and normally they'd be in the same half. As to the reasoning about the little bits remaining, the logic is that when there is a reasonable chance of the whole area being covered the specks would tend to be far apart relative to the size of the shapes. The big assumption is that a speck is likely to either be completely covered or completely missed, my guess was this doesn't make a big difference. I had a quick look at paint coverage and a couple of other things but didn't see anything close but I agree with Bo Jacoby someone has probably looked at something similar. Also you get a nicer formula I think if you define N=A/S, i.e. the number of shapes you'd need if they had no overlap. The formula then is p(N,n)=exp(-K*N*exp(-n/N)) and solving for n you get n=N*log(K*N/log(1/p)) so it gets dominated by N*log(N). So the smaller the drops the more paint you have to spray to be absolutely sure of covering every single point, quite a lot of paint in fact. I appreciate that you think the simulations disagree with the formula but I have no idea from your description what it is that you were looking for or what you saw. Dmcq (talk) 10:16, 26 June 2010 (UTC)

A study of the effects of bombing in [1] might give something, I don't have access Anything which references it might also be useful. percolation is the study of when the area becomes solid, that is independent of n but might I guess relate to the shape effect K. Dmcq (talk) 12:48, 26 June 2010 (UTC)

Counting dust or bacteria is also related [2], but perhaps more to the percolation problem Dmcq (talk) 13:10, 26 June 2010 (UTC)

Thanks for those Dmcq. After a bit of forward and backward citation chasing with the help of the Science Citation Index, this paper looks most promising: Miles, R. E. (1969). "The Asymptotic Values of Certain Coverage Probabilities". Biometrika. 56 (3): 661–680. JSTOR 2334674.. I'm fortunate enough to have access to it, but i don't have time to read it properly right this minute and to be honest its format is a bit mathematical for me so i thought i'd just post the ref in case someone else can 'decode' it first. Qwfp (talk) 15:24, 26 June 2010 (UTC)

On second thoughts this one looks more promising, though more heavily mathematical: Svante Janson (1986). "Random coverings in several dimensions". Acta Mathematica. 156 (1): 83–118. doi:10.1007/BF02399201. {{cite journal}}: Cite has empty unknown parameter: |1= (help). Interestingly his Theorems 1.1 & 1.2 do involve the Gumbel distribution, but the expressions are a lot more complicated than Bo's expression above. Qwfp (talk) 17:43, 26 June 2010 (UTC)

@Dmcq. Thanks for your further explanation about the dumbbells. I get it now.

@Bo Jacoby / Dmcq. The simulation works like this. I have a b x b grid, area A = b^2. Onto this I place c x c squares, each area S = c^2. For each square I place, I randomly choose a cell in the b x b grid, and I align the bottom left of the c x c square with the bottom left of the random cell. Then I mark off all the covered cells in the b x b grid, ignoring any part of the c x c square that falls off the edge. I carry on placing squares until the b x b grid is completely covered (or n reaches some large cutoff point). I repeat this many times, building up a probability distribution that the b x b grid will be completely covered by n c x c squares. Then I draw a graph of this and overlay the graph of the exp(-K*A*exp(-nS/A)/S)) equation with K = 1. When c = 1, the fit is good. With other values of c, the graphs are completely different. 86.185.76.146 (talk) 17:51, 26 June 2010 (UTC).

Thanks. Consider computing coordinates modulo b in order to avoid edge effects. Bo Jacoby (talk) 18:08, 26 June 2010 (UTC).

That is a very good point. I was thinking that edge effects wouldn't be such a big deal, but, of course they are the way I was doing it because the left and bottom edges get many fewer hits than they ought to. Making the change you suggest makes a big difference actually. The results still do not match the proposed formula, but they are quite a bit closer and approaching a situation where I don't feel completely certain that the discrepancy is not just due to other imperfections in the simulation (such as having a granular grid). However, the discussion above seems to be moving away from the idea that the simple formula ought to work? 86.185.76.146 (talk) 18:49, 26 June 2010 (UTC).

Result in Janson 1986

I've had a go at decoding and simplifying a result in the Janson 1986 paper i mentioned above (Svante Janson (1986). "Random coverings in several dimensions". Acta Mathematica. 156 (1): 83–118. doi:10.1007/BF02399201. {{cite journal}}: Cite has empty unknown parameter: |1= (help)). I think i've got the gist of it, but i suggest this is checked by someone with a deeper knowledge of maths than me.

It appears the relevant result is (1.3) in Theorem 1.1 on p84, the second page of the paper. He allows the shapes ('small sets' in his terminology) to be random convex sets. The result depends on the ratio of the area of the region to the average area of the small set; i'll call this ratio r for simplicity (Dmcq called it N above but that clashes with the notation in this paper so would get too confusing). There's also a constant α that depends on the shape of the small set; in Section 9 he shows that α = 1 for the cases of a circle of fixed size and a rectangle of fixed size (he gives other examples too where it's different). In two dimensions, the result is that the distribution of n, the number of small sets required to cover every point of the region at least once, is given by:

$n/r-\log r-2\log \log r-\log \alpha \;{\xrightarrow {d}}\;U,$

where U has a standard Gumbel distribution and ${\xrightarrow {d}}$ denotes convergence in distribution as r → ∞, i.e. as the area of the region becomes much larger than that of the small set. Qwfp (talk) 19:50, 26 June 2010 (UTC)

Wow. I tried plugging this in to my (crucially modified!) simulation and it doesn't seem to be a perfect match-up. However, I'm now within the realm where I'm not sure if the discrepancy might be due to deficiencies in the simulation (granularity, covering squares not vanishingly small compared to covered region). So, I think it doesn't tell me much even if I've got the workings right. Speaking of which, could anyone check my workings?

What I need to check against the simulation is a graph of p versus n, where n is the number of covering squares and p is the probability that these n squares will completely cover the larger region. Based on the formula Qwfp posted above, I'm using

p = Exp(-Exp(-x))

where

x = n / r - Log(r) - 2 * Log(Log(r)) - Log(alpha)

r = A / S (A is area of large region; S is area of covering squares)

alpha = 1

Have I got this part right? 86.185.76.146 (talk) 20:52, 26 June 2010 (UTC).

I believe so. That formula is very like the one we had before but instead of K being a constant just depending on the shape it is (log r)²α. I'll have to have a think what that (log r)² factor means, I'm rather surprised by it but I guess the paper has put a bit more thought into it. It conflicts with the dumbbell argument though the difference isn't great. The formula is equivalent to:

$p(n,r)=e^{-\alpha (\log {r})^{2}re^{-{\frac {n}{r}}}}$

the dumbbell argument implies

$p(2n,2r)=(p(n,r))^{2}$

but using the formula on the two sides gives

$e^{-\alpha (\log {2r})^{2}2re^{-{\frac {n}{r}}}}$

and

$e^{-\alpha (\log {r})^{2}2re^{-{\frac {n}{r}}}}$

log r and log 2r have only a very small relative difference if r is large but they're still different. Dmcq (talk) 22:43, 26 June 2010 (UTC)

Right, thanks. Excuse me if my grasp of this is rather tenuous, but might not the value of alpha be different in the case of the dumbbell? 86.185.76.146 (talk) 22:52, 26 June 2010 (UTC).

Hopefully it should behave as two discs, I believe there must be some problem with saying

$p(2n,2r)=(p(n,r))^{2}$

Perhaps it has something to do with there actually being some variation of the numbers in both halves, it wouldn't divide into exactly n in each half when you put 2n discs in the whole area. I haven't any really good idea what I did wrong yet though. Dmcq (talk) 23:09, 26 June 2010 (UTC)

I think I misunderstood you. Although you called it the "dumbbell" argument, is this $p(2n,2r)=(p(n,r))^{2}$ formula actually using dumbbell shapes at all? Are you not just saying that 2n discs each of area S in a region of area 2A should be equivalent to two independent lots of n discs of area S in two regions of area A? As I understood it, the "dumbell" argument was specifically to show that the formula couldn't be the same for a dumbell and a disc because it led to the contradiction p = p^2, right? In that case, the only thing that gives in the formula is alpha, so alpha should be different for a dumbbell? I have to admit, I find it hard to conceive of this parameter alpha that could be the same for a circle and a square and yet differ for some other unknown shapes. 86.185.76.146 (talk) 23:49, 26 June 2010 (UTC).

Looking further at section 9, Janson says "smaller α corresponds to more efficient coverings". He shows that α=1 for a square of fixed size and fixed orientation, but 4/π for a square of fixed size but random orientation. It's 2 for a triangle of fixed size and fixed orientation, and 3√3/π for a triangle of fixed size but random orientation. (Orientation is irrelevant for circles of course.) So squares of fixed orientation cover better than those of random orientation (which seems intuitively right), but vice-versa for triangles (which seems reasonable when you think about it). Qwfp (talk) 12:39, 27 June 2010 (UTC)

Sorry yes the bit in that formula doesn't depend on anything about dumbbells. What I'm saying is wrong is that 2n discs each of area S in a region of area 2A must not be entirely equivalent to two independent lots of n discs of area S in two regions of area A. In fact we could have n+d in one part and n-d in the other and one would have to sum the combinations with their various probabilities. I assumed that since d would be insignificant compared to n it didn't matter, however it could differ by a few times the square root of n and that must be what allows the (log r)² factor to exist. I'll see if I can do the sums tomorrow, it looks a little nasty, and if that makes the problem disappear. Dmcq (talk) 00:33, 27 June 2010 (UTC)

I'm afraid I haven't properly attempted to follow your dumbbell argument Dmcq, but the Janson result above is valid only for convex sets, and the dumbbell shape clearly isn't a convex set. There are some more recent papers on Svante Janson's website, the latest of which gives a very brief overview of previous work on "covering by translates of a set" in its introduction and then says "In almost all cases above, the set S is taken to be convex". (This latest paper isn't making that assumption but it's about a different problem — what's the minumum number you need when you get to choose where to put them). Qwfp (talk) 09:49, 27 June 2010 (UTC)

The dumbbell argument had nothing to do with that, I was just looking at two lots of circles in two area adjoined. I had a quick go at adjusting it to see what happened if I used probabilities one standard deviation away as in

$p(n+{\sqrt {\tfrac {n}{2}}},r)p(n-{\sqrt {\tfrac {n}{2}}},r)\approx p(2n,2r)$

which I hoped might give me an idea what's happening and I got that one would have to have

${(\log {2r})}^{2}\approx {(\log r)}^{2}\cosh {\sqrt {\frac {n}{2r^{2}}}}$

one can approximate the cosh as $1+n/4r^{2}$ but that gives a result which is even further from what I wanted when I assumed zero standard deviation. It does show though that the exponent should grow faster than the original one with just a constant K which is I suppose heading in the right direction. I think I'll take Janson's word for it :) Dmcq (talk) 13:20, 27 June 2010 (UTC)

Yeah, he seems to be a clever guy, to judge by Google Scholar, which gives him an h-index somewhere around 35—pretty amazing for a mathematician, (assuming there's only one with this name). Perhaps we should have an article on him... Qwfp (talk) 13:41, 27 June 2010 (UTC). Created a stub on Svante Janson. Please expand! This is not really my area. Qwfp (talk) 14:47, 27 June 2010 (UTC)

By the way that factor of 2 in front of the "log log r" is the only part of the formula that depends on the dimension of the space, to which it is equal. He also comments that the coupon collector's problem is a zero-dimensional analogue. The "log log r" term then disappears. Qwfp (talk) 13:41, 27 June 2010 (UTC)

Linear or non-linear time series

How do you tell if a time series is linear or non-linear? Is there a formula? Thanks 92.28.242.52 (talk) 14:12, 25 June 2010 (UTC)

Or, if its an easier question to answer, how do you measure the extent to which a time-series is non-linear or chaotic? Thanks 92.15.15.76 (talk) 09:43, 26 June 2010 (UTC)

A time series that is linear with time will have the form x(t) = at + b. If there are errors or uncertainties in the observations x(t) then there are various methods for estimating the "best" values for the parameters a and b - the simplest method is ordinary least squares, but others are mentioned in our linear regression article. Once you have a linear model, then you can test the goodness of fit between model and observations using various correlation tests. Note that non-linear is not the same as chaotic - a time series of daily mid-day temperatures over the course of several years will be non-linear, but not chaotic. Gandalf61 (talk) 11:44, 26 June 2010 (UTC)

Thanks, you've described a straight line (possibly with noise also) but you have not answered either of the two questions. 92.15.13.228 (talk) 16:12, 26 June 2010 (UTC)

If you compare the goodness of fit of a linear model with that of as quadratic model, which uses up a degree of freedom, and you get a better fit with the linear model then that's a good indication linear is a good model. Dmcq (talk) 14:14, 26 June 2010 (UTC)

Thanks, why? 92.15.13.228 (talk) 16:12, 26 June 2010 (UTC)

Assuming the data contains a random component, there is likely no way to tell for certain -- you can only tell probabilistically. For example, regress y on a constant, x, and x^2. If the coefficient on x is statistically significant while the coefficient on x^2 is statistically insignificant, then you have evidence that the model is more likely to be linear than it is to be quadratic. If you want to test higher-order non-linearities, you can include x^3, x^4, etc. This is a better approach than comparing the goodness of fit for a couple of reasons: (1) without doing additional math, you don't know whether the improvement in the goodness of fit is statistically significant, (2) R^2 is guaranteed to increase when you add the quadratic term, so you must use an adjusted R^2 -- but, there are several (or none) to choose from depending on what you intend to do with the data and whether your dependent variable is discrete or continuous.Wikiant (talk) 16:17, 26 June 2010 (UTC)

Would calculating or estimating the Lyapunov exponent be a way? 92.15.13.228 (talk) 16:20, 26 June 2010 (UTC)

The answer to that is beyond my expertise. However, my limited intuition is that this would not help since you are dealing with data that contains a random component. The Power transform is often used in econometrics to test for linearity vs. several types of non-linearity. Wikiant (talk) 17:46, 26 June 2010 (UTC)

MUSIC INVERSION AS A MATHEMATICAL PATTERN

I have lost a beautiful video, demonstrating the subject material. I recall one example included: Rimsky-Korsakov's Coq d'or, displaying the bird's happy prediction, contrasting with the bird's gloomy prediction - it was the identical notes, just inverted, evoking the contrasting emotion. The video was likely from university lecture.

Any help locating such video would be immensely appreciated. MUSIC INVERSION AS A MATHEMATICAL PATTERN Padraeg Sullivan 23:32, 25 June 2010 (UTC) —Preceding unsigned comment added by PADRAEG (talk • contribs)

June 26

Vector calculus

I found a question in my textbook that really confuses me: Find a C¹ function f: R³ --> R³ such that takes the vector i + j + k emanating from the origin to i - j emanating from (1,1,0) and takes k emanating from (1,1,0) to k - i emanating from the origin.

I am confused because I thought vectors are displacable, so it doesn't matter where it originates from. 173.179.59.66 (talk) 08:35, 26 June 2010 (UTC)

I guess that "i − j emanating from (1,1,0)" means (i − j) − (i + j). Bo Jacoby (talk) 10:06, 26 June 2010 (UTC).

Why is that? 173.179.59.66 (talk) 11:24, 26 June 2010 (UTC)

This is an interesting question. It sounds like we should think of the tangent bundle Tℝ³. We have two vectors based at two points, and a function ƒ : ℝ³ → ℝ³. The first vector, written above as i + j + k is based at the origin, and we want it to be mapped to the vector i - j based at (1,1,0). This means that ƒ(0,0,0) = (1,1,0) and the differential at the origin takes i + j + k to i - j, i.e.

${\mbox{d}}f_{\mathbf {0} }:T_{\mathbf {0} }\mathbb {R} ^{3}\to T_{f(\mathbf {0} )}\mathbb {R} ^{3}\ \ {\mbox{is such that}}\ \ ({\mbox{d}}f_{\mathbf {0} })(1,1,1)=(1,-1,0).$

Practically, you need to evaluate the Jacobian matrix of ƒ, and then evaluate it at x = y = z = 0. You'll also need to do the same for the second vector. You want ƒ(1,1,0) = (0,0,0) and

$({\mbox{d}}f_{(1,1,0)})(0,0,1)=(-1,0,1).\,$

Then you just need to check the differentiability of ƒ, and prove it's C¹. •• Fly by Night (talk) 11:11, 26 June 2010 (UTC)

Oh, I don't really need a solution to the problem...but if you care, there's a slightly easier but much less elegant approach. 173.179.59.66 (talk) 11:20, 26 June 2010 (UTC)

I see... so you mean that you thought that all vectors were unmovable? Well, given a manifold (just think of the plane if you like), there is something called a tangent bundle. This is a fibre bundle whose fibres are tangent spaces. The tangent space at a given point is the space of all vector based at that point that are also tangent to the manifold. So in the case of the plane, the tangent space at (1,0) is the space of all vectors based at (1,0). Let's say we have a map from the plane to the plane. If it carries (1,0) to (0,1) then the differential takes all of the vectors based at (1,0) to the space of vectors based at (0,1). The differential gives a linear map between the tangent space to the plane at (1,0) and the tangent space to the plane at (0,1). So vectors are very much movable. More generally, if f : M → N is a differentiable map between two manifolds, M and N, then the differential

${\mbox{d}}f_{x}:T_{x}M\to T_{f(x)}N$

is a linear map between tangent spaces. So in some sense it actually moves vectors from one manifold onto another manifold. •• Fly by Night (talk) 11:30, 26 June 2010 (UTC)

Contingent events

What techniques do actuaries or others use to determine the probability of contingent events?--220.253.96.217 (talk) 13:53, 26 June 2010 (UTC)

Have you had a search of Wikipedia using the search box at the top of the page? Try sticking in 'contingent events'. Bayes Theorem may also be of use. Dmcq (talk) 14:08, 26 June 2010 (UTC)

I think the concept you're asking for is conditional probability. 75.57.243.88 (talk) 17:19, 26 June 2010 (UTC)

Limits

I want to show that a function tends to zero as x tends to zero. Let

$f(x):={\frac {e^{-1/x^{2}}}{x^{3}}}$

How do I show that ƒ(x) → 0 as x → 0? I've tried L'Hôpital's rule and power series expansion, but I keep falling flat. For example:

$f(x)\sim {\frac {1}{x^{3}}}\left(1-{\frac {1}{x^{2}}}+{\frac {1}{2!\,x^{4}}}-{\frac {1}{3!\,x^{6}}}+\cdots \right)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k!\,x^{2k+3}}}$

But that doesn't exactly tell me much. L'Hôpital's rule just gives higher order denominators with each and every iteration. Any ideas? •• Fly by Night (talk) 14:55, 26 June 2010 (UTC)

I would solve it using the Landau notation. I'm strapped for time at the moment so 'mafraid I can't show you how. 76.229.193.242 (talk) 15:02, 26 June 2010 (UTC)

The substitution y = 1/x² simplifies the limit to $\lim _{x\to 0}{\frac {e^{-1/x^{2}}}{x^{3}}}=\lim _{y\to +\infty }{\frac {y^{3/2}}{e^{y}}}$. You should be able to solve the latter easily with whatever method you are used to.—Emil J. 15:16, 26 June 2010 (UTC)

That is, the manipulation is only valid for $\lim _{x\to 0+}f(x)$, but if you show that to be zero, then $\lim _{x\to 0-}f(x)$ is also zero because f is an odd function.—Emil J. 15:21, 26 June 2010 (UTC)

Easy as that! How foolish I feel now. Thanks EmilJ. •• Fly by Night (talk) 15:33, 26 June 2010 (UTC)

power series

why is it that your article says $f^{\prime }(x)=\sum _{n=1}^{\infty }a_{n}n\left(x-c\right)^{n-1}=\sum _{n=0}^{\infty }a_{n+1}\left(n+1\right)\left(x-c\right)^{n}$ and not $f^{\prime }(x)=\sum _{n=1}^{\infty }a_{n}n\left(x-c\right)^{n-1}=\sum _{n=0}^{\infty }a_{n+1}\left(n+1\right)\left(x-c\right)^{n-2}$

In the second summation you're replacing n everywhere by n+1. It couldn't be more obvious than that. Zunaid 22:22, 26 June 2010 (UTC)

June 27

Speed please

After seeing all those x's and y's and things above, I fell a bit sheepish asking this here, but it will help with an edit I will (may) make to an article. What speed is an aircraft averaging in knots if it covers 10,310 ft in 33 seconds, and (2) if it covers 20,555 ft in 112 seconds? Moriori (talk) 00:03, 27 June 2010 (UTC)

1) 185.106303 knots 2) 108.736648 knots.––220.253.216.181 (talk) 01:55, 27 June 2010 (UTC)

Note that you can do this with the Google calculator built into their search engine: "10,310 ft / 33 s in knots" yields "(10 310 ft) / (33 s) = 185.106303 knots" -- ToE^T 08:01, 27 June 2010 (UTC)

Bayesian Inference?

Suppose I flip a biased coin three times, and the result is H, T, H. I infer from this data that the probability of H when I flip the coin is 2/3. Suppose I flip the coin a fourth time, and the result is H. I now infer that the probability of H when flipping the coin is 3/4.
What I would like to know is if I can use bayesian inference to arrive at my conclusion. Let P(H) be the prior probability that I score H when flipping the coin, in this case 2/3. Let P(H|E) be the posterior probability that I score H when flipping the coin, or the probability the flipping the coin results in H given that I score H when I flip it the fourth time, in this case 3/4.
Bayes theorem states that $P(H|E)={\frac {P(E|H)\;P(H)}{P(E)}}$
What does P(E|H) and P(E) represent, and what are their values in the example I have presented?––220.253.216.181 (talk) 01:50, 27 June 2010 (UTC)

Consider a sample of n=3 flips out of a population of N=4 flips. The following table of 4 rows (for h = 0, 1, 2, 3 heads in the sample) and 5 columns (for H = 0, 1, 2, 3, 4 heads in the population) describe the odds.

                    4 1 0 0 0
                    0 3 2 0 0
                    0 0 2 3 0
                    0 0 0 1 4

As you observed h=2, consider the third row (0, 0, 2, 3, 0). So H=2 or H=3 with odds as 2 : 3. (The impossible outcomes have odds zero). So the probability that the fourth flip will be head (H=3) is 3/5=60%. The odds are computed by the expression $\scriptstyle {\binom {H}{h}}{\binom {N-H}{n-h}}$. If you divide by the column sum $\scriptstyle \sum _{h}{\binom {H}{h}}{\binom {N-H}{n-h}}={\binom {N}{n}}$ you get the hypergeometric distributions $\scriptstyle P(h|H)={\frac {{\binom {H}{h}}{\binom {N-H}{n-h}}}{\binom {N}{n}}}$. The unconditional probabilities are $\scriptstyle P(H)={\frac {1}{N+1}}$ and $\scriptstyle P(h)={\frac {1}{n+1}}$. Bayes rule for conditional probability says that

$P(H|h)={\frac {P(h|H)P(H)}{P(h)}}={\frac {{\frac {{\binom {H}{h}}{\binom {N-H}{n-h}}}{\binom {N}{n}}}{\frac {1}{N+1}}}{\frac {1}{n+1}}}={\frac {{\binom {H}{h}}{\binom {N-H}{n-h}}}{\binom {N+1}{n+1}}}$.

confirming that $\scriptstyle P(H=3|h=2,N=4,n=3)={\frac {{\binom {3}{2}}{\binom {4-3}{3-2}}}{\binom {4+1}{3+1}}}={\frac {3}{5}}=0.6$.

Bo Jacoby (talk) 03:56, 27 June 2010 (UTC).

The inferences you give are wrong, so there's no sense asking if they can be found with Bayesian inference. The correct conclusions can be found with Bayesian inference.

The coin has a property p associated with it. p is the probability that the coin will land heads when tossed, and depends on the bias of the coin. p is unknown, and is thus treated as a random variable with a certain distribution. This distribution is your prior. A uniform prior is usually adequate. The density of the distribution is $f(p)=1$.

After the coin is tossed once and lands H you update your distribution:

$f(p|X_{1}=H)={\frac {\mathrm {Pr} (X_{1}=H|p)f(p)}{\mathrm {Pr} (X_{1}=H)}}={\frac {p\cdot 1}{1/2}}=2p$

Where the denominator is found by integrating over p.

By the time you had 2H, 1T the distribution is $f(p)\propto p^{2}(1-p)$. If you then want to know the probability of heads in the next toss, it's $\mathbb {E} [p]=3/5$. If you toss it again and get a third H the distribution is $f(p)\propto p^{3}(1-p)$, and then $\mathbb {E} [p]=4/6$.

The family of distributions which is best for this kind of problem is the beta distribution, because whenever you update the distribution still belongs to the family. The uniform distribution is a special case. The inferences you suggested correspond to a degenerate beta distribution, which essentially says that p is either 0 or 1 with equal probabilities (which is of course not appropriate). -- Meni Rosenfeld (talk) 08:41, 27 June 2010 (UTC)

What's the justification for using the uniform prior?––220.253.216.181 (talk) 09:20, 27 June 2010 (UTC)

It's the highest-entropy distribution on the unit interval. This can be taken to mean that it makes the least assumptions above and beyond the fact that $0\leq p\leq 1$, so it is appropriate when you really know nothing about p. However, in practice you may have some knowledge about the coin before ever tossing it - for example, you may know that coins of this type usually have $0.4\leq p\leq 0.6$. For simplicity in calculations it is best to encode this knowledge as a beta distribution if at all possible. -- Meni Rosenfeld (talk) 10:59, 27 June 2010 (UTC)

The beta distribution is the limiting case of $\scriptstyle P({\frac {H}{N}}|h,N,n)={\frac {{\binom {H}{h}}{\binom {N-H}{n-h}}}{\binom {N+1}{n+1}}}$ for $\scriptstyle N\rightarrow \infty$ . But it is unnecessary to consider a large population. To obtain the probability that the next flip of a coin is head, it is sufficient to consider a population consisting of the sample supplemented by the next flip: N=n+1. The probability that the next flip is a head is simply $\scriptstyle P(H=h+1|h,N=n+1,n)={\frac {h+1}{n+2}}$. Bo Jacoby (talk) 13:31, 27 June 2010 (UTC).

Domain/Range of this function

What is the domain/range of this function?

$g(x) = \sqrt(16-x^4)$ —Preceding unsigned comment added by 69.230.55.21 (talk) 02:53, 27 June 2010 (UTC)

The domain and range are something you specify, not derived from a formula. For instance the domain and range of 2n could be either the integers or reals or complex numbers. By the way range (mathematics) has two possible meanings, either codomain or image (mathematics). The purpose of them is to allow a set of function (mathematics) to be treated uniformly without going into the details about the individual functions.

For general practical maths purposes I believe they may mean yet another thing where a function is considered as a set of pairs (x,y) where no two pairs have the same x, this is a graph of a function. The domain then is considered to be the set of all x such that there exists a pair with x as the first element and the range is considered to be the set of all ywhich occur in any of the pairs. They implicitly assume x is restricted to the reals unless they say otherwise. I'll assume this is what is meant by the question.

For sqrt there is he additional problem that it does not even form a function in the modern sense if you include both positive and negative square roots. I'm not sure whether the original question meant just the non-negative square root or both. If only the non-negative square root is meant then the 'range' is the non-negative real numbers otherwise it is all the real numbers. The domain is where the argument to the square root is non-negative so 16-x^4 is greater than or equal to 0. so x^4 ≤ 16 so -4 ≤x^2 ≤ 4. For reals x^2 is always greater than or equal to zero so 0 ≤x^2 ≤ 4 so -2 ≤x ≤ 2. So your domain is the reals from -2 to 2 inclusive. Dmcq (talk) 10:07, 27 June 2010 (UTC)

And for the struck through piece, the range probably is the possible non-negative values of the function which are 0 to 4 inclusive. Other meanings of range could include all the reals or just the non-negative reals or -4 to 4 but 0 to 4 is probably what's wanted. Dmcq (talk) 11:35, 27 June 2010 (UTC)

And of course you could always allow complex answers.... -mattbuck (Talk) 10:38, 27 June 2010 (UTC)

According to Square root, ${\sqrt {t}}$ (or sqrt(t)) always means the nonnegative root for nonnegative t. -- Meni Rosenfeld (talk) 11:22, 27 June 2010 (UTC)

Mr. Humanzee, your query is slightly cryptic as you formulated; no surprise, wouldn't you be satisfied with the answers you got. However, I think you mean, first of all, that $x$ is a real number, and that the "domain", is the larger set of real numbers where the expression is also well-defined as a real number, sometimes named "natural domain". This is not implicit in the question as you put it; see the answers above, but it is the most likely interpretation, given the high-school flavour of your question. That said, you need $16-x^{4}$ to be non-negative, that is

$16\geq x^{4}$

that is equivalent to $|x|\leq 2$ (is that clear to you?), and this interval is the domain.

The Range of a function is the set of all values taken by it. In your case, $x^{4}$ takes all values between 0 and 16 as $x$ varies in our "natural domain", $-2\leq x\leq 2,$ so that $16-x^{4}$ also varies between 0 and 16. Taking the square root, you get all values between 0 and 4, and in conclusion the range is therefore the interval [0,4]. As a side remark: not only you should better try to put your question into more precise terms, but it would be nice if you show some attempt to solve the problem. Otherwise, people may think you are just trying to get your homework done, while you are enjoying the panorama. --pma 17:53, 27 June 2010 (UTC)

Natural domain, thanks, I was thinking vaguely there should be a term for that concept. There's no article on it but my guess it is notable. Dmcq (talk) 20:23, 27 June 2010 (UTC)

Yes; I realized later that you had already answered in all detail the OP's question; anyway repeating things is not bad.--pma 22:05, 27 June 2010 (UTC)

The probability that a hypothesis is true before any evidence is presented

Bayesian inference provides a method for updating the probability that a hypothesis is true when presented with some evidence, but it requires a prior probability that the hypothesis is true before the evidence is presented. This prior probability could have been formulated using the same method with different evidence and an even earlier prior probability. Presumably there must be an original probability that a hypothesis is true before any evidence is presented at all, so what is it? ––220.253.216.181 (talk) 09:37, 27 June 2010 (UTC)

Just make a guess and make sure it has a wide margin of error and allows for you being totally and entirely wrong. Then stick to it. Don't do the business of going back and changing your prior to say something is even less likely if you find it actually happens! Dmcq (talk) 11:04, 27 June 2010 (UTC)

"Just making a guess" doesn't always work. If your prior was ill-chosen, the posterior may remain bad even after a lot of evidence. -- Meni Rosenfeld (talk) 11:10, 27 June 2010 (UTC)

The thing I was thinking of about changing priors was for instance the assumption Saddam Hussein had weapons of mass destruction. After not finding any the assumption was changed to that he definitely had them and was even more cunning and evil then previously supposed in hiding them. Dmcq (talk) 11:20, 27 June 2010 (UTC)

Sure, switching to a different prior after collecting data is wrong. My point was that from the start you should spend some time thinking about what prior accurately reflects your current state of knowledge. -- Meni Rosenfeld (talk) 11:28, 27 June 2010 (UTC)

As I should have guessed we have an article on it Prior probability Dmcq (talk) 11:09, 27 June 2010 (UTC)

Possible approaches are the principle of maximum entropy and the Jeffreys prior. Occam's razor can be applied to assign a hypothesis less prior probability the more complex it is. -- Meni Rosenfeld (talk) 11:10, 27 June 2010 (UTC)

If there are just two possibilities, either the hypothesis is true or the hypothesis is false, and you have no way of distinguishing between these two, then prior odds are 1:1 and the probability is 1/2. Bo Jacoby (talk) 13:59, 27 June 2010 (UTC).

Hmmm ? My hypothesis is that if I flip a coin 10 times I will get 5 heads and 5 tails. This is either true or false. There are just two possibilities. I have never flipped this coin before and I have no knowledge of whether it is fair or biased. Does that mean that the prior odds of 5 heads from 10 flips are 1:1 ? And what then are the prior odds of 6 heads from 10 flips ? Or 4 heads from 10 flips ? Are they all 1:1 ? Gandalf61 (talk) 15:51, 27 June 2010 (UTC)

I disagree with Bo. The natural choice of prior probability when you have absolutely no evidence is a uniform distribution of the underlying random variable. Not a uniform distribution of the truth/falsity of the hypothesis. So, in your example, we assume the coin is fair and calculate the probability of the hypothesis accordingly. I'm not quite sure what to do if the random variable is distributed over the real numbers, say, where there is no uniform distribution... --Tango (talk) 16:11, 27 June 2010 (UTC)

I think Bo was speaking literally about a case where there is no underlying random variable, just a hypothesis for which we cannot distinguish its truth from its falsity. Imagine someone you trust giving you an object and tells you that it is either a thingamajig or a doojigger, but not both. You have no idea what a thingamajig or a doojigger are. What is the probability that the object is a thingamajig? It's 1/2.

More generally, the discrete uniform probability is maximum entropy over a finite set of items.

In Gandalf's example the options can be distinguished - one is much more specific than the other. Likewise for "I can either win the lottery or not. Hence, equal odds."

As for unbounded random variables - an improper uniform prior can be used, as can an "almost uniform" distribution like Cauchy. -- Meni Rosenfeld (talk) 17:40, 27 June 2010 (UTC)

But if you were given an object and told it was either a thingamajig, a doojigger or a thingamabob and given a hypothesis "it is a thingamajig" then the natural prior probability of that hypothesis is not 1/2, it's 1/3. A hypothesis is not just an arbitrary sequence of words. Those words have meanings. When we say we have no prior knowledge we don't mean that we don't even understand the language the hypothesis is written in (if that were the case, then yes, our best effort would be to guess 50/50). I'm not sure what you mean by a case without an underlying random variable - if there isn't a random variable, then why are we talking about probabilities? In your example, the random variable is the identity of the object with the sample space {"thingamjig","doojigger"}. --Tango (talk) 18:11, 27 June 2010 (UTC)

Again, the key part is "no way of distinguishing". In your example, one alternative is that the object is a specific type among 3 equally probable types, and the other is that the object is any of 2 types. There is no symmetry.

I can contrive more examples where the active ingredient is not ignorance of the language used. I concede that Bo's original observation might only be usable in contrived examples.

Ignore the part about no random variable. -- Meni Rosenfeld (talk) 19:45, 27 June 2010 (UTC)

I think the lottery example points to the issue of assigning prior probabilities. How does one assign the prior probability of winning an otherwise unspecified lottery? In this situation you aren't told *which* lottery was played (you have no clue how many numbers were picked), or even where it was played (you can't guess which lottery might have been played). You don't even have a price/payout ratio to reason from. However, saying that winning and not winning are equally probable wouldn't be accurate, as most lotteries don't work that way (although there are some (non-governmental) lotteries which do have a 1:1 win:loss ratio). -- 174.24.195.56 (talk) 18:25, 27 June 2010 (UTC)

One thing you can do is assign the system in question to a general reference class, and use data on that. "Ok, this is a lottery, and lotteries usually have winning probabilities of such-and-such, so I'll use a prior based on that." The result will depend on the reference class you choose, so your mileage may vary. -- Meni Rosenfeld (talk) 19:45, 27 June 2010 (UTC)

(edit conflict) If you have an underlying random variable, then you do know something. The assumption was that you had no prior knowledge. Gandalf's coin has an unknown probability p for showing head, and the problem is to find a reasonable prior distribution $\scriptstyle f(p)$ . Then compute the prior probability that your hypothesis is true: $\scriptstyle \int _{0}^{1}f(p){\binom {10}{5}}p^{5}(1-p)^{5}dp$. If you have absolutely no knowledge about the coin, except that it may produce heads or tails when flipped, then $\scriptstyle f(p)=1$. If you can look at the coin and it looks symmetrical, then choose $\scriptstyle f(p)$ according to that. Bo Jacoby (talk) 17:56, 27 June 2010 (UTC).

All you know is what the hypothesis says. It's pointless to speculate about the probability of an unknown hypothesis. --Tango (talk) 18:11, 27 June 2010 (UTC)

Don't complicate simple cases with generality. When flipping a coin with two faces that are only superficially distinguishable, the prior odds heads:tails = 1:1. Knowledge about the performance history of the coin may changes these odds: if 8 flips gave 8 tails then the odds for the 9'th flip is heads:tails = 1:9. Bo Jacoby (talk) 21:11, 27 June 2010 (UTC).

This is a general question. I was disagreeing with your statement that there being an underlying random variable counts as prior knowledge. Prior knowledge means some kind of evidence. The statement of the hypothesis, which is all that is required to determine what the random variable is, does not count. --Tango (talk) 21:21, 27 June 2010 (UTC)

Yes this is a general question, and a difficult one, but many specific cases of the question are easy. What is the disagreeing about? Are we getting different answers to the same questions, or is it just about wording? All you know is what the hypothesis says you say, but what if you don't know what the hypothesis says, but all you know is that there is a hypothesis which may be either true or false? Bo Jacoby (talk) 04:08, 28 June 2010 (UTC).

This is an interesting article. Count Iblis (talk) 00:26, 28 June 2010 (UTC)

Surface area of a slice of a sphere

In a recent question posted to the Math Reference Desk, one answerer mentioned the result (attributed to Archimedes) that the surface area of a slice of a sphere formed by intersecting two parallel planes with the sphere depends only on the distance between the planes but not their locations. I thought that was a very interesting result and wondered how Archimedes might have derived it. Regardless of whether it is Archimedes' method, is there a proof for the result that uses only techniques understandable by elementary school students? --173.49.17.92 (talk) 16:03, 27 June 2010 (UTC)

I expect Archimedes derived the result by The Method of Mechanical Theorems and proved it by the method of exhaustion. Algebraist 16:07, 27 June 2010 (UTC)

The area is liek the volume between two spheres as shown in [3]. You can see that the result for spheres is exactly the same as the result that the area of a ring only depends on the length of the tangent to the inner circle.Dmcq (talk) 16:21, 27 June 2010 (UTC)

Glue a stamp to the surface of the earth, and project it unto a circular cylinder touching the earth around the equator, (the projection lines being parallel to the equatorial plane and radial from the axis). The projection is shorter than the stamp in the north-south direction, but equally many times longer than the stamp in the east-west direction, due to some similar triangles. So the area of the stamp equals the area of its projection. And so the area of the sphere equals its projection upon the cylinder, which is the circumference times the diameter. Bo Jacoby (talk) 20:25, 27 June 2010 (UTC).

An estimate to the closed form expression of a function

Let A be the set of all positive even integers.

I have the following function f:A-->N, where N is the set of all natural numbers. The first few values taken by f are:

1,2,3,4,5,6,7,8,9,10,11,11,12,13,14,15,16,17,18

i.e. f(2)=1, f(4)=2, f(6)=3 and so on. What I am interested in is an estimate of what a closed form expression might look like for f. Its known that f(n)<=n/2, and the values indicate a very small shift from n/2, so small that it isnt visible in the begining. Can any give me any pointers. Thanks--Shahab (talk) 17:39, 27 June 2010 (UTC)

How about f(n) = n/2 + [n = 2] – 2[n > 22], where [·] is an Iverson bracket? Qwfp (talk) 17:54, 27 June 2010 (UTC)

Is the Iverson bracket usually allowed in closed form expressions? It's not one of the usual elementary functions. (Of course, you can choose whatever functions you like to consider elementary for a certain problem, but since the OP didn't specify we should assume we're using the usual functions (plus floor and ceiling, perhaps, since we're working with integers).) --Tango (talk) 18:21, 27 June 2010 (UTC)

By the way, f(n)<=n/2 appears to be false for n=2. Qwfp (talk) 17:56, 27 June 2010 (UTC)

yes that was a mistake. Thanks-Shahab (talk) 18:04, 27 June 2010 (UTC)

Corrected. How should I modify it now? Actually the problem is part of a much bigger problem, a similar result gives an estimate dealing with the ceiling function. Thanks-Shahab (talk) 18:08, 27 June 2010 (UTC)

$f(n)={\frac {n}{2}}-\left\lfloor {\frac {n}{24}}\right\rfloor$ works (that's the floor function - I don't think you'll find anything that doesn't require something like floor, except for an 18th degree polynomial, which probably isn't what you want). --Tango (talk) 18:21, 27 June 2010 (UTC)

Thank you. Would you mind telling me how did you deduce it? -Shahab (talk) 18:33, 27 June 2010 (UTC)

The function is n/2, minus a small error term. That error term is 0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1 - a function with a step in it like that can usually be written using the floor function (or ceiling function), so I just worked out what function would cross from being 0.something to 1.something at the appropriate point, and n/24 does so. --Tango (talk) 18:37, 27 June 2010 (UTC)

You haven't really given enough terms, though. The difference between your function and n/2 only changes once. You can't really predict where else it will change from just one data point. I've guessed that it will change by 1 every 12 terms, but that's a complete guess. --Tango (talk) 18:39, 27 June 2010 (UTC)

Thanks for the help and the explanation Tango-Shahab (talk) 19:16, 27 June 2010 (UTC)

A somehow simpler guess is that the error jumps only once. Dear Shahab, have you got a definition for this function, or some more terms? --pma 19:25, 27 June 2010 (UTC)

Hi Pmajer. I hope you are well. I am glad you are around. I'll try to describe the whole problem to you, in fact I will be glad if you (or anybody else here) can give me any general advice regarding the problem.

Consider the equation v+x+y-z = b (b is an even natural number). It is known that there exists a least natural number r(b) such that if {1,2...r(b)} is partitioned into 2 classes arbitrarily, at least one contains a solution to the given equation. My final goal is to find out a formula for r(b). To this end, I have written a computer program (using brute force) to estimate various values of r(b) for different b, and I use the bound r(b)<= b/2 as v=x=y=z=b/2 is a trivial solution. The problem is that the program starts taking an awful lot of time to estimate r(b) as b increases because the number of possible partitions increases as a power of 2.

Is there any general advice on how to approach this problem, any techniques etc you (or anyone) can suggest. I will really appreciate it.--Shahab (talk) 19:13, 28 June 2010 (UTC)

Is this a well-known search algorithm, and, if so, what is it called?

I have some questions about a method I have thought of for finding shortest routes between 2 locations on a map:

1) Does it give the right answer?

2) Is it already well-known (which I expect it will be as it is so simple)? and if so does it have a name?

3) As this method is basically an analogue computer, does it have a digital version, ie one that can be described by a list of steps as in the Graph search algorithms articles?

The method:

a) Physically make a copy of the map with string and knots, where the knots represent locations on the map and the lengths of the bits of string represent the distances.

b) Take hold of the two knots representing the 2 relevant locations and pull them as far apart as possible without breaking anything.

c) The pieces of string that are now taut represent the shortest route.

I have looked at the Graph search algorithms, and it does not seem to be any of those.

FrankSier (talk) 23:31, 27 June 2010 (UTC)

I've no idea about the name of graph search algorithms. However your physical method is limited by the accuracy and size of knot tying and placement and by the flexibility of cotton. IE it works for crude graphs where different paths are not very similar in lengths, but won't distinguish them well. There are mathematically precise algorithms. -- SGBailey (talk) 06:09, 28 June 2010 (UTC)

It's been called the ball and string model in some papers and also used as an illustration of how some problems can be solved easily physically, I forget where I saw it used in an elementary demonstration, I don't think it was called the ball and string model there, a pity as I like those sorts of models. Dmcq (talk) 09:43, 28 June 2010 (UTC)

Google Scholar (and Google web for that matter) search for "ball and string model" finds this academic paper: "New dynamic SPT algorithm based on a ball-and-string model", P Narvaez, KY Siu, HY Tzeng. IEEE/ACM Transactions on Networking 2001. doi:10.1109/90.974525. Neither the model nor the phrase is referenced in the paper so it appears they thought it up, although others may have come up with it too independently of course. Qwfp (talk) 09:57, 28 June 2010 (UTC)

You might be interested in reading about Dijkstra's algorithm. Sorry, it was already in the list that you linked. Dbfirs 12:56, 28 June 2010 (UTC)

I had another google and it looks like George Minty referred to using strings knotted together to find the shortest path in a 1957 paper when describing the problem. Dmcq (talk) 13:17, 28 June 2010 (UTC)

Oh, so he did: . JSTOR 167474. {{cite journal}}: Cite journal requires |journal= (help); Missing or empty |title= (help) It's not really a paper though, just a very short comment. Nevertheless, it's been cited 39 times according to Google Scholar: [4] Qwfp (talk) 13:45, 28 June 2010 (UTC)

I think there should be a measure of citation density given by citations/pages. That would concentrate minds on cutting papers down though it might make some of them even more unintelligible :) Dmcq (talk) 14:41, 28 June 2010 (UTC)

AK Dewdney had a great popular-science-level discussion on analog computers, I think it was in The Armchair Universe, but it may have been Turing Omnibus. He mentions your algorithm and a number of others, such as spaghetti sort. (He has a clever algorithm for finding the two most distant points in a tree, too.) He also has a good discussion of the computational complexity and limitations of such algorithms. Eric. 66.30.118.93 (talk) 02:44, 29 June 2010 (UTC)

June 28

Bayesian inference with very little information

See the subheading "False positives in a medical test" in the article Bayesian inference. How would the problem be solved (i.e. the probability of a false positive) if P(B|A), P(A), P(B) were completely unknown?––115.178.29.142 (talk) 00:52, 28 June 2010 (UTC)

The result $\scriptstyle {\frac {{\frac {1}{2}}{\frac {1}{2}}}{\frac {1}{2}}}={\frac {1}{2}}$ is as unreasonable as the assumption. Bo Jacoby (talk) 04:27, 28 June 2010 (UTC).

Exponents and Fractions

I do not understand how to approach this question: ${\frac {(x^{a-1})^{3}}{(x^{2a})(x)}}.$ The answer is $x^{a-4}.$ Thank you for the help. —Preceding unsigned comment added by 24.86.187.199 (talk) 05:25, 28 June 2010 (UTC)

To simplify this expression, start by getting both the numerator and the denominator as expressions with a single x then see if it becomes obvious. -- SGBailey (talk) 05:57, 28 June 2010 (UTC)

Wikipedia doesn't do your homework. 194.100.253.98 (talk) 05:55, 28 June 2010 (UTC)

OP: note that yours "x^a-1" is very ambiguous! it should be possibly read "$x^{a}-1$" but in fact from the answer you wrote, I understand you mean "x^(a-1)", that is $x^{a-1}$. I've put the formulas in more readable style. Please try to learn how to write maths formulas, that would make easier for you to get answers. In these cases, you can edit similar formulas and see how they did it. On a different note, please don't forget to sign your post. --pma 08:57, 28 June 2010 (UTC)

See Exponentiation#Identities and properties. -- 58.147.53.253 (talk) 12:46, 28 June 2010 (UTC)

Statistical sample

Hello. I have a question concerning statistical samples. I have written a small program that flips two coins each round for 50000 rounds. In order to demonstrate the Boy or Girl paradox, I made it ignore results where both coins turn up to be tails and instead flip the coins again during the same round, so I end up with 50000 legal rolls. Now, I'm wondering about the official sample size of the program - is it 50000 (the amount of legal rolls) or are the ignored rolls also officially part of the sample? 194.100.253.98 (talk) 05:54, 28 June 2010 (UTC)

Isn't this the question at the heart of the paradox? Your choice will determine your answer. Dbfirs 12:51, 28 June 2010 (UTC)

Well, I know that the answer is 1/3. It's just that another person claims that while my answer is correct, officially the double tails sets are also a part of the sample (even though ignored as they don't qualify by the problem's criteria). I chose not to include them in the sample size count, since they are "impossible" by the way the problem is defined. Does the sample by definition include discarded, "impossible" random rolls? 194.100.253.98 (talk) 13:01, 28 June 2010 (UTC)

You're trying to estimate a probability, or equivalently, the proportion of "rolls with at least one head" where "both rolls are heads". So yes your total sample space is only those rolls that were valid - the two-tailed rolls are not counted. Confusing Manifestation(Say hi!) 00:02, 29 June 2010 (UTC)

Thank you! 194.100.253.98 (talk) 05:25, 29 June 2010 (UTC)

General Cardioid Equation

Hey mathletes-

I'm trying to understand cardioid functions in polar coordinate form.

Is the most general form P(theta)=2a(1-cos(theta))? that is to say, all cardioids are of this form?

thanks209.6.54.248 (talk) 15:28, 28 June 2010 (UTC)

See Cardioid, yes they are all of that form though of course the equation can change if the shape is moved around. The term is also sometimes applied to any shape that looks like that. Dmcq (talk) 15:58, 28 June 2010 (UTC)

Limaçon

Limaçon translates to mean snail, but why are the Limaçon curves named after snails? •• Fly by Night (talk) 19:27, 28 June 2010 (UTC)

They look like snail shells. The resemblance may be a bit strained but that's typical in the naming of curves.--RDBury (talk) 21:37, 28 June 2010 (UTC)

Are these formulas known in the litterature of statistics?

In the above thread the discrete probability distribution $\scriptstyle P(H|h,N,n)={\frac {{\binom {H}{h}}{\binom {N-H}{n-h}}}{\binom {N+1}{n+1}}}$ is defined from Bayes' theorem based on the hypergeometric distribution $\scriptstyle P(h|H,N,n)={\frac {{\binom {H}{h}}{\binom {N-H}{n-h}}}{\binom {N}{n}}}$ and the marginal distributions $\scriptstyle P(h|N,n)={\frac {1}{n+1}}$ and $\scriptstyle P(H|N,n)={\frac {1}{N+1}}$. As this computation is very elementary I expected to find the probability distribution $\scriptstyle P(H|h,N,n)$ in the litterature, but so far I have searched in vain. So my question is: is it known, or is it original research on my part? If it is known, then please provide a reference and the name of the distribution $\scriptstyle P(H|h,N,n)$. (I call $\scriptstyle P(H|h,N,n)$ the discrete induction distribution, and the hypergeometric distribution $\scriptstyle P(h|H,N,n)$ is the corresponding discrete deduction distribution).

The mean ± standard deviation of the hypergeometric distribution is

h ≈ f (H, N, n)

where f is defined by f (H, N, n) = $\scriptstyle {\frac {Hn}{N}}\pm {\sqrt {\frac {{\frac {Hn}{N}}(1-{\frac {H}{N}})(1-{\frac {n}{N}})}{1-{\frac {1}{N}}}}}$.

The mean ± standard deviation of the induction distribution is

H ≈ −1− f (−1−h, −2−n, −2−N).

My proof of this result is boring, but the result itself is fascinating, because it says that the transformation

(H, h, N, n) → (−1−h, −1−H, −2−n, −2−N)

translates between induction and deduction. Is this formula known?

In the J (programming language) the deduction and induction formulas are

  deduc=.(*(%+/))([,:%:@*)(*/@}.%{.)@(-.@((1,,:)%+/@]))
  induc=.*&_1 1@(+&1 0)@(-@>:@[deduc~-@(+#)~)
  1 2 induc 4
    1.4      2.6
0.489898 0.489898

The example shows that if you got 1 tail and 2 heads in the first 3 flips, then you should expect 1.4±0.5 tails and 2.6±0.5 heads in a total of 4 flips. Bo Jacoby (talk) 23:18, 28 June 2010 (UTC).

June 29

Vector similarity

I can calculate the cosine between two unit vectors in two and three dimensions. Doing some calculations in pencil, I think that doing it in four dimensions is a trivial jump from three dimensions. I would like to know how complex does calculating the cosine between two unit vectors become as dimensions are added. I measure complexity by how many variables are required and how many mathematical operations need to be performed. Is it a linear progression (ie: 10 dimensions is twice the complexity of 5 dimensions)? Is it an exponential progression? Does it actually become easier as you start piling on the eleventh, twelfth, and thirteenth dimension? -- kainaw™ 03:09, 29 June 2010 (UTC)

Usually we say that the angle between vectors A and B is $\arccos \left({{A\cdot B} \over {|A|\cdot |B|}}\right)$.

In that formula, $A\cdot B$ denotes the vector dot product which is straightforward to compute if the vectors are in coordinate form. When A and B are unit vectors, the denominator is 1, of course. 75.57.243.88 (talk) 04:08, 29 June 2010 (UTC)

I didn't realize that I was performing the dot product. So, using 10 dimensions, it would simply be a matter of 10 multiplications and 9 additions. Each dimension adds one multiplication and one addition. So, it is a linear increase in complexity - unless I am missing something. -- kainaw™ 04:13, 29 June 2010 (UTC)

Yes, that's correct.

At the risk of sounding snide (actually, I will definitely sound snide), I'll remind you that I linked to the Cosine similarity article, which includes the known fact that the cosine is equivalent to the dot product, as early as here. It is of course clear to anyone who uses cosine similarity in practice. If you're only now realizing this basic fact, I think you really are biting more than you can chew, and that you shouldn't try to contribute to subject areas you have no grasp on. -- Meni Rosenfeld (talk) 06:35, 29 June 2010 (UTC)

So, what you are saying is that if you don't use vector-based mathematics in health informatics (my area of study), you should avoid learning about it and pretend that it doesn't exist. I was apparently suffering from the illusion that mathematics was a good subject to study whenever work was slow so work could be improved in the future. -- kainaw™ 18:53, 29 June 2010 (UTC)

Invalid proofs

Important proofs which are later seen to be flawed are common throughout the history of mathematics – early proofs of the four colour theorem are prime examples. I'm interested in prominent examples of the following (and what the repercussions were): a result is proved in a seemingly airtight way; the mathematical community accepts the proof; a large amount of mathematical work follows the proof, and new proofs ground themselves in the original result; and then the first proof is exposed as flawed years later, invalidating everything that had followed. Also, I'd like to know whether this kind of thing has happened even in the absence of an initial proof – where mathematicians implicitly assume a result without decisive proof, and then that result is seen to be false and the house of cards collapses. In a documentary I was watching, Andrew Wiles recounted how people were doing new mathematics based on the Taniyama–Shimura conjecture before Wiles used it to prove Fermat's Last Theorem. Although that particular conjecture turned out to be true, it provides an example of what I mean. Thanks in advance. —Anonymous Dissident^Talk 11:03, 29 June 2010 (UTC)

It's pretty rare for a published "theorem" to turn out to be false, but List of published false theorems mentions a few. 75.57.243.88 (talk) 15:08, 29 June 2010 (UTC)

False proofs get published all the time, but not in peer-reviewed journals. If a proof survives peer-review, then it is extremely unlikely that it's wrong. The mathematical community would not accept the proof without it being in a peer-reviewed journal and wouldn't use the theorem in their own work either. Mathematicians do sometimes assume conjectures to be true (eg. the Riemann hypothesis) and prove other results, but they do so in full knowledge of the fact that the conjecture could end up being false and their result would be invalid. --Tango (talk) 15:33, 29 June 2010 (UTC)

For recent examples of paper published in major peer-reviewed journals (some would say the major math journals: Inventiones and the Annals) see Daniel Biss. I would agree that published, peer-reviewed papers whose main theorems are incorrect are rare. Reviewers often evaluate the general proof strategy and convince themselves that the details could be filled in without too much trouble, so sometimes the exposition or reasoning have minor flaws, or one finds incorrect or redundant intermediate results. Phils 16:33, 29 June 2010 (UTC)

I don't believe any example of your situation "a result is proved...the mathematical community accepts the proof; a large amount of mathematical work follows...then the first proof is exposed as flawed" yet exists. That's not to say it won't ever happen though. The papers making up the Classification of finite simple groups have all been peer reviewed, and a great deal of work assumes it is correct, but the sheer length and complexity (tens of thousands of pages according to our article) is pretty staggering. Our article claims "Jean-Pierre Serre is a notable skeptic" of the proof, but the link given doesn't seem to support that. Tinfoilcat (talk) 18:17, 29 June 2010 (UTC)

Solving for the same exponential variable on both sides of an equation

Those who may know me from the other reference desks will discover that I am a mathematical ignoramus. How do I solve for n in the following equation?

axⁿ − a = byⁿ

Thank you so much. Marco polo (talk) 15:27, 29 June 2010 (UTC)

So a, b, x and y are all known and you want to know n? You can rearrange that equation to $\left({\frac {x}{y}}\right)^{n}={\frac {b}{(1-a)}}$. That can then be solved using logarithms. --Tango (talk) 15:37, 29 June 2010 (UTC)

You seem to have misread the equation as axⁿ − xⁿ = byⁿ.—Emil J. 15:39, 29 June 2010 (UTC)

Oh, yes, I switched the terms when I factorised it... oops! --Tango (talk) 15:53, 29 June 2010 (UTC)

There's probably no elementary way to solve this equation. You can solve it numerically, though. -- Meni Rosenfeld (talk) 15:54, 29 June 2010 (UTC)

It might be possible to make a good approximation if you knew something about the sizes of a,b,x,y Tinfoilcat (talk) 16:07, 29 June 2010 (UTC)

A solution can in lucky cases be computed by the iteration

$n=\lim _{i\rightarrow \infty }f^{i}\left({\frac {\log {\frac {b}{a}}}{\log {\frac {x}{y}}}}\right)$

where

$f(n)={\frac {\log {\frac {b}{a}}-\log(1-{\frac {1}{x^{n}}})}{\log {\frac {x}{y}}}}$

because the original equation is $\scriptstyle n=f(n)$. Bo Jacoby (talk) 19:40, 29 June 2010 (UTC).

This is revealing but somewhat disillusioning to a math dullard such as me. No wonder I struggled with this! I assumed that my weak math skills (and poor memory of algebra class more than 30 years ago) had left me unable to see the solution. I am writing about the global debt crisis and comparing prospects for GDP growth (ax − a, where a represents GDP and x represents the annual rate of growth) with total debt (b) and the rate at which government debt would grow if deficits were eliminated now (or the interest rate on government debt + 1) (y). I am trying to determine the number of years (n) it would take for GDP growth to pay off this debt at different levels of growth. So I have to approximate toward this number by trial and error, then? This destroys my (apparently erroneous) idea that algebra offers an elegant way to solve any equation. Thanks for your responses. Marco polo (talk) 19:54, 29 June 2010 (UTC)

Easier to solve for y, to give the level of growth (y – 1) needed to pay off debt in n years. Qwfp (talk) 20:10, 29 June 2010 (UTC)

It is important to know that x>1, but you did not tell us that. There is no trial and error here. Give me values for a, b, x, y and I will compute n for you by the formula. Bo Jacoby (talk) 21:44, 29 June 2010 (UTC).

Fan Curve

What is the fan curve equation - pressure vs volume flow rate - or how is it derived? In other words how is it related to the specifics of the fan (no of blades, blade angle, blade length etc)? Also how is the efficicency curve of the fan derived (if it can be derived - Im not talking about the friction loss in bearings etc but the hydraulic losses which depend on geometry)? If the answers to the above are not available, can someone tell me how the efficiency varies with fan rpm? Im not talking about ways to calculate/measure the above but get an equation for it. Thanks

Fluid dynamics is tricky business - I suspect that the answers to most of your questions cannot be expressed with a simple formula. However, they should be obtainable from a computerized simulation for any specified system. -- Meni Rosenfeld (talk) 18:31, 29 June 2010 (UTC)

June 30

Admissible decision rule pdf

On the page admissible decision rule, how is pdf F(x|θ) derived?--Alphador (talk) 00:09, 30 June 2010 (UTC)

Inverse scaling

I'm sure this is an incredibly simple problem. I have two identical squares. When the first square is scaled down 43% (any arbitrary amount), the second square copies the first square, so it also reduces 43%. By what percentage would I have to scale up the 2nd square to make it appear to keep its original scale (even though it's being reduced 43%). In other words, by what amount do I need to scale up the 40% to get it back to 100%? It's easy to figure out for 10%, 25%, 50% -- (scale by 1000%, 400%, and 200% respectively). But my brain isn't working for arbitrary amounts. --70.130.58.45 (talk) 01:26, 30 June 2010 (UTC)

From your 10% / 1000% answer, I assume that you are scaling down to 43% and not by 43%. So, you want to know x where 0.43 x = 1? Try x = 1 / 0.43 ≅ 2.3256 = 232.56%. -- 58.147.53.253 (talk) 01:37, 30 June 2010 (UTC)

It appears to me that you are asking what percent you've reduced when you've reduced by 43% twice (so you can increase by the inverse percent and get back to the original size). If the original size is X and you reduce by 43%, the new size is .43*X. If you reduce again, the new size is .43*.43*X which is 0.18*X. So, divide by 0.18 and you get (0.18*X)/0.18 = X. If your scale amount is Y, you just divide by Y squared. -- kainaw™ 01:39, 30 June 2010 (UTC)

nth power...

I am trying to solve a = (b+c)^n for b = . Can't figure out where to even start with this, any hints please?--Dacium (talk) 02:03, 30 June 2010 (UTC)

Take the nth root of both sides. --Tango (talk) 02:07, 30 June 2010 (UTC)

duh... its one of those days where my brain isn't working :) a^(1/n)-c=b --Dacium (talk) 02:11, 30 June 2010 (UTC)