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July 26

Integrals

If you have$\int {u}{v}\,dx$, where u and v are both functions of x, can you split it up into $\int {u}\,dx*\int {v}\,dx$? 92.0.139.45 (talk) 10:39, 26 July 2008 (UTC)

No, not usually. For example, if u(x)=x and v(x)=x then

$\int {u}{v}\,dx=\int {x^{2}}\,dx={\frac {x^{3}}{3}}$

$\int {u}\,dx\int {v}\,dx=\left(\int {x}\,dx\right)^{2}={\frac {x^{4}}{4}}$

See integration by parts for the rule that actually applies here. Gandalf61 (talk) 10:51, 26 July 2008 (UTC)

It does work for addition, though. $\int (u+v)\,dx=\int u\,dx+\int v\,dx$ --Tango (talk) 15:18, 26 July 2008 (UTC)

3d annulus?

What is the name for the 3-d equivalent of an annulus, ie. the region between two spheres, not the same size, with a common centre? 203.221.127.38 (talk) 17:16, 26 July 2008 (UTC)

If there is a name for such, it is not in List of geometric shapes#Curved but probably should be. GromXXVII (talk) 17:52, 26 July 2008 (UTC)

"Thick shell" would be a good name eg see Shell theorem#Thick shells 87.102.86.73 (talk) 17:58, 26 July 2008 (UTC)

"solid of revolution of the annulus" would be a fairly unambiguous interpretable description.87.102.86.73 (talk) 18:41, 26 July 2008 (UTC)

If we are to be rather pedantic, this is somewhat ambiguous, since if the axis of revolution does not go through the center of the annulus, then you get something else. I would use "symmetric difference of two concentric balls". Oded (talk) 00:59, 27 July 2008 (UTC)

"Hollow Sphere" is in fairly common use http://www.google.co.uk/search?ndsp=20&um=1&hl=en&q=hollow%20sphere&ie=UTF-8&sa=N&tab=iw 87.102.86.73 (talk) 18:56, 26 July 2008 (UTC)

I would interpret "hollow sphere" to mean a shape with zero thickness, in other words a 2-sphere. I think "thick shell" is the best term. Solid of revolution is ambiguous if you don't specify the axis - unless you say otherwise, everyone is going to assume you mean a diameter, but it's still ambiguous. --Tango (talk) 23:01, 26 July 2008 (UTC)

As you might expect, it's called an annulus. --C S (talk) 17:41, 27 July 2008 (UTC)

A 3-annulus? I've never heard the phrase before, but I suppose it's pretty clear what it would mean. --Tango (talk) 18:41, 27 July 2008 (UTC)

This term is used in all dimensions, not just 2 and 3. For example, see annulus theorem, one of the most famous results in geometric topology. Sadly, WP seems lacking in this department, but I'll get around to it eventually. (I don't watch this page regularly. Leave a note on my talk page if you have further questions)--C S (talk) 19:11, 27 July 2008 (UTC)

Thanks, folks. You always deliver :) 202.89.166.179 (talk) 11:58, 28 July 2008 (UTC)

In a nutshell, its gotta be a shell, surely

Simultaneous equations

I have three simultaneous equations in three variables. I either have to 'solve the simultaneous equations or prove that they have no solution'. After some investigation, it becomes apparent that the first two equations can only have a solution if they are the same equation. This now leaves me with two equations in three unknowns. I can tell from this that there is no unique solution to these equations, however does it count as solving the equations if I put one variable in terms of another, which I can do? Thanks 92.3.123.172 (talk) 18:45, 26 July 2008 (UTC)

Is this hypothetical or do you have the equations? Because - After some investigation, it becomes apparent that the first two equations can only have a solution if they are the same equation doesn't make much sense to me, are they not linear equations?

If you have two equations with three unknowns then you should do what you suggest, which is as near to a solution as you may be able to get.87.102.86.73 (talk) 19:02, 26 July 2008 (UTC)

'Solving equations' means to find the solutions to the equations. If there is no solution, (which is usually the case when the equations contradicts one another, such as 0x=1), it means stating that fact. If there is only one solution, (which is usually the case when the equations are linear, such as x+y=2; x−y=0), it means to state that solution. If there is a finite number of solutions, (which is usually the case when the equations are of finite degree > 1, such as x²=x+1), it means to list these solutions. If there is a continuum of solutions, (as is usually the case when there are more unknown than equations, such as x=y), it is not possible to list all the solutions. The best one can do is to put one variable in terms of another, so yes, it counts! Bo Jacoby (talk) 19:09, 26 July 2008 (UTC).

Please keep all advice as general as possible, I want to do all the hard work for myself. The equations are:

$x+y+az=2$

$x+ay+z=2$

$2x+y+z=2b$

If you make y the subject of the first equation and sub it into the second, then equality can only hold if either a=1 or y=z. Subbing both of those conditions into equations 1 and 2 reduces both of them to $x+2y=2$.

While I appreciate your advice 87.102.86.73, I find it a bit unhelpful because the questions tells me to either solve the simultaneous equations or prove they have no solution. You seem to be suggesting something that is in between these two things. Does giving the equations make it any clearer? 92.3.123.172 (talk) 19:13, 26 July 2008 (UTC)

x+y+az=2=x+ay+z

y+az=ay+z

(a-1)z=(a-1)y

x=y y=z (or a=1 as you say)

so the equations become

x+(1+a)z=2

2x+2z=2b

Which you can solve as you said - BUT you have 5 unknowns and only three equations.87.102.86.73 (talk) 19:26, 26 July 2008 (UTC)

You must have made a mistake because from the first two I get x=y. Did that make sense? Can you take it from there.87.102.86.73 (talk) 19:23, 26 July 2008 (UTC)

No, you made the mistake; you eliminated x in your first step, leaving everything in terms of y and z, and then somehow added in back in in the last line. Try subbing in x=y and then try y=z. I also forgot to mention that a and b are both constants. I think I should probably just go with putting one variable in terms of another. 92.3.123.172 (talk) 19:28, 26 July 2008 (UTC)

Yes, (corrected) - you still have too many unknowns.87.102.86.73 (talk) 19:31, 26 July 2008 (UTC)

I wrote in my last post that a and b are both constants, not variables. Unless, by 'too many unknowns', you mean 3 variables for two equations, which I mentioned when I started this thread. 92.3.123.172 (talk) 19:33, 26 July 2008 (UTC)

After solving $x+2y=2$ and $2x+y+z=2b$ simultaneously, the only way they can both have a solution is if they are the same equation. So I take it that I just have to prove the three original equations have no solution, which I think I've basically done by showing that the only solutions make them into the same equation. 92.3.123.172 (talk) 19:40, 26 July 2008 (UTC)

(that should have been an edit conflict... somehow it just put my comment below the ones that weren't present when I started editing) Here's your mistake: you proved that a=1 OR y=z, which is correct. Then you assumed that a=1 AND y=z, which is incorrect. When y=z, a doesn't have to be 1. When a=1, y doesn't have to equal z. When you have an "OR" like that, you need to divide the problem into 2 subproblems and solve them separately. First substitute a=1 (leaving y and z as independent unknowns) and solve (or prove unsolvable) the resulting three equations. Then go back to the start, substitute y=z (but leave a unknown) and do it again. Your final answer will be something like "If a=1, there are _____ solutions. If a!=1, there are _____ solutions." --tcsetattr (talk / contribs) 19:51, 26 July 2008 (UTC)

Ah, I see. It does make sense not to solve with a=1 and y=z at the same time. It'll still leave me with three unknowns for two equations but perhaps the equations won't have to be identical for there to be a solution. Thanks. 92.3.123.172 (talk) 19:58, 26 July 2008 (UTC)

You correctly deduced that either a=1 or y=z. What you should do now is split the rest of the solution into two cases, like this: "Case 1: a=1. Then [...]. Case 2: a≠1. Then it follows that y=z and [...]." You shouldn't substitute both a=1 and y=z at the same time, since that will get you only a subset of the solutions. Note that in case 1 you won't be able to solve the equations completely—you'll be able to find y+z but not y or z individually. -- BenRG (talk) 20:03, 26 July 2008 (UTC)

You have as one option x+(1+a)z=2 and 2x+2z=2b (or x+z=b)

Therefor 2-(1+a)z=x=b-z

2-az=b

2-b=az

(2-b)/a=z ?! This is a solution?

After solving $x+2y=2$ and $2x+y+z=2b$ simultaneously I get:

x+2y=2 and 2x+y+z=2b

2-2y=b-y/2-z/2 or 1-x/2=2b-z-2x

4-4y=2b-y-z or 2-x=4b-2z-4x

z=2b+3y-4 or 2z=4b-3x-2

87.102.86.73 (talk) 19:50, 26 July 2008 (UTC)

I think the confusion is because you actually have a family of sets of simultaneous equations. Your final answer is going to something along the lines of "For these values of a and b, there are infinitely many solutions, of the form (whatever), for these other values of a and b, there is a unique solution, (whatever) and for these values of a and b, there are no solutions." It looks like you've made a good start, just keep going like you are but keep in mind that the number of solutions will depend on a and b. --Tango (talk) 23:10, 26 July 2008 (UTC)

Solving n equations in n unknowns is conveniently split into two steps. The first step is to combine the equations, trying to end up with n equations each in one unknown. The second step is to solve these independent equations one by one. When the equations are linear, as in your case, the second step is reduced to making a division. The equations are x+y+az−2 = x+ay+z−2 = 2x+y+z−2b = 0. Eliminate the variable x by subtracting the first expression from the second. As both expressions are 0 the difference is also 0. (x+ay+z−2)−(x+y+az−2) = 0 or (a−1)y+(1−a)z = 0. Consider separately case one: a=1, and case two: a≠1. Case one gives x+y+z−2 = x+y+z−2 = 2x+y+z−2b = 0. The two first expressions are identical, so effectively there are only two equations: x+y+z−2 = 2x+y+z−2b = 0. You eliminate the variable y by subtracting the first expression from the second. You get (2x+y+z−2b)−(x+y+z−2) = 0. You eliminate x by subtracting twice the first expression from the second: (2x+y+z−2b)−2(x+y+z−2) = 0. Having done the hard work you end up with one equation in x and another equation in y and z. In the second equation you may consider z to be an independent variable, and y to be defined as a function of z. Then proceed to case two. Bo Jacoby (talk) 05:04, 27 July 2008 (UTC).

I know it's been a while, but I would like to suggest some answers. Again, if I'm wrong, please don't tell me the answers.

There are only two ways that the first two equations can have a solution; if either a=1 or y=z. In the case a=1, the solution of the whole system is x=2(b-1) and y+z=2(2-b). In the case y=z the solution of the whole system is $x={\frac {ab+b-2}{a}}$ and $y=z={\frac {2-b}{a}}$. In both cases there are infinitely many solutions but in the case y=z a=0 is not a possible solution. How's that? 92.0.60.66 (talk) 17:30, 2 August 2008 (UTC)

Well done. The solution (x,y,z) = ((ab+b−2)/a, (2−b)/a, (2−b)/a) counts as one solution because a and b are constants. What happens if a=0? The answer depends on whether b=1 or not. Bo Jacoby (talk) 21:35, 2 August 2008 (UTC).

Thanks for checking it for me, really appreciate it. I don't see though why b equalling 1 will have any bearing on the solution if a=0; isn't division by zero undefined? 92.0.60.66 (talk) 21:39, 2 August 2008 (UTC)

Yes, division by zero is undefined, so you must insert a=0 in the original equations to solve that case. Bo Jacoby (talk) 08:30, 3 August 2008 (UTC).

Probablity equation?

Is there any probablity equation using which we can assess the probablity of frequent natural events like rainfall on a particular day? If so, what is it?117.201.96.242 (talk) 19:14, 26 July 2008 (UTC)

The short answer is not really.

But if it rains x days a year you can say the probability of rain is x/365 everyday.

If you have many years of data you can assess each day separately for the probability of rain, and get a better guess. But you will need data from previous years etc..87.102.86.73 (talk) 19:29, 26 July 2008 (UTC)

And, of course, the probability of rain in the next few days can better be estimated based on current weather patterns (is a warm front moving in ?) than historical data. StuRat (talk) 05:16, 27 July 2008 (UTC)

Once again you have done an immense favour to be. I'm absolutely surprised by being familier to your depth of knowledge. Thank you for answering. The common people should learn from you.117.201.96.242 (talk) 20:04, 26 July 2008 (UTC)

We ARE the common people. But some of us are more common than the others!

Trigonemtry problem

I've been itching away at this for the post couple hours, but still can't get this figured out. Please help me, and if you can, please tell me how you got to the answer.

tan alpha + cot alpha = 4/2

alpha 's period is (4/7 , pi/1)

cos^2(x) - 2sin(x)cos(x) - sin^2(x)

so

given f(x) = cos^2(x) - 2sin(x)cos(x) - sin^2(x)

Find the min positive period of f(x)

Find the max and min values of f(x) —Preceding unsigned comment added by Benadrill (talk • contribs) 21:35, 26 July 2008 (UTC)

You seem to be sometimes saying alpha is a variable, at other times it's a constant. I'll assume it's a constant, in which case the first equation becomes a simple quadratic (two solutions), and the answers follow easily. The last one, try applying some trig identities. -mattbuck (Talk) 21:58, 26 July 2008 (UTC)

The function you've called ƒ(x) is just cos(2x) − sin(2x), by a couple of standard trigonometric identities. That has period π. Michael Hardy (talk) 18:09, 28 July 2008 (UTC)

July 27

Infinite Series

I have two questions about infinite series. The first one is a concept question. In the all the books I have read, it always says that if $a_{1},a_{2},a_{3},...$ are positive and decreasing ($a_{1}\geq a_{2}\geq a_{3}\geq ...\geq 0$), then the following alternating series $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+...$ converges if an only if the summands $a_{n}$ approach zero. My questions is that if all of these hypothesis are met, then $-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_{6}-...$ would also converge if and only if the summands approach zero as n goes to infinity? This is true right? Because all I am doing is that I am taking the first series, which is converging, and then multiplying it by -1 which will be well defined. And by a similar argument, $a_{2}-a_{3}+a_{4}-a_{5}+a_{6}-...$ will also converge if and only if the summands approach zero as n gets large, right? Because all I have done is multiplied by -1 and then added the first term?

The second question is, how do I find out in which region (in the upper half plane y>0) does the series $\sum _{n=1}^{\infty }(-1)^{n}{\frac {(\ln n)^{x}}{n^{y}}}$ diverges, converges conditionally, or converges absolutely?. Thanks!69.224.231.242 (talk) 07:33, 27 July 2008 (UTC)

First question - no. It's complicated - firstly the series a1,a2,a3 might not converge even if the terms tend to zero eg 1/1,1/2,1/3,1/4,1/5 etc

WRONG. You didn't read the question carefully. The answer is yes. Michael Hardy (talk) 18:05, 28 July 2008 (UTC)

Try reading the fucking question lord hardy - it doesn't make any sense because : quote "it always says that if $a_{1},a_{2},a_{3},...$ are positive and decreasing ($a_{1}\geq a_{2}\geq a_{3}\geq ...\geq 0$), then the following alternating ..."

I've underlined the two bits you need to check on. Hope that helps87.102.86.73 (talk) 20:41, 28 July 2008 (UTC)

Try reading that carefully and making any fucking sense out of it. B doesn't follow A.87.102.86.73 (talk) 20:40, 28 July 2008 (UTC)

Also try 1/2+1/3+1/4+1/5 etc does that converge - the numbers are decreasing??87.102.86.73 (talk) 20:49, 28 July 2008 (UTC)

Also check the bit about "if and only if the summands $a_{n}$ approach zero" - if a(n)=a(n) in the two series mentioned (rather than a1=a1-a2 a2=a3-a4 etc ) then the statement is true if "if and only if the summands $a_{n}$ approach zero". ie the part about the summands is irrelevent?87.102.86.73 (talk) 21:25, 28 July 2008 (UTC)

Second - the series a1,-a2,a3,-a4 etc - no . In fact there are series which oscillate and diverge which can be shown to have a real value (of the sum) at a limit of infinity.

Sorry you said 'summands' - if the 'summands' are a1-a2, a3-a4 etc then yes..

So the third question - no. obviously you are right that if the series A converges then the series -A also converges.

The problem is with your use of the term if and only if - if you remove that then your reasoning is correct. Ignore that - I missed the bit about summands.87.102.86.73 (talk) 12:06, 27 July 2008 (UTC)

You still need to bear in mind that a1>a2>a3>a4 etc does not guarantee convergence of the series.87.102.86.73 (talk) 12:08, 27 July 2008 (UTC)

Please see Convergent_series#Examples_of_convergent_and_divergent_series87.102.86.73 (talk) 12:11, 27 July 2008 (UTC)

I think you may find useful the two articles Alternating series and Alternating series test.87.102.86.73 (talk) 11:45, 27 July 2008 (UTC)

Have you tried combining odd and even terms into a new series and then testing for convergence using the 'ratio test' see Convergent_series#Convergence_tests.87.102.86.73 (talk) 11:31, 27 July 2008 (UTC) You would need to find the boundaries at which the series stops converging obviously.87.102.86.73 (talk) 11:44, 27 July 2008 (UTC)

Everything you say in the first paragraph is correct. For your second question, you'll probably want to use the alternating series test (that's what your first paragraph is about, if you haven't heard the name) for convergence. I haven't thought about absolute convergence yet. Algebraist 11:52, 27 July 2008 (UTC)

I agree. For absolute convergence, the hardest part is to work out the case of y=1 and figure out for which x this converges. After you do this, you get the answer for other y using the comparison test. For the tricky case y=1 (I would guess that there is a WP page on this, but I was not able to find it) you can use the integral test, if you know integration. If not, then it can also be done directly. Let us know and we can give you a few hints. Oded (talk) 16:42, 27 July 2008 (UTC)

The answer to the first question is yes. If

$a_{1}-a_{2}+a_{3}-a_{4}+\cdots \,$

converges, the so does

$-a_{1}+a_{2}-a_{3}+a_{4}-\cdots .\,$

Often the blind lead the blind on this page, it seems. Michael Hardy (talk) 18:03, 28 July 2008 (UTC)

You're not wrong there - because that is the answer to the SECOND question. As you say 'you didn't read the question carefully' well done.87.102.86.73 (talk) 21:06, 28 July 2008 (UTC)

Was I not clear enough? I already said 'Everything you say in the first paragraph is correct'. Algebraist 18:31, 28 July 2008 (UTC)

You were right, 87.102 was wrong. --Tango (talk) 18:35, 28 July 2008 (UTC)

Oh I was wrong - why didn't someone tell me.87.102.86.73 (talk) 20:36, 28 July 2008 (UTC)

I didn't because I could barely work out what you were talking about at all. Algebraist 21:25, 28 July 2008 (UTC)

Bit of politeness wouldn't kill everybody — whether you're pointing out an error or having one pointed out to you, try to keep it down to a dull roar, please, thanks guys! --_{tiny plastic} Grey Knight ⊖ 21:28, 28 July 2008 (UTC)

probability

probabiliti based short cut to solve question speedly —Preceding unsigned comment added by Anujay12 (talk • contribs) 18:42, 27 July 2008 (UTC)

What question? --Tango (talk) 18:55, 27 July 2008 (UTC)

This may be the biggest leap in question interpretation since "suitly emphazi", but I'm going to suggest Monte Carlo method as a way to use random input to help solve a problem. Confusing Manifestation(Say hi!) 23:19, 27 July 2008 (UTC)

Ah, it's been such a long time since I've "suitly emphazied" anything. :-) StuRat (talk) 17:31, 28 July 2008 (UTC)

Exponential Proportions

I was wondering, if i wanted to write 7^4 in cube form, what would it look like. I mean, how 64^1/2=2^3, what value of x makes 7^4=x^3. And how do you do that, so in the future i can do it myself. Simple terms please, just finished Algebra 1 last year.--Xtothe3rd (talk) 22:32, 27 July 2008 (UTC)

The value of x which makes $7^{4}=x^{3}$ is the cube root of $7^{4}$, which can be written as ${\sqrt[{3}]{7^{4}}}$ or $(7^{4})^{1/3}$. Using the rule $(a^{b})^{c}=a^{bc}$ it can be simplified to $7^{4/3}$. --tcsetattr (talk / contribs) 23:11, 27 July 2008 (UTC)

So then, (7^4)^(1 / 3) = 7^(4 / 3) = 13.3905183... more or less -hydnjo talk 01:45, 28 July 2008 (UTC)

You got be very careful.

If you write 7^4 = x^3

Then

x^3 = 2401

x = 13.39 or x = -6.69 - 11.6 i or x = -6.69 +11.6 i

or if you want exact answers

x = 7 (7)^(1/3) or x = -7 * (-7)^(1/3) or 7 * (-1)^(2/3) * (7)^(1/3)

122.107.182.1 (talk) 11:14, 28 July 2008 (UTC)

July 28

Point wise union

If I have two sets of sets; {{1,2},{3}} and {{4},{5,6}} then the cartesian product is {({1,2},{4}),({1,2},{5,6}),({3},{4}),({3},{5,6})}. However I am after the set of sets that I get when I instead of producing ordered pairs take the union. So that I arrive at $\{\{1,2\}\cup \{4\},\{1,2\}\cup \{5,6\},\{3\}\cup \{4\},\{3\}\cup \{5,6\}\}=\{\{1,2,4\},\{1,2,5,6\},\{3,4\},\{3,5,6\}\}$. Does this operation have a name? Taemyr (talk) 14:29, 28 July 2008 (UTC)

This may not be what you're looking for, but you can write the operation compactly as a set comprehension: $\{x\cup y:x\in A,y\in B\}$. -- BenRG (talk) 17:49, 28 July 2008 (UTC)

That set is the range (or image) of the union of those sets. Is that any help? --_{tiny plastic} Grey Knight ⊖ 20:36, 28 July 2008 (UTC)

This operation is useful in the study of set-theoretic ideals. It can be notated by a union sign with an underbar. I just call it pointwise union. --Trovatore (talk) 20:47, 28 July 2008 (UTC)

Then I'l go with that. Thx all. Taemyr (talk) 00:00, 29 July 2008 (UTC)

Notation for multivariate polynomials of at most degree m

I’m looking for a notation for the set of polynomials over a field $F$ in variables $x_{1},x_{2},...,x_{n}$ of total degree at most $m$. Anyone know any? GromXXVII (talk) 19:41, 28 July 2008 (UTC)

It tends to be used for PDEs and mixed partial derivatives, but the Multi-index notation will serve. RayAYang (talk) 20:35, 28 July 2008 (UTC)

$F_{m}[x_{1},x_{2},...,x_{n}]$ or $F^{m}[x_{1},x_{2},...,x_{n}]$ should work, as long as you define it the first you use it. The only issue I can see is that it isn't clear if the notation means degree at most m, or degree exactly m. And, of course, the second choice could be mistaken for meaning polynomials over a vector space, but that would be rather weird. --Tango (talk) 21:38, 28 July 2008 (UTC)

July 30

Probability of extraterrestrial intelligent life

So I've been idly pondering the likelihood of intelligent extraterrestrial life and I'm wondering about a probability question. I don't have much experience in probability (although plenty in other mathematical areas). Suppose an experiment with desirable outcome (say, the formation of a star with a planet on which an intelligent species evolves), which has an unknown, nonzero but presumably low probability, and furthermore, many, many repeated and (assumedly) independent trials (7.22 x 1022, say). Given that it has occurred at least once, is there any way we can quantify the probability that it occurred only once? Is it possible to quantify the probability of a very unlikely event occuring precisely once in a very large number of trials? Cheers! Maelin (Talk | Contribs) 06:09, 29 July 2008 (UTC)

If an event has probability p, then the probability of it occuring exactly once in N independent trials is $Np(1-p)^{N-1}$. For example, if you toss a coin the probability of seeing exactly one heads in 4 tries is 4*(1/2)*(1 - 1/2)^3 = 1/4. For low probability events, such as seeing intelligent life appear, $Np(1-p)^{N-1}\approx Np$ which means that if it only occured once we would expect $p\approx {1 \over N}$. Dragons flight (talk) 06:42, 29 July 2008 (UTC)

Hi. See Drake equation for the problem at hand, and Poisson distribution for the statistics of low-p high-N systems (and maybe Stein's method too, if you feel like some rigour). HTH, Robinh (talk) 07:29, 29 July 2008 (UTC)

But note that if the occurrence of life in each different star system are truly independent events, then the occurrence of life here on Earth tells us absolutely nothing about the probability of life occuring elsewhere in the Universe - otherwise the events would not be independent. Gandalf61 (talk) 09:49, 29 July 2008 (UTC)

Well, it does tell us that the probability is nonzero (assuming the number of places where intelligent life could evolve is finite). Obligatory link to anthropic principle. « Aaron Rotenberg « Talk « 10:50, 29 July 2008 (UTC)

Is possible to calculate x and y such that there is a 95% chance than the probability of life arising on a planet is greater (less) than x (y) given we know it's happened once? That might be more useful than just an expected value. --Tango (talk) 18:34, 29 July 2008 (UTC)

So is there a way of calculating the likelihood that an outcome with unknown probability occurs precisely once in 7 x 10²² trials? Maelin (Talk | Contribs) 01:09, 30 July 2008 (UTC)

Well, no, since it will be different for different probabilities. --Tango (talk) 03:33, 30 July 2008 (UTC)

If you suppose one can quantify a) L the number of planets in the universe, and b) P the probability of intelligent life (whatever that is) arising on a planet, then you certainly have the gall to calculate (1 - P)(L - 1) = the probability of life not arising on the (L - 1) planets that are not this one.Cuddlyable3 (talk) 13:42, 31 July 2008 (UTC)

I suppose you mean (1 - P)^{(L - 1)}. For that you also need the gall to assume life arises independently on different planets. Algebraist 13:51, 31 July 2008 (UTC)

I finally realised that the intuitive feeling I had that some sort of figure was calculable was based on the notion that knowledge of the existence of life on the first planet we examined was evidence to suggest that P couldn't be very low - that is, high values of P suggest other life is probable, and low values of P are themselves unlikely because life arose on Earth. Fortunately, I finally realised that this is anthropic reasoning - since P could just as easily be low as high, and in universes with low P, if intelligent life did arise as a fluke, such reasoning would lead inhabitants to the erroneous conclusion that P on their universe is unlikely to be low, when in fact it's merely they who are unlikely.

This, I guess, spells the death of the "it's such a big universe, it seems unlikely that of all those stars, only ours has life on it" argument for the probable existence of aliens. Thanks to all who helped! Maelin (Talk | Contribs) 12:15, 2 August 2008 (UTC)

Cardinality

Hi. Consider the natural numbers whose factorisation contains, at most, one of each prime. That is, their factorisation is a proper set, not just a bag. The first few of these numbers are 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29. I'll call these the unique factor naturals, or uf-nats. Obviously, being a subset of the natural numbers, the set of all uf-nats has cardinality aleph-zero. Now, the set of all primes (call it P) also has cardinality aleph-zero. All uf-nats are a subset of P, and any subset of P must be a uf-nat. So there is a one-to-one correspondence between uf-nats and subsets of P.

The set of all subsets of P (P's power set) is therefore equal to the set of all uf-nats. P's power set has cardinality 2^P = 2^aleph-zero = aleph-one. How is it possible for a set with cardinality aleph-one (P's power set) and aleph-zero (the uf-nats) to be in a one-to-one correspondence? 79.72.192.135 (talk) 12:02, 29 July 2008 (UTC)

In your mapping, an uf-nat is mapped to a finite subset of P. There are no uf-nats corresponding to infinite subsets of P - for example, there is no uf-nat that corresponds to the set of all primes greater than 2. So the cardinality of the set of all uf-nats does not have to be the same as the cardinality of the power set of P - indeed, as you have shown, it is smaller. Gandalf61 (talk) 12:14, 29 July 2008 (UTC)

Yes, that would do it, thanks. So there are only countably many finite subsets of a countable set, but uncountably many infinite subsets. It also occurred to me to type that sequence into OEIS; it turns out what I called uf-nats are also called square-free numbers. 79.72.192.135 (talk) 12:22, 29 July 2008 (UTC)

Darnit, I was just coming here to post the Sloane index and you've beaten me to it! Oh well, it is A005117 if anyone else happened to want to know. --_{tiny plastic} Grey Knight ⊖ 12:42, 29 July 2008 (UTC)

Yes, they are called square free and we even have an article on them Square-free integer. --Tango (talk) 18:36, 29 July 2008 (UTC)

which of non-linearly related variables is distributed normally?

In high school, we were taught that "most naturally occurring random variables are distributed normally," by which they meant stuff like height, weight etc. within a given population. Later I found out the connection between this and the Central Limit Theorem. Then something occurred to me: if X is normally distributed, then non-linear transformations of X will not be. The problem is that most naturally occurring random variables can be transformed non-linearly into other fairly natural variables (eg. tree trunk radius into cross-section area), so only a fraction of those can be normal. Another good example is height and weight: these are related non-linearly, although not as closely as radius and area, so they can't both have normal distribution curves within a population, for a given gender (as the Normal distribution article confirms). Is there any general way that you can determine which of two related variables will be normally distributed, and is it possible that both will have a roughly normal, but skewed distribution? 202.89.166.179 (talk) 15:56, 29 July 2008 (UTC)

Your question has essentially all the ingredients of the answer. You should expect near normal distribution of a variable $X$ if it is a result of many independent or nearly-independent contributions (where no particular few are dominant). Here, "contribution" should be interpreted in the additive sense. So if $Y$ is a non-linear function of $X$, you would still have $Y$ as a function of many independent events, but in this case the events would not combine by addition. For example, a somewhat reasonable model for the distribution of the price of a stock at a fixed future time is the exponential of a normal random variable. This is based on the explanation that individual events have multiplicative effects on the stock price, which is the same as an additive effect on the log of the stock price. Oded (talk) 16:18, 29 July 2008 (UTC)

Thanks for that answer. But is it possible that two variables, related but not linearly, can both have nearly normal distributions, ie. both skewed? 202.89.166.179 (talk) 17:21, 29 July 2008 (UTC)

That happens a lot, in a way. For example, suppose that $Y=f(X)$ and $X$ is a normal random variable. If $f$ is nearly linear near the mean of $X$, then $Y$ will be nearly normal. This is not quite precise, since I have not said what nearly and near mean, but it is possible to come up with precise versions of this statement. Many useful and naturally occuring functions are smooth, and the smooth functions are close to linear on small scales. Oded (talk) 21:23, 29 July 2008 (UTC)

Thanks again, :) 202.89.166.179 (talk) 05:05, 2 August 2008 (UTC)

Fibonacci Series

Does it start with a 1 or a 0? --117.196.136.3 (talk) 16:29, 29 July 2008 (UTC)

Conventionally it starts with a 0, as in Sloane's OEIS: A000045. --_{tiny plastic} Grey Knight ⊖ 16:35, 29 July 2008 (UTC)

But note that the initial 0 is the 0th term of the sequence. I think everyone agrees that the 1st and 2nd terms are both 1, the only dispute being that some prefer to leave off the 0th term. Algebraist 17:12, 29 July 2008 (UTC)

Okay, good clarification. :-) For an encore, discuss the difference between a geometer's and topologist's n-sphere in 100 words or less. :-) --_{tiny plastic} Grey Knight ⊖ 17:22, 29 July 2008 (UTC)

A topologist's n-sphere is a naked topological space. A geometer's one is equipped with a smooth structure and probably a Riemannian metric. Algebraist 17:29, 29 July 2008 (UTC)

Related to what Algebraist says, I think the important thing is that the element of index 0 is 0, and the elements of index 1 and 2 are both 1. This results in the nice property that $F_{\gcd(m,n)}=\gcd(F_{m},F_{n})$. This normally means that the sequence starts with a 0 (but can actually extended infinitely both forwards and backwards), unless you're one of the lots of people who believe natural numbers start with 1, in which case the sequence starts with two 1s. – b_jonas 16:29, 31 July 2008 (UTC)

Tall tree heights

How do you determine the height fo a tall tree without climbing up with a tape measure —Preceding unsigned comment added by 79.64.140.117 (talk) 20:36, 29 July 2008 (UTC)

Cut it down! No actually, this might not be the best idea. Let's see... Maybe you can use some geometry and the fact that light travels in straight lines? Oded (talk) 21:27, 29 July 2008 (UTC)

Or a rangefinder and the Pythagorean theorem. But without fancy tools I’d probably go with measuring the shadow, finding the angle of the sun, and using similar triangles. GromXXVII (talk) 21:39, 29 July 2008 (UTC)

Gah, ya beat me to it, Grom. Yes, measure the shadow, and then take the measurement of the shadow of a meterstick, and see how many of the meterstick-shadows fit in the tree-shadow. Or something like that. That doesn't count for the roots, though. But I don't think that matters. Excuse my rambling. IceUnshattered (talk) 22:53, 29 July 2008 (UTC)

Usually in a forest you cannot discern the shadow of a tree. Oded (talk) 23:26, 29 July 2008 (UTC)

Measure a known distance from the trunk with a tape measure. Stand at that distance and get some kind of rod and line it up from your eye to the top of the tree. Measure the angle from the rod to the horizontal (a string with a weight on the end hanging from the rod to give you a vertical can help) and then use basic trig to calculate the height - remember to add the distance from the ground to your eyes to the final result. --Tango (talk) 01:13, 30 July 2008 (UTC)

Climb up with a stone and a stop watch. Drop the stone from the top of the tree and measure the time it takes to hit the ground. Use Newton's second law to compute height. Density of branches has pros and cons. Tlepp (talk) 10:55, 30 July 2008 (UTC)

Wait until a bird lands on top of the tree. Measure the angular size of the bird. Shoot the bird. Measure the bird. Do the calculation. McKay (talk) 11:05, 30 July 2008 (UTC)

(This method requires a set of bathroom scales). Weigh yourself at the bottom of the tree (call this $W_{0}$), then climb it and weigh yourself at the top (calling this $W_{1}$) (this method also requires good balance). If you know that the radius of the Earth at the tree's location is $R_{0}$, then the height of the tree is just $R_{0}\left({\sqrt {\tfrac {W_{0}}{W_{1}}}}-1\right)$. --_{tiny plastic} Grey Knight ⊖ 13:12, 30 July 2008 (UTC) more seriously, User:Tango's answer seems best

First, acquire a barometer. Then use one of the methods listed in the barometer question article. Gandalf61 (talk) 13:16, 30 July 2008 (UTC)

I find the tree's custodian and offer to exchange the barometer for telling me the height of the tree. --Random832 (contribs) 13:29, 30 July 2008 (UTC)

I don't know if a dryad would be interested in having a barometer, but I suppose it can't hurt to try. --_{tiny plastic} Grey Knight ⊖ 13:51, 30 July 2008 (UTC)

Climb up the tree with a piece of string, cut the string to be as long as the tree is high. Climb back down the tree, and measure the string. Philc 0780 13:49, 30 July 2008 (UTC)

On a cloudy day, or if the shadow isn't clear for some other reason, use a slingshot to shoot a stone so that it just makes it over the tree at the top. Time its fall to the ground and you can calculate the height: the horizontal travel won't change the time for the stone to fall to the ground. --Slashme (talk) 13:52, 30 July 2008 (UTC)

If you can control the initial velocity and angle of the stone, adjust those so the stone just hits the top of the tree, and you can calculate the trajectory, and thus the height of the tree. --Carnildo (talk) 20:37, 30 July 2008 (UTC)

Come on, people, that's enough. I've given the standard method, the question is answered, let's stop with all the jokes, huh? --Tango (talk) 17:50, 30 July 2008 (UTC)

What jokes? ō_ō hydnjo talk 23:46, 30 July 2008 (UTC)

Assume a spherical tree... « Aaron Rotenberg « Talk « 04:45, 31 July 2008 (UTC)

The real standard method, if the base of the tree is not accessible but everything round about is level, is to take two measurements of the angle to the top of the tree from points a known distance apart in line with the base. If the distance is x and the angles are alpha and beta, then the height is x/(cot beta - cot alpha).…217.43.211.113 (talk) 09:55, 31 July 2008 (UTC)

To Grey Knight: I recommend measuring a weight with a kitchen scale instead of a bathroom scale becasue bathroom scales are often inaccurate if not placed on a flat horizontal surface (they can give wrong results even on a fluffy carpet), and the treetop doesn't provide for such a surface. – b_jonas 16:23, 31 July 2008 (UTC)

I hope I'm not messing up a test, but this question is often used to see if people will ask that very important 'why' question when given a task. Dmcq (talk) 20:09, 2 August 2008 (UTC)

The standard tool used for measuring angles in triangulation is the theodolite, but perhaps a sextant can be used as well. --Bowlhover (talk) 12:01, 3 August 2008 (UTC)

The article on horizon shows how to calculate the height of a tree by walking away from it till the tip just disappears. Just measure how far you walk and how high you are. It has a helpful picture as well as the formula. Dmcq (talk) 19:14, 3 August 2008 (UTC)

Finding all roots to a 4th degree polynomial

The first of many such questions goes like this:

Given: f(x)=x^4-3x^3+2x^2-7x-11 Find all roots to the nearest .001

I tried to separate the groups with parenthesis, so, (x^4-3x^3)+(2x^2-7x)-11. But I didn't think this works when there's an x^0 in the equation. So I moved the 11 to the other side, (assuming f(x) was equal to zero) and tried to factor out from what was left. But I couldn't factor. The next thing I thought was matrices, so I plugged them into a 5X1 Matrix and hit "rref(". But the response was merely the matrix I sent in. I've heard there's a long way, but I can't remember it. Could somebody help? --Ye Olde Luke (talk) 03:04, 30 July 2008 (UTC)

There are several ways of approaching this, but one way might be to iteratively use Newton's method. By hand, that would be considered a "long way", though. Baccyak4H (Yak!) 03:27, 30 July 2008 (UTC)

Since the question says "to the nearest .001" I expect you are intended to do it numerically. If you're allowed to use a computer (as it seems you are), I suggest plotting a graph (you could plot it by hand if you want, but a computer is far easier!) to find out roughly where the roots are and then try an iterative method to find the roots (rearrange f(x)=0 to x=g(x) and then plug a value of x near the root into the new function, g, and keep doing so until you have a precise enough answer). That's not guaranteed to work (the values of x may get further and further away from the root, rather than closer), but it often does. If it doesn't, you'll need some other numerical method - if you're being asked that kind of question, I imagine you've been taught some. --Tango (talk) 03:31, 30 July 2008 (UTC)

The general fourth degree polynomial equation is known to be solvable by radicals (i.e., extracting roots, multiplication, division, addition and subtraction) since 1545. See the Timeline of algebra and quartic equation. The solution on that page looks rather complicated. I plugged your equation into Mathematica, and Mathematica spits the four solutions, two real and two complex. For example, one of the real solutions is 3.32901. For the higher degree equations, it has been proved by Galois that generally no solution by radicals is possible. In such cases, one reverts to numerical methods. To use Newton's method effectively, you need to first find a point that is quite close to a solution. This is a nontrivial problem. See, for example the Acta 1989 paper by Doyle and McMullen. Oded (talk) 05:25, 30 July 2008 (UTC)

In fact, PlanetMath has the quartic formula, http://planetmath.org/encyclopedia/QuarticFormula.html, although for finding the solutions it’s pretty much useless because a method of derivation using the given values would be both involve less computations and be less error prone than plugging everything in. GromXXVII (talk) 10:55, 30 July 2008 (UTC)

I graphed the equation using my graphing calculator, and 3.3201 apears to be one of the zeroes. The other is somewhere close to -.9121094. I'm sorry, I've been on summer vacation for months, and can barely dreg up what I've already stated. I there any way to clarify the Wikipedia article examples? I vaguely recall learning that the funny thing about Complex numbers is that as solutions they always come in pairs, one negative and one positive. So if I can find one, I can find the other, right? (By the way, you're all getting Reference Desk barnstars for your help)--Ye Olde Luke (talk) 05:45, 30 July 2008 (UTC)

Regarding the "pairs" of complex solutions, you are thinking of the complex conjugate root theorem, on which we have an article. Which articles do you think have confusing examples? Perhaps those pages can be improved (you could even help if you like!). --_{tiny plastic} Grey Knight ⊖ 09:17, 30 July 2008 (UTC)

Here is something that works for this particular equation and does not require any advance math and you can do by hand (if you care to). First, if you evaluate the polynomial at the three points $-1,0,4$, you get positive, negative and positive answers, in this order. Therefore, you can use the bisection method to find approximations of two roots, one between 0 and 4, and the other between -1 and 0. (I guess you already did this.) You will need to get these roots with a higher level of accuracy than the target of 0.001, since you will use them to find the others. (I think the answers you wrote are not accurate enough.) So say the root approximations that you found where a and b. Then you divide your original polynomial P by the polynomial $(x-a)(x-b)$, using polynomial division. If a and b where the exact roots, the division would leave no remainder, and you would get a second degree polynomial. However, since they are not quite the exact root, the division would have a remainder, and you would have

$P=Q\,(x-a)(x-b)+R,$

where Q is a second degree polynomial that you find, and R is the remainder, which is a linear polynomial. Now, since a and b are approximate roots, the coefficients of R will be rather small in absolute value. Therefore, if you plug into P the roots of Q, you get something close to zero. In other words, the roots of Q are approximations of the roots of P. You find the roots of Q using the quadratic formula, which gives you approximations of all four roots of P. Oded (talk) 15:30, 30 July 2008 (UTC)

After the polynomial division, I got "x^2-3.9x+3.51+(-26.36x+14.53/x^2+.9-3)" Since you said to find the roots of Q, which I took to be x^2-3.9x+3.51 , I used the quadratic formula, which looked like this:

$x={\frac {3.9\pm {\sqrt {15.21-14.04}}}{2}},$

But, when I reduced that, it didn't give me complex answers, it gave me 2.49 and 1.41 . The only thing I didn't do was make my first two answers more exact. Would that really cause the whole problem to turn out incorrectly? --Ye Olde Luke (talk) 18:54, 30 July 2008 (UTC)

You've done something wrong with the polynomial division. You should end up with the remainder being of the form ax+b, there shouldn't be a 1/x² term. --Tango (talk) 19:01, 30 July 2008 (UTC)

I parrot-copied the way the Wikpedia article did it, and their answer looked like that. Did I pick the wrong section of the article to copy? --Ye Olde Luke (talk) 19:05, 30 July 2008 (UTC)

For one thing, did you multiply out $(x-a)(x-b)$ before dividing? If your approximation is good enough, then the remainder R that you get would have small coefficients. There is something else that might be confusing to you, that I wrote the equation as $P=Q\,(x-a)(x-b)+R$, while on that polynomial division article they would write it instead in the form ${\frac {P}{(x-a)(x-b)}}=Q+{\frac {R}{(x-a)(x-b)}}$. If you write out your steps in greater detail, we'd be able to make sure they are ok. By the way, when you wrote above "x^2-3.9x+3.51+(-26.36x+14.53/x^2+.9-3)", did you mean $x^{2}-3.9x+3.51+{\frac {-26.36x+14.53}{x^{2}+.9x-3}}$?. In any case, the remainder does not look small enough. Oded (talk) 21:01, 30 July 2008 (UTC)

I think I found your mistake. It occured when you multiplied out $(x-a)(x-b)$. Presumably, you got $x^{2}+0.9x-3$, whereas the correct answer is approximately $x^{2}-2.4x-3$. Is that it? Oded (talk) 21:12, 30 July 2008 (UTC)

You're right. I forgot to add -3.3 to .9 . I'm gonna try the problem again, one minute. --Ye Olde Luke (talk) 00:05, 31 July 2008 (UTC)

All right, I've completely writen out the steps I took to solve the problem, but to save Reference Desk space, I've completed it at User:Ye Olde Luke/proof. Please check to see if I did all the steps right. --Ye Olde Luke (talk) 01:08, 31 July 2008 (UTC)

I haven't checked your arithmetic, but the method looks right. Checking the numbers is easy - just plug them into the polynomial and see if you get zero! One thing - you were asked for values accurate to the nearest 0.001, but haven't gone anywhere near that close. --Tango (talk) 01:17, 31 July 2008 (UTC)

Oops. I fixed that now. --Ye Olde Luke (talk) 01:30, 31 July 2008 (UTC)

Now, you didn't - you need to use the more precise real values in the calculation of the complex ones (in fact, you should use slightly more precise values in order to account for any multiplication of errors). --Tango (talk) 01:36, 31 July 2008 (UTC)

(edit conflict) I think you got the long division right. But it seems you made a few errors here and there. At some point you switched the sign of the root near -0.9, and the square root leading to the imaginary part was messed up. Tango is also right. You need more accuracy. Regardless of the target of 0.001, just to feel safe I would not be happy to drop the remainder if its coefficients are larger than 0.1. Once you work more accurately, you can check that you are getting the right answers by plugging in, as Tango suggested. Do you know how to get the first two real roots accurately? Oded (talk) 01:40, 31 July 2008 (UTC)

On a TI-84 graphing calculator? No. I just zoomed in a whole bunch of times.--Ye Olde Luke (talk) 01:47, 31 July 2008 (UTC)

That should work pretty well. It's basically an electronic version of the bisection method. That article will explain how to do it by hand. --Tango (talk) 01:56, 31 July 2008 (UTC)

Oh, okay. I've increased the acurracy to 3.3290176, and -.9112938. I'm going to use the more exact figures to adjust my work. --Ye Olde Luke (talk) 02:37, 31 July 2008 (UTC)

Darnit. With the more precise numbers, one answer worked out great, leaving a remainder of less than .0002. I made a mistake somewhere with the other. --Ye Olde Luke (talk) 03:10, 31 July 2008 (UTC)

You have some difficulties with consistency. (I would not hire you to do my taxes. :-) In the third from the top box on the right, you did not carry the -11 down and made an error in the column with the -7 in it. Oded (talk) 04:00, 31 July 2008 (UTC)

I see what I did wrong! When you add a negative to a negative, it gets further from zero! That was a dumb mistake. All right, I think I've gotten it finished. Both remainder coefficients were well below .1, and Oded or myself have (I think) managed to locate all arithmatic mistakes. Is it correct? --Ye Olde Luke (talk) 06:05, 31 July 2008 (UTC)

One more small mistake. ${\sqrt {-14.\dots }}={\sqrt {14.\dots }}i=3.7\dots i$. Always plug it in to check. Oded (talk) 06:26, 31 July 2008 (UTC)

Mistake fixed. Thanks. One last thing: this time, to check, I plugged .291 + 1.882i into the orignal equation to see if it worked. The answer returned was: 6.3257249E-5+8.2283318E-5i . I'm not sure what that means. Did that happen just because my answer is a little approximated? --Ye Olde Luke (talk) 06:50, 31 July 2008 (UTC)

That's exponential notation for 0.000063257249 + 0.000082283318i, a complex number that's pretty close to zero. Its magnitude (absolute value) is ${\sqrt {0.000063257249^{2}+0.000082283318^{2}}}\approx 0.00010378836$ so you can be pretty confident you've got the right answers. --tcsetattr (talk / contribs) 07:58, 31 July 2008 (UTC)

Congrat's! You got it. To elaborate on tcsetattr's entry, the E-5 means that you need to multiply the previous number by 10^-5. Oded (talk) 15:50, 31 July 2008 (UTC)

Wow, I actually did it! Thanks Oded, Tango, and everyone else who helped me out! And, as per my promise, everyone who helped me out gets a barnstar! --Ye Olde Luke (talk) 17:11, 31 July 2008 (UTC)

Using the J (programming language), (which is downloaded for free from www.jsoftware.com), you write

  p. _11 _7 2 _3 1

where the sign p. solves the equation -11-7x+2x^2-3x^3+x^4=0. The answer is

┌─┬────────────────────────────────────────────────┐
│1│3.32901 0.29114j1.8818 0.29114j_1.8818 _0.911294│
└─┴────────────────────────────────────────────────┘

meaning that the four solutions are approximately 3.329, 0.291+1.882i, 0.291-1.882i, and -0.911 . If you want to know what you are doing, study the Durand-Kerner method. Bo Jacoby (talk) 11:23, 1 August 2008 (UTC).

Adding to a running average

I hope someone can help with this, it's a problem I've come up against whilst coding a game, but I think the problem belongs here rather than on the Computing board.

I have access to a number which is a pre-calculated average of a series of other numbers. I do not know what the individual values making up this average are. All I do know is how many values there were. I want to add a new number to this mystery series, and get a new average value. Is this even possible? If not -- how best can I get a reasonable aproximation of what the value would be?

195.195.236.129 (talk) 18:40, 30 July 2008 (UTC)

Sure. If your old average is A and you had previously n numbers and you get a new number x, then the new average is ${\frac {n}{n+1}}A+{\frac {1}{n+1}}x$. Check it out using the definition of average. Oded (talk) 18:48, 30 July 2008 (UTC)

It's fairly simple to show why this is true, too. Your running average was found by dividing the running total by the number of values, so you can regain that running total by multiplying the average back by the number of values (so $nA$ in Oded's notation). Adding on your new number, your new total is $nA+x$, and you've got one more value so you find the new average by dividing by $n+1$ to get ${\frac {nA+x}{n+1}}$ (which is the same formula as above).--PaulTaylor (talk) 19:06, 30 July 2008 (UTC)

The running average article tells us a more stable formula you shoule use if you don't do calculations with exact numbers: the new average should be computed as $A+{\frac {x-A}{n+1}}$. (I'm not really good in numerical analysis, but I do remember having read this somewhere, probably in Knuth vol 2. There's also a similar formula for computing the population variance incrementally, but I can't tell what it is.) – b_jonas 16:13, 31 July 2008 (UTC)

July 31

Prime Factorisation

While finding out all of the prime factors of a number by the prime factorisation method, do we take the smallest possible prime number or the largest possible prime number, while dividing each time? 117.194.228.194 (talk) 17:17, 31 July 2008 (UTC)

Doesn't matter, either will work. I usually start with small ones since it's easier to see if they divide the number. --Tango (talk) 17:23, 31 July 2008 (UTC)

Right. The small ones are easier and hopefully when you divide by all the small factors your number is small enough that it is easier to find the larger prime factors, if there are any. Oded (talk) 17:46, 31 July 2008 (UTC)

Minor quibble, but prime factorisation is just another name for finding out all of the prime factors of a number. The method I think you're talking about is trial division. As the article mentions, there are faster ways of factoring large numbers than trying every possible factor in whatever order. -- BenRG (talk) 18:07, 31 July 2008 (UTC)

And just to be explicit, the reason why it doesn't matter which prime factor you choose is the fundamental theorem of arithmetic. —Keenan Pepper 03:32, 1 August 2008 (UTC)

117.194.228.194: consider number=a*b*c*d*e*f. As long as you divide abcdef until one factor remains, you will always get the complete list of numbers. For speed, just divide by whatever factor you think of first. --Bowlhover (talk) 11:43, 3 August 2008 (UTC)

Algebra refresher

In the equation, (-2X^3)(-X^2)(2X), do you add the exponents together or multiply them? This sample test I have doesn't explain how I get to the answer of 4X^6. Thanks, Dismas|^(talk) 23:44, 31 July 2008 (UTC)

You add. Xⁿ is just a shorthand for X multiplied by itself n times, so (X²)(X³) (say) is (XX)(XXX)=XXXXX=X⁵. It's clear that multiplying can't be right, since in that case multiplying by X¹=X wouldn't change the exponent. Algebraist 23:50, 31 July 2008 (UTC)

Well, when you put it that way, I can see where I should have been able to see it... Thanks! Dismas|^(talk) 00:11, 1 August 2008 (UTC)

Hope I'm not giving you a usless answer or making you confused, but there are cases where you multiply the exponents too. In (X^2)^3 would become X^(2*3) or X^6. --Wirbelwindヴィルヴェルヴィント (talk) 16:03, 1 August 2008 (UTC)

You can also regard (X^2)^3 as (X^2)(X^2)(X^2)=(XX)(XX)(XX)=XXXXXX=X^6 --Iamunknown 03:55, 3 August 2008 (UTC)

August 1

Statistics

Hello. I'm a grad student who's been asked to look at some data and come up with some statistics for possible publication, but the data set is small and limited to "present/not present", so I'm not sure how to approach it. To give a little more detail, I have been provided two collections of items, each of which has been attributed with specific features, as follows:

	Collection A	Collection B
Feature 1	0/30	49/50
Feature 2	5/30	45/50
Feature 3	15/30	40/50
Feature 4	0/30	10/50
Feature 5	20/30	0/50

My experience with statistics is very limited, and I'm having trouble working with the various statistics articles, so the questions I have are as follows:

1. How would I go about determining the degree of significance of individual features with this data, considering that it's limited to present/absent data as opposed to a range of values?
2. How could I figure out the likelihood and error that, given the value of three features, an item belongs in a particular set?
3. What kind of other useful statistics can I even generate regarding this data?

Honestly, I'm not even sure where to start, so any help you can provide would be extremely helpful. Many thanks. - But I Played One On TV (talk) 15:18, 1 August 2008 (UTC)

For Q3, one possibility might be odds ratio. For Q1 and Q2, you would need to apply a reasonable statistical model. The top of the odds ratio article might offer a hint. Baccyak4H (Yak!) 16:07, 1 August 2008 (UTC)

Thanks for the odds ratio pointer! I've been reading all day and definitely learning, albeit slowly. Now then, would I be completely out to lunch to try to calculate the p-values for each feature by doing the following?:

• Define the null hypothesis as OR=1.0
• Run the natural log of the odds ratio through the standard logistic function to get a range 0<=f(OR)<=1
• Use the probability mass function where n = sum of collection members (80), k = n * f(OR), and p = 0.5 (which is f(1.0))

Thanks again! - But I Played One On TV (talk) 22:11, 1 August 2008 (UTC)

Further, if you find those articles too dense, call someone in your school's stats (or in lieu of that, math) department and ask them. Since you mention a publication is possible, you will probably get someone's attention. :-) Baccyak4H (Yak!) 16:21, 1 August 2008 (UTC)

Hello. Assuming that the collections, A and B, are samples from a big population, you want to know about the population based on knowing the sample. The probability of each feature can be estimated with some uncertainty from the sample data. A collection contains n items of which i have some feature. If you knew the probability, x, that a random item of the population has the feature, then you could compute the probability p_i(x) that the collection contains i = 0, 1, 2, ... , n items having that feature. This probability is given by the binomial distribution

$p_{i}(x)={n \choose i}\cdot x^{i}\cdot (1-x)^{n-i}.$

This distribution is summarized by it's mean value ± standard deviation:

$i\approx n\cdot x\pm {\sqrt {n\cdot x\cdot (1-x)}}.$

This means that if you know the population frequency, x, of some feature, then you can estimate the sample frequency of the feature, which is

${\frac {i}{n}}\approx x\pm {\sqrt {\frac {x\cdot (1-x)}{n}}}.$

This formula for estimating i knowing x and n is however not what you want. You want a formula for estimating x knowing i and n. The distribution function for x, knowing i and n is still

$p_{i}(x)={n \choose i}\cdot x^{i}\cdot (1-x)^{n-i}$

apart from an unimportant normalization factor. This distribution function is known as the beta distribution. The mean value of the beta distribution is not ${\frac {i}{n}}$, but rather ${\frac {i+1}{n+2}}$, and the standard deviation of the beta distribution is not ${\sqrt {\frac {{\frac {i}{n}}\cdot (1-{\frac {i}{n}})}{n}}}$ but rather ${\sqrt {\frac {{\frac {i+1}{n+2}}\cdot (1-{\frac {i+1}{n+2}})}{n+3}}}.$ So the formula you want is

$x\approx {\frac {i+1}{n+2}}\pm {\sqrt {\frac {{\frac {i+1}{n+2}}\cdot (1-{\frac {i+1}{n+2}})}{n+3}}}.$

Substituting your data into this formula gives the following estimates for the population frequencies:

	Collection A	Collection B
Feature 1	0.03±0.03	0.96±0.03
Feature 2	0.19±0.07	0.88±0.04
Feature 3	0.50±0.09	0.79±0.06
Feature 4	0.03±0.03	0.21±0.06
Feature 5	0.66±0.08	0.02±0.02

Now you want to know if the two collections can be believed to come from the same population. Is a number from a distribution 0.03±0.03 likely to be equal to a number from a distribution 0.96±0.03? Compute the difference

$(0.96\pm 0.03)-(0.03\pm 0.03)=(0.96-0.03)\pm {\sqrt {0.03^{2}+0.03^{2}}}=0.93\pm 0.05$

Is this difference likely to be zero? Zero is 0.93/0.05=18.6 standard deviations away from the mean value of the distribution. This difference is highly significant.

Be warned that different statisticians use different approximations and thus may reach different results. The above approach may not be considered standard by your teacher. Have fun! Bo Jacoby (talk) 23:16, 1 August 2008 (UTC).

Hi again! I just wanted to ask you a quick followup question to my original question. How would I go about determining the probability that an element is part of "Collection B" if it has the three most significant features, in this case Feature 1, Feature 2, and an absence of Feature 5 (with individual probabilities of 0.96±0.03, 0.88±0.04, and 0.66±0.08 respectively)? Many thanks for your help! User: But I Played One On TV (talk) 15:47, 11 September 2008 (UTC) (The question is moved to here from my talk page. Bo Jacoby (talk) 17:56, 11 September 2008 (UTC).)

This question has nothing to do with the populations but only with the samples. So the probabilities above are not relevant. You need to go back to the original data material and count how many elements in each of the collections A and B that has the new Feature 6, which is "Feature 1 and Feature 2 and not Feature 5". Then you get four numbers:

A11 = number of elements in collection A having feature 6 (= 0)

A12 = number of elements in collection A not having feature 6

A21 = number of elements in collection B having feature 6

A22 = number of elements in collection B not having feature 6

and the sums

A10 = A11+A12 = number of elements in collection A (= 30)

A20 = A21+A22 = number of elements in collection B (= 50)

A01 = A11+A21 = number of elements having feature 6

A02 = A12+A22 = number of elements not having feature 6

A00 = A10+A20 = A01+A02 = total number of elements

The probability you ask for is A21/A01. Bo Jacoby (talk) 05:01, 12 September 2008 (UTC).

Fantastic, thank you! If I wanted to include error in this calculation, could I use the values we got form the beta function and propagate those deviations? - But I Played One On TV (talk) 19:57, 15 September 2008 (UTC)

Thanks for the nice words. No, there is no error of calculation. In this case the probability is known. What does it mean? Out of the elements having feature 6 you pick one element at random. The probability that this element is in collection B is exactly A21/A01. Bo Jacoby (talk) 18:48, 16 September 2008 (UTC)

Thank you Bo. You're my statistical hero. - But I Played One On TV (talk) 19:34, 16 September 2008 (UTC)

Perhaps you want to consider a random element from the population? Then rename the variables. You must distinguish between the number of elements in the population, A₀₀₀, and inside the sample, A₁₀₀, and outside the sample, A₂₀₀, and the corresponding known sample counts, A₁₁₁, A₁₁₂, A₁₂₁, A₁₂₂, and the unknown counts outside the sample, A₂₁₁, A₂₁₂, A₂₂₁, A₂₂₂. The probability, that a random element having feature 6 is in collection B, is A₀₂₁/A₀₀₁=(A₀₂₁/A₀₀₀)/(A₀₀₁/A₀₀₀). (A zero in a digit position in the index indicates a summation). Now A₀₂₁/A₀₀₀ is the probability that a random element of the population is in collection B and has feature 6, and A₀₀₁/A₀₀₀ is the probability that a random element of the population has feature 6. These probabilities are not known but can, (assuming A₀₀₀>>1) , be estimated by the beta distribution

${\frac {A_{021}}{A_{000}}}\approx {\frac {A_{121}+1}{A_{100}+2}}\pm {\sqrt {\frac {{\frac {A_{121}+1}{A_{100}+2}}\cdot (1-{\frac {A_{121}+1}{A_{100}+2}})}{A_{100}+3}}}$

and

${\frac {A_{001}}{A_{000}}}\approx {\frac {A_{101}+1}{A_{100}+2}}\pm {\sqrt {\frac {{\frac {A_{101}+1}{A_{100}+2}}\cdot (1-{\frac {A_{101}+1}{A_{100}+2}})}{A_{100}+3}}}$

Division gives

${\frac {A_{021}}{A_{001}}}\approx {\frac {{\frac {A_{121}+1}{A_{100}+2}}\pm {\sqrt {\frac {{\frac {A_{121}+1}{A_{100}+2}}\cdot (1-{\frac {A_{121}+1}{A_{100}+2}})}{A_{100}+3}}}}{{\frac {A_{101}+1}{A_{100}+2}}\pm {\sqrt {\frac {{\frac {A_{101}+1}{A_{100}+2}}\cdot (1-{\frac {A_{101}+1}{A_{100}+2}})}{A_{100}+3}}}}}={\frac {(A_{121}+1)\cdot {\sqrt {A_{100}+3}}\pm {\sqrt {(A_{121}+1)\cdot (A_{100}-A_{121}+1)}}}{(A_{101}+1)\cdot {\sqrt {A_{100}+3}}\pm {\sqrt {(A_{101}+1)\cdot (A_{100}-A_{101}+1)}}}}$

I don't know any exact formula for the mean and standard deviation of the quotient between two random variables. Bo Jacoby (talk) 21:23, 16 September 2008 (UTC).

Geometry :convex quadrilateral

let ABCD be a convex quadrilateral .Consider the points E and F such that C is the mid point of the line segment AE and D is the mid point of line segment BF. Evaluat with proof ,the ratio of [ABEF] and [ABCD]

---

ACtually I am not sure what is meant by [ABCD]. —Preceding unsigned comment added by Khubab (talk • contribs) 22:33, 1 August 2008 (UTC)

The brackets mean "the area of". I'm thinking about what the best solution to this problem may be (I already have one solution, but it's almost surely not what you're looking for). But remember, do your own homework -- that's the policy of the Reference Desk. You should also be a little more specific when you ask for help. 76.238.91.100 (talk) 04:11, 2 August 2008 (UTC)

Au contraire — I thought the brackets meant the cross-ratio, as used for example here. —Keenan Pepper 04:15, 2 August 2008 (UTC)

Hmmm, I didn't know about that. Anyway, if the brackets mean area of, then the answer would be very simple; just use the formula

${\mbox{Area(ABCD)}}={\frac {1}{2}}{\overline {AC}}\;{\overline {BD}}\sin \theta$

where $\theta$ is the angle between the diagonals ${\overline {AC}}$ and ${\overline {BD}}$. This formula works for all quadrilaterals.

EDIT: Concerning the cross-ratio, I'm reading into it.

EDIT2: I looked at this page, and I believe there are typos in their formulas. Cut-the-knot has the correct ones. Also, does anyone know how Wolfram defined the cross ratio of a square to be 2 here (or the values for any of the other shapes for that matter)? 76.238.91.100 (talk) 04:25, 2 August 2008 (UTC)

(Wolfram) Well AC and BD are Lsqrt(2) whereas AD and BC are just L ? (the points on the square go ABCDA etc) is that it?87.102.86.73 (talk) 22:32, 2 August 2008 (UTC)

That was for coplanar points - it made sense to me? (except the ratio 1/2 for a square when they already have 2 for a square)87.102.86.73 (talk) 22:36, 2 August 2008 (UTC)

There is an EDIT to this below. Please explain. The Wolfram article just offhandedly states that it is possible to define a cross ratio for any 4 coplanar points, but it doesn't really elaborate how to perform the calculation. Elsewhere I found out that one could create an arbitrary point P, draw lines from P to the 4 points in question (in this case, the corners of the square), and then find the cross ratio of the 4 intersecting lines thus formed (for which there is an unambiguous definition for the value of the cross ratio). However, the value depends on the placement of P, meaning that with this method a shape cannot be assigned a cross ratio value without specification of the reference point.

EDIT: Alright, I think I've got it figured out. On the Wolfram page, they are treating the points as complex numbers to calculate the cross ratios for the shapes like the square. Multiple values are possible for each shape (explaining why a square has a cross ratio of 2 and 1/2), although these values are all related by the equations given earlier in the page. However, the cross ratio of a quadrilateral in the complex plane is a real number only if the points are concyclic. The OP just says a "convex quadrilateral," so I assume now that what I said earlier about using a reference point P is the only reasonable method to use here. I haven't determined whether the OP is properly solvable (i.e., yields an answer that doesn't depend on the placement of P) this way, though. But then again, there is always the small chance that the answer is supposed to be a complex number. 75.4.141.243 (talk) 00:42, 3 August 2008 (UTC)

August 3

Markup syntax for a mathematical sequence and \lbrace \rbrace not documented

I was trying to edit markup syntax for a mathematical sequence containing braces {a_i} i.e. this syntax: ${a_{i}}$

The braces kept getting silently suppressed when appearing inside the math markup tag. Yes I know that in general outside math tag, (double-)braces would signify a Wikipedia reference; but inside the math tag, that has no meaning, hence braces should be treated as normal characters (or else, rejected and flagged).

I checked the Editing help and Reference and eventuallly deep down after much hunting I found the correct syntax is http://en.wikipedia.org/wiki/Help:Formula#Parenthesizing_big_expressions.2C_brackets.2C_bars i.e. you still need to use lbrace rbrace inside math tag even though that's meaningless. This is not well documented, such documentation as is is buried, and the fact that braces are silently suppressed is also bad.

Suggestions:

• add \lbrace \rbrace to the standard Editing Insert bar

Insert: – — … ‘ “ ’ ” ° ″ ′ ≈ ≠ ≤ ≥ ± − × ÷ ← → ·

• have the markup editor detect and flag braces {} inside a math (or other) markup tag and suggest 'you probably meant \lbrace \rbrace ?'
• document this more prominently, with examples

Smcinerney (talk) 12:31, 3 August 2008 (UTC)

This is documented on the page documenting Wikipedia's implementation of TeX. Where else do you think it should be documented? As indicated at Help:Displaying a formula, you don't actually need \lbrace \rbrace, you can just escape them with backslashes, thus: $\{a_{i}\}$. It would be a very bad idea to warn people using braces in TeX, since the reason they need to be escaped is there importance in TeX markup, e.g. <math>\frac{1}{2}</math> produces ${\frac {1}{2}}$. Algebraist 12:47, 3 August 2008 (UTC)

Braces are necessary in <math> \frac{11+1} 2 </math> displaying ${\frac {11+1}{2}}$ while <math> \frac 11+1 2 </math> displays ${\frac {1}{1}}+12$. Braces are displayed in <math> \{11+1\} </math>: $\{11+1\}$ but not in <math> {11+1} </math>: ${11+1}$. Bo Jacoby (talk) 13:58, 3 August 2008 (UTC).

But you also need to use \left\{ and \right\} to allow the braces to grow to the size of the expression between them. Michael Hardy (talk) 16:29, 4 August 2008 (UTC)

My comment was about the editing process, as experienced by a newbie user (e.g. me). Your first instinct is to use raw braces\{\}. Then, you realise they're silently suppressed. But when you look at the Insert bar for special symbols, all you see is: Insert: – — … ‘ “ ’ ” ° ″ ′ ≈ ≠ ≤ ≥ ± − × ÷ ← → · and below it you see Symbols: ~ | ¡ ¿ † ‡ ↔ ↑ ↓ • ¶ # ½ ⅓ ⅔ ¼ ¾ ⅛ ⅜ ⅝ ⅞ ∞ ‘ “ ’ ” «» ¤ ₳ ฿ ₵ ¢ ₡ ₢ $ ₫ ₯ € ₠ ₣ ƒ ₴ ₭ ₤ ℳ ₥ ₦ № ₧ ₰ £ ៛ ₨ ₪ ৳ ₮ ₩ ¥ ♠ ♣ ♥ ♦ m² m³ You might typically take this to imply that braces are not special symbols. The Insert and Symbols bars certainly have odd and rare selections, since braces are going to be far more common than (say) '≈' or '§' or '№'. Hence I'm suggesting braces should be added on the Insert bar.

Please replicate this exercise for yourselves - just go take a look at the Insert and Symbols bars as shown to an editing user. This is not user-friendly.

Also, I'm saying it is seriously Bad News that the editor silently strips out braces inside a math tag without saying something like 'Braces stripped from math tag - did you mean to use \{\} ?' How else would a user become aware that (single) braces are treated specially?

Algebraist - it would not be hard to automatically distinguish between braces occurring inside a math tag but not part of a TeX expression like \frac, e.g. my <math> {a_k} </math> example, and your TeX expression example where they are part. (If you disagree, can you show a counterexample?)

[I was not aware of the alternative \left\{ syntax instead of \lbrace which Michael H shows.] But the various syntaxes for braces need to be more accessible from the editing page - which they currently aren't. How do you propose this might be done? Smcinerney (talk) 17:11, 4 August 2008 (UTC)

Your problem is in the phrase "inside a math tag bot not part of a TeX expression". Everything inside <math></math> is a TeX expression. Once you've entered that math tag, you're now required to write in TeX. One of the basic syntactical features of TeX is that unescaped braces are used for grouping and do not appear as braces in the output. You won't get far until you learn this. You might as well ask for a "warning" whenever a caret appears in the math exprsesion.... "Warning: you used a caret. Because it means something in TeX (superscript) it will not be appear as a caret in the output!"

The "Symbols" list you mention shouldn't even be used if you're entering an expression using <math></math> tags, since there are TeX equivalents for those symbols.

A separate cheat-sheet below the edit box, containing commonly used TeX keywords... maybe. And as long as we're making unreasonable demands, I'd rather use eqn --tcsetattr (talk / contribs) 20:07, 4 August 2008 (UTC)

It is not apparent from the Editing Help that everything inside a <math> tag will have single-braces stripped since it must be TeX. Please reread the Editing Help and verify for yourself that that connection isn't made. You won't get far until you learn this. You won't get far in teaching people this, unless you actually document it in the relevant place - which it currently isn't - which was my issue. The "Symbols" list you mention shouldn't even be used if you're entering an expression using ${\displaystyle }$ tags, since there are TeX equivalents for those symbols. Again, how on earth could anyone be expected to know that, based on the Editing Help? again this is not documented, and needs to be. Currently you can freely click on Symbols and insert them inside TeX - which is undesirable. A separate cheat-sheet below the edit box, containing commonly used TeX keywords... maybe. I emphatically second that thought. At minimum, how to clearly document (on the Editing Help page, not buried deep in the TeX documentation) that normal markup doesn't apply inside <math> tags? As per Oliphaunt's suggestion, at minimum some link that says "the contents of <math> tags require TeX syntax".

And as long as we're making unreasonable demands I very strongly object to your tone.

So, what is our consensus on what needs documenting and where?Smcinerney (talk) 00:55, 6 August 2008 (UTC)

I'm not sure which help pages you're referring to. Help:Editing doesn't mention math at all, so you couldn't have learned that the <math> tag exists if that was the only help page you read. If you looked at Help:Math (also known as Help:Displaying a formula, Help:Formula and a bunch of other names), the first thing it says is "MediaWiki uses a subset of TeX markup"; a bit later, under "Syntax", it says "Math markup goes inside <math> ... </math>"; perhaps it would be better to change the words "Math markup" to "TeX markup" there but TeX is mentioned several times in the next few sentences so it's not exactly deeply hidden. If there are any other help pages that mention the <math> tag but don't also mention TeX, tell us where they are. --tcsetattr (talk / contribs) 05:12, 6 August 2008 (UTC)

I don't think it would be easy to tell when the braces were intended to be literal. I can't give a counter example unless you give a method of telling, though. If you give provide an algorithm for distinguishing, then I'll try and find an example that it doesn't work for (if I can't, I'll implement it for you!). --Tango (talk) 04:21, 5 August 2008 (UTC)

How about the simple 'Raw braces are illegal inside <math> tags unless associated with a TeX operator'? (Hence if they occur without any operator, that is always obviously wrong)Smcinerney (talk) 00:57, 6 August 2008 (UTC)

To respond to the suggestions of improving this made by the original poster, the only viable option is to "document this more prominently, with examples". <math> tags are a bit of a specialist thing, in that they require knowledge of TeX, whereas standard wiki syntax doesn't. However, I don't quite see how the documentation can be improved to help here. TeX just has a lot of features, resulting in a long page at WP:MATH. Oliphaunt (talk) 10:58, 5 August 2008 (UTC)

Derivative of the generalized factorial function.

Yes, the generalized factorial function, extended to all complex numbers (validly) by Euler, save for negative integers. But, just to keep the convoluting complex analysis stuff out, let's consider the factorial function x! extended to all real numbers save for negative integers.

Now, I propose, as I'm sure some other people have, that:

d/dx(x!) = ∫ e^(-t) * ln(t) * t^x dt

with lower and upper limits 0 and infinity, respectively.

Confirm/deny? I was just playing around with the limit definition and reached that. Point out any errors I may have made, if that's okay. —Preceding unsigned comment added by 124.191.116.239 (talk) 14:51, 3 August 2008 (UTC)

The generalized factorial function is normally called the gamma function. Your formula is equivalent to the formula for the derivative of gamma in that article. Algebraist 15:26, 3 August 2008 (UTC)

August 4

Raffle winning strategy

We're having a raffle at work. Each prize has a jar associated with it. You buy tickets and put your tickets in the jar for the prize that you'd like to win. Four of the prizes are vacation time. So, which option gives me better chances of winning at least one of the vacation time prizes, A) putting all my tickets into a single jar or B) spreading my tickets over all four jars? Let's assume that I can only win one of the vacation prizes and let's assume that all four jars are equally filled with other people's tickets. Dismas|^(talk) 02:56, 4 August 2008 (UTC)

Say each of the jars has N tickets, and you have 4. If you put all four in the first jar, the probability you win is 4/(N+4). If you put one in each jar, the probability that you lose is (N/(N+1))^4, so the probability that you win is 1 - (N/(N+1))^4. For any positive N, the latter probability is greater, so you should spread your tickets over all the jars. However, as N increases, the difference between the two strategies decreases, and in the limit of large N, they are the same. So if there are a lot of other people, it would probably be a better strategy to guess which jar has the fewest tickets, and put all your tickets in that one. —Keenan Pepper 04:21, 4 August 2008 (UTC)

(Edit conflict, and we've said almost exactly the same thing, but I'll post mine anyway!) Ok, let's assume you buy four tickets, and there are n other tickets in each jar. If you put all your tickets in one jar, your chance of winning is simply ${\frac {4}{n+4}}$ (you have 4 tickets and there are n+4 tickets in the jar). If you put one ticket in each jar, your chance of winning at least one prize is one minus the chance of winning no prizes, in other words $1-\left({\frac {n}{n+1}}\right)^{4}$ (the chance of losing each jar is ${\frac {n}{n+1}}$, and there are 4 jars, so the chance of losing all of them is $\left({\frac {n}{n+1}}\right)^{4}$). I then plugged those formulae into a maths computer package and found that putting 1 in each jar is better for any value of n, although the difference becomes very small as n gets very large. The difference is at most 14% for n=1 and drops below 1% for n>20. So, my advice would be to put them in different jars if there are only a few tickets in each, otherwise put them all in the one with the least tickets (obviously, that requires you to play last, but I'm sure you can manage that if you try!). --Tango (talk) 04:24, 4 August 2008 (UTC)

Tango's maths computer may be replaced by the binomial series

$1-\left({\frac {n}{n+1}}\right)^{4}=1-(1+n^{-1})^{-4}=1-\sum _{i=0}^{\infty }{{-4} \choose i}n^{-i}={\frac {4}{n}}-{\frac {10}{n^{2}}}+{\frac {20}{n^{3}}}-\cdots$

${\frac {4}{n+4}}=1-(1+4n^{-1})^{-1}=1-\sum _{i=0}^{\infty }{{-1} \choose i}4^{i}n^{-i}={\frac {4}{n}}-{\frac {16}{n^{2}}}+{\frac {64}{n^{3}}}-\cdots$

So

$1-\left({\frac {n}{n+1}}\right)^{4}-{\frac {4}{n+4}}={\frac {6}{n^{2}}}-{\frac {44}{n^{3}}}+\cdots$

confirming the behavior for big values of n. Bo Jacoby (talk) 08:35, 4 August 2008 (UTC).

Thanks! I split the tickets amongst the jars. We'll see how I did tomorrow... Dismas|^(talk) 09:58, 4 August 2008 (UTC)

Absolutely, it just seemed quicker to get the computer to do it! Obviously not quite quick enough, though, by 3 minutes! Also, the computer told me the two formulae didn't switch places at any point before n got large enough to make it all simple - that could also be done analytically, of course, but I didn't see the point. --Tango (talk) 21:59, 4 August 2008 (UTC)

Thanks for the help all! I didn't any of the vacation time. Though I did put a ticket in the jar for a pair of hand knitted mittens and hat. I'm going to save them for winter and give them to my wife.  :-) Dismas|^(talk) 14:11, 7 August 2008 (UTC)

Poisson Distribution

I am trying to do this question, but getting bogged down in sums of factorial fractions.

Y is the number of calls received in one hour, and it has a Poisson(20) distribution. Given Y=y, each of the y calls is redirected with probability 0.2. If X is the number of redirected calls, show that X~Poisson(4).

I have taken exp(-20) out the front of an infinite sum over i of 1/(i-j!)i! multiplied by 16^j multiplied by (1/4)^i. I can see where I need to get - the infinite sum of (16^j)/j! needs to come out the front, and then we need to somehow wind up with a 4^i/i! out of what is left. I don't really understand how to manipulate the factorial fractions to get there. help, please! —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 06:19, 4 August 2008 (UTC)

For a fixed j, you need to calculate that sum. It would help if you make the substitution $k=i-j$. Oded (talk) 11:33, 4 August 2008 (UTC)

Thankyou! Now I see how easy it actually is. I'll remember that tip for sure. —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 01:08, 7 August 2008 (UTC)

Looking for Simon Plouffe's magical formulae

I'm looking for Simon's magical formulae that produces the digits of Pi in base 10 without first calculating the preceeding digits.

I know for sure that this formulae exists because it said so in Wikipedia.

On Simon's page Simon Plouffe, it says "Plouffe discovered an algorithm for the computation of π in any base in 1996. He has expressed regret for having shared credit for his discovery of this formula with Bailey and Borwein.".

Any idea where is the formulae for base 10? 122.107.146.22 (talk) 09:13, 4 August 2008 (UTC)

An algorithm that can compute digits of a mathematical constant without using preceeding digits is called a spigot algorithm. There is a well-known spigot algorithm for π called the Bailey–Borwein–Plouffe formula, but this calculates hexadecimal (or, equivalently, binary) digits of π, not decimal digits. I am not aware of a decimal spigot algorithm for π. As far as a decimal spigot algorithm is concerned, this page says "Plouffe, for one, has found a way to compute individual decimal digits of pi without calculating preceding digits. That makes it possible to compute a particular decimal digit of pi using a pocket calculator, Plouffe says. However, his algorithm is fairly slow and clearly impractical for determining the millionth or billionth decimal digit of pi." It gives another link here, which I haven't checked. I don't understand the comment about efficiency, because I would have thought a spigot algorithm should be more or less equally efficient whether it is calculating the millionth, billionth or just the tenth digit - that is kind of the point of spigot algorithms. Gandalf61 (talk) 10:21, 4 August 2008 (UTC)

Hmm. The latter article appears to have moved to http://www.lacim.uqam.ca/~plouffe/Simon/articlepi.html, the link in the first is broken. GromXXVII (talk) 11:01, 4 August 2008 (UTC)

Fabrice Bellard lists this approach for any base, from 1997. It's a slight improvement over Plouffe's 1996 formula in that it runs in O($n^{2}$) time rather than his O($n^{3}\log \left(n\right)^{3}$). --_{tiny plastic} Grey Knight ⊖ 16:22, 4 August 2008 (UTC)

confusing problem

explanation: ok, so I guess I should make this clear first off, this isn't homework. Thankfully I stopped having to do math homework quite a few years ago... yet for some reason a part of my brain keeps coming up with these problems that the rest of me is usually completly mentally unequipped to answer. They're just little thought experiments that I come up with by accident that bug me until I work them out (or more usually get someone else to do it for me). This is one of those problems, and I would be most grateful if someone could help me out here before I go completely insane.

The only way I can think about the problem without my brain melting is in high-school esque math problem terms, imagining actual physical objects and properties (balls, boxes, colors). I don't know if that's clear enough but seeing as I know nothing about actual mathematics this is the best I can do. Hopefully it can be translated into proper math speak.

problem: You have a number of balls of different colors. The number of colors is X. There an equal number of balls of each colour. This amount of balls of each color is Y. There is also an indeterminate amount of boxes that can hold the balls. However each box can only contain a certain number of balls, no more and no less. The number of balls each box can contain is Z. In addition, no box can contain more than one ball of the same colour; and no two boxes can ever have the same combination of coloured balls.

The questions that are bugging me are: for what combinations of values for X, Y and Z is it possible to put all of the balls into boxes without any left over balls? And for these combinations, how many boxes are needed?

The answer can be in an algebraic equation, I think I can get my head around that. However if this problem unknowingly steers into some really complex math stuff, a little explanation might be necessary. Thanks :)

--86.146.162.73 (talk) 15:12, 4 August 2008 (UTC)

I don't know the answer and I haven't attempted to think about it, but start by looking under combinatorial design both here and on Google. And Google Scholar and Google Books. Michael Hardy (talk) 16:20, 4 August 2008 (UTC)

Why not start with small cases to get an idea? What happens with (X,Y,Z)=(1,1,1)? (2,1,1)? (2,2,1)? - Rainwarrior (talk) 16:37, 4 August 2008 (UTC)

The problem sounds equivalent very closely related to the generation of balanced incomplete block designs. The key word here is "incomplete": not every color will be in every box. That page seems somewhat obtuse but it might be a good start. I would also recommend Google searches, per MH above. Baccyak4H (Yak!) 17:30, 4 August 2008 (UTC)

Thanks for the help but as I suspected, this goes into territory that is completely beyond my ken. Perhaps I should have been more clear about my complete lack of aptitude; I can barely do multiplication in my head (I suspect I suffer from dyscalculia, but I've never been diagnosed). I don't understand anything in the above article past "in". I really don't know why I torture myself by coming up with questions like this :p I think I'm going to try Rainwarrior's idea instead and work it out graphically in MS Paint or with colored boxes in Excel or something. --86.146.162.73 (talk) 17:57, 4 August 2008 (UTC)

I know for a fact that there are tables of sets of {X, Y, Z} out there (I used them in grad school...), but I do not have any refs on hand. They are likely older ones—this topic has lost some of its research luster, as the main reason it was studied was to avoid doing more complicated matrix operations by hand before cheap computing was available.

That said, per RW it should not be too hard to enumerate the smaller examples. Keep in mind the necessary restrictions: you have assumed that Z<X, since otherwise some boxes would have more than one ball of a given color. Also, from that article (rewritten slightly):

(# boxes) × Z = X × Y (equiv to the bk=vr expression)

The other expression in that article states a restriction which I am not sure is required by your conditions.

Baccyak4H (Yak!) 18:33, 4 August 2008 (UTC)

(unindenting) Is this so hard as all the posters indicate? I get X*Y/Z<=X!/Z!(X-Z)! and in addition X*Y/Z must be integer. Taemyr (talk) 04:10, 5 August 2008 (UTC)

That doesn't work. Try X=1, Y=2, Z=2. It satisfies your condition, but still fails because of the "no two boxes can ever have the same combination of coloured balls" condition. --Tango (talk) 04:16, 5 August 2008 (UTC)

Actually, your counterexample isn't a counterexample (provided X!/Z!(X-Z)! is understood as a binomial coefficient). The maximum number of boxes used is the binomial coefficient X choose Z, denoted $\scriptstyle {\binom {X}{Z}}$, since this is the number of possible distinct combinations of X colors into sets of size Z. The total number of balls XY must not exceed $\scriptstyle {{\binom {X}{Z}}Z}$, since otherwise there will be balls left over. In addition, XY/Z must be an integer since the total number of balls must be divisible by the box size (each box, if used, must be completely filled). Now, these two conditions are obviously necessary, but I can't see a proof that they are also sufficient, although it seems quite plausible. siℓℓy rabbit (talk) 13:24, 5 August 2008 (UTC)

Good point - my example is a degenerate case and doesn't really prove anything. I agree necessity is obvious, sufficiency is more of a problem, though. I suspect there may be a tighter bound on the total number of balls, although I haven't even tried looking for one yet. It won't be anything simple, since that bound can be attained in some trivial cases (eg. (1,1,1)). --Tango (talk) 09:22, 6 August 2008 (UTC)

Is there a formula for the probability of rolling a sum? given a pair of 6-sided dice?

In other words, is there some neat little algebra formula that I can just plug in?

Like the odds of rolling a 7 is .166(repeating) percent and the odds of rolling a 2 or 12 are both .0277(repeating)

in other words, is there an equation like f(x) = y

The odds of rolling a 10 can be either 6+4, 5+5, 4+6 thus f(10) = .0833333

Is there a algebraic or statistical formula that I can relate the input variable to the output formula?

I know that the answer is yes, if the function is continuous, but this dice problem is a stepwise function. (the derivative doesnt exist, thus you can't use calculus to come up with the equation)

Is there a catchy formula for this simple phenomenon? I've checked the dice page and the binomial page and used google search exclusively on en.wikipedia.org Sentriclecub (talk) 16:11, 4 August 2008 (UTC)

The probability that the sum is x is

${\frac {6-|x-7|}{36}}.$

I don't find that very "catchy", the way most identities in probability that you see displayed here are. That's probably why you don't see it. Michael Hardy (talk) 16:17, 4 August 2008 (UTC)

You might be able to derive a catchy expression by viewing the problem as one of counting partitions of x into n parts (in your case n=2) where the partition sizes are restricted to be no greater than six. That would be the numerator and 6ⁿ would be the denominator. The restriction to be no greater than six might make the general derivation tricky, however. Baccyak4H (Yak!) 17:51, 4 August 2008 (UTC)

Excellent, I saw that too, that the sum goes 1,2,3,4,5,6,5,4,3,2,1 all over 36. I was squaring and then taking the sq-root, but using simple absolute value operators is much more aesthetic. Sentriclecub (talk) 19:07, 4 August 2008 (UTC)

August 5

Integral of an Absolute Value

What technique should I use to integrate $\int (|x^{2}-4|,x,0,3)$? I tried u-substitution but the $2x$ does not cancel out.

Thank you!

Matt Saladin (talk) 00:41, 5 August 2008 (UTC)

Do you mean $\int _{0}^{3}|x^{2}-4|\,dx$? If so, just split it as $\int _{0}^{2}|x^{2}-4|\,dx+\int _{2}^{3}|x^{2}-4|\,dx$. Algebraist 00:44, 5 August 2008 (UTC)

(edit conflict)Do you mean $\int _{0}^{3}\left|x^{2}-4\right|dx$? I think the only reasonable way is to use the definition $\left|x\right|={\begin{cases}x{\mbox{ if }}x\geq 0\\-x{\mbox{ if }}x<0\end{cases}}$ and from there split the integral into two parts. Confusing Manifestation(Say hi!) 00:48, 5 August 2008 (UTC)

That's weird. Why does your use of \left and \right force the small space before dx? Algebraist 00:53, 5 August 2008 (UTC)

Haven't a clue. Confusing Manifestation(Say hi!) 00:10, 6 August 2008 (UTC)

I think when you do \rigth Tex interprets this as a right side deliminator, which has different spacing than a verticle bar (which I think is normally interpreted as a binary operator). Oded (talk) 01:33, 6 August 2008 (UTC)

Alright, I get it now. Thank you very much. Sorry for the poor notation, I haven't used LaTeX before. —Preceding unsigned comment added by Syst3m3rr0r (talk • contribs) 00:52, 5 August 2008 (UTC)

I messed around trying to see if you could do a substitution and ended up having to do integration by parts repeatedly and getting a power series: which is not quite desirable. The only two ways I know of to antidifferentiate an absolute value are what they’ve done above, or to use the fact that ${\frac {d}{dx}}|x|={\frac {|x|}{x}}$ which is essentially the same thing just more cumbersome but avoids the use of piecewise functions. When combined with the chain rule for more complicated things, it gets very cumbersome quite fast, for instance ${\frac {d}{dx}}|x^{2}-4|={\frac {|x^{2}-4|}{x^{2}-4}}\cdot 2x$ GromXXVII (talk) 11:53, 5 August 2008 (UTC)

You think that's bad, try using the alternative definition $\left|x\right|={\sqrt {x^{2}}}$. Confusing Manifestation(Say hi!) 00:10, 6 August 2008 (UTC)

Doesn't that method fail because the region of integration includes a non-differentiable point? Doesn't integration by parts require the things you're differentiating to be differentiable? --Tango (talk) 00:39, 6 August 2008 (UTC)

I think yes and no. If you were to break it into two(or more) different integrals at the non-differentiable point(s) you’ll get the same expression for both pieces – and so that single expression holds for all other points. However, because of the introduced the denominator to do the antidifferentiation, the expression doesn’t exist at that point(s) anyway. So yes I should have been more careful, but I think it is still valid. GromXXVII (talk) 17:59, 6 August 2008 (UTC)

Permutations

how many different groups of six can i make using numbers 1 through 54 without using the same number twice —Preceding unsigned comment added by 99.207.12.77 (talk) 09:43, 5 August 2008 (UTC)

Do you mean without using the same number twice in each group (for example {[1,2,3,3,4,5];[1,2,3,4,4,5]} would be illegal but {[1,2,3,4,5,6];[1,2,3,5,6,7]} would be legal) or without using the same number twice at all (for example {[1,2,3,4,5,6];[6,7,8,9,10,11]} would be illegal)? --Slashme (talk) 09:51, 5 August 2008 (UTC)

Assuming you mean the first of these (equivalent to the question 'how many different hands of six cards are there from a standard deck of cards with two (different) jokers included), the answer is ${\frac {54!}{6!(54-6!)}}={\frac {54\times 53\times 52\times 51\times 50\times 49}{6\times 5\times 4\times 3\times 2\times 1}}$. See Combination.--130.88.123.142 (talk) 11:40, 5 August 2008 (UTC)

Small order of operations fix: It should be ${\frac {54!}{6!(54-6)!}}$. StuRat (talk) 13:08, 5 August 2008 (UTC)

${\frac {54!}{6!(54-6!)}}=25827165$. These answers assume the order of the numbers doesn't matter, for example [1,2,3,4,5,6] and [2,1,6,4,5,3] are not different. Somebody other than the original poster added the headline "Permutations", but order usually does matter in a permutation, and then the answer would be ${\frac {54!}{(54-6)!}}=18595558800$. The original poster used the word "groups". Group (mathematics) has a special meaning which was not intended here. Set would be a better word if the order doesn't matter. PrimeHunter (talk) 13:27, 5 August 2008 (UTC)

Again, the parens seem to be in the wrong place (in the first equation, but correct in the second). I believe that (54-6!) would mean (54-6×5×4×3×2×1) = (54-720) = -666, which, while an interesting number (the anti-antichrist ?), is not what was meant here. I believe it should be written as (54-6)! or 48!. StuRat (talk) 01:26, 6 August 2008 (UTC)

Yes, the "permutations" was me; I should have said "combinations". Very sloppy, sorry ;-) --Slashme (talk) 13:55, 5 August 2008 (UTC)

NP-completeness of 3-colouring

I'm trying to figure out the proof of the NPCness of 3-colouring (this isn't homework). Karp's paper is very terse/cryptic. Garey and Johnson leave it as an exercise. Sipser also leaves it as an exercise, but gives the hint to use an OR-gadget like this:

   *
  / \
 *---*
 |   |
 *   *

OK, but how do I use this gadget? I'm having trouble figuring out which are the inputs and which is the output. Also, graph-colouring isn't "biased" towards any colour but OR-gates are "biased" towards true --- how does one resolve this assymetry? --Taejo|대조 15:32, 5 August 2008 (UTC)

It’s in Algorithms by Johnsonbaugh and Schaefer, although [possibly] different from the approach you have. They reduce 3SAT to 3-colorability by constructing a [fairly complicated] graph which is 3 colorable iff the corresponding expression is satisfiable. GromXXVII (talk) 20:34, 5 August 2008 (UTC)

I don't know how it was originally solved, but I think this will work. The graph has

• nodes labeled T and X with an edge between them;
• a node for each variable;
• a node for the negation of each variable, forming a triangle with the variable and X;
• a node connected to X for each internal wire connecting an OR gate to an OR gate in the CNF circuit; and
• one of those OR-gadgets for each OR gate in the CNF circuit, with the bottom nodes being the inputs and the top node being the output. If the output wire goes to an AND gate then the output node is T.

The OR-gadget doesn't implement an OR gate, but I don't think it matters. The important thing is that the output can't be T unless at least one input is T. It can be F when one input is T, but that will never lead to spurious solutions. In effect you're replacing every $\vee$ in the CNF expression with a variable ranging over $\{\vee ,\wedge \}$, but it's never helpful to set it to $\wedge$. -- BenRG (talk) 21:25, 5 August 2008 (UTC)

Which graphical calculator is best for me?

Hi, I'm considering getting a graphical calculator for my statistics, and would like to know which to get. I want one that allows the greatest number of statistical functions (multiple regression, ANOVA, etc.) and also, just as importantly, has the greatest capabilities for solving equations and systems of equations (eg. being able to solve non-linear simultaneous equations in 2 or more variables, exactly or by approximation). Ease of use is also important, but less so. Are there any I should look at? It's been emotional (talk) 16:42, 5 August 2008 (UTC)

To be able to solve equations exactly you’ll want something with a Computer Algebra System – which essentially puts you at a TI-89/92 or the competing models made by other brands. As for statistics – if you’re going to doing anything beyond a class you might consider a computer package (there are some available for free). Some of the more advanced calculators will do a lot of statistics, but it can be cumbersome especially if you have a lot of data. GromXXVII (talk) 23:10, 5 August 2008 (UTC)

Price support example: utter nonsense, or not?

The example in the Price support article makes no sense to me. It seems to be comparing things that can't logically be compared. That's a vague complaint, but one specific issue I have with it is that the government wouldn't just throw away the 200 units of goods it bought for $1200. Is there a good reason for neglecting the value of those goods, or is the example just nonsense? 128.165.101.105 (talk) 19:02, 5 August 2008 (UTC)

I can't comment on the maths in the example but the government is quite likely to store the surplus in a huge warehouses until it is about to go off, and then sell it in third world countries at a price that undercuts local producers and puts them out of business. Who this benefits I'm not at all sure. Dmcq (talk) 19:28, 5 August 2008 (UTC)

In the case of agriculture at least I think they normally pay not to produce as opposed to acting as a retailer. GromXXVII (talk) 20:23, 5 August 2008 (UTC)

I don’t think the value of the goods is neglected, the example makes note about how the cost to the consumers is actually 450+1200, the 1200 being the cost to keep the goods off the market. GromXXVII (talk) 20:23, 5 August 2008 (UTC)

The US government has bought and/or destroyed perfectly good food crops to prop up prices. This seems illogical, however, since they're unwilling to allow competition and capitalism to work, this is the only way to keep prices high during periods of overproduction. StuRat (talk) 01:19, 6 August 2008 (UTC)

My question was actually supposed to about the logical consistency of the argument. At the end it just says "For this reason, price supports are considered inefficient.". I don't see where that conclusion comes from at all. There's all this arithmetic going on that doesn't make sense, and doesn't support the conclusion.

Also, even if I believe that price supports are bad, that doesn't change the fact that the article doesn't mathematically prove that. 128.165.101.105 (talk) 14:10, 6 August 2008 (UTC)

I don’t think the article is claiming they’re bad, but rather saying their inefficient meaning that the benefit is monetarily less than the cost. That is, the direct cost to the consumers is only 450, and the benefit to producers is 550. But the indirect cost to consumers adds another 1200, which together with the previous 450 is larger than 550. Also note I’m not an economist, I’m just trying to interpret the what the article is saying GromXXVII (talk) 15:04, 6 August 2008 (UTC)

I think inefficient may refer to Pareto efficiency, its inefficient because somebody is worse off. It wouldn't be my first objective. Dmcq (talk) 18:32, 6 August 2008 (UTC)

Sorry it could be Pareto efficient because the producers get more money, I believe the article is saying it isn't efficient because the consumers could give the money directly to the producers so the producers go the same amount but the consumers lost less. Dmcq (talk) 18:43, 6 August 2008 (UTC)

In other words, it's more efficient for the government to just pay farmers to produce less, rather than buying the surplus and throwing it away. Algebraist 18:48, 6 August 2008 (UTC)

This would certainly be a saving because the government could pay less to the producers but the producers would still get the same amount because they don't have to pay to produce the surplus. You have to wonder what the whole point of keeping the price high is though. If it is to support the producers through lean periods it would be better just to subsidize them in such years - or even just have a basic subsidy anyway. I believe this is the way the EU is going with agriculture. A bit of a pity really as set aside protected lots of wild space. Dmcq (talk) 20:22, 6 August 2008 (UTC)

Ray-tracing in more exotic non-euclidian spaces?

I've seen ray-traced views of elliptic space and hyperbolic space, but I was wondering about some of the stranger non-Euclidian spaces, like projective space and taxicab space, if there were a way to draw ray-traced images in those spaces. --Zemyla^t 21:58, 5 August 2008 (UTC)

Well, to do ray tracing you need at least some notion of straight line and a way of embedding the viewer in the space. If you want directional lighting and reflection and refraction and such you need some notion of angle also. A Riemannian manifold gives you all of those automatically, and you can find ray-traced images made with more complicated Riemannian manifolds than the elliptic and hyperbolic planes, like these movies. The projective space RP³ with the usual geometry is just a model of elliptic geometry. I'm not sure what would count as an angle or even a straight line in the taxicab space. -- BenRG (talk) 23:07, 5 August 2008 (UTC)

A natural definition of 'straight line' is 'shortest path' (or something similar). For taxicab geometry, the shortest path between two points is massively non-unique, so I doubt you can get very far. Algebraist 23:14, 5 August 2008 (UTC)

Taxi-cab might be possible - assuming there are opaque objects in a transparent field. Then replacing the rays with a 'project plane' eg a X x Y array of points. Then there are two options:

a. Each path is equally likely (and contributes) - therefor the 'illumination' of a point in the array by an object is proportional the fraction of paths un-obscured.

b. If there is a unblocked path to the object then it is seen (black & white)

because multiple 'objects' would be visible from a given 'screen point' some sort of summable transparent representation of the 'projection' would be needed eg The screen point intensity is proportional to the sum of the 'colours' of each point visible.

It's not really a good geometry to do ray tracing in.87.102.5.5 (talk) 02:14, 6 August 2008 (UTC)

For hyperbolic geometry specifically, see negative curves and other blog entries on the same blog.

August 6

Ramsey's Theorem Proof

The section on Proof of Ramsey's Theorem for 2 colors contains the following sentence:

Now |M| ≥ R(r − 1, s) or |N| ≥ R(r, s − 1), again by the pigeonhole principle.

Can someone please explain how the inequalities follows from the pigeonhole principle. Thanks--Shahab (talk) 04:44, 6 August 2008 (UTC)

The graph contains $|M|+|N|+1=R(r-1,s)+R(r,s-1)$ vertices. Therefore one or the other of the stated inequalities must hold. Eric. 89.51.105.61 (talk) 08:25, 6 August 2008 (UTC)

(After edit conflict( I believe it follows from the fact that vertex v lies in a complete graph with R(r-1, s) + R(r, s-1) vertices, so

$|M|+|N|=R(r-1,s)+R(r,s-1)-1$

but if |M| < R(r − 1, s) and |N| < R(r, s − 1) then

$|M|+|N|\leq R(r-1,s)+R(r,s-1)-2$

which is a contradiction. Invoking the pigeonhole principle doesn't seem to be the most natural way to explain this. Gandalf61 (talk) 08:48, 6 August 2008 (UTC)

That's what I had arrived at. So this means that the reference to the pigeonhole princinple is incorrect. Thanks--Shahab (talk) 09:21, 6 August 2008 (UTC)

I'll clarify that point in the proof given in the article. Eric. 89.51.105.61 (talk) 10:00, 6 August 2008 (UTC)

Trigonometry: Secant to Cosecant

I'm having a lot of trouble understanding this problem from Teach Yourself Trigonometry:

"The value of $\alpha$ is acute, and $sec(\alpha )=k$. Find in terms of k the value of $cosec(\alpha )$."

Now,

$sec(\Theta )={\frac {1}{cos(\Theta )}}$

and

$cosec(\Theta )={\frac {1}{sin(\Theta )}}$

So if I start with

$k={\frac {1}{cos(\alpha )}}$

Multiply both sides by cos(a) and divide by k

$cos(\alpha )={\frac {1}{k}}$

The sine is 90 degrees minus the cosine:

${\frac {\pi }{2}}-{\frac {1}{k}}=sin(\alpha )$

So it seems to me that the answer should be:

${\frac {1}{{\frac {\pi }{2}}-{\frac {1}{k}}}}={\frac {2k}{\pi (k)-2}}=cosec(\alpha )$

But the book says that the real answer is:

${\frac {k}{\sqrt {k^{2}-1}}}$

Can someone tell me where my thinking went wrong, and what the right way to approach this problem is? - 209.242.51.68 (talk) 05:23, 6 August 2008 (UTC)

The error is with "sine is 90 degrees minus the cosine". It's actually $\sin(\theta +{\frac {\pi }{2}})=\cos(\theta )$. --Tango (talk) 05:42, 6 August 2008 (UTC)

Or $\sin({\frac {\pi }{2}}-\theta )=\cos(\theta )$, but not $\sin(\theta )={\frac {\pi }{2}}-\cos(\theta )$. -- BenRG (talk) 11:47, 6 August 2008 (UTC)

What you want is the Pythagorean trigonometric identity. Algebraist 12:49, 6 August 2008 (UTC)

Polynomial that generates ALL primes

I have seen in a book (The Music of the Primes, Marcus du Sautoy, 2004 ed.) a polynomial in 26 variables whose positive outputs constitute the complete set of prime numbers. Can anyone tell me who proved it and when?

Thanks in advance, Fish. (talk) 17:28, 6 August 2008 (UTC)

According to formula for primes, it was Jones, James P.; Sato, Daihachiro; Wada, Hideo; Wiens, Douglas (1976), "Diophantine representation of the set of prime numbers", American Mathematical Monthly, 83: 449–464. Algebraist 18:05, 6 August 2008 (UTC)

In case du Sautoy doesn't makes this clear, you should be aware that this is in no way an interest fact about the primes. It is an interesting fact about the expressive power of integer polynomials. In fact, for any Recursively enumerable set of positive integers, there is a polynomial which takes exactly those positive values. Thus the result for primes is a trivial special case. Algebraist 18:56, 6 August 2008 (UTC)

Thank you very much for a model answer: timely, concise and accurate. As to your second point, the polynomial is not discussed in any detail at all (which I presume is why he does not quote the source himself), so the article you directed me to contained much of interest. I don't understand it all, but it's interesting anyway! Thanks. Fish. (talk) 22:39, 6 August 2008 (UTC)

list methods which can demonstrate ability to perform a given function

Optimal classification is an arrangement of attributes for a set of elements in an attribute-value system which minimizes the number of attribute queries necessary to identify a particular element within an element set. Because there may be more than one arrangement of attributes which minimize the number of queries, the function is termed optimal rather than minimal. By this definition what method(s) in mathematics can clearly demonstrate their capability of performing this function? (Not a homework question BTW, merely doing article research to determine the method(s) which can perform this function such as when one might collect a list of sort routines.) —Preceding unsigned comment added by 71.100.162.249 (talk) 20:17, 6 August 2008 (UTC)

Riiiight! hydnjo talk 22:19, 6 August 2008 (UTC) Acknowledging failure to AGF  :( -hydnjo talk 15:49, 7 August 2008 (UTC)

Let's see if I understand what you're saying. You have a list of attributes for each element. Maybe "Green", or "11010", or something. You want to know which attributes are helpful in telling the elements apart quickly, and you want references for algorithms that are provably effective in finding these attributes. Am I right so far? You say, "a particular element". Are you looking for the fewest attributes necessary to distinguish this from the rest, or to distinguish all elements from each other? Black Carrot (talk) 07:49, 7 August 2008 (UTC)

• List of attributes... correct.
• Which attributes are helpful in telling the elements apart quickly... correct.
• Want references for algorithms that are provably effective in finding these attributes... correct.
• Am I looking for the fewest attributes necessary to distinguish a particular element from the rest or to distinguish all elements from each other... the latter, if I understand your question correctly.

In general, all attributes are assumed to be shared by a common group of elements with the purpose of the algorithm being to distinguish any one element from all other elements in the list of elements. If you threw in a foreign element with a single attribute permitting that element's identification with only one query and added this attribute to the list of attributes being considered for all then the algorithm would treat it no differently than any other attribute nor the new element any differently than the other elements. In other words, if the group of elements were varieties of trees and you threw in a tin can and added the attribute of "totally made of metal" it would not make any difference in terms of the order in which the attributes were placed by the algorithm to permit the fewest queries to be made to identify either the trees or the tin can except that when unrelated elements are combined in such a fashion that identifying the tin can might take a greater, or the greatest, number of queries than if it were part of a list of metal objects with attributes more closely related to identifying them, such as types of metal by percent for each metal object, etc. —Preceding unsigned comment added by 71.100.162.249 (talk) 09:38, 7 August 2008 (UTC)

Note that "Optimal classification" was the subject of a former article and AfD debate. Gandalf61 (talk) 12:15, 7 August 2008 (UTC)

Yes it was and has since survived a second attempt at deletion by the same user when it was moved to the Wikibooks. The purpose of this question is in fact to give those who claimed in the deletion discussion here an opportunity to prove that other methods exist. Unlike here in the Wikipedia, in the Wikibooks project, if other methods do in fact exist then they are easily enough included in the book as other chapters. My feeling is, however, that these were false claims and the nomination for deletion itself was the equivalent of an entomologist seeing a new creature he had never seen before and squishing it into the ground for that reason. —Preceding unsigned comment added by 71.100.162.249 (talk) 12:25, 7 August 2008 (UTC)

If I can stay awake through all the long descriptions... you're looking for a 20Q algorithm? --tcsetattr (talk / contribs) 20:33, 7 August 2008 (UTC)

Similar idea, but no. I'm looking for algorithms intended to address bounded problems, i.e., to find the order of attributes which will in combination define a specific class of elements with the fewest attributes rather than algorithms to find the order of fewest attributes which will in combination define an unbounded class of elements. —Preceding unsigned comment added by 71.100.162.249 (talk) 04:50, 8 August 2008 (UTC)

Offhand I'd guess this is what they call an NP-hard problem like the travelling salesman problem and you won't even get a quick way to check a solution is correct. It would be nice to be proved wrong but I think you'll probably have to make do with heuristics for a good solution. There's probably a degree in it either way for somebody ;-) Dmcq (talk) 22:18, 7 August 2008 (UTC)

Actually, there is at least one algorithm that does this already which has been rejected recently as an article in the Wikipedia on claim that other methods exist which already solve this problem, such as methods for trimming decision trees. Out of scientific curiosity and for the sake of scientific objectivity I seek only to examine such claims to reach an empirical determination for myself. —Preceding unsigned comment added by 71.100.162.249 (talk) 04:50, 8 August 2008 (UTC)

August 7

Natural log integral

Does my working go wrong at any point?

$\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \theta )\ d\theta$

$\theta =\pi /4-\phi$

${\frac {d\theta }{d\phi }}=-1$

$=-\int _{\frac {\pi }{4}}^{0}\ln(1+\tan({\frac {\pi }{4}}-\phi ))\ d\phi$

$=-\int _{0}^{\frac {\pi }{4}}\ln(1+\tan({\frac {\pi }{4}}-\phi ))\ d\phi$

$=-\int _{0}^{\frac {\pi }{4}}\ln(1+{\frac {\tan {\frac {\pi }{4}}-\tan \phi }{1+{\tan {\frac {\pi }{4}}}{\tan \phi }}})\ d\phi$

$=-\int _{0}^{\frac {\pi }{4}}\ln(1+{\frac {1-\tan \phi }{1+{\tan \phi }}})\ d\phi$

$=-\int _{0}^{\frac {\pi }{4}}\ln({\frac {1+\tan \phi }{1+\tan \phi }}+{\frac {1-\tan \phi }{1+{\tan \phi }}})\ d\phi$

$=-\int _{0}^{\frac {\pi }{4}}\ln({\frac {1+\tan \phi +1-\tan \phi }{1+\tan \phi }})\ d\phi$

$=-\int _{0}^{\frac {\pi }{4}}\ln({\frac {2}{1+\tan \phi }})\ d\phi$

$=-\int _{0}^{\frac {\pi }{4}}\ln 2-\ln(1+\tan \phi )\ d\phi$

$=\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )-\ln 2\ d\phi$

$\Rightarrow \int _{0}^{\frac {\pi }{4}}\ln(1+\tan \theta )\ d\theta =\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )-\ln 2\ d\phi$

The whole point of the question is that I have to show that

$\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \theta )\ d\theta ={\frac {\pi }{8}}ln2$.

I could do this if instead of

$\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \theta )\ d\theta =\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )-\ln 2\ d\phi$

I had

$\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \theta )\ d\theta =\int _{0}^{\frac {\pi }{4}}\ln 2-\ln(1+\tan \phi )\ d\phi$

Because both

$\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \theta )\ d\theta$ and $\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi$

bound the same area and therefore have the same numerical value.

So I could say that

$\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \theta )\ d\theta =\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi$

$\Rightarrow \int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi =\int _{0}^{\frac {\pi }{4}}\ln 2-\ln(1+\tan \phi )\ d\phi$

$\Rightarrow \int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi =\int _{0}^{\frac {\pi }{4}}\ln 2-\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi$

$\Rightarrow 2\int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi =\int _{0}^{\frac {\pi }{4}}\ln 2\ d\phi$

$\Rightarrow \int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi ={\frac {1}{2}}\int _{0}^{\frac {\pi }{4}}\ln 2\ d\phi$

$\Rightarrow \int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi ={\frac {1}{2}}[\phi \ln 2]_{0}^{\frac {\pi }{4}}$

$\Rightarrow \int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi ={\frac {1}{2}}[{\frac {\pi }{4}}\ln 2-0\ln 2]$

$\Rightarrow \int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi ={\frac {1}{2}}[{\frac {\pi }{4}}\ln 2]$

$\Rightarrow \int _{0}^{\frac {\pi }{4}}\ln(1+\tan \phi )\ d\phi ={\frac {\pi }{8}}\ln 2$

Since that would all work so nicely, I am convinced that I must have missed out a minus sign at some point but I'm not sure where. Was it when I switched the limits between which I had to integrate?

If my mistake is missing a minus please tell me; if it's some other mistake then could you please tell me that it's not the minus problem but don't tell me what I have actually done wrong. I'd like to try and do as much of this by myself as possible. Thanks 92.1.69.95 (talk) 12:37, 7 August 2008 (UTC)

Near the beginning of your working, just after your substitution, you have

$-\int _{\frac {\pi }{4}}^{0}\ln(1+\tan({\frac {\pi }{4}}-\phi ))\ d\phi =-\int _{0}^{\frac {\pi }{4}}\ln(1+\tan({\frac {\pi }{4}}-\phi ))\ d\phi$

If you swap limits you must multiply the integral by -1 as well. so this should be

$-\int _{\frac {\pi }{4}}^{0}\ln(1+\tan({\frac {\pi }{4}}-\phi ))\ d\phi =\int _{0}^{\frac {\pi }{4}}\ln(1+\tan({\frac {\pi }{4}}-\phi ))\ d\phi$

which gives you the reversal in signs that you are looking for. Gandalf61 (talk) 13:13, 7 August 2008 (UTC)

Exponentially losing viewers

A newspaper story yesterday said a TV show "has, over the past few seasons, been exponentially losing viewers".

The story is HERE, but it is not necessary to read the story in order to understand the question.

I know that the word "exponentially" has come into wide use to mean "very rapid increase", whether or not the increase is exponential in the mathematical sense of the word.

What I'm wondering is whether the idea of "exponential decline" makes any sense mathematically. Can a finite number (such as number of viewers) which can at most decrease to zero, decrease exponentially?

Thanks, Wanderer57 (talk) 13:44, 7 August 2008 (UTC)

It can certainly do so approximately, as radioactive particles do (see exponential decay). It would mean that the rate of loss of viewers was constantly decreasing, in such a way that the half-life remains constant. Of course, this would break down with the last few people, but it could be a sensible approximation until then. I expect that article's just talking nonsense though. Algebraist 13:54, 7 August 2008 (UTC)

Interestingly, the exponential decay of particles also breaks down for the last few particles. The chances that the writer actually ran the data through a curve-fitting program and determined that the resulting curve approximates an exponential rate of decay is almost zero. However, I suppose we could come up with a reason for a non-linear rate of decline, if we suppose that everyone who stops watching tells their freinds it's now uncool to watch, and those friends then stop watching and tell their friends. I would guess that a decreasing S-curve would be more likely to match the rate of decline, as that would start with a plateau, then have a slight rate of decrease, then a large rate of decrease, then again a slow rate of decrease, dropping to a lower plateau of loyal viewers who will stay with the program no matter what. That lower level may or may not be enough viewers for the program to remain profitable. StuRat (talk) 14:12, 7 August 2008 (UTC)

Thank you both for the feedback. If I understand this correctly, a show that was "losing viewers exponentially" could be losing viewers at a very high rate (say 50% per week) OR losing viewers very gradually (1% per year), OR anything in between. To put this another way, "losing viewers exponentially" by itself says very little about the rate of loss.

Did I get that right? Wanderer57 (talk) 17:13, 7 August 2008 (UTC)

Yes. All it says is that the rate of loss varies proportionally to the number of viewers left. Algebraist 17:15, 7 August 2008 (UTC)

It is worth noting that "exponentially" is often used incorrectly by the lay public as a synonym for "fast". Hence when a news reporter says something like "exponentially losing viewers" I tend to assume what they really mean is "rapidly losing viewers". Dragons flight (talk) 18:25, 7 August 2008 (UTC)

The decrease was monitored only the last few seasons, and it is not certain that the rate of decrease will remain contant in the future. It may not be a good idea to write that we lost a certain number of viewers per year, implying a linear decay, because a linearly decreasing function must from a certain point of time become negative, while the number of viewers cannot be a negative number. It may be better to write that we lost a certain fraction or percentage of the viewers each year, implying an exponential decay, although an exponentially decreasing function take fractional values and never becomes zero, while the number of viewers take integer values only and might eventually become zero. But 'linear' or 'exponential' says nothing about the actual decay, only about how to describe it, and about how to forecast the future number of viewers based on it. Bo Jacoby (talk) 22:20, 7 August 2008 (UTC).

Matrices

Can anyone tell me the history of matrices: who invented them and who found out their magical properties regarding application to certain physical problems? —Preceding unsigned comment added by 79.76.225.183 (talk) 23:24, 7 August 2008 (UTC)

Since you linked to the matrix disambiguation page, I will assume you haven't looked at the article: matrix (mathematics). Come back if you still have questions after reading the article. Jkasd 00:55, 8 August 2008 (UTC)

Im interested in the philosophy of matrices and why they appear to be so powerful in soling all sorts of problems. Is it just coincidence that this construct is so universally useful? —Preceding unsigned comment added by 79.76.225.183 (talk) 01:30, 8 August 2008 (UTC)

Matrices are really just a way of representing linear transformations between vector spaces. That's the reason, for example, for the (on its face, rather peculiar!) rule for multiplying them—it corresponds to taking the composition of the corresponding linear transformations (that is, do one transformation, then the other). --Trovatore (talk) 01:33, 8 August 2008 (UTC)

August 8

Gaussian function

I'm trying to prove to myself something at the Gaussian function article. The article says that the Gaussian integral

$\int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}$

and that therefore

$\int _{-\infty }^{\infty }ae^{-{(x-b)^{2} \over 2c^{2}}}\,dx=ac\cdot {\sqrt {2\pi }}$

I understand why the a showed up in the answer: because it's a constant that was inserted after the integral sign. But I'm not sure how you would arrive at the second equation given the first equivalency (in other words, how you would express the left side of second equation in terms of the left side of the first). Could someone lead me through it? (I'm the family member of a scientist who wants to understand this for the purpose of their research.) Thanks! —anon —Preceding unsigned comment added by 70.23.85.94 (talk) 02:44, 8 August 2008 (UTC)

You need to use integration by substitution. Specifically, set $u={\frac {x-b}{c{\sqrt {2}}}}$. Algebraist 02:51, 8 August 2008 (UTC)

By the way, from a geometrical point of view, this is completely obvious. Regarding the first integral as the area under the graph of $e^{-x^{2}}$, we obtain the second by shifting the graph horizontally a distance b (which doesn't change the area), stretching it vertically by a factor a (which multiplies the area by a), and the stretching it horizontally by a factor $c{\sqrt {2}}$ (which multiplies the area by $c{\sqrt {2}}$). Algebraist 03:09, 8 August 2008 (UTC)

Thanks so much for the u-substitution; I understand perfectly now. And thanks for the geometrical insight as well. —Preceding unsigned comment added by 70.23.85.94 (talk) 03:27, 8 August 2008 (UTC)

Sudoku minimum and maximum for fixed outcome

I was wondering what is a minimum number of digits that must be filled in in a sudoku grid before there is only one solution and, considering that it varies depending on where they are placed, what is the maximum number that can be filled in and still leave the puzzle with more than one solution. I was trying to figure it out on paper but got nowhere, and don't have the math to do it except by trial and error.--70.107.9.159 (talk) 04:25, 8 August 2008 (UTC)

The most that can be given without specifying the solution is all but four. Your other question is massively harder, and the answer is not known. The least anyone's come up with is 17 (in thousands of different ways). See mathematics of Sudoku for more. Algebraist 04:29, 8 August 2008 (UTC)

The Dilogarithm

I am trying to evaluate
$\operatorname {Li} _{2}(-1)=\int _{-1}^{0}{\frac {\ln(1-t)}{t}}dt=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k^{2}}}=-{\frac {\pi ^{2}}{12}}$
whose exact value is already well known. So, I picked the integral and tried so solve it by making the substitution u=1-t, du=-dt and transforming the integral to
$\int _{-1}^{0}{\frac {\ln(1-t)}{t}}dt=\int _{1}^{2}{\frac {\ln(u)}{1-u}}du=\int _{1}^{2}\ln(u)-{\frac {\ln(u)}{u}}du=\left[u\ln(u)-u-{\frac {(\ln(u))^{2}}{2}}\right]_{1}^{2}$
which comes out to an exact value of $\ln(4)-1-{\frac {(\ln(2))^{2}}{2}}$. What am I doing wrong?--A Real Kaiser...NOT! (talk) 05:55, 8 August 2008 (UTC)

$\int _{1}^{2}{\frac {\ln(u)}{1-u}}du=\int _{1}^{2}\ln(u)-{\frac {\ln(u)}{u}}du$ is bogus. ${\frac {a}{b+c}}\neq {\frac {a}{b}}+{\frac {a}{c}}$ --tcsetattr (talk / contribs) 06:06, 8 August 2008 (UTC)

UGGGGGHHHHHHHHHHHHHHH, I am so blind. I can't believe that I made an algebra one mistake.--A Real Kaiser...NOT! (talk) 07:28, 8 August 2008 (UTC)

Cosmological constant influencing the moon and pioneer orbits?

The article on the Kepler_problem_in_general_relativity assumes that the cosmological constant is zero. What happens to the field and to the orbit if this is not the case? Is there a connection to the pioneer anomaly? Bo Jacoby (talk) 07:53, 8 August 2008 (UTC).