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September 4

12 people in a family, all with birthdays in different months

Say there's a family of 12 people: 2 parents and 10 children. I'm told all the kids came from single births. Someone claims that none of the 12 have a birthday in the same month as any of the others, i.e. there's exactly one birthday for each of the 12 months. How likely is this claim to be true? -- JackofOz (talk) 21:26, 4 September 2009 (UTC)

As a rough estimate if you took 12 random people and assumed randomly distributed birthdays among the 12 months (which they aren't), the probability of them having one birthday per month is 12!/12¹², which is about 1 in 19000. That's not to say it's a lie. Unusual things do happen some of the time, and when they do there's typically disproportionate attention brought to them, which can make them seem like they happen more often then they rightfully should. I think there's probably a name for that effect but I don't know what it is. Rckrone (talk) 21:52, 4 September 2009 (UTC)

That is "how likely it is to be true" if the family were chosen uniformly from among all families with 12 children.

Rckrone wrote:

assumed randomly distributed birthdays among the 12 months (which they aren't)

Here's another instance of the use of "randomly" to mean "uniformly". This is an oddly persistent meme. I don't know why. Michael Hardy (talk) 22:09, 4 September 2009 (UTC)

What I meant to say was that 1/19000 is a rough estimate of the probability of this happening among 12 random people. I agree that the probability that the claim is true is much much higher. In fact I don't see any reason to think it isn't. As for misusing "randomly" I have no explanation. Rckrone (talk) 22:17, 4 September 2009 (UTC)

Suppose there are n families with 2 parents and exactly 10 children, and the 12 birthdays are uniformly distributed in the 12 months. The probability of at least one occurrence of one birthday for each of the 12 months is 1 - (1-12!/12¹²)ⁿ = 1 - 0.999946276783ⁿ. This passes 50% at n = 12902, and it is 99.999999999999999999999% at n = 1000000. I don't know how many such families there are and how far from uniform distribution the birthdays really are but it seems highly likely that there exists such an occurrence. PrimeHunter (talk) 23:33, 4 September 2009 (UTC)

Oh, I have no doubt it has happened many times, because it's obviously well within the realms of possibility. But if I were a betting man, and someone asserted this particular family had this characteristic, I was wanting to know whether it would be a better bet to lay money they were telling the truth, or they were telling a lie (or, to be charitable, they were mistaken). That may not be exactly what I originally asked, but it's what I meant. -- JackofOz (talk) 23:43, 4 September 2009 (UTC)

It would depend on the reliability of the person making the claim. If they knew the family and I didn't then I would be reluctant to offer a bet to them. If they accept the bet then it may be because they feel fairly confident they are right. Almost everybody knows some things which are rare, and rare things often draw attention and may be told many times. PrimeHunter (talk) 00:13, 5 September 2009 (UTC)

"the use of "randomly" to mean "uniformly". ... I don't know why". The article randomness explains that the word "random" refers to a discrete uniform distribution: "Governed by or involving equal chances for each of the actual or hypothetical members of a population". Mathematicians change the meaning of words to include special cases, such that a coin with two tails still produces a random variable even if the outcome is neither random nor variable, but predictable and constant. Bo Jacoby (talk) 06:43, 5 September 2009 (UTC).

The 12!/12¹² calculation assumes that the birthdays are independent of one another. I'd say that it would be hard work to establish/defend any statement of independence here. Robinh (talk) 09:20, 5 September 2009 (UTC)

That's certainly true (and even the assumption that the distribution is uniform measured by months is weird and not supported by the principle of indifference -- we know, after all, that some months are longer than others). But as a practical matter, to get a first approximation without doing too much work, it's a reasonable guess. --Trovatore (talk) 20:59, 5 September 2009 (UTC)

For a busy family like that, independence is the last thing I'd assume. Mom gets pregnant in month X, kid #1 is born 9 months later, Mom spends 2 months nursing and getting her parts back to normal, and at month X+11 gets pregnant again. Kid #2 is born at X+20, rinse, repeat. A new pregnancy every 11 months, and since 11 (months between pregnancy starts) and 12 (months in a year) are relatively prime, you get all the births in distinct months. See also Cheaper by the Dozen. 67.122.211.205 (talk) 07:49, 6 September 2009 (UTC)

• "Uniformly distributed" does not mean the same as discrete uniform distribution to most readers. Surely "randomly distributed" is less likely to be misinterpreted? Both are unlikely in these circumstances. Dbfirs 09:56, 6 September 2009 (UTC)

Here's a point that nobody mentioned. What is the likelihood of conscious manipulation of the birth months by the parents? Some fertile couples might make it a point to create such a family, and then all purely mathematical considerations are totally out the window (which they probably should have been from the beginning on other grounds). It makes a nice little exercise if you stipulate a uniform distribution, but another question should be put in its place unless you want a very complex mostly non-mathematical discussion.Julzes (talk) 01:06, 8 September 2009 (UTC)

Canonical Forms for PDEs

Okay so working with 2nd order linear PDEs, I cam across the fact that there are canonical forms for different kinds of PDEs. For hyperbolic PDEs, the canonical forms are $u_{xx}-u_{tt}=f(x,t,u,u_{x},u_{t})$ or $u_{xt}=g(x,t,u,u_{x},u_{t})$. My question is why are there two canonical forms for hyperbolic PDEs? Is there a significance to this distinction? Thinking with analogies to conics sections, I think that we just "rotate" one form by 45 degrees to get another form. Why do we do that? Is one easier to solve in some cases than the other?--97.118.56.41 (talk) 22:37, 4 September 2009 (UTC)

This is a very nice question. There are some papers by Farid Tari, and some other co-authors, that address the classifications of implicit differential equations.

1. "Bifurcations of binary differential equations." Proc. Roy. Soc. Edinburgh Sect. A 130 (2000), 485–506. (With J. W. Bruce and G. J. Fletcher)
2. "Duality and implicit differential equations", Nonlinearity 13 (2000), 791–812. (With J. W. Bruce)
3. "Two-parameter families of implicit differential equations", Discrete Contin. Dynam. Systems, 13 (2005), 139–262.
4. "Two-parameter families of binary differential equations", Discrete Contin. Dynam. Systems, 22 (2008), 759–789.

Maybe these will help ~~ Dr Dec (Talk) ~~ 09:57, 5 September 2009 (UTC)

It's nothing special. As you know, both form are very common in applications. E.g. as you know, waves and oscillations are described by the first one. As you know, $u_{tx}=0$ is easier to solve, in fact immediate. And, as you know, one can easily pass from one form to the other. So, if you already know the answer, why are you asking? ;) (@Dec: the OP's eq. are semilinear hyperbolic equations, there's nothing implicit!)--pma (talk) 18:18, 5 September 2009 (UTC)

No, but it's a question of classification. Such classifications always follow similar lines. I thought it might have been of some interest to him/her. ~~ Dr Dec (Talk) ~~ 20:24, 5 September 2009 (UTC)

September 5

The Sorgenfrey plane

Ok I know how the usual proof of non-normality goes, but I've been thinking about something that smells fishy to me. Take the line y=-x. The proof goes by showing the set of all rational points and irrationals can't be separated by open sets using Baire's category theorem. But what if we do this: for every rational p/q, cover the point on the line with [p/q,p/q+1/(q^q)) X [-p/q,-p/q+1(q^q)), where X denotes cartesian product. Notice how this half open square is VERY small. It certainly looks plausible that every irrational has a half open square (with the irrational at the bottom left corner) that's disjoint from the union of the rational squares, because every irrational is surrounded by a cloud of rationals with very large denominator. I've been trying to show this is not possible but it seems to have something to do with approximation of irrationals by rationals, continued fractions maybe (which I don't know much about). What am I doing wrong? Breath of the Dying (talk) 09:00, 5 September 2009 (UTC)

Eh, only quadratic irrationals like square roots can be approximated by continued fractions. (At least if you mean the conventional type with only 1's as numerators.) Professor M. Fiendish, Esq. 09:17, 5 September 2009 (UTC)

That is not true. Algebraist 12:16, 5 September 2009 (UTC)

You were probably thinking about the fact that quadratic irrationals have a repeating continued fraction expansion. -- Meni Rosenfeld (talk) 17:15, 5 September 2009 (UTC)

That particular scheme will fail for irrational points that can be well-approximated by rationals. For example, setting $c=\sum _{j=1}^{\infty }10^{-j!}$, any neighbourhood of (c,-c) intersects your neighbourhood of the rational points. Algebraist 12:22, 5 September 2009 (UTC)

As to the first question, as you recalled: in this topology, the closure of any neighborhood of the closed set $\{(r,-r):r\in \mathbb {Q} \}$, meets the anti-diagonal in uncountably many points. If this is OK for you, I really don't see why it is not OK for you that in particular, the nbd you wrote has this property.--pma (talk) 17:58, 5 September 2009 (UTC)

Great retrosnub icosidodecahedron

Great retrosnub icosidodecahedron.

I keep seeing conflicting vertex figures for this one.

1. Some sources say the vertfig is 3.³/₂.3.⁵/₃.3.
2. Other sources say it is ³/₂.³/₂.³/₂.⁵/₃.³/₂.
3. And still others say it is ^{(3.3.3.3.5/₂)}/₂.

Which is right?

On Wikipedia, the vertfig shown is ^{(3.3.3.3.5/₂)}/₂, but application of the Wythoff symbol |³/₂ ⁵/₃ 2 gives 3.³/₂.3.⁵/₃.3. No idea where ³/₂.³/₂.³/₂.⁵/₃.³/₂ came from. Professor M. Fiendish, Esq. 09:04, 5 September 2009 (UTC)

Just going by the vertex figure image there, which I'm assuming is correct, it looks to be a pentagram with the faces being four triangles and one pentagram so I guess the problem is how do you represent that with the notation? Just listing the 5 faces would be ambiguous with the case where they're arranged in a pentagon, so wikipedia handles that with the "/2" at the end to represent 5/2 instead of 5. It looks like another way is to specify the orientation of the faces relative to the vertex. 3/2 is a triangle but connecting the vertices in the opposite order, and similarly 5/3 is an inverted pentagram. I'm not really sure on how the orientations dictate the shape of the vertex figure, but it could be that those other two versions you mentioned both represent the same thing. Vertex configuration seems to explain. Rckrone (talk) 18:04, 5 September 2009 (UTC)

Wouldn't ³/₂.³/₂.³/₂.⁵/₃.³/₂ (all numbers <2) be indistinguishable from 3.3.3.3.⁵/₂? That is, if all the faces are backward, isn't it the same as if all are forward? —Tamfang (talk) 07:08, 10 September 2009 (UTC)

I agree with you there. Unless it turns the polyhedron inside-out or something :-). I think the first has a problem because 3.3/2 would have 2 Δs over each other. Professor M. Fiendish, Esq. 03:18, 13 September 2009 (UTC)

Uniform Continuity of sin(x³)/(x+1)

Hi again guys, sorry about the influx of analysis from me on here, I'm working through a bundle of questions and saving the ones which really stump me for here! I'm trying to work out whether ${\frac {\sin(x^{3})}{x+1}}$ is uniformly continuous over $[0,\infty )$ and I have no idea how to begin. The fact sin(x³) has a derivative tending to infinity leads me to think it might not be, but then the function decreases in magnitude to 0 too which may affect things, and I'm unsure even what answer to expect, let alone how to proceed.

Any help would be greatly appreciated!

Thanks a lot, Spamalert101 (talk) 14:32, 5 September 2009 (UTC)

It's pointwise continuous, and tends to a limit as x tends to infinity, so it's uniformly continuous. This is a variant of the Heine–Cantor theorem. This is very similar to your last question: in both cases we have a function with good local behaviour that is badly behaved at infinity, so multiplying it by a function that decays to 0 at infinity gives good global behaviour. Algebraist 14:54, 5 September 2009 (UTC)

And to make it quantitative, you may try and prove that your function is Hölder continuous of exponent 1/2: for instance, $\scriptstyle |f(x)-f(y)|\leq 5{\sqrt {|x-y|}}$ holds true for all $x$ and $y$ in $\scriptstyle [0,\infty )$. --pma (talk) 20:50, 5 September 2009 (UTC)

The restriction to any bounded interval [0, a] is a continuous function on a compact set and so is uniformly continuous, and the restriction to [a, ∞] only takes values near 0, so it can't vary by more than ε if a is big enough. Michael Hardy (talk) 20:58, 5 September 2009 (UTC)

Properties of an isosceles triangle

Let ABC be triangle with AB=AC=6 cm . If the circumradius of the triangle is 5 cm, find the length of BC . —Preceding unsigned comment added by 117.98.100.98 (talk) 16:44, 5 September 2009 (UTC)

This looks a lot like a homework problem. I will give you this hint, though: if you draw the triangle with its circumscribed circle, you will notice that the line between the vertex A of the triangle and the center of the circle is perpendicular to the side BC of the triangle. When you get your answer and punch it into your calculator (which you shouldn't do until the last step), the 3rd and 4th digits after the decimal should be 6 and 9 respectively, assuming I did my math correctly.

Note that my approach was thrown together from very faint recollections from my geometry class many years ago, and may not be the way your teacher is expecting you to solve it. --COVIZAPIBETEFOKY (talk) 18:12, 5 September 2009 (UTC)

"...the 3rd and 4th digits after the decimal should be 6 and 9 respectively, assuming I did my math correctly." Of course, I found a mistake in my work. The answer after correcting my work still includes both of those dirty digits, though; just not in the same place. --COVIZAPIBETEFOKY (talk) 18:20, 5 September 2009 (UTC)

A hint: Use Heron's formula. Bo Jacoby (talk) 18:22, 5 September 2009 (UTC).

There certainly may be something I'm missing here, but I'm pretty sure this leads nowhere. The area by Heron's formula and the area given by (1/2)base*height are exactly the same expressions, so setting those two equal certainly doesn't really help at all. Was there something else you had in mind? --COVIZAPIBETEFOKY (talk) 18:36, 5 September 2009 (UTC)

You can use law of sines and the area formula Δ=ab*sinC/2 to relate the circum-diameter to the area. Anyway, Circumscribed circle lists a formula that can be applied directly to this problem. Rckrone (talk) 18:54, 5 September 2009 (UTC)

Here is a simple approach to the problem, based on properties of triangles you should have already seen in school.

• Calling the center of the circumcircle O, note that the line AO bisects BC at a right angle, at say point D (see isosceles triangle).
• Note that triangle AOB is also an isosceles triangle, and the angle bisector of angle AOB bisects the line AB at right angles, at point E. Can you determine the lengths of sides AO, OB, AE, EB and OE ?
• Now what can you say about triangles AOE and ABD ? How do you use that to calculate length of side DB and hence, BC ?

Hint: Look up similar triangles and pythagoras theorem. Also, no calulator is required to solve this problem! 98.220.252.228 (talk) 18:46, 5 September 2009 (UTC)

Triangle ABO has sides 5,5,6. s=(5+5+6)/2=8. Area according to Heron is √(8·3·3·2)=12. This is = (1/2)base*height = (1/2)·5·(|BC|/2). Solve to get |BC|=48/5=9.6 . This, COVIZAPIBETEFOKY, is what I had in mind. Bo Jacoby (talk) 20:54, 5 September 2009 (UTC).

I'm guessing that by "isoscale" you mean "isosceles". Remember that if angles α and β are opposite sides a and b respectively, then

${\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}={\text{circumdiameter}}$

(see law of sines). So you have

${\frac {6}{\sin \beta }}=10,$

and hence

$\sin \beta ={\frac {3}{5}}.$

Since two angles are β and one is α and they add up to a half-circle, that gives you α. Then you've got

${\frac {a}{\sin \alpha }}=10$

so you can find a. Michael Hardy (talk) 21:11, 5 September 2009 (UTC)

...and I just noticed the arithmetic is really simple: you don't need numerical approximations from a calculator, since sin β = 3/5. Consequently cos β = √(1 − (3/5)²) and as luck would have it, we've got a Pythagorean triple and so cos β = 4/5 exactly. Therefore sin(β + β) = 2 sin β cos β = 2(3/5)(4/5) = 24/25. Since α + β + β = half-circle, we must have sin α = sin(β + β). So that sine is 24/25. SSo we get a = 48/5 exactly. Michael Hardy (talk) 21:23, 5 September 2009 (UTC)

How nice that Michael's and mine seemingly different roads lead us to exactly the same result. Bo Jacoby (talk) 06:09, 6 September 2009 (UTC).

35 should have been 25. -- Meni Rosenfeld (talk) 07:14, 6 September 2009 (UTC)

In my elementary school times I formed the opinion that almost all rectangular triangles have edges in the ratio (3:4:5). --pma (talk) 21:33, 6 September 2009 (UTC)

Little did you know that in time, you would learn that this is indeed true - for an appropriate measure on the set of all triangles, of course. :) -- Meni Rosenfeld (talk) 19:44, 7 September 2009 (UTC)

Dense in C

Resolved

--Shahab (talk) 19:41, 6 September 2009 (UTC)

I want to show that $\mathbb {Z} [{\frac {i}{2}}]$ is dense in $\mathbb {C}$, the complex plane. By $\mathbb {Z} [{\frac {i}{2}}]$ I mean the set of all evaluations of polynomials with integer coefficients at ${\frac {i}{2}}$ (where $i=(0,1)$). How can I show that any open ball around an arbitrary z contains infinitely many such evaluations?--Shahab (talk) 19:48, 5 September 2009 (UTC)

Let r be an arbitrary positive real number. We need to show that you can find a polynomial that evaluates to a number within a distance of r from z. Choose integer n such that ${\frac {1}{2^{n}}}<{\frac {r}{\sqrt {2}}}$ and n=0 (mod 4). Then choose integers p and q such that ${\frac {p}{2^{n}}}+{\frac {iq}{2^{n+1}}}$ is a close to z as possible. That number is $pX^{n}+qX^{n+1}$ evaluated at i/2 and is less than r from z. --Tango (talk) 20:34, 5 September 2009 (UTC)

Thanks. I have a doubt though. How do we choose p and q such that ${\frac {p}{2^{n}}}+{\frac {iq}{2^{n+1}}}$ is as close to z as possible? I understand choosing n by the Archimedian property. Similarly what property do we use when choosing p and q?--Shahab (talk) 20:55, 5 September 2009 (UTC)

To get p, say, we take the integer part of the real part of z, call it a. Then test a2ⁿ+b for b in {0,1,...,2ⁿ-1} and choose the closest. It is a finite number of trials, so you will definitely find a closest. Do the same for q and the job is done. --Tango (talk) 21:01, 5 September 2009 (UTC)

Nice. You may also say: p and q such that 2p+iq is as close as possible to 2ⁿ⁺¹z (a minimum distance that is certainly not greater than $\scriptstyle {\sqrt {5}}/2$, &c) However as soon as you've observed that Z[i/2] contains all numbers p/2ⁿ+iq/2ⁿ it's done.--pma (talk) 21:12, 5 September 2009 (UTC)

Yes, that is pretty much equivalent. --Tango (talk) 21:42, 5 September 2009 (UTC)

Thank you both.--Shahab (talk) 19:41, 6 September 2009 (UTC)

Divisor of a Riemann surface

Can someone please explain what is meant by a divisor on a compact connected Riemann surface? I cannot find the relevant Wikipedia article (rather unusual). 86.210.207.133 (talk) 23:17, 5 September 2009 (UTC)

Divisor (algebraic geometry) is the relevant article, I think. --Tango (talk) 23:44, 5 September 2009 (UTC)

Grrrrr, no mention of Riemann surfaces... 86.210.207.133 (talk) 01:15, 6 September 2009 (UTC)

Any compact Riemann surface is a projective variety. --PST 03:57, 6 September 2009 (UTC)

In this case (because Riemann surfaces have complex dimension 1) a divisor is just a function $D\colon X\to \mathbb {Z}$ with discrete support, ie for a compact Riemann surface it's just a finite (formal) linear combination of points on the Riemann surface. --Xedi^Talk 04:54, 7 September 2009 (UTC)

Function elements

Is there a Wikipedia page concerning function elements?

Mathworld has one... http://mathworld.wolfram.com/FunctionElement.html —Preceding unsigned comment added by 86.210.207.133 (talk) 23:33, 5 September 2009 (UTC)

Have you tried searching Function elements? Intelligentsium 02:27, 6 September 2009 (UTC)

September 6

Evaluating a probability estimator

Hi. I have random variables X and Y, where X is nominal and Y takes values of 0 and 1. I have n observations of the form $(X_{i},Y_{i})\,\!$, where several observations may share the same X. I denote $p=\mathrm {Pr} (Y=1|X)\,\!$ (so p is a random variable which becomes determined when given X), and ${\hat {p}}=f(X)$ is an estimator for p. I want to know how good this estimator is - specifically, I want to approximately know $\mathbb {E} ((p-{\hat {p}})^{2}).$ Is there a standard way of doing this?

I think I have reduced the problem to that of estimating $\mathbb {E} (p^{2})$, in case this is any simpler. Thanks. -- Meni Rosenfeld (talk) 16:42, 6 September 2009 (UTC)

Actually I had an idea for this, but originally I thought it would be computationally intractable. After looking into it I see it's not that bad. The idea is to assume that p follows the beta distribution, find the MLE of the distribution parameters given the data, and calculate the mean and variance for those parameters. Does this sound reasonable? Is there a better way? -- Meni Rosenfeld (talk) 18:06, 6 September 2009 (UTC)

You might be interested in the following article: http://psychology.wikia.com/wiki/Inferential_statistics . Bo Jacoby (talk) 20:08, 6 September 2009 (UTC).

Ok, thanks. -- Meni Rosenfeld (talk) 18:39, 8 September 2009 (UTC)

Automorphisms of Z[x]

What are all the ring automorphisms of $\mathbb {Z} [x]$? I believe Z has only one: the identity. What is the general procedure of approaching the problem of determining Aut(R)? Thanks.--Shahab (talk) 22:01, 6 September 2009 (UTC)

Yes, the (unital) ring automorphism is determined by where you send x. x needs to be in the image, so x has to go to something of degree 1. I suspect it can go to ±x+n for any n in Z, and to no other place. If so, this is the infinite dihedral group, represented as the matrices { [ ±1, n ; 0, 1 ] : n in Z } under matrix multiplication. In general finding the automorphism group of fundamental, innocuous rings can be the source of long-standing open problems, such as (the strong) Nagata conjecture and the description of the Cremona group. I wrote a short thing on this a while back (the enwiki article is just a sentence). JackSchmidt (talk) 22:47, 6 September 2009 (UTC)

Well, it's clear the leading coefficient of the image of x must be ±1. If it is ax + n, then a poly in the image of the automorphism can be written as aQ(x) + constant, so it can't be surjective unless a is a unit.John Z (talk) 00:30, 7 September 2009 (UTC)

September 7

Latex align* multiple points of alignment.

Resolved

I am trying to type up a displayed equation for which I want to have multiple points of alignment, one at equals and one at the point where I give reasons for each line. I have tried doing it like this:

  \begin{align*}
     \frac{x}{y} + \frac{a}{b} &= xy^{-1} + ab^{-1}  & ( \text{by definition})\\
        &= xy^{-1}1 + ab^{-1}1  & (\text{where 1 is the multiplicative identity})\\
        &= xy^{-1}bb^{-1} + ab^{-1}yy^{-1}  & (\text{inverse property of mulitplication } bb^{-1} = yy^{-1} = 1) \\
  \end{align*}

But the problem is that the reasons get right aligned whereas I want them to be left aligned, preferable such that there is at least a \quad spacing between the equation and the reason. Is there any way to do this? AMorris (talk)●(contribs) 04:41, 7 September 2009 (UTC)

The default alignment for the align environment is left right left right... so it alternates left and right. You can just put two & before the text instead of one, making it align to the left. --Xedi^Talk 05:56, 7 September 2009 (UTC)

To expand on Xedi's answer:

  \begin{align*}
     \frac{x}{y} + \frac{a}{b} &= xy^{-1} + ab^{-1}  && ( \text{by definition})\\
        &= xy^{-1}1 + ab^{-1}1  && (\text{where 1 is the multiplicative identity})\\
        &= xy^{-1}bb^{-1} + ab^{-1}yy^{-1}  && (\text{inverse property of mulitplication } bb^{-1} = yy^{-1} = 1) \\
  \end{align*}

The idea is that the &s alternate between being /alignment markers/ and /column separators/. So the first & means "align the first column at the = sign". The second & delimits the end of the first column and the start of the second column. The third & means "align the second column at the (". And so on. Eric. 67.169.125.37 (talk) 07:57, 7 September 2009 (UTC)

Perfect - that sounds like it will do exactly what I wanted. It was frustrating because I couldn't find this documentation anywhere when I looked up how align works. Thanks guys! AMorris (talk)●(contribs) 12:05, 7 September 2009 (UTC)

Yeah, Latex is a pain. Sometimes you just have to know the right magic incantation, or know about some obscure behavior that you'd never think to look up. Eric. 67.169.125.37 (talk) 19:46, 7 September 2009 (UTC)

Oh, now that I looked at the align again, I seem to recall that having a \\ on the last line gives you some extra whitespace after the equations that you probably don't want. You can see for yourself whether you like it. Eric. 216.27.191.178 (talk) 23:06, 7 September 2009 (UTC)

Reduction to Canonical Form

Hello, I have been trying to reduce $2u_{xx}+u_{xy}+yu_{yy}=0$ in the region y>1 for the last couple of hours but not getting anywhere. In this region, the PDE is elliptic. Letting a=2, b=1, and c=y, the discriminant is negative. So I set

${\frac {dy}{dx}}={\frac {b+{\sqrt {b^{2}-4ac}}}{2a}}={\frac {1+{\sqrt {1-8y}}}{4}}={\frac {1+i{\sqrt {8y-1}}}{4}}$

and (I think) that I am supposed to solve this differential equation and then let $\phi (x,y)$ be the real part of the solution and let $\psi (x,y)$ be the imaginary part and then I will have my change of coordinates which will reduce this PDE to the canonical form (the Laplacian equals some function). But the problem is that when I try to do this integral I get

$-i{\sqrt {8y-1}}+\ln(-i+{\sqrt {8y-1}})$

which I can't separate into the real and imaginary part. And then when I tried to integrate the real and the imaginary part separately (which is correct I hope) and use the results as the transformation, it doesn't work. I let ${\frac {dy}{dx}}=1/4$ giving me $\phi (x,y)=4y-x$ and then ${\frac {dy}{dx}}={\frac {\sqrt {8y-1}}{4}}$ giving me $\psi (x,y)={\sqrt {8y-1}}-x$. The mixed partials don't go away and also it is a nightmare when I try to invert it. What am I doing wrong? Is there an easier way of doing this or am I making some stupid mistake? Thanks! 97.118.56.41 (talk) 07:24, 7 September 2009 (UTC)

You said that you can't seperate into the real and the imaginary part. Well, if

$z:=-i{\sqrt {8y-1}}+\ln(-i+{\sqrt {8y-1}})$

as you say then we can split this into it's real and imaginary parts:

$\Re (z)=\ln \left(\left|-i+{\sqrt {8y-1}}\right|\right)+\Im \left({\sqrt {8y-1}}\right)\ ,$

$\Im (z)={\mbox{Arg}}\left(-i+{\sqrt {8y-1}}\right)-\Re \left({\sqrt {8y-1}}\right)\ .$

Now, if we assume that y > 1 (in fact y > ⅛ would do) then we get

$\Re (z)=\ln \left(2{\sqrt {2y}}\right)\ ,$

$\Im (z)=-{\sqrt {8y-1}}-\arctan \left({\frac {1}{\sqrt {8y-1}}}\right).$

(I've ignored the multi-valuedness and just worked with principal values) ~~ Dr Dec (Talk) ~~ 09:59, 7 September 2009 (UTC)

FRACTALS

210.212.239.181 (talk) 11:24, 7 September 2009 (UTC)harshagg

I want to know about fractals and their application and how to design them.I am just a beginner in this field even in wikipedia itself i can know about only their theory not use and i don't know can be that i couldn't understand what they say or mean to say.Pls help me.If its not against rule i can give my e-mail id so that u can contact me at for longer period of time.

There are many different types of fractals. Here at Wikipedia we have a general article on fractals, articles on certain types of fractals such as Julia sets, De Rham curves and iterated function systems, and even articles on individual notable fractals, such as the Mandelbrot set and the Koch snowflake. Our list of fractals by Hausdorff dimension gives a list of about 70 different individual fractals. The index of fractal-related articles gives a longer list of our articles about fractals. Gandalf61 (talk) 11:50, 7 September 2009 (UTC)

One nice fractal with a fairly elementary construction and plenty of (mathematical) applications is the Cantor set. -GTBacchus^(talk) 12:11, 7 September 2009 (UTC)

If you have MS Windows you might also like to try out the Fractint program. Dmcq (talk) 14:06, 7 September 2009 (UTC)

Irreducible representations of signed permutations (double young diagrams?)

Hello,

Young tableau describes how one can use Young tableaux to get all irreducible representations of the symmetric group. Another good text on that can be found here [1]

The symmetric group on $n+1$ symbols is precisely the finite Coxeter group $A_{n}$.

The Coxeter group $B_{n}$ consists of the $2^{n}n!$ signed permutations on the set $\{1,....n\}\cup \{-1,...,-n\}$ I heard that that group has a similar construction for its irreducible representations, using something like "double young diagrams". I can find very little on that on the internet. Can anyone tell me a bit more about this (even a few names that might help my search could be useful).

I am also interested in which irreducible characters appear in the decomposition of the characters of $B_{n}$, induced by parabolic subgroups. For $A_{n}$ this is completely known, and the coefficients are known as Kostka numbers.

Many thanks, Evilbu (Talk) unsigned by Evilbu

You seem to have made a series of statements; without posing a question. Are we to assume that your question is "Which are the irreducible characters appearing in the decomposition of the characters of $B_{n}$, induced by parabolic subgroups?" ~~ Dr Dec (Talk) ~~ 12:27, 7 September 2009 (UTC)

My sincere apologies. My main question is: how to describe the irreducible representations of that group $B_{n}$? My secondary question is: which of these appear (and if possible: how often) in the decomposition of the characters of $B_{n}$, induced by parabolic subgroups? Evilbu (Talk)

I don't know of any particularly special treatment of Bn, but the chapters on wreath products in Kerber's Representations of Permutation Groups is where I learned how to deal with Bn.

• Kerber, Adalbert (1971), Representations of permutation groups. I, Lecture Notes in Mathematics, Vol. 240, Berlin, New York: Springer-Verlag, doi:10.1007/BFb0067943, MR0325752
• Kerber, Adalbert (1975), Representations of permutation groups. II, Lecture Notes in Mathematics, Vol. 495, Berlin, New York: Springer-Verlag, doi:10.1007/BFb0085740, MR0409624
• Young, Alfred (1930), "On Quantitative Substitutional Analysis 5", Proceedings of the London Mathematical Society. Second Series, 31, doi:10.1112/plms/s2-31.1.273, ISSN 0024-6115, JFM 56.0135.02

The chapters in question are (Kerber 1971, Ch. 1 doi:10.1007/BFb0067945, Ch. 2 doi:10.1007/BFb0067946) and (Kerber 1975, Ch. 1 doi:10.1007/BFb0085742). The Coxeter group of type Bn is also known as the hyperoctahedral group, and is originally discussed in (Young 1930). JackSchmidt (talk) 16:35, 7 September 2009 (UTC)

A^n*B^n ≠ (A*B)^n for matrices (get around)

In the following series. F=Fo(A^n*B^n*(1-B)^0+Bicof*A^(n-1)*B^(n-1)*(1-B)^1......Bicof*A*B*(1-B)^(n-1)+(1-B)^n)

Bicof =(Binomial coeficient)

I would like to simplify this to Fo(AB+(1-B))^n. Using the binomial theorem

However A and B are both Matrices so A^n*B^n ≠ (A*B)^n. A is a upper triangular matrix and B is a diagonal matrix, with all elements less than 1.

Once that is done the function can be further simplified using the limit form of the exponential function and manipulating B.

Any ideas would be useful. The problem arose from tedious crusher breackage formulas. Startibartfast —Preceding unsigned comment added by 41.240.93.114 (talk) 15:09, 7 September 2009 (UTC)

This question has me totally blown away. What does Fo mean? What exactly does the binomial coefficient of a matrix-binomial power expansion mean? For example, for square matrices M and N we have: (M + N)² = M² + MN + NM + N². The binomial coefficients don't appear, correct me if I'm wrong, until we have a commutative algebra. ~~ Dr Dec (Talk) ~~ 16:05, 7 September 2009 (UTC)

I'd guess the OP wants something for binomials like the exponential map formula for Lie groups, but that's fairly hairy. There's an inversion formula for a sum of two matrices also which is probably somewhere on wiki but I don't think it would help either. Dmcq (talk) 16:53, 7 September 2009 (UTC)

Diagonal matrices commute with everything, don't they? Doesn't that mean that, in this case, A^n*B^n = (A*B)^n? --Tango (talk) 16:58, 7 September 2009 (UTC)

I thought that for a moment too, but it's not true:

$\left({\begin{array}{cc}1&2\\3&4\end{array}}\right)\left({\begin{array}{cc}1&0\\0&2\end{array}}\right)=\left({\begin{array}{cc}1&4\\3&8\end{array}}\right)$

$\left({\begin{array}{cc}1&0\\0&2\end{array}}\right)\left({\begin{array}{cc}1&2\\3&4\end{array}}\right)=\left({\begin{array}{cc}1&2\\6&8\end{array}}\right)$

~~ Dr Dec (Talk) ~~ 17:06, 7 September 2009 (UTC)

Scalar matrices commute with everything. Algebraist 17:13, 7 September 2009 (UTC)

Ah, yes. Sorry. --Tango (talk) 17:20, 7 September 2009 (UTC)

To the OP: if you rewrite your question in a more readable and understandable way, then I will think about it.--pma (talk) 18:45, 7 September 2009 (UTC)

A topological counterexample

Does there exist a compact T₂ first-countable separable space which is not second-countable? I would especially like it if the example was totally disconnected (i.e., does there exist an uncountable Boolean algebra B such that every ultrafilter in B is countable generated, and B − {0} is the union of countably many ultrafilters?) The Sorgenfrey line almost works, but it's not compact, only Lindelöf. — Emil J. 17:24, 7 September 2009 (UTC)

Counterexamples in Topology gives the weak parallel line topology as fulfilling all your conditions, including total disconnection. Algebraist 17:36, 7 September 2009 (UTC)

Excellent, thanks. This is what I was looking for. — Emil J. 09:45, 8 September 2009 (UTC)

I was about to suggest this book myself. ~~ Dr Dec (Talk) ~~ 17:41, 7 September 2009 (UTC)

debeat? —Preceding unsigned comment added by 92.8.193.46 (talk) 01:44, 8 September 2009 (UTC) debeat,debate,debeaten no? --pma (talk) 05:48, 8 September 2009 (UTC)

Base Conversion

I was wondering what happens when you convert to a base beyond 36. From what I understand after base 10 you use letters to fill in for the numbers (i.e in hexadecimal A is for 11, so on). But after base 36 you run out of letters. It wouldn't make much sense to express the decimal number 100 in base 37 as 2 26 (2*37 + 26*1). Thanks in advance 66.133.196.152 (talk) 18:26, 7 September 2009 (UTC)

You could make up more symbols, perhaps using another alphabet, or using both lowercase and uppercase letters. Or you could do what you say wouldn't make much sense, and represent each digit in decimal, with spaces (or some other non-numerical character) to separate the digits. Algebraist 18:32, 7 September 2009 (UTC)

[2] Base 64

Also Ascii85

Both from computer science for human input of big numbers..

They're common in URL's too eg youtube http://www.youtube.com/watch?v=wc8bLHznswA (probably base 64)

83.100.250.79 (talk) —Preceding undated comment added 18:57, 7 September 2009 (UTC).

(I fixed your base 37 number.) --Tardis (talk) 14:57, 9 September 2009 (UTC)

Presentation of dihedral group

I took abstract algebra 2 years ago but am sitting in on it this year. On Friday, the Professor said that $\langle r,s|r^{n}=1,sr=r^{-1}s\rangle$ is a presentation for $D_{2n}$, the dihedral group on an $n$-gon. A student asked if he needed also $s^{2}=1$ and he said he did not. The book, Dummit and Foote, lists it with $s^{2}=1$, as does the article Dihedral group, and the professor OFTEN makes mistakes, so I am just wondering if he is right. Thanks StatisticsMan (talk) 22:46, 7 September 2009 (UTC)

The professor is wrong. s has infinite order in his version. Algebraist 22:48, 7 September 2009 (UTC)

Indeed. Take, for example, the dihedral group D₆. This Mathworld article shows us that

$D_{6}=\langle x,y:x^{6}=y^{2}=1,xy=yx^{-1}\rangle .$

(In fact, if you're interested, the article gives you the conjugacy classes, the centre of D₆, the commutator subgroup, the abelianisation, the left cosets, and the character table of D₆) ~~ Dr Dec (Talk) ~~ 00:22, 8 September 2009 (UTC)

As far as I understand, a presentation is not unique. It is simply a set of generators and relations that imply the group MUST be the given group. So, that article says the given presentation must lead to D_6. But, it does not rule out that some presentation with less relations also must give D_6. I know that presentation leads to D_6 already. My question is, does the one with one less relation, given above, also lead to D_{2n}.

As far as Algebraist's comment, I do not understand that. Again, based on my understanding of presentations, it is not possible that the presentation I first listed gives that s has infinite order. The reasoning is simple, D_{2n} satisfies those generators and relations. So, that first presentation certainly does not imply that s has infinite order. It may be that there does exist some group with s having infinite order that satisfies those generators and relations. This would then prove that presentation does not give D_{2n}. But, that presentation certainly does not imply s has infinite order. StatisticsMan (talk) 00:38, 8 September 2009 (UTC)

You don't seem to understand what a group presentation means. A group presentation does not describe a family of groups, it describes a specific group.

By definition, the group G with presentation $\langle r,s|r^{n}=1,sr=r^{-1}s\rangle$ is the quotient of the free group on two generators (r and s) by the (normal subgroup generated by) the given relations. That is, it is the freest group on two generators satisfying those relations. So, if any group H generated by r and s and satisfying those relations has s of infinite order, then s must have infinite order in G, because H is the homomorphic image of G (with r going to r and s to s) and the order of an element can't go up when you take a homomorphic image. H=$\langle r,s|r^{n}=1,sr=r^{-1}s,r=1\rangle$ is an obvious example of such a group. Algebraist 00:44, 8 September 2009 (UTC)

It is possible that I don't understand it exactly :) I will think about this and check out the book for more info on the topic. Thanks. StatisticsMan (talk) 00:51, 8 September 2009 (UTC)

To put it another way: the group G is the freest group on two generators satisfying those relations. The only relations holding in G are those which must hold in any group in which the given relations hold. Because (for any n), there is a group satisfying the relations given for G with sⁿ≠1, this relation need not hold, and hence does not hold in G. So s has infinite order in G. Algebraist 01:01, 8 September 2009 (UTC)

You are correct that a given group can (in fact, will) have many different presentations. Proving that a given presentation has the minimal number of generators and relations can be rather difficult, as can proving that two presentations represent the same group. Algebraist's comment about the order of s is sufficient proof that the presentation your professor gave is a different group, though. --Tango (talk) 01:06, 8 September 2009 (UTC)

Indeed, even the problem of whether a given presentation represents the trivial group is undecidable in general. Algebraist 01:30, 8 September 2009 (UTC)

I think StatisticsMan's objection to Algebraist's response is fine (Algebraist's reasoning is fine too). SM probably objects to the argument "there is no order relation for s given, so s has infinite order", which is not a very good argument. SM specifically mentioned preferring an explicit group satisfying those relations where s has infinite order; this is a very good attitude to have, as it can be difficult or impossible to say anything about a particular finitely presented group based solely on its presentation. However, it is very easy to give a concrete, explicit example of a group satisfying those relations in which s has infinite order (and presumably so easy that Algebraist didn't see the need to spell it out). The example is just the semi-direct product of an infinite cyclic group ⟨s⟩ with a normal subgroup ⟨r⟩ in which s acts as inversion. Such a group in fact has the presentation ⟨r,s:r^n=1,sr=r^-1s⟩, but it is also quite easy to see it at least satisfies that presentation while s has infinite order. In case semi-direct products have not yet been covered:

To give an even more concrete realization of the group consider the matrix group generated by r = [ z,0,0,0 ; 0,1/z,0,0 ; 0,0,1,0 ; 0,0,0,1 ] and s = [ 0,1,0,0 ; 1,0,0,0 ; 0,0,1,1 ; 0,0,0,1 ], where z is a primitive n'th root of unity in the complex numbers. It is easy to check that r has order exactly n, that sr=r^-1s, and that s has order infinity, since s^(2n) = [ 1,0,0,0 ; 0,1,0,0 ; 0,0,1,2n; 0,0,0,1 ] and s^(2n+1) = [ 0,1,0,0 ; 1,0,0,0 ; 0,0,1,2n+1 ; 0,0,0,1 ]. JackSchmidt (talk) 01:38, 8 September 2009 (UTC)

The example I gave (Z, with 1=s, 0=r) is even easier. Algebraist 01:46, 8 September 2009 (UTC)

Ah, too true. I missed it in your second response. Your example is also a good introduction to a much more widely applicable technique called "quotient methods". You find some nice quotient group where you can work, like the abelianization, ⟨r,s:r^n=1,sr=r^-1s,r=1⟩≅Z, and answer the questions there instead of in the unknown finitely presented group.

Well, at least he now has two concrete realizations of the most general group satisfying the presentation his professor described. Both the semi-direct product and the matrix group are actually isomorphic to ⟨r,s:r^n=1,sr=r^-1s⟩. JackSchmidt (talk) 02:05, 8 September 2009 (UTC)

That's only the abelianization if n is odd. Algebraist 02:14, 8 September 2009 (UTC)

September 8

set

can the infinite set r = {r₁, r₂, ... r_n, ...} where $r_{k+1}={\frac {k+3}{k+2+r_{k}}}$ and $r_{1}={\frac {3}{2}}$ be described by some function $f$ where $f(k)=r_{k}$? —Preceding unsigned comment added by 92.8.193.46 (talk) 01:43, 8 September 2009 (UTC)

You are looking for a closed formula for the function, as the function itself can simply be defined as the mapping f:k-->r_k.Julzes (talk) 01:48, 8 September 2009 (UTC)

Defining s_k = 1/(r_k-1) you get a much nicer recursive relationship, s_k+1 = -(k+3)s_k - 1, with s₁ = 2. You can construct an explicit (but awful) formula for r_k from that:

$r_{k}=1+(-1)^{k-1}\left[(k+2)!\left({\frac {1}{3}}+\sum _{n=2}^{k}{\frac {(-1)^{n}}{(n+2)!}}\right)\right]^{-1}$

Rckrone (talk) 04:04, 8 September 2009 (UTC)

Trivially so: you have already provided an inductive definition of the function you desire. — Carl (CBM · talk) 04:12, 8 September 2009 (UTC)

can all sets of that form where $r_{k}=r_{k}(r_{k-1},r_{k-2},...r_{k-n+1},r_{k-n})$ be written as a function of $k$ in the same manner provided the first $n$ terms are given? —Preceding unsigned comment added by 92.8.193.46 (talk) 11:10, 8 September 2009 (UTC)

Clarify what you mean by 'of that form' in the more complex situation.Julzes (talk) 17:58, 8 September 2009 (UTC)

You may invent ad hoc notations signifying any solution you need. But the sequence (1, 2, 4, 16, 65536, ...) satisfying r _k+1 = 2 ^{r _k}, r ₀ = 1 is not written in terms of elementary functions. Bo Jacoby (talk) 12:38, 8 September 2009 (UTC),

Does it help you to know that

$r_{k}=1+{\frac {(-1)^{k-1}e}{\Gamma (k+3,-1)}}\ ?$

Where Γ is the incomplete Γ-function and e is Euler's constant? (N.B. the incomplete Γ-function is related to the Γ-function which generalises the factorial function to take a complex variable z with Re(z) > 0.) ~~ Dr Dec (Talk) ~~ 15:06, 8 September 2009 (UTC)

Can the choice of numbers and the sequence be changed and still have a closed form like that, Declan? It looks like a special case.Julzes (talk) 02:21, 9 September 2009 (UTC)

I'm not sure, I just though evaluating Rckrone's expression for $r_{k}$ might help some how. I don't really understand the problem in its totality. ~~ Dr Dec (Talk) ~~ 15:38, 10 September 2009 (UTC)

The sorgenfrey plane 2

I've done some more thinking to my question above (the sorgenfrey plane). Let A be all the rational points on the diagonal and B be a {r+sqrt2 | r is rational}. Obviously A and B are both dense and disjoint. Now if you recall how the proof of "regular+countable basis -> normal" goes, we can apply the following theorem: if A and B can be covered by two countable collections {Un} and {Vn} respectively, and that the closure of each Un is disjoint from B, and the closure of each Vn is disjoint from A, then A and B can be covered by disjoint open sets. The above A (rationals) and B (rationals+sqrt2) satisfy this property, hence they can be covered by disjoint open sets. I believe my intuition was right: it is possible to separate two disjoint dense sets, provided both are countable. I don't know what's up with the irrationals; there's more to them than meets the eye. The irrationals are a Baire space in the order topology, but rationals aren't, even though both are densely ordered. My solution is: there's something fishy with the irrationals that our intuition fails to capture. Anyone agree?

To Algebraist: I tried using your example but cant see how it'll work for p/q+1/(q^q). The page Liouville number just says |x-p/q|<1/(q^n), for any n. But I guess it's good enough, it must be that there's always a real number that can be approximated extremely closely by a rational p/q with a low denominator q, no matter how we rig the half open square.

To pma: you must have had one those those moments when two totally unrelated things give exactly the same result right? This and other "miracles" as I call them. I don't believe in miracles, so when I see something like this I tend to give it a deeper thought and try to bring more intuition into it. So please, if you accept logic without question that's fine by me, but things like Godel's incompleteness theorem and Skolem's paradox really puts my faith down. Breath of the Dying (talk) 02:18, 8 September 2009 (UTC)

I agree, in principle; but let's go back: the main thing was very simple: no matter how one fixes an Euclidean nbd I_r for each rational r, the set of all points that belong to infinitely many I_r,

$\cap _{F\in {\mathcal {F}}}\cup _{r\in \mathbb {Q} \setminus F}I_{r}$

is an intersection of countably many dense open subsets of $\mathbb {R}$, hence an uncountable set by the Baire's category argument (${\mathcal {F}}$ being the family of all finite subsets of $\mathbb {Q}$). This somehow is counter-intuitive, since it seems it tell us that a very large part of the irrational numbers are stuffed somewhere very very closely to the rationals. For instance, the trascendental number that Algebr. provided, is approximable by rationals as fast as a lightning. On the contrary, one would expect that being so "close" to the rationals would leave very small room. I think in this case one's intuition is just too much attached to finite or smooth analogies. I'm confident that your doubts will soon disappear. --pma (talk) 06:49, 8 September 2009 (UTC)

The intuition that "being so close to the rationals would leave very small room" is actually correct to the extent that the set of irrational numbers with approximation exponent greater than 2 has Lebesgue measure zero. — Emil J. 11:59, 8 September 2009 (UTC)

Good remark --so maybe what is counterintuitive is just any second category sets with null Lebesgue measure --pma (talk) 14:11, 8 September 2009 (UTC)

mathematics

what is the solution of (integration(differentiation of dx)) —Preceding unsigned comment added by 122.168.71.208 (talk) 10:30, 8 September 2009 (UTC)

Your question as posed does not entirely make sense, but you might find something useful in our article on the fundamental theorem of calculus. Gandalf61 (talk) 10:59, 8 September 2009 (UTC)

Your question is not specific enough to be meaningful, but I'd guess that you're taking integration and differentiation to be inverse operations, as in (square(square root of x)). So what do you think the solution to your question is?86.155.186.70 (talk) 11:06, 8 September 2009 (UTC)

(ec) If you're asking what the indefinite integral of the derivative of f(x) is, it's f(x). If you're asking what $\int f(x)dx$ means, see integral. Where we're calculating the integral over a range (see definite integral), the product of f(x) and dx (see differential) points towards the notion of considering the area under the curve of y=f(x) as the infinite summation of rectangles with height f(x) and width dx (see Riemann sum). In that way, the long-s could be intutitively related to the capital sigma notation. That's the brief overview of the generic notation. If what I've said doesn't help, you'll probably need to come back and be more specific. —Anonymous Dissident^Talk 11:10, 8 September 2009 (UTC)]

Isn't $\int f'(x)\ dx=f(x)+c$ for some constant c? So the antiderivative of the derivative is unique up to an addative constant. ~~ Dr Dec (Talk) ~~ 15:13, 8 September 2009 (UTC)

I might be wrong here, but is +c really necessary when we're considering the antiderivative of the derivative of a function we already know? +c notes an ambiguity – but if we already know the original function, there is no ambiguity. Speaking from a purely theoretical and objective reference frame, the indefinite integral of the derivative of f(x) is f(x), without the +c – because c is already known and is part of the function, if it exists. Contrast this to the position of a person who is trying to integrate the derivative of a function they do not know. Naturally, I'm only assuming that we do know f(x) here. —Anonymous Dissident^Talk 02:23, 9 September 2009 (UTC)

Derivatives don't have to be integrable, in the Lebesgue theory - so it's not always the case that $\int f'(x)\ dx$ even makes sense. But in suitably nice circumstances, this does work. Tinfoilcat (talk) 16:42, 8 September 2009 (UTC)

A derivative of a differentiable function does not have to be Lebesgue integrable, but it is always Henstock–Kurzweil integrable, and its indefinite integral is indeed the original function up to an additive constant. — Emil J. 17:12, 8 September 2009 (UTC)

Although if a function f is only differentiable a.e. then $\int f'(x)\ dx$ may still be still meaningful in the Lebesgue sense, but not necessarily equal to the original function unless f is absolutely continuous. Rckrone (talk) 17:22, 8 September 2009 (UTC)

Piecewise functions

I am getting confused on where to shade...

$h(x):={\begin{cases}3,&{\mbox{if }}-1\leq x\leq 1\\4,&{\mbox{if }}1<x\leq 4\\x,&{\mbox{if }}4\leq x.\end{cases}}$

and how do you graph y=x? Accdude92 (talk) (sign) 13:30, 8 September 2009 (UTC)

I suggest you start by graphing points, and see if you detect a trend. Ray^Talk 14:10, 8 September 2009 (UTC)

(ec) When you have a function outputting a constant, it's as easy as graphing a line like y=10. When the function outputs different constants for different ranges, which line you're drawing differs based on x value. For instance, for "3 if -1≤x≤1", you're graphing y=3 from x = -1 to x = 1. Another way of thinking about it is that you're graphing y=3 for every x value that meets the criterion "x is greater than or equal to -1 and less than or equal to 1". y=x is a line that cuts the origin diagonally – every x value is matched to every y value: (1,1), (2,2), (3,3) and so on. —Anonymous Dissident^Talk 14:19, 8 September 2009 (UTC)

Dihedral group of a polygon

Resolved

Someone asked me to explain the dihedral group of a polygon to them. This is what I said: Dihedral group is the group of symmetries of a polygon by reflection and rotation. If n is odd then besides the n rotations there are n reflections around the n lines originating from the n vertices and bisecting each interior angle. If n is even then besides the n rotations there are n reflections around the perpendicular bisectors of the n edges. So in either case the dihedral group has order 2n.

What I am not sure is whether in case n is odd the reflections are taken around the angle bisectors or not. Can someone clarify this. I searched on the net but couldn't find anything. Thanks--Shahab (talk) 15:35, 8 September 2009 (UTC)

The dihedral group of a regular n sided polygon is generated by the reflections in the perpendicular bisectors of its sides and the bisectors of its interior angles. If n is odd each axis of symmetry connects the mid-point of one side to the opposite vertex. If n is even there are n/2 axes of symmetry connecting the mid-points of opposite sides and n/2 axes of symmetry connecting opposite vertices. In either case there are n axes of symmetry altogether and 2n elements in the symmetry group. Reflecting in one axis of symmetry followed by reflecting in another axis of symmetry produces a rotation through twice the angle between the axes. Gandalf61 (talk) 15:48, 8 September 2009 (UTC)

Thanks for the quick response. This information should also be in the Dihedral groups article.--Shahab (talk) 16:09, 8 September 2009 (UTC)

Metric completion of a field

This question should be elementary, but it's giving me the blues. Suppose we have a field, $\left(K,+,\cdot \right)$ endowed with a absolute value $|\cdot |:K\rightarrow \mathbb {R} _{+}$ that satisfies the three properties:

• AV1: $\left|x\right|=0\iff x=0$
• AV2: $\left|x\cdot y\right|=\left|x\right|\cdot \left|y\right|$
• AV3: $\left|x+y\right|\leq \left|x\right|+\left|y\right|$

Now, suppose $K$ is not metrically complete, i.e., there exist sequences $\left(x_{n}\right)_{n=1}^{\infty }$ that are Cauchy in $K$ with no limit in $K$. We define ${\overline {K}}$ to be the set of Cauchy sequences in $K$ modulo Cauchy sequences that converge to zero. Then, it is natural to define an absolute value, $\left|\cdot \right|^{\star }$ by $\left|\left(x_{n}\right)_{n=1}^{\infty }\right|^{\star }=\lim _{n\rightarrow \infty }\left|x_{n}\right|$. This is all very nice.

My question: how can I prove that $\left|\cdot \right|^{\star }$ is unique, i.e., that it is the only absolute value that we can define on ${\overline {K}}$ that is well-defined, satisfies the axioms AV1-AV3, and restricts to $\left|\cdot \right|$ on $K$ (i.e., $\left|\iota \left(x\right)\right|^{\star }=\left|x\right|$, where $\iota \left(x\right)=\left(x\right)_{n=1}^{\infty }$, the constant sequence)?

I think there's something subtle going on here that's eluding me. I've read Complete metric space, but it didn't quite ease my mind. The first paragraph under the header "Completion" asserts this claim, but doesn't say how it's proven. Can someone give me a hint as to what I'm missing? Thanks in advance. -GTBacchus^(talk) 15:41, 8 September 2009 (UTC)

An absolute value defines a topology, and it is continuous with respect to this topology. Let $|\cdot |'$ be another absolute value on ${\overline {K}}$ which satisfies the properties above. If $x=(x_{n})_{n=0}^{\infty }$ is a Cauchy sequence, then $x=\lim _{n\to \infty }x_{n}$ in ${\overline {K}}$, thus $|x|'=\lim _{n\to \infty }|x_{n}|'=\lim _{n\to \infty }|x_{n}|=|x|^{*}$. Now, the catch with this argument is that I am tacitly assuming that the topology induced by $|\cdot |'$ coincides with the natural topology on ${\overline {K}}$, and I don't quite see why this should be true in general. — Emil J. 16:59, 8 September 2009 (UTC)

Also, I've taken a look at Complete metric space#Completion as you suggest, and I do not see anything related to absolute values there. The generic extension method described there only implies that there exists a unique (uniformly) continuous function $|\cdot |^{*}$ which extends $|\cdot |$. It does not assert that $|\cdot |^{*}$ is an absolute value (though this should be easy enough to prove), and more importantly, it does not assert anything about uniqueness wrt the class of not-necessarily-continuous absolute values. Are you sure that the claim you are trying to prove is even true? — Emil J. 17:07, 8 September 2009 (UTC)

No, I'm not sure. The claim I really want to prove involves a non-Archimedean absolute value (featuring the strong triangle inequality), but I think it might be true in general. Actually, the assertion is simply that the absolute value extends "naturally" to the completion. Uniqueness would be bonus. This comes from a set of lecture notes, and may very well be an ill-posed question. If it's not true, I suppose there's a counter-example?

I've already proven that $|\cdot |^{\star }$ is an absolute value satisfying the same AV1-AV3 axioms. -GTBacchus^(talk) 18:51, 8 September 2009 (UTC)

(ec) What you need is the universal property of the completion $\iota :K\to {\overline {K}}$. For any metric space $K$ and any uniformly continuous map $f:K\to M$ to a complete metric space $M$ there exists a unique uniformly continuous ${\overline {f}}:{\overline {K}}\to M$ such that $f=\iota \circ {\overline {f}}$ (you may say that ${\overline {f}}$ is the unique u.continuous extension of $f$ over ${\overline {K}}$, if you like to think $K$ as a subset of ${\overline {K}}$).

This situation may be described in term of an adjunction, if you like the language of category theory. In any case the universal property allows to extend the absolute value and the algebraic operations from $K$ to ${\overline {K}}$ together with their identities. There are just some small detalis to work out; in case just ask. --pma (talk) 17:07, 8 September 2009 (UTC) PS: as to the unicity matter: precisely, ${\overline {f}}$ is the unique continuous extension, and it turns out to be uniformly continuous with the same modulus of continuity of $f$. --pma (talk) 17:13, 8 September 2009 (UTC)

Ok, so it's the unique continuous extension, but there may be other extensions that aren't continuous? This makes sense to me, and reminds me of a function that appears in Counterexamples in Analysis. There's an additive but non-homogeneous function on $\mathbb {R}$ that sends each number to the sum of the coefficients of its representation in terms of the basis of $\mathbb {R}$ as a vector space over $\mathbb {Q}$. That function, as I recall, is additive, but not remotely continuous. Its graph is dense in the plane, again IIRC. -GTBacchus^(talk) 18:51, 8 September 2009 (UTC)

Sure. So, the whole matter is extremely simple; should something be still unclear to you, we can help you to clarify it. --pma (talk) 19:56, 8 September 2009 (UTC)

A good algebraic number theory book should handle this. I recall that Janusz's Algebraic Number Fields had a very readable and careful discussion, and Lang's Algebraic Number Theory probably covers this adequately as well. Unfortunately I have neither of these books with me and I don't remember the answer! I'll check Milne (available here) if I remember to do so later.

Regarding proving the claim, I got stuck at the same place as Emil, in proving that $(x_{n})_{n=0}^{\infty }=\lim _{n\to \infty }\iota (x_{n})$ in ${\overline {K}}$. I think that if you drop requirement 2 of an absolute value, that $|xy|=|x||y|$, then you get a counterexample in $\mathbb {R} /\mathbb {Q}$ following the plan you mentioned. I think the best bet for finding a counterexample without dropping requirement 2 is in the completion $\mathbb {C} _{p}$ of ${\hat {\mathbb {Q} }}_{p}$. Eric. 216.27.191.178 (talk) 00:56, 9 September 2009 (UTC)

Here's a fairly trivial counterexample. Pick K such that Aut(K/K) is nontrivial (for example, we may take K = Q(i) with the usual complex absolute value, so that K = C is algebraically closed). Fix a nonidentical field automorphism σ of K which is identical on K, and define |x|' = |σ(x)|^*. On the one hand, it is easy to see that |·|' is an absolute value extending |·|. On the other hand, since σ is nonidentical, and K is dense in K, σ is discontinuous, and a fortiori it is not an isometry. Thus, there are x, y such that $|x-y|^{*}\neq |\sigma (x)-\sigma (y)|^{*}=|\sigma (x-y)|^{*}=|x-y|'$, i.e., |·|' ≠ |·|^*. — Emil J. 10:45, 9 September 2009 (UTC)

Interesting. I'm looking at Algebraic Number Theory, by Jürgen Neukirch, page 124, from the section on "Completions". Ahem:

Finally, one proves the uniqueness of the completion $\left({\overline {K}},|\cdot |\right)$: if $\left({\overline {K}}',|\cdot |'\right)$ is another complete valued field that contains $\left(K,|\cdot |\right)$ as a dense subfield, then mapping $|\cdot |-\lim _{n\rightarrow \infty }a_{n}\mapsto |\cdot |'-\lim _{n\rightarrow \infty }a_{n}$ gives a $K$-isomorphism $\sigma :{\overline {K}}\rightarrow {\overline {K}}'$ such that $|a|=|\sigma (a)|'$

How does this jibe with your counterexample, which looks perfectly reasonable to me? (By the way, what's the LaTeX for the angle-shaped "hat"? I'm using \overline in place of it now.) -GTBacchus^(talk) 15:49, 9 September 2009 (UTC)

What he claims is uniqueness up to an isomorphism (preserving the field structure, absolute value, and elements of K). It's perfectly consistent with the example, in fact, his σ is my σ. A hat is \hat in TeX: ${\hat {K}}$. — Emil J. 16:09, 9 September 2009 (UTC)

Ah, so the isomorphism preserves algebraic relations, but non necessarily the topology, because the absolute value in continuous in one case but not in the other? Is that right? -GTBacchus^(talk) 19:11, 9 September 2009 (UTC)

Let me summarize and state the whole thing as I see it: a list of simple facts:

1. Let K be a AV field (field with absolute value, i.e. AV1 AV2 AV3 hold). Then, $d_{K}(x,y):=|x-y|$ makes $K$ a metric space, and a topological field (the field operation are continuous).
2. There exists an AV field ${\hat {K}}$ such that $K$ is a sub AV field of ${\hat {K}}$ (i.e., it is a subfield of ${\hat {K}}$ and its AV is the restriction of the one of ${\hat {K}}$); moreover ${\hat {K}}$ is complete as a metric space and $K$ is dense in ${\hat {K}}$. Any such ${\hat {K}}$ is called an AV field completion of $K$.
3. The AV field ${\hat {K}}$ may be constructed starting from a metric completion $M,{\hat {d}}$ of the metric space $K,d_{K}$; then the AV and the field operations of $K$ extend uniquely to $M$ in such a way to make it an AV field ${\hat {K}}$ whose distance $d_{\hat {K}}$ coincides with the distance ${\hat {d}}$.
4. If ${\overline {K}}$ is another AV field completion of K, there exists a unique homeomorphism $\sigma :{\hat {K}}\to {\overline {K}}$ such that $\sigma _{|K}=id_{K}$; moreover, $\sigma$ is an isomorphism of AV fields.
5. Remark. (EmilJ example) If K is an AV field and H is a super-field of K, then H may admit several AV that make it a completion of K (essentially due to the fact that if K is not complete there is a lot of trascendence in the completion extension).

Note that there is an analogous situation in the context of completion of normed linear spaces, including the remark (that becames much simpler: there is a lot of linear independence passing to the completion.) Do you agree? There are possibly typos here and there.--pma (talk) 20:13, 9 September 2009 (UTC)

Is "homeomorphism" a typo? If not I'm confused about that since Emil J gave a convincincing argument that a map like that would be discontinuous, but maybe there's something I'm not getting about the topology involved here. Rckrone (talk) 23:19, 9 September 2009 (UTC)

From Merriam-Webster: "a function that is a one-to-one mapping between sets such that both the function and its inverse are continuous and that in topology exists for geometric figures which can be transformed one into the other by an elastic deformation".Alan (talk) 23:27, 9 September 2009 (UTC)

I think the question was not, "what does homeomorphism mean", but rather "is homeomorphism in this context a typo for homomorphism, in this case, of valued fields?" A homomorphism (no 'e') preserves algebraic structure, but is not necessarily continuous. Or, if it is necessarily continuous in the case of valued fields, that's a theorem. I'm also wondering about the continuity issue. -GTBacchus^(talk) 05:00, 10 September 2009 (UTC)

Not a typo: I wrote "a unique homeomorphism $\sigma$" to emphasize that $\sigma$ is unique even in the class of all homeomorphism that extend the identity of K forgetting the algebraic structure; I could have even written "a unique continuous $\sigma$": this unique continuous element $\sigma$ is then automatically an isometry of metric spaces and an isomorphism of AV field (in particular, a homomorphism). On the contrary, if we try to characterize $\sigma$ algebrically, we meet difficulties, exactly because (in general) there are a lot of non-continuous automorphisms of ${\hat {K}}$ that extend the identity of K, as EmilJ's remark showed. The completion of an absolute valued field is primarily a completion of a metric space: the algebraic extension follows the metric extension. There are other mixed topological-algebraic structures, where a topological property is characterized in a purely algebraic way (first example that cames to my mind: a complex-valued homomorphism of a Banach algebra is necessarily continuous); but this is not the case. Still I do not understand why you seem to be unsatisfied with a unicity property stated in terms of the topological structure. Here the natural thing is to think in terms of completion of metric spaces: in this more general and simpler category, completion is unique up to a unique isometry (or if you like, up to a unique homeomorphism that is automatically an isometry). Once this is clear, take it to the context of valued fields: the same $\sigma$ will be automatically a valued field isomorphism also. --pma (talk) 06:23, 10 September 2009 (UTC)

I'm sorry if I'm being dense about this (no pun), and I thank you for your patience. In your numbered points above, here's my understanding:

1. Start w/ an AV-field K, on which the AV and field operations are continuous.
2. There exists some metric completion of K that is an AV-field (with AV and field ops continuous) that contains K as a dense sub-AV-field.
3. You can get such a completion by completing K as a metric space to obtain K-hat, and then uniquely extending the AV and operations - all of which presumably remain continuous.
4. If there's some K-bar (different from K-hat) that does what K-hat does (completes K metrically, and extends the AV and algebra so they're still continuous), then K-bar and K-hat are equivalent via a unique homeomorphism that respects all the structure.
5. Instead of starting with a metric completion, we could take a super-field, which extends the algebra first, and add topological structure to it afterwards. Then we can get away with a discontinuous AV. Even though it restricts to the old AV on K, and gets along w/ the field operations, it's not a metric completion of K, so point #4 doesn't apply. That's why #3 is a more natural way to do it.

Is that all correct? -GTBacchus^(talk) 07:29, 10 September 2009 (UTC)

I understand where I was confused (thanks pma) and I think it's the same place, so I will give it a shot. Referring to Emil J's example, the map σ as a regular-old field isomorphism from K to itself is discontinuous with the topology induced by |.|*. However as AV fields, K with metric |.|* is distinct from K with metric |.|'. Call them K₁ and K₂. σ as a map between AV fields from K₁ to K₂ is continuous since open sets according to the topology induced by |.|* get mapped to open sets according to the topology induced by |.|' (instead of using |.|* at both ends like before), which can be seen by the fact that |x|* = |σ(x)|'. In other words σ preserves the AV as the transition is made from the one metric to the other. Rckrone (talk) 08:13, 10 September 2009 (UTC)

One way to think about it is that there are different ways to embed the metric completion of K into the superfield K, two of them being K₁ and K₂, and σ is the AV field isomorphism that relates them. They both contain the same sets of elements, so merely as fields they are identical, but with the added structure of a metric they are distinct. Rckrone (talk) 08:38, 10 September 2009 (UTC)

agreed. Also, they have the same set of elements, and they are isomorphic, but not via the identity map.--pma (talk) 09:24, 10 September 2009 (UTC)

(ec) Correct. As to point 1, note that the continuity of the structure mappings (sum, product, multiplicative inverse, absolute value) follows as a consequence of the AV field axioms and the choice of the topology on K (the metric topology of the distance d_K(x,y):=|x-y|), so it doesn't need to be assumed. In particular, the distance d_K makes

• $+:K\times K\to K$ a Lipschitz function;
• $\cdot :K\times K\to K$ a locally Lipschitz function (it's Lip on every disk); also, $x\to ax$ is Lipschitz (of constant |a|) from K to K;
• $(\cdot )^{-1}:K\setminus \{0\}\to K$ (multiplicative inverse) a locally Lipschitz function: precisely, Lipschitz on the complement of every nbd of 0 (just by AV2, because |1/x-1/y|=|x-y|/|xy|);
• $|\cdot |:K\to \mathbb {R}$ (the abs value) a Lipschitz function, just by AV3.

As to point 3, if M is any completion of K as metric space (hence a plain metric space with no algebraic structure) the above structure maps (+, $\cdot$, inverse, $|\cdot |$), thanks to the universal property of metric completion, have unique continuous extensions respectively to maps:

• ${\hat {+}}:M\times M\to M$ ;
• ${\hat {\cdot }}:M\times M\to M$ ;
• ${\hat {(\cdot )^{-1}}}:M\setminus \{0\}\to M$ ;
• ${\hat {|\cdot |}}:M\to \mathbb {R}$ ,

that a priori are only continuous maps: but then all the AV field axioms holds for them by continuity, since they hold on the dense subset K. In particular the distance of M is, of course, just ${\hat {|x-y|}}$. Technical detail: notice that, while the continuous extension of the sum and of the AV is an immediate application of the universal property of metric completion, it is a bit less immediate for the product and the inverse, because they are only locally uniformly continuous; the latter is possibly not even defined on a complete domain, M\{0}. But that difficulty is very easily bypassed: first extend uniquely by continuity the product map (respectively, the multiplicative inverse map) in the closed sets of MxM (resp., of M\{0}) where it is Lipschitz, then glue together the extensions using the unicity property, and observe that the result of the gluing is still locally Lip, hence continuous.

As to point 4, we can make a stronger statement putting a comma pregnant with meaning in the middle (making it a non-restrictive clause): K-bar and K-hat are equivalent via a unique homeomorphism , which respects all the structure. That is, the identity map on K extends to a unique homeomorphism between them, and, this unique homeomorphism turns out to be an AV field isomorphism. Is that OK? (note: I used everywhere the term AV field just because I'm not sure about the exact term, maybe you know). I think the other people above also agreed with this picture: maybe they may correct or add something. --pma (talk) 09:18, 10 September 2009 (UTC)

That's awesome. Thank you so much. Thank you all, who have commented here. I'm, like, smarter now. :) I'll check back for any further notes or corrections here, but I really feel comfortable with what we've just walked through, including the technical points about the product and multiplicative inverse maps not being uniformly continuous on the whole field.

Neukirch, in the book I consulted, calls the structure we're talking about a "valued field"; I don't know how standard that terminology is. -GTBacchus^(talk) 11:07, 10 September 2009 (UTC)

Poker and math

What problem, if any, do computers have in playing poker? Could a computer playing online poker win against human players? --Mr.K. (talk) 16:12, 8 September 2009 (UTC)

The combinatorial work necessary for optimal estimates of the strengths of others hands is hard, probably intractable. Coming up with good heuristics is also difficult. Dealing with the non-mechanistic issues (bluffing, modeling the other players) is also highly non-trivial. But yes, they can, and they do, regularly. My success at Sam Fox Strip Poker was limited (as was, to be honest, my motivation, due to the 320x200 greyscale images). --Stephan Schulz (talk) 16:34, 8 September 2009 (UTC)

Limit poker is much easier to do than no-limit, I believe. Have a look at Computer poker players Tinfoilcat (talk) 16:38, 8 September 2009 (UTC)

Yeah that's right. It's all about the pot odds (and more importantly the implied pot odds). If you need to put in x% of the pot to call and you expect to make a winning hand more than x% of the time then you've got good poot odds and you should call. (Obviously you might not make you hand and lose, but if you made the same call 1,000 times then you'd expect to make a profit). Like Stephan says: it's the bluffing that is hard to get around. You can make a really big bet so that the pot odds are so bad that the player wouldn't normally call, e.g. if you bet the twice the pot then the pot odds are will be so bad that you'd need to be able to make a winning hand 2 times in 3 (or better) to be able to call. But that's the bluff: I could have rubbush cards, but make a big bet so that the other player doesn't have the pot odds to call. ~~ Dr Dec (Talk) ~~ 17:13, 8 September 2009 (UTC)

Yes, limit is easier for a computer. There are far fewer options to consider (you can fold, call or raise by a fix amount, rather than fold, call or raise by any amount of your choice [as long as it is over the minimum raise]). --Tango (talk) 20:49, 8 September 2009 (UTC)

Hermite polynomials and Hermite Interpolation

Does anyone know if there is a connection between Hermite polynomials and Hermite interpolation? ...Other than the association with Charles Hermite.

When interpolating a polynomial it can help to use a basis other than simple powers of x so I thought the Hermite polynomial might be useful for Hermite interpolation in the same way as the Newton basis polynomials are used in the Divided differences algorithm.

Neither article in Wikipedia states such a relationship but there is a link to an external article on Hermite interpolation from the article on Hermite polynomials... but that could just have been places in error.

Thanks,

Yaris678 (talk) 19:38, 8 September 2009 (UTC)

Well, I do not see any connection at all between Hermite polynomials and interpolation problems. --pma (talk) 20:10, 8 September 2009 (UTC)

Derivative of a function between matrices

Hi there refdesk!

I've been completing a couple of old exercises in analysis and I've come upon the following:

Define f: M_n → M_n by f(A) = A³. Show that f is differentiable everywhere, then find its derivative.

Now I've realised I haven't a clue how to differentiate a function between matrices (M_n the set of real n x n matrices), nor even how the derivative is defined. Could anyone point me in the direction of an explanation or give me a very brief rundown of how to solve a question like this, which sounds like it's probably very simple once you know how!

Thanks a lot,

Typeships17 (talk) 20:38, 8 September 2009 (UTC)

Matrix calculus might help. --Tango (talk) 20:44, 8 September 2009 (UTC)

Doesn't the space of n × n matrices have a manifold structure? The space of n × n matrices would be an n²-dimensional manifold. So write a general matrix as M := (m_i,j), and then compute M³ = (m_i,j)³. Then you just need to check that each of the entries of M³ is differentiable as a function of (m_1,1,…,m_i,j,…,m_n,n). This is clearly true since each entry with just be a homogeneous polynomial of degree 3. ~~ Dr Dec (Talk) ~~ 20:50, 8 September 2009 (UTC)

Yes, it can be thought of as a family of n² functions each defined on $\mathbb {R} ^{n^{2}}$ and differentiated accordingly. --Tango (talk) 21:33, 8 September 2009 (UTC)

Thanks a lot guys, I'll try that link out now. In addition, I was wondering if anyone could explain to me why the derivative $Df\vert _{x}(h)$ of a function between Rⁿ and R^m is orthogonal to x in not-too-technical terms: I can get my head around the concept where m=1 and you can look at level sets, but I'm not sure what geometrical argument you can apply when mapping to more than 1 dimension. Thanks very much, Typeships17 (talk) 23:39, 8 September 2009 (UTC)

sorry but it is absolutely not clear what you mean. What's x? What's $Df\vert _{x}$? --pma (talk) 09:48, 9 September 2009 (UTC)

I think I know what s/he might be getting at. If f : R^m → Rⁿ is a submersion with x₀ a regular point of f and y₀ a regular value of f then the tangent space to {x | f(x) = y₀} at x₀ is given by the kernel of the differential of f evaluated at x₀ (i.e. the Jacobian matrix evaluated at x₀). In the case when n = 1 then the differential if just the gradient vector of f and its kernel is just the orthogonal space of the gradient vector. ~~ Dr Dec (Talk) ~~ 22:37, 9 September 2009 (UTC)

I've tried to answer what I think the question was talking about. See here. Let me know if I've made any errors. ~~ Dr Dec (Talk) ~~ 12:10, 10 September 2009 (UTC)

Consider the big laugh expansion: $(A+H)^{3}=A^{3}+(HAA+AHA+AAH)+o(\|H\|)$: your $f$ is differentiable at any $A$ and the Fréchet differential is $Df(A)H=HAA+AHA+AAH$. This holds for $A$ in any Banach algebra of course.--pma (talk) 07:04, 9 September 2009 (UTC)

thanks for the correction... ehm, I made an error counting to 3 ;-) --pma (talk) 13:51, 9 September 2009 (UTC)

Is this a hyperbola or...

Draw line segments connecting (a,0) to (0,1), (a-1,0) to (0,2) etc. up to (1,0) to (0,a). Is the curve that is approximated a hyperbola (well, half of one), or is it something else? Donald Hosek (talk) 21:57, 8 September 2009 (UTC)

You mean an envelope like this? It certainly looks like a hyperbola to me, although I can't guarantee it. --Tango (talk) 22:08, 8 September 2009 (UTC)

As the very article you linked states, it's a parabola, not a hyperbola. Algebraist 22:22, 8 September 2009 (UTC)

Don't stop when y gets down to 0; keep going. Likewise keep going in the opposite direction. If you do that, you'll see that the curve does not have asymptotes. So it cannot be a hyperbola. Michael Hardy (talk) 02:09, 9 September 2009 (UTC)

Indeed, to get an equilateral hyperbola one has to take all line segments whose X and Y intercepts have constant $product$, not $sum$. BTW, using a ruler, this fact gives a more precise way of drawing a hyperbola, compared with drawing by points. --pma (talk) 15:13, 9 September 2009 (UTC)

(ec) I think you're talking about the envelope of the family of lines given by

$y=\left({\frac {1+t}{t-a}}\right)x+(1+t)\ ,$

where, for each value of t with 0 ≤ t ≤ a – 1, we get the equation of a line. Well, let

$F((x,y),t):=y-\left({\frac {1+t}{t-a}}\right)x-(1+t)\ .$

Then for a fixed t_{0 the corresponding line is given by}

$\{(x,y)\in \mathbb {R} ^{2}:F((x,y),t_{0})=0\}.$

In this case, the envelope is given by solving F = F_t = 0 in terms of x and y. Doing this gives the envelope, or discriminant:

${\mathcal {D}}_{F}(t)=\left({\frac {(a-t)^{2}}{a+1}},{\frac {(1+t)^{2}}{a+1}}\right).$

This curve has the property that all of the lines are tangent to it. The question is now: is this curve part of a hyperbola? ~~ Dr Dec (Talk) ~~ 22:25, 8 September 2009 (UTC)

One way to visualize that it's a parabola rather than a hyperbola without finding the explicit function is to consider what happens when you continue drawing lines past where you stopped: (0, 0) to (0, a+1), (-1, 0) to (0, a+2),... (-n, 0) to (0, a+1+n),... The slope of these lines go through the vertical and approach 1. You can also continue in the other direction to see a similar trend. Rckrone (talk) 22:49, 8 September 2009 (UTC)

Yes, that's a(n arc of) parabola. Let's call s>0 and t>0 respectively the X and the Y intercept of a segment in the family. If c is the constant value of s+t in the family of segments, following Declan's approach you'll get

$x^{1/2}+y^{1/2}=c^{1/2}$;

and if you rationalize and change coordinates after a π/4 rotation (i.e. x=(u+v)/2, y=(u-v)/2) you get a normal form of a parabola.

BTW, you can generalize this picture a bit. Given any $\alpha >0$ and $c>0$, the envelope of the family of all segments within the first quadrant with x and y intercepts satisfying

$s^{\alpha }+t^{\alpha }=c^{\alpha }$

is a curve of equation

$x^{\frac {\alpha }{\alpha +1}}+y^{\frac {\alpha }{\alpha +1}}=c^{\frac {\alpha }{\alpha +1}}$

(yes, $\alpha =2$ gives 2/3).

You do not need any differential calculus; it's a plain consequence of the Hölder inequality in $\mathbb {R} ^{2}$ (and for your particular case $\alpha =1$, it's just a Cauchy-Schwarz).--pma (talk) 08:28, 9 September 2009 (UTC)

hey!now that's all in the first link, too! ;-) --pma (talk) 13:54, 9 September 2009 (UTC)

September 9

Effect of type I vs Type II error cost difference

What work has been done on how an appropriate p-value for null hypothesis testing changes when type I errors are more or less costly than type II? (What piques my interest about this question is its implications for understanding the evolution of apophenia.) NeonMerlin[3] 05:21, 9 September 2009 (UTC)

To be precise, p-values are not affected by such considerations: those values are merely calculated from statistics taken from the sample; what type I/II error considerations can do, is affect confidence levels: Bear in mind that p-value can be defined as the least significance level under which your null hypothesis is rejected (here significance level is 1-α, where α stands for confidence level).
In regard with type I/II error issues, set any other considerations aside (such as sample size, which has an inverse relation with both error probabilities), the thumb rule is that the more relatively costly type I error is, the less significance level you are willing to set, or in other words, the more probable is that the statistic from a sample will fall into the non-rejection interval (or the more probable is that the p-value is greater than the (prefixed) significance level. Pallida  Mors 18:21, 10 September 2009 (UTC)

Why angle side angle?

http://imgur.com/MpXTh.jpg

It's given that angle 1 is equal to angle 2, angle 3 is equal to angle 4, point D is the midpoint of line BE, and that line BC is equal to segment DE. BC is equal to BD via substitution, thus triangle ABD is congruent to triangle EBC via ASA, but it's bugging the hell out of me. Shouldn't it be AAS, seeing as how BE is outside of B and C? How would BC help me anyway? Ugh. I hope I phrased that in an understandable way. --Glaesisvellir (talk) 22:05, 9 September 2009 (UTC)

Once you have two angles, you can easily calculate the third using the fact that the sum of all the angles in a triangle is 180 degrees. That means it really doesn't make any difference whether you have ASA or AAS. --Tango (talk) 22:34, 9 September 2009 (UTC)

Thank you. The only problem is, what if the measurements of the other two angles are not listed, and measuring them is not an option? --Glaesisvellir (talk) 00:50, 10 September 2009 (UTC)

If you're trying to prove two triangles are congruent, such as the ones in the image you linked, you can argue that since two of the angles of each triangle are equal, then the third angles of each triangle must be equal. Referring to the image, that argument is $m\angle A=180^{\circ }-m\angle 1-m\angle 3=180^{\circ }-m\angle 2-m\angle 4=m\angle E$. Rckrone (talk) 03:53, 10 September 2009 (UTC)

September 10

Weaker variant of Schur for sum of two = irreducibles: two eigenvalues?

Hello,

I was considering endomorphisms of \mathbb{C} G- modules, and something funny happened with the eigenvalues. I was wondering if this were true:

If V is the \mathbb{C} G-module V with V=A+B, with A and B two isomorphic irreducible modules, then every endomorphism f has at most two complex eigenvalues. Moreover, every vector v\in V is either an eigenvector of f, or is in an invariant subspace with respect to f of dimension two.

It looks like it true, but I've never seen it before so that is why I am in doubt.. I was also wondering if there were more general things like this that are known and very important.

Many thanks, Evilbu

Homomorphisms behave nicely with respect to direct sum decompositions. In your case, where V is the direct sum of two copies of A, End(V) is isomorphic to the 2x2 matrix ring with entries in End(A). Since you work over C, this is really the 2x2 matrices over C. Now at least it's obvious that there can be at most two eigenvalues. Tinfoilcat (talk) 16:46, 10 September 2009 (UTC)

maths?

what is maths —Preceding unsigned comment added by 59.96.142.2 (talk) 17:05, 10 September 2009 (UTC)

math is using numbers to get new numbers.Accdude92 (talk) (sign) 17:07, 10 September 2009 (UTC)

Mathematics is the science and study of quantity, structure, space, and change. Mathematics is both the queen and the handmaiden of all science. --LarryMac | Talk 17:42, 10 September 2009 (UTC)

I'd say she's also science's drinking buddy, confidant, and occasional tennis doubles partner. -GTBacchus^(talk) 17:22, 11 September 2009 (UTC)

Mathematics is an esoteric discipline whose higher reaches are guarded jealously by those in the know so that it is hard for new entrants to join them. Thus in some ways mathematics, like other learned professions, provides a good living for its practitioners.86.152.79.120 (talk) 22:33, 10 September 2009 (UTC)

That is the opposite of the truth. Lots of popularizations get written by mathematicians, but far fewer people are willing to read them than mathematicians might wish. Michael Hardy (talk) 21:00, 11 September 2009 (UTC)

Is it possible for "maths" not to use numbers and still be properly considered as maths? 202.147.44.84 (talk) 22:30, 10 September 2009 (UTC)

Yes. Algebraist 22:33, 10 September 2009 (UTC)

Yes. Topology and geometry don't involve a great deal of numbers (they pop up from time to time, but mostly just to index things). --Tango (talk) 22:43, 10 September 2009 (UTC)

Someone has to properly ingrain the idea in humans' minds that mathematics is not solely about numbers, equations or formulas. Although number theory is a branch of mathematics, it is not concerned with "the arithmetic of big numbers" - such operations may be done with a computer (some cannot but that is irrelevant to mathematicians). Furthermore, although mathematics does deal with equations in a vague sense (see algebraic geometry, field theory and algebraic number theory), a mathematician is not really concerned with "solving equations" but rather how the solution sets behave. In algebraic geometry, one is concerned with the "geometric" properties of solution sets; in field theory, one is sometimes concerned with algebraic extensions of fields (an algebraic extension of a field (F) is simply another field containing F whose every element is the solution of a non-zero polynomial with coefficients in F). For instance, a non-mathematician may ask which numbers can be a solution to an equation. A mathematician will then formalize this: Define a number to be algebraic over the rational numbers if it is the solution to some non-zero polynomial with rational coefficients. The mathematician will now investigate this in greater depth by analysing some non-trival (first) examples of algebraic numbers. For instance, $x^{2}-2=0$, has one solution equating to ${\sqrt {2}}$; this is "non-trivial" since it "cannot be written as a fraction" (again, this idea must be formalized). After this, a mathematician will attempt to modify this example and investigate what one may obtain - what about, $x^{2}+1=0$, he/she will ask. I urge the OP to attempt this - it is remarkable how this question leads to a totally new branch of mathematics. Lastly, "formulas" are not of interest in mathematics - if at all they serve a purpose it is how one obtains them and not what they state. In differential geometry, these so-called "formulas" exist but the way in which they are obtained leads to new insights in differential topology, for instance.

I have told you what mathematics is not but I have not told you what it is. I would love to do so but I think that the best basic examples of this may be found in Group (mathematics) and Vector space - wonderfully written articles which may be understood by even someone who cannot count (by the way, one need not know how to count to do mathematics - I would not know how to count if my reputation as a mathematician (as viewed by "society") were not at stake). Ideas relating to numbers are basic to mathematics and existed more than 2000 years ago. If I had to give a "definition" for mathematics (which numerous people have attempted to do), I would define it as: "The logical deduction of truth from a set of axioms." The more experience I have with the subject, the further this definition changes in my mind - I am sure that I will never be able to find a definition which suits me perfectly. --PST 05:06, 11 September 2009 (UTC)

Isn't your description a bit extreme in the opposite direction? I'd say that some formulas really are of interest, et cetera. Also, would a person who cannot count know the meaning of "two vectors" and "17th century"?

My personal definition for mathematics used to be along the lines of "The logical deduction of truth from a set of axioms", but lately I'm of the opinion that "mathematics" is a name for two distinct things - one is a method of obtaining truths (using axioms, proofs and the like), and one is a body of knowledge in subject matters which are traditionally explored with this method (quantity, structure, ...). Theoretical computer science is explored with the mathematical method, but is not usually considered to be included in the mathematical body of knowledge. -- Meni Rosenfeld (talk) 06:05, 11 September 2009 (UTC)

By my comment I implied that "counting" is simply an assignment of symbols and sounds to commonsense. In particular, it is not essential that mathematics be based on an agreement upon universal terminology; each and every person may define his own truth and do mathematics. It is just a question of mathematical interest - can a deep theory emerge from this, for instance. Thus, if a student finished high school without knowing how to count, I would not form any opinion of his/her mathematical ability but I would ask him/her for a substitute that he/she has developed.

With regards to formulas, I do not know whether they are really of interest except for the method that obtained them. For instance, one can give a formula for the area of a circle but that does not mean much except for the applications that it may produce. The general ways in which one may obtain that formula (which I found interesting the first time I saw them) are of actual interest. Although a formula can sometimes reflect the method which produced it, the method will have deep insights which the formula cannot encapsulate. --PST 02:19, 12 September 2009 (UTC)

Factor in a load of things, then factor out what it's all about. That's maths by my WP:OR, though it doesn't say what it's all for. Dmcq (talk) 10:55, 11 September 2009 (UTC)

I'd say mathematics is the study of abstract structures. And two abstract structures are the same precisely if they're isomorphic. Hence mathematics is the study of truths that are preserved by isomorphism. Michael Hardy (talk) 21:01, 11 September 2009 (UTC)

I think that Michael's definition above concurs with what Cantor once famously quoted - "Mathematics is not the study of objects but the relations between them." I agree that mathemaics is the study of truth preserved by isomorphism, as well. Nevertheless one runs into the point that Meni Rosenfield stated above - "Theoretical computer science is explored with the mathematical method, but is not usually considered to be included in the mathematical body of knowledge." If one thinks about it, all fields require the logical deduction of truth but the nature of the "axioms" change. For instance, scientists will logically deduce truth from experimental evidence to build a theory. Although my next comment will be debatable, I think that it is the case - all bodies of knowledge are in some sense mathematical in nature. This is perhaps explained by the fact that there are many mathematicians in history who have explored other fields (to name a few, Emanuel Lasker and John Nash, but there are far more). In effect, every field is mathematics once the "truth" has been decided. Deciding the truth is what is unique to the field. --PST 02:19, 12 September 2009 (UTC)

Paper towels

If you know the radius of the roll and the radius of the inner cylinder, what is the formula for the length? Thickness is harder to measure, so I guess what I really want is a formula for how much is left when the radius is half what it started out to be.Vchimpanzee · talk · contributions · 22:23, 10 September 2009 (UTC)

The length of the paper towel on the roll will depend upon the thickness of an individual sheet of paper; there is no way around that. If you find it difficult to measure the thickness of one sheet, you could try measuring the thickness of ten or so sheets (say, wrapped around the roll to make it easier) and divide by ten.

Once you know the thickness of an individual sheet, there are a few ways to estimate the length. One way might be to measure the volume of the paper towels (by measuring the volume of a cylinder with the larger radius and subtracting the volume of the inner cylinder), and use that and the thickness to get the length. If you are careful, you will notice that this method does not depend upon the height of the roll (if you leave the height h as a variable when computing the volume, you will notice the h cancels out at the end when computing the length of the paper towel).

As for your second question, "how much is left when the radius is half...", I don't understand what you are asking. Do you mean, what proportion of the paper towel has not been used when the outer radius has been halved? Eric. 216.27.191.178 (talk) 23:02, 10 September 2009 (UTC)

Usually they tell you on the package how many sheets are in a full roll, and the size of each sheet. That lets you figure out the length of a full roll (call this L). The length will also be (approximately) proportional to the area of the annulus that you see when you look at the roll from the end. Let R be the radius of the full roll and r be the radius of the inner cylinder. Then the area of the annulus is $A_{1}=\pi (R^{2}-r^{2})$. The thickness of the annulus is of course R-r. When enough towels are used up that the outer radius of the roll is half what it started at (i.e. it is R/2), the area of the annulus is of course $A_{2}=\pi \left(({R/2})^{2}-r^{2}\right)$. The length of that partial roll will be ${A_{2} \over A_{1}}L$. If what you are really asking is the length once the thickness is half what it started as, then just find the new outer radius $R_{2}=r+{{R-r} \over 2}$ and similarly calculate the new area and new length. —Preceding unsigned comment added by 70.90.174.101 (talk) 23:21, 10 September 2009 (UTC)

A similar question was posed here. Clearly, a common mathematical thought induced by the closet --"When I'll get out, I'll post on the RD/M" ;-) --pma (talk) 06:32, 11 September 2009 (UTC)

September 11

Matrices and equations

I know that one may use matrices to solve linear equations of the form

${\begin{cases}a_{1}x_{1}+a_{2}x_{2}=c_{1}\\a_{3}x_{1}+a_{4}x_{2}=c_{2}\\\end{cases}}$

Where the c and a are constants. There are a number of methods, but the one I know best is Cramer's rule, which makes the solving of x₁ and x₂ a relatively simple matter:

$x_{1}={\frac {\begin{vmatrix}c_{1}&a_{2}\\c_{2}&a_{4}\end{vmatrix}}{\begin{vmatrix}a_{1}&a_{2}\\a_{3}&a_{4}\end{vmatrix}}}={{c_{1}a_{4}-a_{2}c_{2}} \over a_{1}a_{4}-a_{2}a_{3}}$

$x_{2}={\frac {\begin{vmatrix}a_{1}&c_{1}\\a_{3}&c_{2}\end{vmatrix}}{\begin{vmatrix}a_{1}&a_{2}\\a_{3}&a_{4}\end{vmatrix}}}={{a_{1}c_{2}-c_{1}a_{3}} \over a_{1}a_{4}-a_{2}a_{3}}$

This is all well and good, but Cramer's rule hits a stumbling block as soon as we have an x₁ or x₂ with a degree ≠1. My question is this: can matrices be used to solve systems of equations that are not linear, such as those containing polynomials and more complex expressions? For example, could a matrix be used to solve this?

${\begin{cases}3x^{2}-2y+5z=356\\4x+y-10z^{\frac {1}{2}}=145\\6x+11y+15z=512\\\end{cases}}$

Thanks for the help. 58.168.51.217 (talk) 07:41, 11 September 2009 (UTC)

To a great extent, no. Solving polynomial systems involves much, much more mathematics, both in the theory and algorithms. Special systems, of course, can be reduced to linearity, e.g. (changing your example a little) you can solve this

${\begin{cases}3x^{2}-2y+5z^{\frac {1}{2}}=356\\4x^{2}+y-10z^{\frac {1}{2}}=145\\6x^{2}+11y+15z^{\frac {1}{2}}=512\\\end{cases}}$

first as a linear system in x² ,y , z^1/2. --84.221.209.213 (talk) 09:14, 11 September 2009 (UTC)

Can you point me toward that mathematics? And I think I'm seeing what you've done here. What we find for z (for example) through Cramer's is really the square root of the z we're looking for. Is that correct? 58.168.51.217 (talk) 10:54, 11 September 2009 (UTC)

Correct. But are you more interested in explicit computations, algorithms, approximation of solutions, or more in the qualitative aspect: what is the shape and the structure of the set of solutions, in particular, existence results and so on. The two aspect are linked, of course. Another point is, what kind of functions a have you in your mind: polynomials, analitic functions, differentiable functions, continuous. Each has a whole theoried behind: commutative algebra algebraic geometry differential geometry nonlinear functional analysis degree theory fixed point theory...

Maybe the most simple interesting result, which is satisfactory both theoretically and for the practical point of view, is the contraction principle. --131.114.73.84 (talk) 12:09, 11 September 2009 (UTC)

If you don't put in additional restrictions then even if all the coefficients are integers and there is only one equation there is no general solution. This is Hilbert's tenth problem. Dmcq (talk) 12:25, 11 September 2009 (UTC)

...yes: talking about integers solutions. That's another main issue: does the OP want integer, real or complex solutions? That changes completely theory and methods.--131.114.73.84 (talk) 13:23, 11 September 2009 (UTC)

There is also the issue that even single-variable polynomial equations may have no easy solution, e.g. x⁵−x−1 = 0 has no solution in radicals. — Carl (CBM · talk) 13:31, 11 September 2009 (UTC)

As an applied mathematician, I would point you to Newton's Method. If you just want numerical values for x, y, z that solve the nonlinear system, that is you best bet. If you want to know if you can express the solution algebraically (i.e. something like x = sqrt(5)+log(4)), then you'll have to ask the pure mathematicians...

BTW: Cramer's rule is a pretty ineffcient way to solve a linear system of equations. Gaussian Elimination is much better. Cramers rule requires you to find n+1 nxn determinants - the most efficient way to calculate a determinant is ... Gaussian Elimination! So better to use that straightaway.195.128.250.134 (talk) 23:25, 11 September 2009 (UTC)

Irrational perfect set

How can I find a nonempty perfect set in $\mathbb {R}$ which has no rational numbers in it. The book in which I found this problem has not discussed measure theory as yet, and so I do not want to use it in any way. Thanks--Shahab (talk) 07:45, 11 September 2009 (UTC)

for instance play with the binary expansions: fix an irrational number c, and consider all numbers 0<x<1 whose binary expansion has x_2k=c_k for all k, the other digits being free.--pma (talk) 08:03, 11 September 2009 (UTC)

Re OP: what do you mean by "perfect"? If you mean perfect set, the set of all irrationals is an example. If you mean perfect closed set, you can do it by a Cantor-set like construction, where you add an extra technique that uses an enumeration of the rationals. I don't want to say more in case this is a homework problem; once you see how the hint works you will be fine. — Carl (CBM · talk) 13:34, 11 September 2009 (UTC)

Perfect sets are always closed. The set of all irrationals is not perfect. — Emil J. 13:50, 11 September 2009 (UTC)

You're right, of course; I am so used to saying "perfect closed set" that I have to think about whether the terms are redundant. In any case, we have three solution techniques now. — Carl (CBM · talk) 13:53, 11 September 2009 (UTC)

More abstractly, you could begin with the inclusion of the Cantor space 2^ω into the Baire space ω^ω and compose this with a homeomorphism between Baire space and the irrational numbers. — Carl (CBM · talk) 13:40, 11 September 2009 (UTC)

No Carl this isn't homework. (Unfortunately I don't go to school). Anyway I thought about pma's hint and came up with the following reasoning. I want to confirm whether this is correct. Let c be an irrational number in [0,1] and $E:=\{x\in [0,1]:x_{2k}=c_{k}\}$. Clearly as the 2kth digit of any $x\in E$ doesn't follow a pattern so x is irrational. Moreover any neighborhood of such an x is bound to contain a $y(\neq x)\in E$ as other then the even placed digits all others may be freely chosen. So all points of E are its limit points. Also if y is any limit point outside E then let 2t be the first position from the left in the decimal expansion of y such that y_2t doesn't coincide with c_t, and let e be the rational number in (0,1) with 1 at the 2t position and 0's elsewhere. Then (y-e,y+e) doesn't contain any element of E. Hence E is closed. --Shahab (talk) 19:05, 11 September 2009 (UTC)

Correct! (Alternatively, you can prove that E is homeomorphic to the Cantor space 2^ω quoted by Carl, certainly a compact space with no isolated points.) --pma (talk) 20:41, 11 September 2009 (UTC)

Life expectancy of the oldest

Today's passing away of Gertrude Baines made me think of the kind of question my old physics prof liked to ask: A general question that requires an understanding of what is relevant, which assumptions to make, and how to apply the correct mathematical methods. Maybe I shouldn't post this here, because what I'm looking for is not a mathematically precise result, but rather an estimate of the order of magnitude. I wanted to figure this out myself, but I need to do other things now; and because it is connected to a current event I rather posted it here for others who might enjoy it as long as it's fresh:

What is the life expectancy of someone who just became the oldest person? Or, asked differently, how long is the expectation value for the time a person holds the title "oldest person"? Of course, you could just go to Oldest people#Oldest living people since 1955 and divide the observed time by the number of people, but how would you go about it if you didn't have that information? If you don't like ignoring what you already know, you might go the other way and ask: Given the observed life expectancy of the oldest, what conclusions can we draw about Senescence § Theories of aging? — Sebastian 20:23, 11 September 2009 (UTC)

The fact that they are the oldest person isn't relevant. You would calculate their life expectancy in the same way as with anyone else, using a life table. Life tables aren't particularly accurate at the upper end due to the very small sample size. If someone is the older person ever, then there aren't any data points to base the expectation on. You could try and extrapolate based on the last few years you do have data for, but that wouldn't be very reliable at all. --Tango (talk) 20:35, 11 September 2009 (UTC)

What the conditional expected lifetime is, given that the person is the oldest person, but not given the age, is a perfectly valid question in the right context. But I think the "right context" for me might involve knowing more than we're given here. Michael Hardy (talk) 20:57, 11 September 2009 (UTC)

I think the events are independent. How does someone dying 1000 miles away affect the chances of another person dying? Even if they were both in some kind of cohort initially. An insurance company might revise it's estimate of the probability based on the new evidence, but I think the actual probability remains the same.--RDBury (talk) 21:23, 11 September 2009 (UTC)

Well, yes, you could work out the probability distribution of the age of the current oldest person and use that, combined with the life tables, to get the life expectancy of a randomly selected oldest person, but I'm not sure why you would want to. --Tango (talk) 21:41, 11 September 2009 (UTC)

Jeanne Calment's answer at age 120 was "A very short one!".John Z (talk) 22:41, 11 September 2009 (UTC)

I guess she already said that at age 90, when she negotiated the reverse mortgage.  ;-) — Sebastian 07:45, 12 September 2009 (UTC)

The general statistical methods for such problems are described in extreme value theory. Robinh (talk) 06:23, 12 September 2009 (UTC)

Yup, that's basically what I've been looking for. I'm a bit surprised that it takes its own theory and that it was discovered so late - I expected it to be rather simple exercise. — Sebastian 07:45, 12 September 2009 (UTC)

Using discrete crime data points to estimate relative rates by geography

I have a set of datapoints that represent individual murders-- each datapoint has a longitude and latitude. How can I produce an estimate for any geographic point, such that the map could be painted with an overlay representing the estimated murder rate for a given area.

Obviously, there will be limits to how accurate the estimates can be, just trying to find a good ballpark sort of measure.

I had thought about doing something along the lines of:

Estimated rate at a point ~= Sum over all data points(1/distance to murder)

But this seems sort of arbitrary, in that, it's something I just made up. Perhaps murder rate should use an inverse square rule instead. Or maybe there is some more advanced procedure for estimating a location-based-rate using a random sampling of data points.

Advice greatly welcomed.

(also know, murder rate isn't the actual issue being explored, it's just easier to describe) --Alecmconroy (talk) 20:47, 11 September 2009 (UTC)

Since when did this forum get so morbid? Anyway, you probably want to find some sort of data analysis/data mining software. This kind of problem occurs in business all the time, though more people who might want to by a product than people who get murdered, and there is commercial software out there to help with it. I don't know if I'm allowed to mention a specific product but try Googling "mapping software" as a first step. It's an interesting problem but it's already being worked by people who are actually getting paid to do it.--RDBury (talk) 21:37, 11 September 2009 (UTC)

Yes, that is an interesting Rorschach isn't it-- when asked to come up with a meaning for a binary discrete samples, I just think murder. Murder has a certain binariness to it in a way that "crime" or "disease" or "profit" just lack. lol I guess I am morbid.

Any way, thanks to the commenters, I think I'm on the right trail. RDBury (or others)-- do we know offhand of any such software that would be easily available to do this? I actually already have the data in google maps, so if it can do it, that would be extra awesome-- but I can't find any mention it in the documentation right now. --Alecmconroy (talk) 09:20, 12 September 2009 (UTC)

Sounds like you want some kind of 2D histogram. Or, more sophisticatedly, density estimation. Yaris678 (talk) 21:47, 11 September 2009 (UTC)

Differential equation

Just wondering if anyone could give a hint (or can solve)

d2[fn(x)]/dx2 = -k/[fn(x)]2

Starts at f(0)=real number, f'(x)=0. Any ideas? (not advanced student)83.100.250.79 (talk) 22:45, 11 September 2009 (UTC)

It's a one dimensional motion with an inverse square law force. Multiply by d[fn(x)]/dx and integrate. You'll get the conservation of energy, that gives a first order autonomous equation, that you can integrate again. --78.13.143.41 (talk) 23:18, 11 September 2009 (UTC)

Got stuck at "Multiply by d[fn(x)]/dx and integrate." - left hand side integration by parts is giving me more dⁿ[f(x)]/dxⁿ ...83.100.250.79 (talk) 23:38, 11 September 2009 (UTC)

Use substitution. --COVIZAPIBETEFOKY (talk) 02:52, 12 September 2009 (UTC)

Substitute what? There's not a lot there.83.100.250.79 (talk) 11:00, 12 September 2009 (UTC)

I'm not going to do all your homework for you. Both integrals can be evaluated using the technique of substitution. Figure out what u should be in each integral, such that the expression inside the integral is of the form f(u)du. --COVIZAPIBETEFOKY (talk) 15:23, 12 September 2009 (UTC)

It's not homework, it's curiosity. And I'm looking for a link to an answer, or more helpful answer, I'm familiar with integration by substitution, but I'm stuck. Is this a well known differential. Can someone link to a method to solve it. Thanks.83.100.250.79 (talk) 15:34, 12 September 2009 (UTC)

If you do the multiplication suggested then on the LHS you will have a function (d(fn(x))/dx) and its derivative - that kind of integral is generally done by substituting u=(the function) (in this case, u=d(fn(x))/dx). On the RHS you have an expression involving a function (fn(x)) and the derivative of that function, so you should substitute u=fn(x). I haven't actually tried it, but that ought to work. --Tango (talk) 16:24, 12 September 2009 (UTC)

I still don't understand the multiplication method... I found a substitution: I found this [4] which gives the answer - I suppose that's what user:78... meant when they said "its one dimensional motion with inverse square law" - I found it by using that as a search term.

It'll probably make sense in a few moments once I've worked it through...83.100.250.79 (talk) 16:51, 12 September 2009 (UTC)

September 12

Simplify integral

Can this be simplified? ${\frac {d}{dv}}\int _{-\infty }^{\infty }\phi _{1}(v_{1})\left(\int _{-v_{1}}^{v-v_{1}}\phi _{2}(v_{2})dv_{2}\right)dv_{1}$

Probably some bad choice of variables here, I'll rewrite it first.

${\frac {d}{dv}}\int _{-\infty }^{\infty }f(y)\left(\int _{-y}^{v-y}g(x)dx\right)dy$

Thanks in advance--Yanwen (talk) 03:54, 12 September 2009 (UTC)

I get f*g where * means convolution. I'm kinda rusty with this type of problem so take for what it's worth.--RDBury (talk) 05:05, 12 September 2009 (UTC)

This appears to be an exercise (I'm guessing homework) in using the Leibniz integral rule and the first fundamental theorem of calculus. 67.122.211.205 (talk) 19:09, 12 September 2009 (UTC)

what is the value of k that must be added to 7,16,43,79 so that they are in proportion?

what is the value of k that must be added to 7,16,43,79 so that they are in proportion? —Preceding unsigned comment added by 122.168.68.145 (talk) 09:32, 12 September 2009 (UTC)

5.--RDBury (talk) 14:16, 12 September 2009 (UTC)

only if 16=19...Tinfoilcat (talk) 16:19, 12 September 2009 (UTC)

In proportion to what? --Tango (talk) 16:20, 12 September 2009 (UTC)

My interpretation was that you're supposed to add a constant k to each term and end up with a bit of a geometric series (i.e. al, al^2, al^3, al^4). This problem does not have a solution Tinfoilcat (talk) 16:28, 12 September 2009 (UTC)

Perhaps they don't have to be consecutive terms from a geometric series? Or could it just be that one must divide the next, which divides the next, etc.? --Tango (talk) 16:41, 12 September 2009 (UTC)