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June 18

Order of polynomial / Gaussian Numerical Integration

Is sqrt(x^7) + x^2 a polynomial of order 7 or 3.5? I want to know this because I want to find the minimum number of Gauss points I need to exactly integrate it, and minimum number of Gauss points = ceiling ( (order of polynomial +1) /2 )

122.57.166.246 (talk) 10:08, 18 June 2010 (UTC)

Neither. ${\sqrt {x^{7}}}+x^{2}$ is not a polynomial. This expression is not difficult to integrate, but I don't think you can integrate it exactly with Gaussian quadrature. Gandalf61 (talk) 10:24, 18 June 2010 (UTC)

The exam question is to find the minimum number of Gaussian points needed to integrate it exactly...is there a mistake in the question then? I would just like to know if the expression is considered to have an order of 3.5 or 7. Thanks

122.57.166.246 (talk) 10:54, 18 June 2010 (UTC)

Or it's a trick question. -mattbuck (Talk) 11:27, 18 June 2010 (UTC)

Try substituting x=u^2. Then you have to manage u^7+u^4. -- SGBailey (talk) 12:47, 18 June 2010 (UTC)

Good idea. But don't forget the extra factor of 2u from dx = 2udu. Gandalf61 (talk) 13:05, 18 June 2010 (UTC)

Usually the word degree rather than order is used. Michael Hardy (talk) 16:37, 18 June 2010 (UTC)

Getting help from refdesk for an exam question doesn't seem appropriate. 75.57.243.88 (talk) 19:11, 18 June 2010 (UTC)

For what it's worth, it is a previous exam question. Thanks everyone for their help. 122.57.169.173 (talk) 23:40, 18 June 2010 (UTC)

Hahn fields

What is a "Hahn field"? While Googling the phrase to get information about the Hahn Field Archeological District, I keep finding websites that refer to Hahn fields or Hartman-Hahn fields; here is one for you with JSTOR access — I don't have it, so I can't read it. This is another one, but it's too in-depth for me to understand. I've searched; we don't appear to have an article on the topic. Nyttend (talk) 12:59, 18 June 2010 (UTC)

See Hans Hahn and Hahn series. I don't know who Hartman is. -- 58.147.53.127 (talk) 19:10, 19 June 2010 (UTC)

How do you call this number?

If you take four numbers, which are not all equal, i.e. 4277, and reorder them from bigger to smaller - 7742 - and re-order them from smaller to bigger - 2477 - and subtract this second number from the first, and go on with the same algorithm with the resulting number, you'll always obtained a specific number. I forgot the name, but it is the name of an Indian mathematician. Does anyone know what it is called?--Quest09 (talk) 17:39, 18 June 2010 (UTC)

Kaprekar's constant Dmcq (talk) 17:54, 18 June 2010 (UTC)

Why don't you do those calculation yourself? You would easily find that the number is 6174. Then you'd find an appropriate Wikipedia article. --CiaPan (talk) 17:59, 18 June 2010 (UTC)

No biting there CiaPan, it was a legit question :-( hydnjo (talk) 02:33, 19 June 2010 (UTC)

I don't bite. I didn't know the answer either, so just tried it with a simple four-operation calculator, and got the result witin two minutes, with no advanced maths or even any special efforts. I just started with the proposed 4277 and got following results:
7742–2477 = 5265
6552–2556 = 3996
9963–3699 = 6264
6642–2466 = 4176
7641–1467 = 6174
and saw 6174 has same digits as 4176, so it loops back to the same subtraction it came from, so this is the number, which Quest09 asks for. Then I simply entered it into the Wikipedia's search box and got it: '6174 is known as Kaprekar's constant'. It definitely didn't require lots of work. Or did it?
Finding it ourselves is much faster than waiting for the answer, and with the algorithm given anyone is able to do it. One only needs to try.... --CiaPan (talk) 06:58, 21 June 2010 (UTC)

CiaPan, no I wouldn't have reached the number. Since I didn't know how many repetitions were possible. I'd have gone through it certainly, but how could I recognize it?--Quest09 (talk) 10:32, 19 June 2010 (UTC)

You know you have reached it, because at the next repetition it gives itself again. --84.220.119.125 (talk) 12:19, 19 June 2010 (UTC)

How about trying it out and having look instead of giving up before you start? Dmcq (talk) 10:53, 19 June 2010 (UTC)

The Kaprekar's_constant article and the D. R. Kaprekar#Kaprekar constant section both state that this property (all non-trivial orbits leading to a single fixed-point) is unique in the decimal system to 3 and 4 digit numbers, but the reference given only states that this "appears" to be the case "only for three and four digit numbers", and examples are given through ten digits. Has this assumption been proven? -- 58.147.53.127 (talk) 13:50, 19 June 2010 (UTC)

No, this is not unique to 3 or 4 digit numbers in base 10. The Kaprekar mapping has one or more non-zero fixed points for almost any number of digits. The reference you gave has a table (near the end) of fixed points or "kernels" of the mapping up to 10 digits - for example, with 6 digits we have:

$182738\rightarrow 763533\rightarrow 431766\rightarrow 631764\rightarrow 631764\rightarrow \dots$

OEIS Sequence A099009 lists fixed points up to 14 digits long, and also gives patterns for generating 3 different infinite sequences of fixed points with increasing numbers of digits. One of these patterns generates a fixed point for any even number of digits ≥ 4; another generates a fixed point for any odd number of digits ≥ 9; and the third generates a fixed point for any number of digits that is a multiple of 3. So the only digit lengths for which a non-zero fixed point does not exist in base 10 are 2, 5 and 7. The Kaprekar Routine page at MathWorld lists fixed points for other bases as well. Gandalf61 (talk) 08:55, 20 June 2010 (UTC)

Thank you; that is very interesting. My question is, however, about the unreferenced assertion that three and four digit numbers are unique (in base 10) for having a single fixed point to which all non-trivial orbits lead. -- 58.147.53.145 (talk) 12:09, 20 June 2010 (UTC)

Oh, yes, I see. Well, looking at the much longer list of fixed points here, we can see other families of fixed points which provide at least two fixed points for each even length ≥ 8 and each odd length ≥ 17. So the only digit lengths with a single fixed point are 3, 4, 11 and 13 (6, 9 and 15 have a second fixed point because they are multiples of 3). And a quick spot check of 11 and 13 shows that there are longer loops for these digit lengths. So 3 and 4 are indeed the only digit lengths for which the Kaprekar mapping sends all numbers eventually to a single fixed point.

A related question is whether are other digit lengths for which the mapping has only fixed points (even though not unique) and no longer loops. I doubt this, as I expect there are extendible patterns for generating loops for any digit length, in the same way as there are for fixed points - but it would make an interesting investigation. Gandalf61 (talk) 13:38, 20 June 2010 (UTC)

Excellent. Thank you. -- 58.147.53.145 (talk) 06:12, 21 June 2010 (UTC)

Mathematicians by country

Which countries provides the world with the best mathematicians?--Quest09 (talk) 17:41, 18 June 2010 (UTC)

Before anyone attempts to answer this question I'd like to ask Quest09 to define what they mean by "best". Zunaid 18:38, 18 June 2010 (UTC)

And what they mean by "provide". The country which provides the best (for pretty much any definition of best) mathematicians with jobs and money for their research has got to be the US. If the meaning is where the "best" mathematicians are born, for example, now that's a rather different question.—Emil J. 18:48, 18 June 2010 (UTC)

On a quick look at [1] from The MacTutor History of Mathematics archive I see that Scotland has had almost as many as the US and Ireland is getting up there with China and India, and England has the most by far. Dunno why but I get this niggling feeling that it might be just a teensy weensy bit biased ;-) Dmcq (talk) 19:04, 18 June 2010 (UTC)

My experience has always been that the Russians, and more generally former USSR countries turn out the most brilliant mathematicians. The problem is that they have never been recognised in the past because much of their work was secretive and even if it hadn't have been; they wouldn't have been given any credit in the West because of political pressures. I think the Russian education system has a lot to do with it; IMHO it is vastly superior to the American education system (and indeed many of the European systems too). The late, great V. I. Arnold was very critical of the American system (he didn't like the simplification of the curriculum and the lowering of standards required to accommodate ill-prepared students, he made the criticism that they could graduate with a jazz history course, but not necessarily with a history of algebra course ). •• Fly by Night (talk) 10:44, 19 June 2010 (UTC)

I think saying that Soviet mathematicians "have never been recognised in the past" or "given any credit" in the west is a bit of an exaggeration. Izvestiya: Mathematics was translating Soviet mathematics into English from 1937 1967, Sbornik: Mathematics from 1967, Russian Mathematical Surveys from 1971. There were many Soviet mathematicians known by their reputation and their work to western colleagues. Tinfoilcat (talk) 12:31, 22 June 2010 (UTC)

June 19

Estimating a minimum from discrete points

Suppose I have a set of discrete values sampled from a function that has no slope discontinuities. I don't know the underlying function, all I have are data. The function looks roughly like a skewed hyperbolic curve with a minimum.

The minimum value sample I have isn't necessarily the minimum of the function. I would like to estimate the minimum (x,y) coordinate of the function somehow, without inventing a model underlying function to curve-fit the sampled values.

One way I can think to do this is to fit a cubic spline to the minimum 4 values and find the minimum of that. I was hoping to do it by projecting the slopes on either side of the minimum and seeing where they intersect, but this seems to work only when the sample values are symmetric around the minimum.

Is there some accepted way of doing this? ~Amatulić (talk) 07:06, 19 June 2010 (UTC)

I suppose you mean: Find (x,y) such that f(x,y) is minimum. But "without inventing a model underlying function to curve-fit the sampled values" makes the problem undefined. Bo Jacoby (talk) 11:28, 19 June 2010 (UTC).

By analogy with the German tank problem, you could try:

${\hat {Y}}=m-{\frac {m}{k}}+1$

to estimate the lowest population y value, where m is the lowest y value observed from k such values. Not that this would give the corresponding x value, nor may the underlying distribution of y values be appropriate - but it's easy.→81.147.3.245 (talk) 12:41, 19 June 2010 (UTC)

Fitting a cubic function to the points in the region of the minimum sounds reasonable to me. Can't see any reason for a cubic spline though if you're only interested in the minimum. It you only use four points you can fit a cubic exactly. If there's error (uncertainty) in the function values you might want to fit to more points by least squares. The choice of how many points is then a trade-off between the amount of error and the size of the region over which the function is well approximated by a cubic—some external info on the amount of measurement error would probably help. Qwfp (talk) 14:19, 19 June 2010 (UTC)

A cubic spline may be needed just to bias the slope of the cubic section that makes the minimum. I think my underlying function is well-behaved enough that I won't need it, though. ~Amatulić (talk) 19:11, 19 June 2010 (UTC)

Continuity of the inverse function

Let M, N ⊆ ℝⁿ open sets, and f: M → N a bijective continuous function. My question: Is the inverse function f^-1: N → M always continuous? --84.62.192.52 (talk) 16:02, 19 June 2010 (UTC)

Yes. This is essentially the theorem of invariance of domain. Algebraist 17:12, 19 June 2010 (UTC)

OK, you said open sets. The proposition is of course not true in more general situations. Michael Hardy (talk) 05:44, 20 June 2010 (UTC)

Equilateral planets

Hi, sorry if this question sounds dumb to you (or even homework-ey), but I'm not entirely sure how to do it myself (and it's definitely not homework). First of all, Planet Earth is ~27,000±1,000 ly away from the galactic center. Now, imagine there are other two planets which are exactly the same distance from the Galactic Center as Earth, and are also exactly apart from each other and Earth (i.e. they would make up the vertexes of an equilateral triangle). My question would be, what is the distance between each planet? I know there's an easy way to know this, but I couldn't come up with it... Thanks in advance! :) 186.80.207.31 (talk) 18:04, 19 June 2010 (UTC)

The distance between a planet and the center is the radius of the circumscribed circle of the triangle; the distance between each planet is its side. Therefore, the latter is ${\sqrt {3}}$ times the former. -- Meni Rosenfeld (talk) 18:26, 19 June 2010 (UTC)

Thank you, but could I ask why? I don't get how you got that number from what you said. 186.80.207.31 (talk) 19:05, 19 June 2010 (UTC)

In Equilateral triangle, it is said that when the side is a and the circumscribing radius is R you have $R={\frac {\sqrt {3}}{3}}a$ (this can be found with some trigonometry). Then $a={\frac {3}{\sqrt {3}}}R={\sqrt {3}}R$. -- Meni Rosenfeld (talk) 19:13, 19 June 2010 (UTC)

I understand now. Thanks again! 186.80.207.31 (talk) 19:58, 19 June 2010 (UTC)

Don't just take the article's word for it -- when MR says if can be found with some trigonometry, he means very simple trigonometry. See 30-60-90 triangle for a diagram showing how applying the Pythagorean theorem to a bisected equilateral triangle give you the length of the side opposite the 60° angle as √3/2 time that of the hypotenuse. (That is, sin 60° = √3/2.) Then make a diagram of your planets and form a triangle with one vertex at the galactic center and the other two at two of your planets. Note that the central angle is 1/3 of a circle, or 120°, and that if you bisect that angle you get a 30-60-90 triangle with a hypotenuse of length R = 27,000 ly and the side opposite the 60° angle of length half the distance between planets. But you know that this latter length is √3/2 R, so the distance between planets is √3 R. -- 58.147.53.127 (talk) 01:17, 20 June 2010 (UTC)

"when MR says if can be found with some trigonometry, he means very simple trigonometry", I wouldn't bet on that always being true though ;-) Dmcq (talk) 11:56, 20 June 2010 (UTC)

It should always be obvious to even the most dim-witted individual who holds an advanced degree in hyperbolic topology. :) -- Meni Rosenfeld (talk) 12:12, 20 June 2010 (UTC)

Is the integral of 1/x = ln(x)+C or ln|x|+C ?

What is the integral of ${\frac {1}{x}}$
$\ln x+C$
or
$\ln {|x|}+C$
? I get conflicting advice ––220.253.96.217 (talk) 22:42, 19 June 2010 (UTC)

Well, it depends on how you look at it. Indefinite integrals are a little bit hokey anyway; the things that make more sense are definite integrals and antiderivatives. If you're thinking of indefinite integrals as definite integrals for which the limits are not specified, then the version with the absolute-value signs work, provided the unspecified limits have the same sign. --Trovatore (talk) 22:45, 19 June 2010 (UTC)

Obviously it doesn't matter when x is real, but what if x is complex? That would cause ln(x) and ln|x| to have very different values.--220.253.96.217 (talk) 22:58, 19 June 2010 (UTC)

Oh, it's completely wrong for any complex-valued calculation. ("It" being the one with the absolute-value signs.) --Trovatore (talk) 23:02, 19 June 2010 (UTC)

So the correct integral is ln(x)+C then? Does that still work when x is negative? ––220.253.96.217 (talk) 23:11, 19 June 2010 (UTC)

In a complex-valued context, it "works", subject to the hassles regarding picking a branch of the log function. In a real-valued context it doesn't work, because the log of a negative number is undefined. --Trovatore (talk) 23:13, 19 June 2010 (UTC)

OK suppose I want to find the area underneath $y={\frac {1}{x}}$ between –2 and –4. That would be ln(–2) – ln(–4) = ln(2) + (2k-1)iπ - (ln(4) + (2k-1)iπ) = ln(2) – ln(4) = ln(1/2), this assumes you choose the same value for k when finding the ln of –2 and –4. Is this a fair assumption to make?––220.253.96.217 (talk) 23:33, 19 June 2010 (UTC) —Preceding unsigned comment added by 220.253.96.217 (talk) 23:32, 19 June 2010 (UTC)

Well, let me put it this way: Did you get the right answer? (In a picky sense, no, you didn't, because you got a negative number, and area can't be negative, but with luck a minute or two's reflection will sort that out for you — you did get the right answer for the definite integral.)

Now, was it just luck, or can you prove it happens in general? --Trovatore (talk) 08:23, 20 June 2010 (UTC)

For reals the |x| is correct but the C isn't the same on both sides of x=0 in that you can't integrate over that point. For complex numbers you can get from one side to the other by going one way or the other round the singularity and use the complex logarithm everywhere. Dmcq (talk) 23:25, 19 June 2010 (UTC)

It's log|x| + constant if x is real and not zero; and of course if x is positive then the absolute value sign can be dropped. Allowing x to be complex opens other cans of worms; as a function of complex x, the absolute value function is not differentiable anywhere.

And as someone noted above, the constant isn't "constant" on the whole line; it's one constant when x is positive and another when x is negative, and the two constants may or may not be equal. Michael Hardy (talk) 23:53, 19 June 2010 (UTC)

The correct answer is: For every simply connected domain D which doesn't contain the zero, the anti-derivative of the function $f={\frac {1}{x}}$ (where f is defined in D) is: Log(x)+C (where Log is any continuous function satisfying exp(Log(z))=z for every z in D), that is: ln|x|+iArg(x)+C (where ln is the real function satisfying exp(ln(x))=x for every positive x, and Arg is any continuous function satisfying tan(Arg(z))=Im(z)/Re(z) for every z in D which has a non-zero real part).

For example, the area underneath $f={\frac {1}{x}}$ (where f is defined in D) along a continuously differentiable path between 6 and -2 is:

(Log(6)+C)-(Log(-2)+C) = (ln(6)+ 2kiπ + C) - (ln(2) + (2k-1)iπ + C) = ln(6)-ln(2) + iπ = ln(3) + iπ.

HOOTmag (talk) 08:16, 20 June 2010 (UTC)

As I said above, the whole notion of an "indefinite integral" (that is, the things with the +C at the end) is a little hokey. It works if you know what it means and why, but if you apply it blindly you can get into trouble.

That's a bit what you're doing when you compute $\int _{-2}^{6}{\frac {dx}{x}}$ and come up with the value $\log 3+i\pi \,\!$. What exactly do you mean by that? It certainly isn't "the area under the curve"; obviously area cannot have a nonzero imaginary part. It isn't the Riemann integral or the Lebesgue integral of that function on that interval; neither of those is defined here (they do have a Cauchy principal value, but it isn't what you calculated).

Now, your answer, $\log 3+i\pi \,\!$, isn't completely wrong. It is the path integral of ${\frac {dz}{z}}$, along a path in the complex plane from −2+0i to 6+0i that travels below 0+0i and never wraps around it. I think that's right; I had to draw a picture in my head. If I drew it wrong, maybe change "below" to "above". --Trovatore (talk) 08:39, 20 June 2010 (UTC)

The quantity of area is a complex number if the area is defined underneath a curve of complex function. HOOTmag (talk) 10:34, 20 June 2010 (UTC)

I thought I had a good imagination about such things but that eludes me. Dmcq (talk) 16:13, 20 June 2010 (UTC)

LOL :) HOOTmag (talk) 16:30, 20 June 2010 (UTC)

What I hear you saying here is basically that you want to define the phrase "area under a curve" to mean the definite integral. I suppose you can do that if you want, though it isn't really standard, and not literally correct (whereas the "area under a curve", when the curve is a nonnegative real-valued continuous function, is really literally an area).

So then forget about areas, and look at what I wrote about the definite integral. Your computation of the definite integral is unambiguously wrong, unless you mean a path integral in the complex plane. But even then you have to take the right path to get the answer to come out the same as the one you got. --Trovatore (talk) 19:06, 20 June 2010 (UTC)

What do you mean by "the right path"? Is there a "wrong" path? Remember that we're talking about a function 1/x defined in a simply connected domain D which doesn't contain the zero, so what kind of "wrong" continuously differentiable path can there be inside D? HOOTmag (talk) 20:54, 20 June 2010 (UTC)

Say what? No we're not. When you take out zero, it's no longer simply connected. --Trovatore (talk) 20:56, 20 June 2010 (UTC)

Oh, wait, I think I see. You want to fix a simply-connected set to work in. Fine, if you pick the right one, then yes, you get the answer you got. But sorry, you can't call it "the area under the curve". No no no, that's just not going to fly at all. --Trovatore (talk) 21:00, 20 June 2010 (UTC)

What do you mean by "if you pick the right one"? Is there a "wrong" one - as long as a simply connected domain D (which doesn't contain the zero) is concerned? What I'm saying is that For every simply connected domain D which doesn't contain the zero, the area underneath $f={\frac {1}{x}}$ (where f is defined in D) along a continuously differentiable path between 6 and -2 is: ln(3) + iπ. Note that just as the quantity of area can be a real non-positive quantity (when the function is real and unnecessarily positive, e.g. check the two areas above and under the curve of f(x)=x between -3 and 2), so the quantity of area can be a complex non-real quantity (when the function is complex and unnecessarily real). HOOTmag (talk) 22:56, 20 June 2010 (UTC)

Then your statement is just wrong again, ignoring your curious phrase "area under the curve". The path integral in question can take any value of the form ln(3)+(2n+1)iπ (where n is any integer), depending on what domain D you choose (and thus what paths are available to integrate along). Algebraist 23:03, 20 June 2010 (UTC)

Look again: For every simply connected domain D which doesn't contain the zero, the anti-derivative of the function $f={\frac {1}{x}}$ (where f is defined in D) is: Log(x)+C (where Log is any continuous function satisfying exp(Log(z))=z for every z in D), that is: ln|x|+iArg(x)+C (where ln is the real function satisfying exp(ln(x))=x for every positive x, and Arg is any continuous function satisfying tan(Arg(z))=Im(z)/Re(z) for every z in D which has a non-zero real part).

For example, the area underneath $f={\frac {1}{x}}$ (where f is defined in D) along a continuously differentiable path between 6 and -2 is:

(Log(6)+C)-(Log(-2)+C) = (ln(6)+ 2kiπ + C) - (ln(2) + (2k-1)iπ + C) = ln(6)-ln(2) + iπ = ln(3) + iπ.

HOOTmag (talk) 23:08, 20 June 2010 (UTC)

Look again yourself. Suppose your domain D is a thin curved strip which goes n-and-a-half times anticlockwise around 0 on its way from -2 to 6. Then any curve from -2 to 6 in D will also go n-and-a-half times anticlockwise around 0, and so Arg will go up by (2n+1)iπ en route. Algebraist 23:11, 20 June 2010 (UTC)

Ah, nice. I hadn't been able to picture any way to get the full general-n case. I think you can also draw a wiggly branch cut that forces you to get this answer, but I can't quite nail it down in my mind's eye — I might actually have to pick up a pencil. --Trovatore (talk) 23:18, 20 June 2010 (UTC)

Then the area underneath $f={\frac {1}{x}}$ (where f is defined in D) along a continuously differentiable path between 6 and -2 is:

(ln(6)+ 2niπ + C) - (ln(2) + (2n+1)iπ + C) = ln(6)-ln(2) - iπ = ln(3) - iπ.

Note that whenever I wrote iπ I intended to include also the option of -iπ. Note that mathematics can't define the very distinction between i and -i, i.e. if you take every true/wrong mathematical statement, and replace i by -i and vice versa, then you get again a true/wrong mathematical statement (respectively).

HOOTmag (talk) 23:41, 20 June 2010 (UTC)

My answer was correct for the definition of "area under a curve" you gave above, which corresponded to the standard path integral, and I sketched a proof of this, which your comment here in no way refutes. I would ask what your comment here is supposed to do, but since you seem to have no interest in engaging with other people's mathematical arguments, and you also seem never to have studied elementary complex analysis, I will instead leave this conversation. Algebraist 23:52, 20 June 2010 (UTC)

My comment was about your claim that any curve from -2 to 6 in D would go n-and-a-half times anticlockwise around 0, and so Arg would go up by (2n+1)iπ en route. so I responded that in your case the area underneath $f={\frac {1}{x}}$ (where f is defined in D) along a continuously differentiable path between 6 and -2 would be (ln|6|+ 2niπ + C) - (ln|-2| + (2n+1)iπ + C) = ln(6)-ln(2) - iπ = ln(3) - iπ. HOOTmag (talk) 00:16, 21 June 2010 (UTC)

(ec) If you pick the right one, you get the answer you got. If you pick a different one, you might not. I think with this formulation there are two possible answers; namely, the one you got, $\log 3+i\pi \,\!$, which corresponds (I think) to a branch cut extending through the top half-plane, and $\log 3-i\pi \,\!$, which corresponds to putting the cut through the bottom half-plane.

As for area being possibly complex, don't be silly. Area is area. It cannot be negative, and it is always real. --Trovatore (talk) 23:07, 20 June 2010 (UTC)

When I wrote iπ I meant also the option of -iπ. Note that mathematics can't define the very distinction between i and -i, i.e. if you take every true/wrong mathematical statement, and replace i by -i and vice versa, then you get again a true/wrong mathematical statement (respectively).

Can't the area be negative? check the two areas above and under the curve of f(x)=x between -2 and 2. According to your attitude, the sum of these two areas would be a positive quantity, as opposed to what's accepted everywhere in Maths. HOOTmag (talk) 23:25, 20 June 2010 (UTC)

No, area cannot be negative. Don't be silly. A definite integral can be negative; area cannot.

It is true that there is a symmetry between i and −i, and your claim about mathematical statements and changing between them is true, given sufficient qualifications, which you haven't given. Nevertheless it is not possible for the same quantity to be both i and −i. --Trovatore (talk) 00:00, 21 June 2010 (UTC)

Check the two areas above and under the curve of f(x)=x between -2 and 2. Adopting your attitude, the sum of these two areas would be a positive quantity, as opposed to what's accepted everywhere in Maths.

Who said that i and -i are the same quantity? I just said that when I wrote iπ I intended to include also the option of -iπ. HOOTmag (talk) 00:16, 21 June 2010 (UTC)

As to your second paragraph, I'm starting to see a pattern here — you write false things, then claim later that you meant something else. Of course in this case, as Algebraist has explained to you, the thing you are now claiming is also false.

As to the first paragraph, yes, the sum of the two areas is positive, and this is accepted everywhere in mathematics; your claim to the contrary is simply wrong. The definite integral is zero, but the definite integral is not the area. --Trovatore (talk) 00:37, 21 June 2010 (UTC)

What "pattern"? which "false" things? I've never said that i is equal to -i, have I? I just said that when I wrote iπ I intended to include also the option of -iπ, because "mathematics can't define the very distinction between i and -i, i.e. if you take every true/wrong mathematical statement, and replace i by -i and vice versa, then you get again a true/wrong mathematical statement (respectively)". That's what I said. where do you see here "pattern" or "flase" things?

You assume that any area can't be negative, but it's not what's accepted in maths. For example, the area (not only the definite integral) of f(x)=x between -2 and 2 is considered in Maths to be zero.

HOOTmag (talk) 01:21, 21 June 2010 (UTC)

Did you also mean 3iπ, 5iπ, and 7iπ, or are you ignoring my point above? Algebraist 23:27, 20 June 2010 (UTC)

No, I'd referred to Trovatore's argument. As to your last argument, I have already referred to it just beyond it. Look above. HOOTmag (talk) 23:45, 20 June 2010 (UTC)

Could you explain, per my comment below, how your argument would go wrong for $x^{-2}$? — Carl (CBM · talk) 23:45, 20 June 2010 (UTC)

If you explain in a rigorous manner how you intend to apply my technique of $x^{-1}$ in $x^{-2}$ then I'll try to show the error. HOOTmag (talk) 00:16, 21 June 2010 (UTC)

Parallel to your argument: In any simply-connected domain excluding the origin, the antiderivative of $x^{-2}$ is $-x^{-1}$, so $\int _{-1}^{1}x^{-2}dx=(-1)-(1)=-2$. But, actually, that integral diverges. — Carl (CBM · talk) 00:28, 21 June 2010 (UTC)

Actually, Carl, I'm having a little trouble following this one too, given that $\int _{-1}^{1}x^{-1}dx$ also diverges. What distinction are you making here? --Trovatore (talk) 00:33, 21 June 2010 (UTC)

I was just picking a different integrand that avoids the logarithm issue, might be more well-known as a divergent integral, and is always positive on the real line minus the origin. The technique would work equally well for $x^{-3}$, etc. — Carl (CBM · talk) 00:41, 21 June 2010 (UTC)

Oh, now I've figured out what you mean. So, my technique is intended to deal with the function f(x)=1/x areas defined by path integrals only (the path being continuously differentiable). HOOTmag (talk) 01:21, 21 June 2010 (UTC)

You have not figured out what I mean. What I mean is that the definite integral you are talking about diverges. — Carl (CBM · talk) 01:42, 21 June 2010 (UTC)

I have figured out. I'm not talking about definite integrals at all, but rather about areas defined by path integrals (the path being continuously differentiable). HOOTmag (talk) 02:04, 21 June 2010 (UTC)

But as Algebraist indicated, the value of this path integral depends on the exact shape of the path, in particular on how many times it winds around the origin. It does not depend only on the endpoints. — Carl (CBM · talk) 02:33, 21 June 2010 (UTC)

Algebraist claimed that any curve from -2 to 6 in D would go n-and-a-half times anticlockwise around 0, and so Arg would go up by (2n+1)iπ en route. so I responded him that in his case the area underneath $f={\frac {1}{x}}$ (where f is defined in D) along a continuously differentiable path between 6 and -2 would be (ln|6|+ 2niπ + C) - (ln|-2| + (2n+1)iπ + C) = ln(6)-ln(2) - iπ = ln(3) - iπ. HOOTmag (talk) 20:13, 22 June 2010 (UTC)

And you're wrong. --Trovatore (talk) 20:54, 22 June 2010 (UTC)

The fact that I'm wrong is equivalent to the fact that you are the smartest person in the world. HOOTmag (talk) 21:32, 22 June 2010 (UTC)

That's also wrong. Does my saying you're wrong qualify me as well? As to specifics the difference would depend on the layer you take for each side so in your terms it is more like (ln|6|+ 2niπ + C) - (ln|-2| + (2m+1)iπ + C) = ln 6 - ln 2 +2(n-m)-1)iπ Dmcq (talk) 21:47, 22 June 2010 (UTC)

He's not the smartest one, so there's no contradiction in the equivalence.

Anyways, adopting your terms, I claim that n=m. Had it not been the case, the path-integral-function (in the simply connected domain D which doesn't contain the origin) wouldn't have been continuous, and consequently it wouldn't have been a path-integral-function (in D) at all.

HOOTmag (talk) 22:09, 22 June 2010 (UTC)

If you're thinking of a branch cut, they don't have to be straight. It can form a spiral out from the origin for instance. Dmcq (talk) 22:29, 22 June 2010 (UTC)

I'm talking about a "regular" (i.e. bi-dimensional) path integral (the path being continuously differentiable). Spirals need more than two dimensions. HOOTmag (talk) 23:06, 22 June 2010 (UTC)

You just keep digging.... --Trovatore (talk) 00:58, 23 June 2010 (UTC)

digging or not, my point (which hasn't been refuted yet) is that for every simply connected domain D which doesn't contain the zero, the anti-derivative of the function $f={\frac {1}{x}}$ (where f is defined in D) is: Log(x)+C (where Log is any continuous function satisfying exp(Log(z))=z for every z in D), that is: ln|x|+iArg(x)+C (where ln is the real function satisfying exp(ln(x))=x for every positive x, and Arg is any continuous function satisfying tan(Arg(z))=Im(z)/Re(z) for every z in D which has a non-zero real part).

My second point is that the "regular" (i.e. bi-dimensional) path integral of $f={\frac {1}{x}}$ (where f is defined in D the path is continuously differentiable) between 6 and -2, can be computed - using the familiar technique of anti-derivatives, this way:

$\int _{-2}^{6}{\frac {dx}{x}}$ = (Log(6)+C)-(Log(-2)+C) = (ln(6)+ 2niπ + C) - (ln(2) + (2n ± 1)iπ + C) = ln(6)-ln(2) ± iπ = ln(3) ± iπ.

HOOTmag (talk) 07:26, 23 June 2010 (UTC)

You should look at the articles Cauchy's integral formula and Residue theorem-- these are topics in complex analysis you seem to be missing. The function f(z) = 1/z has a pole at z=0 (since the denominator is z) with a residue of 1 (since the numerator is 1). This means that any path integral of f(z) that goes counterclockwise around z=0 once has the value 2πi.

Since line integrals are additive over paths, if the path goes counterclockwise around z=0 n times, the value is 2nπi, if it goes clockwise around z=0 n times, the value is -2nπi. (This n is the Winding number.) When you choose a path between z=6 and z=-2, as you know, you can't go straight across the origin because of the singularity at z=0. You can go above or beneath z=0, and it changes the final answer: ln(3)+πi or ln(3)-πi. You can loop around z=0 as many times as you want, which allows us to get ln(3)+(2n+1)πi for any integer n. The answers above that you reject really are correct, and it's important to understand why, or you'll miss something basic about the line integral you propose.

With line integrals, you should expect that a different path can give you a different answer. You need Cauchy's integral theorem to show that in this case, all we need to know about the path is how many times it loops around z=0. You should work through the proof of this theorem and the Residue Theorem to really understand your question; I'm sure that those who responded above have personally worked through the proofs of these theorems at some point. 98.235.80.144 (talk) 15:34, 23 June 2010 (UTC)

It looks like you haven't read all of what I've written, because I did refer to all of that many times, and once again my readers ignore the fact that I'm talking about a function which is defined in a simply connected domain which doesn't contain the zero, the path being inside this domain. So please read now our article Cauchy's integral theorem, and see that for every function f which is holomorphic in a simply connected domain D, F being f's anti-derivative, the integral of f between a and b, along a continuously differentiable path γ (in D), is: $\int _{\gamma }f(z)\,dz=F(b)-F(a).$. Conclusion: the path integral is path independent. HOOTmag (talk) 18:58, 23 June 2010 (UTC)

No-one is ignoring this fact. Everyone agrees that if you fix a domain D (simply connected, does not contain 0), then the value of the integral is independent of the choice of the path in D. But if you choose a different domain E (again simply connected and not containing 0), the value of the integral for paths in E may be different from the value obtained from paths in D. Example: if you choose a "fat semicircle" above the real axis the itegral is log(3)+iπ, and if you reflect this domain in the real axis the integral is log(3)-iπ. Tinfoilcat (talk) 19:14, 23 June 2010 (UTC)

You have ignored my computation, which coincide with yours. So let me quote myself:

$\int _{-2}^{6}{\frac {dx}{x}}$ = (Log(6)+C)-(Log(-2)+C) = (ln(6)+ 2niπ + C) - (ln(2) + (2n ± 1)iπ + C) = ln(6)-ln(2) ± iπ = ln(3) ± iπ.

How many times should I have to repeat this "±" ?

HOOTmag (talk) 19:27, 23 June 2010 (UTC)

I have not ignored your computation. Firstly you contradict yourself by saying 1) "the path integral is path independent", and 2) there are two different possibilities for the path integral (the ±). Secondly, there are infinitely many other possible values of the integral, depending on different choices of domain. Let me add another example: take a path that starts at 6, travels below the real axis then crosses it between -2 and 0, curves back again above the real axis so that it meets it to the right of 6, then proceeds below the real axis to -2. Then the value of the integral is...? Tinfoilcat (talk) 19:50, 23 June 2010 (UTC)

(ec) No, the different possible values have nothing to do with the fact that "i" and "-i" are algebraically indsitinguishable over the real numbers (i.e., the fact that complex conjugation is a ring automorphism of the complex numbers that fixes the real numbers).

Choosing a domain D and choosing a branch cut for the complex logarithm are different ways of saying the same idea, and it seems to us that you're missing this idea.

Below, I have drawn a path from -2 to 6 in the complex plane. This path is a simply connected domain D, not containing 0; and on this path, the integral is ln(3)-3πi.

Notice that $e^{-3\pi i}=-1$, so -3πi is one of the possible values for ln(-1). But you need to specify the branch cut or the domain D to know which value to use. On the domain D below, the only correct answer is ln(3)-3πi. It is not ln(3)+3πi or ln(3) ± πi.

                                             _____|_________             
                                            /     |         \          
                                           /   ___|__        \         
                                          /   /   |  \        \        
                                   ------(---|----|---)--------|-------
                                          \ -2    |  /         6       
                                           \______|_/                  
                                                  |                    

If you want to see the details of this example explicitly, the path I've drawn can be parametrized by $z={\frac {2+4t}{3}}e^{-\pi it}$ for t from 1 to 4.

Then the differential dz is $dz={\frac {4-\pi i(2+4t)}{3}}e^{-\pi it}dt$.

So we have $\int _{-2}^{6}{\frac {dz}{z}}=\int _{1}^{4}{\frac {4-\pi i(2+4t)}{2+4t}}dt=\int _{1}^{4}{\frac {4}{2+4t}}dt+\int _{1}^{4}-\pi idt=\ln(3)-3\pi i$.

98.235.80.144 (talk) 20:25, 23 June 2010 (UTC)

HOOTmag: Let us consider the following simply connected domain D which does not contain the origin: It contains all points except those on the spiral $\{\theta \exp(100i\theta )|\theta \geq 0\}$. Any path from -2 to 6 in this domain circles many times around the origin, and the integral along such path will have a large coefficient for the imaginary part.

Your mistake is that you think $\arg(z)$ must be between $-\pi i$ and $\pi i$ (or likewise). In fact, for a domain like the one I described, it takes a wide range of values. -- Meni Rosenfeld (talk) 09:17, 24 June 2010 (UTC)

No, this is not my mistake, because in my first post here I made it clear that the anti-derivative of the function $f={\frac {1}{x}}$ is: ln|x|+iArg(x)+C, and then I claimed that the area underneath $f={\frac {1}{x}}$ along a continuously differentiable path between 6 and -2 is: (ln(6)+ 2kiπ + C) - (ln(2) + (2k±1)iπ + C). So how could I have thought that $\arg(z)$ must be between $-\pi i$ and $\pi i$ (or likewise), after my clear words about the constant k involved in describing the function Arg?

If you really want to know about my thoughts, look at my response on 98.235.80.144's talk page.

HOOTmag (talk) 09:53, 24 June 2010 (UTC)

That was the "likewise" part - you thought that arg was always restricted to a range of $2\pi$, or something similar.

It looks like you understand now. People have tried to explain to you that there are domains like this as early as the post by Algebraist 23:11, 20 June, it's unfortunate that you have resisted for this long. -- Meni Rosenfeld (talk) 10:11, 24 June 2010 (UTC)

A well known proverb, in an ancient language, goes like this: LO HA-BIE-SHAHN LA-MED...

I resist as long as I feel that other people's arguments don't relate to my own argument. e.g. Algebraist's (and others') argument about the spiral was not intended to refute my thought about the constant k, but rather to refute other opinions which were mistakenly ascribed to me, while 98.235.80.144 did refer to my position.

HOOTmag (talk) 10:32, 24 June 2010 (UTC)

Just as a heuristic: any solution for $\int _{-1}^{1}x^{-1}dx$ that would work in the same way and give a finite value for $\int _{-1}^{1}x^{-2}dx$ cannot possibly be correct, because the latter integral diverges. But HOOTmag's technique would appear to work inthe same way for both of these integrals. The problem is that these integrals correspond to a path along the x axis, right through the singularity. — Carl (CBM · talk) 23:15, 20 June 2010 (UTC)

June 20

Fields medal

How come no women has ever got the fields medal? —Preceding unsigned comment added by 58.109.119.6 (talk) 08:16, 20 June 2010 (UTC)

Many people haven't got the Fields medal. Some of them were not qualified. The rules discriminate against age, but not against gender. Bo Jacoby (talk) 11:09, 20 June 2010 (UTC).

Also, mathematics is very much a male dominated subject. In my faculty there was a single female member of the academic staff amongst 50 or so male members. Why haven't any women won Fields Medals? Well, it's just a statistical thing more than anything else. •• Fly by Night (talk) 20:48, 22 June 2010 (UTC)

A-invariant subspaces

Let V a finite-dimensional vector space over a field K, A ∈ Hom_K(V,V), V = ⊕_i=1^k W_i with A-cyclic subspaces W_i ≠ {0}, and A_i := A_|Wi for 1 ≤ i ≤ k. Furthermore, the characteristic polynomials f_Ai are pairwise coprime. My question: Is v = ∑_i=1^k v_i with v_i ∈ W_i for all 1 ≤ i ≤ k an A-generator of V iff v_i is an A-generator of W_i for all 1 ≤ i ≤ k? --84.62.192.52 (talk) 13:26, 20 June 2010 (UTC)

If you don't get an answer here, you might try taking this question to www.mathoverflow.com . However if you post it there, you're going to need a little more motivation for the question. Read the FAQ - questions that aren't presented so as to interest mathematicians will be closed very quickly. 24.6.2.115 (talk) 22:37, 23 June 2010 (UTC)

Right-angled triangles in a quadrant

The "number of right triangles with nonnegative integer coordinates less than or equal to N and one corner at the origin." (A155154) seems to have a very regular graph- is there an explicit formula for this sequence? 70.162.12.102 (talk) 19:40, 20 June 2010 (UTC)

It seems very unlikely to me. There are n² triangles with the right angle at the origin, but then for the ones with the right angle elsewhere it gets pretty messy pretty fast. Any right triangle on the lattice is going to have its legs in a rational ratio so we can think of all the triangles as scaled up versions of the triangles with leg lengths a, b, with a, b positive and relatively prime. There are two possible orientations, so just consider the one where the right angle is closer to the x-axis than the other vertex. For given a, b there is the triangle with right angle at (a, 0) and the other vertex at (a, b) but we can also replace our "x unit" (0, 1) with any (c, d) where c, d are non-negative and not both zero. Then the "y unit" becomes (-d, c), so the vertices of the triangle become (ac, ad) and (ac - bd, ad + bc). For these points to be inside the region we want they have to satisfy bd ≤ ac ≤ n and ad + bc ≤ n. So the problem is now counting all the values of a, b, c, d that satisfy these conditions for a given n, which is not straight forward. Rckrone (talk) 21:31, 20 June 2010 (UTC)

I agree, there probably isn't a nice formula (it is, of course, almost impossible to prove that). There may, however, be an asymptotic formula that is very accurate even for small values of n. There must be some explanation for the graph looking so regular. --Tango (talk) 00:27, 21 June 2010 (UTC)

a_n ≈ 3.1735n^2.14937±7.65 for 0≤n≤44. Bo Jacoby (talk) 12:44, 23 June 2010 (UTC).

Or perhaps a_n ≈ 3.1356n^2.152891.00303^±1. Bo Jacoby (talk) 08:33, 24 June 2010 (UTC).

June 21

Books on Randomness

Could anyone recommend some good books on randomness? More specifically, how randomness affects everyday lives and is often interpreted by people as having patterns. It needs to be at a level that a non mathematician can understand. 'A Random Walk Down Wall Street' is a good example of what kind of book I am looking for though it does not need to be financially based. Also, I would be interested if there are any books that really delve into how randomness and perceived patterns can be utilized/exploited, for example Casinos using the house edge against gamblers thinking they have patterns and systems that beat the odds.

I searched the archive to see if there were any previous questions along this line but could not find any, although searching for randomness is a tough search. 63.87.170.174 (talk) 16:26, 21 June 2010 (UTC)

You might want to try The Black Swan by Nassim Nicholas Taleb. -mattbuck (Talk) 16:36, 21 June 2010 (UTC)

Based on the reviews at Amazon, it looks like The Black Swan is significantly worse than Taleb's earlier book, Fooled by Randomness, which I think is also more aligned to what the OP is looking for. I haven't read either yet. -- Meni Rosenfeld (talk) 17:35, 21 June 2010 (UTC)

The Drunkard's Walk: How Randomness Rules Our Lives by Leonard Mlodinow is a good non-mathematician's introduction, covering a lot of the areas you mention such as casinos. --OpenToppedBus - Talk to the driver 13:59, 22 June 2010 (UTC)

Grand Piano lid prop angles

Most modern grand pianos' lid props appear to form a 90º angle where they meet the underside of the piano's lid. It seems logical to me that the lid prop is less likely to slip at that angle because there would be a direct load transfer of the weight of the piano's lid to the support stick. That is, grand piano manufacturers intentionally use a 90º angle for safety reasons. Could someone show me the mathematics, perhaps using vector analysis, to prove my hypothesis?

The reader may want to visit http://en.wikipedia.org/wiki/Grand_Piano to see a couple of pianos that do not appear to use the 90º angle. Note the Louis Bas grand piano of 1781 and Walter and Sohn piano of 1805.Don don (talk) 16:52, 21 June 2010 (UTC)

Thinking about it I'm surprised any of the props meet the lid at a right angle. That means the prop would have to be supported on all sides to stop it from slipping after being knocked. If it is at an angle it could just fit into an angle, sounds easier to me. I don't think one need worry much about the strength of the prop and the force will always go straight down it since the ends aren't held firmly in place. Dmcq (talk) 19:57, 22 June 2010 (UTC)

Which Statistics Test?

I'm trying to determine the effect taking a particular class has on student retention rates to the next grade (sophomore year). I have the number of freshman students who started and the number who went on to sophomore year (about 66% of 1600). I also have the number of those same freshman who took this class, and how many of this subset went on (about 90% of 300). Obviously the class made a difference, but what test do I use to prove it (with significance)? It's been over five years since my last statistics class. I'm not really dealing with samples, these are the official numbers. Do I still use the z-score, right tailed hypothesis test... even though the z-score is 9.19 before fpc, 10.18 after? Do I use fpc even though it's not really a sample? I've got to run similar tests for ten other years.160.10.98.34 (talk) 20:47, 21 June 2010 (UTC)

Pearson's chi-square test seems appropriate here. But it looks like you have made an observational study, so keep in mind that correlation does not imply causation. -- Meni Rosenfeld (talk) 07:49, 22 June 2010 (UTC)

McNemar's test. HTH, Robinh (talk) 07:51, 22 June 2010 (UTC)

I can't see any pairing here, so I can't see why McNemar's test would be appropriate rather than Pearson's chi-square test. As for finite population correction (fpc), you use that if you're trying to estimate uncertainty in your estimate of a population parameter when your sample is a substantial fraction of the size of the population. Here, you sample is the entire population, so the fpc should be 0 as you know the proportion in the population exactly, i.e. its standard error is 0. However, i think you can still interpret statistical tests of a null hypothesis, such as Pearson's chi-square test, without assuming variability is due to sampling from a larger population. Another matter to consider as you're repeating this for ten other years is multiple comparisons. And Meni is right to remind you about not reading causation into this - you say "obviously, the class made a difference", but that's not obvious at all from what you've told us. Were the students randomly allocated to take this class? If they weren't, you can't assume those who took the class are comparable to those who didn't, e.g. maybe the more motivated students were more likely to choose this class. See self-selection. Qwfp (talk) 08:26, 22 June 2010 (UTC)

You meant "but that's not obvious at all". -- Meni Rosenfeld (talk) 08:34, 22 June 2010 (UTC)

Fixed, thanks! Qwfp (talk) 08:41, 22 June 2010 (UTC)

Thanks to all. Can I still do a Chi Square test even though the 1600 number includes the 300 students who took the class? Most of the examples I've come across deal with two independent, non-overlapping groups. Just so you know, I appreciate it's correlation, not causation. We're planning on testing whether the major indicators of academic success are different for these groups too. —Preceding unsigned comment added by 160.10.98.106 (talk) 13:06, 22 June 2010 (UTC)

The first step before you conduct the test is to construct a 2×2 contingency table of non-overlapping groups, i.e. "took class", "didn't take class" vs. "went on to next year", "didn't go on to next year". That's just a couple of straightforward subtractions. Qwfp (talk) 13:45, 22 June 2010 (UTC)

The probability P₁, that a student having taken the class went on, has a beta distribution with α=1+0.90·300=271, β=1+0.10·300=31, α+β=302, mean = μ = $\scriptstyle {\frac {\alpha }{\alpha +\beta }}$ = 0.897351 . standard deviation = σ = $\scriptstyle {\sqrt {\frac {\mu -\mu ^{2}}{\alpha +\beta +1}}}\!$ = 0.0174356 . So P₁ ≈ 0.897351±0.0174356 . The probability that a student not having taken the class went on, is P₂≈ 0.646837±0.0131106 . The difference P₁ − P₂ = 0.250514±0.0218149. Zero is μ−11.4836σ. This difference is highly significantly different from zero. Bo Jacoby (talk) 14:47, 23 June 2010 (UTC).

Huh? How do beta distributions come into this? This is just a straightforward comparison of two binomial probabilities, i.e. Pearson's chi-squared test for a 2×2 contingency table. There's no need to give the probabilities themselves a distribution. Granted in reality the probability of going on to the next grade may vary between individuals within each group (class takers and non-takers) depending on all sorts of other factors, but there's not the information here to say anything about that; all you know are the overall probabilities for each group. Qwfp (talk) 23:55, 23 June 2010 (UTC)

If you know a probability, P, and a sample size, n, then the number of successes in the sample, i, has a binomial distribution. But our situation is that we know n and i, while P is unknown. The distribution of the continuous parameter P is not binomial, but beta, with parameters α=i+1 and β=n−i+1. Bo Jacoby (talk) 09:15, 24 June 2010 (UTC).

Oh, i see now, sorry; you're taking a Bayesian approach, while Meni and I (and, i think, the orginal poster) were being frequentist. Either is fine and going to come to essentially the same conclusion here (though don't usually associate Bayesian inference with phrases like "highly significantly different"). I'm still more concerned about (over)interpretation. 21:38, 24 June 2010 (UTC)

June 22

Euclidean Geometry

A friend asked me to prove that the shortest distance bewteen two points is covered by a straight line. I told him that I couldn't, because this was an axiom of Euclidean geometry. Am I right? 173.179.59.66 (talk) 05:54, 22 June 2010 (UTC)

No, actually, that has been used as an axiom or definition (for example, by Archimedes), but not by Euclid. In his 20th proposition in The Elements, Euclid proves that any two sides of a triangle add up to more than the third side. It follows from this that a straight line is shorter than a route made up of a sequence of straight lines in different directions. I don't know whether or not Euclid also proves that a straight line is also shorter than a curved path; I don't think he would have been able to do it rigorously. --Anonymous, 06:41 UTC, June 22, 2010.

Okay, much thanks. 173.179.59.66 (talk) 06:52, 22 June 2010 (UTC)

Modern mathematicians use the calculus of variations to show that straight lines have the shortest path property. This also allows one to show that, for example, the shortest path along a sphere between two points is a great circle. HTH, Robinh (talk) 07:46, 22 June 2010 (UTC)

If you take the length of a smooth arc $\textstyle \alpha :[0,1]\to \mathbb {R} ^{n}$ to be $\textstyle \int _{0}^{1}|\alpha '(t)|dt$, then the inequality is just $\textstyle \int _{0}^{1}|\alpha '(t)|dt\geq \left|\int _{0}^{1}\alpha '(t)dt\right|=|\alpha (1)-\alpha (0)|.$ The analogous inequality follows if you use the more general notion of total variation of the arc $\alpha$ (then no regularity assumption is needed on $\alpha$). --pma 19:35, 22 June 2010 (UTC)

Kepler's First Law (Antiderivative)

In the process of deriving Kepler's First Law, the antiderivative $\int \!{\frac {ldr}{r{\sqrt {2\mu Er^{2}+2\mu Cr-l^{2}}}}}$ needs to be evaluated. By book gives the solution as an arcsine of something, but I've been having trouble reproducing the result. It seems that every math program I try (as well as my own attempts) yield a messy combination of imaginary numbers and logarithms. Can anyone help me solve this antiderivative? —Preceding unsigned comment added by 173.179.59.66 (talk) 10:45, 22 June 2010 (UTC)

I haven't checked where that equation comes from but are you sure about that r on its own in the denominator? That would be where your log comes from. Dmcq (talk) 11:22, 22 June 2010 (UTC)

Are you sure it isn't arcsec? ${\frac {1}{x{\sqrt {x^{2}-1}}}}={\frac {d}{dx}}\mathrm {arcsec} \,x$ 129.67.37.143 (talk) 11:28, 22 June 2010 (UTC)

Sorry my comment about logs, expression is okay. Try differentiating arcsin(a+b/r) and you should get a similar expression that you can change around to get that what you want. Dmcq (talk) 12:31, 22 June 2010 (UTC)

World Cup group stages - how likely is it that drawing of lots would be required?

I've been following the 2010 FIFA World Cup and there's been some discussion about scenarios that might lead to some of the groups being decided by drawing lots. This seems to me to be an unfair way to make the decision, and I'd like to know the prior possibility (i.e. not from where we are now but from the beginning of the tournament) that this might happen.

There are four teams in each group, each playing each of the others once, and the first two qualify for the next round. FIFA uses the following criteria to rank teams in the Group Stage.

1. greatest number of points in all group matches;
2. goal difference in all group matches;
3. greatest number of goals scored in all group matches;
4. greatest number of points in matches between tied teams;
5. goal difference in matches between tied teams;
6. greatest number of goals scored in matches between tied teams;
7. drawing of lots by the FIFA Organising Committee.

I guess the other thing we'd need to know is the general distribution of scores within the group stages, which we can find by looking at the group stage results from the last three world cups in 2006, 2002 and 1998 (before that there were fewer teams, which might possibly affect results - best to eliminate any variables we can).

Anyone care to have a go at working out the likelihood that one or more groups would have to be decided by drawing lots? Presumably you'd do it by running some kind of simulation, but I know my statistical knowledge isn't up to it. --OpenToppedBus - Talk to the driver 12:07, 22 June 2010 (UTC)

If every game in the group resulted in a equal draw - eg all 1-1 or all 0-0 - then whatever happens you get a 4 way tie. You either draw lots or start using fewest yellow/red cards or choose alphabetically or ... -- SGBailey (talk) 13:16, 22 June 2010 (UTC)

Of course, whatever criteria you have there's a possibility of a tie. My concern is that there's too high a possibility of a tie using the current criteria.

But that's only based on my perception. What I'm looking for is someone to say, "actually, based on a normal distribution of results, that'll only happen once every 50 years" - in which case I'd probably think it's fine. Or alternatively, "statistically, you're likely to have to decide something by drawing lots once in every other tournament", in which case I would think FIFA were remiss in not adding an extra criteria based on number of cards, number of corners, number of times teams have hit the woodwork, etc, any of which would be fairer than just picking names out of a hat. --OpenToppedBus - Talk to the driver 13:48, 22 June 2010 (UTC)

To the best of my knowledge its happened once in the history of the World Cup, in 1990 for Group F to be precise. Republic of Ireland and the Netherlands came 2nd and 3rd respectively after a draw, though both qualifed as that was back in the days of 6 groups where the 4 best 3rd placed teams qualified. Its never actually affected if someone qualified before at World Cup level, though may have happened in other competitions. - Chrism would like to hear from you 19:45, 22 June 2010 (UTC)

I disagree that any of those methods would be better than drawing lots. The object in soccer is very clear: score more goals than the other team and you win. If you use other criteria to decide who advances like how many corners a team takes, that adds new incentives. Then it become somewhat of a different game: a contest to score goals and not allow corner kicks. That's not really what the game is supposed to be. As it stands, it's more valuable to score a goal than to prevent the other team from scoring a goal, which seems a little bit questionable to me but whatever. I guess they want to encourage more aggressive play.

As for drawing lots, I don't see what's unfair about it even it's a really crappy way to lose. Also sorry this doesn't really address the question at all. Rckrone (talk) 22:43, 23 June 2010 (UTC)

June 23

hint needed for apparently difficult problem

Hello all... I have a problem which I have been grappling with for some time. Let b be a positive integer and consider the equation z = x + y + b where x,y,z are variables. Suppose the integers {1,2,...4b+5} are partitioned in two classes arbitrarily. I wish to show that at least one of the classes always contains a solution to the equation.

I have tried using induction on b. The case b = 1 has been solved entirely by me. But I cannot understand how to use the induction hypothesis to prove the result. The more I think of it, the more I feel that a different approach to the problem is needed, but I cant figure out what. It is sort of a special case of a research problem, which has been solved in a more general way. I have little experience of doing research on my own, and so will be glad if anyone can offer me any advice or hints. Thanks - Shahab (talk) 07:06, 23 June 2010 (UTC)

Do x, y and z need to be different? -- Meni Rosenfeld (talk) 07:26, 23 June 2010 (UTC)

No.-Shahab (talk) 07:34, 23 June 2010 (UTC)

It looks like a pigeonhole principle problem to me but I haven't figured it out yet. Dmcq (talk) 08:33, 23 June 2010 (UTC)

The other thing I can see is that if you consider x+x+b you tend to need to have the 0 mod 3 and 1 mod 3 ones into separate sets and then the 2 mod 3 ones have to be stuck in and cause problems. Dmcq (talk) 09:48, 23 June 2010 (UTC)

Here is an outline of a solution. Let's call the two partitions A and B and let's try to split the integers from 1 to 4b+5 between A and B so that neither A nor B contains a solution to z=x+y+b.

Let's put 1 in A. Then if b+2 is also in A we have x=1, y=1 and z=x+y+b all in A. So we must put b+2 in B.

Now, if 3b+4 is also in B, we have x=b+2, y=b+2 and z=x+y+b all in B. So we must put 3b+4 in A.

Now we have 1 and 3b+4 both in A. If 4b+5 is in A then we have ... what ? Or if 2b+3 is in A then we have ... what ? And if we put 2b+3 and 4b+5 both in B, along with b+2 which is already in B, then we have ... what ?

(There may be a more elegant solution that uses induction on b and/or the pigeonhole principle, but I can't see it.) Gandalf61 (talk) 14:45, 24 June 2010 (UTC)

That seems plenty elegant to me, thanks, I wish I'd spent a bit more time on it and got something like that. Dmcq (talk) 21:22, 24 June 2010 (UTC)

dy

In one of my textbooks it says $\int dy=[function]dx$ What is meant by the $\int dy$ —Preceding unsigned comment added by 76.230.251.114 (talk • contribs) 17:01, 23 June 2010

Generally the notation $\int dy$ means the same thing as $\int 1\cdot dy$. --Trovatore (talk) 17:50, 23 June 2010 (UTC)

I bet your textbook says $\int dy=\int f(x)dx$ meaning $y=\int f(x)dx$. Bo Jacoby (talk) 19:33, 23 June 2010 (UTC).

Solving equations like this

How do you solve equations like this for y:
${\frac {dy}{dx}}=xy$
Where the derivative of y with respect to x is equated to some function of y and x? —Preceding unsigned comment added by 203.22.23.9 (talk) 22:55, 23 June 2010 (UTC)

See separation of variables. --Tango (talk) 23:00, 23 June 2010 (UTC)

I'm not sure how rigorous this is, but we can multiply both sides by dx and then divide both sides by y. This gives a dx on one side and a dy on the other side, so we integrate to give

$\int {\frac {1}{y}}\,{\mbox{d}}y=\int \!x\,{\mbox{d}}x$

Evaluating these two indefinite integrals gives ln(y) = ½x² + c where c is the constant of integration. We can solve for y by taking the exponentiation of both sides to the base e. Writing k for e^c to simplify things we have

$y=e^{{\frac {1}{2}}x^{2}+c}=ke^{{\frac {1}{2}}x^{2}}.$

We can check this solution, and show that it is right:

${\frac {{\mbox{d}}y}{{\mbox{d}}x}}=kxe^{{\frac {1}{2}}x^{2}},\ \ xy=x\left(ke^{{\frac {1}{2}}x^{2}}\right)=kxe^{{\frac {1}{2}}x^{2}}.$ •• Fly by Night (talk) 13:48, 24 June 2010 (UTC)

From what I remember, the method that is normally taught is the integrating factor method. Rewrite the equation as

${\frac {{\mbox{d}}y}{{\mbox{d}}x}}-xy=0$

Now, our integrating factor is $e^{\int -x\,dx}=e^{-{\frac {1}{2}}x^{2}}$. Multiply through and we get

$e^{-{\frac {1}{2}}x^{2}}{\frac {{\mbox{d}}y}{{\mbox{d}}x}}-e^{-{\frac {1}{2}}x^{2}}xy=0$

This is actually the derivative of a product. Simplifying, we get

${\frac {\mbox{d}}{{\mbox{d}}x}}\left(ye^{-{\frac {1}{2}}x^{2}}\right)=0$

This can be easily integrated and rearranged to get

$y=ce^{{\frac {1}{2}}x^{2}},\,c\in \mathbb {R}$

Readro (talk) 14:16, 24 June 2010 (UTC)

See also separation of variables. 75.57.243.88 (talk) 21:45, 24 June 2010 (UTC)

June 24

Constrained solutions of a linear equation

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

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

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

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

Simple formula

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

Simplify the equation to 41s₁−36s₂−5p=0. The inequality s₁>p implies that s₂>p. The inequality p>0 implies that 41s₁−36s₂=5p>0, so 0<s₂/s₁<41/36. Choose 0<p<s₁ freely and compute s₂=(41s₁−5p)/36. Bo Jacoby (talk) 13:18, 24 June 2010 (UTC).

Probability Distributions

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

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

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