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June 15

Data analysis help

Suppose there is a row of 8 posts; the distances between the posts are the same. I take a measuring tape and precisely measure the distance from goal post 2 to every other goal post. Now, I need to calculate the distance between the posts as precisely as possible.

How should the data be analyzed? Should I do a linear regression, plotting distance against "m", where "m" is the mth post from post #2? If so, m=1 has two measurements (one from the second post to the first, another from the second to the third); how do I handle that? Also, when plotting the graph in Excel, should I make the y-intercept 0?

If a linear regression isn't the most accurate approach, what is? BTW, I'm actually trying to analyze some diffraction data I've taken. This isn't homework; the diffraction experiment was done to find the wavelength of a laser, but the different analysis methods I've tried give different results, and worse, this difference is well outside experimental error! --Bowlhover (talk) 06:30, 15 June 2010 (UTC)

I think linear regression is probably the best approach. Certainly better than taking differences between successive measurements and averaging these differences, which would amount to using only the distance from the first to the last. I'd call your reference post from which you took the measurements post 0, then the other posts are numbered –1, 1, 2, 3, 4, 5, 6. Plot the displacement (vector) from the reference post against the post number, i.e. the displacement to post –1 is negative. You can probably force the y-intercept to be zero, but i think i'd want to know more about exactly how you took the measurements to be completely sure (presumably the lines have finite width — did you measure to and from the centre of each line or the edge? Do they all have the same width?) --Qwfp (talk) 07:36, 15 June 2010 (UTC)

When the precisions dy_i are proportional to the measurements y_i, you should take the logarithm before you do the linear regression: log(|y_i|) = log(|i|)+k. Then you naturally have the y-intercept to be zero. Bo Jacoby (talk) 08:17, 15 June 2010 (UTC).

If you were careful to always measure from the same point of post 2 then you are trying to get the m in c+m*x+error, which is what linear regression is good at. The y intercept doesn't matter, the slope is what you want since c is the negative of the error for x=2. Dmcq (talk) 09:18, 15 June 2010 (UTC)

I think it's better if you describe the diffraction problem you have. The best course of action will crucially depend on a model for the measurement errors.

If you have several measurements of a quantity, each equal to the quantity plus an error which is normally distributed and independent of the other errors, then the optimal estimate will be a weighted average of the measurements, each weighted by the inverse of its error's variance. -- Meni Rosenfeld (talk) 09:30, 15 June 2010 (UTC)

To Meni & Qwfp: here's a pretty detailed description of my experiment. I shone a 532 nm laser onto a human hair so that the diffraction pattern appeared on a wall that's roughly perpendicular to the laser beam. I put a piece of paper on the wall so that the diffraction pattern was on the paper, and used a pencil to indicate where the minima were, and where the central bright region was. I repeated the experiment with a red laser whose wavelength I didn't know (and incidentally, isn't on the label). Then I used a plastic ruler to measure the distances of all the minima from the central region for both lasers. I measured from the center of the line I drew for the central minima to the center of the line for the minima in question. If this isn't the best approach, I can go back and remeasure the minima in a different way, though repeating the experiment itself would take quite a while. --Bowlhover (talk) 17:48, 15 June 2010 (UTC)

Instead of using a human hair, wouldn't it be easier to calibrate the wavelength of a laser by single slit diffraction? The relationships between the nulls and maxima from the central node are quite simple if you know the width of the slit. ~Amatulić (talk) 18:41, 15 June 2010 (UTC)

The outer lines will be further apart because the wall is flat. Dmcq (talk) 19:34, 15 June 2010 (UTC)

Assuming the posts or hairs aren't moving relative to each other, it sounds like your major source of error is in your measurement technique (the plastic ruler and pencil). Try repeating the experiment a few hundred times, then compute the mean and standard deviation as per usual. Science is fun sometimes - usually it's just hard yakka like anything else. Zoonoses (talk) 14:35, 18 June 2010 (UTC)

Getting gradient of a contour

I have three points on an unknown curve in 2d, and need to calculate the gradient at the middle point. My approach is to do a polynomial interpolation (second order, I think, if I have 3 points), solve the system of 3 equations where the points (xi,yi) are used in the equation of the parabola (ax^2 + bx + c = y), and get the gradient from the derivative (2ax+b=0) However, this assumes the parabolas used for the interpolation have to be symmetrical to the y axis.

(1) Should I just rotate everything so that the two side points will be on a horizontal line?

(2) Or are there simpler better methods, maybe a direct equation to get a gradient or a normal vector at one point just by giving its neighbors?

(3) If I have 5 consecutive points, a 5th degree polynomial would result, I think, in overfitting (Runge's phenomenon). Is there a better way to use the points which are only second neighbors of my middle point, to get a better interpolation/approximation? --131.188.3.21 (talk) 11:15, 15 June 2010 (UTC)

I'm just considering that the parabola might not be a good idea at all:

                    x (p2)
 
 x (p1)               x (p3)

In this case I'll get a very inaccurate solution, the tip of the parabola will be very far from the middle point p2, and even a linear interpolation and getting the median of the angle at p2 would do better (the median has the same direction as the "normal", the gradient would be just its rotation by 90 degrees). --131.188.3.21 (talk) 11:38, 15 June 2010 (UTC)

How do you know it's "inaccurate" if you don't know what the true curve is?

I can have reasonably many and dense samples, but having them too dense, I feel noise will have a very great effect. --131.188.3.21 (talk) 12:40, 15 June 2010 (UTC)

So there's noise! This completely changes the problem. You need to use as many points as possible and do least squares fitting. -- Meni Rosenfeld (talk) 12:48, 15 June 2010 (UTC)

I think that if you choose the minimum-curvature parabola as I suggest below, rather than the one with a vertical symmetry axis, the result will be more natural-looking (in particular, with a vertex closer to p2).

By "median" did you mean angle bisector? -- Meni Rosenfeld (talk) 12:04, 15 June 2010 (UTC)

Yes, I do. I'm sorry, but I'm not that good in English terminology. --131.188.3.21 (talk) 12:40, 15 June 2010 (UTC)

[ec] The problem is hopelessly underspecified. Any additional prior information you have about the curve or the points can completely alter the results.

It seems plausible that under some reasonable assumptions, a good interpolation will be the parabola for which the curvature at the vertex is minimal (a general parabola in the plane has 4 degrees of freedom; the points constrain 3; you can minimize the curvature over the fourth).

For 5 points, I'd use the conic section which passes through all points. -- Meni Rosenfeld (talk) 11:42, 15 June 2010 (UTC)

Thanks for the quick answer. The problem itself is that a number of sample points are given which should constitute a contour of some kind. I just need the gradients, normals etc. at those points. I think, for a reasonably smooth curve and samples close enough, 3 or 5 consecutive points would be enough, more would just result in overfitting. Yeah, a neural network to fit it for many points, with higher importance at the middle one and lower at the far ones would be a "perfect" solution, but I need a reasonably fast and simple one (at it has to be calculated maybe millions of times).

Do you mean by your suggestion with the parabola to rotate the parabola in the plane so that the "tip" of it will be at the middle point? --131.188.3.21 (talk) 11:50, 15 June 2010 (UTC)

hmm, and I think the conic section will not work if the curve is not convex. --131.188.3.21 (talk) 11:53, 15 June 2010 (UTC)

Is the position of each point known precisely, or are they measured with error? Error in one coordinate or both, in which case with equal errors or different errors? Are the points equally spaced in one coordinate or the other or in distance along the curve? Qwfp (talk) 11:58, 15 June 2010 (UTC)

They are not necessarily equally spaced, but by having many samples close to each other, I can choose points that are. I think I'll try out the exact parabola fitting for 3 points, and if it does not works adequately, I'll just approximately fit it to more than 3 points. --131.188.3.21 (talk) 12:40, 15 June 2010 (UTC)

[ec] For the oblivious setting, I think I have a simpler solution than my original suggestion: Interpolate with a circle passing through the 3 points. The center should be easy to calculate, and then the normals are just vectors from the center.

If you want something more advanced, splines is what you probably want. For more advice we do need some more details about your problem:

• Where do the contours come from? Are they completely general, do they follow some simple algebraic formula, are they smooth (how smooth)?
• How many points do you have per curve?
• How fast is "fast"? Is that millions of calculations per second, per millisecond, per hour? Modern computers can do quite a lot millions of times per second.
• How accurate should the results be?
• Are the points known exactly or approximately?
• Is there a significance to orientation, location and\or scale? Is an upright parabola as likely a priori as a lying down one? Is a circle as likely as a larger circle, or one that is centered on a distant point?

-- Meni Rosenfeld (talk) 12:44, 15 June 2010 (UTC)

The contours come from real world shapes, so there it not much a priori information about their orientation.

I think I'll try the following methods, in this order, if the accuracy of first one will not suit my needs:

1. Fit a circle through 3 points

2. Fit a parabola with minimum curvature through 3 points. (have 3 equations for the 3 points, + one for the curvature (what do you mean by that? Max radius of a circle at a point and two close neighboring points? I could not find how the term "curvature" is used for parabolas), but 5 unknowns from (Ax + By)^2 + Cx + Dy + E ... did I understood something wrong? )

3. Some numerical minimization for many points and a parabola or spline

Thanks for the ideas. --131.188.3.20 (talk) 13:06, 15 June 2010 (UTC)

Regarding parabolas: I mean Curvature in a non-parabola-specific sense. The curvature at the vertex of the parabola is inversely proportional to the distance between the vertex and the focus. I can't think of a reason the minimum-curvature parabola will necessarily have the middle point as a vertex, but I can't rule it out either. Anyway, finding this parabola is complicated and should give similar results to the circle, which is much simpler.

A parabola actually has 4 unknowns, because multiplying A, B, C, D and E by a constant gives the same parabola. -- Meni Rosenfeld (talk) 14:02, 15 June 2010 (UTC)

It sounds to me you are talking about a normal graph one value of y for each x like the way a hill goes up and down in vertical section rather than a 2d contour round a hill which forms a loop. For the graph a straightforwrd parabola would be the usual choice for three points though you might know better - for instance a sandpile might form a sharp triangle or you might know there is a lot of local variability in which case a flatter parabola which didn't go through the points might be an appropriate smoothening. There are articles on curve fitting and polynomial interpolation and others plus lots of packages to do it, but saying one interpolation is better than another is a difficult thing. Dmcq (talk) 14:51, 15 June 2010 (UTC)

Riddle (which I invented):

Resolved

Try to define the constant-symbol "i" as an imaginary unit, using merely: the very constant-symbol "i" (having no interpretation given in advance), quantifiers (universal/existential), conjunction (no other connectives), parentheses, identity, addition, multiplication, and two (bound) variables only. The formula is to be true if and only if the constant-symbol "i" is interpreted as an imaginary unit. The answer will be given here within 2 days if nobody has preceded me by Thursday midnight. HOOTmag (talk) 21:35, 15 June 2010 (UTC)

Working on the basis that 1 - 1 + 1 - 1 + ... can be defined as 1/2, I say that i = (-1) * (1/-1) * (-1) * (1/-1) * ... -mattbuck (Talk) 21:45, 15 June 2010 (UTC)

Who has let you use the symbols "1", "/", "-" ? HOOTmag (talk) 21:49, 15 June 2010 (UTC)

It doesn't make sense anyway... A divergent series can be made to equal anything you like, pretty much, and I don't see how that relates to a divergent product. --Tango (talk) 22:03, 15 June 2010 (UTC)

Of course. HOOTmag (talk) 22:15, 15 June 2010 (UTC)

Your challenge isn't particularly clearly defined, but is something like this acceptable: $\mathbb {C} =\mathbb {R} (i){\text{ such that }}i\cdot i=x\in \mathbb {R} {\text{ where }}\forall y\in \mathbb {R} \,(x\cdot x\cdot y=y){\text{ and not }}\forall y\in \mathbb {R} \,(x\cdot y=y)$? --Tango (talk) 21:53, 15 June 2010 (UTC)

So why not simply: (y)(iiiiy=y) and not (y)(iiy=y)? HOOTmag (talk) 22:15, 15 June 2010 (UTC)

That is slicker, but of course it doesn't distinguish i from −i. Neither does Tango's, of course. In fact i cannot be defined in the language of the real numbers with plus and times, if the only constant symbols you're allowed are to be interpreted as reals. That's because there's an automorphism of the complex numbers, in that language, that takes i to −i. --Trovatore (talk) 22:28, 15 June 2010 (UTC)

Sorry, but I didn't talk about the "number" i, but rather about the "constant-symbol i", which should be defined as "an imaginary unit", i.e. you should define the constant-symbol "i" as an either imaginary unit. HOOTmag (talk) 22:44, 15 June 2010 (UTC)

If you have a constant symbol whose intended interpretation is the imaginary unit, then there's nothing left to do. The answer to your riddle is just i. --Trovatore (talk) 22:46, 15 June 2010 (UTC)

I didn't talk about a constant symbol whose "intended interpretation" is the imaginary unit, but rather about a constant symbol "to be defined" as an either imaginary unit. i.e. you have to find a formula (containing just the symbols mentioned in the riddle), which is a true formula iff the constant symbol "i" is interpreted as an imaginary unit. HOOTmag (talk) 22:54, 15 June 2010 (UTC)

Well, you weren't very specific about that. I don't think it's standard to refer to −i as "an imaginary unit". What you have is the imaginary unit, which is only i, not −i. --Trovatore (talk) 22:56, 15 June 2010 (UTC)

When I talked about "an imaginary unit", I simply meant "a root of the equation: xx+1=0". If it hadn't been clear enough, then I've made it clear now. HOOTmag (talk) 23:06, 15 June 2010 (UTC)

You are, of course, right. You can't distinguish between i and -i. Any distinction is entirely arbitrary. I don't see why -i can't be considered an imaginary unit. It's a unit (in the number theory sense) in the Gaussian integers and it's imaginary. That seems like as good an interpretation of "imaginary unit" as any. --Tango (talk) 23:41, 15 June 2010 (UTC)

Depends on what you mean by "can't distinguish". There are precise senses in which you can't distinguish between i and −i, but they assume some sort of limitation on the expressive power of the language you're using. If you have a language that has a constant symbol for i, then it's easy to define i — it's already done.

By the way, HOOTmag, that's part of the problem with the way you expressed your riddle. If you have a constant symbol, then you don't need a definition for it. If you want a definition, then what you want is a formula that accepts one free variable, not constant.

Tango, I think you have in mind some more philosophical issue about whether i is really well-specified when we talk about it in natural language. That's an interesting question, but doesn't really bear on whether −i is "an" imaginary unit. As far as I know, when people talk about "the" imaginary unit, they mean i. --Trovatore (talk) 05:10, 16 June 2010 (UTC)

I've been talking about the constant-symbol "i" not yet interpreted. Note that a given language is not necessarily associated with any interpretation, untill you determine the very interpretation. Actually, I'm talking about a language in which the constant-symbol "i" is the only symbol which has no interpretation given in advance, and the formula should "define" the meaning of "i", so that the formula is to be true if and only if one interprets the constant-symbol "i" as an imaginary unit. Anyways, if you are not satisfied with constant-symbols having no interpretation given in advance, then I let you replace the constant-symbol by a free variable (which would be the third variable in the formula, since the formula is allowed to have also two bound variables). HOOTmag (talk) 08:03, 16 June 2010 (UTC)

What I'm explaining to you is the standard usages in mathematical logic. You can go ahead and make up your own if you want, but it will impede communication. --Trovatore (talk) 08:11, 16 June 2010 (UTC)

Our controversy over the "standard usages" (or over whether a language is necessarily associated with any interpretation) is not the issue, because I've already let you replace the constant-symbol by a free variable (which would be the third variable in the formula, since the formula is allowed to have also two bound variables). HOOTmag (talk) 08:38, 16 June 2010 (UTC)

It's not about whether a language is ipso facto associated with an interpretation. It's about what constitutes a definition. If you have a constant symbol i, then i is definable via the one-free-variable formula $x=i$, with free variable x. That's really all there is to it. I understand your rephrasing, and it's fine once you explain it. But it isn't standard usage, period. --Trovatore (talk) 08:44, 16 June 2010 (UTC)

Again, our controversy over the "standard usages" is not the issue, because I've already let you replace the constant-symbol by a free variable (which would be the third variable in the formula, since the formula is allowed to have also two bound variables). HOOTmag (talk) 08:48, 16 June 2010 (UTC)

It is not "our" controversy. The usage I have explained is standard. --Trovatore (talk) 08:52, 16 June 2010 (UTC)

Our new controversy over whether the old controversy is ours or not - is not the issue, because I've already let you replace the constant-symbol by a free variable (which would be the third variable in the formula, since the formula is allowed to have also two bound variables). HOOTmag (talk) 09:17, 16 June 2010 (UTC)

Hmm? No, there's no controversy there. I already said that your rephrasing, along with the explanation of what you meant by "an imaginary unit", was fine. --Trovatore (talk) 09:20, 16 June 2010 (UTC)

If there's no controversy, then my riddle is looking forward to your solution... HOOTmag (talk) 10:00, 16 June 2010 (UTC)

You could go with that, but I felt it was clearer to use x, and I was allowed two variables, so why not use them? I think it helps to make it explicit that i² is real. --Tango (talk) 23:41, 15 June 2010 (UTC)

You've used three variables, because you have quantified the "y" twice, i.e. your y's are two different variables, and x is the third variable. Anyways, you have used the negation, which is not a legitimate connective. To make it clear, I let you only use: the very constant-symbol "i", quantifiers (universal/existential), conjunction (no other connectives), parentheses, identity, addition, multiplication, and two (bound) variables only. HOOTmag (talk) 08:03, 16 June 2010 (UTC)

You've edited the question. It did not say that. It said "connectives". Logical connective says (in the 2nd paragraph): "Also commonly, negation is considered to be a unary connective." You can't change the rules half-way through. (As for y being two different things, you have a point, but it's easy enough to re-write it to get around that. I'm not inclined to do that, though, after you've edited the original question.) --Tango (talk) 18:22, 16 June 2010 (UTC)

You're right, so you've solved the riddle according to its old (wrong) version. I was wrong and I apologize. Anyways, how would you re-write your suggestion to get around the two y's? HOOTmag (talk) 19:52, 16 June 2010 (UTC)

$\mathbb {C} =\mathbb {R} (i){\text{ such that }}i\cdot i=x\in \mathbb {R} {\text{ where }}\forall y\in \mathbb {R} -\{0\}\,(x\cdot x\cdot y=y{\text{ and not }}x\cdot y=y)$ --Tango (talk) 20:36, 16 June 2010 (UTC)

Who has let you use the zero? The permitted symbols don't contain it. HOOTmag (talk) 21:39, 16 June 2010 (UTC)

You didn't explicitly say I could use $\mathbb {R}$, either, but if I'm using bound variables you've got to let me bind them. If you like, I can remove the -{0} bit and add "or y+y=y" to the bit in brackets. --Tango (talk) 22:23, 16 June 2010 (UTC)

I've provided a detailed list of symbols, none of which is the symbol $\mathbb {R}$, so you couldn't have used this symbol either. Anyways, now I understand how you can use only one "y" (provided that the list of permitted connectives is unlimited). HOOTmag (talk) 23:33, 16 June 2010 (UTC)

Take i to be the Pauli matrix sigma_x and take the identity to be the 2 by 2 identity matrix. Count Iblis (talk) 02:48, 16 June 2010 (UTC)

I'm looking for a formula which uses very specific symbols, but you haven't supplied such a formula yet. HOOTmag (talk) 08:03, 16 June 2010 (UTC)

I see! Also I made a stupid error anyway, as sigma_x^2 = 1. I should ave writen down the matrix which has a 1 and a minus 1 on the anti-diagonal. Count Iblis (talk) 14:04, 16 June 2010 (UTC)

As a Wikipedian my answer is

constant i = Imaginary unit

The problem I see with the question is that one has to assume the imaginary numbers and the various axioms relating to them in the first place if one is going to define i. In which case one might as well write something like i = (0,1) in the complex numbers. Dmcq (talk) 09:34, 16 June 2010 (UTC)

I've been talking about the constant-symbol "i" not yet interpreted. Note that a given language is not necessarily associated with any interpretation, untill you determine the very interpretation. Actually, I'm talking about a language in which the constant-symbol "i" is the only symbol which has no interpretation given in advance, and the formula should "define" the meaning of "i", so that the formula is to be true if and only if one interprets the constant-symbol "i" as an imaginary unit. Anyways, if you are not satisfied with constant-symbols having no interpretation given in advance, then I let you replace the constant-symbol by a free variable (which would be the third variable in the formula, since the formula is allowed to have also two bound variables). So, the premitted symbols will be as follows: the free variable "i", quantifiers (universal/existential), conjunction (no other connectives), parentheses, identity, addition, multiplication, and two bound variables only. Note that no usage of "1"/"0" is permitted, Nor is the comma-symbol "," permitted. HOOTmag (talk) 10:00, 16 June 2010 (UTC)

Here's one with no connectives and only one quantifier: $\forall x\,(i\cdot i\cdot x+x+x=x)$.—Emil J. 10:04, 16 June 2010 (UTC)

Great! I haven't thought about that! Thank you Emil. HOOTmag (talk) 10:12, 16 June 2010 (UTC)

And this one does not meet the specs because of negation, but it is quantifier-free: $i\cdot i\cdot i+i+i=i\land i+i\neq i$.—Emil J. 10:28, 16 June 2010 (UTC)

I don't think either of those is sufficient. They don't specify that i² is real, which I think is a key point. Otherwise we could be talking about any number of mathematical objects. There are matrices that satisfy those equations, for instance. $1\in \mathbb {Z} _{2}$ satisfies them. I could probably think of plenty of others given some time. --Tango (talk) 18:31, 16 June 2010 (UTC)

The domain of discourse is the set of complex numbers (I forgot to indicate that), so "i" must be a complex number. HOOTmag (talk) 19:52, 16 June 2010 (UTC)

That's a big thing to forget. If you've already defined the complex numbers, I'm not sure what the point is of defining i. You must have already done it, or you wouldn't know what "complex numbers" means. --Tango (talk) 20:30, 16 June 2010 (UTC)

Oh, I'm sure you haven't figured out what I'd meant by my original riddle (you're innocent of course; it's me who'd had to put things clearer). Given the set of complex numbers as the domain of discourse, I'd been looking for a formula which uses the permitted symbols only, and which is true if and only if the constant-symbol "i" is interpreted as an imaginary unit. Note that Emil did solve this riddle. HOOTmag (talk) 21:39, 16 June 2010 (UTC)

Nice answers to a nice riddle -but the challenging problem now is how to put the question so as to be pedantic-proof ;-) --pma 22:09, 16 June 2010 (UTC)

You must be new here: Welcome to the Wikipedia Mathematics Reference Desk! --Tango (talk) 22:25, 16 June 2010 (UTC)

:-) --pma 20:42, 18 June 2010 (UTC)

A proof? Oh dear no, that would be WP:OR ;-) Dmcq (talk) 22:32, 16 June 2010 (UTC)

That's not particularly challenging;) The question was to find a formula with one free variable in first-order logic with equality using at most two quantifiers, no parameters, and no connectives besides conjunction, which defines {i, −i} in the structure $(\mathbb {C} ,+,\cdot )$.—Emil J. 13:19, 17 June 2010 (UTC)

well, now the challenging problem is finding a pedantic objection to this --pma 20:42, 18 June 2010 (UTC)

June 16

Could someone please look at this supposed solution to Collatz conjecture

Could someone please look at this edit and determine if it's legitimate. Seems like a radical change, something that would be obvious to everyone in the field. Shadowjams (talk) 04:20, 16 June 2010 (UTC)

It's bogus. See [1] and identify relevant points as an exercise. And a valid proof would be in Annals of Mathematics, not sites.google.com. 75.57.243.88 (talk) 05:21, 16 June 2010 (UTC)

As far as my superficial understanding of the problem can tell, this paper doesn't trigger any of Scott's heuristics. Also, I don't know if it's a contradiction that a discovery will be published informally before we can see it in a journal. -- Meni Rosenfeld (talk) 08:17, 16 June 2010 (UTC)

It triggers #1 in that it's a PDF and not TeX, that's a bit iffy in that he might have used TeX to generate the PDF. From what I read there is a definite hit on #7 and #8 though. In any case, there are reasons why WP has criteria for reliable sources and keeping it from being a forum for the discussion of dubious claims is one of them. We tend to be lax on this in math articles in general, but an sharp eye should be kept on articles about well known unsolved problems.--RDBury (talk) 12:28, 16 June 2010 (UTC)

That paper was obviously written in LaTeX. — Carl (CBM · talk) 12:38, 16 June 2010 (UTC)

I don't even know how to parse "it's a PDF and not TeX". What would you expect from something that is TeX? A source *.tex file? DVI? That would be suspicious. -- Meni Rosenfeld (talk) 13:50, 16 June 2010 (UTC)

Whatever, that wasn't my main point anyway.--RDBury (talk) 03:21, 17 June 2010 (UTC)

I had a quick look and the end with its cardinalities looks wrong to me but it should be left to a professional to check. The main point here though is that Wikipedia cannot accept WP:Original research even if it is correct so it should not be in the article. It needs to be properly checked and published first especially as it is a long standing conjecture. Dmcq (talk) 08:34, 16 June 2010 (UTC)

In general, it simply isn't our place here to evaluate these things. If the proof is correct, it will be discussed in public relatively soon on professional mailing lists, and we can use those to add a (qualified) statement about the solution to our article. Until then, we shouldn't link to a self-published preprint that claims to solve a famous open-problem. One sign that the proof is correct: someone else will rewrite it in a clearer way. For an example of this in history, look at the "PRIMES is in P" proof, which was recognized very quickly as correct and rewritten by Dan Bernstein like so [ [2].

That being said, this paper is not obviously bad. The author clearly has some mathematical background and the exposition is clear enough that the results can be checked in principle by someone with the background and patience to do so. And FWIW I am a professional mathematician. — Carl (CBM · talk) 12:58, 16 June 2010 (UTC)

The abstract rings some alarm bells though: "The proof of Collatz theorem uses an astonishing involutive monoid automorphism... " (my emphasis) and "Have fun !" AndrewWTaylor (talk) 13:39, 16 June 2010 (UTC)

If someone had proved it I'd be quite happy for them to drop anal-retentive mode for a moment. Personally I'd prefer a bit more of the tongue in cheek and personal stuff like you get in Maths Intelligencer. Dmcq (talk) 15:10, 16 June 2010 (UTC)

FWIW, I was indeed rather astonished that the thing makes a well-defined automorphism.—Emil J. 16:32, 16 June 2010 (UTC)

I'm afraid that in the middle of page 5, $\sim _{\tau }\,$ is not, in fact, compatible with the monoid structure of M (i.e., it is not a congruence): for example, $\varphi \sim _{\tau }\psi$, but $\varphi \beta \not \sim _{\tau }\psi \beta$, as $[\varphi \beta ]_{\tau }=\{\varphi \beta ,\psi \gamma \}$.—Emil J. 16:26, 16 June 2010 (UTC)

Partitions of n

Is there an explicit formula for the number of partitions of n, i.e. for p(n)? I've read the article and understand that p(n) is given by the coefficient of xⁿ in

$\prod _{k=1}^{\infty }{\frac {1}{1-x^{k}}}\ .$

Let's say I wanted to know p(n) for all n ≤ 49; would I only to need to compute the Taylor Series of

$\prod _{k=1}^{49}{\frac {1}{1-x^{k}}}\ ?$ •• Fly by Night (talk) 13:59, 16 June 2010 (UTC)

Yes, because the rest of the product is 1 + terms of exponent greater than 49.—Emil J. 14:48, 16 June 2010 (UTC)

However, I think the more efficient way is the one described in the "Intermediate function" section. -- Meni Rosenfeld (talk) 15:06, 16 June 2010 (UTC)

Maybe, maybe not. It depends on the method being used. I've got a computer algebra package, so it takes half a second to calculate the coefficients of my last product. I just wanted to make sure I wouldn't miss any terms by truncating the product prematurely. •• Fly by Night (talk) 15:12, 16 June 2010 (UTC)

Partitions of n with a conditions

How would I work out the partitions of n given the assumption that there are exactly six numerically distinct summands and each of the summands is less than a given number? The motivation for this question is the lottery. We have balls numbered from 1 through to 49, and six balls are drawn at random from the 49. I would like to know how many different draws would give balls whose face values sum to a given number. Clearly there is only one combination that sums to 21, namely 1 + 2 + 3 + 4 + 5 + 6 = 21. Similarly, there is only one combination that sums to 279, namely 44 + 45 + 46 + 47 + 48 + 49 = 279. So, given 21 ≤ n ≤ 279 how many partitions of the form n = a₁ + a₂ + a₃ + a₄ + a₅ + a₆ are there given that 1 ≤ a_k ≤ 49 and a_k ≠ a_l if k ≠ l? •• Fly by Night (talk) 14:14, 16 June 2010 (UTC)

Well, you can get a simple upper bound by considering the sum (eg X) divided into n sections. We have n-1 "separators", so the supremum would be ${\binom {X+n-1}{n-1}}$ (choosing n-1 elements from a list of X+n-1 objects). However, this does allow repeated values, so is not the least upper bound. -mattbuck (Talk) 14:43, 16 June 2010 (UTC)

Thanks, but I'm looking for an explicit value; not a bound. One key premise was that the values can't repeat. •• Fly by Night (talk) 15:05, 16 June 2010 (UTC)

There's simple solutions with either condition relaxed, either allowing repeats or no limit like 49, but I don't know an easy solution of your problem. For something like this just running a straightforward program might be best. You'd need a compiled language like C to get the answers quickly rather than using Basic if it is done straightforwardly way with these numbers. So if you're not into programming a bit of help might be a good idea. Dmcq (talk) 15:41, 16 June 2010 (UTC)

[ec] Just getting the obvious out of the way: For sufficiently small numbers, you can brute-force it. For 6 and 49, it took my naive implementation less than a minute to get results for every n. -- Meni Rosenfeld (talk) 15:45, 16 June 2010 (UTC)

I thought about using a program, but there are far too many cases to check. I'll give it a go, but I think my computer will just freeze. I'll report back ASAP. •• Fly by Night (talk) 17:03, 16 June 2010 (UTC)

Make sure that your variables are ascending, that is, each starts at a value one more than the previous. This gives you 20 million cases (each being a simple addition) which is well within the abilities of even a weak computer in an unoptimized programming environment, and will give you results for every n.

Also, if you're content with an approximate solution, you don't have to enumerate all possibilities - you can choose the variables randomly, and repeat enough times to get the required error bounds. -- Meni Rosenfeld (talk) 18:32, 16 June 2010 (UTC)

Here's a naive C# implementation (that is probably suitable for similar languages) that took 30 milliseconds on my machine. I've tried optimizing it a bit but that didn't have a practical effect.

int[] results = new int[280];
for (int x1 = 1; x1 < 50; x1++)
{
 for (int x2 = x1 + 1; x2 < 50; x2++)
 {
  for (int x3 = x2 + 1; x3 < 50; x3++)
  {
   for (int x4 = x3 + 1; x4 < 50; x4++)
   {
    for (int x5 = x4 + 1; x5 < 50; x5++)
    {
     for (int x6 = x5 + 1; x6 < 50; x6++)
     {
      results[x1 + x2 + x3 + x4 + x5 + x6]++;
     }
    }
   }
  }
 }
}

-- Meni Rosenfeld (talk) 19:11, 16 June 2010 (UTC)

Why not simply use generating functions? Evaluating

$f(x)=\sum _{n_{1}<n_{2}<\ldots n_{6}}x^{n_{1}+n_{2}+\ldots n_{6}}={\frac {x^{21}}{\prod _{k=1}^{6}\left(1-x^{k}\right)}}$

is quite easy. This is without the restrictiong that the numbers are smaller than 50, but with that restriction you can still exactly evaluate the summation and extract the answers using series expansions. Count Iblis (talk) 17:52, 16 June 2010 (UTC)

This is what I get when I work out the correct generating function, taking into account the restriction that the numbers be strictly increasing and not be larger than N, using Mathematica

$-\left({\frac {x^{6}\,\left(-x^{15}+x^{5\,N}-x^{6\,N}+x^{3\,\left(1+N\right)}-2\,x^{4\,\left(1+N\right)}+x^{5\,\left(1+N\right)}+3\,x^{3\,\left(2+N\right)}-x^{4\,\left(2+N\right)}-x^{2\,\left(3+N\right)}+3\,x^{3\,\left(3+N\right)}-2\,x^{2\,\left(4+N\right)}+x^{3\,\left(4+N\right)}-3\,x^{2\,\left(5+N\right)}-2\,x^{2\,\left(6+N\right)}-x^{2\,\left(7+N\right)}+x^{10+N}+x^{11+N}+x^{12+N}+x^{13+N}+x^{14+N}+x^{15+N}-x^{7+2\,N}-2\,x^{9+2\,N}-2\,x^{11+2\,N}-x^{13+2\,N}+x^{4+3\,N}+2\,x^{5+3\,N}+3\,x^{7+3\,N}+3\,x^{8+3\,N}+2\,x^{10+3\,N}+x^{11+3\,N}-x^{1+4\,N}-x^{2+4\,N}-2\,x^{3+4\,N}-3\,x^{5+4\,N}-2\,x^{6+4\,N}-2\,x^{7+4\,N}-x^{9+4\,N}+x^{1+5\,N}+x^{2+5\,N}+x^{3+5\,N}+x^{4+5\,N}\right)}{{\left(-1+x\right)}^{6}\,{\left(1+x\right)}^{3}\,\left(1+x^{2}\right)\,\left(1-x+x^{2}\right)\,{\left(1+x+x^{2}\right)}^{2}\,\left(1+x+x^{2}+x^{3}+x^{4}\right)}}\right)$

Putting N = 49 and expanding to order 279 gives:

$x^{21}+x^{22}+2\,x^{23}+3\,x^{24}+5\,x^{25}+7\,x^{26}+11\,x^{27}+14\,x^{28}+20\,x^{29}+26\,x^{30}+35\,x^{31}+44\,x^{32}+58\,x^{33}+71\,x^{34}+90\,x^{35}+110\,x^{36}+136\,x^{37}+163\,x^{38}+199\,x^{39}+235\,x^{40}+282\,x^{41}+331\,x^{42}+391\,x^{43}+454\,x^{44}+532\,x^{45}+612\,x^{46}+709\,x^{47}+811\,x^{48}+931\,x^{49}+1057\,x^{50}+1206\,x^{51}+1360\,x^{52}+1540\,x^{53}+1729\,x^{54}+1945\,x^{55}+2172\,x^{56}+2432\,x^{57}+2702\,x^{58}+3009\,x^{59}+3331\,x^{60}+3692\,x^{61}+4070\,x^{62}+4494\,x^{63}+4935\,x^{64}+5426\,x^{65}+5940\,x^{66}+6506\,x^{67}+7097\,x^{68}+7748\,x^{69}+8423\,x^{70}+9163\,x^{71}+9933\,x^{72}+10769\,x^{73}+11637\,x^{74}+12579\,x^{75}+13552\,x^{76}+14603\,x^{77}+15690\,x^{78}+16856\,x^{79}+18059\,x^{80}+19349\,x^{81}+20673\,x^{82}+22087\,x^{83}+23540\,x^{84}+25082\,x^{85}+26663\,x^{86}+28340\,x^{87}+30051\,x^{88}+31860\,x^{89}+33706\,x^{90}+35648\,x^{91}+37625\,x^{92}+39703\,x^{93}+41809\,x^{94}+44016\,x^{95}+46253\,x^{96}+48586\,x^{97}+50944\,x^{98}+53402\,x^{99}+55875\,x^{100}+58446\,x^{101}+61031\,x^{102}+63706\,x^{103}+66388\,x^{104}+69161\,x^{105}+71928\,x^{106}+74781\,x^{107}+77624\,x^{108}+80542\,x^{109}+83440\,x^{110}+86412\,x^{111}+89348\,x^{112}+92350\,x^{113}+95311\,x^{114}+98324\,x^{115}+101285\,x^{116}+104295\,x^{117}+107235\,x^{118}+110215\,x^{119}+113119\,x^{120}+116048\,x^{121}+118889\,x^{122}+121751\,x^{123}+124507\,x^{124}+127274\,x^{125}+129930\,x^{126}+132581\,x^{127}+135109\,x^{128}+137629\,x^{129}+140008\,x^{130}+142370\,x^{131}+144587\,x^{132}+146771\,x^{133}+148800\,x^{134}+150794\,x^{135}+152617\,x^{136}+154397\,x^{137}+156004\,x^{138}+157554\,x^{139}+158923\,x^{140}+160236\,x^{141}+161354\,x^{142}+162410\,x^{143}+163273\,x^{144}+164062\,x^{145}+164654\,x^{146}+165176\,x^{147}+165490\,x^{148}+165732\,x^{149}+165772\,x^{150}+165732\,x^{151}+165490\,x^{152}+165176\,x^{153}+164654\,x^{154}+164062\,x^{155}+163273\,x^{156}+162410\,x^{157}+161354\,x^{158}+160236\,x^{159}+158923\,x^{160}+157554\,x^{161}+156004\,x^{162}+154397\,x^{163}+152617\,x^{164}+150794\,x^{165}+148800\,x^{166}+146771\,x^{167}+144587\,x^{168}+142370\,x^{169}+140008\,x^{170}+137629\,x^{171}+135109\,x^{172}+132581\,x^{173}+129930\,x^{174}+127274\,x^{175}+124507\,x^{176}+121751\,x^{177}+118889\,x^{178}+116048\,x^{179}+113119\,x^{180}+110215\,x^{181}+107235\,x^{182}+104295\,x^{183}+101285\,x^{184}+98324\,x^{185}+95311\,x^{186}+92350\,x^{187}+89348\,x^{188}+86412\,x^{189}+83440\,x^{190}+80542\,x^{191}+77624\,x^{192}+74781\,x^{193}+71928\,x^{194}+69161\,x^{195}+66388\,x^{196}+63706\,x^{197}+61031\,x^{198}+58446\,x^{199}+55875\,x^{200}+53402\,x^{201}+50944\,x^{202}+48586\,x^{203}+46253\,x^{204}+44016\,x^{205}+41809\,x^{206}+39703\,x^{207}+37625\,x^{208}+35648\,x^{209}+33706\,x^{210}+31860\,x^{211}+30051\,x^{212}+28340\,x^{213}+26663\,x^{214}+25082\,x^{215}+23540\,x^{216}+22087\,x^{217}+20673\,x^{218}+19349\,x^{219}+18059\,x^{220}+16856\,x^{221}+15690\,x^{222}+14603\,x^{223}+13552\,x^{224}+12579\,x^{225}+11637\,x^{226}+10769\,x^{227}+9933\,x^{228}+9163\,x^{229}+8423\,x^{230}+7748\,x^{231}+7097\,x^{232}+6506\,x^{233}+5940\,x^{234}+5426\,x^{235}+4935\,x^{236}+4494\,x^{237}+4070\,x^{238}+3692\,x^{239}+3331\,x^{240}+3009\,x^{241}+2702\,x^{242}+2432\,x^{243}+2172\,x^{244}+1945\,x^{245}+1729\,x^{246}+1540\,x^{247}+1360\,x^{248}+1206\,x^{249}+1057\,x^{250}+931\,x^{251}+811\,x^{252}+709\,x^{253}+612\,x^{254}+532\,x^{255}+454\,x^{256}+391\,x^{257}+331\,x^{258}+282\,x^{259}+235\,x^{260}+199\,x^{261}+163\,x^{262}+136\,x^{263}+110\,x^{264}+90\,x^{265}+71\,x^{266}+58\,x^{267}+44\,x^{268}+35\,x^{269}+26\,x^{270}+20\,x^{271}+14\,x^{272}+11\,x^{273}+7\,x^{274}+5\,x^{275}+3\,x^{276}+2\,x^{277}+x^{278}+x^{279}$

How much time did computing the expansion take? -- Meni Rosenfeld (talk) 19:42, 16 June 2010 (UTC)

I asked Mathematica this by evaluating Timing[Series[%,{x,0,279}]] and it says 2.69 Seconds. Not bad for my antique 400 MHZ processor :). Also, generating the generating function is very easy, you just iterate this process. Sum the geometric series x^k from k = r to N. Then replace r by k + 1, multiply the result by x^k and then sum again from k = r to N. At this stage you have summed over n6 and n5, so you have to repeat this process until you have summed over n1. At the last step you, of course, set r = 1. Count Iblis (talk) 21:05, 16 June 2010 (UTC)

I was using Maple, and have 2 GB of RAM and a Pentium Dual Core at 2.16 GHz and 2.17 GHz and I stopped the procedure after two and a half hours and 6 MB use of memory. Although I was just using Boolean commands: for a from 1 to 49 do etc. I don't have the faintest idea how to use C, C+, C#, or anything link that. •• Fly by Night (talk) 21:40, 16 June 2010 (UTC)

It's amazing what you can do with generating functions, I'm impressed. Anyway you've got the complete solution in one rather long line there. Dmcq (talk) 22:28, 16 June 2010 (UTC)

It really shouldn't have taken that long. Can you post the input you gave to Maple? -- Meni Rosenfeld (talk) 06:38, 17 June 2010 (UTC)

This shorter line is sufficient to show the point that the coefficient to $x^{21}$ is 1 and that the coefficient to $x^{279}$ is also 1

$\scriptstyle x^{21}+x^{22}+2\,x^{23}+3\,x^{24}+5\,x^{25}+7\,x^{26}+11\,x^{27}+\cdots +7\,x^{274}+5\,x^{275}+3\,x^{276}+2\,x^{277}+x^{278}+x^{279}$

Bo Jacoby (talk) 06:44, 17 June 2010 (UTC).

My Maple input, well, it's a little embarrassing. I did this:

Total := 0:
for a from 1 to 49 do
for b from 2 to 49 do
for c from 3 to 49 do
for d from 4 to 49 do
for e from 5 to 49 do
for f from 6 to 49 do
if a + b + c + d + e + f = 100 
and a < b
and b < c 
and c < d
and d < e 
and e < f
then Total := Total + 1:
fi: od: od: od: od: od: od:

•• Fly by Night (talk) 23:07, 17 June 2010 (UTC)

The innermost expression here is computed approximately N⁶ times and is complicated compared to Meni Rosenfeld's which is computed about N⁶/6! times and is quite simple. If the expression takes 5 times longer then whole program would take about 3600 time longer - a second becomes an hour. So if his naive implementation was in Maple too on a comparable machine this algorithm would take two and a half days. Comparing to the C# taking 30 milliseconds if it was coded using this algorithm would take two minutes. It looks to me like the Maple is 1000times slower. That seems a bit high to me, I'd have though they'd have done a bit more optimisation for speed, but it wouldn't be wholly unreasonable. Dmcq (talk) 07:57, 18 June 2010 (UTC)

For the record, my 1-minute implementation was in Mathematica. Yeah, I was also surprised how slow it was compared to C# (the two implementations were conceptually the same). -- Meni Rosenfeld (talk) 11:33, 18 June 2010 (UTC)

As I suspected, you didn't use my advice to start each variable at a value one higher than the last. As Dmcq says this requires x720 cases and makes each longer. You should have done

Total := 0:
for a from 1 to 49 do
for b from a+1 to 49 do
for c from b+1 to 49 do
for d from c+1 to 49 do
for e from d+1 to 49 do
for f from e+1 to 49 do
if a + b + c + d + e + f = 100 
then Total := Total + 1:
fi: od: od: od: od: od: od:

Also, this will give you the result for just one value of n. You can get them all with same effort by defining an array of values for each n, and each time incrementing the array member in the index of the sum.

Anyway, I remind you that if you just want the results, use Count Iblis' polynomial. -- Meni Rosenfeld (talk) 11:33, 18 June 2010 (UTC)

Fastest access to the values would be to preload an array and get value from that in a language like C# or even assembler. Slowest is to use an interpreted language and an unoptimised algorithm for each value. Here the difference in speed would be about 100,000,000,000,000 to 1. It just shows how easy it is to lose a few factors of a thousand and not notice. You'd never have done that with an old visible record computer! Dmcq (talk) 13:13, 18 June 2010 (UTC)

re: "As Dmcq says this requires x720 cases ..."; note that Dmcq wrote, "about N⁶/6! times". 49⁶ / ₄₉C₆ ≃ 990. -- 58.147.53.173 (talk) 15:17, 18 June 2010 (UTC)

Dmcq said "approximately N⁶" vs. "about N⁶/6!", which is x720. In fact, two wrongs make a right here, as it's really $N!/(N-6)!$ vs. $N!/(6!(N-6)!)$, which is still x720. -- Meni Rosenfeld (talk) 18:42, 19 June 2010 (UTC)

Here is a perl one liner which uses recursion both for obfuscation and for ease of changing the number of sumands without adding additional loop variables:

perl -le 'sub f{if(@_<6){f(@_,$_)for($_[-1]+1..49)}else{$i=0;$i+=$_ for@_;++$s[$i]}}f();print"$_ $s[$_]"for(1..$#s)'

Runtime (98 seconds on my netbook) is about 4 times longer than with the nested variables.

An improvement to the nested loop approach would be for a to range from 1 to 44, b from a+1 to 45, ..., f from e+1 to 49.→81.147.3.245 (talk) 11:38, 19 June 2010 (UTC)

Wolfram Alpha

Can I get Wolfram Alpha to calculate the area of a quadrilateral given its vertex co-ordinates? I can't seem to get it to compute using the syntax "quadrilateral vertex coordinates {(x,y),(x,y),(x,y),(x,y)}

Thanks in advance,

PerfectProposal 14:47, 16 June 2010 (UTC)

Works for me. You can also start with "area of a" to remove the unneeded stuff. -- Meni Rosenfeld (talk) 15:16, 16 June 2010 (UTC)

If your polygon is a quadrilateral, you may well do that yourself with a simple calculator... For the solution see article section Polygon#Area and centroid (formula for A). --CiaPan (talk) 18:44, 18 June 2010 (UTC)

June 17

Map pinpont problem

Say a friend chooses a secret point on a map, and gives you a random starting point. When you choose successive points, he gives you the hints "warmer," "colder," and "neither" to indicate if you've moved radially closer to the secret point or not. Given that you want to find the secret point as quickly as possible, what would be your optimal scheme? One possibility would be to choose points until you are neither "warmer" nor "colder" with respect to your random starting point, and thus form a curve of constant radius about the secret point; you could then extrapolate the center of the circle this begins to form. Is there a more efficient way? —Preceding unsigned comment added by 24.117.105.163 (talk) 05:18, 17 June 2010 (UTC)

Let point B be warmer/colder/neither than point A. The segment bisector between A and B divides the map into a warm region and a cold region, the line itself being neither. Once the warm region is the inside of a polygon, you should choose your next point such that the polygon is divided into (almost) equal areas, finding X by the bisection method. Bo Jacoby (talk) 06:21, 17 June 2010 (UTC).

The problem only talks in terms of 'warmer' and 'colder' not 'warm' or 'cold', one can only directly compare with the last point. I'm wondering what 'neither' means and I don't think it is necessary and it may cause problems if it is a range, also there must be an area which is counted as the target rather than it being a point. Anyway it sounds an interesting problem, somebody else very possibly has investigated it before but I'll not look at google yet! ;-) Dmcq (talk) 13:39, 17 June 2010 (UTC)

You can always (except maybe pathological cases) find a new point so that its segment bisector with the last point will halve the area. "Neither" probably means just that, that the new and last points are equally distant from the target. This won't happen very often, but if it does you're very lucky because the search will then be restricted to a single line. -- Meni Rosenfeld (talk) 13:49, 17 June 2010 (UTC)

Anyway my first solution is to use grid steps one way until I get colder, then do this at right angles, then choose a smaller grid and do it from the new point I'm at till I'm sufficiently close. Dmcq (talk) 13:43, 17 June 2010 (UTC)

I must obviously explain more carefully to Dmcq. Let the given point be A=(0,0). Choose the point B=(1,0). If B is warmer than A then the unknown point X=(x,y) satisfies x>1/2, and if B is colder than A then x<1/2, and if B is neither warmer nor colder than A then x=1/2. Next choose the point C=(1,1). Now you also know if y>1/2 or y<1/2 or y=1/2. The 'warm' region is the convex area in which the unknown point is known to be located. Now choose a point D far away in the warm region, hoping that it is colder than C. Then the warm region is a finite triangle, otherwise it is infinite. Bo Jacoby (talk) 15:57, 17 June 2010 (UTC).

Sorry Bo Jacoby, silly me, you're quite right. I really should stop myself when I have the urge to rush something off before I have to go somewhere. It just means I keep thinking how silly I am till I get back :) Dmcq (talk) 16:57, 17 June 2010 (UTC)

Don't torment yourself. Communication is a joint effort. Bo Jacoby (talk) 11:07, 18 June 2010 (UTC).

Thanks for the responses everyone! The bisection method is definitely a good one. Is it also the most efficient? It seems that perhaps some clues may be gleaned from previous hints that allows you to cut out more than half of the remaining area. 24.117.137.23 (talk) 22:59, 18 June 2010 (UTC)

No, all information is coded in the 'warm' polygon.Bo Jacoby (talk) 20:02, 23 June 2010 (UTC).

Proportional_navigation may be helpful. Zoonoses (talk) 14:45, 18 June 2010 (UTC)

It seems to me that if we are really talking about mathematical points, then the expected number of steps will always be infinite no matter what the algorithm. The best you can do is bisect the remaining region, but there are as many points in half of it as there were originally (i.e. an infinite number). Therefore there isn't really a "best" algorithm, or to put it another way, "The only way to win is not to play". --Anonymous, 19:33 UTC, June 18, 2010.

True, but surely there are more efficient ways to get "close enough." 24.117.137.23 (talk) 22:59, 18 June 2010 (UTC)

Arithmetic sum question

The question says: $S\sim Geo\left({\frac {1}{12}}\right)$ so $P\left(S=s\right)={\frac {1}{12}}{\left({\frac {11}{12}}\right)}^{s-1}$ where $1\leq s<\infty$ and so $P\left(S\leq s\right)=\sum _{i=1}^{s}{\frac {1}{12}}{\left({\frac {11}{12}}\right)}^{i-1}=1-{\left({\frac {11}{12}}\right)}^{s}$, but I'm having trouble following the logic of that last summation. So far I've got: $\sum _{i=1}^{s}{\frac {1}{12}}{\left({\frac {11}{12}}\right)}^{i-1}={\frac {1}{12}}\left[{\left({\frac {11}{12}}\right)}^{0}+{\left({\frac {11}{12}}\right)}^{1}+\dots \right]={\frac {1}{12}}\left[{\left(1-{\frac {11}{12}}\right)}^{-1}\right]=1$ because ${\left(1-x\right)}^{-1}=x+x^{2}+\dots$; where am I going wrong in this? Thanks in advance. It Is Me Here ^{t / c} 11:11, 17 June 2010 (UTC)

You have confused $\sum _{i=1}^{s}{\frac {1}{12}}{\left({\frac {11}{12}}\right)}^{i-1}$ with $\sum _{i=1}^{\infty }{\frac {1}{12}}{\left({\frac {11}{12}}\right)}^{i-1}$. For the summation formula of a finite geometric series, see Geometric series. -- Meni Rosenfeld (talk) 11:26, 17 June 2010 (UTC)

Sorted it now, cheers. It Is Me Here ^{t / c} 12:00, 17 June 2010 (UTC)

Given a family of sets, I can generate a partition of their union set, by looking at their overlapping...

... as an example, let us consider 3 sets A, B and C; this sets generate 7 disjoint nonempty subsets, that is: A.B.C (elements belonging to the intersection of the 3 sets) A.B.notC (elements belonging to A and B, but not to C) A.notB.C (... and so on) A.notB.notC notA.B.C notA.B.notC notA.notB.C Now, given a family of sets, I can always build such a partition. Which is the name of one of such part? In a partition, one of these disjoint subset is called "part" or "block" or "cell"; but it is a more general concept, it does not imply that it comes from a family of sets. Does a name already exist? —Preceding unsigned comment added by 130.88.195.163 (talk) 18:08, 17 June 2010 (UTC)

Eight - you missed out the set excluding all 3. Dmcq (talk) 18:20, 17 June 2010 (UTC)

And you missed "partition of their union set". Algebraist 18:23, 17 June 2010 (UTC)

True, seven then. I think I call them clause (logic), except they don't have to contain every literal so A on its own would be a clause. I havve comae across a term from programmable logic devices but I couldn't find it now. Canonical form (Boolean algebra) uses 'minterm' to refer to them but that's not what I was trying to think of. Dmcq (talk) 18:38, 17 June 2010 (UTC)

Well, although I need a definition like this for decomposing a set of logical formulae, I would like to have a name in set theory; something like "the partition of the union set given by the family". —Preceding unsigned comment added by 130.88.195.163 (talk) 18:45, 17 June 2010 (UTC)

I think full disjunctive normal form in Disjunctive normal form where are the clauses contain every literal is the closest I'm going to get but that refers to the full logical formula. Dmcq (talk) 18:50, 17 June 2010 (UTC)

The following notation is not described in wikipedia because it is original research on my part, but I guess I can confidently inform you. The numbers are called ordinal fractions. The first set A=100, and notA=200. B=010, and notB=020. C=001, and notC=002. This defines the 27 sets 000 001 002 010 011 012 020 021 022 100 101 102 110 111 112 120 121 122 200 201 202 210 211 211 220 221 222. The 8 sets 111 112 121 122 211 212 221 222 are mutually disjoint. (For example 121=A.notB.C). Bo Jacoby (talk) 23:56, 17 June 2010 (UTC).

Moment-generating function of the Binomial distribution

You are told that $X\sim B\left(n,p\right)$ and so $M_{X}\left(\theta \right)={\left(q+p{e}^{\theta }\right)}^{n}$ where $q=1-p$; then you are told that $Z={\frac {X-\mu }{\sigma }}$ and are asked to use the linear transformation result to show that $M_{Z}\left(\theta \right)={\left(qe^{-{\frac {p\theta }{\sqrt {npq}}}}+pe^{\frac {q\theta }{\sqrt {npq}}}\right)}^{n}$. Now, I have managed to get to $e^{-\theta {\sqrt {\frac {np}{q}}}}{\left(q+pe^{\frac {\theta }{\sqrt {npq}}}\right)}^{n}$, but don't know how to turn that into the final answer (or 'break open' that final ${\left(a+b\right)}^{n}$ bracket). Any help would be appreciated. It Is Me Here ^{t / c} 19:48, 17 June 2010 (UTC)

$-\theta {\sqrt {\frac {np}{q}}}=n\left(-{\frac {p\theta }{\sqrt {npq}}}\right)$

Dmcq (talk) 20:10, 17 June 2010 (UTC)

Ah, OK, thanks. It Is Me Here ^{t / c} 20:19, 17 June 2010 (UTC)

some controversial questions

We have some controversial questions on Talk:Exponentiation#0.E2.81.B0 .E2.80.94 limits and continuity.

1. Is the set of integers a subset of the set of real numbers?
2. Is 5+0i>3+0i ?
3. Is 0=0.0 ?

You comments are appreciated. Bo Jacoby (talk) 23:37, 17 June 2010 (UTC).

To which my answer in all three cases is no. Dmcq (talk) 23:43, 17 June 2010 (UTC)

The answer to the first question is no, strictly speaking, although the point is pedantic. The second is really a matter of notation and convention. I would be interested in how you justify a negative answer to the last question. "0.0" is just notation. For all you know, it could be a placeholder I use for the "integer" 0. Phils 23:49, 17 June 2010 (UTC)

Not that I agree with it, but I guess the justification is that 0 is an integer and 0.0 is a real number. I think that's hypercorrect, personally... --Tango (talk) 23:55, 17 June 2010 (UTC)

Well I'd only distinguish between them when I though it was important to distinguish between integers and real numbers (or in fact any other thing like a complex number or zero matrix). Dmcq (talk) 23:59, 17 June 2010 (UTC)

I would distinguish them in a clearer way, though. $0_{\mathbb {Z} }$ and $0_{\mathbb {R} }$, perhaps. --Tango (talk) 00:02, 18 June 2010 (UTC)

Mathematics is not programming or engineering (and besides, many high level programming languages would interpret the number as a real or as an integer correctly, depending on context). Unless the authors specified it explicitly, nobody reading "0.0" in a math journal would assume that it means the zero element of the reals as opposed to an integer. Phils 00:04, 18 June 2010 (UTC)

Could someone expand on the answers for the curious few in the peanut gallery? Why the distinction between "real zero" and "integer zero"? Does this imply that the combined length of three identical line segments is technically not equal to the length of the original line segment after being scaled by a factor of three (e.g. 3 × 2.6 ≠ 3.0 × 2.6) ? -- 140.142.20.229 (talk) 01:01, 18 June 2010 (UTC)

Well, technically you can't multiply elements in different rings/groups, so you couldn't multiply the integer 3 by the real number pi, but you could multiply the real number 3 by the real number pi, and you can set up an injection from Z to R by f(x) = x. In practice there's no difference, because by convention we assume that such an injection is made, so in 3*pi we're multiplying f(3) by pi. -mattbuck (Talk) 01:27, 18 June 2010 (UTC)

For question 2, it's again a technicality - 5+0i = 5, and 3+0i = 3, but when working in the complex plane (which is implied by the i) you have the issue that C is not a well-ordered set - you have numbers a,b where a !< b and b !< a and further a != b.

For question 3, 0 multiplied by itself is technically undefined, but by convention we say that anything multiplied by 0 is 0 (or at least for finite values) and so we say that 0.0 is 0. -mattbuck (Talk) 01:32, 18 June 2010 (UTC)

Well, R isn't a well-ordered set either... but presumably you meant a total order.

For question 3, I think the OP was referring to the real number 0.0, where the period is interpreted a decimal point. So then the question is just whether we equate the integer 0 with the real number 0.0, to which we would usually answer yes, unless we have a good reason to be pedantic about the distinction. --COVIZAPIBETEFOKY (talk) 01:42, 18 June 2010 (UTC)

What is the distinction to be pedantic about? Bo Jacoby (talk) 07:59, 18 June 2010 (UTC).

As others have said, it depends on our specific constructions of the integers and real numbers, but usually, the real numbers that correspond to integers are distinct objects from the integers. So the distinction is between the set of integers, which are often constructed, eg, as an equivalence class on ordered pairs of finite ordinal numbers, or the set of real numbers that correspond to some integer. --COVIZAPIBETEFOKY (talk) 11:24, 18 June 2010 (UTC)

You can multiply an integer, specifically, by any member of an additive group. It is defined as repeated addition in that group.

If you consider Z not to be a subset of R, then you can't have $f(x)=x$. Writing this just muddies the issue. What you can have is "$f(x)$ is the real number corresponding to x".

Of course 0 multiplied by itself is defined. That is, in any structure where there is a member called "0" and an operation called "multiplication", 0*0 is defined (and in all such structures I can think of, it is equal to 0). -- Meni Rosenfeld (talk) 08:46, 18 June 2010 (UTC)

Claiming that the sentence the integers are a subset of the real numbers is false with no definition of integer and real numbers around makes little sense. If you look at the "construction" sections of Integer and Real number, you will notice that they are formally disjoint sets. They consist of different objects, whatever approach you take to construct the real numbers. Of course, in almost every context, the integers are identified with their image inside the reals. Regarding the three segments, the only physically reasonable answer is that the two lengths are the same. Formalism and abstraction exist to make complicated things intelligible, not obfuscate the obvious. Phils 01:36, 18 June 2010 (UTC)

I have a better question.

• Is 1+1=0?

The answer is "it depends on which 1, 0 and + we mean". If we mean the integers 1, 0 and integer addition, then this is false. If we mean the $\mathbb {Z} _{2}$ elements and $\mathbb {Z} _{2}$ addition, this is true. We use integers more often than $\mathbb {Z} _{2}$, so by convention, without a context explicitly specified, the notation refers to integers and we say the statement is false. But this does not mean that "1+1=0" is really, truly false, only that conventionally we use it to mean something that is false.

People speak of "the set of integers", but there are actually many sets which can be referred to as "the set of integers". There's a set of equivalence classes of ordered pairs of finite ordinals with the obvious relation; to this set I shall refer as "Integers1". Then we go on to construct the set of real numbers, and in it we identify elements which correspond to the elements of Integers1. We call the set of those elements Integers2. Integers1 and Integers2 are of course disjoint. The question "Is the set of integers a subset of the set of real numbers?" reduces to the question "which set of integers do we mean?". If we mean Integers1, no. If we mean Integers2, yes.

In most cases it doesn't matter which set we mean when we say "integers". I suspect that usually people actually think about Integers2, but maybe that's just me. When it comes to questions where it does matter, since apparently we don't have a clear convention on what we mean by default, we just need to specify what we mean. For 0^0, if the exponent is considered an element of Integers1, it follows from the natural definition of integer exponents that the answer is 1. If the exponent is considered an element of Integers2 - a real number which happens to be an integer2 - it follows from the natural definition of real exponents that this is indeterminate.

For question 2 - it's again a question of Reals1 vs. Reals2 (a subset of Complexes). But even if we do mean Reals2, we can and should define a relation > on the complexes which is not total, and holds only for unequal real2 numbers.

Question 3 - Like I said, I think 0 and 0.0 both refer to the Integers2-0, so they're equal. If it turns out that the convention is that 0 refers to Integers1 and 0.0 to Integers2, then no. -- Meni Rosenfeld (talk) 08:36, 18 June 2010 (UTC)

Of course, 0.0 could represent the zero element in the field of rational numbers, and "set of integers" could mean algebraic integers (in algebraic number theory texts it is common to use "integer" to mean any algebraic integer, and to distinguish the integers in $\mathbb {Z}$ or $\mathbb {Q}$ by calling them "rational integers"). As several editors have already said, the answers to all three questions depend on context and interpretation. Gandalf61 (talk) 09:40, 18 June 2010 (UTC)

Apart from $f(x)=0^{x}$, are there any examples in mathematics where $\scriptstyle f(0)\neq f(0.0)$ ? Bo Jacoby (talk) 11:18, 19 June 2010 (UTC).

There is a similar issue when you move from the real numbers to the complex numbers; ln(2.0) is a single real number, but you have to choose a branch cut to define ln(2.0 + 0i). This is because "ln" has different meanings in the two contexts. — Carl (CBM · talk) 11:36, 19 June 2010 (UTC)

It sounds unlikely that there's any other straightforward thing like that, but here's a similar conundrum - what's the difference between zero and nothing? In fact in some definitions of 0 there's no difference but it's a good one to ponder. Dmcq (talk) 11:48, 19 June 2010 (UTC)

Can I be an idiot and ask why the set of integers is not a subset of the set of real numbers? I would have no problem writing ℤ ⊂ ℝ. After all, n ∈ ℤ ⇒ n ∈ ℝ and so ℤ is a subset of ℝ. In fact, don't we have a sequence of nested proper subsets given by ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ ⊂ ℂ ⊂ ℍ? •• Fly by Night (talk) 12:25, 19 June 2010 (UTC)

With the usual definitions of those sets, none of your proposed chain of subsethoods holds. Each set does contain a canonical isomorphic copy of the previous one, though, so you can if you wish redefine ℕ to be its canonical copy in ℤ (and so on for the rest), and then ℕ will indeed be a subset of ℤ. Meni has already explained this above. Algebraist 12:31, 19 June 2010 (UTC)

Are there industry standard definitions for the integers, real numbers, etc? The definitions of ℂ and ℍ are built on ℝ. (I didn't see Meni's post; there was far too much thread to read. Sorry.) •• Fly by Night (talk) 13:04, 19 June 2010 (UTC)

P.S. The more I think about it, the less it makes sense. If the integers are not, by defintion, a subset of the real numbers then it must mean that there is a definition of the real numbers for which not all integers are considered as real numbers. What possible definition of the real numbers could there be for which not every integer is considered to be a real number? Let's say there exists an integer, n, which is not a real number; then what is n? •• Fly by Night (talk) 14:11, 19 June 2010 (UTC)

Let me give you some examples. The natural number 2_N is defined as the ordinal $\{\varnothing ,\{\varnothing \}\}$. Depending on the construction, the integer 2_Z may be the ordered pair $\langle \varnothing ,2_{\mathbb {N} }\rangle$ (if I choose to represent the + sign by the empty set), or the set $\{\langle a,b\rangle \mid a,b\in \mathbb {N} ,a=b+2_{\mathbb {N} }\}$, or something else. The rational number 2_Q may be the ordered pair $\langle 2_{\mathbb {Z} },1_{\mathbb {N} }\rangle$ or the set $\{\langle a,b\rangle \mid a,b\in \mathbb {Z} ,a=2_{\mathbb {Z} }b\}$. The real number 2_R may be the set $\{a\in \mathbb {Q} \mid a<2_{\mathbb {Q} }\}$ or the set of all rational sequences which converge to 2 etc. The complex number 2_C can be the pair $\langle 2_{\mathbb {R} },0_{\mathbb {R} }\rangle$. All of these 2's are different objects, even though 2_Z plays in Z the same role as 2_R plays in R; there is a canonical embedding of Z into R which maps 2_Z to 2_R. There are no industry standard definitions of these mathematical structures, or rather, there are too many of them, but essentially none of the usual constructions make, say, the integers a subset of the reals. Pretty much the only reasonable way how to achieve that would be to make explicit exceptions in the construction, such as: a real number is either a Dedekind cut on the rationals with no rational limit, or a rational number itself. This is awkward to work with, and it's much easier to go with the usual strategy where the canonical embeddings are not literal inclusions.—Emil J. 15:58, 19 June 2010 (UTC)

I'm sorry but, to me, this seems to be totally over the top. You have the real numbers, then most people would, in your notation, make the following definitions: ℕ = {0_R, 1_R, 2_R,…}, ℤ = {…, –2_R, –1_R, 0_R, 1_R, 2_R,…}, ℚ = {x/y : x, y ∈ ℤ ∧ y ≠ 0_R}, and ℂ = {x + iy : x, y ∈ ℝ}. Thus ℕ ⊂ ℤ ⊂ ℝ ⊂ ℂ. I guess the problems come from the fact that when you try to construct the integers or the reals from nothing then you end up having to use different techniques because of the different nature of the integers and the reals. The results are then incompatible. So as sets ℕ ⊂ ℤ ⊂ ℝ ⊂ ℂ, but the construction of one doesn't fit with the construction of the next. •• Fly by Night (talk) 18:37, 19 June 2010 (UTC)

The problem is that there's no good way to construct real numbers without using rational numbers. And there's no good way to construct the rational numbers without integers, and so on. So you have to start with an elementary definition of natural numbers and move up from there.

Of course, you can also leave the real numbers undefined, and add some axioms to give them the properties you want. Then you can define integers as specific real numbers. But most people prefer to have the amount of undefined concepts and axioms at a minimum. Since you can do all standard mathematics with just sets, and you can't do much without sets, the standard approach is to axiomatize set theory and define everything else in terms of sets.

Until now I thought it was also standard to discard the initial definitions of integers etc. once they're no longer needed, and redefine everything as subsets of complex numbers. -- Meni Rosenfeld (talk) 18:55, 19 June 2010 (UTC)

Yeah, that's not common in set theory at least. The number 2 will almost always mean {0,1} to a set theorist, where 0 = {} and 1 = {0}. — Carl (CBM · talk) 19:00, 19 June 2010 (UTC)

But isn't this all a problem with the construction and formalisation of the different number systems? If you treat the real numbers intuitionally and geometrically (see the the number line picture above) then there really is not problem at all. Aren't all of these problems coming from our own inability to formally construct and define the different number systems in a cohesive way? It all feels like it's been cobbled together like a patchwork quilt. If there are so many technical hurdles to overcome when formalising the real numbers then how on earth can we make sense of Riemannian manifolds? (Never mind more exotic creatures like Kähler manifolds.) •• Fly by Night (talk) 19:10, 19 June 2010 (UTC)

This is getting like the argument amongst educationalists over whether to introduce multiplication as repeated addition and what to do when measurements are multiplied together when repeated addition doesn't work. The reals just are different from the integers. Dmcq (talk) 19:55, 19 June 2010 (UTC)

Right. To FlyByNight: the reals are conceptually and qualitatively different from the natural numbers. The natural numbers are about counting things. You can't count things with the reals, not even the ones that correspond naturally to the naturals.

Remember the scene in Terminator 2 where Arnold has been ordered not to kill anyone, and after an intervention of surpassing violence tallies up the deaths, and they come to 0.0? It's funny, not just because of unnecessary precision, but because it's a category error. There aren't any .5 people, and there aren't any .0 people either. --Trovatore (talk) 23:22, 19 June 2010 (UTC)

The reals are indeed conceptually and qualitatively different to the integers; but since when does a subset have to be conceptually and qualitatively the same as one of its supersets? For example, the empty set ∅ is a subset of any given set; but not every set is conceptually and qualitatively the same as the empty set. •• Fly by Night (talk) 22:22, 20 June 2010 (UTC)

There is no bar to including objects of disparate type in the same set. For example, even if you don't consider the natural numbers to be real numbers, you could always take the union of the naturals and the reals, and the naturals would then be a subset of that union.

So it's not a question about the subset relation in general. It's a question of whether the natural numbers are naturally considered a subset of the reals. Are they really the same sort of thing? I think there is a reasonably clear distinction even at an intuitive level. The real number 1.0, for example, is a position along a continuous line; the natural number 1 is more like "what all unique objects have in common". These are different things.

So when it turns out that the most convenient codings turn out quite different (1 being coded as the singleton of the empty set, 1.0 as a collection of sequences of rationals), rather than see that as a defect of the coding, we can embrace that as a clear reflection of the underlying typedness of the objects.

Of course, there are plenty of contexts where the distinction doesn't matter, or even where it's actively useful to ignore it, and in those contexts we happily ignore it without comment, counting on the reader to make the necessary allowances. --Trovatore (talk) 22:50, 20 June 2010 (UTC)

Thank you, everyone, for comments and answers. The article on isomorphism has reference to a relevant article: Mazur, Barry (12 June 2007), When is one thing equal to some other thing? (PDF). The ring of integers is defined abstractly as an axiomatic system. The axioms are sufficient for calculations, but two questions remain unanswered: Are the axioms consistent, such that at least one realization exists? and: are the axioms sufficient, such that two realizations are necessarily isomorphic? In order to settle the first question a set theoretical construction of integers is made, such as described above. When such an example satisfy the axioms, then the axioms cannot be contradictory. To me this is the sole purpose of the construction. The ring of integers is defined abstractly modulo isomorphism by the axioms, and a concrete realization is defined by the set theoretical construction. In the abstract sense, the ring of reals contains a unique subring of integers, and we write ℤ ⊂ ℝ and 0=0.0 . In the concrete sense, different, (albeit isomorphic), constructions can be made, and the zero of one construction is not the same mathematical object as the zero of some other construction and then 0≠0.0 . So it seems to me that the controversy is about differentiating between the abstract and some concrete interpretation of the words 'integer' and 'real'. In the abstract sense the three original questions are answered affirmatively, as Fly by Night and I myself would do. Bo Jacoby (talk) 05:55, 21 June 2010 (UTC).

No one doubts that the integers and, for lack of a better term, "integer reals", are isomorphic as rings. So for the purposes of ring theory, you can consider them the same, at least when not considering them as embedded in some larger ring.

But that's only for the purposes of things that can be expressed in the language of rings. Your original question was motivated by exponentiation, which is not a ring operation, and is not definable in the language of rings. There is no notion of raising an element of a ring to the power of another element of the ring. --Trovatore (talk) 06:29, 21 June 2010 (UTC)

Rings don't even have greater than and less than so that lets out question 2. Dmcq (talk) 14:16, 21 June 2010 (UTC)

Abstract ring elements do have well defined powers to nonnegative integer exponent. So if x is a real number and n is a nonnegative integer, then xⁿ (including 0⁰ = 1) is defined by ring operations. Abstract integers also have a well defined total order, which is the restriction of the order of the reals. Bo Jacoby (talk) 15:46, 21 June 2010 (UTC).

Abstract ring elements do indeed have well-defined powers to natural-number exponents. But that just makes my point for me! When you're defining ring-element-to-natural-number, the ring can be anything you want, but the natural numbers are always the same. The natural numbers are therefore something external to the ring. --Trovatore (talk) 18:27, 21 June 2010 (UTC)

Is the set of integers a subset of the set of real numbers? It depends on exactly how real numbers are defined, and so can only be answered in the context of a particular construction of the real numbers. What is true is that there is a subset of the real numbers Z' = {..., -3.0, -2.0, -1.0, 0.0, 1.0, 2.0, 3.0, ...} which is isomorphic to the integers Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}. Because Z' behaves exactly like Z does, and because it satisfies the usual axioms, most mathematicians will say "the integers are a subset of the reals", although in certain technical contexts the difference between Z and Z' may become important. -- Radagast3 (talk) 10:13, 22 June 2010 (UTC)

Why say that 1.0 is different to 1? I was always under the impression that 1 = 1.0 = 1.00 = 1.000 = 1.0000 = …; after all, for any integer n

$1.0\dots 0=1+\left(\sum _{k=1}^{n}{\frac {0}{10^{k}}}\right)=1+\left(\sum _{k=0}^{n}0\right)=1+0=1.$

If 1 and 1.0 are different then we're in real trouble. Why is there a focus on using decimal notation anyway? I don't think we can write every real number as a decimal, can we? Only the rationals can be written using decimals and bars (finite or repeating decimal expansions). I mean, how do we write √2 in decimal notation (base 10)? (But maybe that's a different question for a new thread) •• Fly by Night (talk) 19:03, 22 June 2010 (UTC)

Of course the real number 1.0 is the same as the real number 1. Apparently people use "1" to mean the integer 1, and "1.0" to mean the real 1. This is because decimal expansions are suitable for real numbers, but not at all for integers. Any real number has a (possibly infinite) decimal expansion. But the focus is not on the decimal expansions, here they're just used as a notational shortcut to distinguish integers from reals. -- Meni Rosenfeld (talk) 07:39, 23 June 2010 (UTC)

Is 5+0i>3+0i? It depends on what comparison operator is intended. The notation indicates that two complex numbers (3 and 5) are being compared, so it's not clear what the term "5+0i>3+0i" means. -- Radagast3 (talk) 10:18, 22 June 2010 (UTC)

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 [3] 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)

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 integral 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) 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" 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) 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 integral 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) 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) 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) 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. 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. 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)

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).

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)

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)

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)