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June 11

Simple Type 1 error problem

I've got a statistics homework problem that I'm not sure how to approach. Here it is:

1) In most industrial procedures quality control is routinely performed. Imagine the following situation: a machine is used to fill two liter soda bottles. Even when the machine is calibrated to put two liters of soda in each bottle, the actual content varies a little from bottle to bottle. The content has a normal distribution with mean 2 liters and standard deviation 0.1 liter. Each day a random sample of 9 bottles is taken from the daily production. If the sample mean is below 1.9 liters or above 2.1 liters, the production process is stopped to check the machine.

a) What is the probability that the sample mean is either below 1.9 liters or above 2.1 liters?

b) What is the probability that the machine is stopped even though it is within working according to specifications? Explain in complete sentences.

Part a is simple; I got .0027 for being +/- 3 standard deviations from mean. I'm not even sure what part b is asking, though. If it's asking the probability that a machine has a mean within [1.9, 2.1], then isn't it the same as part a? Anyone know what's going on here? --151.141.244.19 (talk) 01:53, 11 June 2010 (UTC)

I won't say if your answer to part 'a' is correct (why is ±3 standard deviations the same as a sample mean of ±1 standard deviation from the desired mean?), but it looks to me like 'b' has the same answer as part 'a'. If the sample mean exceeds the required bounds, the machine is stopped and checked. ~Amatulić (talk) 04:42, 11 June 2010 (UTC)

Because a sample consists of 9 bottles, and the square root of 9 is 3. Bo Jacoby (talk) 19:59, 12 June 2010 (UTC).

I think part (b) is just asking you to put your answer from part (a) into a complete sentence, e.g. one beginning, "The probability that the machine is stopped even though ..." --Qwfp (talk) 19:39, 11 June 2010 (UTC)

Proof that Euclid’s fifth postulate cannot be proven.

How did Bolyai and Lobachevski prove that Euclid’s fifth postulate cannot be proven?91.104.181.154 (talk) 22:24, 11 June 2010 (UTC)

The fifth postulate was shown independent of the others by the demonstration of a model of hyperbolic geometry which satisfied the other postulates but not that one. Dmcq (talk) 23:11, 11 June 2010 (UTC)

By the way I recently went along to an exhibition of crocheted models of the hyperbolic plane. Here's the person who started it all off Daina Taimina. Dmcq (talk) 23:14, 11 June 2010 (UTC)

2.8651496649...

I've found this to be the "other" positive value of x for which $\Gamma (x)={\sqrt {\pi }}$. Does anybody know anything else about this number? --Lucas Brown 23:22, 11 June 2010 (UTC)

Well I tried out Plouffe's inverter and it didn't have it. Dmcq (talk) 23:56, 11 June 2010 (UTC)

You can do more tests using integer relation algorithms if you compute more significant digits (at least a few dozen). Count Iblis (talk) 03:13, 12 June 2010 (UTC)

Here are a few more digits:

2.8651496649764734274885554227037096412511096062528695651871023239515553871026286151412171720144439329280321651530519737995653560...

-- Meni Rosenfeld (talk)

June 13

Meromorphic invertible functions

The Möbius transformations are the only invertible meromorphic functions on the complex plane. In the more general case with n complex functions of n complex variables, are there any invertible functions with only n-1 complex dimensional singularities where $\lim _{w,z,\cdots \to singularity}{\frac {1}{f(w,z,\cdots )}}$ exists other than those of the form $f(w,z,\cdots )={\frac {aw+bz+\cdots +c}{jw+kz+\cdots +l}}$? 74.14.111.223 (talk) 06:06, 13 June 2010 (UTC)

Limits / infinite series

To abridge the question somewhat:

${\text{Show that }}{\frac {\theta }{e^{\theta }-1}}<1{\text{for all }}\theta {\text{. Show also that it is near }}1{\text{ if }}\theta {\text{ is small and near }}0{\text{ if }}\theta {\text{ is large.}}$

Now, I'm afraid I could not really get anywhere with this, trying an approach which began $e^{\theta }>0\implies e^{\theta }-1>-1$, but then did not know where to go from there. The answer says, verbatim:

$e{\text{.}}g{\text{. Expression is }}{\frac {\theta }{\theta +{\frac {\theta ^{2}}{2!}}+...}}$ (1)

${\text{always }}<1$ (2)

${\text{and this is }}\approx 1{\text{ if }}\theta {\text{ is small}}$ (3)

$\approx 0{\text{ if }}\theta {\text{ is large}}$ (4)

Now, I understand how they got to (1), but (2), (3) and (4) are something of a mystery to me at present. Could anyone enlighten me, please? It Is Me Here ^{t / c} 13:54, 13 June 2010 (UTC)

Surely theta > 0 is implied? If you understand how they got to (1), then (2) is obvious: the denominator is numerically larger than the numerator. If theta is small, then the higher power terms in the denominator are dominated by the linear term so your expression is approx theta/theta, this gets you (3). For (4), divide numerator and denominator through by theta and consider whats happens as theta tends to infinity. Zunaid 14:34, 13 June 2010 (UTC)

I think that you want θ ≥ 0, because as θ grows negatively then e^θ becomes very small, and your function behaves like y = –x for very large, negative θ. In fact, the line y = –x is an asymptote as θ tends to negative infinity. When θ = 0 your function is indeterminate (we get zero divided by zero). We can use L'Hôpital's rule to evaluate the limit of your function as θ tends towards 0, and we can show that this limit is 1. Also, it's clear that your function tends to 0 as θ tends towards positive infinity. This half-answers your question. You just need to show that your function doesn't misbehave away from those values. Personally, I would work out the derivative, and try to show that it is always negative. If you draw a graph of your function, you see that it decrease constantly for all θ. So let's calculate the derivative of your function:

$f'(\theta )={\frac {(1-\theta )e^{\theta }-1}{(e^{\theta }-1)^{2}}}.$

Clearly (e^θ – 1)² ≥ 0 for all θ and it follows that the sign of the derivative is given by the sign of g(θ) = (1 – θ)e^θ – 1. Well, that's not strictly true. When θ = 0 then the derivative is indeterminate (we get zero divided by zero). We can use L'Hôpital's rule to evaluate the limit of g(θ) as θ tends towards 0, and we see that this limit is ½(1 – e)^–1 (which is negative!). Now that we know what happens at θ = 0; to show that the derivative of your function is negative for all θ, we simply need to show that g(θ) < 0 for all θ ≠ 0. To do this, we can look for the stationary points of g(θ). We see that $g'(\theta )=-\theta e^{\theta },$ and so $g'(\theta )=0$ if and only if θ = 0. Is this a maximum or a minimum? Well, $g''(\theta )=-(1+\theta )e^{\theta }$ and so $g''(0)=-e<0$ meaning that θ = 0 is a maximum of g(θ). In fact, it is a global maximum since it is the only stationary point. Notice that g(0) = 0 and so g(θ) < 0 for all θ ≠ 0. It follows that $f'(\theta )<0$ for all θ. So your function has value 1 when θ = 0 and constantly decreases towards 0 as θ increases towards positive infinity. •• Fly by Night (talk) 15:42, 13 June 2010 (UTC)

P.S. The sample solution is a bit hand-wavy. What it is saying is that close to zero, the lowest order terms in the Taylor Series dominate. So your function will behave like θ/θ = 1 for very small θ. For very large θ, the highest order terms will dominate. What this means though is unclear. What is the highest terms in an infinite Taylor Series? Your function behaves very differently as θ tends towards negative infinity than it does when θ tends towards positive infinity. Personally, the sample solution isn't at all rigorous; like I said: it's a bit hand-wavy. •• Fly by Night (talk) 15:59, 13 June 2010 (UTC)

For $\theta >0$, the denominator is $>\theta +\theta ^{2}/2$. This is enough to show that the limit is 0 - no need to muse over the meaning of "highest terms". Also, after you factor out the θ all terms are increasing, so the function is strictly decreasing. -- Meni Rosenfeld (talk) 17:27, 13 June 2010 (UTC)

Fly by Night, surely if $g\left(\theta \right)=\left(1-\theta \right)e^{\theta }-1=e^{\theta }-\theta e^{\theta }-1$ then $g'\left(\theta \right)=e^{\theta }-\theta e^{\theta }$? It Is Me Here ^{t / c} 19:16, 13 June 2010 (UTC)

No, I'm afraid not. Try using the product rule. In particular:

${\frac {d}{d\theta }}(\theta e^{\theta })=\left({\frac {d}{d\theta }}\theta \right)e^{\theta }+\theta \left({\frac {d}{d\theta }}e^{\theta }\right)=e^{\theta }+\theta e^{\theta }$

So you will find that $g'(\theta )=-\theta e^{\theta }$ as stated. •• Fly by Night (talk) 19:23, 13 June 2010 (UTC)

Silly me, thanks! It Is Me Here ^{t / c} 19:42, 13 June 2010 (UTC)

Meni Rosenfeld, could you elaborate on the last sentence of your reply, please? It Is Me Here ^{t / c} 19:19, 13 June 2010 (UTC)

As Zunaid suggested, you can write ${\frac {\theta }{e^{\theta }-1}}={\frac {\theta }{\theta +\theta ^{2}/2!+\cdots }}={\frac {1}{1+\theta /2!+\cdots }}$. For $\theta >0$, Every term in the denominator is increasing (gets larger when θ gets larger), so the sum is also increasing. The function is therefore decreasing. This is not really necessary for the questions asked, it was more of a reply to Fly by Night. -- Meni Rosenfeld (talk) 19:31, 13 June 2010 (UTC)

Oh, I see now. Thanks, everyone. It Is Me Here ^{t / c} 19:42, 13 June 2010 (UTC)

(edit conflict) I like your solution Meni; it's a nice simple solution... I was using a sledgehammer to crack a walnut. Full marks Meni! •• Fly by Night (talk) 19:44, 13 June 2010 (UTC)

Er sorry, one more thing. Meni Rosenfeld, if the function is strictly decreasing, does that necessarily mean that it tends to y=0? Could one have a strictly decreasing function tending to y=-2 or y=3 or something? It Is Me Here ^{t / c} 19:47, 13 June 2010 (UTC)

Since the denominator grows without bound whilst the numerator is constant, the fraction as a whole tends to zero. (The proof of this fact is one of the elementary results demonstrated when limits are introduced in under-grad maths. The proof would typically use a squeeze theorem and/or a suitable limit comparison test to simplify the denominator, then prove that the simplified fraction converges to zero by the usual epsilon-delta definition. Once the course moves on to more difficult problems you can typically take those elementary results as a given without further proof.) Zunaid 21:47, 13 June 2010 (UTC)

No and yes, respectively. A function which is strictly decreasing can tend to 0, or to -2, or 3, or any other real number, or negative infinity. My proof that the function is decreasing was just to show that what Fly by Night did could have been done simpler (in particular, that the sample solution was rigorous, in the sense that once you write the Taylor series, everything else follows trivially). My proof for what you were actually asked - that the limit is 0 - was in the first sentence: After θ is factored out, the denominator is $1+\theta /2+\cdots >1+\theta /2\to +\infty$, so the function tends to 0. -- Meni Rosenfeld (talk) 08:15, 14 June 2010 (UTC)

Not all strictly decreasing functions have a limit of zero. The function ƒ(θ) = k + e^–θ is a strictly decreasing function; i.e. if θ > φ then ƒ(θ) < ƒ(φ), and its limit as θ tends to positive infinity is k. Since k can be chosen arbitrarily, we can give ƒ any limit we choose. •• Fly by Night (talk) 19:04, 14 June 2010 (UTC)

Taylor series

In one of your articles there is:${\begin{aligned}(x+y)^{n}&={n \choose 0}x^{n}y^{0}+{n \choose 1}x^{n-1}y^{1}+\cdots \end{aligned}}$ <extended content>

What is meant by ${n \choose 0}$? —Preceding unsigned comment added by 76.229.192.126 (talk) 21:05, 13 June 2010 (UTC)

${n \choose 0}={\frac {n!}{0!(n-0!)}}={\frac {n!}{n!}}=1,$ meaning that there is one way to choose no element out of n elements. See article N choose k. DVdm (talk) 21:14, 13 June 2010 (UTC)

It's rather ${n \choose 0}={\frac {n!}{0!(n-0)!}}={\frac {n!}{1\cdot n!}}={\frac {n!}{n!}}=1,$ but the conclusion is correct. --CiaPan (talk) 21:33, 13 June 2010 (UTC)

Of course, a typo induced by correcting a typo. Sorry. Silly DVdm (talk) 21:44, 13 June 2010 (UTC)

If you tell us which article you are talking about we can insert the appropriate link. Bo Jacoby (talk) 21:51, 13 June 2010 (UTC).

Naturally, one place where it appears is Binomial theorem, but there there is already a link to Binomial coefficient. -- Meni Rosenfeld (talk) 08:20, 14 June 2010 (UTC)

I had pointed to article N choose k, which redirects to Binomial coefficient. DVdm (talk) 09:57, 14 June 2010 (UTC)

What are you replying to? Bo asked where did the OP encounter the formula. -- Meni Rosenfeld (talk) 10:53, 14 June 2010 (UTC)

Bo's indentation suggests he was asking me. DVdm (talk) 13:34, 14 June 2010 (UTC)

I see. Since Bo was clearly replying to the OP, I've taken the liberty to correct his indentation. -- Meni Rosenfeld (talk) 14:49, 14 June 2010 (UTC)

June 14

Proof that infinity equals -1

x=1+2+4+8+16+...

2x=2+4+8+16+...

2x-x=(2+4+8+16+...)-(1+2+4+8+16+...)

x=-1

But x is clearly infinity, so infinity equals -1. Where's the error with this? --76.77.139.243 (talk) 13:26, 14 June 2010 (UTC)

Interesting, but I believe the error is the step after

2x-x=(2+4+8+16+...)-(1+2+4+8+16+...)

Because infinity - infinity = infinity, the next step should actually be

x = infinity, which is consistent with your premise -- Rick Van Tassel ^{user|talk|contribs} 13:30, 14 June 2010 (UTC)

We have an article 1 + 2 + 4 + 8 + · · ·. Algebraist 13:35, 14 June 2010 (UTC)

You can't rearrange the order of summation in an infinite sum. Readro (talk) 13:41, 14 June 2010 (UTC)

One error is that you can't subtract series in that way unless they converge to a finite number.

Infinity minus infinity is an indeterminate form. That means that if ƒ(x) and g(x) can both be made as large as desired by making x close enough to some specified point (which may be infinite) then ƒ(x) − g(x) may approach anything, depending on which functions ƒ and g are. Michael Hardy (talk) 13:45, 14 June 2010 (UTC)

The error is to assume that "x is clearly infinity". Bo Jacoby (talk) 14:18, 14 June 2010 (UTC).

Indeed! minus 1 is the most reasonable answer. Infinity does not really exist, so if you have a divergent expression, it must be the result of some illegal formal manipulations applied to an expression that does correspond to a finite number. You can then guess what the correct expresson is, but any formal method for doing that, must itself contain at least one illegal manipulation, as you have to undo the assumed illegal manipulation that led to the divergent expression in the first place.

It turns out that errors produced by formal manipulations, like interchanging Taylor expansions with integrations when that is not allowed, using series expansion outside the domain of convergence etc. etc., tend to be universal in the sense that they can usually be corrected by some generic resummation technique. Count Iblis (talk) 14:29, 14 June 2010 (UTC)

There is only one error here. The expression is not a sum. It is shorthand for a limit. Treating it as a sum and attributing some of the standard properties of sums (like distributivity, commutativity and/or associativity) creates the trouble. DVdm (talk) 14:33, 14 June 2010 (UTC)

Indeed. Those properties only hold for convergent series, so by using them you are assuming x isn't infinity. Once you've assumed that, it is hardly surprising that you get an answer that is consistent with it. --Tango (talk) 14:45, 14 June 2010 (UTC)

Even for a series which sums to a limit okay you can do some wierd things by rearranging the terms if it isn't absolutely convergent. See the Riemann series theorem. Dmcq (talk) 15:17, 14 June 2010 (UTC)

The rules for resumming divergent series are discussed in this article. Count Iblis (talk) 15:21, 14 June 2010 (UTC)

True, but in this case all the terms are positive, so it isn't an issue. --Tango (talk) 15:25, 14 June 2010 (UTC)

See formal power series. You do not need to assume convergence. The algebraic rules apply nicely. However you are only allowed to insert or remove a finite number of parentheses, so 1−1+1−1+1−1+1−1+... = 1/2 is not equal to (1−1)+(1−1)+(1−1)+(1−1)+... = 0. Also you are only allowed to change the order of a finite number of terms, so 1−1/2+1/3−1/4+1/5−1/6+...=ln(2) is not equal to 1+1/3−1/2+1/5+1/7−1/4+..., even if these series are convergent and contain the same terms.

By the way in p-adic numbers where p is 2 what you have is correct. ...1111111.0 does equal -1. Dmcq (talk) 17:28, 14 June 2010 (UTC)

Program for creating mathematical graphs

Cross-posted to the computing desk.

Hi everyone. I'm looking for a program to draw graphs easily (I do mean graphs not charts - essentially circles connected by lines). I'd like a program which has the following:

Graphical User Interface

Ability to drag/drop vertices around.

Preferably ability to resize the vertices, alter colours, alter thickness/colour of the edges, etc.

Does anyone know any software which can do this? -mattbuck (Talk) 17:53, 14 June 2010 (UTC)

You might try Inkscape. It has it's faults but it's free and you can upload the .SVG files to Wikipedia.

Perhaps one of the packages in the List of mind mapping software or List of concept mapping software… looking through the former now, how about Graphviz? Not used it myself though. Qwfp (talk) 09:07, 15 June 2010 (UTC)

Try yEd. Barsamin (talk) 14:36, 18 June 2010 (UTC)

You can try Cytoscape too, especially useful for overlaying data on the graphs.

How do you write integration limits of the solution in TeX?

How do you write limits of integration after finding what the integral is. For example, let's say if the question was to integrate x^2 from 0 to 1, the (partial) solution to this is x^2/2 evaluated from 0 to 1, how would you write this in TeX? 142.244.151.247 (talk) 21:07, 14 June 2010 (UTC)

You use the square bracket \left[ and \right]. Make sure you use them in pairs or it won't compile. If you type \left[ \frac{1}{2}x^2 \right]_0^1 you get

$\left[{\frac {1}{2}}x^{2}\right]_{0}^{1}$

Putting \left before bracket makes LaTeX adjust the size of the bracket to fit the text. You can use \left( \right), \left{ \right}, etc... But make sure you put them in pairs. If you don't want a bracket then use \left. or \right. to omit the left or right bracket. For example \left. \frac{1}{2}x^2 \right]_0^1 gives

$\left.{\frac {1}{2}}x^{2}\right]_{0}^{1}$

You can even get more exotic brackets, say pointy brackets, by typing \left\langle \right\rangle:

$\left\langle {\frac {1}{2}}x^{2}\right\rangle$

Without the \left and \right these expressions compile as:

$[{\frac {1}{2}}x^{2}]_{0}^{1}\ {\mbox{ and }}\ \langle {\frac {1}{2}}x^{2}\rangle$

•• Fly by Night (talk) 21:16, 14 June 2010 (UTC)

The one I've seen the most is similar to the second one but with a vertical line instead of a bracket. Apparently there is no universal standard though.--RDBury (talk) 02:31, 15 June 2010 (UTC)

As in \left. \frac{1}{3}x^3 \right|_0^1 ?

$\left.{\frac {1}{3}}x^{3}\right|_{0}^{1}$

-- 58.147.52.243 (talk) 06:53, 15 June 2010 (UTC)

That's right. However, if the general integral form contains more than one term (sometimes also when there's a single term, that contains more than two factors, one of which is constant), then the vertical bar alone is not clear enough. The same is even more important if the integral appears as a term in a sum, say something like
$f(x)+\int _{2}^{3}y(x)\,dt=7x^{2}+{\tfrac {3}{4}}\ln x+2\cos(x^{2}){\Big \vert }_{2}^{3}$
—it's hard to say which terms should be taken at 2 and 3 and then subtracted, and which should not. Putting brackets makes things clear:
$f(x)+\int _{2}^{3}y(x)\,dt=7x^{2}+{\Big [}{\tfrac {3}{4}}\ln x+2\cos(x^{2}){\Big ]}_{2}^{3}$
CiaPan (talk) 07:46, 15 June 2010 (UTC)

The standard is square brackets on either side:

$\int _{0}^{1}x^{2}\ {\text{d}}x=\left[{\frac {1}{3}}x^{3}\right]_{0}^{1}={\frac {1}{3}}.$

I only mentioned the one-sided brackets so the person setting the question could use the functionality in other settings. The single line on the right is usually used as a short hand for evaluation. So ⅓x³ evaluated at x = 0 can be written as

$\left.{\frac {1}{3}}x^{3}\right|_{x=0}$ •• Fly by Night (talk) 23:04, 15 June 2010 (UTC)

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)

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)

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) 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.

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).

Proportional_navigation may be helpful. Zoonoses (talk) 14:45, 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)

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)

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)

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)

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)