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March 1

Am I smarter than a 5th grade math textbook?

I am perfectly mortified to have to be asking this at all, but I'm really confused and have no one else to consult. My husband teaches 5th grade math, and I'm helping him out by grading some papers. One of them has the following question: "Yolanda ate 2/5 of the carrots on the plate, then her brother ate 9 of the remaining carrots. Which expression shows how to find how many total carrots there were?" It's a multiple-choice question, but none of the provided answers makes any sense. The teacher's edition gives the answer as "(2/5)c - 9," but this doesn't make sense to me. Is this answer actually correct? If so, why? If not, what IS the correct answer? — Preceding unsigned comment added by 172.3.129.135 (talk) 19:46, 1 March 2013 (UTC)[reply]

Should be (2/5)c-9 = 0 , then c can be worked out by rearranging the formula to be 22.5 carrots.....

Yolanda eats (2/5) x 22.5 carrots which leaves 9 carrots for her brother to eat.

I agree with you its a badly written question and answer.

Ap-uk (talk) 22:28, 1 March 2013 (UTC)[reply]

The problem does not state that no carrots were left at the end. If you want an expression for the number of carrots that remain, it would be (3/5)c - 9, assuming c is the number at the beginning. So either you copied something down incorrectly, or the textbook contains a typo. Looie496 (talk) 22:49, 1 March 2013 (UTC)[reply]

Yes assuming no carrots left at the end this gives 15 carrots to start but the wording implies there are several remaining.--Gilderien Chat|List of good deeds 23:22, 1 March 2013 (UTC)[reply]

Indeed. There's not enough information to work out the number of carrots we began with, because "9 of the remaining carrots" means there were still some left over, and we're not told how many. If c is the starting number, then the simplest expression for the starting number is, surprise, surprise, c; and the other data is just herrings in various shades of red. Unless we assign a value, d, to the number left over at the end. Then d = (3c-45)/5, and c = (5d+45)/3. But if none of these are acceptable answers, then the paper contains a typo and students should not be burdened with the responsibility for that. -- Jack of Oz ^[Talk] 01:06, 2 March 2013 (UTC)[reply]

Thanks all. Actually I came to many of the same conclusions JackofOz did, but "c" wasn't a possible answer, nor was any equation or any expression with multiple variables. (And the problem was definitely asking the students to create an expression of the state of things using numbers and mathematical symbols rather than solving for an unknown value.) After posting this I was able to get in touch with my father, who went farther in math than I did (not that it should matter for a fifth-grade problem; he's just my go-to math expert) and actually enjoys this type of problem, and he didn't know what to make of it either. I know I copied it down correctly, so the problem was with the materials - and not for the first time, either. I just threw the question out altogether and gave them a score based on eleven problems instead of twelve. And now my husband's on his way home from work and we can spend the evening griping about the wretchedness of teaching materials in the state of California. — Preceding unsigned comment added by 172.3.129.135 (talk) 01:18, 2 March 2013 (UTC)[reply]

They just needed to write the last sentence differently. Instead of "Which expression shows how to find how many total carrots there were ?" they should have asked "Which expression shows how many carrots remain, as a function of the original number of carrots, c ?". Your husband can have the students correct the books, either in this class or the next, to make it a reasonable practice question. StuRat (talk) 04:10, 2 March 2013 (UTC)[reply]

Apart from mathematical infelicities, the language of the original question also leaves a lot to be desired. "How many total carrots" - are we to assume that some of these objects are total carrots but some are only partially carrots (crosses between carrots and hedgehogs, perhaps)? Reminds me of Eric the Half Bee. -- Jack of Oz ^[Talk] 08:21, 2 March 2013 (UTC)[reply]

Yes, the expression "total carrots" makes the whole question meaningless, or at least so ambiguous that you have to guess at the meaning. Even if we interpret it way StuRat suggests, the answer should be (3/5)c - 9, not (2/5)c - 9. Gandalf61 (talk) 09:59, 2 March 2013 (UTC)[reply]

If the question had been "total carrots eaten" then the expression would have been "(2/5)c + 9" (where c is the original number of carrots). Was this one of the options? Dbfirs 17:52, 2 March 2013 (UTC)[reply]

Not to mention that anyone eating 9 whole carrots in a sitting, on a regular basis, may very well turn orange. StuRat (talk) 17:56, 2 March 2013 (UTC) [reply]

I have seen that happen to someone who tried to diet by filling up on carrots. Gradually going orange from the fingers onwards.Manytexts (talk) 06:16, 6 March 2013 (UTC)[reply]

March 2

Correlation =//= Causation... Ever?

If there is a strong positive correlation, then that is still a correlation, right? That can't be considered a causation, because it is still a correlation. My reasoning is that any correlation, no matter how high or low, can be tampered with by selecting a critical r value and comparing the critical r value to the obtained r value (Pearson product-moment correlation coefficient) in order to achieve a desirable result? For example, there may be a lot of evidence (very strong correlation) to suggest a statistically significant change, but because the statistician chooses a ridiculously high critical value, the statistician fails to reject the null hypothesis and maintains that there is no change, because there is a conflict of interest in the experiment that may influence the statistician's own life in some bad way, which the statistician obviously denies in order to preserve his academic credentials even though lying is liable to being accused of academic dishonesty and thus can greatly ruin his career. Are there real examples of such things happening where statisticians intentionally manipulate the data in order to achieve a desirable result and report the result to a scientifically and mathematically illiterate general public? Sneazy (talk) 06:39, 2 March 2013 (UTC)[reply]

I can't make sense of all that, but there is a famous book called How to Lie with Statistics that is probably the sort of thing you are looking for. Looie496 (talk) 06:47, 2 March 2013 (UTC)[reply]

Actually, the first sentence is the question. Sneazy (talk) 06:55, 2 March 2013 (UTC)[reply]

Actually there are two questions and one statement with an appending question mark. Looie answered your second question. As for the first, indeed correlation is correlation and not causation. Your reasoning is somewhat convoluted and the agency of a statistician need not be invoked to show that they are fundamentally and conceptually different. — Preceding unsigned comment added by 202.65.245.7 (talk) 08:51, 2 March 2013 (UTC)[reply]

(edit conflict)The article on Statistical hypothesis testing explains how hypothesis testing is normally done. It's not just comparing r-values, though a high r-value suggests that there is something to investigate (which might or might not be a causal link). I assume you've read the article Correlation does not imply causation. Dbfirs 08:59, 2 March 2013 (UTC)[reply]

In the real world, we assume causation all the time, when all we can really prove is correlation. For example, observing the correlation between smoking and lung cancer deaths was taken as evidence that smoking causes lung cancer deaths, before the specific mechanisms were discovered. It might have turned out that people with a certain gene which causes lung cancer also, as a result of that gene, craved cigarettes, but this seems unlikely. StuRat (talk) 18:00, 2 March 2013 (UTC)[reply]

... but causation was widely denied (especially by tobacco companies and smokers) even when the correlation was shown to be strong. Dbfirs 08:24, 3 March 2013 (UTC)[reply]

Yes, and we're still with the problem that no specific case of lung cancer can be blamed on smoking, which is a problem for those wishing to file lawsuits (of course, one could also argue that anyone who started smoking in the last few decades had fair warning of the danger, but that's another issue). StuRat (talk) 16:26, 3 March 2013 (UTC)[reply]

Restricted polyabolos

Does anyone know the number of possible restricted polyabolos?? Please see Talk:Polyabolo for what this means. Georgia guy (talk) 20:41, 2 March 2013 (UTC)[reply]

The sequence begins 1, 2, 2, 6, 8, 22, 42, 112, 252, 650, 1584, 4091, 10369, 26938, 69651, 182116, 476272, 1253067, 3302187, 8733551… (sequence A197465 in the OEIS). —Bkell (talk) 01:02, 3 March 2013 (UTC)[reply]

March 3

Human-made mazes (complex branching passages)

What are some of the most complex mazes (complex branching passages) made by humans? They can be in buildings or tunnels or any other structures made by humans. (In planning this question about three months ago, I decided that it would be best if I first asked a more basic question. That question is archived at Wikipedia:Reference desk/Archives/Mathematics/2012 December 7#Mazes (complex branching passages).)
—Wavelength (talk) 03:30, 3 March 2013 (UTC)[reply]

I'm not sure I understand what you would classify as a "maze." Does something like the National Highway System (United States) count? —Bkell (talk) 04:07, 3 March 2013 (UTC)[reply]

Yes, it does. I am looking for examples in various categories: (1) buildings, (2) tunnels, (3) urban streets, (4) highways, and (5) other structures.

—Wavelength (talk) 15:36, 3 March 2013 (UTC)[reply]

How about the electrical wiring of a Boeing 747, or the world's telecommunication network, or the UK's electricity supply system from generation to domestic consumer (much wider than the UK, as there is an interconnector with France, who may well have physical links with other countries)?←86.186.142.172 (talk) 16:21, 4 March 2013 (UTC)[reply]

For a building example, see Kowloon_Walled_City, which was essentially one building at the end. It was built without plan, and famous for the complicated routes and "hacked" connections. SemanticMantis (talk) 00:20, 5 March 2013 (UTC)[reply]

Thank you.—Wavelength (talk) 03:34, 6 March 2013 (UTC)[reply]

"Large" microchips such as CPUs are easily more complex than a large city's road network. Roger (talk) 12:29, 8 March 2013 (UTC)[reply]

Limit points in the cartesian plane

Am I correct in thinking that in the Cartesian plane when any Set A (for example A = { x | x = ( 0 , 1 / n ) where n ∈ ℕ } is assigned a set A' of limit points these points are all those an infinitesimal distance from the points in set A (e.g. (0,0) is a limit point of the example above?--Gilderien Chat|List of good deeds 16:08, 3 March 2013 (UTC)[reply]

I think you get the idea, but perhaps you should rephrase it in terms of open sets. In the Cartesian plane you can replace the open sets with open metric ε-balls in the formulation (which are open sets). So a point p is a limit point of A if every ε-ball, ε > 0, around p contains a point of A other than p. See Limit point. So if a point q is at a positive distance of at least δ from every point of A (except perhaps q itself), no matter how small, it is not a limit point of A because you can chose ε = δ/2 for the ε-ball around q. That open set contains no points in A except possibly q itself. YohanN7 (talk) 18:07, 3 March 2013 (UTC)[reply]

Thank you, that goes a long way to alleviating my confusion. So does that mean the example I gave above has 0 as its only limit point? The definition you gave seems to imply that every point in a continuous set is itself a limit point for example in x^2 + y^2 > k^2 would this imply that every point on the plane except those in or on a circle of radius k around (0,0) is a limit point of the set? --Gilderien Chat|List of good deeds 18:58, 3 March 2013 (UTC)[reply]

Yes, limit points can be in the set too. So the limit points of

\{\,(x,y):x^{2}+y^{2}>k^{2}\,\}

are all points except those strictly inside the circle of radius k around the origin—the points on the circle are limit points. —Bkell (talk) 19:02, 3 March 2013 (UTC)[reply]

And therefore whether x is a limit point of A is independent of whether x ∈ A? Thank you.--Gilderien Chat|List of good deeds 19:17, 3 March 2013 (UTC)[reply]

I am under impression that the YohanN7's formulation excludes from the A' set separated points of A (although it allows inner or border-but-not-separated points), while Gilderien's formulation allows all of them into A'. Am I right? For example let's take the set A defined as union of a square without one side, a line segment without one of its end plus a single point:

A = [0, 1)×[0, 1] ∪ [2, 3)×{0} ∪ (4, 4)

What is A' then (according to both formulations, if not identical)? --CiaPan (talk) 21:42, 3 March 2013 (UTC)[reply]

From what I understand, YohanN7's formulation would include internal points in the set as well, for example for the set S = { x | 0 < x < 1 } there would exist a set S' = { x | 0 ≤ x ≤ 1 } which would essentially be { 0 , S , 1 }. --Gilderien Chat|List of good deeds 21:50, 3 March 2013 (UTC)[reply]

Okay, but my question relates to separate points. For a finite set A, say containing 3 points, is A' equal A or is A' empty? --CiaPan (talk) 22:05, 3 March 2013 (UTC)[reply]

A - whether q is part of A is independent of whether it is part of A' .--Gilderien Chat|List of good deeds 22:34, 3 March 2013 (UTC)[reply]

The subtle point here in the definition of a "limit point" is whether a neighborhood of the point p must contain points of the set other than p. There is no general agreement on this matter. So there are various other terms that are often used in this setting. For example, here are definitions from Tom Apostol, "Mathematical Analysis":

A point p is an adherent point of the set S if, for all ε>0, the ε ball around p contains points of S.
A point p is an accumulation point of the set S if, for all ε>0, the ε ball around p contains points of S\p.

By "limit point" some authors mean an "adherent point", and others mean an "accumulation point" (in this sense). Wikipedia (AFAIK) generally means "accumulation point". Sławomir Biały (talk) 23:18, 3 March 2013 (UTC)[reply]

Sometimes it just happens that a relatively easy concept becomes confusing because there are slightly different conventions and different terminology applies, or worse, the same terminology for slightly, but distinctly, different concepts. It's annoying, but unavoidable. Just look before you leap!

My formulation was meant to be a Cartesian plane version of "Let S be a subset of a topological space X. A point x in X is a limit point of S if every neighbourhood of x contains at least one point of S different from x itself.", which is the formulation in the wikipedia article Limit point. YohanN7 (talk) 11:29, 4 March 2013 (UTC)[reply]

March 4

What did von Neumann's variant of von Neumann–Bernays–Gödel set theory look like?

The "History" section of our article on von Neumann–Bernays–Gödel set theory starts with "The first variant of NBG, by John von Neumann in the 1920s, took functions and not sets, as primitive," without further elaboration, and the set theory books I have seen usually do not go into von Neumann's original conception at all, leaving me with the question, what did von Neumann's variant of NBG look like? — Tobias Bergemann (talk) 13:11, 4 March 2013 (UTC)[reply]

Doubling the cube

Can anyone explain me in simple maths that the neusis construction at http://en.wikipedia.org/wiki/Doubling_the_cube#Using_a_marked_ruler why equals cuberoot2? 117.233.117.95 (talk) 13:34, 4 March 2013 (UTC)[reply]

(Image copied here for ease of reference.) There's a proof here ("Construction 3", starting on page 7 of the PDF) of what appears to be the same construction, though I haven't worked through it yet. AndrewWTaylor (talk) 17:10, 4 March 2013 (UTC)[reply]

Penrose tiles puzzle pieces

Does anyone have any suggestions on a set of Penrose tiles jigsaw puzzle pieces. They're for my nephew. (Well, really they're for me... :-) Sławomir Biały (talk) 23:31, 4 March 2013 (UTC)[reply]

A classic interlocking jigsaw puzzle would not follow the boundaries of Penrose tiles. You could do that, but why not get actual Penrose tiles and assemble them yourself into what ever size Penrose pattern you want ? (Instead of buying the tiles, you could also cut them out of construction paper.) Note that patchwork quilts are sometimes made in such a way, so, if either of you are interested in sewing, that's a more practical application, starting from fabric "cells": [1]. StuRat (talk) 01:43, 6 March 2013 (UTC)[reply]

Yes, I'm looking for a good set of tiles that fit together, maybe with some kind of edge matching as explained in the article. I don't want to cut them out myself, as I lack the patience for it. Sławomir Biały (talk) 16:47, 6 March 2013 (UTC)[reply]

What about http://www.gamepuzzles.com/pentuniv.htm#TD? Maybe that contains something? -- Toshio Yamaguchi 12:57, 7 March 2013 (UTC)[reply]

March 5

A maximum/minimum value question

Dear Wikipedians:

A question calls for the finding of the absolute maximum and minimum value of $f(x,y,z)=x^{2}+y^{2}+z^{2}$ subject to the constraints $x-y=1$ and $y^{2}-z^{2}=1$

Using Lagrange multipliers, I set up the following function:

$L(x,y,z,\lambda ,\mu )=x^{2}+y^{2}+z^{2}+\lambda (x-y-1)+\mu (y^{2}-z^{2}-1)$

Setting the first partial derivatives of L to zero, we obtain

$2x+\lambda =0$	(1)
$2y-\lambda +2\mu y=0$	(2)
$2z-2\mu z=0$	(3)
$x-y-1=0$	(4)
$y^{2}-z^{2}-1=0$	(5)

From (3) I find $\mu =1$ , substituting into (2):

$2y-\lambda +2y=0$

$\lambda =4y$ (6)

Substituting (6) into (1):

$2x+4y=0$

$x=-2y$ (7)

Substituting (7) into (4):

$-2y-y-1=0$

$-3y=1$

$y=-{\frac {1}{3}}$

Substituting $y=-{\frac {1}{3}}$ into (5):

$\left(-{\frac {1}{3}}\right)^{2}-z^{2}-1=0$

$z^{2}=-{\frac {8}{9}}$

This means that there are no real solutions to the original problem. However the official solutions clearly indicate the existence of real solutions. I am wondering what is going on?

Thanks for all your help.

74.14.60.239 (talk) 02:42, 5 March 2013 (UTC)[reply]

Equation 3 has an alternate solution. There is no absolute maximum subject to those constraints, but there is certainly an absolute minimum. Looie496 (talk) 03:49, 5 March 2013 (UTC)[reply]

I'm going to do it a bit differently. I notice that we can get the function in terms of y only:

$x-y=1$

$x=1+y$

$y^{2}-z^{2}=1$

$-z^{2}=1-y^{2}$

$z^{2}=y^{2}-1$

$z=sqrt(y^{2}-1)$

$f(x,y,z)=x^{2}+y^{2}+z^{2}$

$f(y)=(1+y)^{2}+y^{2}+(sqrt(y^{2}-1))^{2}$

$f(y)=(1+2y+y^{2})+y^{2}+(y^{2}-1)$

$f(y)=2y+3y^{2}$

From here, set the derivative equal to zero:

$0=2+6y$

$y=-1/3$

Now find x and z:

$x=1+y$

$x=1+(-1/3)$

$x=2/3$

$z=sqrt(y^{2}-1)$

$z=sqrt((-1/3)^{2}-1)$

$z=sqrt((1/9)-1)$

$z=sqrt(-8/9)$

So, I get the same results as you. What does the book list as the solution(s) ? StuRat (talk) 04:35, 5 March 2013 (UTC)[reply]

What you have demonstrated with perfect clarity is that that method doesn't work. The problem is that the equation

y^{2}-z^{2}=1

limits the range of y -- it can never be between -1 and 1. So when looking for the extrema, in addition to the point where the derivative is zero you also have to consider the edge points of the range. Looie496 (talk) 04:53, 5 March 2013 (UTC)[reply]

I may have missed something when solving for

(sqrt(y^{2}-1))^{2}

. This should also have a negative root, I think:

$f(y)=(1+y)^{2}+y^{2}+(sqrt(y^{2}-1))^{2}$

$f(y)=(1+2y+y^{2})+y^{2}+-(y^{2}-1)$

$f(y)=(1+2y+y^{2})+y^{2}-y^{2}+1$

$f(y)=2+2y+y^{2}$

From here, set the derivative equal to zero:

$0=2+2y$

$y=-1$

Now find x and z:

$x=1+y$

$x=1+(-1)$

$x=0$

$z=sqrt(y^{2}-1)$

$z=sqrt((-1)^{2}-1)$

$z=0$

So, if we plug in (0,-1,0), we get:

$f(x,y,z)=x^{2}+y^{2}+z^{2}$

$f(x,y,z)=0^{2}+(-1)^{2}+0^{2}$

$f(x,y,z)=1$

Not sure if that's the min or max, though. I'd have to try a few nearby values to test it out. Does this match the book answer ? StuRat (talk) 05:01, 5 March 2013 (UTC)[reply]

The minimum and the maximum are to be found amongst the roots of the first derivative. All you have to do is reduce the function in question to a single-variable function. And you do precisely that by replacing x and y with expressions in z. From y² - z² = 1 we deduce that y² = 1 + z². And from x - y = 1 we deduce that x = 1 + y = 1 ± √(1 + z²) ⇔ x² = 2 + z² ± 2√(1 + z²). Our function thus becomes f(x, y, z) = F(z) = 3 + 3z² ± 2√(1 + z²). All we have to do now is calculate F'(z), find its roots, and see which of them has the smallest, respectively the largest value. F'(z) = 2z(3 ± 1 / √(1 + z²)), whose roots are z₀ = 0 and z_{1, 2} = ±2i√2/3. Since complex solutions are apparently not acceptable, F(0) = 3 ± 2, meaning your minimum and maximum value are 1 and 5, respectively. — 79.113.208.129 (talk) 05:04, 5 March 2013 (UTC)[reply]

We agree on the minimum, but not the max. How about the case of

f(3,2,sqrt(3))=16

? That's more than 5, and seems to be consistent with the constraints. StuRat (talk) 05:23, 5 March 2013 (UTC)[reply]

Yes, stupid mistake: a point (in this case, 0) cannot be both maximum and minimum at the same time ! And -apart from that silly blunder-, I forgot to add that one also has to take into account the values of the function at the extreme points of its interval of definition, in this case

\pm \infty

, both of which are

+\infty

. So the minimum and the maximum are the smallest and largest values from among 1, 5, and

\infty

. Ergo, the minimum is 1 and the maximum is

+\infty

. — 79.113.208.129 (talk) 05:47, 5 March 2013 (UTC)[reply]

Very good, we are now in complete agreement. StuRat (talk) 05:53, 5 March 2013 (UTC)[reply]

EXCEPT you don't need z here and the whole mess with possibly complex square roots. Just leave

z^{2}=y^{2}-1

and plug that z² directly into the f formula. --CiaPan (talk) 17:03, 5 March 2013 (UTC)[reply]

That's what I did, although I did need to take into account the negative root. StuRat (talk) 02:42, 6 March 2013 (UTC)[reply]

Additionally I would exclude +infinity form the final answer. The maximum value of a real function would be a real number and infinity is not a real number. The answer 'maximum of f is infinity' is incorrect IMHO, it should be 'there is no global maximum, f is not bounded above'. --CiaPan (talk) 10:04, 6 March 2013 (UTC)[reply]

Yea, that's how I'd say it, but I'm flexible. StuRat (talk) 17:42, 6 March 2013 (UTC)[reply]

March 6

Gamma ~= 1 / √3

Is there any particular reason for why $\gamma \approx {\frac {1}{\sqrt {3}}}$ , or is it merely a simple coincidence ? — 79.113.237.34 (talk) 00:59, 6 March 2013 (UTC)[reply]

Probably a coincidence 130.76.64.120 (talk) 01:58, 6 March 2013 (UTC)[reply]

Follows from Gaussian quadrature approximation of gamma, using two-point quadrature on each of the intervals [n,n+1]. Note that you get the easier approximation of gamma as 1/2 by using the trapezoid approximation instead. So it's not surprising GQ gives a better (irrational) approximation that is also fairly simple and involves the square root of 3.... Sławomir Biały (talk) 02:07, 6 March 2013 (UTC)[reply]

Thanks ! :-) — 79.113.237.34 (talk) 03:52, 6 March 2013 (UTC)[reply]

A strange relationship to seeing in art drawing

Talking to my art class about the simple method of using a pencil to "sight" at arm's length, a figure for proportions in order to put it on the page. There must be a formula to explain why this can be done and is a pretty good guide. You hold the pencil out against the subject, say his or her head, and put the thumbnail at that point. You check how many times, say it goes into the body length. Then, closer, on your drawing, you check the headlength in the same way, and divide it into the body length to make corrections. I'm not asking about how the division happens, but how it's translated so simply from a full human figure into the sketch you're making. If this isn't math, please recommend a ref desk that could handle it. Thanks in advance, Manytexts (talk) 02:23, 6 March 2013 (UTC)[reply]

Yes, it's math/optics. See vanishing point and perspective drawing. StuRat (talk) 02:35, 6 March 2013 (UTC)[reply]

Ah. Though I went to these articles & almost all the "see also" related ones, I am left no wiser. Too techie for me, Manytexts (talk) 06:20, 6 March 2013 (UTC)[reply]

Are you a student or the teacher ? If you are a student, you might benefit from learning the basics of perspective. Try making a picture of a room, using one point perspective. Here's a video to guide you through it: [2]. They used a computer, but you can do the same thing with paper, pencil, and a ruler. StuRat (talk) 06:37, 6 March 2013 (UTC)[reply]

Note that a person standing won't be distorted noticeably in a perspective view (unless the "eye point" is really close). That is, their head is the same size in proportion to their legs, as it is in real life. This would be different if they were lying down, with either their head or feet near us. One interesting detail from the world of art, is that Michelangelo's David is actually sculpted with perspective in mind. That is, the feet are actually proportionally smaller, since the viewer will presumably be closer to those than the head, and the feet would seem huge otherwise. StuRat (talk) 06:56, 6 March 2013 (UTC)[reply]

If I understand the question correctly, the answer is just the fact that things that are proportional to each other remain in the same proportion as they get bigger or smaller. For example, if a painting contains two objects, one of which is twice as large as the other, then those two objects will have the same proportion between them (2:1) in any print of the painting, regardless of the size of the print. The absolute sizes of the two objects may differ (the objects in the painting will be tiny on a postcard and huge on a billboard, for example), but the proportion between them will stay the same.

When you realistically paint a scene from real life, your painting may be larger than the real-life scene, or it might be smaller, but your aim is to keep all of the proportions correct. The pencil-measuring technique is simply calculating proportions to ensure that they match the scene. (I don't think it has anything to do with vanishing points or perspective.) —Bkell (talk) 16:54, 6 March 2013 (UTC)[reply]

Calculating RGB values for images

I came across a specific problem I don't know how to solve. I want to create a game using the M.U.G.E.N. engine. For that, I need to create images to later be used as sprites to be used in-game. In order to create those images, I need to first define the colors to be used for those sprites. Those colors are RGB colors representing a palette to be used for those sprites. I can use exactly 256 colors for this palette. I know that I already need three specific colors: (255, 60, 255), (0, 0, 0) and (255, 255, 255). I want the other colors to be equally distributed in my color model. That is, I want to define the remaining 253 colors in a way such that I increment the RGB parameters one-by-one by a constant value. Now my question is, how do I calculate the constant c with which I have increment x, y, z in the RGB triple (x, y, z) such that I come from (0, 0, 0) to (255, 255, 255) in 253 steps? (Note that it doesn't matter for me that I probably don't hit (255, 60, 255) and (255, 255, 255) exactly, as they are already defined). I hope the problem description is understandable, if something is unclear, I will try to clarify. -- Toshio Yamaguchi 14:03, 6 March 2013 (UTC)[reply]

To clarify: the steps would be (0, 0, 0) (0+c, 0, 0) (0, 0+c, 0) (0, 0, 0+c) (0+2c, 0, 0) (0, 0+2c, 0) (0, 0, 0+2c) (0+c, 0+2c, 0) (0+c, 0, 0+2c) (0+c, 0+c, 0) (0+c, 0+c, 0+c) (0+2c, 0, 0+c) (0+2c, 0+c, 0) etc -- Toshio Yamaguchi 14:26, 6 March 2013 (UTC)[reply]

First, I'm going to ignore the fact that you've predefined three colors; you'll see why at the end. What you want to do, conceptually, is to define

{\sqrt[{3}]{256}}

equally spaced points along each of the three dimensions (the red dimension, the green dimension, and the blue dimension), so that when you take all possible combinations you get

({\sqrt[{3}]{256}})^{3}=256

colors. Of course, there's a problem here, because

{\sqrt[{3}]{256}}

isn't an integer; it's about 6.35. You have a hard limit of 256 colors, so you can't round that number up to 7 (because 7³ = 343). So you need to round it down to 6. Now, 6 equally spaced points along a dimension that goes from 0 to 255 should include the 0 point and then 5 more points that end with 255: so these points need to be placed at multiples of 255/5 = 51. So that's the value of c you're looking for.

This defines only 6³ = 216 colors, which leaves room for 40 other colors of whatever you'd like. You've predefined three colors, but two of those—(0,0,0) and (255,255,255)—are already included in the 216. So add (255,60,255) to the 216 defined by the multiples of c, and then do whatever you like with the remaining 39 colors. (See also Web-safe color.)

If you really want to define all 256 colors and have them be "as equally spread out as possible" in the color space, there are several ways to do that, but they won't give you a perfectly spaced out set of colors like you ask for. —Bkell (talk) 16:36, 6 March 2013 (UTC)[reply]

Agreed, but note that the red, green, and blue increments don't need to be identical. You could use 7 R, 6 G, and 6 B, for example, for 252 total colors. This would give you (0,0,0), (255,255,255) and (255,51,255). That last one isn't quite (255,60,255), so you could pick one or the other, depending on whether that exact color is more important, or having the colors around it be evenly spaced. I wouldn't recommend using both, though, as they would be difficult to distinguish. If you stick with (255,60,255), then I suggest redistributing the remaining red spacing, as 0, 60, 115, 166, 213, 255. Also note that evenly spaced numerically doesn't always mean evenly spaced visually. Unfortunately, this may vary by display device and settings, though, so evenly spacing them numerically is the best we can do. StuRat (talk) 17:27, 6 March 2013 (UTC)[reply]

If you're going to use differing increments for the three colors, I would suggest adding the "extra" precision to the green channel (so 6 points along each of the red and blue dimensions and 7 points along green). This is because the human eye is more sensitive to green than to red or blue. This same compromise is commonly made with 16-bit color, which often uses 5 bits each for the red and blue components and 6 bits for green. —Bkell (talk) 17:11, 6 March 2013 (UTC)[reply]

However, that makes the closest color to (255,60,255) become (255,43,255), which is even worse spacing. StuRat (talk) 17:34, 6 March 2013 (UTC)[reply]

So? Just add in (255,60,255) if you need it. You have four extra colors, after all. —Bkell (talk) 18:17, 6 March 2013 (UTC)[reply]

Yes, but it won't be evenly spaced from (255,0,255), (255,43,255) and (255,85,255), (255,128,255), ..., which was also a goal. StuRat (talk) 18:34, 6 March 2013 (UTC)[reply]

No, reread the original question. We are supposing that three colors, including (255,60,255), have already been assigned, and the question asks how to evenly distribute the remaining colors. Quote: "Note that it doesn't matter for me that I probably don't hit (255, 60, 255) and (255, 255, 255) exactly [with the remaining 253 colors], as they are already defined." —Bkell (talk) 19:32, 6 March 2013 (UTC)[reply]

Yes, evenly distribute the remaining colors around those already assigned. In a worst case scenario, what would be the point in having colors (255,60,255) and (255,61,255) ? They realize they aren't likely to hit it exactly, but would still presumably like to be close. StuRat (talk) 20:20, 6 March 2013 (UTC)[reply]

First, the question asks for an even distribution of 253 colors throughout the color space. It does not condition this distribution on the values of the colors that have already been assigned. It specifically asks for a scheme in which the remaining 253 colors are equally spaced in each of the three dimensions; there is no part of that that takes into account the three predefined colors, and the question specifically says those colors don't matter in the distribution of the remaining 253.

Second, why does it matter if you have two colors that are close together? You're going to have 256 colors in your palette anyway, whether you use them all or not. So why not evenly distribute 216 or 252 colors, and then add in the other colors that you've decided you definitely need? Alternatively, you could fix three colors and then, taking those colors into account, use a complicated heuristic or a force-based algorithm or solve a linear program or use a color quantization algorithm or something in an attempt to optimally distribute the other 253 colors throughout the color space, but that isn't what the original question asks for. That's why I said, "If you really want to define all 256 colors and have them be 'as equally spread out as possible' in the color space, there are several ways to do that, but they won't give you a perfectly spaced out set of colors like you ask for." —Bkell (talk) 20:31, 6 March 2013 (UTC)[reply]

Well, we disagree on what the OP wants, so we need them to clarify it for us. StuRat (talk) 20:54, 6 March 2013 (UTC)[reply]

Wow, first of all, thanks for the lot of input. Yes, as Bkell notes, the distribution of the 253 colors is in no way dependent on the three colors already defined (as such, those three colors can actually be ignored for this problem, I just mentioned them in order to present the problem exactly). So what I need is a way to calculate c. It isn't a problem if some of the values come close to the predefined ones, as in this case I can just change them to any arbitrary value that might be useful. But the others should follow the desired distribution, in order to give a harmonic and broad pool of colors. Also it might as well be that some of the colors will be changed from the colors following the distribution if necessary, but I need a rich and harmonic starting point. I think that if I just choose the colors by feeling or something I might get an inbalanced pool of colors which might be very hard to fix later. -- Toshio Yamaguchi 21:36, 6 March 2013 (UTC)[reply]

OK, Bkell has your answer then, at 6 possible values for each of red, green, and blue, which makes the value of C = 51. This gives values of 0, 51, 102, 153, 204, and 255 for each color. Or, if C can be different for the different colors, leave it at 51 for red and blue, and set it to 42.5 for green (remembering to round off the actual values), which gives you 7 possible values: 0, 42, 85, 127, 170, 212, 255. You get a total of 216 + 3, or 219 colors in the first scheme, and 252 + 3, or 255 colors, under the second scheme. Can we mark this Q resolved ? StuRat (talk) 22:09, 6 March 2013 (UTC)[reply]

Yes, this can be marked as resolved. Thanks. -- Toshio Yamaguchi 23:04, 6 March 2013 (UTC)[reply]

Will do. Can I ask whether you intend to use the (6,6,6) scheme or (6,7,6) ? StuRat (talk) 23:09, 6 March 2013 (UTC)[reply]

I think I will use the (6, 6, 6) scheme and adjust that later as necessary. The colors resulting from this might not be the colors I will need exactly, but it provides me with a spectrum I can fiddle with later and I think will be much easier than defining all values from scratch. To give a bit more information, I plan to make a clone of a computer game using the M.U.G.E.N. engine and I need a basic palette for the game sprites. -- Toshio Yamaguchi 23:26, 6 March 2013 (UTC)[reply]

Resolved

Errors in measurments and estimations

Hello, I am interested in some biological quantity (let's say the height of penguins). I can measure several independent values (different penguins) $m_{i}$ . However, my measurments are imperfect. Each one has its own error (variance) $e_{i}$ . This can also be seen as the unknown value $v_{i}$ being drawn from a known distribution $N(m_{i},e_{i})$ . Methods described in Weighted_mean can now help me to estimate the distribution of my values $v_{i}$ (distribution of heights among my penguin population) with mean $M$ and variance $V$ .

My question is how exact is this estimate? Ideally I would want a 2x2 covariance matrix on $M,V$ . I cannot imagine to be the first one with this problem, but I did not find a solution. Do you now a solution or some more search terms?

many thanks, -- 95.208.114.213 (talk) 17:07, 6 March 2013 (UTC)[reply]

My first thought is to do a parallel weighted mean, using the same weights for each data point, but instead of using the actual data (penguin heights), use the variance at each point, and apply the same weighting factor to that. This should give an upper-bound for the variance, with the law of large numbers bringing it down from there. StuRat (talk) 18:15, 6 March 2013 (UTC)[reply]

I generally don't find it useful to think in such abstract terms. Once you figure out what biological question you are trying to answer, it will focus you on the methods appropriate for answering that question. Looie496 (talk) 22:35, 6 March 2013 (UTC)[reply]

That gives a parsing error for me.--Gilderien Chat|List of good deeds 22:47, 6 March 2013 (UTC)[reply]

Posting here is really slow today

I've been having trouble posting here to the math reference desk for several hours now. I repeatedly get the Wikimedia Foundation error page. I've posted a notice about it at Wikipedia:Village pump (technical)#Posting to the math reference desk is really slow if anyone else is experiencing the same thing. —Bkell (talk) 20:14, 6 March 2013 (UTC)[reply]

Same here, and not just this page. This Q above was posted a dozen times, over a couple hours, because the OP apparently kept getting that error and reposting, even though it really worked. Most annoyingly, sometimes that error means it failed to make the change, and sometimes it doesn't. And, if the change was made, it may not show up right away. StuRat (talk) 20:27, 6 March 2013 (UTC)[reply]

March 7

Another integral formula for the Gamma/Factorial function

$x!=\Gamma (x+1)=\int _{0}^{\infty }{\frac {dt}{e^{\sqrt[{x}]{t}}}}$ . At the same time, we already know that $x!=\Gamma (x+1)=\int _{0}^{\infty }{\frac {t^{x}}{e^{t}}}\ dt$ .

I guess my question would be four-fold: (1) did anyone notice this alternate expression before ? (2) are there other meaningful or interesting properties attached to the function $e^{\pm {\sqrt[{n}]{x}}}$ ? (3) are there any other integral expressions for the factorial/Gamma function ? and (4) what are we to make of the fact that the integrals of these two different but related exponential functions have the exact same value over R₊ ? — 79.113.241.192 (talk) 09:09, 7 March 2013 (UTC)[reply]

(1) Probably

(2) I can't see that this one is particularly meaningful but the function will surely have some unique properties.

(3) There are as many integral representations as substitutions you care to make.

(4) Maybe it's worthy of a mildly interested 'huh' noise. That's probably about it.

Can you use propositional logic/calculus to construct mathematical models for quantitative research?

I would like to construct a model depicting the influence of the majority over the minority in election. I would argue that the majority can influence the minority groups. How can I use a simple propositional model for this? Can you give some advice or examples? — Preceding unsigned comment added by Joshua Atienza (talk • contribs) 09:25, 7 March 2013 (UTC)[reply]