Welcome to the mathematics section
of the Wikipedia reference desk.

Select a section:

WP:RD/MA

Want a faster answer?

Main page: Help searching Wikipedia

How can I get my question answered?

Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
Post your question to only one section, providing a short header that gives the topic of your question.
Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
Don't post personal contact information – it will be removed. Any answers will be provided here.
Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
Note:
- We don't answer (and may remove) questions that require medical diagnosis or legal advice.
- We don't answer requests for opinions, predictions or debate.
- We don't do your homework for you, though we'll help you past the stuck point.
- We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

May 6

What kind of function does this data represent? How do I represent the formula in Excel?

I have a set of data,and I typed it into an Excel spreadsheet. See below.

Inches	Gallons	Difference	Average
1	3	3	3
2	8	5	4
3	14	6	5
4	22	8	6
5	31	9	6
6	40	9	7
7	50	10	7
8	61	11	8
9	73	12	8
10	85	12	9
11	97	12	9
12	110	13	9
13	123	13	9
14	136	13	10
15	150	14	10
16	164	14	10
17	178	14	10
18	193	15	11
19	207	14	11
20	222	15	11
21	237	15	11
22	252	15	11
23	267	15	12
24	282	15	12
25	297	15	12
26	312	15	12
27	327	15	12
28	348	21	12
29	356	8	12
30	371	15	12
31	386	15	12
32	400	14	13
33	414	14	13
34	427	13	13
35	441	14	13
36	454	13	13
37	467	13	13
38	479	12	13
39	491	12	13
40	502	11	13
41	514	12	13
42	524	10	12
43	533	9	12
44	542	9	12
45	549	7	12
46	556	7	12
47	561	5	12
48	564	3	12

The table represents some data that came with my oil furnace. The first column reflects how many inches of oil are in the furnace. The second column converts the inches into how many gallons of oil are present in the furnace. I myself created the third and fourth columns. So, my question is: why would the relationship between inches and gallons be so odd? I was expecting a simple linear function. Why would it not be linear? What would be the correct name for this function? How can I determine the equation (function) with which I can input "inches" and determine "gallons"? Thanks. Joseph A. Spadaro (talk) 02:20, 6 May 2016 (UTC)[reply]

Are you sure about the line (28", 348 gal)? That value looks out of place. I'd have expected (28", 342 gal) instead. -- ToE 03:11, 6 May 2016 (UTC)[reply]

Thanks. Yes, I double checked the "28 inches = 348 gallons" line. It was actually the one below that that raised my eyebrows the most (the "29 inches = 356 gallons" line). These numbers make no sense. I don't see a rhyme or reason. But, then again, I was expecting a purely simple linear function. Joseph A. Spadaro (talk) 03:18, 6 May 2016 (UTC)[reply]

Then that line (28", 348 gal) is most likely in error in your furnace's documentation. Perhaps whoever typed up the table for the furnace or oil tank manufacturing company misread a "342" as a "348" from the list of values handwritten by whoever measured the actual capacity of that model of tank. Note that if it were (28", 342 gal), then the differences on the the (29", 356 gal) would be repaired as well. -- ToE 03:33, 6 May 2016 (UTC)[reply]

OK. Let's assume that was a typo in the documentation. Still, the rest of the numbers seem odd to me. I guess I see a pattern, "sort of". But not really. Joseph A. Spadaro (talk) 04:07, 6 May 2016 (UTC)[reply]

Calculating a "first difference" column was a smart move. Notice that when you are more than a foot and a half from either end of the tank, the volume increases at a rate of 14 to 15 gallons per inch, meaning that the cross sectional area of the internal volume of the tank is approximately 15 gal ÷ 1 inch = 15 · 231 cubic inches ÷ 1 inch = 3465 square inches. (Assuming the US liquid gallon.) If you measure the length of the tank and multiply it by its maximum width, you should get a number close to but a bit larger than this, as the external dimensions also include the thickness of the tank walls. (Internal structures such as baffles or pipes will also throw the numbers off a bit.)

Also notice that the difference column is (to within one gallon) symmetric top to bottom, meaning that the top shape of the tank is the same as the bottom shape of the tank. -- ToE 07:05, 6 May 2016 (UTC)[reply]

Yes, I did notice that the "difference column" (column 3) had numbers that appear to be in a (semi-)circular or "bow" pattern. Joseph A. Spadaro (talk) 07:16, 6 May 2016 (UTC)[reply]

Is the oil tank rounded at the top and bottom, as with this typical tank? -- ToE 03:35, 6 May 2016 (UTC)[reply]

Not sure. Have to double check. But I think so. Why? Joseph A. Spadaro (talk) 04:05, 6 May 2016 (UTC)[reply]

The relation is linear only if the horizontal cross sectional area of the tank is constant with height. See Cavalieri's principle.--Jasper Deng (talk) 05:12, 6 May 2016 (UTC)[reply]

So, with the numbers above, that clearly cannot be the case. I assume. What would be the correct name for this function? How can I determine the equation (function) with which I can input "inches" and determine "gallons"? I want to create a modified version of that chart. In my modified version, I want the inches (first column) to increase by 1/4 inch each time (as opposed to the current increase of 1 full inch). So, I'd want the first column to be 1 inch; 1.25 inch; 1.50 inch; 1.75 inch; 2.00 inch; and so on. With the correct corresponding gallon values. Thanks. Joseph A. Spadaro (talk) 05:41, 6 May 2016 (UTC)[reply]

You need to know the horizontal cross-sectional area of the tank as a function of height for an exact answer. Your problem is one of interpolation, for which there are many possible approaches. For now all I can say is that the volume is an increasing function of the water level.--Jasper Deng (talk) 05:57, 6 May 2016 (UTC)[reply]

If you really wanted to, you should be able to generate a three part piecewise function to model the volume as a function of fill height. (And if I were doing so, I would first see how well a semicircular top and bottom of the ends worked, and if they matched closely, then I'd have a good prediction of the overall dimensions of your tank just from your table.) But that doesn't seem necessary if you just want to increase the resolution of your table to the quarter inch. Linear interpolation should work well enough -- even for the top or bottom inch (where the true function will be the least linear) as your volumes were given only to the nearest whole gallon anyway. -- ToE 07:25, 6 May 2016 (UTC)[reply]

I was thinking of doing some type of interpolation. And I was thinking of this: I would take all of the gallon values between, say, inch 12 and inch 13. So, 12 inches equals 110 gallons; 13 inches equals 123 gallons. That is a difference of 1 inch and its corresponding 13 gallons. So, for each half-inch, there is an approximate increase of 6.5 gallons. And for each quarter-inch, there is an approximate increase of 3.25 gallons. Is that a correct interpolation? I don't need the numbers to be 100% precise and exact. A decent "ball park" figure is fine. I assume my interpolation method is "accurate enough" (for my purposes). But, here is my problem, since this function is not linear. That specific interpolation will only work in that specific interval of inches (12 to 13 inches). It has nothing to do with the other intervals in the chart. Do I have to create an interpolation 48 separate times? One interpolation that occurs between inch #12 and inch #13. (Which I did above as my example.) Then a totally different interpolation for inch #13 and inch #14? And so on, for all 48 inches? Is that what I need to do? There is no "one" model that will work for all 48 inch intervals? Thanks. Joseph A. Spadaro (talk) 15:44, 6 May 2016 (UTC)[reply]

Here is a quick and dirty fit using a logistic function as a base:

{\text{Gallons}}={786.38 \over 1+e^{0.0785*\left({\text{inches}}-23.834\right)}}-113.31

Compared to your table the average error is about +/- 2 gallons, with the max error of a little more than 6 gallons. Perhaps not ideal, but you didn't specify how accurate a solution you wanted, and of course increasing accuracy will presumably mean more complicated expressions. Dragons flight (talk) 07:53, 6 May 2016 (UTC)[reply]

Thanks. But I am confused. Shouldn't there be an exact and precise function for something like this? I mean, the shape of the tank is not random. It is indeed some very specific geometric shape. Joseph A. Spadaro (talk) 15:37, 6 May 2016 (UTC)[reply]

Your linear interpolation is fine between about fourteen inches and 36 inches. Outside this range, I would just draw a graph and sketch the smoothest curve going near to the points, and use this for interpolation. Remember that the given data has been rounded to a whole number of gallons, so your interpolation could be more accurate than the printed values. The tank might not be an exact geometric shape, and internal structure might distort any data derived from a function. Dbfirs 15:56, 6 May 2016 (UTC)[reply]

You are saying that my linear interpolation is fine (between a certain section of the chart). So, I have to essentially do 48 separate interpolations? Joseph A. Spadaro (talk) 16:08, 6 May 2016 (UTC)[reply]

No, I'm saying that the graph is a straight line between these limits because the cylinder has a constant cross-section, so you can use a very simple formula using equal increments. My claim is that this formula might be more accurate than the data published, because of rounding. Dbfirs 08:41, 7 May 2016 (UTC)[reply]

@Joseph A. Spadaro:

{\frac {\Delta h}{\Delta v}}

against h looks a lot like a semi ellipse to me, which would make sense if your tank is a cylinder with a horizontal axis (imperial units (I think) make it an ellipse rather than a circle). If you take that to be true, then

{\frac {dh}{dv}}={\sqrt {\frac {900(49x-x^{2})}{2401}}}

(based upon the vertices being (0,0), (49,0) and (24.5,15)). This gives the rather untidy equation for v in terms of h as

{\sqrt {\frac {900}{2401}}}({\frac {x^{\frac {3}{2}}}{2}}{\sqrt {49-x}}-12.25{\sqrt {x(49-x)}}+600.25\arcsin {\frac {\sqrt {x}}{7}})

. I will admit I used wolfram alpha for the integral and an automated tool for the ellipse. I'm lazy. — crh ₂₃ (Talk) 17:45, 6 May 2016 (UTC)[reply]

@Joseph A. Spadaro: No. There are infinitely many functions that interpolate your data. So unless you know more about the shape of the tank, no function will necessarily be better than another. The reason is that if you give me any function that interpolates the data, I can add to it any other function that is zero at the integers and obtain another valid interpolating function.--Jasper Deng (talk) 18:01, 6 May 2016 (UTC)[reply]

Your tank is a simple cylinder with a radius of $r\approx 24.01$ inches. If h is the height in inches:

${\text{Gallon}}=14.9630*{1 \over 2}\left(\left(h-r\right){\sqrt {1-\left({h-r \over r}\right)^{2}}}+r\sin ^{-1}\left({h-r \over r}\right)+{\pi r \over 2}\right)$

Except for the h = 28 value that you already noted appears weird, this formula agrees with your table for all values within less than 1 gallon. Dragons flight (talk) 19:15, 6 May 2016 (UTC)[reply]

@Dragons flight: Thanks. But, are you sure that formula works correctly? When I apply that formula, I get different results. For example, when I input 8 inches, I get a value of negative 502.5414 gallons. (When it should be positive 61 or so.) And other input numbers for other inch values also yield incorrect results for gallons. See my related question, posted here: Wikipedia:Reference desk/Computing#How do I translate this equation into an Excel formula?. Thanks. Joseph A. Spadaro (talk) 04:51, 7 May 2016 (UTC)[reply]

I made a sign error with the last term, now fixed. Sorry about that. Dragons flight (talk) 07:51, 7 May 2016 (UTC)[reply]

@Dragons flight: Are you sure that you are not also missing another factor of r? I assume you rolled it into the constant (14.9630), but without units I can't be sure.--Jasper Deng (talk) 08:11, 7 May 2016 (UTC)[reply]

Holy smoke. How on earth did you come up with that formula? (Generally, speaking. I don't need/want all the specifics.) So, that formula will work for any value that I plug in for "inches"? Even if the inches are 1/4-inch or 1/2-inch increments? Thanks. Joseph A. Spadaro (talk) 20:21, 6 May 2016 (UTC)[reply]

Also, do you have the Excel version of how that formula is typed? Thanks. Joseph A. Spadaro (talk) 21:06, 6 May 2016 (UTC)[reply]

(edit conflict) He made an educated guess about the shape of your tank, namely that it is a cylinder lying on its side with radius r. Then the cross-sectional area at height h is given by

A(h)=2l{\sqrt {r^{2}-(h-r)^{2}}}

where l is the length of the cylinder. The volume contained below height h is given by integration (roughly speaking, summing up all the little slices below height h, each of which has area given by the cross-section and thickness dh), i.e.

V(h)=\int _{0}^{h}A(t)dt=2l\int _{0}^{h}{\sqrt {r^{2}-(t-r)^{2}}}dt

. Using a trigonometric substitution to evaluate the integral should yield what Dragon's flight has above. --Jasper Deng (talk) 21:08, 6 May 2016 (UTC)[reply]

Addendum: For Excel, use this. I also strongly encourage you to empirically verify the correctness of this formula because it is based on the assumption that the tank forms a perfect cylinder lying perfectly on its side. For posterity I also encourage you to learn the mathematical derivation of formulae like this so you can solve similar problems on your own.--Jasper Deng (talk) 21:10, 6 May 2016 (UTC)[reply]

You stated: For posterity I also encourage you to learn the mathematical derivation of formulae like this so you can solve similar problems on your own. No, I don't understand any of this. It's all over my head. Thanks. Joseph A. Spadaro (talk) 21:15, 6 May 2016 (UTC)[reply]

@Joseph A. Spadaro: Well, you asked for how the formula was derived. You won't be able to do that (not even at a basic level) without at least an intuitive understanding of integral calculus. Think about slicing the tank into infinitely many infinitesimally thin sheets of volume and adding up the volumes of all of them. Note that when

h=2r

the formula reduces to just the volume of a cylinder.

Not to be nitpicky, but it is also wrong to assume that we could have found the formula just from the data points, because like I said, there are many ways to interpolate a given set of data. The formula here was based on what turned out to be quite a good guess at the shape of the tank, which is more information than just the data points alone.--Jasper Deng (talk) 21:34, 6 May 2016 (UTC)[reply]

I love calc as much as the next bloke, but a bit of trig will do this job quicker. Our article Circular segment unfortunately does not include area as a function of sagitta h, but it does include the easily derived area as a function of central angle θ. A=(r²/2)(θ-sinθ). It also includes the as easily derived h=r[1-cos(θ/2)]. Solve for θ, substitute into the formula for A, apply a trig identity or two, and Bob's your uncle: A=r²acos(1-h/r)-(r-h)sqrt(2rh-h²). -- ToE 04:29, 7 May 2016 (UTC)[reply]

@Thinking of England: My calculus-related comment was meant for a general tank where you do not necessarily have the nice properties of a cylinder, which I had to assume because the OP had not confirmed that it was indeed a cylinder.--Jasper Deng (talk) 08:03, 7 May 2016 (UTC)[reply]

I don't understand that formula. It is giving me "A", which is "area"? The area of what? I want a formula where I can enter "H" (height in inches) and get a result of "G" (gallons). Joseph A. Spadaro (talk) 05:03, 7 May 2016 (UTC)[reply]

@Joseph A. Spadaro: Just multiply by the length of the tank, which for a cylindrical tank is constant with respect to the height. The area given here is the vertical cross-sectional area (distinct from the horizontal cross-sectional area I used above).--Jasper Deng (talk) 08:13, 7 May 2016 (UTC)[reply]

Thanks. Does anyone know the exact formula that one would type into Excel to match the formula above by User Dragons flight? This link doesn't make any sense to me at all. Thanks. Joseph A. Spadaro (talk) 21:14, 6 May 2016 (UTC)[reply]

Taking a step back from the above discussions, I roughly plotted the "difference" data above, and they appear to be a good fit for a horizontal cylinder with a radius of 24 inches and a length of 72 inches. This gives us a volume of about 130,000 cubic inches which equates to about 564 gallons. A web search yielded several tank volume calculations, including this one, which will calculate the liquid volume from the tank dimensions (Length = 72, Diameter = 2 x Radius = 48 inches) and fill depth. A spot check of the results shows a very good match. As for the formula, this page is among several that give variants of what I've adapted as:

{\text{Gallon}}={1 \over 231}L\left(R^{2}\cos ^{-1}\left({R-h \over R}\right)-\left(R-h\right){\sqrt {2Rh-h^{2}}}\right)

where R = radius = 24 inches, L =length = 72 inches, and h is the measured fill depth. The first part of the inner calculation calculates the "pi slice" sector cross-sectional area that includes the measured liquid, and the second part subtracts out the triangular portion above the liquid, leaving only the liquid cross section. Multiplying by cylinder length and converting to gallons yields the desired result. The first of the above referenced web pages has a good illustration of this. An equivalent Excel formula would be:

=(1/231)*A2*(A1*A1*ACOS((A1-A3)/A1)-(A1-A3)*SQRT(2*A1*A3-A3*A3))

where A1 is R, A2 is L, and A3 is h. -- Tom N talk/contrib 07:50, 7 May 2016 (UTC)[reply]

~~I found the glitch in Dragon Flight's equation - the sin^-1 term should be subtracted, not added:~~

{\text{Gallon}}=\color {Red}14.9630*{1 \over 2}\left(\left(h-r\right){\sqrt {1-\left({h-r \over r}\right)^{2}}}{\color {Red}{-}}r\sin ^{-1}\left({h-r \over r}\right)+{\pi r \over 2}\right)

, ---(Update: This was wrong. Dragon Flight's equation above is correct.)

or in Excel:

~~=14.963*(1/2)*((A12-A1)*SQRT(1-((A12-A1)/A1)^2)-(A1*ASIN((A12-A1)/A1))-((PI()*A1)/2)) (as requested on computing board)~~

=14.963*(1/2)*((A12-A1)*SQRT(1-((A12-A1)/A1)^2)+(A1*ASIN((A12-A1)/A1))+((PI()*A1)/2)) (corrected)

It wasn't obvious at first,but Dragon Flight's corrected equation and the one I posted just above are equivalent.

{\begin{aligned}{\text{Gallon}}&={1 \over 231}L\left(R^{2}\cos ^{-1}\left({R-h \over R}\right)-\left(R-h\right){\sqrt {2Rh-h^{2}}}\right)\\&={1 \over 231}L\left(R^{2}\left({\pi  \over 2}{\color {Red}{-}}\sin ^{-1}\left({R-h \over R}\right)\right)-\left(R-h\right){\sqrt {R^{2}-(R-h)^{2}}}\right){\color {Red}{\text{corrected}}}\\&={1 \over 231}L\left({\pi R^{2} \over 2}{\color {Red}{-}}R^{2}\sin ^{-1}\left({R-h \over R}\right)-\left(R-h\right)R{\sqrt {1-\left({R-h \over R}\right)^{2}}}\right){\color {Red}{\text{corrected}}}\\&={1 \over 231}LR\left({\pi R \over 2}{\color {Red}{+}}R\sin ^{-1}\left({h-R \over R}\right)+\left(h-R\right){\sqrt {1-\left({h-R \over R}\right)^{2}}}\right){\color {Red}{\text{corrected}}}\\&=7.48\left(\left(h-R\right){\sqrt {1-\left({h-R \over R}\right)^{2}}}{\color {Red}{+}}R\sin ^{-1}\left({h-R \over R}\right)+{\pi R \over 2}\right){\color {Red}{\text{corrected}}}\\\end{aligned}}

-- Tom N talk/contrib 02:23, 8 May 2016 (UTC)[reply]

My apologies to Dragon Flight and any who tried to make sense of my last post. Dragon Flight had already corrected his equation, but I was working off an old version when I tried to reconcile the differences and ended up introducing my own error (another flipped sign). The corrections have been annotated above. -- Tom N talk/contrib 00:14, 11 May 2016 (UTC)[reply]

Let x be the number of inches and y=f₁(x) be the number of gallons.

f₁(24)=282, f₁(36)=454, f₁(48)=564.

The table shows that f₁ is symmetric around the point (x,y)=(24,282).

f₂(x)=f₁(24+x)−282 is an odd function. f₂(x)=−f₂(−x).

f₂(0)=0, f₂(12)=172, f₂(24)=282.

f₃(x)=f₂(x)−282x/24 =f₁(24+x)−282−47x/4

f₃(0)=0, f₃(12)=31, f₃(24)=0.

f₄(x)=(1−(x/24)²)x/24

f₄(0)=0, f₄(12)=3/8, f₄(24)=0.

f₅(x)=31*8/3 f₄(x)=31(1−(x/24)²)x/9

f₅(0)=0, f₅(12)=31, f₅(24)=0,

f₅(x) interpolates f₃(x).

f₆(x)=f₅(x)+47x/4 =31(1−(x/24)²)x/9+47x/4 interpolates f₂(x)

f₇(24+x)=282+31(1−(x/24)²)x/9+47x/4 interpolates f₁(24+x).

This third degree polynomial reproduces the table within a few gallons.

Bo Jacoby (talk) 20:28, 11 May 2016 (UTC).[reply]

Thanks, all. Joseph A. Spadaro (talk) 03:20, 12 May 2016 (UTC)[reply]

That interpolation is clearly not as good as the ones previously given, which match even more closely and are derived from the (assumed) geometry of the tank.--Jasper Deng (talk) 06:39, 12 May 2016 (UTC)[reply]

May 7

how can lines/curves of infinite length have positive area?

came up in editing an article...someone described that "There exist nonrectifiable curves of nonzero 2-dimensional Lebesgue measure" and therefore "curves with infinite length can have positive area." this makes no intuitive sense to me...can this be explained in any kind of intuitive manner? (I assume it's true as the editor seems genuinely knowledgeable)..68.48.241.158 (talk) 16:21, 7 May 2016 (UTC)[reply]

See Osgood curve. Sławomir Biały (talk)

Also, do there exist curves/lines of finite length that have area?68.48.241.158 (talk) 16:56, 7 May 2016 (UTC)[reply]

No, in general a set can only have one finite nonzero Hausdorff measure. See Hausdorff dimension. Sławomir Biały (talk) 17:46, 7 May 2016 (UTC)[reply]

The articles cited don't even explain technically why these curves have area, as far as I can tell...and they certainly don't offer any kind of intuitive explanation.....?????68.48.241.158 (talk) 18:01, 7 May 2016 (UTC)[reply]

You are asking for an intuitive explanation, but it might help if you said what point you are starting from. For example, are you already familiar with the Lesbegue measure? Do you know calculus? Or do you mean something really elementary like high school math? Dragons flight (talk) 18:45, 7 May 2016 (UTC)[reply]

I took calculus in high school, studied humanities/philosophy in college..so not too advanced as far as math proper...and it's been a while since I've done/studied any math proper...I'm wondering if there's any even conceivable way to explain it intuitively...obviously, we're talking about an abstraction here..so such a thing could never exist in the real world/be measured for area in the real world...do all infinitely long curves have area? or only certain kinds? and how/at what point to they become imbued with area?? is this approaching it in a philosophical way that doesn't make much sense??68.48.241.158 (talk) 19:05, 7 May 2016 (UTC)[reply]

I encourage you to look at the concept of an improper integral first. --Jasper Deng (talk) 19:08, 7 May 2016 (UTC)[reply]

don't see how helpful as far what I'm asking for...granted, I may be asking the impossible..68.48.241.158 (talk) 19:16, 7 May 2016 (UTC)[reply]

But that's the most elementary way there is to calculate areas under infinite curves. I won't entertain any "philosophical" discussions of this subject because such discussions are necessarily hand-wavy (i.e. vague and not rigorous).--Jasper Deng (talk) 19:21, 7 May 2016 (UTC)[reply]

but we're not talking about calculating areas under infinite curves but the area of the curves themselves....?? does the area under the curve somehow become mixed up with the curve itself in particular infinite curves????68.48.241.158 (talk) 19:24, 7 May 2016 (UTC)[reply]

Then what you are talking about makes no sense. (Also, please stop using excessive question marks, I get it that you're confused, you don't need to have so many). I think the editor probably meant area enclosed, since the area of a one-dimensional curve itself is zero.--Jasper Deng (talk) 19:46, 7 May 2016 (UTC)[reply]

it now appears you don't understand it either and will be awaiting an explanation as I am! we're talking about one-dimensional infinite curves having area!68.48.241.158 (talk) 19:51, 7 May 2016 (UTC)[reply]

I agree with the OP - it seems you are unfamiliar with the concept of a Space-filling curve. That is, you can have a continuous function

f(x):[0,\ 1]\to \mathbb {R} ^{2}

such that its image

f([0,\ 1])

, a subset of

\mathbb {R} ^{2}

, has positive area. -- Meni Rosenfeld (talk) 20:11, 7 May 2016 (UTC)[reply]

To say that a line or curve (whether finite or infinite) "has" an area seems wrong to me, and I suspect that's just a sloppy way of saying that the line or curve "bounds" an area. Of course a finite curve can bound an area, as in a circle, but also an infinite curve may, as in many fractals, such as the Koch snowflake. StuRat (talk) 20:05, 7 May 2016 (UTC)[reply]

Stu, what you may be missing is that these "curves" are allowed to self-intersect. Think of taking a blue pen, and coloring in a square entirely blue, by moving it back and forth at random over the area of the square until you've scratched out a solid area. It's not exactly like that, but it'll do for a start. There are some nice pictures at space-filling curve that might help. --Trovatore (talk) 20:40, 7 May 2016 (UTC)[reply]

Yes, that's very much like my illustration down below, filling in a rectangle with an infinite number of infinitely thin V's (which doesn't require self-intersecting curves, BTW). However, the difficulty the OP is having in conceptualizing this is that, unlike the pen, the line segments have no width, so even a huge number of lines should not color in the rectangle even the slightest percentage. It's difficult to make the jump from this to an infinite amount of lines completely filling it in. StuRat (talk) 21:09, 7 May 2016 (UTC)[reply]

seems wrong to me too...which is why I asked..but apparently it's somehow true...may need very expert math types to weigh-in (I believe the person who suggested it is a very expert math type)...68.48.241.158 (talk) 20:10, 7 May 2016 (UTC)[reply]

If we define "curve" as "the image of a continuous function from

[0,1]

to a space (the Euclidean plane in our case)", then a curve can actually have a positive area. -- Meni Rosenfeld (talk) 20:11, 7 May 2016 (UTC)[reply]

John von Neumann famously said, "Young man, in mathematics you don't understand things. You just get used to them.". Many mathematical objects are pathological and defy our existing intuitions. If we study them hard enough, we develop new intuitions which can accommodate them.

I think the best way to gain intuition for space-filling curves is to study in detail the construction of a particular one, such as the Hilbert curve. -- Meni Rosenfeld (talk) 20:17, 7 May 2016 (UTC)[reply]

thanks, we're finally on to something with the "space-filling curves"...still don't intuitively understand it, but as you suggest may not be possible to....68.48.241.158 (talk) 20:20, 7 May 2016 (UTC)[reply]

Also, they might mean that an infinite curve approximates an area. Consider the following (poorly drawn) rectangle:

      
|    |

Now what happens if we start filling it in with V shapes ? Here's just 2:

      
|\/\/|

But, if you had an infinite number of infinitely narrow V's, couldn't you say that this approximates filling in the area of the rectangle, much as 0.9999 repeating approximates, and in a sense is, 1.0 ? StuRat (talk) 20:18, 7 May 2016 (UTC)[reply]

I have a feeling you're off here, Stu, but would have to defer to more expert math folk..68.48.241.158 (talk) 20:22, 7 May 2016 (UTC)[reply]

Well, if we consider the area as "the set of all points within a boundary", and my example above, with an infinite number of V's, does in fact occupy all points within the (rectangular) boundary, doesn't that meet the formal definition ? StuRat (talk) 20:50, 7 May 2016 (UTC)[reply]

what I don't understand is how the curve becomes imbued (so to speak) with area...as each distinct part of the curve would be said to have no area, wouldn't it??68.48.241.158 (talk) 20:30, 7 May 2016 (UTC)[reply]

I don't believe you can break it down into finite pieces like that, as is often the case in calculus. Also see Zeno's paradoxes and emergent property. And to go to the next dimension, consider a solid of revolution, composed of an infinite number of cross sections, each of which have area, but no volume themselves. StuRat (talk) 20:56, 7 May 2016 (UTC)[reply]

Such curves tend to have a fractal structure. So, while you certainly can zoom in, the magnified curve will appear qualitatively very similar to the original curve. The irregularities never smooth out, as they do for differentiable curves. Sławomir
Biały 21:30, 7 May 2016 (UTC)[reply]

If you think of the curve as having a width of 0, then the area of a curve of any finite length ℓ is 0 × ℓ = 0, but the area of a curve of infinite length is 0 × ∞, which is an indeterminate form. Getting an indeterminate form for an answer doesn't mean there is no answer; it means you can't tell by that technique what the answer is. A curve of any finite length has to have area zero, but a curve of infinite length doesn't have to, at least not by this line of argument. And it turns out that some actually don't. -- BenRG (talk) 21:41, 7 May 2016 (UTC)[reply]

Or, if there is no other method to obtain a definitive solution, perhaps we just declare a solution and use that, by convention. StuRat (talk) 21:58, 7 May 2016 (UTC)[reply]

If these curves in fact have an area, then what is it? shouldn't areas be measurable? the area can't be infinite as these fractals can fit inside a small finite area...?68.48.241.158 (talk) 02:10, 8 May 2016 (UTC)[reply]

OK, so first of all, technically, it's not the curve that has nonzero area. The curve is a map. What has nonzero area is the range of the curve; that is, the set of all points that the curve passes through at least once.

For a true space-filling curve that lives inside the unit square, [0,1]×[0,1], the range is just the entire square, and its area is 1. --Trovatore (talk) 04:08, 8 May 2016 (UTC)[reply]

In advanced mathematics, the concepts of length, area, volume, etc. are usually defined by the Lebesgue measure. That article is probably not very useful to you, but let's try and discuss some examples and perhaps that will help. First of all, one of the important things to understand about the Lebesgue measure is that the measure of an open set is defined to be equal to the measure of the closure of that set. For example, the length of the open interval (0,1) is equal to the length of the closed interval [0,1] is equal to 1. This is true, by definition, even though the open interval doesn't include the points 0 or 1. Similarly, the area of an open disk (e.g. all points less than R distance from the origin) is equal to the area of the comparable closed disk (e.g. all points less than or equal to R distance from the origin). Roughly, the closure of a set means all points in the set plus any points that are outside but "infinitely close" to the set. Specifically, a point is infinitely close if for any distance greater than 0 there is always a point in the original set less than that distance away. In other words, closing the set adds in the edges and fill in any simple holes. For example, if you take the interval [0,1] minus the point 0.5 = [0,0.5) + (0.5,1], then the closure is still [0,1]. Now consider a more complicated example, the interval [0,1] minus all rational numbers. That removes an infinite number of points but still leaves all the irrational numbers which is also an infinite number of points. For every rational number there is always an irrational number infinitely close to is. Hence the closure of all irrational number in the interval [0,1] is the whole interval [0,1]. Even though the irrational set has an infinite number holes, it's length is still defined to be the same as the whole interval, i.e. 1. Now, let's go a step further. What is the area of a fractal or space-filling curve? Well first, we have figure out what is the closure of that curve. So, we count not only the curve itself, but also any points that are infinitely close to the curve itself. Even though the curve itself has no "width", an curve that winds infinitely densely through a space can be infinitely close to every point in that area. When that occurs, the closure of the curve becomes whole area. At that point the area of curve, according to the Lebesgue measure, is the whole area of the space it fills. Hope that helps. Dragons flight (talk) 07:55, 8 May 2016 (UTC)[reply]

What now? No no no. We're not taking the closure. You can't do that. The rational numbers in the unit interval have measure 0, but their closure has measure 1. Anyway, any continuous image of the closed unit interval is compact, and therefore itself closed. So taking the closure, in this case, doesn't do anything.

No, the curve really traces out every single point in the unit square. It can't do that without self-intersection. --Trovatore (talk) 08:06, 8 May 2016 (UTC)[reply]

Yeah, I'm approximating a bit. For the purpose of an "intuitive explanation". The measure of a set is usually the same as the measure of the closure of a set. There are exceptions, e.g. your rationals example (which is why I gave the irrationals in the unit interval). The Peano curve, our classic space filling curve, is of course the limit of an iterative process. At any finite step in that process you have a non-self-intersecting curve of finite length. However, we can imagine a curve in the infinite limit that fills the space (and also happens to be self-intersecting). In particular, if I choose an irrational point in the plane (e.g. (pi/4, e/3) or something), there is no curve in the Peano generation process that reaches that point. Such points are only reached in the infinite limit. The points reached by the limit curve are "roughly" the closure of the set of points reached by the set of Peano generating curves. Of course if you want to try a more precise explanation, please go ahead. Dragons flight (talk) 08:34, 8 May 2016 (UTC)[reply]

I'm sorry, I think your explanation is worse than useless. You're saying things that just aren't true, and not even on point. The space-filling curve quite literally traces out every point. It doesn't just get arbitrarily close. Getting arbitrarily close doesn't help at all. --Trovatore (talk) 08:45, 8 May 2016 (UTC)[reply]

It seems that lines/curves are normally thought of as a kind of boundary/demarcation that don't actually exist in reality, but are abstractions..it makes intuitive sense that they have zero width and no area...to suggest there exist such things that do have an area flies in the face of intuition...I'm not even sure expert mathematicians can be said to have a genuine intuition about this (as an earlier poster referred to mathematical pathology)...it still doesn't quite make sense to me even as an abstraction...it would seem that for these curves to have an area then they should have a measurable width, which they don't seem to...otherwise one is simply measuring the area enclosed by a fractal...68.48.241.158 (talk) 13:17, 8 May 2016 (UTC)[reply]

No, it's not the area enclosed by a fractal that we mean here. There exist Jordan curves whose two-dimensional Lebesgue measure is not zero. By the Jordan curve theorem, these are actually boundaries of open subsets of the plane. So, yes, the boundary of an open domain can have positive area. As to whether it "exists" in reality, all of mathematics represents an idealization which cannot be said to "exist". However, I disagree with the (apparently widely-held) belief that pathologies are not present in nature. Dynamical systems appearing in many applications exhibit pathological chaotic behavior. Indeed, the "pathologies" studied by mathematicians in ergodic theory are fairly tame from the perspective of real-world dynamics. Even geometrical pathologies exist in nature, such as corals, which exhibit isometric immersions of flat and hyperbolic surfaces, whose mathematical existence was only very recently proven. Life imitates art; art imitates life.

The issue here is apparently how one defines concepts such as area and length in the first place. For planar regions, area is easier to define. We know how to compute the area of a rectangle. So we also know how to compute the area of an infinite collection of disjoint rectangles: we add their areas. Now if a region is given, we may not know precisely how to calculate its area, but we can at least say that the area is not greater than a certain amount. Indeed, all we need to do is cover the region with a collection of rectangles. If a region is contained in a collection of rectangles, then the area of the region cannot be greater than the sum of the areas of the rectangles. That seems like a very intuitive idea. Now, the area of a region R can be defined as the smallest value of the area (actually the infimum) over all areas A such that the sentence "the area of R is not greater than A" holds. This is the basic idea of the Lebesgue measure.

It is a standard theorem in analysis that there exist Jordan curves (that is, boundaries of bounded, connected, and simply connected open sets in the plane) whose area A, in this sense, is not zero. That is, any covering of the curve by rectangles has area is bounded below by the number A.

Length is actually much harder to define, because it is not possible to cover a curve with line segments. Instead, we cover a curve with rectangles, and (effectively) sum their perimeters rather than areas. It is not at all obvious how to make this actually work in practice; it is the basis for the Hausdorff measure. Sławomir
Biały 13:49, 8 May 2016 (UTC)[reply]

Really, you can get positive measure with a non-self-intersecting curve? I was wondering about that. Do we have an article on that? The examples in space-filling curve don't seem to help.

What does such a curve look like in a neighborhood of a point of density? --Trovatore (talk) 21:07, 8 May 2016 (UTC)[reply]

Wasn't my example with the infinite V's filling a rectangle a case of non-intersecting lines completely filling the rectangle ? StuRat (talk) 02:58, 9 May 2016 (UTC)[reply]

It has to be a curve; that is, you have to find a continuous map from the interval [0,1] onto your set. I don't see that you specified a way to do that. (To be fair, several of our articles don't, either, and I think that's a flaw — but the difference is that it can be done, whereas I don't think yours can. But feel free to prove me wrong.) --Trovatore (talk) 09:48, 10 May 2016 (UTC)[reply]

I take it you mean [0,0] to [1,1]. My method would work for that or any other rectangular range you wanted in 2D. Here's 3 line segments connecting [0,0] to [1,1]:

1+---+
 |/\/|
0+---+
 0   1

You can use any odd number of line segments at increasing steep angles to start from [0,0] and end at [1,1]. When you go to an infinite number of line segments, all points in the unit square are then covered, and none intersected, unless you consider touching to be intersecting. StuRat (talk) 22:34, 10 May 2016 (UTC)[reply]

You need to prove that the sequence of curves you have described converges uniformly. However, it clearly does not, because arbitrarily small subintervals of [0,1] must get stretched out to intervals of length greater than 1. So, while in some sense the limit of this construction does cover the square, it is not a continuous image of the unit interval. It's actually worse than this, because the construction you have given does not even converge pointwise, except at the endpoints. So it doesn't even define a discontinuous function from the interval to the square. Sławomir Biały (talk) 22:51, 10 May 2016 (UTC)[reply]

I'm not sure we're at the right stage of the discussion to bring in pointwise v uniform convergence. Let's just focus on getting across what needs to be accomplished, at a high level.

Stu, no, I really did mean the interval [0,1]; that is, the set of all x with 0≤x≤1. You have to find (or at least, show that there exists) a continuous function f that takes in values between 0 and 1 inclusive, and spits out points in your rectangle. You haven't described such a function, not even in vague terms. That's what needs to be accomplished, first. --Trovatore (talk) 23:17, 10 May 2016 (UTC)[reply]

I anticipate the retort as "pass to the limit", which was what seems to be intended here. The functions are the graphs of a sequence of triangle waves whose union has a closure that fills out the square. I think the observation that the limit does not exist as a function is very relevant to this discussion. Sławomir Biały (talk) 23:34, 10 May 2016 (UTC)[reply]

Yes, for example the boundary of the Mandelbrot set is believed to have this property. But, more generally, according to the Denjoy–Riesz theorem, every compact totally disconnected planar set is a subset of a Jordan arc. It's easy to construct such "fat" Cantor sets in

\mathbb {R} ^{2}

. Sławomir Biały (talk) 21:26, 8 May 2016 (UTC)[reply]

Added after the fact, but lost in an edit conflict: Up to sets of measure zero, these latter examples probably just resemble fat 2-dimensional Cantor sets near each point of density. (Measure cannot "see" the connectedness of the set.) Sławomir Biały (talk) 21:32, 8 May 2016 (UTC)[reply]

Oh, so I saw your link up above to Osgood curve. The picture is nice; it's clear that you can get positive measure. What's less obvious to me is that what's left, after you take out the wedges, is actually (the range of a) curve. Same thing for the boundary of the Mandelbrot set. Can you say something about that? (It would be nice to add it to the article, as well. Also, the Hilbert curve and Peano curve articles could say something about what the actual curve is, as opposed to the stages.) --Trovatore (talk) 21:29, 8 May 2016 (UTC)[reply]

It seems like an interesting question in general: is the boundary of a bounded connected and simply connected domain homeomorphic to the circle? I believe the answer is yes, but regardless of the answer to this general question, for the curve depicted at Osgood curve (due to our own David Eppstein), the boundary is already a uniform limit of curves. For each n, there is a compact connected simply-connected set

K_{n}

that is a union of

4^{n}

mutually congruent triangles,

T_{n}

. Eppstein's Osgood curve is the intersection of the

K_{n}

. It is hopefully obvious what the

T_{n}

are in the picture. For instance,

T_{0}

is the largest triangle that contains the picture.

T_{1}

is any of the four smaller triangles, and so forth. Naturally, the triangles

T_{n}

and

T_{m}

are not similar unless

n=m

. However, we still have

\operatorname {diam} (T_{n})=2^{-n/2}\operatorname {diam} (T_{0}).\qquad (1)

Now, let

\phi _{n}

be a sequence of continuous homeomorphisms so that

\phi _{n}

maps each of the

4^{n}

equal subintervals forming a partition

[0,1]

into each of the

T_{n}

that comprise

K_{n}

. Then, owing to (1), any such sequence

\phi _{n}

is uniformly Cauchy. It is easy to show that the limit function is one-to-one (hence a homeomorphism). Sławomir Biały (talk) 23:03, 8 May 2016 (UTC)[reply]

you state some interesting things; some of which make some sense to me...I do think at least some of the concepts raised by the question are of a philosophical nature as opposed to of a strict mathematics nature (of course it's difficult to ever entirely separate out the two)..but I don't doubt the mathematical truth of what's being discussed...68.48.241.158 (talk) 14:19, 8 May 2016 (UTC)[reply]

To clarify my earlier comment: Expert mathematicians do have an intuition for things like this. It's just a different intuition than the one they started out with. Sometimes you need to adapt mathematical concepts to your intuition, sometimes you need to adapt your intuition to the mathematical concepts. -- Meni Rosenfeld (talk) 14:30, 8 May 2016 (UTC)[reply]

Some comments in Alexander Grothendieck's Esquisse d'un Programme may be of interest on "intuitive" versus "modern":

"I would like to say a few words now about some topological considerations which have made me understand the necessity of new foundations for “geometric” topology ... (256)

After some ten years, I would now say, with hindsight, that “general topology” was developed (during the thirties and forties) by analysts and in order to meet the needs of analysis, not for topology per se, i.e. the study of the topological properties of the various geometrical shapes. That the foundations of topology are inadequate is manifest from the very beginning, in the form of “false problems” (at least from the point of view of the topological intuition of shapes) such as the “invariance of domains”, even if the solution to this problem by Brouwer led him to introduce new geometrical ideas. Even now, just as in the heroic times when one anxiously witnessed for the first time curves cheerfully filling squares and cubes, when one tries to do topological geometry in the technical context of topological spaces, one is confronted at each step with spurious difficulties related to wild phenomena.

This [stopgap way to avoid difficulties] is a way of beating about the bush! This situation, like so often already in the history of our science, simply reveals the almost insurmountable inertia of the mind, burdened by a heavy weight of conditioning, which makes it difficult to take a real look at a foundational question, thus at the context in which we live, breathe, work – accepting it, rather, as immutable data. It is certainly this inertia which explains why it took millennia before such childish ideas as that of zero, of a group, of a topological shape found their place in mathematics. It is this again which explains why the rigid framework of general topology is patiently dragged along by generation after generation of topologists for whom “wildness” is a fatal necessity, rooted in the nature of things.... The only thing in all this which I have no doubt about, is the very necessity of such a foundational work, in other words, the artificiality of the present foundations of topology, and the difficulties which they cause at each step. ...

... It need (I hope) not be said that the necessity of developing new foundations for “geometric” topology does not at all exclude the fact that the phenomena in question, like everything else under the sun, have their own reason for being and their own beauty. More adequate foundations would not suppress these phenomena, but would allow us to situate them in a suitable place, like “limiting cases” of phenomena of “true” topology. [1]

On Grothendieck’s tame topology comments on recent developments & has more quotations. I would say that Robert Ghrist's Elementary Applied Topology is an accessible text in a similar spirit and neighborhood.John Z (talk) 02:25, 9 May 2016 (UTC)[reply]

That is an interesting quote. I would associate this philosophy also with Michael Gromov, particularly the h-principle. Sławomir
Biały 16:45, 9 May 2016 (UTC)[reply]

May 8

statistical difference

Hi all,
It says here (page 99) that if we have a 2-universal hash functions family $H$ , mapping $\{0,1\}^{n}$ to $\{0,1\}^{m}$ and $T\subseteq \{0,1\}^{n}$ , then the following distributions have statistical difference of $(2^{m}/|T|)^{\Omega (1)}$ :

Choose $h$ uniformly from $H$ , and then choose $y$ uniformly from $T\cap h^{-1}(0)$ . Output $(h,y)$ .
Choose $y$ uniformly from $T$ , and then choose $h$ uniformly from $\{h\in H|h(y)=0\}$ . Output $(h,y)$ .

I couldn't find any explanation why is it guaranteed to have this statistical difference. Can someone please explain it to me?
Thanks a lot! — Preceding unsigned comment added by 109.186.49.82 (talk) 21:41, 8 May 2016 (UTC)[reply]

Rotational Invariants

I have a 2-sphere and I want to define 2 rectangular areas on it. They are "spherical rectangles" with borders that are some meridians and parallels. Sorry for not using Greek letters all the way through-they refused to take subscripts, superscripts. Let angle θ <t> be the inclination (polar) angle, and angle φ <f> the azimuthal angle. I will use Greek and Latin letters interchangeably. I need three angles t: $0<t_{1}<t_{2}<t_{3}<90^{o}$ and two angles f: $0<f_{1}<f_{2}<360^{o}$ It is clear they define two contiguous (adjacent) spherical rectangles. The rectangles touch each other on $t_{2}$ parallel. A function f(θ,φ) is defined on the area that is the sum of both rectangles. The portions of the function f(θ,φ) on both rectangles are different. Let $f_{1}$ (θ,φ) be the portion of the function f(θ,φ) on one rectangle and $f_{2}$ (θ,φ) the corresponding portion on the other rectangle. A basis of fully normalized Spherical Functions

\mathrm {Y} _{l}^{m}(\theta ,\phi )=\mathrm {N} _{l}^{m}e^{im\phi }\mathrm {P} _{l}^{m}(cos\theta )

(1)

is defined on the whole 2-sphere but I will consider only the portion that is covered by the above two rectangles. Each function $f_{1}$ (θ,φ), $f_{2}$ (θ,φ) and $f$ (θ,φ) will be expressed as

{\mathfrak {f}}_{1}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}f_{l}^{m}Y_{l}^{m}(\theta ,\phi )

(2)

{\mathfrak {f}}_{2}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}g_{l}^{m}Y_{l}^{m}(\theta ,\phi )

(3)

and

{\mathfrak {f}}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}q_{l}^{m}Y_{l}^{m}(\theta ,\phi )

(4)

I then fix one particular index ${l}$ which defines a subspace in the functional Hilbert space and compute these expressions (asterisk marks complex conjugate):

{\mathfrak {I}}_{1}={\sum \limits _{m=-l}^{m=+l}}f_{l}^{m}f_{l}^{m*}

(5)

{\mathfrak {I}}_{2}={\sum \limits _{m=-l}^{m=+l}}g_{l}^{m}g_{l}^{m*}

(6)

{\mathfrak {I}}={\sum \limits _{m=-l}^{m=+l}}q_{l}^{m}q_{l}^{m*}

(7)

It is very important for me to know if additivity is preserved and

{\mathfrak {I}}={\mathfrak {I}}_{1}+{\mathfrak {I}}_{2}

(8)

Thank you. --AboutFace 22 (talk) 22:54, 8 May 2016 :(UTC)

In order to expand your function into spherical harmonics you need to know it on the whole sphere. Ruslik_Zero 03:10, 9 May 2016 (UTC)[reply]

NO, I don't need to know if it is on the whole sphere. It is NOT on the whole sphere. It is a locally defined function and when the coefficients are computed integration has local limits. It can be assumed that the function is defined on the two rectangles but on the rest of the sphere the amplitude of the function is zero. --AboutFace 22 (talk) 12:25, 9 May 2016 (UTC)[reply]

I think Ruslik's point is valid, but you can get around this by assuming the function is 0 outside the domain you're interested in. Otherwise, while the expansion may exist, it won't be unique. As a simple example using Fourier series instead of spherical harmonics, you can expand the function f(x)=1 on the interval [0, π] as a sin series and get a square wave, or you can expand it as simply 1. If you are assuming 0 outside the two rectangles then yes, the expansions of the two separately will match what you get for their union. But the two functions you're talking about have to match on the common boundary, otherwise it the function f isn't defined on the whole rectangle. Presumably if the functions did not agree then the series would converge to the average of the two values, I know this is true at least for Fourier series. It should be noted that, unless your functions are 0 on the boundaries, convergence will be slow and not uniform, so the series may not be practical. --RDBury (talk) 18:29, 9 May 2016 (UTC)[reply]

Expansion in spherical Bessel functions would be more appropriate for domains of this kind. Alternatively, you can conformally map to the disk via a Schwarz-Christoffel map and take a Bessel expansion in the disk. Sławomir Biały (talk) 18:34, 9 May 2016 (UTC)[reply]

Thank you, @RDBury and @Sławomir Biały. Yes it is a physical situation, function $f$ is a well behaved function with no singularities anywhere and it is smooth and the two parts $f_{1}$ and $f_{2}$ match on their common boundary. It is possible that there is an "abrupt drop" at the outside boundary of the combination of two rectangles $f$ . I am sorry, I don't understand your (@RDBury) sentence "yes, the expansions of the two separately will match what you get for their union." I am sorry, I would appreciate if you confirm the expression (8). I am sorry about it. I need yes or no. Otherwise I feel your sentence is ambiguous.

I just meant to say additivity is preserved, since you're assuming the function 0 outside the given domains. Not sure about

{\mathfrak {I}}={\mathfrak {I}}_{1}+{\mathfrak {I}}_{2}

, but f₁ and f₂ are orthogonal so I would think so. --RDBury (talk) 22:36, 9 May 2016 (UTC)[reply]

@Sławomir Biały, Thank you for the suggestion about Bessel Functions. I have thought about it before but now I need what I posted about. --AboutFace 22 (talk) 21:56, 9 May 2016 (UTC)[reply]

@RDBury, thank you for the confirmation. --AboutFace 22 (talk) 22:48, 9 May 2016 (UTC)[reply]

May 9

Preservation of precious Science and Mathematics knowledge

To whom it may concern,

This question has been really bothering me for quite some time, and has really made me pessimistic, maybe you can answer it.

With all this mathematics and Scientific research being done every day, how could we be sure that all this research and discoveries will be preserved for future generations? How could we know that research findings and discoveries in Mathematics and the Sciences today, will be available to humanity, in, say, 1,000 years? I asked this question to many Mathematicians, and they unanimously said there is no good system that preserves our discoveries today.

What could we learn from the contributions of Euclid, Thales, and Archimedes, in which their discoveries are still available until today even though most of the discoveries from ancient Greece and the Library of Alexandria have been lost and/or destroyed? Is it true that many mathematical and scientific discoveries are being lost today because there is no good way of preserving and archiving new and old Scientific and Mathematical discoveries? I heard the science is being lost today at an unprecedented level due to poor archiving and/or not properly preserving data and discoveries throughout the world.

Links:

http://unesdoc.unesco.org/images/0010/001055/105557e.pdf

http://www.unesco.org/bpi/eng/unescopress/2002/02-fea10e.shtml

http://www.un.org/apps/news/story.asp?NewsID=3902#.Vvhn4PsrLIX

What can we do to retrieve all of the research that we have lost already? — Preceding unsigned comment added by 98.113.174.114 (talk) 21:55, 9 May 2016 (UTC)[reply]

Don't you worry. Google will digitize everything :-). Actually there is no need to preserve anything. Studies of scientific citations have shown that science grows on its skin, in short only the most recent papers are referenced. Old discoveries if useful are incorporated in contemporary knowledge and will be passed to the next generation, if there is a next generation - it is another question. --AboutFace 22 (talk) 22:06, 9 May 2016 (UTC)[reply]

All is transitory, nothing is permanent. Sławomir Biały (talk) 23:10, 9 May 2016 (UTC)[reply]

May 10

Finding a bijection between a permutation of n digits and the range [1,n!]

Hello,

I am given a permutation of the digits 1-9, representing an ID number. I try to find a bijection that would link each number to an integer in the range 1-9!, according to its relative order. For example:

f(123456789) = 1, f(123456798) = 2, ....,f(987654312) = 9!-1 f(987654321) = 9!

Any hints or suggestions regarding the finding of such a function?

Thanks! — Preceding unsigned comment added by 212.179.21.194 (talk) 07:44, 10 May 2016 (UTC)[reply]

I think it is slightly easier to do this for g(N)=f(N)-1. You can write

g(N)=a_{1}8!+a_{2}7!+a_{3}6!+\dots +a_{9}

,

where

a_{1}

is one less than the first digit of N (and it may be easier to do this with digits 0..8 instead of 1..9). To find

a_{2}

, you map the remaining digits to 1..8 and continue recursively. But I have no idea whether this is the most efficient way of going about this. —Kusma (t·c) 09:54, 10 May 2016 (UTC)[reply]

Its not particularly pretty, but you can construct the answer recursively by noting that the first 8! numbers start with 1, the next 8! with 2 etc... and applying the same logic to the remaining digits. — Preceding unsigned comment added by 128.40.61.82 (talk) 10:08, 10 May 2016 (UTC)[reply]

Thank you both for your answers, why did you choose to represents the number as a linear combination of factorial terms? 212.179.21.194 (talk) 11:11, 10 May 2016 (UTC)[reply]

It seems natural, as the first digit is 1 for 8! times, the second digit is 2 for 7! times, the third digit is 3 for 6! times, ... (and similar things with other digits). Really, as 128.X has said, you just look at how many numbers start with what digits. —Kusma (t·c) 14:48, 10 May 2016 (UTC)[reply]

It doesn't add much to what's already here but you might be interested to read the section of the article Permutation#Permutations in computing which discusses this problem. Dmcq (talk) 15:05, 10 May 2016 (UTC)[reply]

Relation vocabulary

WP:RDL#Fill in the blanks includes a question on mathematical vocabulary. (Answers would be best placed there, on the Language Desk.) -- ToE 14:14, 10 May 2016 (UTC)[reply]

Railroad tracks

I stand on a bridge over railroad tracks, directly above the symmetry axis between the two rails, and take a picture of the (totally straight) rails all the way to the horizon, while pointing the camera along the symmetry axis. The images of the rails get closer and closer the higher you look on the photo. What is the equation for the curve of the image of one of the rails? And how about in the special case where the camera is no higher than the rails? Loraof (talk) 17:03, 10 May 2016 (UTC)[reply]

It is a straight line. The articles Graphical projection and vanishing point have a few details. Sławomir Biały (talk) 17:09, 10 May 2016 (UTC)[reply]

Sławomir Biały's answer is correct for a rectilinear projection, which is what you get with a typical camera and lens (modulo some distortion). In the case of a vertical cylindrical panorama, like this one, the image is a sinusoid. (If the camera is in the plane of the rails, then it's a straight line, which in this case is a sinusoid of amplitude 0.) -- BenRG (talk) 19:02, 10 May 2016 (UTC)[reply]

Thanks. Now if I'm on a small asteroid with a very near horizon, presumably both rails seem to just end at the height of the horizon, without merging. But what if I'm on a flat planet of infinite extent? Do the two straight lines intersect at a place on the photo representing the same elevation as the camera, with neither continuing beyond there? Loraof (talk) 19:31, 10 May 2016 (UTC)[reply]

Assuming a flat planet and an ideal pinhole camera, the horizon is the intersection of the film with the plane parallel to the ground and containing the aperture (so any object at the same height as the camera will also project to the horizon). The image of a track is the intersection of the film with the plane containing the track and the aperture. The track images intersect at a point on the horizon, namely the intersection of the film with the line parallel to the tracks and containing the aperture.

On a curved asteroid there are no straight lines (unless you mean great circles on a sphere, in which case there are no parallel lines), so you'd have to be more precise about how the tracks are laid out, but most likely they will not intersect anywhere in the image. -- BenRG (talk) 20:24, 10 May 2016 (UTC)[reply]

May 11

Maximum coverage of plane by identical regular polygons...

Looking for the maximum coverage of the plane by each regular identical N-polygon. For N=3,4,6 the answer is 1.00. For 8, I think it is the Truncated square tiling where the part covered = 2sqrt(2)-2. Any ideas on 5 ,7 and other numbers? The limit as N gets large is equal to the $\eta _{h}={\frac {\pi }{2{\sqrt {3}}}}\approx 0.9069.$ . Which is worst?Naraht (talk) 18:26, 11 May 2016 (UTC)[reply]

I suggest starting here and following the links backwards. (An earlier version of this post had a bunch of other links, but I think this is the best one and will answer most of your questions.) JBL (talk) 19:43, 11 May 2016 (UTC)[reply]

May 12

System of Linear Eequations

Is the following statement true? This seems to me wrong, but I can't find any counterexample...

Statement: Let a system of linear equations such that each equation has exactly k terms $y_{i}\in \{x_{i},1-x_{i}\}$ where $\{x_{i}\}$ are the variables of these equations . That is, we have a system of linear equations, each of the form: $\sum _{y_{i}\in \{x_{i},1-x_{i}\}}{y_{i}}=1$ . For example, the system may consist of the equation $x_{1}+(1-x_{2})=1$ (here, $y_{1}=x_{1},y_{2}=1-x_{2}$ ).

This system of equations has some solution over $\mathbb {Z}$ iff it has a 0-1 solution (i.e, it has some solution over $\{0,1\}\subset \mathbb {Z}$ ). 213.8.204.30 (talk) 06:43, 12 May 2016 (UTC)[reply]