Wikipedia:Reference desk/Archives/Mathematics/2011 September 23

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September 23

360 sq centimeters

I have been asked to paint a picture which measures that exact dimension above, that is to say if the picture is comprised of drawn squares, (say each of a different colour) this non-mathematician would assume that the external dimension should be 19 x 19 cm. However that produces 361 actual different colour squares, why should this be and how do I arrive at 360 exactly?85.211.209.246 (talk) 07:09, 23 September 2011 (UTC)

That's because 19 × 19 = 361. If you want 360 exactly, try, for example, 36 × 10, or 18 × 20, or another pair of integers that multiply together to make 360. See Factorization. —Bkell (talk) 07:28, 23 September 2011 (UTC)

Or you could start with a 19 x 19 square and remove the central square to leave 360 squares. Gandalf61 (talk) 08:16, 23 September 2011 (UTC)

This is what I have done, but I put Gold Leaf in on the odd square at the Golden Mean.--85.211.209.246 (talk) 06:24, 24 September 2011 (UTC)

Speed of Light

Why is it that the ability of an object to exceed the speed of light means that things could 'arrive before they leave'? Thanks. asyndeton talk 10:36, 23 September 2011 (UTC)

This probably belongs on the science desk, but this has to do with Relativity of simultaneity. Call the point in space-time where the particle starts it's journey A, and the point where it ends B. If the particle is going faster than the speed of light in your frame of reference then B is outside the light cone of A (the pair is "spacelike"), so there's some other frame of reference where the two events are simultaneous, and yet others where B occurs before A. Rckrone (talk) 12:52, 23 September 2011 (UTC)

And that can then be used to construct a set up allowing you to send signals into your own past, see here. Count Iblis (talk) 16:59, 23 September 2011 (UTC)

Could someone please check this comment on Balanced ternary?

Resolved

Someone has queried the addition table on the Balanced ternary talk page. I believe that the table as given in the article is correct, but another comment makes me think that I have misunderstood the meaning. Please could someone check it. -- Q Chris (talk) 14:45, 23 September 2011 (UTC)

Your best bet for this sort of thing is WikiProject Mathematics. In particular, try posting on Wikipedia talk:WikiProject Mathematics. WikiProject Mathematics deals with the upkeep, addition and improvement of articles. The reference desk is more like, well, a reference desk. — Fly by Night (talk) 18:08, 23 September 2011 (UTC)

Thanks, I'll try there.

The cartesian plane as a sphere - theoretical maths

Hi. I understand that the cartesian plane is typically used as a flat grid starting with $x$ and $y$ grid coordinates. As a mathematical representation, what would be the result of adding curvature to the grid itself (ie. geodesics) so that the ±x coordinates meet at the back of the sphere now formed by this curvature, and so that the ±y coordinates similarly meet at the back of this sphere? The point opposite the origin (mathematics) would be (±∞, ±∞) in terms of x, y. It would make no difference whether functions on this grid were on the outside or inside surface of the sphere. Now, suppose that the anti-origin is in itself a type of origin (think of the North/South poles on Earth as an example). Let the grid ignore the problems associated with addition and subtraction, which would inevitably result from mixing numerical numbers and infinities. My observations of such a framework include:

• A simple hyperbolic base function, for example f(x) = 1/x, would intersect both the x and y intercepts at the anti-origin, rather than having asymptotes.
• An odd function (linear, cubic...), including those with zero or infinite slope, would complete a full circle, as would sinusoidal functions of continuous sort, with infinite domain/range (question: what would happen to a helicoid parabola?).
• A parabola (quadratic function, quartic...) may exist in the form where its lines converge on being parallel as its non-vertex approaches the anti-origin, and may either meet the anti-origin in parallel, converging lines (think closed universe), or have a rounded shape similar to the vertex.
• A circular equation may or may not have a mirror reflection of itself about the anti-origin, assuming for instance that the main circle is centered about the origin.
• A cross-section across this sphere may in fact reveal aspects of the function in 3-D space that are uncovered only in spherical use of the cartesian grid, similar to a chord in a circle, but instead in a sphere (actually, I don't know how to make sphere-chord volume calculations; a similar but unrelated thought is: is a chordal graph a possible analogue?).

Of course, there are likely many more possibilities associated with this type of cartesian reasoning. My question is, has this type of application already been used in standard mathematics, or did I somehow invent the concept? Iff it is already in use, for what practical or scientific applications has it been used? Is there any connection between this cartesian plane-sphere and fractal geometry? What are some existing Wikipedia articles underlying my concept? Finally, what other properties would exist from this cartesian sphere, and could I possibly call it a 3-sphere? Thanks. ~AH1 ^(discuss!) 14:46, 23 September 2011 (UTC)

A concrete construction of what you describe is called the stereographic projection, which has many applications indeed, for instance to cartography. This has been known in the area of astronomy for thousands of years. Sławomir Biały (talk) 15:09, 23 September 2011 (UTC)

maximum traffic flow

I was watching an old episode of Futurama where they are on a robot planet. In once scene, a bunch of robots travel through an intersection for exactly 1 minute, (running over the delivery team without notice), after which the intersection is empty. That got me thinking. What are the formulas for figuring out the maximum number of robots that can go through a road intersection in 1 minute based on the following constraints...

• Each robot is identical and fully occupies a space of 1'x1'
• Both roads are width n in feet, and are perpendicular to each other running N/S and E/W
• Robots must travel at a constant 3' per second in a straight line
• An equal number of robots must cross the road from each direction (1/4 are going north, 1/4 are going south, 1/4 east and 1/4 west)
• Robots may not collide
• At the end of 1 minute, the intersection must be empty

Any ideas?

Googlemeister (talk) 18:30, 23 September 2011 (UTC)

They go 3' per second, so at most 3 can go through in 1 second, which means that an upper bound for the maximum number that can go through is 60*3*2=360. (The "*2" since you can have two opposing lanes running at the same time, right?) This only allows two opposite lanes to go through, so this plan violates your 4th condition. But if we let the N/S lanes go for 30 seconds, then let the E/W go for 30 seconds, we should be able to get (almost) 360. Maybe you'll miss out on a few cars when you do the switch from N/S to E/W.

This all assumes that each lane is wide enough for one car. If the lanes are bigger, multiply by the appropriate factor. Staecker (talk) 21:50, 23 September 2011 (UTC)

You seem to have assumed that the intersection is only 1' square. However, it would actually be the widths of the two intersecting roads. So, the wider the roads, the longer it would take to cross the intersection, and the longer it would take to switch directions. StuRat (talk) 03:34, 25 September 2011 (UTC)

Also note that this type of multi-lane switchover is best:

   ^^^
   |||
   |||
   |||
   ||<-----
   |<------
   <-------

While this type of switchover is worse:

   ^^^
   |||
   |||
   |||
     <-----
    <-----
   <-----

This only represents 2 directions, so you'd need to cycle through the other 2 directions, too, within the minute. This assumes that the bots can move both north and south in the same lanes and east and west in the same lanes. This might create traffic problems outside the intersection, but I'm going to consider that issue beyond the scope of the question. StuRat (talk) 03:38, 25 September 2011 (UTC)

Yes, we can just handwave away traffic going into and out of the intersection, but perhaps I was unclear that 1/4 of robots must be going in each individual direction simultaneously, not alternating (even though alternating is almost certainly more efficient). Googlemeister (talk) 15:59, 26 September 2011 (UTC)

Derivative of Sine

I'm trying to find a purly geometrical proof for the fact that

${\frac {\text{d}}{{\text{d}}\theta }}\sin \theta =\cos \theta \,.$

The proof I want to find uses the "first principals" definition:

$\lim _{\delta \to 0}\left({\frac {\sin(\theta +\delta )-\sin \theta }{\delta }}\right)=\cos \theta \,.$

The proof is pictorial, and uses a diagram of a triangle with other sub-triangles and chords. Please don't give an analytic proof: I already know that proof. I need the graphic that I once saw as a student. I've been asked to give a high school "master class" for some talented young students. I wanted to get them to work it out for themselves from the picture… but then I realised that I couldn't remember the picture myself, and that I couldn't find it on Google. (Oh, the shame!) — Fly by Night (talk) 23:40, 23 September 2011 (UTC)

Something like this? Sławomir Biały (talk) 23:47, 23 September 2011 (UTC)

Thanks Sławomir, but that seems far too complicated. It starts with "As a large square of side curls into a cylinder, an uncurled unit square is kept tangent to it, so that the unit square's diagonal always lies tangent to the helix formed by the larger square's diagonal. Projecting the helix into three mutually perpendicular planes…". Like I said: there's a picture of a triangle with sub-triangles and chords. It's really very simple, but also very fiddly. — Fly by Night (talk) 00:01, 24 September 2011 (UTC)

I don't know if this is what you're going for, but if you draw the unit circle with the angles $\theta$ and $\theta +\delta$, then the quantity $\sin(\theta +\delta )-\sin \theta$ is the length of an interval that you can draw next to the (x,y) coordinates intersected by $\theta +\delta$, and $\delta$ is approximated by its secant line between the two angles, which, combined with the third line segment you can draw to represent $\cos \theta -\cos(\theta +\delta )$, make a right triangle. (I'm too lazy to draw a picture.)

This is easiest to think about when $0<\theta <\theta +\delta <\pi /2$; to turn it into a full-fledged proof, you probably have to consider cases, and you also have to either prove or take on faith that the ratio between the secant line and the corresponding angle goes to one as the angle goes to zero. --COVIZAPIBETEFOKY (talk) 04:23, 24 September 2011 (UTC)

Thanks COVIZAPIBETEFOKY, I think I get the picture (without a picture)! I think I'll use the formula sin(α + h) = sin(α).cos(h) + sin(h).cos(α), and then use the geometrical argument (see the image) along with the sandwich theorem to show that

$\lim _{h\to 0}{\frac {\sin h}{h}}=1\,.$

From that the result follows; but there will be a case-by-case progression as you say. I think it's much nicer to try and get them to work it out for themselves, using geometrical arguments, than just giving them a rule to learn. Thanks for the advice. — Fly by Night (talk) 17:34, 25 September 2011 (UTC)