Wikipedia:Reference desk/Archives/Mathematics/2011 February 25

= February 25 =

Limit monstrosity
$$\lim_{x \to 0+} \frac{-x^3}{2\sqrt{x^2-\frac{x^4}{4}}-2x}$$

I tried dividing the top and bottom by x.

$$\lim_{x \to 0+} \frac{-x^2}{2\sqrt{1-\frac{x^2}{4}}-2}$$

But the limit of the denominator is still zero, and I can't find any other common factors between the numerator and the denominator. Widener (talk) 00:21, 25 February 2011 (UTC)


 * Multiply and divide the original limit by the "conjugate": $$2\sqrt{x^2-\frac{x^4}{4}}+2x$$. It will simplify the denominator.   Sławomir Biały  (talk) 00:23, 25 February 2011 (UTC)
 * Thank you, the answer's 4. By the way, I also tried multiplying and dividing my second equation by it's "conjugate" $$2\sqrt{1-\frac{x^2}{4}}+2$$ and that didn't seem to work. Do you know why that might be? Widener (talk) 00:34, 25 February 2011 (UTC)
 * That should also work.  Sławomir Biały  (talk) 00:38, 25 February 2011 (UTC)
 * Oh, I just made a mistake. Never mind and thank you for your help! Widener (talk) 00:48, 25 February 2011 (UTC)
 * Another thing you could try is, after getting the key expression in the denominator into the form $$\sqrt{1-\frac{x^2}{4}}$$, expand that in a power series and keep as many terms as you need (won't be many). You know what $$\sqrt{1-u}$$ is, approximately, when $$u\,\!$$ is small? --Trovatore (talk) 00:41, 25 February 2011 (UTC)
 * Like a Taylor series? and $$\lim_{u \to 0+} \sqrt{1-u} = 1$$ doesn't it? Widener (talk) 01:58, 25 February 2011 (UTC)
 * Well, yes, a Taylor series, but you don't really have to get that involved &mdash; it's actually just the binomial theorem. The limit you wrote down is correct, but when I said you wouldn't need very many terms, I didn't actually mean you could get by with just one :-)  --Trovatore (talk) 02:18, 25 February 2011 (UTC)
 * If differentiation is needed then why not use L'Hôpital's rule? It is a weaker differentiablitity condition han having a well defined jet. — Fly by Night  ( talk )  02:38, 25 February 2011 (UTC)
 * I had been pleasantly surprised, up to now, that no one had brought up l'Hôpital. In my view l'Hôpital is a very nice thing to have around in the relatively rare instance that numerator and denominator are both easy to differentiate, but unfortunately students tend to use it as a substitute for understanding what's going on.  If only there were some way to keep them from finding out that the rule exists, until they've learned how to do it in ways that require a little more thought and a lot less algebraic manipulation.  Then and only then you could tell them, hey kids, here's a handy shortcut in certain relatively uncommon cases, and they could have it in their toolkit for those cases, but would have a better understanding. --Trovatore (talk) 03:08, 25 February 2011 (UTC)
 * Sorry to disappoint you Trovatore. I agree with what you said. I never much enjoyed analysis, and avoided it as an undergraduate. My calculus education was just that: plain old calculus and not much analysis; which was at the hands of applied mathematicians. That might explain my fondness for such underhanded tricks. I promise to try harder next time :-) — Fly by Night  ( talk )  14:52, 26 February 2011 (UTC)


 * Yet another trick is to set $$y=\sqrt{1-\tfrac{x^2}{4}}$$ whence $$x^2 = -4(y^2-1)$$. Then we just need to evaluate $$\lim_{y\to 1}\tfrac{4(y^2-1)}{2y-2}$$, and limits of rational functions are easy. A systematic method is to reduce by $$y-y_0$$ with long polynomial division repeatedly until either numerator or denominator becomes nonzero at $$y_0$$; faster in this case is to remember that $$y^2-1 = (y-1)(y+1)$$ so the entire fraction immediately reduces to just $$2(y+1)$$. –Henning Makholm (talk) 21:25, 26 February 2011 (UTC)

Vertices of a hyperbolic tiling
Does anyone know of an easy and fast way to generate the coordinates of the vertices of a hyperbolic tiling? (The order-7 triangular tiling, for instance, or any other convenient one). I need something suitable for doing numerical integration. Sławomir Biały (talk) 00:53, 25 February 2011 (UTC)


 * Use complex numbers as 2D-coordinates. Construct a 7-gon within the unit circle. The otehr 7-gons in the order-7 triangular tiling are related to the first 7-gon by a mobius transformation. Simply choose the parameters of the mobius transformation such that it maps 2 the original 7-gon onto an adjacent 7-gon. Keep doing this until you have the desired amount of 7-gons. See also Poincare disk model 213.49.109.45 (talk) 07:27, 25 February 2011 (UTC)


 * With "7-gon" replaced by "equilateral triangle", that would be more or less the definition of a hyperbolic tiling. I was hoping for something that did not involve repeatedly reflecting in circles.  It's very slow this way, and doesn't give the rapid grid generation I would hope for.  But thanks anyway.


 * I have more or less already decided that there is no easy way to do this, so I have already adopted a different (although less satisfactory way) to do the integration.  Sławomir Biały  (talk) 12:52, 25 February 2011 (UTC)


 * Ask commons:User:Tamfang, he has generated lots of hyperbolic tilings, so he must have some method. You can see he's created most of commons:Category:Hyperbolic_tilings.  &#x2013; b_jonas 14:12, 25 February 2011 (UTC)


 * Thanks, I'll do that.  Sławomir Biały  (talk) 14:16, 25 February 2011 (UTC)


 * My method probably is not what you want. I find the coordinates of one vertex (and the edges incident to it), and use mirrors to convert each pixel's coordinates to those of an equivalent point within the fundamental region. —Tamfang (talk) 06:04, 1 April 2011 (UTC)


 * Don Hatch does it in Java —Tamfang (talk) 06:08, 1 April 2011 (UTC)

Pumpkin weight problem
A colleague gave me this problem today:

"A 400 lb. pumpkin is 98% water by mass. If by dessication, the water content drops from 98% to 96%, how much does the pumpkin now weigh?"

And this is supposed to be so surprising because, he explained, that if done correctly, one will be astonished that the pumpkin after said dessication weighs 200 lbs.

He gave me the following formula: .96xWater + .96x8 = Water, then 0.04xWater = .96x8 and I just don't understand how losing water in the amount of 2% of the original mass calculates to so much mass lost for the pumpkin overall. If every 2% was that much mass, the pumpkin would have to originally weigh thousands of pounds, not just 400.  DRosenbach  ( Talk 02:53, 25 February 2011 (UTC)


 * The water content drops from 98% to 96% of the mass at the time, not from 98% to 96% of the original mass. It may be easier to see what happens by considering the non-water mass. If we assume the non-water mass is constant then it's 2% of 400 lb. You can compute how much that is, and then what the new total mass must be if the constant non-water mass is suddenly 4% of the new mass. PrimeHunter (talk) 03:07, 25 February 2011 (UTC)


 * You are given:
 * $$m_{pumpkin_1}= 400lbs,$$
 * $$m_{water_1}=0.98m_{pumpkin_1}= 392lbs,$$
 * $$m_{water_2}=0.96m_{pumpkin_2}$$
 * Now, since the overall weight lost is entirely due to weight of water lost:
 * $$m_{pumpkin_1}-m_{pumpkin_2}=m_{water_1}-m_{water_2}$$
 * Substituting from above:
 * $$400lbs-m_{pumpkin_2}=392lbs-0.96m_{pumpkin_2}$$
 * and rearranging terms:
 * $$400lbs-392lbs=m_{pumpkin_2}-0.96m_{pumpkin_2}=(1-0.96)m_{pumpkin_2}$$
 * $$8lbs=0.04m_{pumpkin_2}$$
 * So:
 * $$m_{pumpkin_2}=\frac{8lbs}{0.04}=200lbs$$
 * WikiDao   &#9775;  17:16, 25 February 2011 (UTC)


 * KISS !!! A dry mass is 2% of 400 lb = 8 lb. After the described partial drying it is 4% of the new mass, so the new mass is 8 lb : (4/100) = 200 lb. End of calculation. No water. --CiaPan (talk) 18:26, 25 February 2011 (UTC)
 * I was mostly just playing around with using mark-up.
 * It is a "surprising" result, though, presented the way it is! :) WikiDao    &#9775;  20:28, 25 February 2011 (UTC)
 * But if I would like the explanation in words rather than in math, is that possible? Just because the math works out doesn't mean I can understand how this makes any sense.  I'm still lost, even though you guys are thoroughly surprised and excited :)  DRosenbach  ( Talk 21:56, 25 February 2011 (UTC)
 * Do you mean something along the lines of "If the non-water content is doubled in percentage without changing in actual weight then the total weight must be halved"? The maths is more elegant.    D b f i r s   22:32, 25 February 2011 (UTC)
 * How about this: At time zero the pumpkin is 2% dry solid and the rest is water. If it loses water to the point where 4% (twice as much as before) of it is now solid, that means that the total mass has been halved. It sounds much less surprising if you phrase it in terms of the non-water mass. Staecker (talk) 18:22, 27 February 2011 (UTC)
 * Maybe an analogy with Achilles and the tortoise will help? Achilles is the water mass, which starts at 392. The tortoise is the 96% water mark, which starts at 384. We start reducing water to close the gap, but by the time we get to 384 water, the 96% mark reduced to 376.32. We continue, but when we get to 376.32 water, the elusive mark is now at 368.9... . Note that a reduction of 8 in the mass caused a reduction of 7.68 in the target - the tortoise is not very slow! Eventually we do close the gap, but only after a very long chase. To find exactly how long, you can treat it as a geometric series, a simple linear equation, or any other method. -- Meni Rosenfeld (talk) 07:37, 28 February 2011 (UTC)

Asymptotes
Can someone explain the following line from the aarticle on Asymptotes to me please: "Generically, by Bezout's theorem, a curve will intersect its asymptote at exactly this number of other points, as the intersection at infinity is of multiplicity two." Why is the intersection at infinity of multiplicity two? Thanks-Shahab (talk) 06:05, 25 February 2011 (UTC)


 * Because the asymptote is tangent to the curve at infinity.  Sławomir Biały  (talk) 12:31, 25 February 2011 (UTC)


 * Actually the multiplicity is two or more so the statement in the article isn't strictly true. For example the tangent to the Folium of Descartes does not intersect the curve at a finite point. Also, the finite points of intersection may have multiplicity higher than one which will throw off the count. It's seems a bit of overkill to invoke Bezout anyway since for a line and a curve you're really only using the Fundamental Theorem of Algebra. I'll try to reword the section when I get a chance, if no one beats me to it.--RDBury (talk) 18:51, 25 February 2011 (UTC)
 * It's true generically, as stated. Sławomir Biały  (talk) 19:05, 25 February 2011 (UTC)
 * If asymotote is tangent to the curve at infinity, why isn't its multiplicity one?-Shahab (talk) 04:36, 26 February 2011 (UTC)
 * An intersection of multiplicity one would an ordinary crossing point. The definition of multiplicity is complicated but the intuition is to think of a a secant line approaching the tangent line. The secant has two points of intersection that merge to form a double point of intersection when the line becomes tangent to the curve.--RDBury (talk) 06:54, 26 February 2011 (UTC)


 * Take a look at line at infinity. This might help you understand what "at infinity" means. — Fly by Night  ( talk )  15:26, 26 February 2011 (UTC)


 * The intersection can have multiplicity greater than 2, of course -- such as when x²y=1 intersects meets its horizontal asymptote with multiplicity 3. I've edited the article accordingly. –Henning Makholm (talk) 21:35, 26 February 2011 (UTC)


 * My understanding was that it was referring to a curve in general position (Generically...), in which case the intersection at infinity does have multiplicity exactly two. But your edit is an improvement anyway.   Sławomir Biały  (talk) 21:48, 28 February 2011 (UTC)

problem with the algorithm for programing b-spline
I have a elliptical contour shaped curve made of by joining 57 points on it (not equally spaced).and i want to find new position of the curve if i give 6 or 7 new coordinates of the points which are located previously at the contour.for this i am trying to approximate the curve passing through these new given points called control points by using b-spline method of 4rth order.that is fine,but the problem is how could i find all 57 points again on this new approximated curve.here is the problem is of knot vectors.how could i find that between two new point how many old point are situated.is there any problem i want to use b-spline for this? Is there any other technique by which this problem could be solved? after that i want to write c program for this.124.124.247.141 (talk) 10:44, 25 February 2011 (UTC) 124.124.247.141 (talk) 11:33, 25 February 2011 (UTC)


 * As far as I can make out you talk about an old curve and a new curve and finding the points on the old curve somewhere on the new curve. The problem I see is that the old points may not be exactly on the new curve so how does one define a point which corresponds to the old point. When you just use points to define a Bezier curve you can say that the point corresponding to a control point is the one which has that as the most important one in the bezier curve near it. I can't see what your question is asking exactly though. Dmcq (talk) 22:54, 25 February 2011 (UTC)


 * If this really is an elliptical curve, then 5 points should define it fully, as with any conic section. So, then, why are you using 57 ?  If the issue is that these points aren't exact, and you want to run an elliptical curve through a cloud of approximate points, then other methods should be used, such as running an ellipse through each combination of 5 and picking the curve which is "in the middle", mathematically.  To make a b-spline from 57 points is risking arriving at a rather lumpy curve.


 * Now, as to mapping points from one curve onto a second curve, a normal (perpendicular) projection of each point onto the new curve is one method. Here, again, a b-spline could complicate things, as you don't necessarily get a single normal projection for each point, while a truly elliptical curve would only allow one normal projection (unless projecting from the center point, or if you count the projection through the center point to the far side of the ellipse).  This gets especially ugly if it's a 3D b-spline.  By contrast, all elliptical curves are planar.


 * If you need to stick with the b-spline, then perhaps rather than doing a normal projection, the best method would be to find the closest point on the b-spline to each point. For every old point, you would need to do this for each arc in the b-spline, and take the point that's closest overall, from the closest point on each arc.  To reduce computation cycles, you might want to just go with the endpoints of each arc, initially, and use the arcs on either side of the closest endpoint for the more in-depth calculations. Note that it's still possible to get two points on the new b-spline which are both the same (minimum) distance to the old point.


 * Here's an example (greatly exaggerated, of course), where the old point (+) projects as 2 new points (°) on the new curve:

\ +  /  \   /   \_/

\ +  /  °   °   \_/


 * StuRat (talk) 01:19, 1 March 2011 (UTC)

Spiral staircase
Hello. I want to know how to calculate the total length of a spiral staircase's outer edge (not homework, I'm just curious about how much distance I can "save" if I take the inner edge). I know (or rather, have access to) the staircase's radius (it's a circle if viewed from top-down), height and the steepness. 212.68.15.66 (talk) 13:47, 25 February 2011 (UTC)


 * Check out Helix. You can get the length of the inner and outer rails that way. Let us know if you have problems applying the formula. However, if you have physical access, you could just use string to measure directly. SemanticMantis (talk) 13:57, 25 February 2011 (UTC)


 * Just think of wrapping a huge sheet of paper round the staircase to make a cylinder and mark the path of the outer rail. When you unwrap the sheet starting at the start of the rail you'll find the mark makes one or more straight lines along the sheet, You need to find the length of the line, which is given by figuring out a rectangle given by the height at the start and the end and the horizontal length given by how many times it went round the cylinder. Dmcq (talk) 17:03, 25 February 2011 (UTC)


 * Just an aside, you do mean a helical staircase, right? Otherwise, it sounds pretty Escheresque.  DRosenbach  ( Talk 17:12, 25 February 2011 (UTC)


 * `Spiral staircase' may indeed be a bit of a misnomer, but it is nevertheless the standard nomenclature for helical stairs. SemanticMantis (talk) 18:43, 25 February 2011 (UTC)


 * Sorry, that was meant as more of a joke. Good weekend to all.  DRosenbach  ( Talk 21:57, 25 February 2011 (UTC)


 * The outer edge of the staircase is a curve on a cylinder. Cut the cylinder vertically and roll it out into a plane. Then the path turns into a straight line having the same length, which is computed using Pythagoras. Bo Jacoby (talk) 12:28, 26 February 2011 (UTC).


 * Some physics comments:


 * 1) It's not actually possibly to walk on the outside edge, assuming a handrail is there. Same for the inside edge.


 * 2) The amount of work is the same, theoretically, as you go up the same height, in either case. However, in the real world, we burn energy when walking on level ground, so it should take slightly more calories to walk near the outside edge. StuRat (talk) 01:00, 1 March 2011 (UTC)