Wikipedia:Reference desk/Archives/Mathematics/2013 April 16

= April 16 =

Curve with even distances
Hi everybody,

I'm looking for a one -dimension curve which the distance between each point on it and to a half circle is the same.

The curve cannot be inside the circle; it must be outside of it.

The curve is not a circle.

Thank you.

Exx8 (talk) 08:43, 16 April 2013 (UTC)


 * Indeed, such curve cannot be a circle. However it can be a union of some circular circular arcs... CiaPan (talk) 09:09, 16 April 2013 (UTC)


 * You might be looking for a Parallel curve (offset curve).--Salix (talk): 09:25, 16 April 2013 (UTC)


 * I'm confused by the req that the distance between each point on the curve be the same. The would seem to mean a series of points, not a curve, since a curve requires zero distance between adjacent points.  I think you are asking about an offset curve, but the way you stated it is confusing.


 * I think you meant "I want a curve with a constant offset outside of a semi-circle". If you don't curve back around the endpoints of the semi-circle, that just gives you another semicircle, with the same center and degrees swept, but with a radius of R+D, where R is the original semi-circle's radius, and D is the offset distance.  If you do want to curve back around the endpoints, then you get two smaller circular arcs appended to the larger offset curve, centered on the endpoints of the original semi-circle, with radius D, such that they have point and tangency continuity, but not curvature continuity, with the main offset curve.  StuRat (talk) 18:58, 16 April 2013 (UTC)


 * I understand Exx8 means a curve C such that for points p on C a point-to-set distance of p to the semicircle S is constant d(p, S) = const. --CiaPan (talk) 05:49, 17 April 2013 (UTC)

Identitites Containing both Pi and e
Are there any other identities out there expressing a relationship between Pi and e, other than Euler's identity $$e^{i\pi}+1=0$$ and the Gaussian integral $$\int_{-\infty}^{\infty}{e^{-x^2}}dx=\sqrt{\pi}$$ ? — 79.113.234.168 (talk) 14:06, 16 April 2013 (UTC)


 * There are a whole host of approximate formula like Ramanujan's constant $$e^{\pi \sqrt{163}} =\ $$$262,537,412,640,768,743.999 999 999 999 25…$ --Salix (talk): 21:01, 16 April 2013 (UTC)


 * BTW, why does $$e^{\pi \sqrt{58}}$$ work too? Reasons are not given in the article you link for this... Double sharp (talk) 13:37, 17 April 2013 (UTC)


 * See here. Count Iblis (talk) 15:57, 17 April 2013 (UTC)


 * eπ − π = 19.999099979... Double sharp (talk) 13:40, 17 April 2013 (UTC)


 * Salix and Double Sharp: I'm not interested in that kind of approaches. — 79.113.234.93 (talk) 14:23, 17 April 2013 (UTC)


 * There is also $$i^{2i}=e^{-\pi}$$. Ruslik_ Zero 18:28, 17 April 2013 (UTC)


 * Which is also based on Euler's identity $$e^{i\alpha}=\cos \alpha + i \sin \alpha$$ where $$\alpha={\pi\over2}$$. — 79.113.213.221 (talk) 05:52, 18 April 2013 (UTC)


 * $$\int_{-\infty}^{\infty}\frac{\cos(x)}{1+x^2}dx=\frac{\pi}{e}$$ . — Count Iblis (talk) 22:38, 16 April 2013 (UTC)


 * $$\sum_{n=0}^{\infty}\frac{1}{1+n^2} = \frac{1}{2} + \frac{\pi}{2}\frac{e^{2\pi}+1}{e^{2\pi}-1}$$ . — Count Iblis (talk) 23:16, 16 April 2013 (UTC)


 * The e in the first formula comes from Euler's identity $$e^{i\alpha}=\cos \alpha + i \sin \alpha$$, and the Pi from the fact that $$arctg'(x) = {1 \over 1+x^2}$$ , along with $$arctg(\pm\infty) = \pm{\pi \over 2}$$ , and $${\pi \over 2} - (-{\pi \over 2}) = \pi$$. — 79.113.234.93 (talk) 14:23, 17 April 2013 (UTC)


 * And the second one is related to the fact that $$\zeta(2) = \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$$ . — 79.113.213.221 (talk) 16:25, 18 April 2013 (UTC)
 * Also, $$\sum_{n=0}^{\infty}\frac{1}{n^2 + a^2} = \frac{a\pi\cdot\coth(a\pi)-1}{2a^2} = \frac{\pi}{2a}\cdot\frac{e^{2a\pi}+1}{e^{2a\pi}-1} - \frac{1}{2a^2}$$ . — 79.113.220.118 (talk) 03:39, 20 April 2013 (UTC)


 * If you want something much less trivial, you can look at
 * $$\int\limits_0^{\infty}\frac{x^3}{1+e^x}dx=\frac{\pi^4}{15}$$ . — Ruslik_ Zero 08:27, 18 April 2013 (UTC)


 * This is ultimately related to the Riemann zeta function by means of the polylogarithm:
 * $$\int\limits_0^{\infty}\frac{x^n}{1+e^x}dx\ =\ (1-{1\over2^n})\cdot\zeta(n+1)\cdot\Gamma(n+1)$$ . — See here.
 * To an odd n corresponds an even argument of Zeta, which is always of the form $${\ p \over q}\cdot\pi^{\ 2m}$$ . So I guess the real question would be: Why are the even Zeta Constants related to Pi ? — 79.113.213.221 (talk) 15:40, 18 April 2013 (UTC)

Surface perimeter of a sphere

 * Any formula for the surface perimeter of a sphere?? I know the following formulas for the surface perimeter of the Platonic solids make sense:


 * Tetrahedron = 6 times edge length
 * Cube = 12 times edge length
 * Octahedron = 12 times edge length
 * Dodecahedron = 30 times edge length
 * Icosahedron = 30 times edge length

But how about a sphere?? Georgia guy (talk) 22:07, 16 April 2013 (UTC)


 * You seem to be summing the lengths of the edges of your polyhedra. Spheres have no edges; you could say that the sum of their lengths is 0, then.  --Tardis (talk) 02:03, 17 April 2013 (UTC)


 * Or, the most facet-nating answer is to say that it has an infinite number of infinitely small edges, and thus has an infinite "perimeter". StuRat (talk) 02:29, 17 April 2013 (UTC)


 * it has an infinite number of infinitely small edges, and thus has an infinite "perimeter" - This is tantamount to saying that $$0 \cdot \infty = \infty$$ — 79.113.234.93 (talk) 14:08, 17 April 2013 (UTC)


 * As noted below, it's actually say that $$ab $$ approaches $$ \infty$$ as $$a $$ approaches $$ \infty$$ and $$b $$ approaches $$ 0$$, which is not the same thing at all. Whether true or not depends on how quickly each approach occurs.  The examples the OP gives, where "perimeter" goes up, at an increasing rate, as the number of polygons increases, leads me to infer that it would go to an infinite ratio/perimeter, in the case of infinite polygons. StuRat (talk) 17:15, 17 April 2013 (UTC)


 * See Method_of_exhaustion. SemanticMantis (talk) 17:05, 17 April 2013 (UTC)


 * Along the same lines, the only sensible way to formulate the question would be to look at the limit at infinity, and that can depend on exactly how you pose it. You would need a way to show what the sequence of shapes is that has a sphere as the limit at inifinity. Presumably, it will involve progressively dividing the shapes up into smaller and smaller polygons, which, assembled, form a polyhedron with an increasing number of faces. You would only need to show that there is a systematic way of doubling the total length in a finite number of steps, to show the limit of the "perimeter" is infinity. For an example of this approach, in a different context, see the proof of Nicole Oresme that the harmonic series goes to infinity. IBE (talk) 08:34, 17 April 2013 (UTC)


 * You could use geodesic domes as an approximation to a sphere. You can construct these with an ever increasing number of polygons.--Salix (talk): 09:40, 17 April 2013 (UTC)


 * For example, here is a 14-frequency icosahedral geodesic dome and its dual.
 * 14-frequency icosahedral geodesic sphere.png14-frequency icosahedral geodesic sphere dual.png
 * Double sharp (talk) 13:26, 17 April 2013 (UTC)
 * (Larger size: ) Double sharp (talk) 13:28, 17 April 2013 (UTC)
 * There are four possible answers: (1) undetermined; (2) zero; (3) the sphere's actual surface; (4) infinity. The first one is the general answer. — 79.113.213.221 (talk) 05:57, 18 April 2013 (UTC)
 * I feel that the Hausdorff dimension of a particular space-filling curve on the surface of the sphere would have some relevance - the variety of possible curves would suggest a variety of possible answers.←86.128.43.105 (talk) 10:03, 18 April 2013 (UTC)


 * Maybe the perimeter tells the total length of the seams used if you sew a sphere-shaped football from flat pieces of cloth. &#x2013; b_jonas 12:52, 19 April 2013 (UTC)