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# March 20

## What is the ith root of 2

I am looking for x where x^i = 2 Oh and how many roots are there? 148.182.26.69 (talk) 02:12, 20 March 2017 (UTC)

A countably infinite number, I do believe. --jpgordon𝄢𝄆 𝄐𝄇 04:14, 20 March 2017 (UTC)
The values are e2kπ − i ln 2, for any integer k. Double sharp (talk) 05:32, 20 March 2017 (UTC)
x=2i. Bo Jacoby (talk) 07:31, 20 March 2017 (UTC).
Did you mean 2(1/i) ? --CiaPan (talk) 09:58, 22 March 2017 (UTC)
The two expressions are equal, --JBL (talk) 12:46, 22 March 2017 (UTC)
Oh, you're right! Just missed it's an imaginary number. :) --CiaPan (talk) 12:55, 22 March 2017 (UTC)

# March 21

## Ellipsoid Example

I am looking for an example for an ellipsoid that centered at (0,0,0), and contains ${\displaystyle [0,1]^{n}}$ (the unit cube), but does not contain the circle defined by the points (1,0,0) and (0,1,0) and (0,0,1)? Thanks :) 212.179.21.194 (talk) 17:17, 21 March 2017 (UTC)

Circle? You probably meant the sphere? Ruslik_Zero 20:18, 21 March 2017 (UTC)
Three points determine a circle, so probably not. Anyhow, the idea is not really different than your previous question: a sufficiently long and skinny ellipsoid will be a counter-example. (Probably any ellipsoid centered at (1/2, 1/2, 1/2) that passes through the vertices of the cube will do the trick, except for the circumsphere itself.) --JBL (talk) 22:48, 21 March 2017 (UTC)
The ellipsoid is supposed to be centered at (0, 0, 0), so that example doesn't work. The circle in question is the intersection of the unit sphere with the plane x+y+z=1. --RDBury (talk) 08:43, 22 March 2017 (UTC)
Ugh, thanks. --JBL (talk) 12:44, 22 March 2017 (UTC)
You won't find such ellipsoid. (At least if you seek an ellipsoid with axes along the coordinate system axes - I haven't analyzed other orientations close enough to come to a general conclusion.)
The three vertices define an equilateral triangle T. Extend the triangle by adding its three copies along its sides. You have a new triangle T1, which is twice as big as T, and midpoints of T1 sides are the vertices of T.
The circle circumscribed on T is inscribed in T1.
The vertices of T1 are (1,1,–1), (–1,1,1) and (1,–1,1).
So T1 is inscribed in a bigger cube [–1,1]3, whose center is (0,0,0).
Due to central and planar symmetry of the big cube and the ellipsoid with the common point of symmetry and cardinal planes (XY, XZ and YZ), if the ellipsoid contains the [0,1]3 cube, then it contains also the [–1,1]3 cube, which in turn contains the T1 triangle, which contains the circle.
Hence a requested ellipsoid does not exist. Regards, CiaPan (talk) 09:54, 22 March 2017 (UTC)
Nice proof :) Gandalf61 (talk) 10:14, 22 March 2017 (UTC)
THX :) I have just realized the central symmetry doesn't suffice and added a note about planar symmetries. --CiaPan (talk) 10:25, 22 March 2017 (UTC)
There is an ellipsoid if you allow oblique axes. Take the ellipsoid to be 12(x-y)2+x2+y2+z2=15. The form on the lhs is positive definite so this is an ellipsoid and not a hyperboloid or some other conic surface, and it's easy to verify that the cube is contained in it. The point
(1/3+1/√3, 1/3-1/√3, 1/3) lies on the circle in question; it's easy to check that it satisfies the equations x+y+z=1 and x2+y2+z2=1. But for this point x-y=2/√3, so (x-y)2=4/3, 12(x-y)2+x2+y2+z2=17>15 so the point is outside the ellipsoid. --RDBury (talk) 04:09, 24 March 2017 (UTC)

Thank you for your answers! BTW, actually, I guessed that such an ellipsoid does exist. 09:19, 24 March 2017 (UTC) — Preceding unsigned comment added by 212.199.149.142 (talk)

The similarity to my previous question, raises another question:

Is that true that whenever n-dimensional ellipsoid E contains a cube B and centered at a vertex of B, then E also contains the minimal (w.r.t containment) sphere S that contains B? 31.154.81.43 (talk) 13:00, 24 March 2017 (UTC)

Again, I believe the answer is no. In general, let P be any finite set of points. Then the intersection of the ellipsoids centered at the origin containing P is the convex hull of P∪−P. To see this, let L(x)≤1 be a plane bounding the convex hull. Then L(x)2≤1 also bounds the convex hull. L(x)2=1 is not an ellipsoid but you can tweak it by a small amount to make it one which still bounds the convex hull. For example if P lies within the sphere ||x||<M then L(x)2+ε||x||2=1+εM2 is an ellipsoid which contains the points. As ε→0 the ellipsoids approach the pair of planes L(x)2=1, so any point not between those planes will be outside the ellipsoids with ε smaller than some value.
So the fact that you're throwing in ellipsoids is really a red herring, you just need to find the convex hull of the points P∪−P. If it does not contain the sphere (or any set) then there is an ellipsoid which does not contain the sphere. --RDBury (talk) 15:34, 24 March 2017 (UTC)
Could you please write down an equation for such an ellipsoid explicitly? 31.154.81.43 (talk) 07:41, 25 March 2017 (UTC)

# March 23

## Mathematical equations do not display correctly

The image at right is a screen capture of part of the article Cauchy–Schwarz inequality. Notice that the last few symbols of the equations are not displayed. Can someone help me correct this problem? Thanks very much for your help!--老陳 (talk) 05:32, 23 March 2017 (UTC)

It displays correctly on my ipad screen. Loraof (talk) 15:34, 23 March 2017 (UTC)
I am using Windows 10 with Microsoft Edge browser on my PC.--老陳 (talk) 23:29, 23 March 2017 (UTC)
I just tried using that setup, and it renders fine for me. Don't know what to tell you. Deacon Vorbis (talk) 04:54, 24 March 2017 (UTC)
Is here the right place for this kind of questions?--老陳 (talk) 05:13, 24 March 2017 (UTC)
The question has the word "Mathematical" in it. I guess it is related to the expression of Mathematical concepts. 148.182.26.69 (talk) 05:22, 24 March 2017 (UTC)
@老陳: Per WP:?, questions about Wikipedia editing should be placed at Wikipedia:Help Desk (or a more specialized venue if you know where to go); reference desks are for questions about the world outside Wikipedia edition. For that particular question, the best venue is either Wikipedia:Reference desk/Computing if the problem is on your end (as it seems to be), i.e. your computer does not display the equation, or the help desk if the problem is that the Wiki-code on the page is incorrect and causes display problems; it is not a big deal either way, I think. TigraanClick here to contact me 13:08, 24 March 2017 (UTC)
@老陳: I have no idea what the reason might be, but try going to the Preferences page Appearance tab (Special:Preferences#mw-prefsection-rendering) and choose 'PNG images' if you use 'MathML'. --CiaPan (talk) 14:25, 24 March 2017 (UTC)
@CiaPan:I tried to change it and the display problem went away. So, I think there's a bug in the 'MathML' option. Thanks for the help.--老陳 (talk) 00:00, 25 March 2017 (UTC)

## Reflection formulas for the Gamma function

We have

{\displaystyle {\begin{aligned}&\Gamma {\Big (}1~+~x{\Big )}~\cdot ~\Gamma {\Big (}1~-~x{\Big )}~=~{\frac {\pi x}{\sin \pi x}}\\\\&\Gamma \left({\tfrac {1}{2}}+x\right)~\cdot ~\Gamma \left({\tfrac {1}{2}}-x\right)~=~{\frac {\pi }{\cos \pi x}}\\&\Gamma \left({\tfrac {1}{4}}+x\right){\bigg /}\Gamma \left({\tfrac {1}{4}}-x\right)~=~\pi ^{2x}\cdot {\frac {\zeta \left({\tfrac {1}{2}}-2x\right)}{\zeta \left({\tfrac {1}{2}}+2x\right)}}\end{aligned}}}

where the latter is nothing else than the functional equation of the Riemann zeta function. I was wondering whether yet another similar relation might also exist, involving ${\displaystyle \Gamma \left({\tfrac {1}{8}}\pm x\right)}$ or ${\displaystyle \Gamma \left({\tfrac {1}{3}}\pm x\right).}$79.113.251.53 (talk) 11:12, 23 March 2017 (UTC)