Wikipedia:Reference desk/Archives/Mathematics/2011 May 15

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May 15

Ellipse

The equation $ax^{2}+bxy+cy^{2}=1$ describes an ellipse, but it is not a standard ellipse because the ellipse's axes are not necessarily parallel to the x and y axes, i.e. it has been rotated. How do you read the angle of rotation from the equation? Widener (talk) 03:13, 15 May 2011 (UTC)

If the major axis forms an angle of $\theta$ with the x-axis, then θ minimizes the value of $ax^{2}+bxy+cy^{2}$ with the substitution $(x,y)=(\cos \theta ,\sin \theta )$. Substituting and differentiating gives $(c-a)\sin(2\theta )+b\cos(2\theta )=0$ which means $\theta ={\tfrac {1}{2}}\tan ^{-1}\left({\frac {b}{a-c}}\right)$. This gives the result up to a multiple of $\pi /2$. -- Meni Rosenfeld (talk) 08:17, 15 May 2011 (UTC)

Did this reply help you? -- Meni Rosenfeld (talk) 15:56, 16 May 2011 (UTC)

Yes. Why wouldn't it? You explained the answer to my question. Widener (talk) 11:48, 17 May 2011 (UTC)

It's common that even an answer as straightforward as this leaves an OP with something to be desired. I don't presume to divine what an OP would think upon seeing an answer of mine. I appreciate having explicit closure for the exchange. -- Meni Rosenfeld (talk) 10:01, 18 May 2011 (UTC)

Multiplication of Cardinals and Order Preservation.

See Multiplication#Properties. If a, b, and c are cardinal numbers, does the following still hold?

" Multiplication by a positive number preserves order: if a > 0, then if b > c then ab > ac. Multiplication by a negative number reverses order: if a < 0 and b > c then ab < ac."
Thanks in advance. voidnature 08:24, 15 May 2011 (UTC)

[ec]Are you referring to cardinal numbers? I don't know of a way to multiply a cardinal number with a negative number. For multiplication by a cardinal number a, this will hold if either $a<b$ or b is finite. -- Meni Rosenfeld (talk) 08:34, 15 May 2011 (UTC)

Yes, I am referring to cardinal numbers. My question is actually concentrated on this bit :"if a > 0, then if b > c then ab > ac". Thankyou. voidnature 08:36, 15 May 2011 (UTC)

Ok, so you need either $a<b$ or b finite. Otherwise, a counterexample is $a=\aleph _{0},b=\aleph _{0},c=1$. However, it will always be true that $ab\geq ac$. I think the axiom of choice might be needed for some of these results. -- Meni Rosenfeld (talk) 08:49, 15 May 2011 (UTC)

Oops, sorry, c has to be zero. So I don't need to worry about $a=\aleph _{0},b=\aleph _{0}$. So my question should be If a and b are cardinal numbers and c is 0, does "if a > 0, then if b > c then ab > ac" hold? Thankyou. voidnature 08:58, 15 May 2011 (UTC)

Every cardinal multiplied by 0 is 0, so you're basically asking, "if a>0 and b>0 are cardinal numbers, is ab>0?" The answer is yes, because the cartesian product of two nonempty sets is nonempty. -- Meni Rosenfeld (talk) 09:05, 15 May 2011 (UTC)

"the cartesian product of two nonempty sets is nonempty": can you give a proof please? voidnature 09:09, 15 May 2011 (UTC)

$a\in A,\ b\in B\Rightarrow (a,b)\in A\times B$. -- Meni Rosenfeld (talk) 10:17, 15 May 2011 (UTC)

Why is S5 a modal companion of CPC?

Why is S5 a modal companion of CPC? It seems like this should imply that S5 implies the translation of excluded middle, which seems to be $\Box p\lor \Box \lnot p$, which seems to say there are no contingent propositions - but surely S5 allows for contingent propositions? 88.104.173.35 (talk) 19:27, 15 May 2011 (UTC)

Plane partitioning algorithm

Resolved

Oh, this is on the tip of my tongue: I hate when that happens. Begins with "L", I think... I'm trying to remember the name of an algorithm which takes an array of points on a plane and partitions the plane such that each point is surrounded by a polygon - I think the margins of which fall equidistantly with another point (or is it some other definition of "influence"?). The resulting diagram looks like a honeycomb made by tipsy bees. What is that algorithm? (it's not a BSP or its ilk)-- Finlay McWalter ☻ Talk 22:22, 15 May 2011 (UTC)

Voronoi diagram perhaps?--RDBury (talk) 23:39, 15 May 2011 (UTC)

Yes, that's it (and no L in sight)! Thanks. -- Finlay McWalter ☻ Talk 23:45, 15 May 2011 (UTC)

“L, I KNOW it begins with L!” – b_jonas 18:46, 16 May 2011 (UTC)