Wikipedia:Reference desk/Archives/Mathematics/2012 February 9

= February 9 =

Unique Factorization Domain
Are Z[sqrt(2)], Z[cuberoot(2)], and Z[sqrt(2),cuberoot(2)] unique factorization domains? Black Carrot (talk) 06:43, 9 February 2012 (UTC)


 * No. For instance, $$119=7\times 17=(11-\sqrt{2})(11+\sqrt{2})$$ and $$123=(5-\sqrt[3]{2})(5^2+5\sqrt[3]{2}+(\sqrt[3]{2})^2)=3\times 41$$.   Sławomir Biały  (talk) 12:23, 9 February 2012 (UTC)
 * But don't $$7=(3-\sqrt 2)(3+\sqrt 2)$$ and $$3=(1+\sqrt[3]{2})(1-\sqrt[3]{2}+(\sqrt[3]{2})^2)$$?
 * Plus fixed a minus to a + in your bit above hope okay Sławomir. Dmcq (talk) 12:42, 9 February 2012 (UTC)
 * Sorry, I was just being stupid. The class number of $$\mathbb{Q}(\sqrt{2})$$ and $$\mathbb{Q}(\sqrt[2]{2})$$ is one.  So both of rings of integers are UFDs.   Sławomir Biały  (talk) 13:17, 9 February 2012 (UTC)
 * How do you calculate the class number? Black Carrot (talk) 00:03, 10 February 2012 (UTC)
 * Well there's articles Ideal class group and class number formula here but if you really want to get into all that I can warn you that entire books have been written on the subject. Dmcq (talk) 09:33, 10 February 2012 (UTC)
 * The first, Z[sqrt(2)] (see Quadratic integer) is an Integrally closed domain (necessary for it to be a UFD, you have to remember to check the Z [?] guy is the whole ring of integers of the number field, as indicated above), and is even a Euclidean domain, using the norm, and hence a UFD, so everything is as nice & tangible as you could want there.John Z (talk) 11:08, 12 February 2012 (UTC)

Offsetting an ellipse
It's been a long time since I thought much about geometry, so bear with me, I'm sure this is rather elementary.

I am drawing an ellipse using a rather straightforward function that takes x,y as the center, two radius lengths (a and b), and a rotational angle as its arguments. The ellipse draws perfectly fine under such conditions. For a rotational angle of 0, a ends up being the horizontal radius and b ends up being the vertical radius, like this:



The function spits out a bunch of coordinates (the number depends on how many steps there are, but that isn't very important for this question) that I can then plot on a grid.

Here's the thing. I need to adjust the plotting of the ellipse so that x,y isn't the center, but coincides with one of the edges of the ellipse. So in the above example, I would shift the ellipse over so that what is now -a would be at point C. No problem for a single axis, of course (just offset every x coordinate by a), but how do I do this for arbitrary angles, keeping the -a point in the above ellipse centered on x,y? It smells like trigonometry to me and I'm very rusty on that.

Another way to do this would be to plot the ellipse coordinates at 0 degrees rotation, then rotate the ellipse with the rotational point being one edge of it. But I've no idea how to do those sorts of transformations on a bunch of x,y coordinates.

Any suggestions? --Mr.98 (talk) 17:06, 9 February 2012 (UTC)


 * As to rotating points, the general (computer guy, at least) thing is to apply a rotation matrix to each point. Slightly more generally, affine transformations (shears, translations, dilatations, and rotations) can be composed ad-nauseum by taking their corresponding transformation matrices and multiplying them together (in order of application).  When applied to a point, the resulting matrix applies all those transformations in one go. So for example if you wanted to rotate something by 0.1 radians about (3,-10). First you'd translate to move the coordinate system so the centre of rotation is at the origin) - so you'd apply the translation matrix for x-3, y+10.  Then you'd apply the rotation matrix. And then you'd translate back to where you came from (x+3, y-10). -- Finlay McWalterჷTalk 20:11, 9 February 2012 (UTC)
 * To be a bit more specific, let's say your basic ellipse is given by
 * $$x=a \cos t,\,y=b \sin t,$$
 * then the rotation counterclockwise by θ is
 * $$x=a \cos \theta \cos t - b \sin \theta \sin t,\,y=a \sin \theta \cos t+ b \cos \theta \sin t$$
 * and the translation so the center is at (h, k) is
 * $$x=h+a \cos \theta \cos t - b \sin \theta \sin t,\,y=k+a \sin \theta \cos t+ b \cos \theta \sin t.$$
 * --RDBury (talk) 21:57, 9 February 2012 (UTC)


 * (ec) I remember you do VB, so this may be helpful. Here's R (programming language) code that does what you want:


 * If you want to try it out in R, create the function using "fix(elip)", and substitute the empty function with the code above. To see the result, after defining the function type


 * --NorwegianBluetalk 22:48, 9 February 2012 (UTC)


 * Thanks! I will puzzle over this a bit further. --Mr.98 (talk) 20:55, 10 February 2012 (UTC)

Stationary functions of a functional
What kind of techniques would a mathematician employ when looking for the functions which minimise or maximise the functional given by $$ \mathcal{F}[f(x)]= \int_0^1 x^2 f(x) \text{d}x - \left( \int_0^1 x f(x) \text{d}x \right)^2 $$ subject to a constraint of the form $$ \int_0^1 f(x) \text{d}x = \text{constant} $$ Thanks for your help — Preceding unsigned comment added by 129.67.37.224 (talk) 23:57, 9 February 2012 (UTC)


 * In general, the techniques for functional optimization problems are those of the calculus of variations. Here you have a quadratic functional on an affine hyperspace, which is unbounded both above and below. Consider $$f=c + th$$ where $$c\in\R$$ is the given constant for the constraint, $$\scriptstyle t\in\R$$ and $$\scriptstyle \int_I h= 0$$. Then $$t\mapsto \mathcal{F}[c+th]$$ is polynomial in $$t$$:


 * $$\mathcal{F}[c+th]= \big(c/3 - c^2/4\big) + \bigg( \int_Ix^2 h -c\int_I xh\bigg)t - \bigg(\int_I xh\bigg)^2 t^2.

$$
 * Now, if for instance we fix an $$h$$ with $$\scriptstyle \int_I h = \int_I x h= 0$$ and $$\scriptstyle \int_I x^2 h \neq0$$, e.g. $$h=x^2-x+1/6$$, the above quantity is a non-constant affine function of $$t$$, hence unbounded from above and from below (homework?)--pm a 08:50, 11 February 2012 (UTC)