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# April 18

## Bayes' Theorem

The wiki article about it has an illustration with a decision tree that I couldn't understand. Is the following an ok way to think of it and explain it?

1. Think of a big square with area 1 representing a probability space. Draw two filled circles A and B (maybe unequal sizes) inside the square for the events A and B, so they partly overlap like a Venn diagram. Their areas are the probabilities Pr[A] and Pr[B]. Let C be the intersection of A and B.
2. Pr[A|B] just means you've picked a point in B and now you want to know the likelihood that it's also in A. That is, it's in the intersection C. So this likelihood is Pr[C]/Pr[B]. Similarly Pr[B|A] is Pr[C]/Pr[A]. Rearranging terms you get Pr[A|B]=Pr[B|A]*Pr[A]/Pr[B] which is Bayes' Theorem.

If yes, is this somehow a wonderful or amazing fact? Why is it usually written as that equation, instead of thinking of intersecting events like the circles above? Thanks! 50.0.136.56 (talk) 03:52, 18 April 2017 (UTC)

Let's add a link to the article: Bayes' theorem. StuRat (talk) 04:32, 18 April 2017 (UTC)
Thanks. There's also a separate article Bayes' rule whose purpose I couldn't discern. 50.0.136.56 (talk) 04:37, 18 April 2017 (UTC)
That's just a redirect to the same article. --76.71.6.254 (talk) 10:53, 18 April 2017 (UTC)
That was not a redirect until today's edit by Trovatore (special:diff/775983852). --CiaPan (talk) 12:44, 18 April 2017 (UTC)
It's possible I jumped the gun a little on that one, but I figured it was easy to undo if someone didn't like it. Looking on the talk page, apparently there's some concern that the rule might be useful even in situations where the hypotheses of the theorem are not strictly met. It seems to me that that clarification could be made within a single article, but it may be that it's content that ought to be merged. --Trovatore (talk) 18:12, 18 April 2017 (UTC)
Oh. You might've mentioned it here, since it was being discussed. Oh well, not important. --76.71.6.254 (talk) —Preceding undated comment added 18:44, 18 April 2017 (UTC)

# April 20

## Rubik's Revenge all checkerboard

Is there a way to make all sides of a solved Rubik's Revenge cube be checkerboard? If you could, please show how with this style of notation: example r ² B ² U ² l U ² r ' U ² r U ² F ² r F ² l ' B ² r ². Many thanks! Anna Frodesiak (talk) 06:58, 20 April 2017 (UTC)

Not certain this will work, but you could try 180 degree rotations of alternate slices in each dimension i.e. 180 degree rotation of columns 1 and 3, rows 1 and 3 and layers 1 and 3. Gandalf61 (talk) 09:22, 20 April 2017 (UTC)
Hi Gandalf61. You are wise, but that only makes four out of six sides checkerboard. Anna Frodesiak (talk) 10:46, 20 April 2017 (UTC)
You could achieve your checkerboard if you could "swap front left middle with rear right middle". Achieving this swap requires some form of parity/entropy to be maintained and if that could be done as a matching "swap front right middle with rear left middle", then you have done the swaps for one of the three axes. Do it three times on the various axes. My guess is that it is doable, but I don't know how. -- SGBailey (talk) 14:02, 20 April 2017 (UTC)
Hi SGBailey. I just can't get my head around that. I'm so sorry. Anna Frodesiak (talk) 18:40, 20 April 2017 (UTC)
I hadn't realized you meant a 4*4*4 cube. -- SGBailey (talk) 06:45, 21 April 2017 (UTC)

An odd thing is that with a normal 3x3x3 Rubik's cube, one can easily checkerboard the whole thing. For some reason this doesn't work with Rubik's Revenge (4x4x4). It should because once the edges and center 4 are matched in colour, the 4x4x4 essentially becomes a 3x3x3. Mystifying! Anna Frodesiak (talk) 18:40, 20 April 2017 (UTC)

I watched through the following Youtube video https://www.youtube.com/watch?v=lnZWBq3C4Rs and given that someone who appears to be a very experienced cuber (or at least owns more types of these cubes that I'd ever seen before (up to 7x7x7, 2x3x3, 2x2x3,1x3x3 and a gear cube) and when he makes checkerboards for the even cubes it is only on four sides, I doubt it is possible. (Certainly not proof though)Naraht (talk) 19:11, 20 April 2017 (UTC)

It's not possible for any even x even x even, the reason being that the cubes aren't painted that way. So you can't even do by disassembling the cube and putting it back together.
Suppose you could. Then looking at the corner pattern from the top it might look like.
  G   B
O Y   W R

R W   Y O
B   G

But now you have RWB in the upper right as well as the lower left, putting it in two places at once. --RDBury (talk) 22:39, 20 April 2017 (UTC)
PS. you could do it by disassembling four cubes and and reassembling them to create four different checkerboards. --RDBury (talk) 22:45, 20 April 2017 (UTC)
I think RDB is right, but it requires a more elaborate proof. The above assumes a particular way of constructing the checkerboard, but there are other ways to consider. For example, in the above diagram one of the blue/green faces might have the two colors the other way around. Or the two colors checkerboarding a side might not be colors that were on opposite faces of the solved cube. You have to prove that any such solution is impossible. --76.71.6.254 (talk) 23:14, 20 April 2017 (UTC)
You're right and I had a more detailed proof, but then I mistakenly thought it was more than needed. I think though, if you start with RWB in the lower left corner and try to fill in the remaining corners using a checkerboard pattern, the result would have to look like what's shown. The other possibilities you mention can only occur if you change the orientation of some of the corners. It is possible to disassemble a cube and a mirror image cube and put them together to form two checkerboards in three ways. --RDBury (talk) 01:53, 21 April 2017 (UTC)

I am still having trouble understanding, but the bottom line is what matters: I cannot be done. Fair enough. Many thanks to all for the thoughtful feedback. I am grateful. Best wishes and happy cubing! Anna Frodesiak (talk) 06:40, 21 April 2017 (UTC)

# April 21

## Projections in Hilbert spaces

For each closed subspace K of a Hilbert space H, let ${\displaystyle P_{K}}$ be the self-adjoint projection onto K. Suppose I have a sequence ${\displaystyle H_{0}\subset H_{1}\subset \dots \subset H}$ of closed subsets of H, and let K be the closure of the union of these subspaces. In what operator topology (if any) does the sequence of projection operators ${\displaystyle P_{H_{n}}}$ tend to the projection operator ${\displaystyle P_{K}}$? 66.66.228.95 (talk) 00:01, 21 April 2017 (UTC)

# April 23

## Coiled Phone Cord

I have tried this experiment repeatedly. If I pick up the handset of a landline phone cord, make a call, and then hang up, I invariably find the cord has become tangled by several loops. Even if I let the cord hang free, so that it lies untangled, after the next call, it will have multiple loops in it. At most I might change hands once or twice, but that does not in my mind account for it then having six or more loops in it when I hang up. Any material that addresses this? Thanks. μηδείς (talk) 02:43, 23 April 2017 (UTC)

## Classifying n-gons

I could have asked this question at any time, but it came to my attention when I saw recent edits to the polygons template.

Triangles can be classified as:

1. Equilateral = 3 lines of symmetry, all sides are equal and all angles are 60 degrees.
2. Isosceles = 1 line of symmetry, 2 sides are equal and one side is different; 2 angles are equal and one angle (the angle formed by the legs) is different.
3. Scalene = no lines of symmetry, no congruent sides and no congruent angles.

To generalize these 3 classifications of triangles into n-gons for n > 3, we can do so as follows:

The generalization of the equilateral triangle is clearly the regular polygon. This is the square for n = 4.

But how about generalizing the isosceles triangle?? An isosceles polygon (a generalization of an isosceles triangle) would be a polygon that is not regular but that has at least one line of symmetry. The non-square rectangle, rhombus, kite, and isosceles trapezoid are all examples of isosceles quadrilaterals.

Likewise, a scalene polygon is a polygon with no lines of symmetry. I don't know whether to categorize a polygon with no lines of symmetry but rotational symmetry (a parallelogram that's not a rectangle or a rhombus) is correctly classified as scalene.

These categories can continue for n-gons for any value n. Do isosceles polygons in general have special properties?? How about scalene polygons in general?? Georgia guy (talk) 12:09, 23 April 2017 (UTC)

Properties of scalene polygons in general would be properties of all polygons. These would be in the article Polygon. As for all polygons with a line of symmetry, you could look at Isosceles trapezoid or Rhombus, and see if any of the properties there generalize. Loraof (talk) 18:59, 23 April 2017 (UTC)

# April 24

## derivative of constrained function is sum of unconstrained partials?

Hello, while studying neural networks I came upon what was described by the lecturer as a "math trick" to solve a particular type of optimization problem using gradient descent. Basically, when optimizing a neural net that requires two parameters to be equal, you can replace the partial derivative for each constrained parameter by the sum of the partials with respect to each constrained parameter. So if you have ${\displaystyle f(x_{1},x_{2})}$, then the derivative of ${\displaystyle f(x,x)}$ with respect to ${\displaystyle x}$ is the same as the sum of the partial derivatives of ${\displaystyle f(x_{1},x_{2})}$ with respect to ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$. So far I have not been able to find a counter example, but I also do not know how to prove it. If anyone has pointers, clues, or proofs please help! Sorry if its obvious, and thanks! Brusegadi (talk) 01:09, 24 April 2017 (UTC)

It's a particular case of the multivariable chain rule; it's easier to see what's going on from the more general form ${\displaystyle {\frac {d}{dt}}(f(x(t),y(t))={\frac {\partial f}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}}$. --JBL (talk) 01:52, 24 April 2017 (UTC)
Thank you so much! Brusegadi (talk) 02:08, 24 April 2017 (UTC)

## Johnson's SU-distribution

Is Johnson's SU-distribution and "shepherd's crook" and Johnson Curve all the same thing? Thanks. Anna Frodesiak (talk) 06:25, 24 April 2017 (UTC)

I want to know because of [3] and [4] that I did. Anna Frodesiak (talk) 10:24, 24 April 2017 (UTC)

## Face value numbers

When we count, for example, coins or banknotes by their face value rather than actual quantity expressed by natural numbers, are those still natural numbers or some other kind? Thanks.--212.180.235.46 (talk) 16:59, 24 April 2017 (UTC)

Natural numbers don't include decimals, and most currencies have a sub-denomination (like cents for dollars), which makes it a decimal number. Yen is one that doesn't, so, for that case, I suppose you could use natural numbers for prices, in most cases (with exceptions for buying in quantity, where they might break the price down by tenths of a yen, etc.) StuRat (talk) 17:38, 24 April 2017 (UTC)