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March 1

Function wanted, sigmoids on a unit square

I'd like a function $f:(0,1)^{2}\to (0,1)^{2}$ . I have two sigmoid functions $sig_{j}:(0,1)\to (0,1)$ . I need to have $f(x,y)\geq sig_{1}(x)$ , and the idea is that f should combine these two sigma functions in a reasonable way that also has an interaction parameter.

What I'm currently using is $f(x,y)=(1-sig_{1}(x))\cdot sig_{2}(y)+sig_{1}(x)$ . This is fine, but there's very little interaction (meaning that that the size of f is mostly just determined by the larger input), and I can't control that interaction. Here's an example of what I have [1], and I'd like a way to basically pull that top right corner of the right panel toward the center.

Question: how can I write f in terms of sig_j so that 1) the end result is broadly sigmoidal for any (horizontal, vertical, or radial) slice, 2) I preserve the inequality condition above, and 3) I'm also able to increase or decrease the interaction strength with some additional parameter $I\in (0,1)\;\mathrm {or} \;I\in \mathbb {R_{+}}$ ? If f(x,y) can be written as f(sig(x),y) as my example above, that would be nice, but not necessary. Thanks, SemanticMantis (talk) 19:42, 1 March 2016 (UTC)[reply]

Here's something I've been playing with. You're given σ₁ and σ₂, increasing on [0, 1] with σ₁(0) = σ₂(0) = 0, σ₁(1) = σ₂(1) = 1. Define

f_{k}(x,y)=1-(1-\sigma _{1}(x))(1-\sigma _{2}(y))(1-\sigma _{1}(x)\sigma _{2}(y))^{k}

where k≥0. For k=0 you get the function f defined above. Also

f_k(0,y)=σ₂(y), f_k(x,0)=σ₁(x)

f_k(1,y)≡1, f_k(x,1)≡1

f is increasing in both the x and y directions.

I not sure how you define interaction or sigmoidalness but this family has at least some of the properties you're interested in. --RDBury (talk) 19:32, 2 March 2016 (UTC)[reply]

@RDBury: Thanks! That will do nicely; I think I can adapt it to my needs. I probably should have seen that family myself, but I got distracted by things that were too simple or too complicated :) Still curious to hear any other suggestions, or any general references on ways to make functions on squares out of functions on intervals. SemanticMantis (talk) 20:26, 2 March 2016 (UTC)[reply]

Expressing how different two quantities are

I came across this video on YouTube and the person describes a situation where two data entries are supposed to be the same but aren't. He tries to describe how different the two quantities are by taking the difference and by taking the %diff. But he points out that the %diff is not symmetrical. Is there a "clean" way of showing how different two quantities are in a symmetrical fashion? ScienceApe (talk) 20:29, 1 March 2016 (UTC)[reply]

You can take the absolute value of the difference divided by the midpoint (average) of the two values. This is typically done in arc elasticity calculations in economics--see Arc elasticity#Formula. Loraof (talk) 20:58, 1 March 2016 (UTC)[reply]

There is no generic good answer to a question like this, as it depends on assumptions about your data. For example, Loraof's suggestion is very reasonable for data that you think of as roughly linear and that only takes positive values; but it works very poorly if your data includes positive and negative values. --JBL (talk) 21:30, 1 March 2016 (UTC)[reply]

Right, it's clean and symmetric, no guarantee of it being "right", "best", or even "reasonable" for a specific application. For example String_metric lists many choices for ways to say how different two strings are. Metric_(mathematics) doesn't have a nice section that lists examples for notions of distance on the real numbers, I suppose because there are just too many. SemanticMantis (talk) 22:31, 1 March 2016 (UTC)[reply]

log

Many possibilities here, the best one depends on the application. If you like the %diff but want it symmetric, you can try

|\log(x/y)|

. Equivalent to %diff for small differences, symmetrical, and gives meaningful results for things which differ by orders of magnitude. -- Meni Rosenfeld (talk) 00:06, 2 March 2016 (UTC)[reply]

To make it equivalent to %diff you need (1) to use the natural logarithm and (2) to multiply by 100. That means that you use logarithm with base e^0.01, approximately 1.01. I am not aware of a conventional name for the unit of logarithm with base e^0.01, so I suggest the name degree and the symbol °. Increasing by 1° is nearly the same thing as increasing by 1%. Increasing by 69° is doubling, and 230° is tenfold. A rate of interest should be measured in degrees per year, rather than in percent per year, because debt increases exponentially, rather than linearly, when subject to interest. Bo Jacoby (talk) 22:37, 2 March 2016 (UTC).[reply]

Of course it's the natural logarithm. As for multiplying by 100 - no, that's not needed. The %diff between 101 and 100 is 1%, or in other words, 0.01.

|\log(101/100)|\approx 0.00995\approx 0.01

, the same.

As for the introduction of new notation... I'm sure we've had this argument before. Your suggestion makes no sense to me. -- Meni Rosenfeld (talk) 23:43, 2 March 2016 (UTC)[reply]

Thanks. You get 1% by multiplying 0.01 by 100%. How do you express a rate of interest of say 8° per year? Bo Jacoby (talk) 07:53, 3 March 2016 (UTC).[reply]

I would simply say "interest of 8% per year, compounded continuously".

The "8% per year" is the number, and "compounded continuously" is what I do with this number. From the latter part I know that the total worth is

e^{at}

after time t. From the former part I know that

a={\frac {0.08}{\mathrm {yr} }}

.

Compare "interest of 8% per year, compounded monthly". Here the same number appears, but I'm using it differently - the formula is now

\left(1+a\cdot \mathrm {mo} \right)^{t/\mathrm {mo} }

. Likewise for any other interest rate and compounding period.

But it doesn't make sense to have a single piece of notation which tells me both the number and which formula to use with it.

Of course, it so happens that "69.3% per year continuously" is the same as "100% per year annually". But the relation is not linear, it's a function applied to the number rather than a part of the number itself. -- Meni Rosenfeld (talk) 00:01, 4 March 2016 (UTC)[reply]

Thanks. The new notation "69.3% per year continuously is the same as 100% per year annually" is very awkward. Unit names indicate both concepts and units. 1 foot = 30 centimeter is a length. 1 doubling = 69° is a logarithm. Bo Jacoby (talk) 08:24, 4 March 2016 (UTC).[reply]

Units are only multiplicative factors. 1 foot = 30 cm and 2 feet = 60 cm (roughly, a foot is actually a bit more), so it works out. But logarithms/compounding is not linear so you can't use it as a unit. 69.3%/yr continuously is equivalent to 100%/yr annually, but 138.6%/yr continuously is not the same as 200%/yr annualy.
It's not a "new notation", it's a well established notation in common use among finance people. I had half a mind to call it by a different name to be more mathematical and less finance, but it wouldn't be different in the concepts so I just stuck with the original.
It's not awkward at all. It's just two things that are equivalent. In general, an interest of a compounded continuously is equivalent to an interest of $(\exp(a\ \mathrm {yr} )-1)/\mathrm {yr}$ compounded annually. -- Meni Rosenfeld (talk) 10:13, 4 March 2016 (UTC)[reply]

Debts grow exponentially, and so the logarithm of a debt grows linearly, and so a unit of logarithm per time is the proper unit for rate of interest. 8° for one year gives 16° for two years, but 8% for one year gives 16.64% for two years, so the unit %/yr for interest rates is misleading and dishonest.
"per year" and "annually" means the same thing.
Claiming that "69.3% per year is the same as 100% per year" cannot be helped being confusing.

Bo Jacoby (talk) 21:12, 4 March 2016 (UTC).[reply]

I'm sorry but I disagree on all counts.

For continuously compounded interest, the total worth after time $t$ is $W_{0}\cdot \exp(at)$ , where $W_{0}$ is the initial worth. The formula has a rate parameter a. The argument of the exponential function is a pure number, so $at$ is a pure number. t has dimension [T] (time) so a must have dimension 1/[T]. So it can be, for example, 0.24 / yr. Since 1 yr = 12 mo, you have 0.24 / yr = 0.02 / mo. The "exponentialness" comes from the function in which this parameter is used, not from the parameter itself.
When I said "100% per year annually" I meant "100% per year compounded annually". I thought it should have been clear from context I merely omitted a word for brevity. As I explained at length, different compounding periods lead to different functions. To have a complete picture you need to know both the function and the parameter. The "100% per year" describes the parameter, the "compounded annually" (or "annually" for short) describes the function. In contrast, "100% per year compounded continuously" uses a different function and thus has different behavior. "100% per year compounded monthly" has different behavior still, and so on.
Why would it be confusing? Is it confusing that a radius of 2 is equivalent to a diameter of 4? Of course not. If you're describing a quantity, you need to specify both the quantity you are referring to (radius or diameter) and what is its numerical value. Using different quantities leads to different descriptions of the same thing. Likewise, "69.3% per year compounded continuously" is equivalent to "100% per year compounded annu∓ally". You can't mention a quantity like "100% per year" and be done without mentioning what kind of compounding takes place. Of course, it's possible that the compounding mechanism can be inferred from context.
It doesn't make sense for logarithms to have different units from other numbers. Units only make sense for homogeneous quantities. For example, you could say area has a different unit than length, since area is homogeneous of degree 2 in length. But exp and log are a power series and are not homogeneous, thus they must accept and return pure numbers. -- Meni Rosenfeld (talk) 23:52, 5 March 2016 (UTC)[reply]

Let me focus on your objection against giving logarithms units. 10 is said to be an order of magnitude bigger than one. 10ⁿ is n orders of magnitude bigger than one, and n=log₁₀(10ⁿ). So order of magnitude is the unit of the base 10 logarithm.

Multiplication by 2 is doubling. So doubling is the unit of the base 2 logarithm. 10 doublings is about 3 orders of magnitude because 2¹⁰≈10³.

The factor 100^−1/5 is called a magnitude when estimating the brightness of stars. So five magnitudes corresponds precisely to a factor of 0.01 in brightness. Magnitude is the unit of the base 100^−1/5 logarithm.

A doubling of frequency is called an octave. A semitone is the unit of base 2^1/12 logarithms, so 12 semitones is one octave.

Special 5 dimensional Antiprism...

To have create an N+1 dimensional Anti-prism, you need two copies of the same N dimensional polytope an arrange the two so that the vertices of one line up with the middle of the N-1 space on the other. These would fall into 3 categories:

The standard 3-dimensional anti-prisms, formed from 2 polygons
The antiprisms formed from two N dimentional simplexes, which becomes the N+1 dimensional cross polytope. (Two Tetrahedra become the 8 cell)
The antiprism formed from two 24-cells. This would have 2 24 cells and 48 5 cells.

This last seems interesting, but I can't find it on Wikipedia. It does have a high level of Regularity (all vertices, edges and faces are the same, but all cells aren't). Is there an article for it?Naraht (talk) 20:30, 1 March 2016 (UTC)[reply]

I could be wrong but since the faces of the 24-cell are octahedra, the "side" faces would be octahedral pyramids, not 5-cells. (Here I'm calling a pyramid formed in 4 dimensions using an octahedron as a base an octahedral pyramid.) WP does have an article Uniform 5-polytope but if I'm right about the side faces then the definition would exclude the figure you're talking about. I'm not convinced the figure is edge and face transitive since there would be some edges/faces that are part of the two 24-cells and some that are not. --RDBury (talk) 22:18, 1 March 2016 (UTC)[reply]

Just a few more computations: The figure has vertex transitive symmetry but the two 24-cells form a system of two blocks for the permutation group. A 2 element set with both elements in the same block, and a two element set with elements in different blocks must fall in different orbits under the action of the group on pairs of elements. This implies that the figure is not edge transitive. Similarly it's not transitive on the dimension 2 faces, even though all these faces are triangles. There are 48 vertices total. There are 192 edges which are part of a 24-cell (96 in each), and 144 edges going from one 24-cell to the other, making a total of 336. For dimension 2 faces, there are 192 in the 24-cells and 576 with a vertex in one 24-cell and 2 in the other. This makes 768 triangles altogether. For dimension 3 faces there are 48 octahedra in the 24-cells, 384 tetrahedra with a vertex in one 24-cell and three in the other, and 288 tetrahedra with two vertices in with 24-cell, 720 total. For dimension 4 faces there are the 2 24-cells, 48 octahedral pyramids as mentioned above, and also 192 5-cells with two vertices in one 24-cell and three in the other. This makes a total of 242. Checking Euler's formula you get 1-48+336-768+720-242+1=0 as expected. If the 24-cells have edge length 2 then they would be distance 2√(√2-1)≈1.287 apart to make all the crossing edges have edge 2 as well. --RDBury (talk) 05:20, 2 March 2016 (UTC)[reply]

You are right, Octahedral Pyramids, not 5 cells for the connections between the two 24-cells. (with one vertex in one and a 3-d figure in the other). I hadn't considered the triangle -line connections for the 4-D faces. So similar to the 3-D antiprisms & prisms in that for them, the Vertices are Transitive and the Edges aren't since edges between the 24-cells are different from those within. Can't find much online about 5 dimensional polyhedra that are neither Uniform or products of (triangle/square/circle) like the cubspherinder. Certainly in the 5 dimensional equivalents of the Johnson Solids, but at this point looking for the 4 dimensional equivalents is a large field with few involved....Naraht (talk) 21:18, 2 March 2016 (UTC)[reply]