/sci/ - Science & Math

Today I discovered a (to me) perplexing relation:

[math]X:=\left(
\begin{array}{cc}
x & 0 \\
0 & x_0 \\
\end{array}
\right)[/math],

[math]H:=\left(
\begin{array}{cc}
0 & h \\
0 & 0 \\
\end{array}
\right)[/math]

then, at least for analytic f,

[math]f(X + H) = f(X) + \dfrac{f(x) - f(x_0)}{x - x_0} H[/math]

(* Some Mathematica: X = {{x, 0}, {0, x0}}; H = {{0, h}, {0, 0}}; X . H - H . X == (x - x0) H // Simplify; MatrixPower[X + H, m] == MatrixPower[X, m] + H (x0^m - x^m) / (x0 - x) // Simplify; *)

In existing language, this family (in, [math]x, x_0, h[/math]) are possible bases of 2x2 representation of the affine algebra (the smallest Lie-algebra), as evidenced by

[math][X, H] = (x - x_0)\, H[/math]

For [math]x_0=-x[/math] that's also a subalgebra of [math]sl(2)[/math] (note that my X is often called H in that context.)

The formula generalizes the standard matrix computation of its exponential map, [math]f=\exp[/math], but from the standard representation (which has [math]x_0=0[/math]), it's usually not clear that a finite difference quotient shows up there.
In the limit [math]x\to x_0[/math] it's somewhat clear that

[math]f(X + H) = f(X) + f'(x) H[/math]

because H is nilpotent, but it's baffling me that the expression also holds on the finite difference level.

It essentially mirrors the idea

"[math]\dfrac{f(x + h) - f(x)}{h} = \dfrac{f(x) - f(x_0)}{x - x_0}[/math]"

but holds exactly.
Has anybody seen this before?

>What am I missing something here?
In the standard model of things, Cauchy reals are equivalence classes (each real being of set cardinality [math]|{\mathbb R}|[/math], mind you) of Cauchy sequences.

A disadvantage of the Cauchy reals is that the completeness property of them as a set is harder to get.
(On the flipside, to even make the Dedekind reals a set, you need stronger powerset-like axioms.)

Btw. I came across of probably very nice papers recently, in those directions.

>Real Analysis in Reverse
https://arxiv.org/pdf/1204.4483.pdf
Now classically discussing in detail all the nice properties (TM) of the reals, or rather their relationship to Dedekind completeness (and I'm sure it's a good ref for having all those at hand formally)

>Completeness of Ordered Fields
https://arxiv.org/pdf/1101.5652.pdf
Classically discussing how far away Dedekind completeness is from the vantage point of having an ordered arithmetical field

The rest is from a constructive angle:

>On the Cauchy Completeness of the Constructive Cauchy Reals
https://arxiv.org/pdf/1510.00639.pdf
Looking at the pains of indeed talking about the Cauchy completness of set collection of Cauchy reals in the strong-as-possible-without-LEM set theory
Keyword: Modulus of convergence

Same author + Rathjen discssing the Dedekind reals here
>On the Constructive Dedekind Reals
http://math.fau.edu/lubarsky/Dedreals.pdf

The Dedekind variant opens up some (arguably hard formalist) theories, locales etc. (https://ncatlab.org/nlab/show/locale))
I think Bauer generally prefers to allow for rich such frameworks and he even has some code going in that direction
https://github.com/andrejbauer/dedekind-reals

There's more angles, though, I know of a few recent type theory PhD theses who are basically about formalization, and some are inbetween those variants.

>>10970700
>>10971335
The incompletness of PA is damning generally.

But once that insight is digested, I don't find the undecidability of the consistency in particular to be all that important. Three points on that:
* If PA where inconsistent (rules out by the premise of the result of course), then it would prove its own consistency (wrongly)
* If you accept some of the countable ordinals, then Gentzen gives you consistency of PA
* Gödels result tells us that there is no PA proof (Con_PA) expressing that all of PA's proofs aren't one of an inconsistency. However, afaik you can prove that for all of PA's proofs, you can find a (constructive, even, I think) proof that they aren't of an inconsistency.
Gödel speaks about consistency in terms of one PA proof with an ∀.
>"PA does not derive its own consistency as an arithmetic statement."
On the flipside, the second version reassures us of a similar notion of consistency, but the ∀ is in the logic we use and not internalized to PA.
>"PA proofs permit us, by conservative means, to prove that they aren't a proof of an inconsistency.".
Formalization: https://arxiv.org/pdf/1902.07404.pdf

So I guess we have to accept undecidability, but not fear for inconsistencies of arithmetic showing up anytime soon.

>>10970700
>>10971335
The incompletness of PA is damning generally.

But once that insight is digested, I don't find the undecidability of the consistency in particular to be all that important. Three points on that:
* If PA where inconsistent (rules out by the premise of the result of course), then it would prove its own consistency (wrongly)
* If you accept some of the countable ordinals, then Gentzen gives you consistency of PA
* Gödels result tells us that there is no proof saying that all of PA's proofs aren't one of an inconsistency. However, afaik you can prove that for all of PA's proofs, you can find a (constructive, even, I think) proof that they aren't of an inconsistency.
Gödel speaks about consistency in terms of one PA proof with an [math] \forall [/math].
>"PA does not derive its own consistency as an arithmetic statement."
On the flipside, the second version reassures us of a similar notion of consistency, but the [math] \forall [/math] is in the logic we use and not internalized to PA.
>"PA proofs permit us, by conservative means, to prove that they aren't an inconsistency.".
Formalization: https://arxiv.org/pdf/1902.07404.pdf

So I guess we have to accept undecidability, but not fear for inconsistencies of arithmetic showing up anytime soon.

>>10918657
To the extent that others find it helpful (and I think that's the case) if he'd copy from MO he'd still roughly need to grasp it.
I'd like to know if the anime posters are one guy of if there are generally several, and also whether it was always the same guy.
I've been on and off posting on /sci/ since 2008, so I'm sure there's a few more PhD's around. The site is definitely more social and fun than reddit of SO/SE. On that note, let me point to a call to read into Artin:
>>10918608

>>10899024
Internal energy, enthalpy or also Gibbs free energy are related by Legendre transforms, just like the Hamiltonian relates to a Lagrangian. A reparametrization leading to other relevant quantities of the system.
https://en.wikipedia.org/wiki/Legendre_Transformations#Definition

>>10899038
Practice and good books? I don't know if there's a better answer to a general question like that.

>>10898946
I guess it also rules out prices given away for "life's work" kind of honors

>>10897641
https://en.wikipedia.org/wiki/Bessel%27s_correction

>>10896579
bumping 4u

https://youtu.be/Q53GmMCqmAM

>>9319899
The -1/12 also exists in classical analysis and already Euler had the most general result on it.

Consider the difference between integral and sum,

[math] \int _m^n f(x)\, {\rm d}x=\sum _{i=m}^n f(i)-\frac 1 2 \left( f(m)+f(n) \right) -\frac 1{12}\left( f'(n)-f'(m)\right) + \frac 1{720}\left( f'''(n)-f'''(m)\right) + \cdots [/math]

where the coefficients are in simple way computed from Bernoulli numbers. A child of this would be e.g.

[math] \frac{1}{3}(4^3-2^3)=(2^2+3^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1) [/math]

but using the above I can generate infinitely many rigid identities that will all have that magic factor.
It's only Ramanujan who studies the same things and make the identification. A classical analogon would be

[math] \sum_{n=0}^\infty n \, z^n - \dfrac {1} { \log(z)^2} = - \dfrac{1}{12} + {\mathcal O}((z-1)) [/math]

You can type that into wolfram alpha for confirmation. The limit z to infinite gives an expression for 1+2+3+..., however, classically there's the log.
>this kills the sum

I bought a mic and am going to start a long term project going through foundations up to statistical physics. You may subscribe here

https://www.youtube.com/channel/UCcrSMnEYhIPX_p127jI23qw

Ignore any test videos, I'll start in the mid of June. I have the first 30 videos of so outlined, but I'll go on for a long time here, no hurry. First Curry-Howard, the logic, relevant foundational constructions and then on the proper math eventually...

Hey guys,

I'll produce some original content and why not post it here also offer for folks to participate.

I'm preparting for my PhD exam and so I searched the web and copied together some general questions. I'll use the to work out what I might say and apply em to the more technical topics for my reseach and thesis sections.
Here's the raw stuff

https://pastebin.com/sB99q5M7

I'll assemble them tomorrow and next week and might re-post a cleaned up version (there's lots of duplications) and if you're super bored you can do it too, or add questions and so on. Or make an interactive document.
Or just wait for the end result.

Hope that helps,
I'm off to the cinema

Enumerate your language and denote by [math] h(n,m) \in \{0, 1\} [/math] whether the n'th Turing machine halts before m steps (1 for halts, 0 for doesn't halt). That's a function
[math] h : {\mathbb N} \time s{\mathbb N} \to \{0,1\} [/math]

Let's define the sequence
[math]s_k := \sum_{n+m=k} \frac{1}{2^n} h(n,m) [/math]

Compute
[math] s := \lim_{k\to \infty} s_k [/math]

Let me make up a hand-wavy analogy - we can get down to business later too.

You just moved into a house. But your gf has a different idea of how to do things with this you space and so you two constantly moving stuff around.

Let h(X,Y) denote the ways in which you can move stuff from X into Y.
You have a kitchen shelf A with a lot space to put stuff (it has some number of free spots), a storage room B (this also has some number of free spots) and a boxes V1, V2, V2... with stuff in it.

So h(V1, A) are the ways in which you can put the things in the box V1 onto the shelf.
(Say V1 contains 5 toys the shelf has 7 spots, well now you can do some combinatorics to see how many possibilities h(V1, A) captures)
Similarly, h(V1, B) are the ways in which you can move stuff from the box V1 into the storage room.
Now say you placed all your stuff on the kitchen shelf A but you guys decide you instead want to move it to the storage room B, in some way. Clearly, the different ways in which you can do this correspond to h(A,B), which are the ways to move stuff from the kitchen to the storage room.
Said again, the ways in which you can change ALL possible strategies h(V1, A) to the strategies h(V1, B) correspond to h(A,B).
And this doesn't depend on the things (the particular box you chose). For all boxes V, the ways in which you can change ALL possible strategies h(V, A) to the strategies h(V, B) correspond to h(A,B). And this is true for kitchen shelves and basement shelves just alike, i.e. holds for all sorts of A,B.

In symbols:
forall V, A, B
h(V, A) -> h(V, B) <=> h(A,B)

A more geometric realization: Consider the Cartesian spaces R^n and R^m. Then e.g. h(R^2,R^3) are all the ways to put a n=2dim sheet into a m=3dim space. Or h(R^1,R^4) are all the ways how to embed a k=1dim line into 4dim space?
How to change a given embedding in h(R^1,R^2) to one in h(R^1,R^4)?

>>8357140
Ha! I posted that list years ago.

There is also
>Making the most out of zero branes and a weak background

Here's a recent paper
http://arxiv.org/pdf/1404.0799v3.pdf

Also, see this list:
http://www2.tcs.ifi.lmu.de/~jjohanns/cute.html

>>>>8359454
Yes, in the sense that the Joy of Cats is a reference to the former. I think.
Btw., from a didactic perspective, I consider that to be a horrible piece of notes.

Completely unrelated question, does anyone know how to approach
log(x+d)==x^r
in x, when d != 0 and r != 1?
Graphically there seems to be one neat solution, always.
I'm trying to gain some intuition comparing powers of x and powers of log of x (or x+1).

In

https://upload.wikimedia.org/wikipedia/commons/d/d2/RiemannPrim1859.djvu

Riemann observed that a integration variable substitution x->n·x in the definition of the Gamma function

<span class="math">\Gamma(s) := \int_0^\infty x^{s-1} {\mathrm e}^{-x}\,{\mathrm d}x [/spoiler]

let's you write

<span class="math">{\frac {1} {n^s}} = {\frac {1} {\Gamma(s)}} \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x [/spoiler]

and then, recognizing the geometric series

<span class="math">\sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} = \frac{1} {x} - \frac{1}{2} + \frac{1}{12}x+O(x^2)[/spoiler]

give you

<span class="math">\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1} { \Gamma(s) } \int_0^\infty \frac{x^{s-1}} { {\mathrm e}^x-1} \, {\mathrm d}x [/spoiler]

He takes the integral into the complex plane, where he <span class="math"> \frac{1} { {\mathrm e}^x-1}[/spoiler] diverges periodically in steps of <span class="math">2\pi\,i[/spoiler].
He discovers that the function obeys a reflection formula

https://en.wikipedia.org/wiki/Riemann_zeta_function#The_functional_equation

<span class="math"> \sum_{n=0}^\infty \frac {1} {n^x} = \frac {(2\pi)^x} {2\cos( (\pi/2)\,x )\,\Gamma(x)} \sum_{n=0}^\infty n^{x-1} [/spoiler]

And obviously,

<span class="math"> {\frac {2^2} {2\cos( \pi )}} = -2 [/spoiler]

<span class="math"> \sum_{n=0}^\infty n = - \frac{1}{12} [/spoiler] :^)

<span class="math"> -2 · - \frac{1}{12} = \frac{1}{6} [/spoiler]

In

https://upload.wikimedia.org/wikipedia/commons/d/d2/RiemannPrim1859.djvu

Riemann observed that a integration variable substitution x->n·x in the definition of the Gamma function

<span class="math">\Gamma(s) := \int_0^\infty x^{s-1} {\mathrm e}^{-x}\,{\mathrm d}x [/spoiler]

let's you write

<span class="math">{\frac {1} {n^s}} = {\frac {1} {\Gamma(s)}} \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x [/spoiler]

and then, recognizing the geometric series

<span class="math">\sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} = \frac{1} {x} - \frac{1}{2} + \frac{1}{12}[/spoiler]

give you

<span class="math">\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1} { \Gamma(s) } \int_0^\infty \frac{x^{s-1}} { {\mathrm e}^x-1} \, {\mathrm d}x [/spoiler]

He takes the integral into the complex plane, where he <span class="math"> \frac{1} { {\mathrm e}^x-1}[/spoiler] diverges periodically in steps of <span class="math">2\pi\,i[/spoiler].
He discovers that the function obeys a reflection formula

https://en.wikipedia.org/wiki/Riemann_zeta_function#The_functional_equation

<span class="math"> \sum_{n=0}^\infty \frac {1} {n^x} = \frac {(2\pi)^x} {2\cos( (\pi/2)\,x )\,\Gamma(x)} \sum_{n=0}^\infty n^{x-1} [/spoiler]

And obviously,

<span class="math"> {\frac {2^2} {2\cos( \pi )}} = -2 [/spoiler]

<span class="math"> \sum_{n=0}^\infty n = - \frac{1}{12} [/spoiler] :^)

<span class="math"> -2 · - \frac{1}{12} = \frac{1}{6} [/spoiler]

In

https://upload.wikimedia.org/wikipedia/commons/d/d2/RiemannPrim1859.djvu

Riemann observed that a integration variable substitution x->n·x in the definition of the Gamma function

<span class="math">\Gamma(s) := \int_0^\infty x^{s-1} {\mathrm e}^{-x}\,{\mathrm d}x [/spoiler]

let's you write

<span class="math">{\frac {1} {n^s}} = {\frac {1} {\Gamma(s)}} \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x [/spoiler]

and then, recognizing the geometric series

<span class="math">\sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} {{\mathrm e}^{-x}-1} = \frac{1}{x} - \frac{1}{2} + \frac{1}{12}[/spoiler]

give you

<span class="math">\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{e^x-1}\,{\mathrm d}x [/spoiler]

He takes the integral into the complex plane, where he <span class="math">\frac{1}{e^x-1}[/spoiler] diverges periodically in steps of <span class="math">2\pi\,i[/spoiler].
He discovers that the function obeys a reflection formula

https://en.wikipedia.org/wiki/Riemann_zeta_function#The_functional_equation

<span class="math"> \sum_{n=0}^\infty \frac {1} {n^x} = \frac {(2\pi)^x} {2\cos( (\pi/2)\,x )\,\Gamma(x)} \sum_{n=0}^\infty n^{x-1} [/spoiler]

And obviously,

<span class="math"> {\frac {2^2} {2\cos( \pi )}} = -2 [/spoiler]

<span class="math"> \sum_{n=0}^\infty n = - \frac{1}{12} [/spoiler] :^)

<span class="math"> -2 · - \frac{1}{12} = \frac{1}{6} [/spoiler]

If you squeeze a box of gas by the small volume dV, the energy you have to apply is whatever the system gains in energy
dU = ∂U/∂V · dV
and we call
p := -∂U/∂V
the pressure. It’s an energy density (consider the units), mediating energy and volume.
The product of energy density and volume is an energy.
Thus,
\int_1^2 p dV
and
\int_1^2 V dp
are energies associated with the squeezing.

If you squeeze a helium balloon, it will get denser: Obviously, as the volume V goes down, the density N/V goes up. If you let go, it will return back to normal.
For ideal gas, the energy density is proportional with the density N/V. The other factor must be an energy, and this mean energy is called k T. Together, you have the ideal gas law
-∂U/∂V = p = k T · N/V

The further you go with the squeezing of the box, the harder it resist.
In fact, for the ideal gas, the pressure diverges as you go to very small volumes
p proportional to 1/V. Here’s the crucial part: Squeezing the ballon from 58 cm3 down to 57 cm3 is easier than squeezing it from 48 cm3 down to 47 cm3. And it’s virtually impossible to squeeze it from 1 cm down to 0 cm. This nonlinearity of effort for different states is where the non-linear "log" comes from.

\int_1^2 V dp = \int_1^2 (k T N / p) dp = k T N log(p_2/p_1)

The fraction of energy densities can be expressed as fraction of particle densities and you get a log of concentrations.
The Nernst equation arises from considerations just like that. Except you move particles along potentials, you don't necessarily press stuff together.

The fact that the energy is proportional to the characteristic energy T shouldn’t be surprising.
The ugly function (logarithm) arises because you talk about an extensive quantity, the energy, in terms of a conjugate variable, the pressure, which is a density (or expressed in terms of the density/concentration directly). Density always means per volume, and there you have that multiplicatively inverse relation.