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# July 19

## Integrating over countable set

As a note, this is not an assigned problem, just something I was thinking about. Suppose ${\displaystyle f(x)}$ is strictly positive and continuous over an open interval ${\displaystyle S=(a,b)}$, so that ${\displaystyle \int _{S}f(x)dx>0}$. We commonly encounter integration over these uncountably infinite sets like ${\displaystyle S}$ in evaluating integrals. Note that if we create the finite set ${\displaystyle D=\{x_{k}:x_{k}\in S,k=1,...,n\}}$, for example, some ${\displaystyle n}$ integers lying in ${\displaystyle (a,b)}$, it appears that ${\displaystyle \int _{D}f(x)dx=\sum _{k=1}^{n}\int _{x_{k}}^{x_{k}}f(x)dx=0}$. So my question is, is it ever possible to have a finite, or countably infinite, subset ${\displaystyle S^{*}\subset S}$, such that ${\displaystyle \int _{S^{*}}f(x)dx>0}$? 50.101.93.214 (talk) 00:09, 19 July 2017 (UTC)

Not in the usual sense. The Lebesgue measure of any countable (including any finite) set is zero.--Jasper Deng (talk) 05:51, 19 July 2017 (UTC)
But counting measure will do the trick.
By the way, the formula for computation you suggest (the sum of integrals) does not make sense. YohanN7 (talk) 07:29, 19 July 2017 (UTC)
Not for infinite sets it wouldn't.--Jasper Deng (talk) 07:36, 19 July 2017 (UTC)
Why not? Integrals are generally allowed to assume the values ${\displaystyle \pm \infty }$. In the present case, since strictly positive functions are considered, only finite values or ${\displaystyle +\infty }$ can occur, and the integral is never ill-defined. YohanN7 (talk) 08:25, 19 July 2017 (UTC)
I (personally) usually don't tend to consider an integral to exist with infinite value, rather that it exists or doesn't.--Jasper Deng (talk) 08:39, 19 July 2017 (UTC)
I have no problem with that. But even if the set is infinite, the integral may be finite. In the present case one has
${\displaystyle \int _{S^{*}}fd\mu =\Sigma _{x\in \mathbb {R} }x\cdot \mu (f^{-1}(x)).}$
(So ${\displaystyle f}$ is actually a step function or simple function according to some definitions (e.g. Herbert Federer's Geometric measure theory) of the term. The seemingly uncountable sum needs its proper definition of course, see Federer again.) The r h s is a countable sum, with strictly postitive terms. It may well be finite. YohanN7 (talk) 09:02, 19 July 2017 (UTC)
Yes, it can indeed be finite since for some functions, ${\displaystyle S^{*}}$ can be chosen to produce an (absolutely) convergent series. But it won't work for every function and every ${\displaystyle S^{*}}$ meeting the OP's definition.--Jasper Deng (talk) 09:11, 19 July 2017 (UTC)
There's no point in arguing over conventions and terminology, but I dare say that in measure and integration theory, infinite values are allowed more often than not, both for the range of functions and the value of integrals. This is probably because it simpliefies the exposition. The only things not allowed (i.e. left undefined) are expressions of the type ${\displaystyle \infty -\infty }$. Whenever ${\displaystyle 0\cdot \infty }$ occurs, it is defined to be zero (at least, I can't think of an exception atm). (Division may also cause problems, but rarely pops up.) YohanN7 (talk) 09:17, 19 July 2017 (UTC)
Yes, this is all very standard. Excluding the value ∞ serves no purpose. Measures take values in the range [0, ∞]. Integrals can be infinite, or they can fail to exist, and these are importantly different cases; excluding the value ∞ would collapse them together, to no benefit. --Trovatore (talk) 07:50, 23 July 2017 (UTC)

Questions: The OP says that "it appears that ${\displaystyle \int _{D}f(x)dx=\sum _{k=1}^{n}\int _{x_{k}}^{x_{k}}f(x)dx=0}$" but I wonder if this is meant to be ${\displaystyle \int _{D}f(x)dx=\sum _{k=1}^{n-1}\int _{x_{k}}^{x_{k+1}}f(x)dx}$ because for any constant ${\displaystyle c\in (a,b)}$ where f(x) is continuous and differentiable, ${\displaystyle \int _{c}^{c}f(x)dx=0}$. Also, is it meant to be that ${\displaystyle a? EdChem (talk) 07:41, 23 July 2017 (UTC)

# July 20

## Goldman and G-domain

I've seen definitions of both Goldman domain and G-domain, but do those two pages describe formally equivalent structures? I understand that G-domain has been used as a synonym for Goldman domain, but is this the standard use? Klbrain (talk) 21:23, 20 July 2017 (UTC)

Not really an expert on this but the definitions given in the article look equivalent. There's already a merge tag in place, perhaps the articles should be merged regardless since the concepts are certainly similar enough that they could be placed in a single article. --RDBury (talk) 15:15, 24 July 2017 (UTC)
Thanks for your view. That was my feeling too, but note that no-one has been willing to commit an opinion on the merge in almost 3 years. I've done some pure maths, but was rather (25 years ...) too rusty to be sure. Klbrain (talk) 17:27, 24 July 2017 (UTC)

# July 22

## The Fibonacci-like sequence x(1)=x(2)=1, x(n+1)=2*agm(x(n), x(n-1))

Hi, as you know, the Fibonacci sequence has f(n+1)=f(n)+f(n-1) = 2*(the arithmetic mean of f(n) and f(n-1)), and lim as n-->+oo of the ratios f(n)/f(n-1) is M = Golden mean, and the lim as n-->-oo is -1/M. I've played with sequences like y(n+1) = 2*(harmonic mean [and similarly rms mean] of y(n) and y(n-1)), but I'm really wondering about the arithmetic-geometric mean(agm) variation: what are the numbers that the ratios x(n)/x(n-1) approaches as n-->+-oo, where {x(n)} is defined in the title above? Are the ratios well known in another context, like perhaps pi/2? Numerical approximations would also be helpful. Thanks. 64.134.223.214 (talk) 00:55, 22 July 2017 (UTC)

Writing a quick one-liner (almost) in Haskell, the ratios of consecutive terms of your sequence seem to be converging to around 1.60207620749888.... Checking the OEIS for this number gives no hits, so no idea if it's expressible in terms of known constants. --Deacon Vorbis (talk) 04:07, 22 July 2017 (UTC)
Thanks. Does your program tell what the ratio x(n)/x(n-1) looks like for large negative n?64.134.223.214 (talk) 04:14, 22 July 2017 (UTC)
Never mind. I'm such an idiot!64.134.223.214 (talk) 03:17, 23 July 2017 (UTC)

It is not too difficult to find an equation satisfied by the limit ratio. I think the following gives 100 digits of your constant using Mathematica (assuming I got the argument of the elliptic function right):

In[30]:= FindRoot[(1+x)/x^2 == 2/Pi*EllipticK[(1-x)^2/(1+x)^2],{x,1.6},WorkingPrecision->100]
Out[30]= {x->1.602076207498882854166181178384113469249890744070779427538587439268519411756346198948331083813417404}


No idea whether this number is related to anything else. —Kusma (t·c) 10:01, 24 July 2017 (UTC)

[edit conflict] This looks like a job for isc rather than OEIS. But I didn't find a match there either.
FWIW, I've put more digits of the ratio at https://pastebin.com/qSkERhCF, it should be correct to the sf specified. -- Meni Rosenfeld (talk) 10:23, 24 July 2017 (UTC)
I guess the process should converge for any "reasonable" mean and give a limit ratio between ${\displaystyle {\sqrt {2}}}$ and 2. —Kusma (t·c) 11:23, 24 July 2017 (UTC)

# July 24

## Associative property and composite functions

I posted a question a bit back. I have developed some conclusions.

Let ${\displaystyle L_{n}=f(T_{n-1})}$ and ${\displaystyle T_{n}=g(L_{n-1})}$, where n is a point in time. Hence,${\displaystyle L_{n-1}=f(T_{n-2})}$ and ${\displaystyle T_{n-1}=g(L_{n-2})}$. Thus, using the substitution property, the composite functions ${\displaystyle h_{L}}$ and ${\displaystyle h_{T}}$ can be defined, such that ${\displaystyle h_{L}=f\circ g=f(g(L_{n-2}))=L_{n}}$ and ${\displaystyle h_{T}=g\circ f=g(f(T_{n-2}))=T_{n}}$. Finally, due to the associative property, ${\displaystyle h_{L}=h_{T}=h}$.

I am not certain this line of thinking is correct. I appreciate any insight anyone could provide. Schyler (exquirere bonum ipsum) 17:44, 24 July 2017 (UTC)

I think this doesn't quite work without the assumption that f and g are commutative as well. ${\displaystyle g\circ f\circ g\neq f\circ g\circ f}$ in general.--Jasper Deng (talk) 18:05, 24 July 2017 (UTC)
Yes, you are confusing associative property and commutative property. Ruslik_Zero 19:55, 24 July 2017 (UTC)