replication – Replica identity value on PostgreSQL read-only replica

In PostgreSQL will the replica identity of a table in a read-only replica database be the same as the replica identity in the primary? I.e., if I run the following query on primary and replica databases will I always get the same result (assuming replication is caught up)?

SELECT CASE relreplident
          WHEN 'd' THEN 'default'
          WHEN 'n' THEN 'nothing'
          WHEN 'f' THEN 'full'
          WHEN 'i' THEN 'index'
       END AS replica_identity
FROM pg_class c
WHERE oid = 'mytable'::regclass;

How to show this integral limit identity?

Let I be a generalized rectangle and let $f: I to mathbb{R}$. Show that $$lim_{ptoinfty}left(int_I|f|^pright)^{1/p} = max|f|.$$ I found it straightforward to show that these integrals in the limit sequence are properly defined and I showed the LHS $leq$ RHS by using the definition of the integral using partitions. I am confused about the approach to showing that $RHS leq LHS$ to give equality. Thanks for any hints and suggestions.

Does it make sense to split two factor authentication between identity providers

Authentication with SAML uses password in the corporate directory. This password is one authentication factor. The TOTP uses key (password) that has nothing to do with password in the corporate directory. Thus these are two independent factors. For authentication an attacker will need to know both.

Both factors are based on the knowledge of some secret. Normally it is better to use factors of different types, e.g. one is based on knowing of something and one based on possessing of something (like token, card, smartphone) or being something (like fingerprint).

But in the reality TOTP does not require user to enter the key each time. Usually the key is generated randomly and stored in the device, e.g. in the token. This toke generates codes based on the key and current time. To compromise this factor often it is easier to compromise the token (e.g. steal it) than to extract the key from it or to retrieve the key from the security system. Thus the security of TOTP is based effectively not on the knowledge of the key, but on the possession of the token.

Thus 2FA in your case is based on knowledge (of the password in corporate directory) and possession (of the TOTP hardware token).

This approach is reasonable and is more or less common. Of course the real strength of the security depends on many things:

  • The entropy of the passwords in the corporate directory (if it is low, such passwords can be easier brute-forced)
  • How is password management implemented in your company (e.g. how is allowed to reset passwords for users, in what cases)
  • How is TOTP organized (e.g. who can get access to the TOTP key on the server side)
  • How disciplined are users (if they keep their passwords secret or write them down and make then accessible to the others)
  • How well user care of TOTP tokens (e.g. to they inform your company immediately as they notice that token is lost/stolen)
  • Etc.

But the approach itself is reasonable.

alter table – Why can’t you add an Identity to an existing colum in SQL Server?

What is so special about an Identity column?


I am pretty sure there is a valid reason, but I’m trying to get a better understanding of SQL.

So what is so special about an Identity column that this can’t be easily added in retrospect?

Why is this not just a single int/bigint flag somewhere that is being incremented every time you insert a new row?

lie algebras – Exponential map generates the identity component of closed linear groups

In page 15 of “Lie groups beyond an introduction” by Prof. Knapp, Corollary 0.20 states that the exponential map (of linear Lie algebra $mathfrak g$) generates the identity component $G_0$ of a closed linear group $G$. The proof goes like

1 by continuity of the exponential map $exp(mathfrak g)$ is connected

2 $exp(mathfrak g)$ contains a neighborhood of 1 in the group (local diffeomorphism)

3 the smallest subgroup containing a nonempty open set in $G_0$ must be $G_0$

3 doesn’t look obvious to me though, I don’t know how to prove that $exp(mathfrak g)$ is a subgroup. In addition, if we can show it’s open, then it’s an open closed connected subgroup, so it must be $G_0$ since $G_0$ is connected.

fa.functional analysis – Obtaining identity from Pokhozhaev formula

From the classical Pokhozhaev formula, how can I obtain that the following identity holds for $u,v in C^2(bar Omega)$?
int_Omega (Delta u(x,nabla v) + Delta v(x,nabla v)) dx = int_{partial Omega} ( frac{partial u}{partial nu}(x,nabla v) + frac{partial v}{partial nu}(x,nabla u) + (nabla u, nabla v) (x,nu) )ds + (N-2) int_Omega (nabla u, nabla v) dx

simplifying expressions – Applying the identity $sin^2 + cos^2=1$ without any other triginometric simplifications

Mathematica does not automatically simplify $sin^2 + cos^2 =1$, but applying any kind of trigonometric simplification results in not only that identity being applied but also all kind of trig identities, usually resulting in the trig functions being of multiple summed variables. I want to just apply the Pythagorean identity without any other ones. Is there a way to do this? So far I have tried using replacement rules but that is very cumbersome.

How can we make the header responsive which can aline with Site Identity Block (generatepress)

if we want to make the header responsive which can aligned with Site Identity Block with or without (generatepress)?
Found that generatepress is the “best” free theme for someone who want to use to develop their “own theme”
I plan to use wp to share some tutorial/personal note. most of them are text, only include image when really needed. Should I just use wp default themes which should be “better” than generatepress in term of SEO, and performance ? (this is my first site ever, when do we need an theme for wp ?)

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Identity involving binomial coefficients and partitions

Working on a problem in the symmetric group I have stumbled upon the following equation:

$$sum_{substack{pi=(1^{c_1},2^{c_2},ldots,n^{c_n})\textrm{partition of }n}}(-1)^{n-sum_{i=1}^nc_i}frac{n!}{prod_{i=1}^ni^{c_i}c_i!}left(sum_{substack{eta=(1^{b_1},2^{b_2},ldots,k^{b_k})\textrm{partition of }k}}prod_{j=1}^k{c_jchoose b_j}right)^ell=0,$$
where $n,k$ and $ell$ are positive integers and $1le k,ellle n-1.$ (Here, $pi=(1^{c_1},2^{c_2},ldots,n^{c_n})$ means that $n=1cdot c_1+2cdot c_2+cdots +c_ncdot n$.)

I have two questions. First, whether this equality has appeared somewhere. (I have checked the book(s) of Stanley on Enumerative combinatorics, but with no luck.)
Second, given $n$ and $k$, I am intersted on the smallest value of $ell$, where the equality above is not satisfied. For instance, when $n:=43$ and $k:=13$, with a computer computation one can check that the equality above is satisfied FOR EACH element in ${1,2,3,4,5,6}$ and $ell:=7$ is the first time where this equality is not satisfied. However, I have no clue in how to get my hands on this value!

functional analysis – Let A and B be a Banach algebra with $1in Asubset B$ (1 is the identity). Let $ain A $ then show that $Spectrum Bsubset SpectrumA$

Let A and B be a Banach algebra with $1in Asubset B$ (1 is the identity). Let $ain A $ then show that
$$1).Spectrum_a Bsubset Spectrum_aA$$
$$2). Boundary(Spectrum_aA)subset Boundary(Spectrum_aB)$$

First, let’s take $xin Spectrum_a B$, I’m going to show that is in $Spectrum_aA$.
since $xin Spectrum_a B$
$a-x*1$ is not invertible. Since $ain A$ , $𝑆𝑝𝑒𝑐𝑡𝑟𝑢𝑚_𝑎 𝐴$ contains x.
This implies $Spectrum_a Bsubset Spectrum_aA$.

Is this correct?
And how we show the second part?