## real analysis – Scaling of double convolution

I am interested in the scaling of

$$F(x_1,x_4)=int_{mathbb R^2} e^{-vert x_1 -x_2 vert -varepsilon vert x_2 -x_3 vert- vert x_3 -x_4 vert } dx_2 dx_3$$

In particular, I suspect that

$$F(x_1,x_2) le C varepsilon^{-n} e^{{varepsilon}vert x_1 -x_4vert}$$
for some universal $$C>0$$ and $$n ge 0$$.

But this is really only based on pure heuristic and I do not know which $$n$$ could be optimal here.

## lie groups – The Double cover in the classical limit of \$U_q(frak{sl}_2)\$

I am trying to learn about Drinfeld–Jimbo quantum groups and I am having trouble with the classical limit of $$U_q(frak{sl}_2)$$. When properly expressed the limit makes sense as $$qto 1$$ – see for example this question.

But we get $$U(frak{sl}_2) otimes mathbb{Z}_2$$. This fact goes to create many problems, such as the effective doubling of the number of representations, half of which are generally regarded as “uninteresting”.

It seems to me that finding a presentation of $$U_q(frak{sl}_2)$$ would be of great benifit. However I can’t seem to find any discussion of this anywhere. Can anybody explain what is going on here? Is the double cover a necessary consequence of quantization? Is there some reason why people felt it best to stick with the double cover approach?

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## How to visualize the double cover of \$SO(4)\$ as two copies of \$S^3\$?

How to visualize the double cover of the rotational symmetry group of $$S^3$$ (which is $$Spin(4)$$, namely the double cover of $$SO(4)$$) as two copies of $$S^3$$?

This is due to $$Spin(4) = SU(2) times SU(2) = S^3 times S^3$$. But how to visualize the above via the perspective of $$S^3$$?

Related question: Implications of \$text{Spin}(3)times text{Spin}(3) cong text{Spin}(4)\$

## c# – Double em comparação

Olá, eu queria saber o porquê de o resultado estar dando como aprovado quando eu escrevo até 6.1, já que o código fala para só dar o aprovado quando o valor da nota for maior ou igual a 7, de 6.1 para cima está dando aprovado mas se eu escrever de 6 pra baixo dá como reprovado.

if(nota >= 7)
{
}

## fa.functional analysis – Energy levels of double well potential

Consider the (quantum) Hamiltonian on the real line
$$H=-frac{hbar^2}{2m}frac{d^2}{dx^2}+V(x).$$

Let us assume that the potential $$V$$ is an even smooth functions with exactly two non-degenerate minima,
and $$lim_{|x|to +infty}Psi(x)=+infty$$. Such a $$V$$ is called double well potential.

Let $$E_1,E_2$$ be the first two minimal eigenvalues of $$H$$. It is known in physics literature (see problem 3 after $$S$$ 50 in Landau-Lifshitz, vol. 3) that under some extra assumption which are not quite specified there one has $$E_2-E_1 approx frac{omegahbar}{pi} exp(-frac{1}{hbar} C),$$
where $$omega, C$$ are positive constant which can be written down explicitly, and the result is understood asymptotically as $$hbarto 0$$.

I am looking for a mathematically more rigorous discussion of this result where, in particular, the assumptions are formulated more explicitly.

## dnd 5e – Can the arcane/druidic focus staff double as quarterstaff?

Probably.

The entries for Staff and Quarterstaff are different in the equipment list. Under the Arcane Focus section you get:

a specially constructed staff (p. 48).

this is distinctly different (and quite a bit more expensive) than a typical quarter staff (the arcane focus version runs 5gp vs 2sp for a quarterstaff). So no, a quarterstaff cannot double for an arcane foci staff, or the other way round (at least without further clarification).

In the Druidic Focus section you get

a staff drawn whole out of a living tree (p 151 PHB)

The druid is much closer to nature and uses a druidic focus rather than an arcane one. We see that the staff for the druidic focus has a far more…utilitarian description. There is nothing here that makes it not a quarterstaff (though it may be rougher than a typical staff). My guess is most Druids would have no problem wielding their staff as a quarterstaff. I mean who is going to argue with you, you have a quarterstaff.

Last thing, there is an argument to be made for allowing implement staves to be treated as quarterstaves under the improved weapon rules on p 46 of BD&D

In many cases, an improvised weapon is similar to an actual weapon and can be treated as such. For example, a table leg is akin to a club. At the DM’s option, a character proficient with a weapon can use a similar object as if it were that weapon and use his or her proficiency bonus.

This means that if your staff focus resembles a quarterstaff (for some value of resemble I guess), then it could be treated as one and your proficiency bonus would apply to attacks with it (and it’d get a d6 for the damage die).

## Double lines division representation

How to make a fraction to display a numerator on top of the denominator in Google Sheet?

## bitcoin core – How to create a double spend to prioritize a new transaction over a previous unconfirmed transaction?

I’d like to generate a double spend transaction and increase the gas fees over the previous, unconfirmed transaction that has been stuck for a week. I know there’s other answers regarding double spend but I’m looking for more of a technical answer specific to my thought process and sample code.

I understand the general premise is to recreate the originating transaction in its entirety. I’ve started by utilizing bitcore-lib to perform a getRawTransaction of my currently stuck transaction. From here, I’m iterating over the vin and performing a getRawTransaction of the vin’s txid to retrieve the parent. I’m then parsing the parent, iterating over each vout and looking for vout.n to match the previous vin.vout. Once I find a match, I’m generating a utxo which effectively looks like this:

const utxo = {
txId: vin.txid,
outputIndex: vin.vout,
script: output.scriptPubKey,
satoshis: output.value
};

I then use these utxos to generate and sign the transaction as follows (sorry for lack of full class, but the actual process of signing transactions is fully functional):

const transaction = new Bitcore.Transaction();

transaction.fee(this.fee());
transaction.from(utxos);
targets.forEach((target) => {
});