4:28 AM
@TheSimpliFire I have added a bounty!!

https://math.stackexchange.com/questions/2888976/the-definite-integral-problem-with-a-twist

5:16 AM
@TheSimpliFire also thinking about $\lim_{n \to \infty}\ \sum_{r=1}^n a_r ( f(\frac{k}{n}r)\frac{k}{n} )$ I realised for us to take the limit we would first have to know the answer at an arbitrary $k$ before taking $k \to \infty$ ... And the limit should exist? I'm curious what happens before I take $k \to \infty$ to the R.H.S of the formula???

Is there some clever limit definition to escape this??

5 hours later…
10:20 AM

2 hours later…
12:08 PM
@MoreAnonymous Brilliant.

@TheSimpliFire I'm just wondering if his "sufficient condition" limits the derivation for the "crazy non-periodic example" we talked about

@MoreAnonymous You mean with $f(x)=\exp(-x)$?

and $g(x) = e^{-x^2}$
1

Background & Motivation I'm trying to verify/disprove the conjectured formula of the weighted integral of $f(x)$: The Definite Integral Problem (with a twist)? $$\lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n a_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbr... But g doesn't feature in the equation It does ...$$(\int_{0^+}^\infty g(x) dx)^2 = \lim_{k' \to \infty} \lim_{n' \to \infty}\ \sum_{i,j=1}^\infty e^{-(i^2 +j ^2) (\frac{k'}{n'})^2} (\frac{k'}{n'})^2 = \lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n a_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{a_r}{r^s}}_{\text{removable singularity}} \int_0^\infty f(x) \, dx $$12:16 PM Nvm Hang on, surely a_r=b_r\sigma(r) When you have your a_r, checking the condition is equivalent to showing that \lim\limits_{n\to\infty}\dfrac{\log^2n}n\sum\limits_{r=1}^n\dfrac{|a_r|}{\sigma(r)}=0 12:37 PM It applies doesn't it? @MoreAnonymous Don't have the time to work through the algebra but if you get zero then your f and g examples hold @TheSimpliFire How do I get 0? The example has \int_{0^+}^\infty g(x) dx)^2 = \text{constant} \times \int_0^\infty f(x) \, dx  ... Me no understand? @MoreAnonymous I meant to check the limit \lim\limits_{n\to\infty}\dfrac{\log^2n}n\sum\limits_{r=1}^n\dfrac{|a_r|}{\sigma(r)}=0 From mathworker21's answer this is needed to show that \sum\frac{b_d}d<\infty ah ... I'm not great at limits (esp. these kinds .. I can only do "black magic physics" math) .. But I'm gonna award him the bounty anyway cause he's done magic! Well yea he deserves the bounty You can only reward it at least 24 hours after you advertise the bounty 12:47 PM @TheSimpliFire I know already tried to award him the bounty be4 :P @TheSimpliFire how good are you with bra-ket notation and Quantum Mechanics? Had a conjecture relating to that one too stuff like this: \vert\psi\rangle? Yup!! Well I can do basic stuff with them (such as linear algebra) but not beyond that as I'm not a physicist I've learnt a bit about them in my chemistry lessons when discussing orbitals/quantum numbers Alright might as well give you the context: So I was planning on trying to put some type of bounds on the S-matrix. But then realise I don't have any other way of talking about the ladder operator than the available in bra-ket notation. So I set out to create a different way of talking about them ... I do (partially) succeed however, it becomes so convoluted I don't know how to apply them to the original problem anymore .. :(((( (the context of what I was trying to do wasn't in the question because putting bounds on the S-matrix is really ambitious) @MoreAnonymous How do you define \vert1\rangle? Is it (1\quad1\quad\cdots\quad1)^\top? 12:59 PM How do you define \vert1\rangle? Is it (1\quad1\quad\cdots\quad1)^\top? Not sure which basis your using? But I would have gone with (1\quad0\quad\cdots\quad0)^\top? Oh, the standard Hilbert space Any way if it helps u can think of them as eigenenergies of a harmonic oscillator Not really, not heard of eigenenergy In quantum computing, a qubit () or quantum bit (sometimes qbit) is the basic unit of quantum informationâ€”the quantum version of the classical binary bit physically realized with a two-state device. A qubit is a two-state (or two-level) quantum-mechanical system, one of the simplest quantum systems displaying the peculiarity of quantum mechanics. Examples include: the spin of the electron in which the two levels can be taken as spin up and spin down; or the polarization of a single photon in which the two states can be taken to be the vertical polarization and the horizontal polarization. In... @MoreAnonymous See the section on Standard representation, it seems that \vert1\rangle=(0\quad0\quad\cdots\quad1)^\top Yea .. I guess you could always do that to bra-ket notation except in QFT (convert it to a matrix) And because we're on the real plane \langle1\vert=\vert1\rangle^\top as conjugacy affects nothing 1:07 PM yups ... yups Two questions In your post you ask for the calculation of ( \sum\limits_{0 < R \leq 1} \hat R)^\dagger | \phi \rangle Though I prefer Hermitian conjugate rather than transpose Is it \left(\sum\limits_{0 < R \leq 1} \hat R\right)^\dagger \vert \phi \rangle or \left\langle\left(\sum\limits_{0 < R \leq 1} \hat R\right)^\dagger \bigg\vert \phi \right\rangle @TheSimpliFire sorry go on? This one: \left(\sum\limits_{0 < R \leq 1} \hat R\right)^\dagger | \phi \rangle OK, and what does \phi denote? 1:10 PM$$\phi any arbitrary row/column vector (in matrix language)
I'm saying if you write these as matrices then:
$$(\hat 1 - A^\dagger)^{-1} = ( \sum_{0 < R \leq 1} \hat R)^\dagger$$

How do you sum the values from 0 to 1? Is it like integration?

Yea... so inutitively I'm thinking I'll order them in descending order?

@MoreAnonymous LHS should be $(I-A^\dagger)^{-1}-I$? Also there's inconsistent use of $<$ and $\le$ in your post...

Ah ... no thats accurately used... Remember $\hat I = \hat 1$
Well if you want$$(\hat 1 - A^\dagger)^{-1} = ( \sum_{0 < R \leq 1} \hat R)^\dagger \implies I - (I + \underbrace{(\sum_{0 < R < 1}\hat R)^\dagger}_{\hat O} )^{-1} = A^\dagger$$ in some sense

I meant what about the extra $\color{red}{-I}$ at the end: \begin{align}I-(I+\hat O^\dagger)^{-1}=A^\dagger&\implies(I-A^\dagger)^{-1}=I+\hat O^\dagger\\&\implies\hat O^\dagger=\left(\sum\limits_{0 < R \leq 1} \hat R\right)^\dagger=(I-A^\dagger)^{-1}\color{red}{-I}\end{align}

1:21 PM
This is the correct equation can't find yours: \begin{align}I-(I+\hat O^\dagger)^{-1}=A^\dagger&\implies(I-A^\dagger)^{-1}=I+\hat O^\dagger\\&\implies\hat O^\dagger=\left(\sum\limits_{0 < R < 1} \hat R\right)^\dagger=(I-A^\dagger)^{-1}{-I}\end{align}
Can you tell me which equation number on that post?

I was just saying you forgot the $-I$ at the end of the LHS

Ah ... I think I know what;s going on ... one moment

Yep, it's the use of $<$ and $\le$
The $\le$ takes into account $-I$, my bad :P

@TheSimpliFire Yup you figured it out :P
Do you think it's an interesting post?

To answer your question, one way would be to evaluate a general $(A^\dagger)^k$ for $k=1,2,\cdots$ and then multiply by $\vert\phi\rangle$
@MoreAnonymous It appears an interesting problem IMO

1:29 PM
@TheSimpliFire Thanks ... I was hoping to make use of it somewhere (bounding the S matrix) and somehow (making the math work out)

1:48 PM
can i ask doubt here?It's very simple question

2:13 PM
Back
@MathGeek What is it

2:27 PM
@TheSimpliFire in reactions like H2 + I2 = 2HI.There is no effect of pressure.I clearly understand this fact by law of Action. But again this reaction Equilibrium gets affected by temperature as it's an Exothermic reaction.And Temperature is proportional to Pressure. Now these two statements are contrary to each other.Please explain where am i wrong?

@MathGeek There's an equal number of moles of gases on each side of the equation
Increasing the pressure has no effect.

@TheSimpliFire yeah i know that fact. but" But again this reaction Equilibrium gets affected by temperature as it's an Exothermic reaction.And Temperature is proportional to Pressure."
by gay lussac law t is propotional to pressure
and if this reaction has effect of temperature. so it must have effect of pressure

I think that's where we need the equilibrium constant
(partial pressures)

@TheSimpliFire ik i'm dumb but can you please explain

@MathGeek You have P1V1/T1 = P2V2/T2. The temperature does not have to increase as pressure increases, when V2 < V1 appropriately
Suppose the pressure is doubled. If the volume is halved, the temperature stays constant.
T is proportional to pressure assuming a constant volume

2:36 PM
since reactions doesn't depends on volume so it can be made to constant

n = cv

ok!! gotcha

@TheSimpliFire in which grade are you ??

last year of secondary

2:38 PM
i'm in class 10
high school

I got taught equilibria at about that time as well

3 hours later…
5:14 PM
@Mathphile Hi. Today in our newsletter : One solution of the equation $$x^3+y^3+z^3=42$$ where $42$ is the "answer to everything" (joke in a science fiction film where a computer should calculate the answer to everything and output 42) :
? print(a)
-80538738812075974
? print(b)
80435758145817515
? print(c)
12602123297335631
? print(a^3+b^3+c^3)
42
?
Finally, after decades of efforts, this solution was found !
in the university of Bristol