Yup, moelcular and orbital symmetries play a huge role in computationa chemistry, which is related to my PhD next year

still yet to get my head around irreducile representations through

Meanwhile, orbits help visualising what a group is doing most of the time

Irreducible representations are just representations that cannot be broken into a direct sum

Where in the case of a group representation (in opposition to representations of lie algebras etc) having any invariant subspace will immediately give you a direct sum decomposition

darn it, I lost most of the code for my Lambert W graph :(

So really we just require that there are no invariant subspaces

It's interesting how important character theory evidently is in chemistry, and Lie group theory is definitely important for particle physics.

Yep, symmetry is amazing right now.

But the condensed matter theory stuff I do doesn't rely on it. We still rely on symmetry, but irreps aren't something we talk about much.

Sure

@Secret so you're a polymath

Just interesting to me how some parts of math are almost-but-not-quite relevant to me.

@DHMO There exists at least one person in each of my stage of life will think I am that, but no I am not I can only confidently said I know to expert level in chemistry since I have studied that through my honours. Anything else I just happened to know enough that most people often fail to found I am actually an outsider and not an expert

@Secret but you know molecular orbitals

@DHMO That's a important prerequiste in studying computational chemsitry, physical chemistry and organic chemistry. They even taught the basics of it in year 1 nowadays

Yeah.

It's how you explain covalent bonding, isn't it?

hello all

@Secret I see

@Semiclassical yes

and also what orientation in order for molecules to react and other stuffs

I mean, Lewis diagrams are the toy explanation for covalent bonding, but MO is what you need to start doing it seriously

@Semiclassical and the toy might be wrong (read misleading) also

Right.

of course, that doesn't mean MO is perfect

Sure.

Organic chemists use qualitative arguments such as frontier MOs to explain why some reactions go and others are hindered. Physical organic chemists goes further by measuring reaction rate to check the transition states, and computational chemists does the hardcore wavefunction treatment where orbitals cobined to form the molecular wavefunction

It's some approximate picture of what's going on with the Schrodinger equation.

and something something DFT

DFT works well for large strutures, where ab initio methods (direct calculation of the waveffunction) beocmes too computationally expensive

those in the material science and solid state chemistry use DFT alot to check the electronic structures of their composites

progress in graphing polar curves!

this isn't what I wanted to graph, but at least I graphed something!

That's definitely progress then :P

I have code in R to plot curves but i'm trying to adjust it to work on polar curves

oh I see the problem

now, how to fix it

much b etter! except this isn't $r = 1 - \sin \theta$...

but it's arguably better than before

success :)

@GFauxPas congraulations

thank you :) only took me two hours

I normally use golden ratio but that doesnt look right

or maybe it does who knows

**** you. Squared zero! (but at the same time, I am glad of you)

so that means, I should try one sided additive identities or throw away one of the distributive laws and see what happens

No, actually I need to throw away exactly one distributive law AND one additive identity. You cannot have two sided additive identities in associative division by zero term algebra

lol I just figured out that it has a built in polar form :)

defaults to clockwise and theta=0 being 12 o clock though

I guess because that's more natural when making polar charts for presentation

Simply put, if Theorem 5 holds, then the no-go theorems follows

@Secret We (me and balarka) discussed the proof of the transcendence of e in my room

I saw, I have not digested completely yet other than how balerka said the intuition is exp being invariant under differentiation

you can always raise question

hi chat

Some main site questions just annoy the hell out of me for how lazy they are

and they banned lmgtfy comments :(

There's one there right now which literally consists of "I know $a=-\text{number }\times x$, what's the frequency?"

well, whats the frequency?

that's not periodic

wait, what

$a(t)$ is acceleration here, i.e. second derivative of $x(t)$.

and there's some trig going on somewhere I guess

and even the slightest amount of inspection of a textbook would give you: For simple harmonic motion, you've got $ma=-k x$ and the resulting angular frequency is $\omega=\sqrt{k/m}$

yes Semiclassical but that's ma

so literally it's just "take the square root of your number"

There can be tricky problems in simple harmonic motion, to be sure; I'd count ones involving initial conditions to be among those.

@GFauxPas Sure. You have to divide through by $m$ to get $a(t)$ by itself, and doing so makes the coefficient of $x(t)$ become $k/m$.

I'm joking semiclassical

ah.

I guess I can't assume things are obvious

It certainly depends on the audience, of course. I wouldn't assume that someone who hasn't been in a physics course recently would remember it.

But if you're asking a question like that, you're probably taking a course on it. And if you're taking a course on it, you have enough tools it -should- be obvious.

Hence why such questions annoy me: The only context in which I can imagine someone wanting to answer that is one in which they should be able to answer it themself.

Can the infinitude of primes be proven by the sieve of eratosthenes?

$4!+0!+5!+8!+5!=40585$

@MatsGranvik how?

@DHMO No I don't know. I am asking.

I don't think I know any proof that uses it

I think you could possibly construct an argument based on the pigeonhold principle?

so: what's the difference between $=$ and $\equiv$?

@AkivaWeinberger i see we were reading the same copy pasted question :P

"If you have $n$ pidgeons and $n+1$ holes there's a pidgeon with at least two holes in it"

@AlessandroCodenotti Yurp.

@heather What's the context? Modular arithmetic?

@AkivaWeinberger, just in general.

Depends on context.

oh.

In modular arithmetic, $“a\equiv b\pmod m”$ means $a$ and $b$ leave the same remainder when divided by $m$.

If $m$ is $10$, it means $a$ and $b$ have the same last digit.

So it's weaker than being equal, they're just similar in a specific way

right, but mod is an operation, like $+$ or $\times$, right? so i'd think $=$ could make sense.

@heather You might want to look up equivalence classes

okay

@heather Not actually.

For a geometric example, consider two congruent triangles.

@AkivaWeinberger oh, it isn't? hmm

You could define an operation,

@Semiclassical but aren't they equal? why congruent and not equal?

like $a\mod m$ is the smallest positive number equivalent to $a$ mod $m$,

Depends on what you mean by equal. For many properties, being congruent is entirely sufficient to say that if it's true for one triangle then it's also true for the other.

but then you'd write it like $a~{\rm mod}~m=b~{\rm mod}~m$ instead of $a\equiv b\mod m$.

For instance, if two triangles are congruent and one of them is acute, then of course so is the other.

@AkivaWeinberger okay, so then my question would be why isn't mod defined as an operation? why is it defined as a "similarity"

On the other hand, I could instead characterize them in terms of how much horizontal range they represent.

The more useful point of view is that $a$ and $b$ fall into the same "category" (more formally, equivalence class), than that they have the same image under that operation.

@AkivaWeinberger, interesting...

why is it more useful?

In that case, one of the triangles may have more horizontal range than the other. So I couldn't just say "what's true for A is true for B" in that case.

The point is that it's all about what properties are preserved under the equivalence. When we say that two things are 'equal' we generally mean that all possible properties are the same, i.e. there's no way to distinguish them.

To be honest, I'm actually not sure. Would make a good MSE question.

Where equivalence comes into play is when you want to preserve some properties while not worrying about others.

:34396120 no, it's alright, I can ask it

Wait no ignore that that's a bad idea

don't ask it?

why not?

seems like a legitimate question to me

'Cause I'd get all the reputation points instead of you

@Semiclassical hmm interesting

that does make sense

@AkivaWeinberger oh, I thought you meant don't ask the question period

@AkivaWeinberger meaningless internet points are pretty awesome =P

For instance, I can make an acute triangle into an obtuse one by stretching it appropriately.

So under an appropriate transformation, I can say that the two are equivalent. Is that a productive instance of equivalency? Maybe, maybe not; it depends on what properties are preserved and to what extent I care about them.

Hi, I am trying to solve a differential equation:

$y''-6y'+9y= 3t-8e^t$

If I care about angles, then that equivalence is useless to me because it doesn't preserve angles. But if I only care about ratios of lengths, then I'm perfectly happy with that equivalence because it doesn't change those.

@Semiclassical thank you, that explanation made a lot of sense

and the prediction method doesn't work and if I try to solve it using Cramers method i get really hard integrals to solve ( $\frac{3t-8e^t}{e^{3t}+3te^{3t}}$

I should also stress that when one talks about \equiv in the context of modular arithmetic, there's not just one such equivalence. There's one for each n>1.

is there any other way to solve this?

Each one preserves some information---for instance, if two numbers are equivalent mod n, then they both have the same remainder when divided by n.

But it doesn't preserve other information: Knowing x=y mod 7 tells me nothing about their remainders when divided by 11.

On the other hand, if I know x=y mod 7, then it must be true that either x=y or x=y+7 mod 14.

So different equivalence relations on the integers will preserve different but inherently consistent information.

But I'm rambling now.

Max, that one is ripe for Laplace transform method

^

do you have an initial condition?

no initial condition

but this is at the beginning of calculus

hmm, then, just put $y(0) = c_1, y'(0) = c_2$

and we didn't have Laplace transforms yet

then undetermined coefficients?

do you know how to solve $y'' - 6y' + 9y = 0$?

yes

then there's a table

the solution is $y=C_1e^t+C_2te^t$

The general rule when it comes to inhomogenous linear ODEs, like the one you have, is that the solution is a general part (which solves the homogeneous ODE) and a particular part (which satisfies the RHS.)

@max Not quite. That'd be true if it were $y''-2y'+y=0$

$y=C_1e^{3t}+C_2te^{3t}$

i mean

Right.

okay, that's the solution to the homogeneous part, homogeneous means equal to zero. call it $y_h$

So you've got the general part of your solution. What remains is to find any particular solution to the inhomogenous ODE.

right, call it $y_p$

ok

There's a number of methods for doing that, but the simplest one is to make a good guess.

and it's a theorem that all solutions are of the form $y_h + y_p$

now i need to find a particular solution that satisfies $ 3t-8e^t$

Or, rather, to make a guess which has enough freedom in parameters that you can make it workk.

$ LEFT = 3t-8e^t$

so the idea is, Max

did you take linear algebra

yes

the space of solutions is an infinite dimensional vector space

you want to find a basis for that space, but you don't know how many basis components you need

so you want to pick all the possible ones, and the extraneous ones will be identically zero

so, let's see

you want to find a function $f(t)$ such that $y,y', y''$ are linear combinations of some "basis functions"

A place to start is to think: I want to differentiate something several times and then take linear combinations thereof to get something linear and something like $e^t$.

So from experience, you know that it's going to be a polynomial

of what degree, Max?

Eh, it should be -partially- polynomial

well right

@GFauxPas could you find other pandigital fractions?

degree 1

You also need something for the e^t.

@GFauxPas such as 2/3 and 3/4 and etc

ax+b

I'm just talking about the $3t$ part

@GFauxPas i'm off now bye

I figured.

bye DHMO

but then $y'' = 0$

So if you take a trial solution $y_p=at+b$, what is $y''-6y'+9y$?

which may be the case, actually

but let's try $y_p = at + b$

@heather I guess the idea is that there's nothing "special" about the set of numbers $0,\dots,m-1$ when we're doing modular arithmetic. Maybe it's more useful to consider $\lfloor-p/2\rfloor,\dots,\lfloor p/2\rfloor-1$ for some problems.

$9at+9-6a= 3t-8e^t$

Max let's just do the $3t$ part for now

but i am missing the e^t part

Ignore the -8e^t part for the moment. We'll add it next.

ok

you got $9at + 9 - 6a = 3t$?

also, it'd be 9b not 9.

So defining a $\rm mod$ operator that gives you the number in the range $0,\dots,m-1$ is kind of artificial. @heather

$9at + 9b - 6a = 3t$

@AkivaWeinberger I'm sorry, I don't know what that notation means in the second sentence

Right. You can then match coefficients to determine a,b uniquely.

so a=1/3 and b=2/9

^ Right.

nice, Max.

So our particular solution should contain a term (3t+2)/9.

and now i need to find a solutaion that satisfied $-8e^x$

now for $-8e^t$

right

so i need to guess another function

and then my specific solution would be the sum of those two functions i guessed?

Right. So, what's a function you'd take derivatives of to get e^t?

Right.

each guess is called an "ansatz" if you want to know the technical term for it. it's german for "attempt"

but i thought of an objection to it, perhaps: if a number x is "equal" to another number y mod some number $n$, that doesn't mean that x=y, but for other operations this is true, like if you had x+n=y+n, x=y @AkivaWeinberger

$Ce^t$

Yup. So let's try that.

What does y''-6y'+9y give in that case?

@heather Floor — largest integer less than or equal to the number. (I might have been off by one, though.) For a specific example, I meant that maybe it makes more sense to consider $-3,-2,\dots,1,2,3$ as our numbers mod 7 rather than $0,1,\dots,5,6$.

$Ce^t-6Ce^t+9Ce^t=-8e^t$

good. now you can equate coefficients

co $4C=-8$, so $C=-4$

C=-2.

yes, -2

so the specific solution is (3t+2)/9-2e^t

Right.

and the theorem is that the sum of the homogeneous solution and the particular solution will give you the solution to the whole thing

linearity and all that

ok, got it

thanks!

Glad to be of help.

sure. you can find tables online of good guesses

this is called the method of undetermined coefficients

There are other methods than this (which amounts to undetermined coefficients, I suppose).

For instance, one approach is to construct the so-called Green function of the ODE subject to the appropriate boundary conditions. That usually shows up when doing boundary-value problems rather than initial value problems.

I never learned that method Semiclassical

Basically, you solve the ODE $Ly=\delta(x-x')$ where $\delta(x)$ is the Dirac delta.

$L$ is a function or a constatn?

operator.

Oh, so, for example, a polynomial in $D_t$

?

Right.

neat, thanks for showing me something new :)

Once you've solved that ODE, you can write any forcing term as $f(x)=\int f(x')\delta(x-x')\,dx'$

aren't these PDEs?

Nah. Though you have to be careful about which variable you're referring to.

I'm being a bit sloppy there.

Wikipedia states it more carefully.

Another approach, which is a bit silly but still works:

we can write the above ODE in operator form as $(D_t-3)^2y=(D_t-3)^2 y_p = 3t-8e^t$

okay

If I want to solve for $y_p$, one thing I can try is to treat $D_t$ as though it's just a variable. In that case, I'd have $y_p = \dfrac{1}{(D_t-3)^2}(3t-8e^t)$.

Going along those lines, I expand the fraction in front in a Taylor series:

This is what Oliver Heaviside did!

I read about this

Yuuup.

Since you've heard of it, I'll allow myself to be lazy and not finish the details :)

no please finish :(

Fiine.

@Max are you still here

yes

$$\frac{1}{(D_t-3)^2} = \frac{1}{9} \frac{1}{(1-\frac13 D_t)^2}$$

after Semiclassical explains this way I'll show you how to do Laplace transforms if you want

that's my personal favorite way

$$= \frac{1}{9}\left(1+\frac{1}{3}D_t+\frac{2}{9}D_t^2+\cdots\right)$$

well

bah, why isn't that last bit rendering.

that requires knowledge about Laplace transform

:/

Okay. So each term is of the form $\frac{2}{3^{n+2}}D_t^{n}$, acting on $3t-8e^t$.

and then you just apply that to the RHS?

Right.

When it acts on 3t, only the first two terms survive. So you get 1/9*3t+2/27*3 = (3t+2)/9---same as before!

neat stuff :)

And when it acts on $-8e^t$, each $D_t$ just acts as the identity. So you get $\sum_{n=0}^\infty \frac{2}{3^{n+2}}(-8e^t) = \frac{2/9}{1-1/3}(-8e^t)=-\frac{8}{3}e^t$...which isn't quite right. Hrm.

How would you solve an equation like: $k = ax^{-2} + bx^{-1} + cx + dx^2$? Solving for $x$.

@jakebird451 Multiply by x^2 on both sides to get a quartic, and then pray it has nice roots.

There is a quartic formula, but it's useless for applications. So unless you can isolate a simple root or two you probably can't say much.

@Semiclassical Hahaha

Oh, I see what I did wrong. Should've been $\sum_{n=0}^\infty \frac{2n}{3^{n+2}}(-8e^t)$.

there we go

in which case it's $\frac{2/9}{(1-1/3)^2}(-8e^t) = -2e^t$.

I can see why Laplace transforms have kind of superceded this method, fewer mistakes

Though the simpler logic is that, since $D_t (-8e^t)=(1)(-8e^t)$, we have $\dfrac{1}{(D_t-3)^2}(-8e^t)=\dfrac{1}{(1-3)^2}(-8e^t)=-2e^t$.

nice

just replace $D$ with $I$ when acting on $e^t$

Right. If you had $e^{\lambda t}$ instead, you'd get $D_t e^{\lambda t}=\lambda e^{\lambda t}$ and so would replace $D_t\to \lambda$ when acting on $e^{\lambda t}$.

So the method is cute, albeit seeming to flaut any interest in rigor :)

Heaviside was a weirdo

I know you can do things to make this more sound, but eh

a smart weirdo

hah

did you know he painted his nails

pink

I think the transform method is nicer for this stuff, but it's nice to see that the formal operator approach can work as well.

Huh.

Are these eigenvalues related to the parity problem?

oeis.org/A275345

In later years his behavior became quite eccentric. According to associate B. A. Behrend, he became a recluse who was so averse to meeting people that he delivered the manuscripts of his Electrician papers to a grocery store, where the editors picked them up.[14] Though he had been an active cyclist in his youth, his health seriously declined in his sixth decade. During this time Heaviside would sign letters with the initials "W.O.R.M." after his name.

Heaviside also reportedly started painting his fingernails pink and had granite blocks moved into his house for furniture.[3]: xx In 1922, he became the first recipient of the Faraday Medal, which was established that year.

maybe it's a false rumor

You can also imagine how this would work if you had $\cos \omega t$ instead. Then we'd have $D_t^2\to -\omega^2$.

Any linear $D_t$ terms would be a tad annoying, but could probably be taken care of by partial fraction decomp of $L(D_t)^{-1}$.

But the Laplace transform has much the same kind of effect: $(Ly)(t)=f(t)\implies L(s)Y(s)-Y_0(s) = F(s)\implies Y(s) = \frac{Y_0(s)+F(s)}{L(s)}$ where $L(s)$ and $Y_0(s)$ are appropriate polynomials in $s$.

and yadda yadda yadda from there on.

You can also couple this nicely with the Green function approach, since taking the Laplace transform of a Dirac delta is trivial.

It all hangs together, as it must.

Is this an okay place for small questions?

Sure. No guarantee you'll get an answer, of course.

Can you have a polynomial over a finite field, of order q, with more then q roots?

It would have to be zero, but I don't understand how you could have more roots then the number of elements the underlying field allows.

I think I'm reading a theorem incorrectly.

What's the degree of said polynomial?

Plus, wouldn't you have to worry about multiplicity as well?

I mean, (x-1)^100 has 100 roots; they just all happen to be the same.

Well, the degree is less then the order of the field.

Okay.

I was under the impression the "reduced polynomial" is uniquely determined, and "sufficient".

So, for polynomials over infinite fields, if they are zero on infinite values, it must be the zero polynomial, because there can only be finite roots.

This old question/answer seems relevant: math.stackexchange.com/a/782777/137524

But, regardless, I don't see how you could ever have a polynomial whose roots over a finite field (not counting multiplicity) would be bigger than the field itself.

This theorem is in Lang, it's supposed to be the analogue for finite dimensional fields. I just was under the impression I didn't have more then q roots to test.

I'm not even sure what that would mean.

What's the theorem you're talking about?

You could have to test if a root shows up more than once, to be sure.

That's exactly what I'm confused about, I think I'm reading it wrong.

Corollary 1.8 in polynomials.

i.e. you might have to test if $\alpha$ is a root not only of of $f(x)$ but also of $f'(x)$, $f''(x)$, etc.

"Let k be a finite field with q elements. Let f be a polynomial in n variables over k such that the degree of f in each variable is less then q. If f induces the zero function on k^(n), then f = 0."

That's a multivariable polynomial?

proof: ..."If n=1, then the degree of f is less than q, and hence f cannot have q roots unless it is 0. ..."

Sure.

@DanielFischer ideas,help or hint here? :) thanks!