Any hints on how to solve $\phi '''(t) + \phi(t) = 0$?

@Lozansky solve $m^3+1=0$

Ah right

finished all my midterms thank fuck

algebra was today and dear god algebra is hard

Analysis exam down

congrats! :)

@DHMO that works if you do an appropriate substitution first

@Semiclassical what do you mean?

well, $\phi(t)$ isn't m^3, is it?

have to assume that the solution is of a particular form to get that

(it is, of course, but one has to do $\phi(t)=e^{i m t}$)

@Semiclassical isn't that the auxiliary equation?

yeah, but one should presumably understand where that comes from :/

and not just say "hey, auxillary equation"

@Semiclassical so you are saying that every time we want to use auxiliary equation we have to prove it first?

if you're doing a course on it, probably yes

if you can just eyeball it in actual practice, fine

The problem was $\phi '(t) + \int_{0}^{t} (t-\xi) \phi(\xi) d \xi = t, \: \phi(0) =0$

but if you're doing an exam then probably it's worth taking the time to say "$\phi(t)=e^{i m t}$"

BTW

@Semiclassical no, you don't need to assume that

to get to the auxiliary equation?

What do you mean?

(probably simpler to do e^{m t} instead but that's hardly different)

Look, let's solve the equation $y''-y=0$

what the heck is an auxiliary equation

without ever assuming that $y=e^{mt}$

@0celo7 What you get when you plug an exponential ansatz into a linear ODE.

I got $\phi(t) = c_1e^{-t}+e^{t/2}(c_2 cos \sqrt{3}/2 t + c_3 sin \sqrt{3}/2 t)$

You mean a characteristic equation?

same thing, yeah

Just me, or won't it be pretty terrible to solve that for $c_1, c_2, c_3$?

@Lozansky Probably easier to write it as $c_1 e^{-t}+c_2 e^{-\omega t}+c_3 e^{-\omega^2 t}$ where $\omega=e^{2\pi i/3}$ is a cubic root of unity.

then you can appeal to stuff like $1+\omega+\omega^2=0$

frikken algebra

how do roots of unity even work

yup

@0celo7 They just be rotaty arrowy things

@Semiclassical Maybe. Would probably simplify the integral at least

@0celo7 ...

$\omega^n-1=(\omega-1)(\omega^{n-1}+\omega^{n-2}+\cdots +1)$

@Semiclassical in which step did I assume that $y=e^{mx}$?

ffs

just because you can do it without it doesn't mean that's how you'd actually do it

@Krijn what does that mean?

@Semiclassical so how do you do it?

@DHMO interesting

@0celo7 You can rotate roots of $X^n - 1$ around

That's all

that seems hard

I assume a solution of the form $y=e^{m t}$. that would require $y''-y = (m^2-1)e^{m x}=0$

I like DHMO's way better

that has two roots, $m=\pm 1$, and those yield distinct solutions

so therefore $e^{x},e^{-x}$ form a basis of solutions

your solution is valid

but my solution needs no assumption

@Semiclassical My advisor has a library book checked out that I want. Is it ok to ask him for it?

Okay. How would your solution look for $\phi'''+\phi=0$, then

@Semiclassical similar

must i type it out?

maybe I'd just type a portion

If you're not willing to type it out, why would I use it as a technique?

@Semiclassical who said I'm not going to type it out?

@s.harp thanks

sigh

@0celo7 Can't hurt. If there's a specific passage you need, he can probably photocopy it

@Semiclassical I have a PDF already

I'll admit, I like that arrangement if I was going to do something like $\phi'''-\phi = \Psi$

I want to read the first chapter or so and see if it's something I want to read more fully

but I can't read that much on a pdf

this is the first part

sure. presumably you'd get two more similar parts

...or you could do $y=e^{m x}$, and not worry about cleverness

that gives $m^3=1$ immediately

somehow we went from $y'''+y=0$ to $y'''-y=0$. probably a typo on my part.

@Semiclassical cleverness? I kind of use the same approach for everything

@Semiclassical That kind of ansatz argument always bother me. I realize most people are ok with it if they want to know some specific solutions; but eh.

If you can guess an answer, and show that it works, then it's perfectly satisfactory.

Ansatz arguments are beautiful

How do you know if there aren't any other solutions?

Well, let me put it like this. Are you really telling me that you'd obtain the auxiliary equation by that chain of reasoning, that takes so long to write out that you don't want to do the whole thing?

@BalarkaSen Because it's a linear ODE

That's the kind of question I usually have in mind.

Now, if you're doing nonlinear stuff, then yeah

ansatz arguments are a bad idea in that case

@Semiclassical then how are you going to do $y''-2y'+y=0$?

Sure, it's a very specific approach, is all what I was saying.

In that particular case, I wouldn't want to use my approach. But it's hardly strange to use a different approach when the general one fails.

@Semiclassical i'm really curious. how would you do it?

Probably I'd do something like $y=ze^{x}$, since the auxiliary equation $m^2-2m+1=0$ has a double root at $m=1$

yeah isn't that what you do for Cauchy-Euler or one of those guys?

$xe^x$ or something

That gives $(z''+2z'+z)e^x-2(z'+z)+ze^x=z'' e^x=0$, so $z=C_1 x+C_2$

Isn't that $x^2y''+axy'+by=0$?

The other way to go at this, of course, is some sort of transform method

Though for that I'd want to be doing $Ly=f$ not $Ly=0$

@DHMO myself I'd put another line between the third and fourth steps, explicitly stating that $\frac{d}{dx}[e^{-x}(y''-y')-e^{-x}(y'-y)]=0$

@Semiclassical fair point

admittedly, things are going to get hairy whatever you do if you're doing something third-order or worse

e.g. $(D-1)^2(D+1)y=0$

what is that?

$D=\frac{d}{dx}$

What does $(D+1)$ mean?

$(D+1)y=Dy+y=y'+y$

wow, notation

but you already factored it for me

yeah.

if it's in factored form, then I can see the advantage of the direction integration route

e.g. $e^{x}(D+1)y=D(e^{x}y)$

actually that's just integrating factors

well, I mean that the next step would be $D(e^x y)=0\implies e^x y=C$

i.e. direct integration

sure

but, i mean, if someone gives me an ODE like $ay'''+by''+cy'+d=0$

i'm going to go directly to the $y=e^{m x}$ ansatz and see if I get distinct roots

if I do, I'm done.

if not, then more labor is required.

Look, we are both using auxiliary equations

just that you assume $x^ne^{mx}$ while I do not

if you have a polynomial $P(\partial_t)\phi=0$ the best thing to do is find the roots of the polynomial and write the equation as $(\partial_t - z_1)^{n_1}\cdot...\cdot(\partial_t - z_m)^{n_m}$. You know that everything in the direct sum of the kernels of the operators $(\partial_t - z_i)^{n_i}$ is a solution and from Picard Lindelöf theorem you get that this solution space has the same dimension as the solution space of your equation

eh. my point is that if you've got something like $y'''-y=0$, then going directly to $y=e^{mx}$ seems far simpler

@Semiclassical no. you solve the auxiliary equation first, or else you won't know if you have repeated roots

UGH

lol sorry

If you try $y=e^{m x}$ and it works, you're done. but regardless, that gives you the auxiliary equation

otherwise you're just pulling an equation out of nowhere and saying "hey, this thing has repeated roots"

and I'd rather pull a generic solution form 'out of nowhere' than an arbitrary equation

fair point

I would say "consider the following equation..."

or even put it to the rough work

I think one could reasonably proceed as @s.harp did, and frame it instead in terms of "how do I write $Ly=0$ as $(D-z_1)...(D-z_m)y=0$"

I am not too familiar with that notation

For example, I can't see this without converting it to "normal" form:

It's handy, especially if you're the kind of person who likes operator methods

$D(e^xy)=(De^x)y+e^x(Dy) = e^x y+e^x Dy = e^x(1+D)y$

you can also write it in the following rather cute way: $e^{-x} D e^x = D+1$

not very different from converting it to $e^xy+e^xy'$

and more generally $e^{-\lambda x}D e^{\lambda x}=D+\lambda$

thanks

which is something you should perhaps recognize, @0celo7: $x$ generates translations in momentum space :>

@Semiclassical how would you use that?

@Semiclassical what is $D$ here? $e^{-i d/dx}$ or something?

just $d/dx$

oh, I'm thinking about the finite translation version

you're talking about the infinitesimal generator?

yeah. i'm being a bit sloppy

@DHMO well, I already have: $0=(1+D)y=e^{-x}D e^x y=0\implies ye^x=C$

where that $D$ notation really becomes fun, though, is if you have inhomogeneous ODEs

for instance $y+y'=e^{2x}$

@Semiclassical sloppiness overload

@Semiclassical let me try

@Semiclassical what is even going on there?

oh, hmm

don't see why that's sloppy

seems clever to me

@0celo7 theorem:

@DHMO yes yes

@0celo7 because you included $y$ in the $D$

the sloppiness was in regards to the finite translations vs. infinitesimal translations business

it is like saying that $fDg=Dfg$

since I don't remember the terminology correctly off the top of my head

I didn't really understand that comment

mine?

what do the exponentials have to do with momentum space

yeah

well, first, let's trade derivatives for momentum i.e. $p=-i D=-i\frac{d}{dx}$

sure

then the above statement becomes $e^{-\lambda x} p e^{\lambda x}=p-i\lambda$

@Semiclassical let me try this...

sure

and if I write $q=i\lambda$, that becomes $e^{-i q x}p e^{i q x}=p-q$

sure

so that's served to translate $p$ by $q$

uh

$p\mapsto p-q$. that's all I'm asserting.

what map is that?

$\hat{A}\mapsto e^{-i q x}\hat{A}e^{i q x}$

although that is really pretty sloppy. I think you're saying $e^{-iqx}p[e^{iqx}f]=(p-q)f$.

sure. but that's pretty common notation in physics.

you typically don't write the thing you're acting on, just the operators themselves

ok

I'm still not sure what you're trying to tell me...

Hello ! Greeting !

@Semiclassical @DHMO Hey !

Can anyone help me verify the solution?

$x$ generates a unitary transformation of $p$ s.t. $p\mapsto p-q$?

I guess I'd put the point like this: just as the momentum operator $\hat{p}$ is the generator of translations in position, so to is the position operator the generator of translations in momentum

@Mahmoud as-salaam

oh. i've also assumed $\hbar=1$ so that $p=-i(d/dx)$.

Are you Muslim @DHMO ? O.O

@Mahmoud no

@Semiclassical of course

as-salaam ??? @DHMO

the cute approach via $D$ for $(1+D)y=e^{2x}$, btw

@Mahmoud I thought "Mahmoud" is an Arabic name lol, seeing its resemblance to Mohammed

is to totally abuse notation :P

Yes it is @DHMO

Poor notation

محمود @DHMO

@Mahmoud and "hello" in arabic is "as-salaam"?

as-salaamo alaykom

sure

Which means May peace be upon you ! :)

السلام عليكم

@Mahmoud you're a muslim?

Ok just ignore me and continue with the problem I shouldn't have interrupted you.

Yes.

namely, $$(1+D)y=e^{2x}\implies y=(1+D)^{-1} e^{2x}= (1-D+D^2+\cdots)e^{2x} = (1-2+2^2+\cdots )e^{2x}=\frac{1}{1+2}e^{2x}=\frac{1}{3}e^{2x}$$

@Mahmoud there's no on-going problem here

@Mahmoud nice to know that

which looks totally absurd, but gives the correct particular solution

@Semiclassical what the hell?

That is the correct reaction, lol

but you forgot $Ce^{-x}$

Hence why I said particular solution

You can't get the homogeneous solution this way.

@Semiclassical $(1+D)y=e^{2x}\implies y=(1+D)^{-1} e^{2x}=\dfrac{1}{1+2}e^{2x}=\dfrac{1}{3}e^{2x}$

Lol

one absurdity less

That's more physics than I can handle

lol

To make that less absurd, you'd want to think of this in terms of some transform method

transform method?

I took the exam @Semiclassical

Laplace?

Laplace transforms, Fourier transforms, etc.

I'm only familiar with Laplace transform

ah.

Fourier transform is similar but not identical

There's probably some way of making that rigorous using pseudo differential operators or something

I know there are relations between them but I forget how

@Mahmoud Ah. How was it?

which transform should I use here?

Don't know, tbh.

I think Laplace would work here

let me try Laplace

Pretty straight forward but I couldn't solve the last problem. @Semiclassical

Okay. What was it asking?

Prove (As always :/) that :

What test?

works perfectly @Semiclassical

yeah

and in that case being able to divide by $(1+s)$ is perfectly legit

$$(a\neq b) \lor(b\neq c) \lor(a\neq c) \implies a^2+b^2+c^2\neq ab+bc+ac$$

@Semiclassical what do you mean?

God I hate Laplace transforms

@0celo7 with a passion?

Almost

I hate them less than linear algebra fwiw

Hmm

For that to hold, you need at least one of $a,b,c$ to be different than the others.

in fact, all of the $\implies$ can be changed to $\iff$

@Semiclassical what does this mean?

oh, I just meant that when I wrote $(1+D)y=f\implies y=(1+D)^{-1} f$ it's rather absurd on the face of it

on the other hand, you also get something like $(1+s)Y=F$ at the level of the Laplace transform

Green's functions.

I see

and in that case it's perfeclty valid to move $(1+s)$ to the other side.

If only one of them is different, then you'd have (without loss of generality) $a=b\neq c$, which would require $2a^2+c^2 \neq a^2+2ac\implies (a-c)^2\neq 0$ which is always true.

so the only case that needs be considered is $a\neq b\neq c\neq a$

but I like DHMO's approach.

@Semiclassical thanks

it basically amounts to rewriting the desired condition as $(a-b)^2+(b-c)^2+(c-a)^2\neq 0$, which is only false when all three are equal.

I'm back sorry.

@Mahmoud as-salaam alaykum

:) @DHMO

@DHMO ,Wa-Alaikum-Salaam :)

Wa alaykum as-salaam =) @DHMO

@MithleshUpadhyay so many arabic xd

:P

I bet noone here speaks my language

Russian is not very rare...

Not russian

@DHMO , you welcome, and Namastey ! :)

Ti govoris po russki, @DHMO?

@Lozansky I'm not Russian

I'm talking about swedish

@MithleshUpadhyay I've seen your question however you need to know the question is whether is about labeled or unlabeled tree

@Lozansky Are you from Russia? where exactly?

Thanks @DHMO for your solution $$=-)$$

@YOUSEFY From the western part of Russia called Sweden ^^

@Mahmoud you are welcome

@YOUSEFY It should be labeled, rt?

@DHMO I had an interesting question on my complex analysis exam

@Lozansky Cam on you are not from Russia!!

@Lozansky what is it?

Is there any good lecture available on the Internet about Set Theory ?

@MithleshUpadhyay According to my knowledge, if the question doesn't specify, then we assume it is about unlabeled tree

@DHMO Find a conformal mapping $w=f(z)$ that maps $|z|<1$ to $|w|<1$ with $f(0)=1/2$

That sounds like a Blaschke product

@YOUSEFY Sorry to disappoint :(

@YOUSEFY , ok , thanks

@Semiclassical What do you think ?

on the question you asked?

I like DHMO's solution; i think it's nicer than what I could've come up with

@Semiclassical I don't know what Blaschke product is

No about Set Theory.

here: en.wikipedia.org/wiki/Blaschke_product

@Mahmoud oh. no idea

@Lozansky what is a conformal mapping?

Okay thanks

From that, I get $z=\frac{w-1/2}{1-w/2}$ up to a multiplicative constant

which gives $w=f(z)=\frac{z+1/2}{1+z/2}$

@DHMO Basically a map that preserves angles

@Semiclassical That's cheating :P We don't know about Blaschke products

lol

yeah