Mathematics - 2017-06-13 (page 1 of 7)

Dodsy

00:08

Reviewing calculus shouldn't be too tough..

Fargle

You have all of us nerds to help you, too, @Dodsy.

Dodsy

for the function $g(x)=8x^2-x+4$ at what tangent point is the instantaneous rate of change equal to -1?

$g'(x)=16x - 1$
$-1=16x-1$
$0=16x$
$x=0$

$g(0)=8(0)^2-(0)+4$
$g(0)=4$

Therefore the tangent point is equal to $(0,4)$

Very trivial.

I think I'll just have to review the chain rule, product rule etc.

Meow Mix

Hey nate have you seen the harmonic series?

or, well, proof that it's divergent

Dodsy

no :)

Meow Mix

Okay so the harmonic series is just

$1 + \frac{1}{2} + \frac{1}{3} + \dots$

we want to prove that this series is divergent

Here's something really cool

Replace every other term with the term after it so everything is duplicated

like this

$\frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \dots$

Now

this is obviously less than our first series

since every other term is replaced with a term smaller than it

However, when you bundle these halves together

you get $1$...

Similarly, when you bundle the $1/4$s together, you get $1/2$

And so on

And so whhat you have, is a series that is less than itself; which could not happen were it convergent

Dodsy

00:15

huh

that's cool.

arctic tern

@MeowMix If you are familiar with calculus, try finding an estimate for $1+1/2+\cdots+1/n$

Meow Mix

Yeah that's easier to read since i dont have mathjax enabled

Dodsy

I can tell it's approaching 2.

:D

Meow Mix

No it's not

It's approaching infinity

As we just proved

Mike Miller

@arctictern let G act on itself by conjugation. can you give a reasonably strong necessary condition that G semidirect G is isomorphic to G x G?

Dodsy

00:19

it's approaching infinity?

Meow Mix

Yeah! we just proved that!

Mike Miller

that's obviously true if G is abelian, and is true for (stupid example...) compact simply connected simple lie groups

Dodsy

I thought it was an infinite series approaching 2.

Akiva Weinberger

@Dodsy Look at the denominators again

Meow Mix

I might have to call Ted in here

Dodsy

00:19

oh

Mike Miller

i would bet it's true for all (finite) simple groups but composition series don't determine the group, so i dunno how to prove that

Dodsy

I see now.

Akiva Weinberger

You're confusing $1+\frac12+\frac14+\frac18+\dotsb=2$ with $1+\frac12+\frac13+\frac14+\dotsb=\infty$.

Dodsy

right.

I was.

Mike Miller

i actually don't have an example for which it's not true

arctic tern

00:20

hmm

how is it true when G is compact simply-connected simple lie group?

Dodsy

I hope I did well on my functions exam.

arctic tern

if the isomorphisms don't have to respect the product structure in any way I think the answer is pretty difficult

Dodsy

so @MeowMix I had a mate in 1 and missed it

I told you I'm rusty.

Mike Miller

i'm a little worried now but the following is what i had in mind

Fargle

@Dodsy RIP

Meow Mix

00:24

Bleghhhh

Akiva Weinberger

@MeowMix Do you know how to find the sum of $\displaystyle\sum_{n\to\infty} \frac{1}{n2^n}$?

Mike Miller

apply the classification of those; a simply connected compact lie group is semisimple, and the number of simple summands is determined by the rank of $\pi_2$. my claim is that one of these factors has to be the obvious normal subgroup of G itself; i probably have to do a bit more work to show this is true. it's certainly true for SU(2)

Meow Mix

I hate doing this

Not your problem Akiva

something else

Mike Miller

then the other factor (by modding out) has to be G as well

Meow Mix

Let me see

Mike Miller

00:24

p gross but i think it works

Akiva Weinberger

@MeowMix Oh, wait, I might have misstated it

Just a moment

Oh, OK, I did misstate it, but you should be able to find the sum anyway :P

Dodsy

when you're evaluating a limit

can you just input the x->n into the equation and solve?

Meow Mix

NO

Dodsy

I think they taught us that in this calculus course.

Akiva Weinberger

I meant $\displaystyle\sum_{n=1}^\infty \frac{n}{2^n}$

Meow Mix

00:26

well then they're bad people

Akiva Weinberger

@Dodsy If the function exists at $x=n$, and if the function is continuous, yes.

Usually the function doesn't exist at $x=n$.

Meow Mix

^

Dodsy

for instance the limit as x approaches -1 for the function $\frac{x^2 + 9x-4}{x^3}$

Akiva Weinberger

Luckily, most functions are continuous (compositions/sums/products of continuous functions are continuous), so the second concern rarely comes up.

Dodsy

this method works.

otherwise, you find the derivative right?

Akiva Weinberger

00:28

Yeah, that's fine. It exists because the denominator isn't zero there, and it's continuous because quotients of continuous functions (where they exist) are continuous.

Meow Mix

What do you mean

Dodsy

I haven't done limits in a while...

Akiva Weinberger

@Dodsy You mean L'Hôpital's rule?

Meow Mix

If it doesn't exist then surely there isn't a derivative

Akiva Weinberger

You only do that when it's an indeterminate form ($\frac00$ or $\frac\infty\infty$).

Otherwise, it will give you the wrong answer.

Dodsy

00:29

oh I see.

here's one that's tricky

Akiva Weinberger

For example, consider $\lim_{x\to0}\frac{x}{x+2}$. It's clearly $\frac02=0$, but wrongly applying L'Hôpital will give you $\frac11=1$.

Dodsy

$f(x)=\frac{x^2-2x-8}{x+2}$ as x approaches -2

Akiva Weinberger

Sure. That doesn't exist at -2 (it's $\frac00$). You have two options.

One is to factor the numerator; the second is L'Hôpital.

Dodsy

Right.

So you wouldn't put the answer as "0/0"

Akiva Weinberger

No.

Dodsy

00:31

you need to rationalize the numerator?

Akiva Weinberger

It's already "rationalized" — that only applies when there are square roots involved.

Dodsy

oh right.

Akiva Weinberger

I mean, rationalization means "getting rid of square roots".

Dodsy

oh I see.

Akiva Weinberger

If you try to factorize the numerator, you want two numbers that multiply to $-8$ and add to $-2$.

("Multiply to $-8$"—that means one's a negative and one's a positive, right?

That's the only way to get two things to multiply to a negative number.)

Dodsy

00:33

oh right, factoring.

-4 and 2

(x-4)(x+2)

so then the answer is -6.

That makes sense.

Akiva Weinberger

Yup.

Just for fun, let's see what L'Hôpital gives us.

Dodsy

okay :)

Akiva Weinberger

You remember how that works, right?

Dodsy

you take the derivative of the numerator and the denom and evaluate?

so it'd be $\frac{2x - 2}{1}$

?

Fargle

L'Hopital still feels like magic to me.

Dodsy

00:35

so it's still -6.

Akiva Weinberger

Yup.

> At $-2$, the numerator is $(-2)^2-2(-2)-8=4+4-8=0$, and the denominator is $(-2)+2=0$. They're both $0$, so it's of the $\frac00$ form and L'Hôpital's rule applies.

Yeah, it gives the same (correct) answer. Which is good.

Dodsy

right, so if the numerator and denominator are equal to 0, then we can use l'hopital

Akiva Weinberger

Right. The same is true if they're both infinity.

Fargle

@Dodsy Or if they would both approach infinity.

Dodsy

okay, great.

Fargle

00:38

Is there a simple-to-explain proof of L'Hopital? After all I know it's still a mystery to me.

Akiva Weinberger

I guess the idea is $\dfrac{f(x)}{g(x)}=\dfrac{f(0)+f'(0)x+\text{stuff}}{g(0)+g'(0)x+\text{stuff}}$

Fargle

I guess that makes sense, but it works for non-analytic functions f,g, too, right?

Akiva Weinberger

I mean, $\text{stuff}$ just equals $f(x)-f(0)-f'(0)x$, really.

Dodsy

hm weird.

Akiva Weinberger

Or the same with $g$.

Dodsy

00:39

now I'm having trouble with a really trivial average rate of change question.

Fargle

Mm.

Akiva Weinberger

I guess the main part of the proof is showing that it's small enough to be ignored.

Dodsy

this is MCQ.

Akiva Weinberger

I don't remember quite how it goes, but it's analysis-y, clearly

Dodsy

$g(x)=-3(x-1)^2+11$ what is the average rate of change between $x=1$ and $x=3$

Fargle

00:40

@AkivaWeinberger I'll look for something, but the intuition is clear. I appreciate it as always.

Dodsy

I'm getting 6

but only -6 is on there

arctic tern

@Fargle have you seen the Geometric Interpretation section on Wikipedia?

Fargle

@Dodsy So the average rate of change is the average value of the derivative, which can be found by taking the integral of the derivative from x = 1 to x = 3, divided by 3 - 1.

Doing that I get -6.

Mike Miller

@arctictern my friend has proved that if G is centerless, it's just G x G

but the symbol-pushing doesn't seem to work if that's not true

Dodsy

oh interesting

This course didn't teach integrals.

but I mean, I could use them still since it's MC

arctic tern

00:41

@MikeMiller Oh, for all G centerless?

Fargle

@Dodsy Well this case is easy, because the integral of a derivative is just the function itself (up to an arbitrary constant).

Mike Miller

yeah

Dodsy

I was using $\frac{y_2-y_1}{x_2-x_1}$

Mike Miller

it's an explicit isomorphism whose center is obviously in the kernel

Dodsy

ohhhhhhhhhhhhh

I was doing it wrong

I put 11 as $y_2$

Fargle

00:42

@Akiva probably has a better idea of how to approach that without invoking integration, my calc is a bit rusty.

Dodsy

so my method works too.

Fargle

transforms into a bad algebraist

Dodsy

lmfao

Mike Miller

i wonder if there's some trick that gives you everything

Dodsy

I have very poor attention to detail, and I'm not using paper.

Fargle

00:43

@Dodsy Yeah, actually, I'm dumb. The average rate of change between 2 points is exactly the slope of the secant line between those two points, so you did it exactly right, -6 is the answer.

Dodsy

lmfao, that's alright :)

Fargle

(Assuming the function is differentiable)

Dodsy

I'm bringing you back to high school right now.

Fargle

lol

tbh I could use it.

Mike Miller

@arctictern he googled a bit and apparently $D_8$ is a counterexample (dunno whether they meen the 8-element group or 16)

Dodsy

00:47

$f(x)=2x^3-4x+7$, find the equation of the tangent line at (2,15)
$f'(x)=6x^2 - 4$
$f'(2)=6(2)^2-4$
$m=20$

$y = m(x-x_1)+y_1$
$y=20(x-2)+15$
$y=20x-25$

right?

otherwise I failed that part of the test today

Akiva Weinberger

Seems right

Yeah, I just graphed it, it's right

Fargle

Nailed it.

Dodsy

Wonderbar

Akiva Weinberger

stares at KIND bar

Dodsy

Sorry if I'm flooding this chatroom with trivialities, guys.

TheGreatDuck

00:51

actually, that isn't a triviality

my question is probably more of a triviality

Akiva Weinberger

'T'sall good.

Dodsy

I love learning math because I find that even years later, you'll be doing some more math and realize you know math that you thought you forgot.

2

Terribly worded, but gets my point across.

Alright, I'm off for a bit guys, thanks @AkivaWeinberger for your help :)

TheGreatDuck

@Dodsy do you know the math behind having a guy walk on an arbitrary surface in space within a videogame?

Fargle

Take care, fam.

TheGreatDuck

Depending on your knowledge of spatial rotations of vectors... you in fact do.

Akiva Weinberger

00:54

Is this like Pacman on a torus? Or what

TheGreatDuck

@AkivaWeinberger heh heh heh

no.

it's pacman on any 3D model

the spaces are just triangular, that's all.

arctic tern

@MikeMiller apparently $G\times\{e\}$ and $D=\{(g,g):g\in G\}$ inside $G\times G$ act like $G\times\{e\}$ and $\{e\}\times G$ inside $G\rtimes G$, but the map $G\rtimes G\to G\times G$ given by $(g,h)\mapsto (gh,h)$ has trivial kernel.

(I didn't actually check it's a homomorphism, but $G\times G$ should be an internal semidirect product of $L=G\times\{e\}$ and $D$ with the conjugation action, since $G\times G=LD$, $L\cap D=1$, and $L$ is normal.)

TheGreatDuck

@AkivaWeinberger how about this on a torus? indiedb.com/games/block-builder

Daminark

01:59

The phrase "In pre-Bourbaki times" just happened...

LegionMammal978

@s.harp The problem is, how would I go about characterizing this set?

Leaky Nun

03:12

@Secret no, they aren't circular, they're just useless.

$a_{0}(0,n+1) = a_{0}(0,n^+)$ is a tautology.

Adeek

03:59

hi @arctictern

I was reading some paper where they defined algebraic geometry $C^{\infty}$ rings

it is crazy haha

check it out

arxiv.org/pdf/1001.0023.pdf

Balarka Sen

@AkivaWeinberger These sort of examples appear in stereochemistry.

Daminark

04:20

O lawd @adeek

O lawd @Balarka as well

Adeek

hi @Daminark

Daminark

How's it going?

Balarka Sen

@Daminark you still messing up your sleep schedule?

Daminark

The bootcamp isn't starting for a couple weeks. In the meantime, relegating as much sleep as possible to sunshine is probably a good idea because fasting

Adeek

good @Daminark

how is boot camp

Balarka Sen

04:23

Ah, right, fair enough.

Daminark

It seems like it's gonna be a lot of fun, I'm excited for it

Until that starts I'll be doing some number theory, some manifolds stuff, and perhaps a bit more on permutation groups

Balarka Sen

I think we proved something new on the NT room. Did you look?

Adeek

NT room ?

@Daminark sounds very exciting

Daminark

Yeah we've got a very secret room nobody knows about where we're doing number theory right here: chat.stackexchange.com/rooms/59893/number-theory-study-group

Adeek

We should have a room for commutative algebra + algebraic geometry :D

Semiclassical

04:26

I'm surprised you didn't call it straight out the Very Secret Number Theory Room

Daminark

Wait that's a good idea

Adeek

@Daminark do you want to create a commutative algebra room + algebraic geometry :D ?

Semiclassical

lol

Daminark

@Adeek Lol, do continue to discuss that on here so I can gawk at it for a while

Adeek

I could share my solutions that I have for AM

Balarka Sen

04:27

There was a commutative algebra/algebraic geometry room wasn't there?

I think it's frozen from inactivity

Adeek

I need to look at the homotopy category of projective R-module.

Daminark

Or that, if it's fine that I be on. Though I'd need to do the stuff mostly from scratch since I didn't learn much ring theory yet

Adeek

oh I didn't know there is one

Semiclassical

it never went anywhere, but I liked the name I had for a chat room I made here

"Semiclassical's methods"

Adeek

@BalarkaSen are you familiar with the homotopy category ?

not the topological one

Daminark

04:30

Fixed!

eyy

Balarka Sen

Superficially.

Adeek

@BalarkaSen I was wondering do you know if there is anything specail with the homotopy category of finitely generated project R-module ?

of bounded complexes

Balarka Sen

I don't.

Adeek

oh ok

Daminark

tries to keep up Uh, it's cool that, like, some ideals are maximal

Balarka Sen

04:37

yeah it's a local ring, and you keep localizing till you die

Adeek

yeah

Daminark

decides to keep to group theory to avoid this dark fate

Adeek

group theory is boring

ring theory in general is much better

Daminark

But this localization to death stuff sounds sp00ky

Also group theory ends up being p dank

(Especially when you get to permutation groups, and apparently representation theory as well)

Adeek

localization is really cool

Balarka Sen

04:43

@Daminark You just locally die, but not globally

Death Sheaf

I like permutation groups. Wish I was better with them

Daminark

Oh OK that's something at least

So, my schtick with group theory was odd. I did a bit last summer, but it was basically trying to reach group actions and Sylow as quickly as possible

Then just before this quarter I was skimming Rotman's Intro to the Theory of Groups, and audited a class on algorithms in finite groups, which focused on permutation stuff

Oh wait

Balarka Sen

lol wrong chat is wrong

Daminark

Rip in chat

Adeek

05:02

nights everyone

Secret

05:58

@AkivaWeinberger one can also imagine performing the improper rotation this way. Take the object, rotate by 90 deg about the vertical axis, then rotate in 4D along the xz axis which is in the plane where the object is standing, project the resulting object back to 3D, and you get the same result

@LegionMammal978 Actually, my aim is want those $a_{\alpha}(b,c)$ to be the integers, so any $a_{\alpha}(b,c)$ should be expressible in terms of applying $f$ repeatedly on the base case $a_0(0,1)$ and nested $a_{\alpha}(b,c)$ s

The idea is that a set membership is imposed so that e.g. $0 \subset a_0(0,1) \subset a_0(0,2) \subset ... \subset a_0(0,\omega)$

thus at least for this sequence, will recover the integers

This is why I don't understand why $a_0(0,2)=f(a_0(0,1))$ is circular, because the $c$ basically count how many times $f$ is applied to $a_0(0,1)$

So if everything is expended out, we should get:

$a_0(0,1), f(a_0(0,1)), f(f(a_0(0,1))), ...$

@LeakyNun Ah right, not paid enough attention to the symbols

Seems my circular reasoning detection is so poor. I wonder if it is because I have been exposed to too many recursive relations recently resulting in my thinking to become circular...

Leaky Nun

@Secret what is the definition of $f$?

@Secret did you say integers?

Secret

$0 \subset a_0(0,1) \subset a_0(0,2) \subset ... \subset a_0(0,\omega)$

Well that looks like integers, except with the $\omega$ at the end, (I can throw away that)

All those $a_{\alpha}(b,c)$ will be numbers expressed in terms of how many times $f$ is acted on the base case

Leaky Nun

1 min ago, by Leaky Nun

@Secret what is the definition of $f$?

Secret

and the number of times $f$ acted is tallyed by the right argument

Ok, question, why is the following real function acceptable?

Domain $\Bbb{R}$
Base case
$f(x)=x$

Inductive case
$f^{n+1}(x)=f(f^n(x))$

So e.g. $f^9(x)=f(f(f(f(f(f(f(f(f(x)))))))))$

Leaky Nun

06:15

@Secret because it's well defined for every positive integer $n$ and real $x$

Secret

How is that differ from $a_{\alpha}(b,c+5)=f(f(f(f(f(a_{\alpha}(b,c))))))$?

Leaky Nun

@Secret because you haven't defined $f$

@Secret here, $f$ is defined as the identity map $x \mapsto x$.

> Base case \ $f(x) = x$

Secret

Ok how about this: $f^{0}(a_{\alpha}(0,1))=1$ will that be a valid base case?

Leaky Nun

Can you copy the rest of the definition again?

Secret

https://a-ta.co/mathjax/!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…

Leaky Nun

06:20

@Secret a base case without an inductive case.

@Secret it's really simple. Pick a value. Ask yourself if the function is well-defined for that value.

Secret

I want $a_0(0,2)$ to be basically behave like $2$ which is why I accept $a_{\alpha}(b,c)$ as valid numbers which is why I don't get $a_0(0,2)=f(a_0(0,1))=f(1)$ is invalid

because I have both the image and the domain specified

here the input value is clearly $1 =\{0\}$

and it outputs $a_{\alpha}(0,2)$ which is defined to be a number and hence a set that contains $1$

and thus $a_{\alpha}(0,3)$ contains $a_{\alpha}(0,2)$ and so on

but anyway, adding the base case I mentioned many lines above in now

Leaky Nun

@Secret what is $f(1)$?

Secret

$f(1)$ is defined to be a set that contains $1$

likewise $f^{n+1}(1)$ is defined to be a set that contains $f^n(1)$

Leaky Nun

"a set that contains $1$" cannot be more vague

Secret

but $2$ is also a set that contains $1$, is it more specific because $2=1^+$?

More generally, I want $a_{\alpha}(0,c)$ to be define by the following relation:

$0\subset a_{\alpha}(0,1)\subset ... \subset a_{\alpha}(0,c)\subset a_{\alpha}(0,c+1)\subset a_{\alpha}(0,c+2), ...$

So if this is not specific enough, what else do I need?

What besides the integers (and more generally a well ordered set) have the above relation?

Leaky Nun

06:42

@Secret so you are still constructing your definition?

Secret

I guess so, ultimately, every $a_{\alpha}(b,c)$ can be mapped back into the natural numbers

and hopefully preserve ordering

Secret

06:59

\begin{align}
K_0 & =\{0,\omega\}\\
K_n & =\{a_{\alpha}(b,c):\alpha \in \textrm{On},b \in K_{n-1},c\in \omega^+\}\\
K & =\bigcup_{n=0}^\infty K_n\\
a_0(0,0) & =_{R00} 1\\
a_0(0,1) & =_{R01} 1\\
f^0(a_0(0,1)) & =_{R02} 1\\
a_0(0,c^+) & =_{R03} f(a_0(0,c)),c>0\\
a_{\alpha}(0,c^+) & =_{R04} f(a_{\alpha}(0,c))\\
a_0(0,\omega) & =_{R05} \sup(a_0(0,c):c\in \omega)\\
a_{\alpha}(\omega,\omega) & =_{R06} \sup(a_{\alpha}(\omega,c):c\in \omega)\\
a_1(\omega,0) & =_{R07} a_0(0,\omega)\\
a_{\alpha}(\omega,1) & =_{R08} a_0(0,\omega)\\

NB, as of this edition, it is still circular, currently fixing R12

\begin{align}
K_0 & =\{0,\omega\}\\
K_n & =\{a_{\alpha}(b,c):\alpha \in \textrm{On},b \in K_{n-1},c\in \omega^+\}\\
K & =\bigcup_{n=0}^\infty K_n\\
a_0(0,0) & =_{R00} 1\\
a_0(0,1) & =_{R01} 1\\
f^0(a_0(0,1)) & =_{R02} 1\\
\underbrace{f\cdots f}_{\textrm{c times}}(a_{\alpha}(b,1))= a_{\alpha}(b,c^+) & =_{R03} f(a_{\alpha}(b,c)),c>0\\
a_{\alpha}(0,c^+) & =_{R04} f(a_{\alpha}(0,c))\\
a_0(0,\omega) & =_{R05} \sup(a_0(0,c):c\in \omega)\\
a_{\alpha}(\omega,\omega) & =_{R06} \sup(a_{\alpha}(\omega,c):c\in \omega)\\

This is the new version, R11 is deleted to prevent the circular thing in Test #01.

Other cases of circularity not yet known. Still testing

and one more:

\begin{align}
\textrm{Ordering R14}\\
\text{ if } \alpha\leq \beta, a_{\alpha}(b,c) & \subseteq a_{\beta}(b,c)\\
\text{ else if }c\leq c', a_{\alpha}(b,c) & \subseteq a_{\alpha}(b,c')\\
\text{ else if }b\leq b', a_{\alpha}(b,c) & \subseteq a_{\alpha}(b',c)
\end{align}

which means the specification of the class $K$ need to be slightly tweaked:

$K_n =\{a_{\alpha}(b,c):\alpha,\beta \in \textrm{On},b,b' \in K_{n-1},c,c'\in \omega^+\}$

JavadocMD

07:14

Just wanted to pop in and express some gratitude. Y'all save me from tearing my hair out when I attempt to math something. Thanks for anything and everything you do to keep the place ticking.

Leaky Nun

@Secret evaluate $f(2)$.

Secret

$=f(f(1))$?

(everything will be expressed in terms of $f$ and 1, once expanded)

o wait a sec..

I need to specify that the domain of $f$ is $K$, not $\omega^+$

then you are right, $f(2)$ by itself is undefined

because I never wrote down a rule to identify $a_0(0,n) \in K$ with $n \in \omega^+$

So... the domain of $f$ will be all the things in $K$, that is, any $a$ terms, and 1, guess I have not missed anything this time...?

which when the $a$s are all expanded, the domain is basically 1

Zee

10:26

It's 7 AM and am gonna go to sleep now...

s.harp

10:48

hi all

Leaky Nun

@Secret so is $a_0(0,2) \in \Bbb N$ true?

MZ97

What do we mean by 'tau' on this site: mathworld.wolfram.com/Convolution.html

On the formula

LegionMammal978

@Secret I'm doing to try and implement this for Test #02

> d τ

In the initial formula

MZ97

If we have given fg, how do I find gf? (Convolution)

$f*g$

Akiva Weinberger

11:05

Isn't convolution commutative?

Fargle

Yeah, convolution is commutative.

MZ97

yeah, but i've been given 'f*g' of a f and g

LegionMammal978

@AkivaWeinberger It would appear so, as you'd just have to switch your reference frame

Fargle

Right, which means that to find $g * f$, you're already done.

Akiva Weinberger

If I'm not too mistaken, it's equal to $\iint_{x+y=t}f(x)g(y)\ {\rm d}x\ {\rm d}y$

which is clearly commutative since you'd just switch $x$ and $y$

Like, you're integrating the function $(x,y)\mapsto f(x)g(y)$ along diagonal lines.

MZ97

11:09

uio.no/studier/emner/matnat/math/MAT2400/v16/mat2400v16bm.pdf

Can you check 4a

'f*g' is given

Secret

@LeakyNun Technically yes, once the identification is made

LegionMammal978

@Secret Once again, give me a minute to run Test #02

Fargle

@MZ97 Substitute $s = u - x$ and find $ds$ and the new bounds of integration.

LegionMammal978

@Secret For R03, does $c>0$ apply for the first or the second equality?

Secret

Take your time, no need to hurry because I am current busy on my supplementary info. writing

R03, it applies to both cause I don't want $a_0(0,0)$ to be touched

LegionMammal978

11:15

got it

MZ97

@Fargle So: ds=-dx?

Fargle

@MZ97 Right, or $dx = -ds$, equivalently.

Now, if $x$ ranged from $-\pi$ to $\pi$, where does $s$ range from?

LegionMammal978

11:37

@Secret Iterative Test #02

(renumbered some of the rules)

Secret

Delete R05, just realised it is basically a duplicate of R04

I have no idea how to fix the conflict between R03 and R04. That is precisely the circular issue Leaky lun and I have been talking about

LegionMammal978

@Secret Ima eat some breakfast now, I'll implement any suggestions for Test #03 when I get back

Secret

The idea is that since e.g. $a_{0}(0,2)$ is basically the number $2$, it will be $f$ acted on $a_0(0,1)$ which is $1$

Think of it this way, initially there are only three things: $f$, $0$ and $1$

Each time $f$ act on $1$ it produces a new thing, similar to how the successor operator acts on $1$ (with suitable 2nd other logic) it produces $2$

MZ97

@Fargle from u+pi to u-pi

Secret

That is, $a_0(0,2)$ does not exist unless it is produced by $f$ act on $a_0(0,1)$ first

I guess that might be why that circular issue keep coming up, because the whole class $K$ literally start building itself by recusive applications of both $f$ and $a$ on existing sets

Put it in another way, $f$ recursively define all higher $a$s starting from the base case $1$

I am not sure if there is any way to phrase that better nor how to express it mathematically

LegionMammal978

11:53

@Secret I still don't see why we need 2 functions $f$ and $a$; the recursive problem in $\mathrm K$ I solved by defining $\mathrm K_0,\mathrm K_1,$ etc. Additionally, this solution for $\mathrm K$ also prevents infinitely deep chains of applications of $a$.

Secret

So $K_n$ will take care of the issue if I want to move from $a_{\alpha}(b,c)$ to $a_{\alpha}(b,c^+)$?

LegionMammal978

@Secret ??? More like from $a_\alpha(b,c)$ to $a_\beta(a_\alpha(b,c),c')$

MZ97

What is the fourier coefficient of a convolution f * g?

given the fourier for f and g

LegionMammal978

@MZ97 Found this with some quick Googling

Might be useful

Secret

O I see, I did not realise that rule already implied that.

Ok in that case I guess $a$ itself is sufficient. Thanks for pointing that out

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