Mathematics

Araske

12:04 AM

seems like you found a glitch in mathematica?

Akiva Weinberger

"Though I do understand it" - meaning you know why this happens?

Araske

what expression falls out of FullSimplify, and where is it not equivalent?

Semiclassical

Well, a clue comes in what mathematica gives you when you sum it up

Namely, it comes out as a combination of modified Bessel functions

(plus coefficients and an overall exponential factor but w/e)

And modified Bessel functions are solutions to ODEs with an irregular singularity at zero.

Hence they in general have series solutions which are only convergent in some sector of the complex plane.

So my rationale for what's happening with Simplify vs. FullSimplify is that the results they give are only equivalent for some range of $x$

Araske

hmm

that would seem to hold up

Semiclassical

Indeed, the results seem to match when $x$ has positive real part. But when you try to analytically continue across the imaginary axis, you get Stokes phenomenon and therefore things get weird

Araske

12:10 AM

its odd that it isn't keeping track of its boundaries in any way

Semiclassical

Yeah, I'm a little puzzled by that.

Daminark

shrug

Araske

rational programs as such would catch that

Semiclassical

My hunch is that Mathematica handles Bessel function relations in such a way where it works out for positive $x$

Araske

i mean if you'd need to make a choice, that's probably the one that's the most useful for the most people

Semiclassical

12:13 AM

Right.

Araske

still like

when you're coding that software to make the choice

make it throw some sorta exception

Semiclassical

Amusingly, if you replace $-x\to x$, you still get an ambiguity.

But the ambiguity occurs when $x=-1$ :)

so flipping the sign inside the summation doesn't swap the location of the ambiguous region

Araske

well I think we've figured out the bug then

Daminark

@Araske bug??? Get the murder stick

Araske

like that's consistent with it choosing a region to work nicely

Semiclassical

12:15 AM

I may post this on Mathematica SE.

Araske

sounds like a plan

Astyx

Semi just broke mathematica

Semiclassical

lol

Daminark

Rude

Araske

rip in mathematica

you will be missed

Semiclassical

12:23 AM

Mathematica is dead, long live Mathematica

Daminark

mlg violin

Araske

how will I ever survive having to calculate things in future analysis courses

Semiclassical

Another funny bit of this. The answer you get contains a few instances of (x^2)^(1/4)

And if you replace that with x^(1/2) you get the same result as FullSimplify

Araske

...wait

that seems real wrong

Semiclassical

well, remember that I'm taking $x\to-1$

Araske

12:27 AM

does FullSimplify work on x = 2

Semiclassical

?

Araske

oh wait nvm i misread that

still seems wrong, but less egregious

Semiclassical

Here are the outcomes I have: 1) Plug in x=-1 directly in the series. 2) Do the sum with symbolic x, then Simplify and evaluate at x=-1. 3) Do FullSimplify instead of Simplify in (2). 4) Before doing the last step in (2), replace (x^2)^(1/4) with x^(1/2).

1) and 3) give one result, but 2) and 4) give a different result.

My suspicion right now is that it's not even the Bessel function that's the problem so much as it is that surd.

Araske

i mean the $(x^2)^{1/4}$ and $x^{1/2}$ are the same on the positive reals

Semiclassical

Sure. It's on the negative reals that it's an issue.

What doesn't quite work with that reasoning, though, is that Mathematica is smart enough that doing either Simplify or FullSimplify on (x^2)^(1/4) doesn't give x^(1/2).

So evidently the bessel functions are doing something.

Araske

12:36 AM

maybe its that it knows its restricting to positive reals, but for some reason isn't telling you, so it's making the sub?

what do the docs for FullSimplify say?

Semiclassical

No idea, haven't looked at them yet.

I think I can narrow the weirdness down to the following:

(x BesselI[1/4, x^2/8]/(x^2)^(1/4) // Simplify) /. x -> -1. returns -0.553357

(x BesselI[1/4, x^2/8]/(x^2)^(1/4) // FullSimplify) /. x -> -1. returns +0.553357i

or, even simpler:

x BesselI[1/4, x^2/8]/(x^2)^(1/4) /. x -> -1. gives -0.553357

x^(1/2) BesselI[1/4, x^2/8] /. x -> -1. returns +0.553357i

oh. and in fact one gets different answers for x^(1/2) versus x/(x^2)^(1/4)

So I think it's two issues. On the one hand, if it ever gets into a situation like the last one it'll give different behavior for the two expressions (first gives 1i, second gives -1)

However, Mathematica is usually smart enough to not simplify the latter expression to the first. Evidently the presence of that Bessel function is enough to throw it off, so that doing FullSimplify convinces it to simplify more than Simplify would.

Daminark

1:44 AM

Hey @Paul

Paul Plummer

Hi @Daminark What is up?

Daminark

Everything's alright, how about you?

Paul Plummer

Pretty good, was failing at ollies a little bit ago.

Daminark

Ollies? I dunno what that is

Paul Plummer

it is like the basic skateboard trick jump onto things. Been trying to learn to skateboard

Daminark

1:58 AM

Ah, neat

Semiclassical

If I remember skateboard video games, you'd do stuff like ollie onto a rail and grind on it. (skateboard video games being such a realistic depiction of things, of course)

Daminark

All actions are effectively and safely learned by playing a video game a few times and then trying to emulate it perfectly

Paul Plummer

Ah maybe that is my problem, didn't play enough Tony Hawk

Daminark

Ah, you see? Go do it and then you'll be infused with the power to do it

Akiva Weinberger

2:55 AM

Bad idea #271: Dropping into the half-pipe on a Segway.

@PaulPlummer

Daminark

Kek

Secret

3:23 AM

Last night dream there is no maths, but there is a chemistry journal article about sigma sigma orbital interactions. Seems my dreams finally knew how to approach in solving problems now. Hopefully in the future, it will help me solve some tricky maths problems without showing me nonsensical maths

Paul Plummer

3:36 AM

Pretty rad hobby @AkivaWeinberger

When I first started skateboarding, a year ago, pokemon go just came out and I was playing and skating, not paying attention and hit a stone causing me to fly off and get a really bad skinned knee

The Raiders of Las Vegas

So much for multitasking :P

Paul Plummer

No kidding. One of the first things you learn is to hate stones/pebbles, poor quality sidewalks and roads

PVAL-inactive

@PaulPlummer I dropped in on a 6 foot quarter, didn't commit enough and chipped 3 front teeth. 15 years later multiple root canals, capping and finally extractions...

Semiclassical

oof. worst I had was going head over handles on my bike in my first semester at grad school

broke my front two teeth and pushed one of the adjacent teeth out of position enough that I couldn't close my mouth fully. not to mention scraping up my face and the impact to my head.

(thank goodness for the helmet)

PVAL-inactive

Your initial injury was probably just as bad as mine.

Semiclassical

3:47 AM

possibly.

PVAL-inactive

It was just it didn't seem like I needed root canal at the time.

and then years later it didn't seem like the teeth would need to be extracted...

Semiclassical

I think it was clear pretty much immediately that I'd need to get them rooted and capped. so my two front teeth in front aren't real :)

PVAL-inactive

Does your dentist think they'll need to be extracted?

If that happened your first semester of grad school, you might be years off that.

But it seems many root canals go that way.

Semiclassical

hasn't gone that way, yet.

PVAL-inactive

good luck.

Semiclassical

3:52 AM

that would've been almost 7 years ago now.

Paul Plummer

@PVAL-inactive Damn. I havn't had anything to bad, but basically when I started trying learn tricks I hurt my knee: my front leg landed in front of the board, back on the board and it slid causing sideways force on my knee. Had stilts for a couple weeks, and whenever I tried to get back into the next day my knee felt awful, until recently

Semiclassical

The thing that has been annoying is that the veneers have broken a few times and had to be replaced, though not recently.

PVAL-inactive

My time between being rooted and needing extraction was somewhat close to that.

I don't know maybe only like 4-5 years

Semiclassical

Gotcha. There's always a wait-and-see element to it, I guess.

Maks

4:41 AM

@Semiclassical

Could you help me find the proof of a theorem ?

Daminark

It's over there!

Maks

Eigen vectors associated to different eigen values are LI

Nevermind

found it

Daminark

Oh maybe I can help with that @Maks

Lol

Maks

Haha

I have another one

Which I found the answer

but I dont understand it

Given $T \in L(V,W)$

$T'$ is a linear transformation

$T^{-1}$ is also a linear transformation

Daminark

Wait so $T$ is invertible?

Maks

4:45 AM

It appears to be

why wouldn't it be?

Daminark

Okay that was a bad typo, you mentioned $T^{-1}$ so I assumed that $T$ was invertible on account of this. I was gonna ask if $T'$ is also invertible

Maks

:38953549 I saw that

@Daminark Oh no no, those were to different statements

$T'$ on one hand, and the inverse on the other

Daminark

Oh I know, I was just saying like, I didn't realize that $T$ was invertible until you mentioned $T^{-1}$

So I wanted to also ask whether $T'$ is assumed to be invertible

Secret

Is it even possible to have a $T(x+y)=Tx+Ty$ but $T^{-1}(x+y)\neq T^{-1}x+T^{-1}y$ and yet $T^{-1}(Tx+Ty)=x+y$ ?

Daminark

@Secret are the sets involved at least groups?

@Maks also just don't bother with what I said, continue asking your question, I'm sure it'll all be clear in due time

Actually wait even a semigroup should do it

Yeah @Secret if you have a bijective homomorphism $T$ between semigroups, then $T(T^{-1}(x) + T^{-1}(y)) = x+y = T(T^{-1}(x + y))$

But then $T$ is a bijection so $T^{-1}$ is a homomorphism

Secret

4:55 AM

Nvm

Maks

That was the question

Its actually a proof

prove that the inverse and the transpose of $T$ both are linear transformations

Daminark

Oh is $T'$ the transpose? I get it now

Okay so I'm gonna phrase slightly more generally

So assume $T$ is a bijective linear transformation, then by the argument above, $T^{-1}$ is also linear

Well, you'll need to handle scalar multiplication as well

$T(T^{-1}(cx)) = cx = T(cT^{-1}(x))$

But with that you're good

Now, let's take an arbitrary linear transformation $S$, might not be invertible

Akiva Weinberger

How do you even define the transpose without matrices?

Daminark

I'm gonna interpret its transpose as its adjoint here

Secret

Yeah transpose is the adjoints in linear maps over the field of reals

Daminark

5:01 AM

Not Hermitian adjoint, that requires $\mathbb{C}$, just a general adjoint via dual spaces

Maks

Yeag

Transpose

In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, that is it switches the row and column indices of the matrix by producing another matrix denoted as AT (also written A′, Atr, tA or At). It is achieved by any one of the following equivalent actions: reflect A over its main diagonal (which runs from top-left to bottom-right) to obtain AT write the rows of A as the columns of AT write the columns of A as the rows of AT Formally, the i th row, j th column element of AT is the j th row, i th column element of A: ...

I dont get why it involves the dual space here

Daminark

Oh, it's a bit tricky

Basically, the transpose is a thing which you'd normally define over matrices

Which is, flip it

Faust7

why doesn't $ \frac {\sin x} {x} = \cos x $

nvm

Daminark

@Faust7 why should it?

Faust7

power series

but i wasnt paying attention the the demoninator

Secret

5:04 AM

You forget the 1 in the cos power series. Everything follows except that 1 becomes 1/x

Maks

@Daminark Go on

Secret

and that prevent them from being equal

Faust7

not sure what secret is talking about

Daminark

But yeah basically on linear transformations you don't quite have a flipping technique a priori. So now you're trying to generalize this notion

You have two ways of doing this

Faust7

but the demonator isnt the same for each term

Daminark

5:05 AM

First thing is to use the dual space

Secret

O I thought all odd powers of sin power series have the same sequence of coefficients as the even powers of cos x power series, nvm then

Daminark

So if $T:V\to W$, then $T^T:W^*\to V^*$

So you have a functional $f:W\to\mathbb{F}$

Then $T^T(f) = f\circ T$

You can check that this is a linear functional on $V$

Now, here's the thing

Finite dimensional vector spaces are isomorphic to their dual spaces

Maks

where did $f : W \to F$ come from ?

Daminark

$f$ is an arbitrary element of $W^*$

Maks

Oh, ok

Daminark

5:10 AM

So $T^T$ acts on the functional $f$ to give a functional $T^T(f)$

But yeah so basically, remember how that isomorphism goes

Maks

where $f$ is any element of $W*$

Will try

thanks !

Daminark

Oh that was rhetorical

I was gonna explain anyway

Choose a basis $(e_1,\ldots,e_n)$ of $V$

Then for $v = \sum c_ie_i$, you let $f(v) = \sum c_if(e_i)$

Okay wait no this isn't what I was going to say, scratch all that

You know they are isomorphic since they have the same dimension

Let's just leave it at that

Anyway, given a basis $(e_1,\ldots,e_n)$ of $V$, you can construct the dual basis, $(e_1^*,\ldots,e_n^*)$, right?

Such that $e_i^*(e_j) = \delta_{ij}$

So what happens is this

Choose a basis for $V$ and a basis for $W$, then there's a unique matrix representation of $T$

Now choose the dual basis for $W^*$ and $V^*$ and express $T^T$ as a matrix

Turns out, those matrices will be transposes of each other in the typical sense

Sounds good @Maks?

Sorry if I was at all confusing

Maks

@Daminark Yep

@Daminark Yep :)

Daminark

Also hey @Liad!

Liad

Hi @Daminark !

Daminark

5:24 AM

How's it going?

Liad

Good, enjoying your free time now?

Daminark

Not totally free, but yeah! :D

Are you done with everything?

Liad

Not yet :P

Daminark

Ah, good luck

Liad

5:59 AM

Thank you

♦ Set Theory Study Group

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♦ $Mathematics$ Set Theory Study Group

$Mathematics$ Mathematics