Nothing in particular. I'm learning about homomorphisms, so was calculating various things relating to them by following the definitions. I found it unsatisfying/wrong that I couldn't reduce the image further.

ahh.

@AkivaWeinberger Counterexample?

@Semiclassical

are you here?

just would like quick mental verification for something

sure

@barrycarter

Consider this infinite grid, consisting of "staircases" going down and to the right; some are dashed, others are not, and the two types alternate.

This is the Cayley graph of my group; $x$ goes down or to the right on a not-dashed staircase, and $y$ goes down or to the right on a dashed staircase.

@akiva just to make sure, is the vertical dashed/solid line along the bottom row supposed to be just dashed?

That one's just supposed to be dashed, sorry

just making sure

@barrycarter In any case, you can tell that $x^2=y^2$, and — after a little thought — you can see that there's no element of order two

if we consider the following

(I believe this is the Cayley graph of $\langle x,y\mid x^2=y^2\rangle$)

@akiva that's a nice example. is there a 'model' for that Cayley graph? (in the same way as the dihedral group $D_{2n}$ is 'modeled' by the symmetries of the $2n$-gon)

let us say we want to compute the delta complex structure on this

ehhhh, i probably can't help

we can consider $RP^2$ as IxI/~

i've never done the guts of simplicial stuff

where the identification is done according to picture

@Semiclassical I have no idea.

$\sigma_1^0 : \Delta^0 \rightarrow RP^2$ where $1 \mapsto [(0,0)]$

kk @akiva

makes sense so far?

@Semiclassical ?

not really. as I said, I haven't done this stuff. so i really am going to have to remain silent.

sometimes, i really want to kick Mathematica. error messages which don't clarify what's going wrong really don't help that much.

doing $1+2\sum_{k=1}^n c_n \sin(n x)$ in your initial condition? sure, that's fine. resumming that series and using that instead? oh no, we can't have that

rant rant rant

...and the fix is to use "2." in a specific place instead of "2"

@#$!@! mathematica

have you tried here?

not lately. but i have it working again now

cool :-)

"By the Chinese Remainder Theorem $Z_{20}$ is isomorphic to $Z_{4}\times Z_5$. Also by the Chinese Remainder Theorem $Z_{20}^*$ is isomorphic to $Z_{4}^*\times Z_5^*$ [If you know rings, you can use directly the CRT for rings, otherwise you can prove the second statement from the standard CRT]" -- what does the second statement of the Chinese remainder theorem say?

it tells you the nationality of the remainder <\joke>

Hehe! It's a rather odd name for a theorem. O_o

I can't find the said second statement that would imply said isomorphism.

Hi!:)

Hello

how are you?

Well, thanks!! And you?

fine thanks :)

@DanielFischer Why are my thanks messages in comments to answers gone?! :O have they been deleted? :o

morning

morning

morning

I am trying to find a,b,c that satisfy the equations, any idea how to do in WA?

Find $a,b,c$.

I am trying to get the symbolic expressions

with radicals

Moved here.

in linear algebra, when someone writes 1 + M what do they normally mean?

@hhh More generally, given $x=A+B$ and $y=A^2+B^2$, you want to find $A-B$? (Here, $y=7$)

is this the all ones matrix added to M?

or just the diagonal?

I'd think just the diagonal

Like, $I+M$

thanks!

@r9m Probably they've been flagged as "too chatty" and deleted. Are you talking about comments you posted, or comments you received, out of curiosity?

@hhh Never mind, that's not the right generalization

@AkivaWeinberger yes, I am bit lost but I was enough curious to silent whether you invented some clever way to solve this with matrices :D

@DanielFischer sounds like the thanks messages he posted

hello btw :)

@hhh What? The stuff about matrices was me talking to Arnold

Although… it's probably doable with matrices @hhh

@DanielFischer Comments I posted here under tired and cody's answers .. I said (+1) Thanks (stuff like that) and they seem to have been removed

@r9m Indeed. Flagged as "too chatty" and deleted. Usually, we tend to let "thanks" comments from the question author stand, especially if there isn't a cluttered thread under the answer. Well. You could comment again. But if they are flagged, they might be deleted too.

@DanielFischer well I think they received my thanks message the first time :-) so no need to comment again ... (seriously though, all I am saying is 'thanks', what's up with all these flagging and stuff :P .. )

Your action of acceptance of a great answer speaks louder than any "thanks" :P

okay ,, now even tired's comment is gone .. so no possibility of awkwardness I presume! :-)

@r9m Self-deleted. Since the comment it answered was deleted, tired concluded that the reply served no purpose any more.

@DanielFischer ah! :( okay ,, that was awkward :P

@r9m is that a skull on your avatar?

hello btw :)

Any thoughts for: if $f$ is continuous on $[1,2]$ and differentiable on $(1,2)$ to prove that exist $c \in (1,2)$ such that $f(2)-f(1)=\frac{c^2}{2} f'(c)$. I suppose it is Lagrange's mean value theorem, but I couldn't find function to apply that theorem.

afternoon, chat

hi, semi

afternoon

hey @mike. what're you up to?

been lazy most o the morning. should start work soon

gotcha

i've been that way with this code i'm trying to work on

trying to figure out the 'right way' to do something in mathematica is sometimes a really tedious affair

though the advantage of putting stuff like that off is that occasionally you think of smarter ways to do things :)

Hello!

@mike what kind've stuff are you trying to work on lately?

generally mathematica seems like a tedious affair

yes, well, to paraphrase Churchill: "it's the most tedious thing except for every other thing"

and there's no great way to avoid tedium when trying to solve PDES numerically

though the fact that my laptop randomly decided to turn off last night and lose work i'd already done didn't help

I'm normally thinking about transversality problems, meaning I've got some equation, which has an associated space of solutions; I would really like that to be a manifold. can I perturb the equation slightly so it is?

i imagine there's 'usually' directions you can perturb that will work, and directions that won't

It's less that - one never really finds an explicit perturbation - and more that you need to demonstrate that there's a big enough class of perturbations that gosh darn it one of them has to work

or more formally, it's Sard's theorem

ah. so you don't really worry about simulating anything.

here's a finite dimensional example. say I have a function f: M -> R, I want the 0 set to be a manifold, and I realize I can actually perturb this very effectively - in fact, I have a vector space of perturbations, V

then I get a map M x V -> R, which I show has 0 as a regular value (basically because of the contribution of the differential from V), this cuts out a "universal zero set"

call that zero set S (a subset of M x V); Thwres a projection map S -> V. Another application of Sard's says some v is a regular value, and hence there's some perturbation v such that the zero set is regular

the hard part here is constructing that really big space of perturbations.

would a (very very simple) example be this? take $H=p^2/2+(x^2-1)^2/2$ mapping $\mathbb{R}^2\to\mathbb{R}$.

Hm, I have a simpler example

Take H=xy

V = R, and the perturbed equation on R^2 x R is xy+t

then the unperturbed zero set is just $(1,0),(-1,0)$ which is not a manifold. but if i perturb $H$ by subtracting some small $\epsilon>0$ then the zero sets become topological circles around those points

well, that is a manifold, but you're right that it's not cut out transversely/of the right dimension

i figured i wasn't saying it quite right, yeah

the better example in that case is if I start with $H+1/2$, since then near $(0,0)$ one has a saddle point

so that perturbing with positive $\epsilon$ gives you a disconnected pair of $S^1$'s and with negative epsilon a single $S^1$

like that. for $H>1/2$ the level set manifold is $S^1$, for $0<H<1/2$ it's a pair of $S^1$'s

and for $H=1/2$ it's not a manifold due to the crossing

sure, that's good

that's a physics-motivated example, as you might guess

and for me the interesting questions are about the integrals, e.g. the classical period $\oint \frac{dq}{p}$ and the classical action $\oint p\,dq$ where the classical trajectories are just the level sets in this case.

things are nice for $H$ near zero or $H$ large, but get weird near $H=1/2$.

for example, if i compute the period at an energy $H=1/2-\epsilon$ then I'll find that it diverges like $\ln\epsilon$

and not coincidentally that's also where things get weird when you try to quantize it. so yeah, that non-manifold behavior at $H=1/2$ is significant

one detail of this, evident in your picture, is thag different choices of perturbations can give completely different answers

if you want to extract data from the solution spaces you should hope there's some common factor.

in practice, being cobordant often turns out to be enough

yeah. in here there's few enough dimensions that the picture is simple. but add in a few dimensions, and yikes.

or infinitely many

i imagine things are still relatively simple if $H$ has global minima. if one takes that as the initial zero set and then perturbs, that should give torii

which sounds rather KAM-ish, but i don't really know that stuff so well

but if i start near a saddle point of $H$...ewwww

in any case, the first step is to show that perturbation is possible.

not quite sure what you mean by that. in the case of the $H$ above, though, I note that I can't perturb $H$ too far in the negative sense because i wouldn't have any classical trajectories.

I mean exactly what I said above... I'm talking about my setting above

the simpler example you quoted?

no, the actual thing I'm working on

ahh.

from context, i imagine the example you're working on is infinite-dimensional?

yeah. it's the space of connections on a bundle on a 3-manifold, mod gauge equivalence

you have the chern-simons functional on this guy, whose critical points are flat connections. That's usually not a dinite set of points so first you need to perturb so that it is; and then you want the spaces of trajectories between these to be manifolds.

does this lead to instanton stuff? i know it does in the stuff i was saying above

as stated this is either classical, impossible, or uninteresting, but I'm working in a modified setup

yeah, this is the beginning of setting up instanton homology.

kk. in the example i plotted above, one has an instanton trajectory in the case of $0<H<1/2$ but not $H>1/2$

you've told me before what it means but recall your instanton is not mine

that's fair. in the case above they would be (at least in the standard story i know) trajectories connecting $x=-1$ to $x=+1$

here one can think of them as gradient flow lines of the Chern simons fubctional in my space above; but one of the most rewarding features is that the notion of an ASD instanton makes sense on any 4-manjfold, not just tubes $Y \times \Bbb R$ (which is where the trajectories would live)

what is $Y$ here?

3-manifold

ah

@AkivaWeinberger What is the counterexample for the first one?

@TheKindCat The group I gave has neither an element of order two nor $xy^{-1}=yx^{-1}$.

@AkivaWeinberger which group?

Um, just a second (sorry)

http://chat.stackexchange.com/transcript/36?m=28316090#28316090

There we go

@TheKindCat

It's the one whose Cayley graph is a bunch of staircases

oh, I see, very nice, thank you very much.

Note that both $x$ and $y$ have infinite order;

as it turns out, if either of them has finite order, then we both have an element of order two and $xy^{-1}=yx^{-1}$ (I think)

Wait, I'm not sure. What if they both have the same order? I need to think about this EDIT: Never mind, got it

Yeah, if either of them has a finite order, then we have an element of order two and also $xy^{-1}=yx^{-1}$

(Also, if one of them has finite order, so does the other. Proof: $x^n=e\Rightarrow x^{2n}=e\Rightarrow (x^2)^n=e\Rightarrow (y^2)^n=e\Rightarrow y^{2n}=e$)

yeah, it is easy to show it

when one of them has finite order

@TheKindCat Here's something interesting and also group-theory related:

Let $F_{\{a,b\}}$ be the free group on two variables, $a$ and $b$.

Consider the subgroup generated by $\{aaa,aab,\dots,bba,bbb\}$, the set of all three-letter combinations

This subgroup is also a free group. (As it turns out, the subgroup of every free group is free.)

man this is like the fifth time I've made sure to bring paper to work with but forgot my pens at home

yeah, this gives an example of two injections between non isomorphic groups

yeah, that sucks

(in both directions=

@TheKindCat What's the rank of that free subgroup?

As in, it's isomorphic to $F_n$, for some integer $n$; what's $n$?

Where $F_n$ is the free group on $n$ generators

It's not $8$, since you don't need all of those generators. $bbb$ can be dropped, for example, since it's equal to $(bba)(aaa)^{-1}(aab)$.

I guess I can still think so it's not a total waste but it's a lot harder to keep my thoughts organized

i remain curious about where that $x^2=y^2\neq e$ group would show up. getting the entire free group on two generators is easy (fundamental group of the doubly-punctured plane)

(Pretty sure the group given above via Cayley graph is just $\langle x,y\mid x^2=y^2\rangle$, by the way, though I haven't proven that it's not smaller.)

how to find a realization where $x^2=y^2$ works is something i'm not seeing

fubdamental group of Klein bottle

that, i can believe

That's abelian, isn't it?

And has elements of order two

Maybe not

I think I confused it with something else

from Wiki's page on the Klein bottle

the fundamental group of the Klein bottle is neither abelian nor has toreion.

"The fundamental group of the Klein bottle can be determined as the group of deck transformations of the universal cover and has the presentation $\langle a,b | ab = b^{−1}a\rangle.$"

Yeah, it is the group given above, my bad

(Define $b'=ab^{-1}$)

i was wondering how one got $x^2=y^2$ from that

i figured it was something simple

the fact that it's the klein bottle means that my hope for a geometric realization was probably doomed since i wouldn't have thought of something non-orientable

How is that not a geometric realization

oh, i didn't mean it wasn't

i meant i wouldn't have thought of it

"since I wouldn't have thought of something non-orientable"

so to the extent that i would've tried to think of it, i would've failed

@Akiva: I did indeed read that.

just because it has a geometric realization doesn't mean i myself would've realized it :)

I'm not even trying to visualize what $x$ and $y$ would be on it

same

(At least, not at the moment; I might end up doodling it later)

That's not hard. Think of the Klein bottle as a quotient of $\Bbb R^2$ by that group of Isometries. I leave it as an exercise.

Evening

By which group of isometries?

i imagine it isn't so bad when you think in terms of the unit square with appropriate identifications, yeah. i just 1) don't remember the identifications, 2) don't know what $a,b$ correspond to on there

That's the exercise.

right.

Ahh

Welcome, Dr. @DanRust

I imagine $x$ is parallel to a side of the square and $y$ is diagonal?

i mean, the Wiki page includes the fundamental polygon

@MikeMiller :D passed my viva with minor corrections

…wait

If we just draw a rectangle, and orient and label the sides such that going counterclockwise reads $“aab^{-1}b^{-1}”$

then that gives us our group as its fundamental group

and it's also the Klein bottle

For those who don't know UK terms: Dan Rust is now Dr. Rust.

Is the minimum polynomial of $\sqrt{i+\sqrt{2}}$ over $\mathbb{R}$ degree 2? I suspect it is because the number is in $\mathbb{C}$. However I only have a degree 16 polynomial for which the number is a root-- $p(x)=(x^4-4)^4+64$ -- and can't seem to reduce it.

(which can be shown by cutting it on the diagonal and rearranging or something like that to get the usual fundamental polygon)

Congratulations!

@rorty Sophie Germain identity perhaps?

Or try to find the real and imaginary parts of that number

@AkivaWeinberger I'm a student in abstract algebra with very limited math background

Basically, it's that $x^4+4=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^4-2x+2)$, except the general identity has more constants

But you could also just try to find the real and imaginary parts of $\sqrt{i+\sqrt2}$ and work from there

I though the minimum polynomial had to be irreducible

Correct

@TheKindCat Yes, that's my problem. I have a reducible polynomial of absurdly high degree...

@rorty's polynomial isn't

$4$ isn't "absurdly high"…

Hey all!

Ohhh

That's degree $16$, ain't it

but if it is irreducible it must have degree 2

@TheKindCat Yes, you're correct. I'm trying to find a degree 2 poly whose root is the number I gave.

I have a quick question about what inner products give us, @MikeMiller! If norms and metrics give us a notion of length and distance, what is it that inner products offer us in the physical sense?

oh ok

but that seems impossible

@Akiva

why do i need the real and imaginary parts?

Well, if you have it in the form $a+bi$ for real $a$ and $b$, it's a lot easier to find the minimum polynomial over $\Bbb R$.

@rorty: You're working way too hard. Verify that the following polynomial has onlh real entries when you multiply it out: $f(x)=(x-i-\sqrt{2})(x+i-\sqrt{2})$.

@MikeMiller Yes, but his number is the square root of that

oh, that makes sense

On the other hand if you want the degree of your number over $\Bbb Q$, you should expect degree 4.

oh :(

Still, same idea works.

Fine. Replace $x$ by $x^2$.

so he gets a degree 4 polynomial?

Then you have something of degree $4$ :P

That's better than degree 16

:)

Good point. You can take over from here.

@Ephemeral: Angles.

(And length, of course.)

who can?

'direction cosines' if you're an engineer, i guess

May I ask how the axioms translate to angles, @MikeMiller?

I'm guessing that the square root of an inner product of something with itself gives a length.

Do you know the formula for the relationship of the angle between two vectors and the dot product?

More straightforwardly, if two things have 0 inner product, tbey're perpendicular.

Oh, yes of course. The inner product is equal to the product of their magnitudes multiplied by the cosine of the angle between them.

i can't figure out whether or not i should be happy that mathematica is giving me something for free :/

mostly because i can't figure out whether it'll still do that when i make things harder

Conjugate symmetry doesn't really ring much of a bell when it comes to angles and lengths, @MikeMiller. $\langle x, y \rangle = \overline{\langle y, x \rangle}$

@Semiclassical I hate wolframalpha because of that

you tell him, what is 2+5 and it say 7 no problem

yeah, it doesn't always help with where you want to go

then you ask him what is 232131+1321321321 and it says, "did you mean banana?"

Of course with something tougher than adition

what i find more annouying is when you ask it for a definite integral with some parameter, and it'll do it just fine for any particular choice

but it won't give you the general formula :/

Sometimes if you ask the integral from 1 to 10 it works, and when you ask from 1 to 1000 it asks if you meant something else

even though the function is continuous

Don't think about xomplex inner products, then.

in this case my issue is that the answer i get when i do NDSolve seems to match three required boundary conditions despite my only having imposed two of them

So complex inner products are a more abstract notion that don't make for intuitive pictures, @MikeMiller?

I'm sure someone will disagree with me, but to me they're just the correct generalization to complex vector spaces.

I mostly think about real inner products which probably significantly influenced my taste.

for good or ill, i'm stuck thinking in terms of wavefunctions on Hilbert spaces when i think of complex inner products

I hear that a lot from you. ;D

hey

@Semiclassical I figured out everything

@DanRust: So what are you up to now?

@MikeMiller yuuuup

whoa, I wish I had everything figured out

if your interested @Semiclassical here is the delta structure on the torus

texpaste.com/n/1ud3bt6b

mmkay.

on $RP^2$

@MikeMiller I have still a long way ahead of me to figure out everything xD

Don't we all, @MikeMiller!

Thanks for your insight into the complex case! ^_^

I would suggest I do not have insight into the complez case. But sure; you're welcome.

So if I want to consider only real valued inner products, I can omit conjugate symmetry as an axiom of my inner product, @MikeMiller?

What does conjugate symmetry say in the real case?

Oh, I think it says that $\langle x, y \rangle = \overline{\langle y, x \rangle} = \langle y, x \rangle$.

hey @skillpatrol how is life

Since real numbers are invariant under complex conjugacy.

Yes. You can't eliminate that, but it's easier to understand. Compare the dot product.

i'm tempted to be all physics-horror here and talk in terms of bra-ket notation, but i think that'd be cruel :)

I've heard of it before. You gotta teach it to me sometime, @Semiclassical!

Maybe when there are less mathematicians in the house :-b

hah

@Agawa001 still kicking around...how about you?

first step: decide to write the inner product as $\langle x|y \rangle$

I follow so far ... :-b

Waiting for catastrophic consequences!

second step: decide to call vectors as $|y\rangle$ and dual vectors as $\langle x |$

Oh lord!

I can see this getting real messy.

Step three?

and therefore, hey, dual vectors act on vectors as $\langle x | y \rangle$ to give numbers

That's awesome.

Wouldn't a vector space then be it's own dual?

huh, interesting fact. homomorphisms $H \to G$ are said to be equivalent if there's some automorphism of G that takes one to the other. they're said to be linearly equivalent if, for every representation of $G$ (homomorphism $G \to GL_n$), the representations are conjugate

not necessarily, I think

If we define an inner product as a map from $V \times V \to F$ that is.

fact: equivalent homomorphisms needn't be linearly equivalent.

as a "special kind of map"

but yes, an inner product gives an isomorphism $V \to V^*$ for finite dimensional vector spaces.

i suspect that isomorphism would also be there for most of the infinite-dimensional cases that a physicist would consider

step three: if i can treat vectors and dual vectors as objects, why not operators? for that reason one sees things like $|Ay\rangle = A|y\rangle$

It's also true for Hulbert spaces where $V^*$ means continuous dual, yes

@MikeMiller So I know that the minimum poly $p(x)$ of $\sqrt{i+\sqrt{2}}$ over $\mathbb{R}$ has to be degree 2; also I know that $p(x)$ must be a multiple of $q(x)=x-\sqrt{i+\sqrt{2}}$ and a divisor of $(x^2-i-\sqrt{2})(x^2+i-\sqrt{2})$. So I tried multiplying $q(x)$ by each degree-1 factor of $x^2+i-\sqrt{2}=(x+\sqrt{\sqrt{2}-i})(x-\sqrt{\sqrt{2}-i})$, but neither product is over $\mathbb{R}$...any tips?

for a lot of purposes, i don't actually think there's any problem with bra-ket notation besides possible aesthetic issues.

on the other hand, it tends to get clumsy fast when doing lots of tensor product stuff

@skillpatrol kickin arround ? dunno what that means ,wheras i m on som kind of rigid time-phase (not kickin nor bein kicked )

@Agawa001 it just means "still alive"

@rorty: Sorry, I'm a little too busy right now.

@MikeMiller No problem, thanks for the help so far.

Maybe @Semiclassical or @AkivaWeinberger?

nah, i'm back to coding

@skillpatrol ofcourse, unless it is ur ghost which is connected

ethernet?

:-D

@rorty $z=a+bi$ is the root of $(x-a-bi)(x-a+bi)=x^2-2ax+a^2+b^2$

That is, $z$ is the root of $(x-z)(x-\bar z)$

Plug in your number into $z$. Note that $\sqrt{i+\sqrt2}+\sqrt{-i+\sqrt2}$ is actually real…

…and is twice the real part of $\sqrt{i+\sqrt2}$.

If we want to look at a Hilbert space, is it necessarily defined over $\mathbb{C}$, @MikeMiller?

No.

@AkivaWeinberger it is not obvious to me at all that $\sqrt{i+\sqrt2}+\sqrt{-i+\sqrt2}$ is actually real. How did you intuit this?

the real part of any complex number is half itself plus half its own complex conjugate

It something plus it's conjugate. (Now, to find out which real number it is — that is, to write it without using $i$ — is another matter.)

(It's not obvious that it's real unless you've seen it before @rorty)

@AkivaWeinberger Embarrassed to say this, but it's not even obvious to me that it's something plus it's conjugate. How do you see this?

I also find it strange that Pinter (author of my abstract algebra text) seems to expect the reader to know this. He does not assume any knowledge of complex analysis.

it's not entirely trivial. it comes from the fact that in this case (and many others) taking the complex-conjugate commutes with the mapping $f(x)=\sqrt{x}$

so you can take the complex-conjugate before or after taking the square-root, and it's definitely easier to do so before

@rorty: I don't think this is as trivial as we're making it out to be. In fact, it's nontrivial to even define the square root in the complex numbers. (An elementary, but simultaneously very deep, fact is that you cannot define a square root function $\Bbb C \to \Bbb C$ that's continuous on the whole complex plane.)

i concur with @mike on that. in fact, strictly speaking what i just said isn't sensible unless i agree to not consider $x<0$.

The way square root here is defined is as follows. Write every complex number that's not a negative real number as $z = re^{i\theta}$, where $r \geq 0$ and $\theta \in (-\pi/2,\pi/2)$. Then (by definition!) $\sqrt{z} = \sqrt{r} e^{i\theta/2}$. This behaves as a square root (when they're both defined, they're inverses to each other).

the domain of definition here (i.e. all complex numbers except for the negative real line) is important

@rorty Alternatively, let $(a+bi)^2=i+\sqrt2$ and solve

Now it might be of some value to try and visualize what this does on the complex plane (again, where it's defined); but most importantly, as @Semiclassical said, it commutes with conjugation. This is because $$\overline{\sqrt{re^{i\theta}}} = \overline{\sqrt{r}e^{i\theta/2}} = \sqrt r e^{-i\theta/2} = \sqrt{re^{-i\theta}},$$ which is what we wanted.

for $a$ and $b$

aaaaa

That is a very interesting discussion, thanks all. So do you all think that this is just not a suitable exercise to find the minimum polynomial of $\sqrt{i+\sqrt{2}}$ in a elementary abstract algebra text?

I do not agree. Akiva's point is right.

While we can talk about fancy math as much as we want, the final answer will come from fiddling with symbols.

i see

agreed. even if there's some subtlety going on with how one thinks about the square-root map in the entire complex plane, it remains the case that there's just not that many ways it could work out

which is to say, there's only so many ways that $(a+bi)^2=i+\sqrt{2}$ can have solutions that are real in $a,b$

@MikeMiller Professional "Symbol Fiddlers", we call them.

is such fiddling instructive?

i wish there were a royal road to geometry

there's a old line by the astronomer Eddington: "a mathematician is never so happy as when he does n't understand what he is talking about." not sure i agree with that, but there's certainly times when the fiddling is all that matters

@rorty It's fairly faithful to the definition, at the very least. What is the definition of $\sqrt z = x$ other than the $x$ such that $x^2 = z$?

You play with the symbols until you get to something you understand better and then you solve that.

sounds reasonable to me!

Are there any stats on the percentage of math postdocs that land a professorship within a few years? How long is a typical postdoc?

Why do you have DFT transform n to n-element vector? I understand that square matrices make things reversible but, at the same time, shorter pulse in time domain should result in wider spectrum and vice versa. So, the length of Fourier input and output must be different, right?

@rorty (And — like in the real numbers — $x^2=z$ has two solutions as long as $z\ne0$.)