so... I'm trying to convince myself of the behavior of matrices as linear maps

Let $S, T$ be linear transformations s.t. $S: U \to V; T: V \to W$; let $T \circ S$ be their composition and let $\mathcal{M}(S), \mathcal{M}(T), \mathcal{M}(TS)$ be their corresponding matrices.

Notice that the kth column vector $s_k$ of $\mathcal{M}(S)$ is the image $S(u_k)$ in a vector space $V$ of the kth basis vector $u_k$ of some basis of a vector space $U$.

But then, notice that a column vector of $\mathcal{M}(TS)$ is the image $TS(u_k)$ in a vector space $W$ of the kth column vector $s_k$ of $\mathcal{M}(S)$. Therefore, a column vector of $\mathcal{M}(TS)$ is the image in $W$ of a composition $TS$ of linear transformations on the kth vector of some basis of $U$, denoted by $TS(u_k)$.

am i a liar?

i've been staring at the definition of matrix multiplication for five hours straight, and I feel like understanding is painfully and slowly seeping in

but I'm not sure if I'm just getting used to it or actually understanding

you're not lying.

this is a sign

thanks, for reading

maybe this is pedantic but it is sometimes helpful to distinguish between the columns of the matrix (which are lists of numbers) and the vectors represented by those columns (which are linear combinations of basis elements).

there's a correspondence between them but it is not an identity.

ah, right. that's actually super relevant, I keep having a hard time distinguishing the two

so the kth column of M(S) is the list of coefficients you'd use in V to write S(your kth basis vector in U) in terms of the chosen basis of V.

right, noted! thanks a lot!

for some reason it's less confusing in an abstract vector space then it is in some R^n or C^n, where the elements of your space actually are lists of numbers

yeah, $A \to B$ instead of $\mathbb{R}^n \to \mathbb{R}^n$ makes it easier to notice the movement of bases

found this graph that makes it somewhat clearer

actually, crystal clear

ah, if this is category theory, it is a nice thing

diagrams like that are a good way to remember whether it's PDP^{-1} or P^{-1}DP or whatever when doing matrix diagonalization.

yeah, but if you actually have calculated a P it looks silly if you do the wrong one.

ok, yeah

i forgot that calculations are something people do

when i taught engineering linear algebra the engineers kept me very honest. no abstract anything. nothing but matrix operations. it was a shame i couldn't assume familiarity with a computing environment, it would have saved a lot of time.

and the idea of teaching engineers to find eigenvalues of 3x3 matrices by hand, i'm not sure i saw the point of that. they could do it though. better than math majors at following algorithms.

If one remembers that change of basis matrix leads to the inverse transformation for coordinate vectors, the formula is a piece of cake, commutative diagram (which I do have in my books) notwithstanding.

@Thorgott Some more than — ahem — others.

"basis matrix"? no... don't tell me there's something I missed

is it just the basis of the codomain of a linear map?

Change-of-basis matrix

Heya (again) Ted

hey again, Sha

ah, right right

I was too lazy to type hyphens on the iPad

From the borsuk ulam thm, the fact that if S^2 is a union of three closed sets then at least one closed set contains a pair of antipodal points. I think it's still true when we replace the statement to two closed sets. But how can I prove this?

What's the difference between unit and dimension?

Dimensions are what we measure, and units are what we measure with?

Can I say degree and radian are units?

If yes, doesn't this imply that the angle is dimensional?

@Wolgwang don't we say that the dimensions of, for instance, a cube, are x units of height, y units of depth, and z units of length?

@Wolgwang in math, "dimension" usually is a lot more rigorous than that would let on.

P.S. @Wolgwang Radians are unitless.

radians are generally regarded as unitless, for example, because a common definition is as a quotient of two lengths.

Oh damn.

disappears into the void

it just never stops anymore.

I think dimensions tend to be some kind of "base" quantities like mass, time, length, and units of measurement are somehow instantiations of these, e.g. acceleration has dimensions $[\ell][t]^{-2}$ and ... standard units $ms^{-2}$, is this fair to say or am I talking b*llocks?

radians may not have units, but they are typically regarded as units themselves

feet per second^2, not "meters" (whatever those are)

@anakhro Ohk

@leslietownes oioi

it's spelt metres, btw

Can anyone help me with this? Thanks!

that's how i think of it. dimensions is somehow getting at the 'data type' of the quantity and then the choice of units is what pins that down to a number.

@TedShifrin Radians are units,yes?

@leslietownes data type is a good description

@EdwardEvans brit detected

retreats to island fortress

Not In the traditional sense. Leslie explained it earlier.

Yeah I've been banned for being too tolerant of other cultures

so, I've always wondered about the labour party in the uk

like, does it openly work with unions?

@Wolgwang perhaps it might be better to say it this way: "radians are units, but their SI base unit equivalent is simply 1".

Angles are one dimensional? Sorry for stupid questions.

@anakhro This is nice.

@shintuku Some do, some don't. Ones that do have an official status as being somehow connected to Labour

The Labour party was good under Jeremy Corbyn. Now it's just the Red Conservative party.

that's flabbergasting

nice word

imagine a government openly supporting an union

that's insane

What's wrong with unionising? lol

no, I mean, unionizing is great

Oh I see, I misinterpreted

Unfortunately the Tories have been in power for over a decade and it's unlikely to change at the next election because Labour got decimated in the last election

I just find it odd that a government would openly support the right to strike

feels like there's a contradiction somewhere there

I think the Labour party is just more open to the idea of a dialogue with the working classes than the Tory party. Also Labour was born out of trade unionists and socialists banding together to sh*t on the Tories so it makes sense

But Starmer is purging the party of socialists as we speak so shrugs

Although the Tories could be described as a weird brand of socialism mixed with nationalism, like "national socialism" or something if you want to give it a name

uh

that's very bad

national socialism is nazi doctrine, literally it's its name

lmao nah i'm kidding, they're just horrible people

hahahahahah

oh well, I guess it was a dream all along. "we're the labour party, we support labour unions!" sounds too good to be true

Labour under Corbyn was moving back towards the original point of the party but he was aggressively smeared by the media during his tenure as leader and it worked. Labour lost something like 80 seats in the Commons at the last election and he was forced to resign, which is very sad.

whew, that's harsh

france seems to be the only western country with a still functional labour party

@EdwardEvans with extreme prejudice?

straight to the Gulag with them

@shintuku The French are very good at protesting. That's not allowed in the UK anymore!

Sorry to ask, but is there by chance an analog of the Hilbert basis theorem for Artinian rings (i.e. if $R$ is Artinian, then $R[x]$ is Artinian)?

well, the antiprotest laws are pretty international at this point no matter the political affiliation

math.stackexchange.com/questions/2987281/… may be of use.

@shintuku That's in response to COVID, but the UK government introduced a bill that restricted the right to free protest based on whether or not the protest "causes a public nuisance".

Sir Cumference is a fun name. well chosen.

@leslietownes haha thanks

quick question. In Lie Derivative formation, we introduce flows to compare Vector Fields in different tangent spaces. In my book, it is say R^n, we don't need this. By R^n, do we include sub manifolds embedded in R^n? For example, any curved surface in R^n.

@EdwardEvans ah, well... that's interesting. what is a protest, anyways

Some kind of civil unrest designed to cause some kind of disruption to initiate a dialogue to provoke change (there's probably a proper definition, idk)

The point is, the UK government introduced a bill that means that protests that are "too loud" or that "block streets" are liable to be broken up by riot police and protesters are liable to a maximum prison sentence of 10 years

hawk, certainly one does not need to work with only R^n, that would be a boring world. it could be the book is trying to convey key concepts without introducing a lot of formalism and coordinate charts.

most US states have similar laws and they are rarely and inconsistently applied. rarely are people incarcerated.

they round you up and you have an arrest on your record, maybe you spend a night in a police station.

@SirCumference no, because artinian = noetherian + dim 0

and dim R[x] = dim R + 1

in fact, take (x) > (x^2) > (x^3) > ...

@leslietownes isn't the word "public nuisance" far too open to interpretation?

phrase*

yes, it is very hard to capture in words what makes an OK protest different from a not-OK one. a lot of laws in the US on this subject are probably unconstitutional, but they're not applied often enough that anybody cares about them.

i mean, people do care about them. i don't mean to be too cynical. one is more likely to be arrested or teargassed at a 'left wing' protest than a 'right wing' one, it would seem. but nobody is jailed for years over speech stuff. the US is still pretty good about that.

my daughter can count up to seventeen..

I think most people are afraid that the UK would not be good at that

a month ago she would stop around a dozen.

That's good, 17 is prime

she also correctly counted the number of mallard ducklings at the duck pond. she used to just start counting and not stop. now she actually stops when she reaches the right number.

lots of people living in tents near the freeway. significantly more than two weeks ago. they must have been moved from somewhere else.

she also knows her age. two and a half. the half is important.

Hello. When two lines intersect, how many regions they create please?

what is the ambient space in which these lines are intersecting?

in R^3 i think i would be hard pressed to say that an intersection of lines creates any regions at all. in R^2 you can fairly say that generally there are four of them.

algebraically if you think of lines in R^2 as zero sets of functions ax + by + c, the four regions are determined by the different possibilities for sign combinations you get if you're not on a line.

a bit like the positive and negative x and y axes now that i think about it.

is there more to the question? is it linked to something else?

What is the definition of a region gometrically?

there are varying definitions. in R^2 i would say, maybe a connected set. connected here means for any two points in the set there is a path connecting them that stays entirely in the set.

so the four 'quadrants' in the usual xy planes are regions because if i have two things in the same quadrant, i can connect them without leaving the quadrant. i can even use line segments to do so, although this would probably not generally be a requirement of a region.

And non-empty interior?

why don't we even insist that they are open sets. that would make my life easier.

very good point.

this is why i was hesitant about R^3. if you have two intersecting lines in R^3, you can connect any points that aren't on these lines to each other without having to cross the lines. there's too much room. an analogue in R^3 would be intersecting planes, where you do get some separation.

look at me, pretending to know geometry.

they're not planes at all, they are level sets of linear functionals. that's all they are. shuts eyes

@leslietownes. So when two lines intersect, a theorem says it has $k-1$ regions although based on what you said, it should be 4. Please correct me if I am wrong

the theorem sounds like it might involve a potentially larger, and potentially variable, number of intersecting lines. with two lines it's just X and you see your four regions right there in the X picture. which isn't a picture.

Tell us the precise statement of this theorem.

$n$ lines separate the plan into $\frac{n^2+n+2}{2}$ regions if no two lines are parallel and no three pass through a common point.

The plane.

Based on that, take any two lines, they should create $k-1$ regions, although based on my understanding to what @leslietownes said, it should be 4

What is $k$?

$k$ stands for number of lines

Where did you get $k-1$?

2^2+2+2 over 2 is indeed 4.

This is the famous pretend-geometric sequence @robjohn mentioned the other day.

It seems either me or the instructor explained it wrong

cut-the-knot.org/proofs/LinesDividePlane.shtml has some arguments for the general case.

You.

there's an n in the formula, and you're talking about a k. i would pin down what k is.

@TedShifrin ah yes...

@TedShifrin. There are then $k-1$ regions between two lines (that is, between 1 and 2, between 2 and 3, ..., and between $k-1$ and $k$). Finally, there is still a region to the left of line 1 and a region to the right of line k, thus there are $k−1+2=k+1$ regions in total.

you really would expect it to be geometric. at least i would.

This is exactly what the instructor mentioned @TedShifrin.

that may be part of an induction proof. an intermediate step where k is playing a role in an inductive argument.

the induction step is not very illuminating for the case of two lines. it's only two lines.

$\binom{n}{4}+\binom{n}{2}+\binom{n}{0}$ is the fourth degree polynomial that I was talking about

$(1,2,3,4,5,\dots)\to(1,2,4,8,16,\dots)$

the cut the knot website includes that formula. cut the knot is a surprisingly good site. it's a nice reminder of how the internet used to be.

That is the number of regions into which the interior of a circle can be divided by connecting $n$ points on the circle

@leslietownes @TedShifrin. Thanks

Isn't that the same question, @robjohn?

it's the same answer, anyway :)

Just dilate the circle to contain all intersection points of the lines.

no, one is a plane by $n$ lines, this is a disk by $\binom{n}{2}$ lines connecting the $n$ points

Oh.

There's a bijection when $n$ is even.

Oh, still not right. “Never mind.”

Yeah, the other is $\binom{n}{2}+\binom{n}{1}+\binom{n}{0}$ There are very similar proofs, as one might expect.

it's not the same answer. who told you that it was?

bailiff...

my dad sent me a photo of zachary's pizza on college avenue today. i guess they aren't doing indoor dining, but the photo seemed to have been taken inside the restaurant, so i hope he got a slice.

8:30 meaning it's time for fighter jets to buzz our house and wake my daughter up.

they are equal, it seems...

as long as $x\ne0$

@robjohn. Thanks. I mean is it okay to multiply and divide by $x$ as long as $x\ne0$?

it is multiplying by $\frac xx=1$

@robjohn. Thanks

uh, how do I refer to the image of $(1 - \delta, 1 + \delta) \in \text{dom} f$? Is $f( (1 - \delta, 1 + \delta) )$ standard? looks ugly

"the image of the inverval $(1 - \delta, 1 + \delta) \in \text{dom}f$ through $f$" sounds redundant, and I'm not sure "through $f$" is standard

does the domain of f actually contain that neighborhood? or are we restricting to things in that neighborhood that are also in the domain of f? subseteq may be preferable to in here, tex-wise.

domain of $f$ actually does contain that neighborhood, and I'm trying to say that the image of that domain through $f$ falls within the $\epsilon$ neighborhood in the image for a limit proof

f((1-delta,1+delta)) is fairly standard, as is f(A) for A a general subset of the domain of f, although it is what people would call an 'abuse of notation' because f is a point mapping and not a set mapping. sometimes books distinguish the two with brackets or other notational machinery.

i would say 'under f' rather than 'through f'

i do prefer the English phrasing of it to the notation, unless the notation is spelled out somewhere else.

it may also be possible to avoid the notation and just assert that if |x - 1| < delta, then something happens.

this would closely parallel a common definition of the limit.

"the image under $f$ of the interval $(1-\delta, 1+\delta)$, falls within the interval $(L-\epsilon, L+\epsilon)$. $\blacksquare$"

works for me. is contained in the interval might be better than falls within. but i'm moderately afraid of heights. falling just spooks me. your mileage may vary.

contains does indeed make a bit more sense

thanks a lot!

Is there a book for topology which is not "boring"?

no.

^

(⌣́_⌣̀)

@leslietownes Why?

i dunno. a lot of the basics of topology are axiomatics that have distilled everything interesting almost completely away. so there's a lot of sludge you just have to wade through. this isn't topology's fault; it's generality is its strength. but it's not fun on every page.

(._.)

it helps to check what you learn against a library of examples. that can be fun. but a decent intro to topology almost always has to be broader than that. there are lots of counterexamples and pathologies you need to be aware of even if they won't come up "in practice."

if you're dealing with topologies on finite sets there are probably some obvious inequalities that shed little light on the actual picture, and maybe harder results. i would include topologies on finite sets in with pathologies. others may differ.

i never had to deal with a topology on a finite set. the topologies i dealt with were always very nice (e.g. induced by norms) or very ugly (e.g. not metrizable, not X countable, not anything, you basically just gave up on topology).

"the basis for a topology" is also a little strange of a phrasing, most topologies have many bases with no obvious choice of one to pick, and maybe no natural choice of a 'minimal' one.

In my textbook, a homogeneous ideal is defined as one that is generated by homogeneous elements. i was wondering whether someone could verify my solution to an exercise:

eek, someone's doing algebra

lol

i'll stand to one side and not distract from people who might actually know what they are talking about.

i'll stand on the other side and wonder really, what is so ideal about an ideal?

For 21,12 and 31,13 square why are the values getting reversed please explain this to me.

try to write it as 2*10+1 and 10+2*1 i suppose

Ok but so what..

Why should this trick work?

@Oxide

$(2\cdot10+1)^2 = 2^2 \cdot 100 + 2\cdot10\cdot1 + 1^2$ and $(10+2\cdot 1)^2 = 1\cdot100 + 2\cdot 10 \cdot 1 + 2^2$

obviously the middle terms are equal, and the hundreds and ones digit just get switched around

if you think about it for a few seconds you'll realise why this doesn't work with 14 and 41 for example

Is there any other reason apart from the fact that the middle terms are not equal?

???

For 14 nd 41??

they are equal

ah, i made a typo earlier

i should have said $2\cdot 2 \cdot 10 \cdot 1$

I coould not get you.

in any case, it is because $4^2 > 10$, so you have to carry the digit

ok..

got you.

@love_sodam immediate corollary. also easy to prove from scratch, though. if S^2=AuB and neither contains an antipodal pair, then inversion defines an involution of S^2 that maps A to B\A and B to A\B. deduce that A and B are disjoint, but then closedness implies one of them is the whole space.

@Thorgott

uh, sorry, i didn't mean to press enter so soon

could you look at my problem above?

Every 60 seconds in Africa, a minute passes

together we can stop that

I couldn't answer single question in SE math thus leaving me with low score

also got not time

uh, suppose $f: \mathbb{R} \to \mathbb{R}$ such that $x \mapsto x^2 + 3$. this function does have an inverse for any element of the domain equal or higher than 3, no? I'm working on my definitions

or, perhaps it's not called an inverse function, but an inverse image?

like, given a subset $A$ of the image of $f$, I want to refer to the section of $\text{dom} f$ of which $A$ is the image. does this have a name?

it would be the inverse of $A$, if the function was bijective and that inverse existed, but clearly the inverse of $f$ in this case doesn't exist

@Oxide it does if you work in base 17 ;-)

lol i knew someone would say this

get over here!

where?

where $x_n=\left(\frac { 1+a_{n+1 }}{ a_n}\right)^n$

and $x^*=\limsup x_n$

How do I get contradiction from here?

It's given that $(a_n)$ is a sequence of positive numbers

where $N\in \mathbb N$ exists by definition of limsup

@shintuku the "preimage of $A$ under $f$" is one way of describing it. "inverse image" is also used although this sounds funny if f isn't invertible.

Hi @leslietownes

often written $f^{-1}(A)$ even if $f$ isn't invertible. here $f^{-1}$ denotes a set mapping and not the inverse of f.

hello koro. what is $a_n$? any sequence of numbers?

Any hint for my question please ?:)

yes @leslietownes: $(a_n)$ is a sequence of positive numbers

Related post: math.stackexchange.com/questions/417551/…

I'm trying to prove by taking log. Why? Because log is changing fractions to +, - and then adding them just seems so easy (see how things get cancelled

i knew i'd seen it somewhere. littlewood.

it might also be in polya and szego.

my instinct is that replacing ln(1+a_n) with ln(a_n), although it gives you a true inequality, loses too much information.

i don't always have the right instincts but that jumped out at me when i looked over your argument.

1+a_{n+1}, rather.

ln(1+a_n) is going to be positive, for example, while ln(a_n) could be very close to -infty.

@leslietownes: In fact I have done that :P just before iteration step

apparently polya initially studied law. we are mirror images.

@leslietownes bless you I was finally able to formulate my proof

you always seem to be very hard at work. you should take a few hours off, maybe study the law or something.

"every 60 seconds in africa, a minute passes" reminds me of something my german teacher always said. "a woman gives birth in the united states every 10 seconds. we must find her and stop her."

like everything he said all the time it was probably from some old radio show.

another one was "next up on the quiz, the topic is the american flag. what flies over city hall?" "pigeons." that was from "it pays to be ignorant" where the contestants were paid for giving dumb answers.

nytimes.com/2021/04/22/science/thomas-brock-dead.html is not mathematical but is a little funny. one of the discoverers of the organism with the polymerase that was first used in PCR because it was stable at high temperatures (close to boiling) is, of course, named Dr. Freeze.

they now know of a lot of polymerases that are stable at higher temperatures than the yellowstone one. but like the yellowstone one they are found in organisms that live in high temperatures. deep sea vents are another place to find them.

i'm trying to write a proof

"the resulting $\delta$ neighborhood will always be contained by the inverse image of the image of $f$ contained in the codomain interval $(4-\epsilon, 4+\epsilon)$"

there has to be a better way to say this, right?

i like to stick fairly closely to the formulations that appear in definitions. "Thus [for reasons just articulated] if |x - a| < delta, then |f(x) - 4| < epsilon."

i guess it's more point-focused than set-focused, if that makes sense. it's how i think.

point-focused in terms of the plane?

talking about individual elements of the domain as opposed to subsets of the domain.

oh, that makes a lot of sense

it's not wrong to do it the other way, you could totally do it either way. it just "feels" conceptually simpler.

maybe i should write a book about this.

but, say you need to map a piece of the codomain to the domain

what do you do if all you have is points

if you give an example, i could probably provide an answer. not sure about mapping 'a piece of the codomain to the domain.' a statement about the preimage of a subset of the codomain under f is also a statement about certain elements of the domain of f, and i'd just pick an arbitrary one of them.

ok, say $f:\mathbb{R} \to \mathbb{R}$, $x \mapsto x^2+3$, and $\lim \limits_{x \to 1} (x^2+3) = 4$, suppose you had to give the preimage of any subset of the codomain of $f$ that actually contains an image of $f$ for an arbitrary $\epsilon$

generally speaking, with limit proofs it is not necessary to identify the best delta, only a delta that satisfies the definition. it's possible to identify the best delta here, although i would not see that as necessary. i'm not sure if this is what you're asking.

i'm just sort of trying to see if I can actually refer

it's possible to identify the preimage under f of any interval using the quadratic formula and some case analysis. but i wouldn't see the proof of the limit as requiring this.

it's really pathetic, but all I'm attempting to do is refer to a piece of the mathematical object

ahh

it isn't clear to me what you're asking. i may be unwittingly changing the subject.

no I think you have it perfectly right

i guess the proof doesn't at all require thinking of the preimage of $f$

epsilon-delta management is good to learn but in practice one would prove the continuity of a function like that by a raft of theorems about algebraic operations on function spaces that preserve continuity. the deltas that arise out of these compound applications of multiple theorems are almost never optimal.

in numerical contexts sometimes you do want to know how good you can get delta, at least up to an order of magnitude relating to that of epsilon, but a raw proof of a limit is not a numerical context.

well, in the specific case of the above $f$, can't we just select the upper bound of the $\epsilon$ neighborhood and use it to form the $\delta$ neighborhood?

i generally wouldn't bother but it's certainly possible to explicitly identify the preimage of (4-epsilon, 4+epsilon) in terms of epsilon.

if someone who wasn't learning analysis asked me to explain why that function had the limit that it did, i would appeal to the continuity of the function g(x) = x (easy proof with delta=epsilon), the continuity of constant functions (easy proof with delta = anything), and the fact that squaring and addition preserve continuity. that's not what you'd see in an introduction to analysis book but it is how people think about it.

in analysis books they always want you to factor polynomials and divide by x and stuff as if you become a blank slate after every problem involving limits.

and so the continuity of a quadratic function would be by analogy with the continuity of a constant function and a linear function, if I understand you correctly

if f is continuous, so is f times f and f times f + anything else that is known or can easily be proved to be continuous. the proofs of these general laws are more widely applicable than anything that engages with a formula for f.

forgot that mathjax does interpret asterisks even if i'm not mathjaxing.

right, makes perfect sense

i once had to teach out of a horrible 'intro to proof' book that introduced epsilon-delta continuity and then discussed a hierarchy of proofs that included "epsilon/2 proofs" and "epsilon/3 proofs" and all of this other stuff. it was like numerology. complete obsession with formalism.

no insight.

this is where I think thinking specifically about the $\epsilon$ neighborhood as an interval on the codomain helps, maybe

you can picture it as an actual neighborhood on the y axis

because ted's not here i can say it. with functions you can draw, it helps to draw the function, imagine the epsilon neighborhood, graph that, and see, yes, this is allowing me to choose a delta.

the picture is more important than the inequalities. if ted arrives, i withdraw all of these remarks.

@leslietownes I think that is pretty much the Ted summoning spell

yeah, i was waiting for it.

how about this:

and think, specifically, of the picture

of the function $x \mapsto x^2 + 3$

Let $f^{-}$ be the preimage of $f$. Notice: either $f^{-}(4+\epsilon)$ defines a distance to $1$ in $\text{dom} f$ while $f^{-}(4-\epsilon)$ doesn't, or, since the function is quadratic and progressively smaller values of the variation $\Delta x$ are needed to achieve the same variation $\Delta y$, the distance of $f^{-}(4+\epsilon)$ from $1$ in $\text{dom} f$ is smaller than the distance of $f^{-}(4-\epsilon)$ from 1.

now, in theory, this designates exactly those parts of $f$ that will be used to choose $\delta$, no?

some of that looks like word salad to me. why do you hate inequalities? write some inequalities. they can be your friends. if the goal is to prove that the limit exists, just deal with the inequalities.

f's kinda stretchy at 1, which suggests we'll need to take delta smaller than epsilon. that's as far as i can get with argument without writing down inequalities.

f'(1) is 2 so we'll probably need delta smaller than epsilon/2. that's numerology but to believe it i need inequalities.

and f'(1) of course is presuming the existence of the limit that we might be trying to prove. so i'm cheating.

well, if we're given a definite limit, 4, we know that by virtue of it being a quadratic equation, two succeeding same-sized intervals on its image will have the second one having a smaller preimage

so no need to cheat with derivatives

careful. i expect that the preimage of [-1,1] is larger than the preimage of [-1000,-998].

exactly

ah

ok, they're not succeeding

they're getting away from 0!

if we're just evaluating the limit at 1, we want to find delta with |x - 1| < delta implying |x^2 + 3 - 4| < epsilon. the latter condition is |x^2 - 1| < epsilon or |x+1| |x-1| < epsilon. we can ensure |x+1| is smaller than, say, fifty million, if delta is less than 50 million - 1. so taking delta = min(50 million - 1, epsilon/(50 million)) the algebra will work out.

that's how i walk through it in my mind. i always give 50 million a starring role.

exaggerating somewhat to make the point that you do not need a quadratic formula when dealing with a problem like this. we are factoring, but that's likely to happen whenever you're dealing with |f(x) - f(a)| and f is a polynomial. x - a is going to drop out of that and there'll be some other stuff that just needs to be bounded in some neighborhood of a.

then you can take delta = min (diameter of neighborhood of a in which that thing is bounded by B, epsilon/B).

It might be helpful to write $x^2+3-4=(x-1)^2+2(x-1)$ and look at the slope near $x=1$

now, I shall spend the next several intervals of my time meditating on this

thanks a lot!

anyone know this notation

whats the difference between blackboard C and a blackboard tilde C

first is complex plain

plane

what is the tilde for?

maybe adding a point "at infinity"? i dunno.

if you think about the riemann sphere, $\mathbb{C}$ is missing the north pole.

sometimes i saw tildes for various kinds of algebraic completions, but C is algebraically closed.

Riemann sphere is usually $\overline{\mathbb{C}}$

i defer to thorgott who actually knows many things. i am only capable of talking bulls*it

@Flows. Context?

he's here! hide the geometry.

talking about the cross-ratio

dw you guys were right its with infinity included I think

Yes, it's the projective line, the Riemann sphere.

Yup, infinity included.

ha! i still get it right some of the time. take that, universe.

me knowing things is merely a farce

i just told my wife i was right about a math thing on the internet. she said, "OK." she has made some very regrettable choices.

I have to agree with you, yet again.

the new thing seems to be that we both agree. i'm a geometer now.

Well, I'm no Banach algebras person.

no, that's what you are now.

I think this is a one-way-only agreement.

still enforceable. thanks to numerous pro-corporate supreme court decisions.

@leslietownes same here