twitter.com/daveinstpaul/status/1412556454926639104 this is wrong.

does latex have an ombudsman. i want to write a letter.

a solution I was happy about got deleted :'(

:(

although I got invited to coach cmb for the pre-imo so im happy about that

@SAJW You might try my videos, linked in my profile. In 3500 I do a first treatment, connecting to solutions of systems of linear equations. Then there is a second treatment in 3510, along with a presentation on manifolds three ways — explicit, implicit, and parametric.

If $f:X\rightarrow Y$ is continuous surjective and $\{U_{\alpha}\}_{\alpha\in A}$ is an open covering of $Y$, then why is $\{f^{-1}(U_{\alpha})\}_{\alpha\in A}$ an open covering of $X$. This is claimed in a proof I'm reading but can't see why. Thanks!

Just write down definitions, Larry.

You don’t even need surjectivity.

Could you give more hint? What if all the U_alpha map to a single open set in X; f is continuous and that would be still possible right?

First, they don’t map. You’re doing preimage..

Take any $x\in X$. Why is it in some set?

Dumb question. After expanding (x+y)^3 in pretend world, is it universal to consider y^3 or x^3 the first term? Is there a universal definition for first term?

I was reading about the Trace theorem in Evans and he talked about how $\partial U$ has measure zero and hence no meaning to a function u restricted to the boundary $\partial U$, but then he defines $L^p(\partial U)$, isn't this space's norma $\int_\partial U $? So isn't this an integral over a set of measure zero?

@Yorch which one?

Hey, @Ted!

@randomgirl because we write left to right it is often to see the leftmost things as 'first' but i would not consider there to be any order involved.

if someone asked me what the first term was, i'd ask, in what ordering. there are various choices in use for polynomials.

i guess x also precedes y in the alphabet. i could see that mattering for at least one order.

That's exactly what I asked. Thanks so much leslie. I just thought it was possibly a dumb question but you reassured me.

you sometimes see this in computational algebra packages. you have to pick an ordering.

@randomgirl I don't know which would be considered first, but I would expand it as $x^3+3x^2y+3xy^2+y^3$, which preserves the order, whether reading right-to-left or not.

Hi @robjohn

how was your holiday?

I just discovered my cat loves peach/nectarine … crazy

@TedShifrin the scent or the fruit?

The fruit!

My cat likes peach. He was begging for my baby's pureed peaches and he ate them right up.

We had a dog that loved the oranges on our tree. The dogs I had as a kid, loved the tangerines from our tangerine tree.

i had to keep my cat from eating mango this morning.

Yes, I just read they love mango.

Good for them, apparently.

some carnivores they are!

she loves fruits and vegetables. if we leave dishes on the table after dinner (we mostly eat vegetarian soups) she will clean them.

My cat also seems very interested in my cappuccino.

i have a vague sense that she should stick to her cat food but in extremely small quantities i am fine with it. it also makes doing the dishes easier.

olivia loves cappuccino.

Never had these experiences with other cats.

do high school teachers not encourage kids to sketch stuff, functions, etc?

i see a lot of questions that could be usefully addressed by a little sketching work.

Yes, or to abuse graphing calculators.

basic transformations of graphs are a standard part of many K-12 curricula, which is not to say it is a well absorbed part. i fear that devices may have played a role in this.

ahh, those corporate calculators.

the persistence of graphing calculators surprised me. i figured they would have gone out with the internet. like CD players. but no.

I think it depends on which country. In Hong Kong, we were not allowed to use graphical calculator in the usual high school curriculum. So in some sense we need to learn how to sketch the graph of a function.

i think plotting devices are a great tool but an awful replacement for sketching

interesting.

i think that makes sense

@TedShifrin I am not sure why any x must be in a set. Could you please help me a bit further?

Any $x \in X$ must be in $f^{-1}(Y)$.

What course are you taking and do you have the prerequisites?

I think there’s not understanding of preimage.

Anyone online to answer another one of my newbie questions? :)

taps the sign.

:)

Q: Show that, if $\tan^2 \theta = 2\tan \theta +1$ then $\tan 2\theta = -1 $

If I was given this question a few weeks ago, I would used the tangent double angle formula on $\tan 2\theta = -1 $ and rearranged it to get it in the desired form. However, I have recently started a course in writing proofs and was wondering if that would be the right course of action? If we let be $A$ be the statement $\tan^2 \theta = 2\tan \theta +1$ and $B$ the statement $\tan 2\theta = -1 $

The question is asking us to show that $A \implies B$ correct? But wouldn't starting with $\tan 2\theta = -1 $ be showing that $B \implies A$?

mm, if you just use identities and the given hypothesis to evaluate tan(2 theta) that sounds fine to me.

it's sometimes helpful to structure a proof of an identity in a way where only one thing changes at any given time. LHS = RHS1. next line, = RHS2 [reason in parentheses], next line, = RHS3 [reason in parentheses]. operating on both sides of an equality is not always illegitimate but it can be very hard to read.

i taught an 'intro to proof' class and that format of writing a proof of an equation may have been the only thing that got through. a prof of an abstract algebra course said all of my students still used it.

he thought they were copying from one another but someone showed him my notes from the class where i said, this is a really legible format for a proof of the form A = B.

you're welcome, universe.

Haha. I see, thanks for the help :)

i am in a food coma because i ate dumplings, a huge plate of nigiri sushi, and then because my daughter rejected them, avocado rolls.

i love sushi, but the usual servings are miserly

when it's high quality the quantity doesn't matter. i like cheaper sushi places where i can load up.

tachibana across from zachary's on college was very good but is now closed. mitama, if it's still open, is good too and cheaper.

the location where mitama is used to be a breakfast restaurant that suze orman of personal finance fame worked at. then a car crashed through it and it went out of business.

there was also a horrific pizza place on that block. i think all of that is gone now.

i miss living in rockridge, i used to be a 5 minute walk from amazing food. now i have to get in a car or order delivery.

i probably ate from cactus taqueria every friday night from 1998 to 2008. the quality dipped near the end of that run.

i still go back when i'm in town.

What is one application of orbit stabilizer theorem without using group actions? @Leslie

@copper.hat I ate veg sushi once at a hotel here :-)

the avocado roll is one of california's more dubious contributions to the cuisine.

i think of the orbit stabilizer theorem as principally about group actions, so i don't know what to make of the question. sometimes you do somewhat arbitrarily choose an action to dig out some result you want from a mess of algebra.

i don't think i ate avocado at all until i was about 35. i found the texture disgusting. now i probably eat two avocados a week. my wife's family has a tree and we get them for free.

Hey, guys I have a basic question about calculus: math.stackexchange.com/questions/4192306/…. I've been thinking about this for about two, three hours and I just rlly wanna know what's going on

I'll appreciate any help I can get

the example you give is not continuous at zero. there are versions of this kind of thing particularly around where derivatives vanish but i do not see that example as one of them.

for example the limit defining F'(0) does not exist. because you can't just look at one side.

ahh i see i see that makes sense

one thing you will see in textbook formulas is primitives not taking into account discontinuities in the function being integrated. for example the integral of 1/x is ln |x| + "C" but you can choose different C's for positive and negative x, and no textbook ever says this.

something something de rham cohomology.

@leslietownes I see. I was thinking that it could be used somehow to prove some result. Anyways, I think I'll understand it more when I learn group actions.

what about the other example with the u-substitution?

i do think that there's quite a lot of algebra where you want something, and you achieve it by picking a group action even if that isn't 'the point' of why you care about that thing.

@lesl

@leslietownes tried it few months back, I didn't like it :'(

i think continuity forces c_1 = c_2 at the origin but i could be wrong about this.

you see examples of this with integrals of inverse trig functions.

koro group actions or avocado? :)

It's always frustrating when the only references you can find for a problem are either way too narrow or way too general.

there's a lot of stuff on the frontier of calculus where it's seemingly too elementary to be regarded as worthy of comment, but too complicated to be worth explaining.

so people just don't talk about it.

the first rule of calculus club is do not talk about calculus club.

lmao thats a shame

I'm trying to find stuff on continuous additive functions, i.e., continuous functions such that f(x+y)=f(x)+f(y). Most sources assume f is just a real-valued function, but I want to allow higher-dimensional domain.

somewhere i have notes on this. i made a long list of stuff that was not justified to any degree of precision in my calculus book.

but i don't want anything this general:

semi, the "cauchy functional equation" may be a good search term.

oh crap you've found it already.

ya

i think usually with mild hypotheses (e.g. measurability) the solutions to the equation are what you would expect them to be, i.e. f(x) = kx for k a scalar.

yeah, and in the higher dimensional case it's linear

@leslietownes avocado :)

and there are obvious 'counterexamples' with crazy stuff if you choose a hamel basis and go crazy.

gross

koro, if you are younger than 35, i recommend trying it again when you are 35. it disgusted me and now i can't get enough of them.

i'm fine with assuming continuity here

mostly i'm just curious what the usual proofs look like. (i have a version of it showing up in a paper i'm reading, and i'm curious how standard their proof is)

you use induction to prove f(p/q x) = p/q f(x) and then continuity to fill in the gaps, i think.

@leslietownes sure, 10 more years for me to try it again then :)

yeah, that's enough for the 1D case

oh

your diet preferences changing in your 30s is not something i anticipated.

so maybe the real question is how to go from homogeneous to linear when it's not 1D

i couldn't stomach pork anymore. i began to eat cooked fish, which i used to find nauseating. and now i eat a lot of avocados.

are continuous homogeneous functions only polynomials?

(generalizing to arbitrary degree here)

there's a result on this too, i forget the name. it is classical.

ah, Euler

hmm

yes, Euler will do

euler's work is amazing because he writes so clearly you can understand it even if you don't read latin.

he's one of the clearest writers who ever existed.

julius caesar is another one.

@Koro i am in my twenties currently, I meant :)

you may be too young to appreciate the merits of the avocado.

it's a weird texture, i don't think i would eat it by itself.

actually it's considered an exotic fruit here and is sold at high price. And dragon fruit also!

they grow like weeds in southern california. my wife's family has more than they know what to do with.

I saw avocado at a mall once then I thought of trying it, I didn't like it though as the taste was so plain not salty, not sweet etc. I liked juice of dragon fruit though :-)

to be clear, the Euler result i have in mind (applied to the 2D case for simplicity) is this: if $f:\mathbb{R}^2\setminus\{0\}\to \mathbb{R}$ is homogeneous of degree one, then $f(x,y)=x\partial_x f(x,y)+y\partial_yf(x,y)$

i guess that doesn't guarantee linearity, tho. take $f(x,y)=(ax+by)e^{x/y}$ for instance.

@leslietownes just to be clear, in the first example in my question, the derivative when you approach 0 from the right is zero but the derivative when you approach 0 from is left is infinity so F’(0) doesn’t exist right?

i would say the limit, not the derivative. the limit is something of a precursor to the derivative and has two sides. because of the jump the limits on either side do not agree.

so the limit without qualification does not exist and the derivative does not exist either.

i don't want to be too formalistic but that is how a lot of professors would want to see that approached.

my cat is roving the halls in search of food. this is not a good sign.

@DavidChoi it doesn't exist simply because it's not continuous at 0

A differentiable function is necessarily continuous.

Could someone help me with a fairly direct real analysis problem?

mathb.in/59573

I've typed it in the link above with some context.

@leslietownes Maybe? Sorry for the tag

i will answer this in the morning if nobody else does. :) it is midnight here.

Cool, thanks! Good night! :)

Why is x/x is not periodic i mean by definition any smallest positive value of T such that f(x+T)=f(x) for all x in domain but book says its not?

Hi all, sorry for a very stupid question: How to properly/usually write the remainder $r$ of $m$ divided $\mod n$? Is it $r = m(\mod n)$ or $r = \mod(n,m)$ or something completely different?

possibly simply $r = m \mod n$, right?

But it always confuses me what is the factor and what the remainder.

@leslietownes Update: I figured it out myself. One of the two directions is false. For your reference, I have written the proof: imgur.com/a/U5TY0P3

that was a good source to read. thanks @robjohn

@Shobhit large exponentials?

Exponentials mod m

that's what I meant. good

I need help with a question. So there is an event which happens with 0.1 probability. If it happens, it will place 3 stones randomly on a 3x5 grid. I am asked what is the probability that there will be 3 stones placed consecutively from the left in the second row. I think it should be 0.1 x (1/15) x (1/14) x (1/13). Am i correct?

So the second row should look like Stone, Stone, Stone, empty, empty

@Shobhit it should be 0.1 x 6 x (1/15) x (1/14) x (1/13)

(1/15) x (1/14) x (1/13) is computing the probability that three stones are placed in the first three squares from the left of the second row, in a particular order

there are 3! orders in which the stones can be placed in those three consecutive squares

or, if you want to think about it orderlessly, the answer needs to be $\frac{1}{ {15 \choose 3}}$ (which is still 6 x (1/15) x (1/14) x (1/13))

oh ok. Got it. Thank you. One more question, so if I was asked what is the probability that none of the stones appeared in the second row then case 1. The event did not happen (which is 0.9) case 2. Stones appeared but none on the second row. So they appeared 3 together in either row or 1 and 2 stones in 1st or 3rd row or vice versa. So for 3 stones appeared together in a row, will it be 0.1 x (5 choose 3) x 6 x (1/15) x (1/14) x (1/13)? @porridgemathematics

the probability that none of the stones appear in the second row is $\frac{ {10 \choose 3}}{ {15 \choose 3}}$ which is $\frac{10 * 9 * 8}{15 * 14 * 13}$

its easier to just count the number of places the stones can appear, and then enumerate the number of 3 -element subsets consisting of those places

rather than try to think about where each stone appears

ok. I'll keep that in mind :)

so the final answer to your follow up question would be $(0.1)(\frac{ {10 \choose 3}}{ {15 \choose 3}}) + (0.9)$

ok. Thank you :) . I have just one more question. I have been stuck on this all day.

If the case was only 2 stones in 2nd row. So would it be (5 choose 2 )/15 choose 3 ?

not quite, i think it would be [(5 choose 2) (10 choose 1)][15 choose 3]

you need to account for the 10 possible places the last stone can be, for each two possible places some two stones can be in the second row

ok. I think I understand. Much appreciated :)

what

wait

how-

why did that happen!!!???!?

hehe

Can someone guide me how to find the curvature and torsion of the curve of intersection of the following two quadric surfaces: a1 x^2+ b1 y^2+ c1 z^2=1, a2 x^2+b2 y^2+c2 z^2=1. I can find curvature and torsion if I can get the parametric. just guide me how to find parametric, then I can solve further on. Thanks in anticipation.

@porridgemathematics The answer you wrote above to my follow up question $0.1 * \frac{10 choose 3}{15 choose 3} + 0.9$. Shouldn't the first term be multiplied by 3! since after choosing 3 places the stones can be placed in any order in those places.

Is there a ring with unity but no units ?

@Shobhit well the denominator is the total number of possibilities, which in this case is a choice of 3 places (regardless of ordering)

if you wanted to count with respect to ordering, both the numerator and denominator would be multiplied by $3!$ as you say, but it would not change the quotient (the probability)

yes ok . thank you.

and so for the configurations Stone, Stone, empty , empty, empty and the configuration Stone, emoty, empty, empty, empty the answers would be $0.1 * 6 * 1/15 *1/14 *10/13$ and $0.1 * 6 * 1/15 * 10/14 * 9 /13$ respectively. Is this correct? @porridgemathematics

on the 2nd row

$\int {\frac{x^2cosx}{1+sinx)^2}}$

From x=0 to $x=\pi$

Any thoughts..?

I am a high-school student and I am not sure if I am ready to learn abstract algebra. I have been reading Logic from "Introduction to Mathematical Logic" by Elliot Mendelson (up to godels completeness theorem but planning to re read again to finish all of the exercises) but I dont know anything about number theory. never studied calculus. Is it enough to even make an attempt to study a subject like abstract algebra? I am planning to read it from "Advanced Modern Algebra" by Rotman.

@gaufler no, the unit is a unit

prithu there is a spectrum of opinion on the best algebra book. i can't think of any reason why a high school student would be unable to get into it. number theory is a common starting point but not the only one.

one distinction you often see is whether a book does ring theory (addition and multiplication) or group theory (just one operation) first. if you had experience with number theory i would recommend a book with ring theory first.

i learned group theory first, it may have been suboptimal but i still remember a lot of it.

@leslietownes I am planning to read abstract algebra in parallel with logic because learning logic alone is getting a bit boring (Thinking of retracing my steps in Logic to finish all the exercises)

For some reason Dummit and Foote seems to be not suitable for me (even though people say it is a classic)

i never took a class on logic. i did absorb some of it by osmosis because my university had a lot of people who worked in that area.

@leslietownes I am just studying logic as a hobby in this lockdown to pass the time. Schools are shut down.

sounds like now is as good a time as any to study logic, then. i hope you and your family are staying safe.

Hello!

I have a question about the meaning of the question rather than solution please.

Solution: the graph with 2 edges is the only one as it has not circuits and thus it's nonisomorphic unrooted tree,

I am confused as to what the solution was compared against to conclude that because I know that isomorphism is based on comparing one graph against another and see if they are isomophic or not.

i don't know what a rooted tree is but it seems like --- and -< and <_ are at a minimum possible and not isomorphic?

hrm

isn't that four vertices?

if the question is, can i count, the answer is no.

@leslietownes. @hyper-neutrino. Thanks for replying. It's $n=3$ vertices

four, three, it's all the same to me.

nice poem :D

@leslietownes. hahaha

i used to tell my students this: if it seems like i'm making some goofy arithmetic mistake, please speak up. there's a 99% chance that you're right about me being wrong and it will save us all some time.

This is the book solution

i didn't mess up too much but it was always really silly stuff, like 4 not being 3.

to answer your question, @Avra, you can try constructing every distinct (non-isomorphic) graph and see that the only two are the tree with two edges and the triangular cyclic graph

Again. I am just confuses as nonisomorphic if two are not isomorphic, so on what basis the solution was built that only graph with 2 vertices is the only nonisomorphic please?

it means that any other unrooted tree with three vertices will be isomorphic, so this is the only distinct graph when comparing by isomorphism

So, once you build a graph with 3 vertices, you will compare it against another 3-based graph for isomorphism ?

Is this how the solution was based onto?

well, there are only 8 possible graphs with 3 vertices (including disconnected ones), so you can just try them all

unrooted i guess means you don't distinguish points on trees and require them to be mapped to one another via isomorphism?

i mean, there only exists one graph with 2 vertices

then, to add a point to a graph of N edges, you just have to consider all 2^N possible extensions, and observe that one is not a connected graph, one is cyclic, and the two that are acylic connected (tree) graphs are isomorphic to each other

(this is a rather tedious and mechanical approach for larger things though, unless you are using a program)

i think hyper neutrinos advice to be exhaustive is good advice. it's not an approach that scales, but it is adequate for the problem.

@hyper-neutrino. Thank you. Pardon me please, but you said, "you can try constructing every distinct (non-isomorphic) graph". So (non-isomorphic) graph to what? Nonisomophic to which graph as this is the part that confuses me?

i also recommend distinguishing 4 from 3. very useful skill in life.

if $z , w \in \mathbb{D}$ (the unit disk), is there a holomorphic map $ f : \mathbb{D} \rightarrow \mathbb{D}$ s.t $f(z) = |z|$ and $f(w) = |w|$?

@Avra like, construct every graph, and notice that if you construct another graph, it's isomorphic to the only solution

@hyper-neutrino. Thanks. So a graph with 2 edges is nonisomorphic to what

To 2 edges enumeration or 3 edges enmeration?

2 edges enumerations are: a->b>c, c->a->b, b->a->c

So we have 3 graphs with 2 edges, why the solution I posed above from the book says this is the only nonisomorphic

wdym 2-edge enumerations?

because all three graphs are isomorphic

This is the solution

How about a->b>c, c->a->b, b->a->c?

two graphs are isomorphic if a permutation of the points' labels results in the same graph

oh crap there isn't always one :(

if you connect AC and BC, then you can relabel A=A B=C C=B and have the same graph I think

since $|f(z)| = |z|$ implies $f$ is a rotation

oh wait no, but for that $f(0) = 0 $ necessarily.. maybe there can be one after all

a and c are not adjacent

Yes, that is a more formal definition of what I described.

So any permuitation of 2 edges result in a nonisomorphic!!

Since c->a->b is nonisomorphic to a->b->c

no, they are isomorphic, because f(c)=a, f(a)=b, f(b)=c is a bijective function from V1 to V2 that satisfies isomorphism

And also a->c->b?

Ohh!

I see now.

f(a)=a, f(b)=c, f(c)=b

you mean there is one-to-one correspondence btw vertices

basically

So, again please why the graph below is the only non-isomorphic?

I guess I get it now:

The solution graphs that are nonisomorphic w/t 3 vertices are with 0, 1 ,2 and 3 edges

but the only one that is tree and unrooted is the one with 2 edges and this is the only one that is nonisomorhic

@hyper-neutrino. This is the solution I found! Thank you very much.

basically, yes. 0 or 1 edge isn't enough for a tree, 3 is too many for a tree

no worries, glad you understood

Since a tree can not have circuits or loops

yep

@hyper-neutrino. So, in those kinds of problems, we have to enumerate all possibilites and pick one that satifies structure: tree, non tree, etc.?

not necessarily

a tree with N vertices will haev N-1 edges. that is guaranteed to be true

Yep!!

(though the inverse is of course not true)

If it has N-1 edges, then it would be the only one!

Otherwise, it's not a tree in case we want a tree

if you have specific problems/questions feel free to ask them again; i'm not sure I can give a general solution for all similar/related problem types

But if we want a graph that can be basically anything, then any that satifies nonisomorphism could be a solution?

enumeration is tedious; construction is faster but requires more intuition I feel

@hyper-neutrino. Thank you very much.

@Avra well, "satisfies nonisomorphism" on its own means nothing because like you said, isomorphism is a property/operator between two graphs, but if you're trying to count non-isomorphic graphs, it just means "the smallest set of graphs such that no two are isomorphic to each other"

@hyper-neutrino. Thanks. You are right

@hyper-neutrino. In case we have rooted trees!

The solution would be two trees

yep.

Here one-to-one correspondence is in terms of length of a path?

well, when taking isomorphism between rooted trees, instead of "a and b are adjacent iff f(a) and f(b) are adjacent", you just have "a is b's parent iff f(a) is f(b)'s parent"

the only thing you really need for rooted trees is to change "a and b are adjacent" to "a is b's parent" (or vice versa), and you have a rooted tree

@hyper-neutrino. Wow! Is it because rooted trees are defined in terms of parent-child relationship?

that's one way of thinking about them / representing

Why we don't take then c (parent) -> a (child of c and parent pf b) - > b (child of a)

that works fine too

but that was not added in the solution!

Maybe it's wrong?

A is the parent of B if and only if B is a child of A, so both are fine, it's just unnecessary to state both

The solution is above with 2 trees only

a (par) - >b (parent of c) - >c

a (par) -> b( child of a) -> c (child of a)

This is how the solution stated it. It was stated in terms of path length not adjacency or parent relationship as I understood

i think that works too

you can list out the path lengths of each vertex to the root

so the first one has [0, 1, 2] and the second has [0, 1, 1]

I am just confused as why the solution did not add the tree I told you about :

c (parent) -> a (child of c and parent pf b) - > b (child of a)

because that's isomorphic to the first graph

isomorhpic here in terms of path lenght?

from the root node?

yeah, sure

so, it's neither adjacency nor parent-child relationship?

they're just different ways of thinking about the same thing

@hyper-neutrino. Confusing, but got it...!

@hyper-neutrino. Thanks again

no worries :)

my relationship with my irritating cat could best be described as adjacency.

my daughter's figured out that the cat doesn't talk. when we talk to the cat, she will say "livvy doesn't talk." but she understands.

@leslietownes. hahah

i'll miss this, it's something of an end of an era. she used to hold the cat in her lap and tell her about her day, and i think that may be gone now.

she still yells at the cat.

hahaha

OMG

Check out the attitude.

tragedy is when i cut my finger, comedy is when you fall into an open sewer and die.

this is comedy.

i stopped short of upvoting the question and those retorts. that's my good deed for the day.

You're a big help.

Of course I know precisely how to answer the question, but won't be bothered to do so.

i don't mean to trivialize this. this is misuse of the site. it's sad that if you put "My Attempt" in there it's suddenly fine when many normal questions without that formatting are not.

but what do i know.

Well, it's not just the lack of attempt. It's lack of clarity on what we know and what we don't know, coupled with rude attitude because I don't cater to his whims.

saying 'google a PDF of the book i'm talking about' is not the way forward.

I do think that three years ago I would have just answered without having gotten so riled up. But the times have changed on this site.

Yes, this is the spoiled generation.

he has an actual professional geometer asking for clarification. resources that did not exist 30 years ago. and we get this.

i mean i am still giggling about it but i don't feel good about why i'm giggling, if that makes any sense.

@Balarka: The math Ph.D. who taught me in high school was surely not in the same league as the one who taught you. Mine did a weak Ph.D. in logic and was really not a good teacher, although she thought she was.

@TedShifrin. Professor, he asks you to look book online!

Wow!

I used to own the book, and I actually can guess exactly what to do, but this person has irritated me beyond normal bounds.

@TedShifrin. Some people might not grow well how to behave properly

I don't think his rudeness is flag-able, though.

it's common-or-garden-variety rudeness.

@TedShifrin. You really don't know how one grow up, this endless I guess

my daughter spent half of the morning yelling at my wife and coming up with bs reasons why she couldn't go to school. i actually asked her 'who raised you?' then i realized, oh.

We could have predicted all this misbehavior based on yours. No-brainer.

she is 1000% my child and i will have to live with that. it's kind of interesting, to be subjected to it. you think, this is what everyone who has ever had to deal with me must have felt like.

she's very kind to animals and offers them her food, which is something i do. we're still figuring humans out.

one time we were trying to trap a groundhog and got a raccoon instead. he had worn off half the fur on his arm trying to get out during the night. i carried him to safety and when i picked him up he just curled up like a kitten. it was a weird interspecies moment.

i assume we're all watching england-denmark.

@TedShifrin.

@TedShifrin. Did the same thing with me!

well, just admit that you don't want to help.

it seems simple enough to me.

i can't even view the question anymore, someone deleted it.

@leslietownes. He/She deleted it

@leslietownes. I am unfamiliar with his problem :(

it strikes me as standard textbook stuff that would have been answered already if they had just stopped to think that they were making a request of other human beings.

@TedShifrin Plenty of people who think highly of themselves, especially in the teaching community.

also me, no longer in the teaching community. i hold myself in the highest regard.

Rightfully so

@Avra @leslie The question amounts to the fact that the hyperbolic metric is conformally equivalent to the euclidean metric.

that's definitely in one of my books. it might be in the garage but it's there.

@BalarkaSen Not to mention plenty of us in every community.