Mathematics - 2021-05-16

12:00 AM

You wrote partial derivatives (and do NOT ever write something like $\partial f/\partial g$ or $df/dg$ — it is meaningless).

copper.hat

i like to think of tangent planes to the graph

not a particularly unusual perspective, of course

Ted Shifrin

How does that elucidate the composition of linearizations?

copper.hat

it does not.

Ted Shifrin

Gosh ... This might be the second time today I am being analytic and ungeometric. I really will lose my geometer's license.

copper.hat

i am arriving late as usual

shintuku

12:03 AM

composition of linear functions multiplies slopes

YES! thank you!

Ted Shifrin

In one dimension, yes.

shintuku

right right

Ted Shifrin

Glad that helped your intuition :)

leslie townes

2:08 AM

asked the daughter what her favorite activity of the day was. expected to hear about the pony again, instead got "i liked to watch the birds hopping around." on our porch, which she sees every day. not the trip to ride a pony, or the trip to the park to swing on swings. such a toddler.

lone student

Good Morning!

leslie townes

hi

lone student

If there are people seeing errors in my solutioncan they please comment (at least)?

@leslietownes Hello!

Thank you.

robjohn

2:38 AM

never mind

Andrew Micallef

2:55 AM

yo

Joseph Sible-Reinstate Monica

I remember a while ago seeing a picture of a pattern on the hyperbolic plane that looked like it took up 1/2 of it, then when you shifted your view it looked like 1/3 of it instead. I think it was in a question here but I'm not 100% sure. Does anyone know what/where this is?

Andrew Micallef

@TedShifrin I think I recall why I kept avoiding Taylor. I can see the utility, but they are super tedious to write out (and with all the sign flipping in cos / sine, a real headache for my dyslexia)

but the expansion of e^x is nice

shintuku

3:19 AM

so, is there any rigorous definition of quotients of differentials?

leslie townes

several of the answers to mathoverflow.net/questions/73492/… are thoughtful and somewhat relevant

shintuku

so that they act independently, and aren't treated as a symbol for a limit

thanks, will look into that!

leslie townes

i like steven landsburg's answer, which characterizes treating dy/dx as a fraction as a 'gateway drug' to more harmful practices. we may not always agree politically but we agree on that.

shintuku

I guess that if you make sure you can always subsitute $dx$ or any differential by $x - x_0$ and add in an error function that goes to zero, you can safely treat it as a real ratio

shintuku

4:27 AM

may the math gods bless Ted Shifrin and the math 3500 lectures

Russian Bot Who Knows Your IP

4:46 AM

yo

mathics is also nice in plotting

Russian Bot Who Knows Your IP

5:06 AM

scilab is also nice and it has many features

idk why i am telling this here lol

Russian Bot Who Knows Your IP

5:20 AM

and these

shintuku

7:20 AM

anyone knows how the author goes from this, to:

$f(x_0 + h) - f(x_0) = f'(x_0)h + \epsilon h$

where $h = x - x_0$

I tried limit algebra, and I died

Koro

shintuku, that follows from definition of limit.

shintuku

I thought about that, but how are they getting an equality from a limit?

so, of course, the initial equation can be written $\lim \limits_{h \to 0} \text{blabla} = 0$

then, by limit algebra, you get something like $\lim \limits_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} = \lim \limits_{h \to 0} f'(x_0)h$

but where the hell is the $\epsilon h$ coming from

is this a canadian limit of some sort

in any case, I've been at this for two hours, I beg for help

Koro

Think about this: if $\lim f(x)=2$ then $f$ can be defined as $f(x)=2+g(x)$ where $\lim g(x)=0$

shintuku

wait, this follows from the epsilon-delta definition?

Koro

So your $\epsilon$ is “something” that tends to $0$ as $h\to 0$.

@shintuku you can prove that using $\epsilon$ delta also

shintuku

7:32 AM

$f(x) = 2 + g(x)$ would be some function that gets away from 2. Right. and as $g(x)$ as a function of $x$ approaches whatever value it is that makes $f(x) = 2$, then $g(x) = 0$

Koro

$g(x)=f(x)-2$ so what happens to $\lim g(x)$?

shintuku

my bad, are we assuming $x$ goes to 0?

nevermind, that's irrelevant

Koro

In my example, it doesn’t matter shintuku. Do you see why?

If this is clear, use similar arguments in your question about derivative.

shintuku

ah, I got it

niiiiice

thank you!

if $\lim f(x) = 2$ holds, then for $f(x) = 2 + g(x)$, whatever $g(x)$ is, it will be $0$ when $x$ goes to whatever it goes to with $\lim f(x)$

Koro

Great!! 😊

shintuku

7:47 AM

wow, finally

I've been at this for too long

thanks a lot

Koro

Morning exercise: Try to prove limit rules using this (by introducing g(x) ) See how simple the proofs become

welcome :)

shintuku

8:36 AM

@Koro quick follow up, if you're up for it (already greatly appreciate the last answer). I've proven the property you mentioned: if $\lim \limits_{x \to x_0} f(x) = L$ exists, assume $| x - x_0| < \delta$, then $-\epsilon < f(x) - L < \epsilon$, which means $f(x) + g(x) - L = 0$ exists, so $f(x) = L - g(x)$ if $|x - x_0| < \delta$. but, would you happen to know why we'd use $\epsilon h$ as a $g(x)$?

looks to me like we can use literally any $g(x)$, as long as $\lim \limits_{x \to x_0} g(x) = 0$

Koro

8:57 AM

Well, in this case please note that we are taking $\epsilon $ as $g(h)$ and not $\epsilon h$.

Also note that $\epsilon h$ in this case can be replaced by $o(h)$.

shintuku

9:20 AM

I'll look into that, thanks!

Pratik Haware

10:04 AM

You know how we read C(m,n) as "m choose n"? Is there a similar way to read P(m,n)?

kunal Ch.

I need guidance with this inequality. consider $|a|·|b|\lt\varepsilon$ where $\varepsilon$ is arbitarily small. If $0\lt|b|\lt10$ then will this $|a|·10\lt\varepsilon$ inequality hold true ?

Balarka Sen

1:24 PM

@sayanGhosh Thanks! I had shared the questions here before: chat.stackexchange.com/transcript/message/57924025#57924025

I had missed your messages earlier, sorry about that.

Peter John

Let $M$ is a finitely generated $A$-module where $A$ is a local ring with maximal ideal $m$. Then $M/mM$ is a $A/m$-vector space so there is a basis $x_1+mM,...,x_n+mM$ of $M/mM$. Now let $L$ be a free $A$-module generated by $x_1,...,x_n$. Define $\phi:L\to M/mM$ by $x_i\mapsto x_i+mM$. Then I want to show the kernel $\{\sum_{i=1}^na_ix_i| \sum_{i=1}^na_ix_i \in mM\}=mL$. $\supset$ is clear but how can I prove the reverse inclusion?

Omni man

1:56 PM

cellular automation

I got problem with it

anyway I am too tired to discuss about it

Thorgott

2:16 PM

@PeterJohn the induced map $L/\mathfrak{m}L\rightarrow M/\mathfrak{m}M$ is an iso, so the kernel can't be bigger

Peter John

2:41 PM

@Thorgott Oh, thank you

sayan Ghosh

2:53 PM

@BalarkaSen Yeah. Got it. Btw, if you wish, can I talk to you through email or via discord bcoz I'm also interested in Differential Geometry and in Algebraic Topology (Homology, Cohomology) as yours afaik.

epsilon-emperor

3:19 PM

Hi, how would you show the countable sub-additivity of the Hausdorff measure? The book I'm reading said it's a particularly tough task

Balarka Sen

3:33 PM

@sayanGhosh Yeah you can send me a message on discord

Russian Bot Who Knows Your IP

ip address

Charlie

4:06 PM

How come if $\det(ABC)=\det(A)\cdot \det(B)\cdot \det(C)$ I can't commute the $\det$'s in the RHS around and obtain arbitrary permutations of $(ABC)$ have the same determinant, unless it's true that a product of matrices has the same determinant regardless of order?

Perhaps the $\det$ function doesn't commute with itself?

epsilon-emperor

It's true @Charlie

Thorgott

you can

epsilon-emperor

det(AB) = det(BA) = det(A)det(B)

Charlie

oh

oh yeah of course I can see how that generalises, ty

epsilon-emperor

Can someone help me here?

H^s(F) is the s-dimensional Hausdorff measure

but I don't understand their explanation

sayan Ghosh

4:32 PM

@BalarkaSen Thank you! Please share the details of your account.

Thorgott

which part is the issue

epsilon-emperor

4:48 PM

@Thorgott So I see how the result follows if the Hausdorff measure is zero for all s > n

I just need help with proving that the Hausdorff measure of bounded sets is zero

After that I can use countable sub-additivity of the Hausdorff measure to conclude the same for all sets in R^n

"By considering covering of \delta mesh cubes, ....., H^s(F) = 0"

this part is the issue

Thorgott

H^s of an n-dimensional cube is 0 if s>n

might be annoying, but this can be computed from the definition

epsilon-emperor

the calculation seems a little crazy

oh but yes that will do the trick

@Thorgott ok so i want delta-covers for the cube --- will it be okay to only consider smaller cubes covering this one?

Thorgott

you just need to partition a cube into smaller and smaller cubes

epsilon-emperor

5:06 PM

i think our definitions of H^s(F) may be different

Thorgott

tell me yours

epsilon-emperor

H^s(F) is lim_{\delta\to 0} H^s_\delta(F)

|U_i| = diameter of U_i

\delta-cover means that every set has a diameter \le \delta

*every set in the cover

Thorgott

that's my definition too

partitioning a cube into small cubes gives you upper bounds for $H_{L/n}^s$ that go to $0$ as $n\rightarrow\infty$ (here, $L$ is the side length of the cube)

epsilon-emperor

then why do you think that only cube covers work?

@Thorgott what's L?

and is n the same n as in R^n

Thorgott

no, ns just a variable

epsilon-emperor

5:11 PM

refers to number of smaller cubes?

Thorgott

no

epsilon-emperor

then? :(

oh okay

so i've to find smaller cubes of diameter atmost L/n

Thorgott

just partition a cube into smaller cubes and see what you get

the L/n I just chose to get actual partitions, but this isn't a significant point

epsilon-emperor

@Thorgott mathb.in/55241

This is what I get

doesn't really go to zero as n goes to infty unless I've missed something

Thorgott

you forgot taking to the $s$-th power in the RHS

epsilon-emperor

5:23 PM

yup i've edited that

Thorgott

now what happens as $m\rightarrow\infty$

epsilon-emperor

the RHS goes to zero!!!!

and delta goes to zero

wow we did it

thanks!

Derivative

For a result that people often call elegant, Conway's actual proof on the look and say sequence is surprisingly computationally intensive link.springer.com/content/pdf/10.1007/978-1-4612-4808-8_53.pdf

I mean knowing that the end result is an algebraic number of degree 71 might give you a clue beforehand, but it was still worse than expected

Thorgott

np, glad you got it

Derivative

@TedShifrin Ooooooh. I read that book. And I really liked it.

leslie townes

5:33 PM

derivative, what is the statement proved in that paper? it's paywalled

Balarka Sen

@sayanGhosh Send me an email on my university email and I'll share it there

You know my ISI email right?

leslie townes

oh, i found a non-paywalled version

Derivative

call the lengths of the terms in the look and say sequence $L_i$. Then $L_{n+1}/L_n\to\alpha$ where $\alpha$ is..

oh

weird, I'm on my home computer, idk why it isn't paywalled for me, but maybe it's because I accessed my university's VPN on it recently

Balarka Sen

Here's a dumb question. I have an ODE $u''(z) + p(z) u'(z) + q(z) u(z) = 0$ where $p, q$ are meromorphic functions on $\Bbb C$. Somehow, $z = 0$ being a "regular singularity" is the only interesting possibility, in which case $p$ has a pole of order $1$, and $q$ has a pole of order $2$ at $z = 0$. Why?

leslie townes

my guess is that it seems elegant because it cleanly cabins the computational aspect away from the concepts, which are simpler and very well tuned to the problem.

it's maybe no coincidence that i found a copy of the paper on one of doron zeilberger's course website. a lot of his work is like that too even if it is frightfully computational.

Balarka Sen

5:38 PM

There is a classical theorem of Fuchs which describes all the solutions when you have such an ODE with a regular singularity.

I guess if you write $v(z) = u'(z)$ then you have the system $(u', v') = A(u, v)$ where $A(z) = (0, 1|-q(z), -p(z))$

So now we can run some monodromy business. I forget how this works

Locally this will always have solutions away from $z = 0$, the problem is to patch, which requires passing to some cover

sayan Ghosh

@BalarkaSen Yes. Sure.

Done.

Balarka Sen

5:56 PM

Sent @sayanGhosh

sayan Ghosh

P.s: I know that since your very childhood you are keen to Mathematics and you are a wonderful Mathematician as I believe. So, I'm very much eager to talk to you 🙂

Balarka Sen

I'm always around and happy to talk about math. But we have met before, haven't we?

sayan Ghosh

Surely not. But I heard about you and your passion for Mathematics few years back from your relative, I think.

Balarka Sen

Oh, I might have confused you with someone else, then. Where do you study?

I wouldn't put my interest in mathematics in lofty terms. It's a good way to pass time, just like playing a good video game

Most of my relatives neither understand video games nor academia so they have a broken perspective of how things are

epsilon-emperor

About the opposite inequality, I'm not sure how the author gets to it

The fact that the s-dimensional Hausdorff measure of the union is 0, only tells us that s > \dim_H \bigcup_{i=1}^\infty F_i

and we have assumed s > \dim_H F_i for every i

but this doesn't allow us to compare \dim_H F_i and \dim_H \bigcup_i F_i right?

Balarka Sen

6:08 PM

The last sentence tells you the Hausdorff dimension of $\bigcup_{i = 1}^\infty F_i$ is less than any number bigger than supremum of the Hausdorff dimensions of each of $F_i$

leslie townes

dave renfro's comment in math.stackexchange.com/questions/3827302/… might help

or balarka's comment right here :)

Balarka Sen

So $\dim_H(\bigcup_{i = 1}^\infty F_i) \leq \sup_i \dim_H(F_i)$, no?

Thanks for verifying, leslie

Thorgott

it's using the characterization of sup as smallest upper bound

epsilon-emperor

@BalarkaSen No wait, how?

Balarka Sen

Fix any $s > \sup_i \dim_H(F_i)$. Then $s > \dim_H(F_i)$ for all $i$.

So $\dim_H(\bigcup_i F_i) < s$ by what you wrote.

epsilon-emperor

6:12 PM

@BalarkaSen Do you want to fix s = \sup_i ?

Balarka Sen

Not really.

epsilon-emperor

@BalarkaSen Agreed but how does this help us compare \dim_H(\bigcup_i F_i) < s and \sup_i \dim_H F_i? both are less than s, that's all I know

wklm

Hi Everyone :) What's the common interpretation of X^+ and X^- where X is a series?

Balarka Sen

@epsilon-emperor By what I said, you have $\dim_H(\cup_i F_i) < s$ for all $s > \sup_i \dim_H(F_i)$. If $a < s$ for all $s > b$, then $a \leq b$.

sayan Ghosh

@BalarkaSen It might be. Great

leslie townes

6:15 PM

wklm: if a_n is the sequence whose partial sums make up the series, a_n^{+} might be the sequence taking the value a_n, if a_n >= 0, and 0 otherwise. similarly a_n^{+} might be the sequence taking the value 0 if a_n >= 0 and a_n (or perhaps even -a_n) otherwise

epsilon-emperor

@sayanGhosh Ah, this makes sense now. Thank you!

leslie townes

so you end up having a_n = a_n^{+} plus or minus a_n^{-} and have notation for just collecting positive and negative terms

epsilon-emperor

How would you prove that statement about a, b and s? Sorry it's probably very elementary

leslie townes

sometimes see this in the proof of riemann's theorem on rearrangements of partial sums of conditionally convergent series

epsilon-emperor

It's intuitive but I'm using it for the first time somehow

Balarka Sen

6:17 PM

@epsilon-emperor Hint: Take inspiration from your username.

OK, better hint. Prove by contradiction.

:)

epsilon-emperor

@BalarkaSen Done! With inspiration from username XD

wklm

thanks a lot @leslietownes

epsilon-emperor

Suppose a > b. Then there exists \epsilon > 0 such that a > a- \epsilon > b. Our hypothesis gives a < a - \epsilon, contradiction! :)

Thank you!

Balarka Sen

@RyanUnger Do you agree that pseudoholomorphic curves should be thought of as quasi-minimal surfaces, where quasi = in the Gromov sense?

epsilon-emperor

@Thorgott hmm we didn't use that in Balarka's proof

Balarka Sen

6:30 PM

@BalarkaSen OK, I think the point is if $\mathbf{u}(z) = (u(z), v(z))$, then $$\mathbf{u}'(z) = \sum_{i \geq 1} \frac{A_i(z) u_i(z)}{z - p_i}$$ where $A_i(z)$ are holomorphic and $p_1, \cdots, p_n$ are the regular singularities.

In this sense, hypergeometric equations are kind of universal because they solve the Riemann-Hilbert problem for all possible representations $F_2 \to \text{GL}_2(\Bbb C)$. OK.

Thorgott

its implicit

epsilon-emperor

@Thorgott Okay cool I'll think about it. Thanks!

Ted Shifrin

7:02 PM

Howdy @Thor and a @Balarka

Hi epsilon and negative @Leslie.

Thorgott

hey Ted

leslie townes

good afternoon

Ted Shifrin

happy drizzly afternoon, @Leslie.

leslie townes

it's only cloudy here. although we could use some drizzle it would cut short the planned afternoon trip to the duck pond. so i hope it stays cloudy.

Ted Shifrin

It drizzled/rained on me for my 2 1/2 hours going to the Farmers Market and other shopping.

leslie townes

7:09 PM

great timing!

Ted Shifrin

Yup. But you know that ducks love rain :)

copper.hat

7:47 PM

lovely ride through tilden to alvarado and back via kensington. super windy at the crest.

back to defcon 5

leslie townes

the PSQs are safe again

Ted Shifrin

ROFL

copper.hat

8:05 PM

i did sneak a look at a few :-)

Ted Shifrin

8:17 PM

This is far from a PSQ, but you both might be interested. I've never seen anything like this.

leslie townes

yeah, wow. i've seen some goofy expressions for the taylor remainder but nothing like that. very weird to see p > 0 in something like this when 1 is a more common dividing point for introducing p.

LandonZeKepitelOfGreytBritn

You can either round numbers, but you can also just discard decimals

What's the exact term for that latter thing again?

not concatenate...

i just can't remember the exact word!!

truncate!

thx guys you are so useful and great

leslie townes

you are very welcome.

if it's true presumably it pops out of applying the mean value theorem for integrals to some exact expression for the remainder involving one or more integrals. i'm trying to imagine a purpose of this. i was thinking an embedding of a function space defined by norms involving differentiation into something simpler, or a renorming of a function space with something simpler.

it drives me crazy when people don't say where the thing came from. i never thought i'd say this, but would it kill him to post a cell phone photo of where he saw it in print?

Andrew Micallef

8:34 PM

morning

speaking of Taylor, I think I am 90% of the way to prooving Eulers formula

I realised that half of the i's in e^ix become -1's and the other half become -i's

leslie townes

around 2001 we had a mystery pdf of something we wanted to cite it if its proof resembled ours (we only had a few pages with a statement of the result and the first two lines of a proof). the guy who found it couldn't remember where he'd gotten it and google didn't index pdfs at the time. we were not able to figure out where it came from.

about 15 years later i googled it and it was from a partially completed chapter of something one of his coauthors had drafted up but never completed. the proof was not complete.

@andrew exactly.

Andrew Micallef

now I'm just scratching my head at what Euler was doing to have noticed this formula in the first place :P

Emma math

Hello everyone, I was wondering whether anyone knows how to test if an exponential distribution is a suitable model for the quakes data in R.

Andrew Micallef

like a goodness of fit?

the folks over at stats.stackexchange.com might know

robjohn

@AndrewMicallef Have you seen this answer?

Andrew Micallef

8:40 PM

is it about euler (i still havent got my proof finished, just thought that was a profitable observation so if it is eular I will decline to click that link)

robjohn

However, plugging $ix$ into the series for $e^x$ and comparing to the series for $\sin(x)$ and $\cos(x)$ is pretty simple.

leslie townes

@andrew, 17centurymaths.com/contents/euler/introductiontoanalysisvolone/… (exponential functions), 17centurymaths.com/contents/euler/introductiontoanalysisvolone/… (trig functions). in translation.

at least in that book, he derives it from the series, as you did.

robjohn

@AndrewMicallef okay, then don't click that link. However, it is a completely different approach than what you are doing.

leslie townes

it's written in an old enough style that it may not be much help :)

"After logarithms and exponential quantities have been considered, circular arcs and the sines and cosines of these must be considered; not only because they constitute another kind of transcending quantity, but also because of the logarithms and exponentials of these that arise when they are involved with imaginary quantities,
which will become clearer below." you said it, euler.

Andrew Micallef

@robjohn no I hadn't seen that and it is totally different to the approach I was taking. Ted sent me off on Taylor series....

robjohn

8:44 PM

which is not a bad way if you can use Taylor Series

Andrew Micallef

also side bar before I run off, any chance chatjax will work on firefox for android in the near future?

@leslietownes yeah I find modern maths hard enough to read

robjohn

@AndrewMicallef It doesn't? I thought it did. If not, have you tried the suggestion for Chrome on Android?

Andrew Micallef

I'll give chrome a look...let you know later today when I have my break

robjohn

I thought there were people here who were using ChatJax running Firefox on Android.

Andrew Micallef

yeah on firefox when I do the bookmark trick it just loads infinitly then craps out (not the best description, but i'm pulling my memory for the symptoms)

robjohn

8:47 PM

You are running Firefox on an Android phone?

Andrew Micallef

yeah

maybe now (5 mins before I have to run) wasn't the best time to bring it up :p

robjohn

okay. See if the suggestion for Chrome on Android helps with Firefox

Andrew Micallef

will do

leslie townes

euler is easier to read than a lot of math of similar vintage but it is not as easy to read as at least some modern books.

that should be the test of a real analysis book. is it easier to follow than euler.

tealing123

Can anyone help me with a financial mathematics question regarding Wiener ProcesseS?

copper.hat

9:00 PM

@TedShifrin very odd Taylor expression. I think I would like the OP to provide more background before I would spend time on that one.

leslie townes

9:26 PM

copper, in comments he says he was 'checking some documents with taylor series' and found that formula. what more context do you want

this was after, not before, he was asked for a source

i'm going to turn this into a best selling novel where the source of the problem is tracked down after deciphering a series of clues in a number of international locations. i'm thinking of calling it The Taylor Code

coming soon to airport bookshops near you

LandonZeKepitelOfGreytBritn

To all my haters in this room:

I hope you lose your watch and are late for an important appointment.

Joe Shmo

oooof 🔥

leslie townes

9:44 PM

what's a watch? that quip must predate the widespread use of mobile phones.

like "here's a [dime, quarter], call someone who cares."

here's my samsung galaxy s5, call someone who cares

Koro

10:06 PM

Hi everybody

Hi Leslie, Ted, Robjohn

Hi copperhat

I have a question on continuity

related to "shadow points"

It is to be proven that for all $\displaystyle x\in \ ( a,b) ,$ we must have $\displaystyle f( x) \leq f( b)$

Here's the Latex version of my proof. mathcha.io/editor/10MG5t0Pto7HvYFl9p8JfyedV3vIXzY0vQTPklOP

My question is: Is continuity really required for proving $f(x)\leq f(b)$?

I think that intermediate value property possession by $f$ is enough.

Am I missing something? Thanks.

Ted Shifrin

@leslie @copper Here!! Whew! I got it.

leslie townes

wow. tiny tex hiccup there, something isn't typesetting.

Ted Shifrin

@Koro: You might want to tell people (other than me, since I know the problem well) what the actual question is. I.e., what is a shadow point and what is the hypothesis?

@leslie to Koro or to me?

leslie townes

to you, after "and now it all falls in place."

Ted Shifrin

Oh, weird. It was OK and then it isn't. Let me check.

leslie townes

10:15 PM

i read somewhere that the preview engine is based on a slightly different flavor of mathjax than the display version once a question is submitted. sometimes the two do not coincide. i ran into this a lot a long time ago. maybe that is still the case.

Koro

@Ted: I saw your comment there and I also tried to solve the Taylor series problem. I got $(x-a)^{p+1}$ and $p$ also but didn't get $(x-a)^p$ 😁

@Ted: I have incorporated the question in Latex version in the link I have shared above.

Ted Shifrin

OK, @leslie. Thanks.

It's fixed. I had so many braces I messed up a few.

Koro

typesetting is fine now.

Any hint/suggestion for my question please?

Ted Shifrin

@Koro: Don't be impatient.

Oh, you have an older version of Spivak's book. The problem was revised because I pointed out to him that one of my wise students had a far easier proof.

Koro

Ahh, I see.

Ted Shifrin

10:23 PM

Sometimes it isn't useful to state an arcane hypothesis if in reality it gets you nothing extra. I mean, the examples of functions that are discontinuous and have the IVP are pretty limited. However, here we want to show that $f(a)=f(b)$. Without continuity, how are you going to get $f(a)\le f(b)$?

Thorgott

are they pretty limited?

I'd think discontinuity is the more generic phenomenon, even amongst Darboux functions

Koro

I'm afraid till part (a), there is no mention of showing $f(a)=f(b)$

Although that comes in later part of the same question.

leslie townes

more generic can still be arcane. :)

in fact, quasi-all functions are arcane. the proof uses the baire category theorem

Ted Shifrin

@Koro It's not always optimal when writing a problem to give something too general for part a, and then make it more specific for part b. The way the problem got rewritten, it says "Suppose $f(a)>f(b)$. Show that the point where $f$ takes on its maximum on [a,b]$ must be $a$." Without continuity we wouldn't even know there's a max.

@Leslie: There was still a TeX error in the proof (a missing fraction), but I have fixed it. I think it's done.

Koro

Ok. I'll think more about the problem.

Ted Shifrin

10:28 PM

Once I remembered Lagrange's trick of varying the point $a$, it wasn't too hard to see how this should go. I wonder why this generalization of the MVT is useful.

@Koro I no longer have the older editions with that statement of the problem. My recollection is that one of my students, having done the exercise, said something like: "Why do all this sup/inf stuff? Just show directly that ..."

Koro

Okay. Noted.

Ted Shifrin

So I no longer remember the original statement that you're attached to.

copper.hat

10:44 PM

@TedShifrin Very nice!

the pointless harassment continues: "Does this answer abide by the EoQS, which you've previously been linked to?. No, it doesn't"

Ted Shifrin

Why are they so hot on your tail, @copper?

leslie townes

Danilo's explanation of the application of that result is not illuminating.

i want a photo of the secret document

copper.hat

it is just one person who seems to track down my 'transgressions'

not quite sure i get the point of tracking down such stuff on questions that have already been closed.

and a 'punitive' downvote. like i give a flying whatever

Ted Shifrin

Have you gone into the mod room to engage in a quiet conversation about this, @copper?

copper.hat

i added a comment to cured

Ted Shifrin

10:54 PM

@Leslie: I just added my two cents' worth about arcane for more arcane.

leslie townes

there's a post on meta which indicates that starting with one post as a jumping off point to browsing through someone's history and tracking down other transgressions is discouraged. at least one mod feels very strongly about this.

LandonZeKepitelOfGreytBritn

@leslietownes I would not wish my biggest enemy to lose his smartphone. That's too cruel. How is he supposed to check his facebook feed while on the bus otherwise?

leslie townes

you are a greater humanitarian than i am.

copper.hat

our dependence on phones is disturbing

LandonZeKepitelOfGreytBritn

@leslietownes Such kind words... You deserve to keep your watch

Ted Shifrin

11:05 PM

@Leslie Have we ducked?

copper.hat

pretty neat answer @TedShifrin

Ted Shifrin