I greet everyone with respect.

Even little comments will be of great help to me. And I am very grateful for that. You can at least tell under the question whether I made a METHODOLOGY error or not. I understand everyone. Your time is precious. So at least you can make a methodological opinion.

Thank you.

If a comment is made under the question, it is most likely that I will receive a notification. Usually there is no notification from here. I haven't figured out how this place works yet.

it isn't clear to me why f has to have extrema when a = b

even if that's true in this example, symmetric functions need not be minimized at fixed points of the symmetry

(a^2 + (b-1)^2)((a-1)^2 + b^2) being a good example

@leslietownes One of my favorite exercises!

I had an example of lower degree.

Sadly, I discovered this after the book was published.

corrected and revised printing. immediately

@leslietownes Thanks. I know this is not correct. But I need deeply understand this fact. I need "meat" of the error. Why this doesn't work..I need rigorous mathematical arguments.

google seems only recently to have begun answering wolfram-like requests to graph something. google.com/…

at least on my version of chrome.

you can fiddle with rotating the graph and the x, y, and z ranges. pretty cool although i wonder why they did it.

lonestudent, i don't have a good sense for constrained optimization problems of the sort in your question. the presence of 2020 suggests it is a contest problem, and while i bear no ill will toward contest problems, they sometimes prioritize being clever over using generally applicable techniques. which is not my vibe.

@leslietownes I Checked in WA. You are right. Now I need $2(a^2+b^2)-(a+b)=2ab$ is cyclic and simmetric. Can we use this fact?

i realize that there are forums full of people who have worked these techniques down to something resembling a science, but i am not among those people. :)

@leslietownes I understand. I need deeply understand the meat of the "error".

The fact $2(a^2+b^2)-(a+b)=2ab$ is cyclic and simmetric..

Do you think this point is irrelevant?

i don't think it's irrelevant. the fact that the constraint is symmetric is probably the beginning of something you need for a solution to a constrained optimization problem involving a symmetric function to be guaranteed to have symmetry.

but nor am i going to work on it. that's a lot of math there and contest math is not my thing. other people may be more likely to dive into the deep end of that.

in a comment, someone's written the constraint in an interesting way. if you write g(a,b) = (a-b)^2 + (a - 1/2)^2 + (b - 1/2)^2 then the constraint is that g(a,b) = g(1,1). which is interesting because not only is g symmetric, but (1,1) is too, and that happens to be where f achieves an extremum.

there is probably a general result along these lines but i'm guessing that isn't what the contest would expect.

Hiya maths friends with precious time.

@lonestudent you might find maa.org/sites/default/files/pdf/upload_library/22/Ford/… interesting. it doesn't appear to be paywalled and is about exactly this. it's a very short article.

i still don't imagine that's how the problem is 'expected' to be done. you're supposed to apply AM-GM to something pulled out of thin air. that's usually how it works.

ha, it even has my example with citation. buniakovsky, 1854.

@leslietownes How rude of them to travel back in time just to steal from your intellect

well, he traveled forward in time to steal from my intellect, and then back in time to cover his tracks.

the russians have always been sneaky.

i wonder if that example has just been passed down from book to book because he chose it. i certainly didn't come up with that idea on my own, although once you see it you remember it.

Historical revisionism with time travel is the absolute worst form of espionage

(What shape is the example?)

Graph no work on firefox mobile

Oh so its got like two pockets where the minima are, but is symetric around a ridge between them, neat

@leslietownes Thank you very much.

@TedShifrin it took me even longer (nearly all week) still to use the substitution method to solve $\int f'(x)\frac1{f(x)} dx$ despite having been given the answer. But I got there!

do contest math problems usually have 'clean' answers?

Good evening Sir.

@copper.hat Usually

copper from what i've been able to gather, contest inequalities / optimization problems might not be as clean as "(1,1)" in terms of numbers, but will be clean in terms of, not drawing upon things that you couldn't get from a small number of standard inequalities

"the smallest positive root of [irreducible quartic]" is unlikely to be the answer to a contest math problem

Thanks to both of you!

he's chipper. something's wrong.

just on that thin edge.

my 17yo son is talking to me as if i am not a complete imbecile, that always improves the mood :-)

maybe that is worse, is an incomplete imbecile worse than one for whom Cauchy sequences converge?

yes. if you're going to be an imbecile you should go all the way.

a trip to ace hardware in rust california for a tub of completion paste is on the cards...

as you may remember, rust was the old name of el cerrito

Good morning everybody. $\ddot\smile$

@copper.hat Good morning Sir.

Good morning @lonestudent!

from rust to ocean view.

personally i prefer both of those names

rust is a good name for a town. i do feel that ocean view is too literal

the irony of me saying that from long beach does not escape me

i agree on both counts, but albany is too east coast

i am sad that i never got to travel the drain tubes underneath el cerrito plaza, they are not inaccessible

i'm still looking for a trick for that minimisation problem

Is it fine to say for the definition of 'entire function' is that if a function is analytic in the domain where domain € complex set ?

What is the end of your sentence?

Do you know the definition or not?

Complex set means the set of all Complex numbers. Yes, i know the defn. A function f(z) is said to be entire if it is analytic at every z€C.

your initial question is not a well formed sentence

Your sentence still makes no sense.

So what precisely are you asking, given the definition.

I am asking is that can we say for a function f(z) to be 'entire function' if it is an analytic function where domain of analytic function is z which belongs to Complex numbers.

So how is your bad sentence different from the actual definition?

Is my definition is correct then?

Your sentence is not correct mathematical language. You should work on using language correctly.

Given a list: 1 2 3 4 5 6 .... => select [1 2], [2, 3], [3, 4], [4, 5], [5, 6] .... what is called? is there any technical term for that?

select pair from a list is obvious not the right term.

Thanks, i will.

@elliptic00 what does that even mean?

any technical term for the selection?

i do not understand what your select notation is supposed to mean.

elliptic, do you want to form a list of pairs of consecutive elements of the list

zip L (drop 1 L) in haskell if you speak that

?

Yep, I think this is the technical term, pairs of consecutive elements from a list

and you want the result to be an ordered list? or just a set?

No, I'm not looking for the code, I think your Haskell does work with zipWith

i am not aware of a technical term for such a thing

i don't think there's a technical term specific to this. only a variety of ways of specifying it precisely

jinx

Ok, I see.

i was being captious

given sequence a_n, n = 1, 2, 3, ..., something like {(a_n, a_{n+1}): n = 1, 2, 3, ...} would do the job nicely

or if you want the ordered list of pairs, maybe you'd have to signal that with something other than curly braces

in py: [(i, i+1) for i in range(1,6)]

Yep, it works,

@copper.hat not again!!

?

Do you see the link?

no, i'm scrolling up to see what i missed...

www dot buylesliecoin dot biz

What you just said. Being captious..

ahh. verrry slow now. just getting ready for bed!

hello lol

almost palindromic

lemme try it

hello lel

hello llel

lello llel

yyyyeeesss lol

bots nowadays are too smart...

yes

the last "implies" in your first set of statements is really just the definition of the ternary relation "_ < _ < _"

x > a or x < -a

not and

you got "and" up above because when x > 0 you also have x > -a and when x < 0 you also have x < a

@leslietownes So for $|x|<a$ it should be $x<a$ or $x>-a\nRightarrow -a<x<a$?

what it is "it" in "it should be"

if |x| < a, then -a < x < a

if |x| > a, then x > a or x < -a

$\implies$ and symbols are not doing anybody any favors here

@leslietownes Ok

up above when you deduced x < a from x being positive and |x| < a, you're also assuming that x is positive. so x > - a also holds in that instance because x > 0 and 0 > -a. it's not just something that you can only deduce in the case that x is negative.

whereas if x > a, you can't deduce x < -a from that. that's maybe part of the vibe of why you get an and in one case and an or in the other.

also helps to draw a picture. i don't know where you are in all of this, whether tihs is being done from order axioms or something else.

@Wolgwang if possible could you stay for sometime in the chatroom that I have been in?

@leslietownes Thanks :-) I got confused after reading this.

i can see how that would be confusing. i don't think it's helpful to think of a function as the same thing as an implication (or set of implications) that might characterize the rule of the function.

even the idea of a 'piecewise' function as a separate class of function, not too helpful. useful in constructing examples, but not really a property of a function.

$\sqrt{x^2}$ probably isn't a piecewise function. $\frac{2}{\pi} \int_0^{\infty} \frac{x^2}{x^2 + t^2} \, dt$ isn't certainly isn't a piecewise function and doesn't even have square roots in it. but $|x|$ is. oh wait they're all the same.

Hello.

Hello

Could someone help me prove this inequality? mathb.in/54947

I intend to use it to prove the Prekopa Leindler inequality and I'm pretty sure it is true

Feel free to assume that all integrals are finite

Ah, got it! Done

Hi @robjohn! How is it going lately? Long time no see! ;)

@user1591719 how are you doing?

@robjohn Not bad, always working on some calculations (lately also being interested in q-analysis, but it takes time to get used to it).

haven't done much there.

@robjohn Let me see what answers on main you have lately. :D

@user1591719 The number of interesting questions that are "acceptable" these days is very thin.

There are some interesting questions, but since they are asked in a "context-free" manner, one cannot answer.

PSQs (problem statement questions) are discouraged, and if someone has no idea what to do, it is hard to include much context.

So helping people who are in the most need of help is discouraged because there are many who have abused the system to get homework questions answered here.

I did come across a nice way to compute $\int_0^{\pi/2}\log^a(\sin(x))\log^b(\cos(x))\,\mathrm{d}x$

@robjohn Good one here math.stackexchange.com/q/4109569

@robjohn Long-standing issues debated on MSE.

yep. You can see both sides.

@robjohn Did you take into account real values for a, b? That integral is a form of Beta function (derivatives).

no, that is complicated by negative bases

@user1591719 derivatives of the Beta function

evaluated at $\left(\frac12,\frac12\right)$

@robjohn Yes, I agree. See (14) here mathworld.wolfram.com/BetaFunction.html

@user1591719 but no logs there

23-26 give some of the derivatives

@robjohn Yes, I know. My point is that one can start from that form of the Beta function to get your integral evaluated.

sure, but the computations can be pretty complicated if you take derivatives of the Gamma function

I suppose that your generalization uses no Beta function (just a guess).

It uses the Beta function, but has a nice calculus to compute the derivatives

@robjohn Absolutely.

@robjohn That's often a big challenge for many integrals alike (especially for higher derivatives).

see this answer

@robjohn Good work to cite.

yeah, I have several answers like that that can be cited elsewhere. It takes a long time to write some of these, but it's worth it in the end.

@robjohn I agree it takes time to get something like that (and a good amount of inspiration, as well).

I'll have to carefully check more of your answers at some point in the future.

@robjohn If you ever make a list with some answers of yours you would highlight, please don't miss me.

(almost no time these days to explore MSE)

I have a long list of answers that I have for this purpose. I haven't really put it out anywhere.

That's great. If you ever care to share, please let me know. Otherwise I'll try the long way to go and check many of them.

just counted them, there are 409 in my list ;-)

The previous one is a very good one I'll consider (it will be cited).

@robjohn Sir, do you help with mathematical logic too?

@robjohn Ohhh, a lot!

last summer I came up with a real only proof of $\sum\limits_{k\in\mathbb{Z}}\frac1{k+z}=\pi\cot(\pi z)$. My previous proof used residues

@robjohn Do you have it as an answer on MSE?

this answer

It's an ancient question, so the votes on that answer are pretty low.

I've been doing a lot of graphics work with Mathematica recently, too.

@robjohn It should deserve many upvotes (I'm sure that many creative ways - speaking in general - can be obtained by reducing everything to simple differential equations).

Maybe did you get the inspiration from some work on differential equations?

That's true. I was always kind of annoyed that I had to cite a residue proof for that when answering a question that was essentially real based.

@user1591719 only very simple work

@robjohn Yeap

@robjohn Simple, but it takes time and a lot of experience to arrive at these ideas.

I agree. Most of my answers are pretty simple, but I would not have been able to come up with them without the experience I've had.

so experience is very useful. I try to tell people not to be discouraged if it takes a long time to come up with an idea. The experience will pay off.

experience and tools in which you are well versed

@robjohn Absolutely. And we often talk about experience gained over many years! :D

Oh, yes. I have many of those.

Nice talk! I'll return around once in a while. A lot of mathematical work to do lately.

@user1591719 keep in touch, I'd like to see what you're up to

I'll be very glad to find your list of answers at any time it is available.

@robjohn Ok. Great!

Have a nice day!

you too!

If f is continuous only [a,b] then its continuous extension g exists that $g(x)=f(x)$ for $x\in [a,b]$ and that g is continuous everywhere on $\mathbb R$

That’s fine

But if we have $(a,b)$ is there instead of $ [a,b]$ then this need not be true

I disagree with this totally because if $f$ is continuous on $(a,b)$ then clearly we can define its continuous extension as $g(x)=f(x)$ for $x\in (a,b)$ and $g(x)=f(a+)$ for $x\le a$ and $g(x)=f(b-)$ for $x\ge b$.

Am I missing something?

Thanks.

How about $(0,1)$ and $\frac1{x(1-x)}$ or even $\sin\left(\frac1{x(1-x)}\right)$

define $f(a+)$ and $f(b-)$ for us

clearly these expressions are what your reasoning hinges on

Ahh I confused continuity of f on (a,b) with “ end points limits should exists”

Also I have one more question: if f is continuous in one set S with metric function $d_1$, then is it true that f is continuous on S under metric function $d_2$ also ?

Background of this question is like this :-

Suppose that E is compact then $f$ is continuous on $E$ if and only if $\{(x,f(x)): x\in E\}$ is compact.

@Koro Consider on any set $X$ the metric given by $d(x,y)=1$ if $x\neq y$ and $d(x,y)=0$ if $x=y$. Can you tell me which functions on $X$ are continuous with respect to that metric and do you see how that answers your question?

I observe that this set on Right side is subset of $E\times f(E)$ and that $f(E)$ is compact by continuity of $f$. There may be metric function on $E\times f(E)$ which makes it a metric space.

And then I plan to prove the function $g(x)=(x,f(x))$ on E continuous proving 1st half of the theorem. @Thorgott I’ll respond to you also.

:57996021: Thorgott, I think that limit at any point in this metric will not exist and that’s because for every $\epsilon \gt 0$ I should have a $\delta \gt 0$ such that $d(x,a)\lt \delta$ such that ….

And $x\ne a$

No I’ll need to think more…

Only constant functions @Thorgott?

That answers my question on continuity on a set under different metrics. The example of this metric which is inherent to every set shows that continuity need not remain as we change metric. Thanks a lot @robjohn @Thorgott

@Koro false

in a sense, this is the opposite of what's true

tell me your reasoning

I think you only made a subtle mistake

For a function to be continuous at any $a\in X$: for $\epsilon \gt 0.02$ (in particular) there exists a $\delta =1.2$ such that $d(x,a)\lt \delta\implies d(f(x),f(a))\lt0.02$ which is possible only when $d(f(x),f(a))=0$ that is $f(x)=f(a)$.

@Thorgott

True for every $x$ in X because of the way we chose delta

?

Given an $\epsilon\gt0$, there is a $\delta\gt0$ so that $d(x,a)\lt\delta\implies d(f(x),f(a))\lt\epsilon$

so you said $\epsilon=0.2$ I say $\delta=0.5$, then the only $x$ so that $d(x,a)\lt\delta$ is $x=a$, and $d(f(x),f(a))=0\lt\epsilon$

@Koro This delta doesn't work, but continuity says that there exist some delta, so you cannot pick one delta that doesn't work and conclude that the function is discontinuous

^

I am new to treatment of continuity in metric spaces.

I know metric spaces though. I know that only finite sets in the metric space( the one you mentioned) are compact

Oh, that metric

If you know some topology it's easier to think about the continuous maps as preimage of open is open here maybe

In $\mathbb R$ @robjohn

@AlessandroCodenotti I know that result in

metric spaces

so, which sets are open in the metric I described?

I’m not able to visualise continuity on this metric space right now.

@Thorgott B(x, r) where $r\lt 1$

@Koro what are the open sets in that metric?

and $x\in X$

any two distinct points are distance 1 apart, so you should visualize this space as a discrete set of points, all uniformly apart from one another

$B(x,r)=\{t\in X: d(t,x)\lt r\}$ where $r\le 1$

so what is that in this space?

That’s an open set.

is $[0,1]$ open in that metric?

@robjohn: intervals will not make sense in that metric?

yes, but I want you to tell me the elements of $B(x,r)$

you can name all of them

The whole set X @Thorgott

Every point in X is an interior point

@Koro why not? it consists of $\{x\in\mathbb{R}:0\le x\le1\}$

Yesss now I guess I’m on track

@Koro no

what are the points in $B(1,0.5)$?

@robjohn: only 1 lies in this ball

now we're getting there

So 1 is an interior point except that we don’t know X is a set of real numbers

My point is that X is a set with some metric defined on it

yes, the reasoning is the same though

@Koro oh, you had been talking about $\mathbb{R}$ in the previous problem, but not this one.

@Thorgott he's talking about the interval I was asking about

there is no $\lt$

Every point in X is an interior point why? Because for every x in X, I can find an r (by taking it less or equal to 1) such that $B(x,r)=\{t\in X: d(t,x)\lt r\}\subset X$

never mind

Sir @robjohn why not?

Sir @robjohn: My apologies if I caused confusion. I responded to Thorgott question wherein he introduced X as a metric space which separates unlike points by 1

@Thorgott: is my reasoning correct now?

In fact every set in X is open

@Koro yes. That is what I was trying to point out with the interval

Because every point is X is interior point.

And also complement of every set in X must be closed

because the arbitrary union of open sets is open

Therefore every set in X is open and closed both

Clopen

Every finite set is compact also

I commented on compactness earlier also

Now we come to continuity of f on X

@Koro so the space is totally disconnected.

I use this definition for disconnectedness: set S is connected if it can not be written as union of two non empty separated sets A and B. By separated I mean: closure of A is disjoint with B and closure of B is disjoint with A.

If not connected then disconnected

And so what are the connected sets here?

If disconnected that would mean no continuous function because continuity preserves connectedness

Hello everyone! :)

why would that imply no continuous functions?

Continuous image of connected is connected, but that does not imply that continuous image of a disconnected space is disconnected. In fact every compact metric space is a continuous image of the totally disconnected Cantor set

@robjohn: I referred to the link of totally disconnected space (I never knew it before) but I have read separated sets and not connected sets (disconnected word was not used). So since there is no connected subset, f would vacuously…

The implication was wrong

let's start with what the connected sets are.

@AlessandroCodenotti I know Cantor set but not Hausdarf space. I think that’s coming from generalised topology

I meant compact metric, sorry (fixed now)

@robjohn no connected set

singletons are always connected.

$\{x\}$ is a connected set

it cannot be broken into two clopen sets

The reason: Let S be a non empty connected set in C then $S=\cup_{s\in S} \{s\}$

@Koro this assumes $|S|\gt1$

@robjohn Ahh yes

it's true that every set in $X$ is open, but your reasoning isn't entirely convincing me

the notion of an interior point is relative and not absolute

@Thorgott: please read interior point as interior point in X

then you're just observing that $X$ is open in itself, which is always true

I have a conundrum. If $a,b,c\in\{0,1\}\subset\mathbb{Z}$, can you find a closed expression using addition, subtraction, max, and min such that if $a=1$, the expression returns $min(b,c)$ and if $a=0$, the expression returns $max(b,c)$? I feel like I've done something similar to this, before, but it is evading me

Let S be any set in X then for any $s\in S$ take $r\le 1$ and then consider $B(s,r)=\{t\in X: d(t,s)\lt r\} =\{s\}$ is open in X and $S=\cup_{s \in S \} \{s \}$ and hence S is open in X because union of open sets in open @Thorgott

(Not sure if I used the term "closed expression," correctly, tbh, but hopefully what I mean is clear: A single mathematical expression using a finite number of $+$, $-$, $max$, and $min$)

@Koro that is correct

so now that you know every subset of X is open in X, what can you say about the continuous functions on X?

Got it, I think: $min(max(1-a,b),max(1-a,c),max(b,c))$

Yeah, that'll work (I translated them into booleans and then worked out a truth table, and then converted them back to max/mins (so I wasn't entirely sure at a first glance))

@Thorgott: That would mean that every function pulls back an open set that is inverse image of every function is open

thereby making every function continuous

Am I correct?

For example: we take the set $S=\{a,b\ }\subset X$ and define $f(a)=c, f(b)=d$ so that $f(S)=\{c,d\ }$, every inverse image of $f$ here is open and that makes $f$ continuous. :)

indeed, every function is continuous

Yay!! I am on track now

Thank you! @Thorgott

On Pg. 33 here, why is it enough to show that vol(\partial\tilde K) \le 2n vol(\tilde K)^{(n-1)/n}? The author hasn't provided any argument for it. library.msri.org/books/Book31/files/ball.pdf

Even in the first sentence after the statement of Theorem 6.2, the author says it is enough to prove vol(K) \le 2^n (which comes from the previous fact)

Why is it so?

Well I've posted it as a question now, because I wanted to type LaTeX lol

the problem with mathematics is that there is very little ground between 'obvious' and 'have no idea'.

day 4 no exercise.

ramps up on exorcise!

its a bit scary how hooked on endorphins i have become

Quick question: Assume a function h(.,.) defined on R^n*R^m, and for any given y, h(.,y) is a convex function of x. Define f(x)=max_{y\in C} h(x,y) where C is a convex set. is the max function here nondecreasing? meaning f(x) is convex?

the function $f$ is convex, but have no idea why you think it would be non decreasing without other hypotheses. take $h(x,y) = -x$, then $f(x) = -x$.

f can be convex without being nondecreasing. e.g. h(x,y) = e^(-x) and C = {0}. then f(x) = h(x,0) is convex but strictly decreasing

or what copper just said

snap

i'll leave you to it, mr. convex

no, no, no, please :-)

my suspicion is that you will need slightly more hypotheses to ensure that f is convex, but i have no proof or idea of what that would be

Got it so if it's pointwise MAX then it would be convex, makes sense, thanks @copper.hat@leslietownes

if each $x \mapsto h(x,y)$ is convex then the $\max$ is.

or $\sup$ really.

the epigraph of the $\sup$ is the intersection of the epigraphs of the constituents.

that makes sense. seems like you just need to visualize it for one hyperplane

or even proof by duh

burned into my brain in cory hall while minimising $\max$ functions.

i like inequalities in convex analysis where you do it for a line or hyperplane and suddenly you have it for all functions

convex analysis has enough intuition that my little cranium can deal with it :-)

there was a time in my life where i calculated that i'd spent what rounded up to a materially nonzero percentage of my life in evans hall

i performed that calculation in evans, of course

OKay I got another one, I seem to have problem with convexity of this MIN and MAX parts! What about a function as f(x)=min_{i=1,...,m}{a_i * x+b_i}

don't tell ted, but draw a picture!

think of what the hypograph looks like

draw $x,-x$ and take the min. you need to experiment a bit.

draw a bunch of random lines, and then color in the region that lies beneath all of them. it's important to color. i learned this from my daughter, who likes crayons.

I've had like 4 years no excercise since I blew out some disks in my back and cervics

sry to hear it.

breathing fresh air helps though

ill stop whining.

whining is my favorite activity no worries

well the ones in the cervics arent really blown I was exagerating

the only ones I completely messed up where l5-s1 and l4-l5

Have you tried cortisol related things?

They work real good but I think they mess u up

so I almost never take them

i am a bit scared of steroids. that is why i am not exercising at the moment, had some hip injection on tuesday.

i have accumulated many injuries & much wear & tear over the years so i really should not complain (but i will :-))

i keep myself in top shape by remaining in front of a computer or in bed at all times

@Leslie Get out and count the ducks.

that's tomorrow's activity. i'm keeping my hip healthy by only doing that once a week.

i also carry my daughter to and from the mailbox and hold her while she opens it (she takes a while because it has a key). upper body strength is important too.