turns out $\dot x = -\beta xy^\alpha$, $\dot y = \beta xy^\alpha - \frac{1}{\sigma}y$ is a dynamical system, not a differential equation. i thought they were the same

They are the same.

Dynamical system is sometimes used to emphasize topological aspects, but they’re the same.

huh, alright then

thanks

dynamical system often includes other systems, difference equations, etc.

and a differential equation with no solutions might not be called a dynamical system, while we're at it.

I would call those discrete dynamical systems, copper, but point taken.

How come the main site is down but chat isn’t?

they're keeping it open so we can nitpick stuff you say.

Who am I? George Santos?

why trump was ok but santos not is beyond me

think of all the good stuff we probably will never know about him

But he was a star on the volleyball team!

i heard he invested in lesliecoin

I’m sure that’s his fake 700K.

:-)

This is the worst of this site.

Just in time for the reopening.

Ugh... four of my five classes were canceled due to low enrollment. As of today, I have been assigned three other classes. I have four preps this semester, one of which is a class I have never taught before. Classes start on Tuesday.

Please, kill me now.

:/

Yikes. Your ratemyprofessor rating too disastrous?

what's the new class?

@TedShifrin No, the high schools kind of f'd us over. It happens a lot.

Certain schools demand large numbers of seats, then fail to fill those seats in the fall.

We can usually squeak out classes in the fall, but then attrition doesn't give us enough students to run the classes in the spring.

Joint enrollment too low?

Attrition?

So what are the four? Maybe leslie and I can kill your rep forever!

@user726941 Second term with not enough finishing first.

@TedShifrin Essentially. For example, one section of the class was supposed to have 30 students last fall. However, one of the high schools asked for 20 seats, but only actually placed 8 students in the class. So the fall started with only 15 students or so.

About half of those failed or dropped.

So, not enough students for the spring section.

They should be held accountable. They lose half next time.

@TedShifrin Heh.

@TedShifrin I have made many suggestions.

For example, I think that the high schools should be responsible for paying some nominal per-student tuition for every seat they request. So they are out the cost of the class, whether or not they fill the seats.

So if they gave you those 4, the part-time faculty who were going to teach them are fired.

They have no budget as it is.

But the college is so afraid of pissing off the high schools, because, in the short term, they could really monkey with enrollment if they decided to completely withdraw from our dual enrollment agreements.

@TedShifrin We have no part-time faculty in the math department.

Perhaps, they haven't been prepared with the right prerequisites.

Oh, so what are the faculty who had those courses teaching?

@TedShifrin Well, I was given three classes from other faculty who where at (or slightly over) load.

Hmm …

Well, no one should agree to overloads.

@TedShifrin We almost always have faculty in overload, here.

Are students given placement tests?

We are very rural. We can't really hire adjuncts or other part-timers.

There just aren't that many qualified people in the region.

So, usually, people do a lot of overload in the fall, and have a smaller load in the spring.

Oh, makes sense.

But, also, people here like to go into overload, because we are horribly underpaid, and the extra income helps.

Got it.

What I would really like is for the college to compute load on an annual basis, rather than semester-by-semester.

Well, it sounds like they’re doing that de facto.

@TedShifrin Nope. We generally get overload pay in the fall, and then are in underload in the spring. But they still have to pay our minimum contract, so we are given make-work to bring us up to load.

So, for example, someone could have 18 load in the fall (with 3 overload hours), then have 12 load in the spring plus 3 load worth of make-work.

In my ideal world, that same person would have 18 load in the fall, and then get a 3 load reduction in the spring.

Or, if they taught more than 12 load in the spring, would be paid overload for those extra hours.

Oh, I see. I agree, but they want the extra money. So is the make-work fake?

@TedShifrin It depends.

Often, yes.

Officially, my job this semester, for 2 whole load, is to rename two courses.

Anyhow, what are the new courses?

Which, in reality, means that I need to fill out two forms, talk to a couple of deans, and make a couple of presentations at the instructional council. But I am a member of the IC, so I'll be there anyway.

Very little actual extra work.

I was associate dept head for 8 yrs but this is nonsense to me.

@TedShifrin I was originally going to teach two sections of "trig", two sections of calc I, and a section of calc II. Of these, only one section of trig made. I have been given a different section of calc I, a section of "college algebra", and a section of "contemporary math".

I've never taught the contemporary math course before.

Is that a math for poets course?

@TedShifrin Essentially, yes.

Those can be cool courses, but not our usual mindless stuff.

If I had time to plan the course, I think it could be a lot of fun to teach. But I don't have the time. :/

you could make it really contemporary by not preparing any lectures.

If I had my druthers, it would be a very projects-based kind of course. Lots of writing, relatively minimal drill-and-kill.

Surely the person who was expecting to teach it can share a syllabus.

@TedShifrin He did. We sat down and hashed it out for four hours today.

So I will be teaching his class.

But his class is not anything like a class that I would teach. :D

Leslie … “y” $\ne$ “aneous”

I get it, Xander. But ask for it again and start to remold it.

Oh, and the best part is that it is an asynchronous class. >:(

Ugh.

Well, covid is expected to get way worse.

@TedShifrin Oh, that class needs work. Comparing the syllabus I was given to the official description of the course in our curriculum system was... uh... eye opening.

Yes, but at least it’s a terminal course, so leeway is ok.

There is a whole section on geometry (including fractal geometry(!)) in the official curriculum which appears nowhere on the syllabus I was handed.

Figures, since people hate geometry.

you should teach it like a prophet of chaos. explain that the math will be so contemporary, every day they will be seeing new math that has never previously existed. and hint that something unpleasant is going to happen to all math that came before.

Well, maybe swap out a section and replace.

@TedShifrin It looks like there might be a week or two at the end. If I cut the section on weighted voting methods (because... who cares...), I'll likely have somewhere between two and two and a half weeks to do some geometry.

I really would like to talk briefly about fractals---and it ties back to exponential growth models in a useful way, so...

It'll happen, one way or t'other.

Actually, the voting stuff is very interesting, particularly in AZ.

@TedShifrin Phone. But yes, voting good. Weighted voting... meh.

@leslie No comment on the explicit “I don’t belong in this course. Do my trivial homework” link I posted?

you described it as the worst of the site, i dunno. it's bad, maybe not the worst.

that's kind of a weird aspect of math, enough of it is broken into pieces bite sized enough that you can get through quite a lot of courses with "just one more nudge, i'm stuck on this one thing." you can't, like, get someone to write an essay for you as easily as you can get a nudge.

enough of an intro class like that, the formulas look like normal calculus, i could see not really focusing on definitions and just nudging around like a pinball. this may have been why my undergrad DG instructor was always in a pissy mood.

you always try a lot more than i would, in the comments, to improve the question.

is cal II proofy or not yet

depends on who is teaching it. if it covers sequences and series, that strikes me as semi-proofy.

at least, that's part of why i think series have a reputation for being so difficult. that and working with inequalities, if you've never done that before, boy does it suck.

Depends on the school, teacher, etc.

but sequences/series is very challenging for most students.

oh it just struck me

first induction proofs and convergence proofs

No induction in standard calc courses.

Just having to decide on integral test versus ratio versus comparison.

What you’re suggesting is real analysis or calc theory. In Europe, calc is real analysis.

i see

i think it's in stewart somewhere although a competent instructor would probably skip that section. and so many things are only in stewart because at one time, someone asked "why isn't [my favorite thing] in the book?" and because he loved selling books he listened every time.

Editors dictate a lot more than you realize. Gotta go with mass rule.

editor who looks like ebenezer scrooge, shouting to an author over the phone: "the witch of agnesi is going in that book"

@TedShifrin I tend to spend about half a lecture on induction in Calc II. Just enough to deal with $\sum n$, $\sum n^2$, and $\sum n^3$.

I say this from experience as both author and calc book reviewer (incl Stewart).

@XanderHenderson We never did that at UGA. No time.

@TedShifrin I make the time by skipping some of the applications at the end.

I like the applications wayyyyy more.

@TedShifrin Yeah, I don't.

Probably a bias resulting from never having taken an actual physics class in my life.

In any event, I need to go to bed.

We’re teaching science/engr students, not many math majors. And the math majors will learn that later.

And copper.hat is here, so I'm leaving. :P

Night.

G'night.

Hi everyone, I would like to solve the least square problem numerically using Matlab which requires the quadratic problem to be in this form.

Taking the expression further, we get

Matlab's numerical solver requires me to multiply the objective function by 1/2 and drop the offset b^Tb, so I get this

In comparison with the analytical solution of the least square problem, the numerical solver provides the same optimal solution. Is it always the case that multiplying by a constant or removing an offset won't affect the optimal solution?

Go back to single-variable calculus and think through your question(s).

Hi!

Is it possible to solve $$\dfrac{\int_0^1(1-x^{2020})^{\frac{1}{2021}}dx}{\int_0^1(1-x^{2021})^{\frac{1}{2020}}dx}$$ ?

Got this while solving

are you really supposed to evaluate it? it might be simpler to decide a simpler question, like, whether the fraction is less than something or greater than something.

(1 - t/n)^(1/n) is pretty close to e^{-t} when n is large. maybe even uniformly in t in any fixed [a,b]. that's one thought.

@leslietownes Yup it's an integer type question

Are you sure of that outside exponent on $n$?

oh fiddlesticks

well i guess i lost the contest

that typesetting is janky

Does anyone know what the "buffalo's way" method is? I cannot search anything related to it?

@leslietownes ouch

@Wolgwang It is, and it is a very nice answer.

BTW, I am back home and recuperating.

@youthdoo Did you check the Wayback Machine?

Any chance there might be a closed form for $I(a,b)=\int_{-\infty}^{\infty}\frac{e^{-x^2}}{\cosh(a)+\cosh(bx)}\mathrm dx$? Mathematica can't find anything.

@robjohn Thanks. But too bad the linked website is blocked for me.

Hi! A random variable X(omega) takes values in an open interval (a,b), so a < X(omega). Does a < E[X] or just a <= E[X]. I'd lean towards the former, am I right?

can anyone please vet my proof?

@robjohn Yippee! Welcome home!

@TedShifrin thanks! Good to relax at home. I have a home health care Nurse Ratched coming at 9:00.

Does the Wayback Machine link I gave above work for you? I didn’t think they required any special privileges.

Works fine for me.

Could be a local blockage (eg censorship). Don’t know from where youthdoo hails.

Yeah, even tictok is getting blocked/censored nowadays.

@robjohn Does it involve the gamma function?

@robjohn I don't know what happened (I hadn't visited the room for months). I hope your recovery is fast :)

@Wolgwang He said it was a very nice answer...

@Wolgwang it does involve the Beta function, but it cleans up very nicely.

No Gamma in the final result

@robjohn :( How can they? It's not even in the syllabus.

Suppose that we have a normed real linear space X, and that $Y\subset X$. If $f$ is a bounded linear functional on Y, then there exists a bounded linear extension $f'$ of $f$ to all of $X$ such that norm of f and that of f' are the same (i.e., $\|f\|=\|f'\|$).

I think that the following proof is problematic: I'll just sketch the outline of it. Take any $b$ in $Y-X$. Then consider the space (let's call it $Y_1$) spanned by $Y$ and $\{b\}$. Suppose that I come up with a bounded linear extension of f to $Y_1$. Suppose that this extension takes the value $b'$ on $b$.I do this process for every $y\in Y-X$, then using these I define $f'$. Is it correct to say that $g$ thus obtained-$g(x)=f(x)$ for all x in X, $g(y)=y'$ is the $f'$?

I think that this is problematic because g that we obtained may not be linear!!

So Zorn's lemma is a must here to prove this. Is my understanding correct? Thanks.

@Wolgwang the ratio can be done without the Beta function. Just some substitutions.

If you want each integral, it involves the Beta function.

How did the old Ratched treat you?

The better question is: Are the lunatics running the asylum?

That's the post-pandemic norm.

koro it's problematic for a lot of reasons, both the nonlinearity issue, and the fact that it doesn't seem to be attempting to maintain any control of the norm. you can definitely prove this by zorning a one-dimensional version of the result but you have to pay attention to the norm. this is a special (maybe classically most fundamental) case of something often called the hahn banach theorem

because there are so many generalizations of it, wikipedia sucks, but my #2 google result for 'hahn banach theorem,' ucl.ac.uk/~ucahad0/3103_handout_6.pdf should cover what you need

Good morning, Munchkin's devil.

hi

Has the rain hit you yet?

it's kinda weird that someone's PDF handout from a 2012 functional analysis class is the second google result for hahn banach theorem, but it's a good handout

yes, falling steadily here but not too hard.

I'm often skeptical about the quality of handouts. Interestingly, someone posted a question about Terry Tao's blog (about differential geometry), and it — surprisingly — had a sloppy error. I don't expect that from him.

it's definitely weird that google's algorithm for one of the most common named theorems in a pretty big subject would select #1, wikipedia, #2, some random handout, and yeah, this definitely doesn't always work

Is that handout any different from the standard treatment of Hahn-Banach in "all" the textbooks?

I guess we can't find "all" the textbooks by googling. Books will be a lost art.

it treats exactly koro's case (i.e. real spaces, and extension only of bounded linear functions and not other objects). sometimes books begin with something slightly more general.

but yes, it's the usual proof.

the wikipedia page is the usual mess of "here is what 30 people at 30 different times thought was important to add about this thing"

Wikipedia is not a good replacement for learning from a text/lecture.

there's more text about the relationship between the theorem and the axiom of choice than there is expressly about koro's fundamental case

admittedly, the set theory thing is interesting, but, haha

Well, isn't this one of the many things equivalent to AOC?

also some scribble scrabble about generalizations that goes on way longer than it needs to

it's not, it's weaker. it's definitely more than ZF though.

@leslietownes thanks Leslie :). Suppose that g obtained in the comment satisfies the 'norm' condition of extension. Still, there will be issue. The link you shared deals with it nicely: they extend the functional to span (YU{a}).

I understand this extension.

score one for the handout.

I think that to extend to all of X, we do need Zorn's lemma.

@leslietownes I also like the new word I see today - Zorning. I like the ring to that.

ask your doctor if Zorning is right for you

Oh, I was mistaken — again.

@Koro Be careful. Zorn means anger or wrath in German.

like Aguirre, der Zorn Gottes

Ohh

Zorn was a German mathematician.

Indeed.

Max Anger, or Max Wrath

what a name

would be a good name for a movie franchise

or at least a video game

ted: the rain picked up, so olivia is howling. just thought you and everyone else should know that she doesn't like the rain.

I'm supposed to go out to a potluck this evening ... in La Jolla. I'm wondering how bad it'll get.

Screech is not particularly scared of water. She gets in the sink and tub and drinks from the tap whenever possible.

Thunder and lightning I am sure is a different matter.

If every closed unit ball in a normed space X is compact, then X is finite dimensional.

It can be proven using Riesz theorem.

Based on the proof, it seems to be that the following proposition is also true: If every unit circle in normed space $X$ is compact, then X is finite dimensional.

we have basically no space between our ceiling and our roof, so even mild rain is really loud. squirrels on the roof will set her off.

What do you mean by "any unit circle"?

@leslie Is it a tin roof?

A hot tin roof, maybe?

$\{x\in X: \|x\|=1\}$

That is the unit sphere. What are you talking about?

Ohh there is only one such set.

@TedShifrin yes, I mean sphere. 'circle'.

A circle must lie in a $2$-dimensional subspace. Hence always compact.

Words matter in mathematics.

you're also not getting a lot out of saying 'all' open or closed balls in a normed space, koro. they're all translates and scaled versions of one another. hence textbooks usually just talk about the unit ball.

hmm, let's only take one unit ball.

normed vector space, i should say. which seems to be the context here.

Then I think that we can replace 'unit ball' by 'unit sphere'.

Yes, and ball and sphere must not be confused, even though calculus students do it all the time.

Question to ponder: Do "spherical coordinates" work in a normed vector space? I.e., if the unit sphere is $S$, is the unit ball homeomorphic to $[0,1]\times S/\sim$, where $(0,x)\sim (0,y)$ for all $x,y$?

Topologically, this is called the cone on $S$, for obvious reasons.

In fact, I think why we are even talking about closed unit balls or spheres. In the statement, we can just replace 'closed unit balls' by a compact set.

that is, if X is a normed space which has a compact subspace, then X is finite dimensional.

What do you mean by subspace?

You might want to reconsider this rash statement.

subset of X with subspace topology induced from norm on X.

When you're working with vector spaces, "subspace" is very ambiguous.

So you're totally wrong.

that is,I was wondering if there is any truth to the statemenet: if X is locally compact normed linear space, then X is finite dimensional.

@TedShifrin yes, I should have made that clear in my comment.

I can always take a compact subset of a finite-dimensional subspace.

What you just said is very different.

@TedShifrin of course. So the earlier statement was not correct.

Ayup.

I hope @robjohn's visit with Nurse Ratched went OK.

So now I rephrase the statement: Suppose that X is a normed linear space. If $Y\subset X$ is compact and contains a neighborhood of $0$ then X is finite dimensional.

hmm, this should be correct because the proof for 'closed unit balls' proceeds as follows:

Do we still know that a closed subspace of a compact space is compact?

So this just reduces to the previous result immediately.

yes, known from topology.

But your neighborhood of $0$ contains a closed ball centered at $0$.

You don't need a new proof.

yes, so suppose its radius is $r$. Now we normalize every vector to have a norm less than r.

The radius is irrelephant.

Multiplication by $r$ is surely a homeomorphism from the unit ball to the ball of radius $r$.

I guess by this point, you should concentrate on the substantial content.

yes, true so the unit closed ball is compact.

(given that some ball centered at 0 is compact)

Right.

@Koro and this also equivalent to being locally compact

Here's a nice question to think about related to this: suppose $K$ is a compact subset of a normed vector space. Must $K$ be contained in a finite dimensional subspace?

@Thorgott Oh nice! Actually, I learnt the proof for closed unit balls. I am now just trying to see what results I get by making some changes in the hypothesis.

this holds not only over $\mathbb{R}$, but over every complete normed field

i learned this claim in algebraic number theory lol

@TedShifrin I'll think about this one. I tried to construct one onto continuous map f from $[0,1]\times S$ to $B$ (B stands for the closed unit ball) such that $f^{-1}\{z\},z\in B$ gives the partition given by the equivalence relation. Then by one of the properties of quotient maps, it will follow that f is a quotient map and then by universal property for quotient maps, the desired homeomorphism will follow. But so far, I have not come up with such f.

I think it can be proven that: Suppose X is compact, Y is Hausdorff and $f:X\to Y$ is onto continuous, then f is a quotient map.

(because f is a closed map.)

@Thorgott I'll try thinking about such f.

thanks all for your help. :-)

@TedShifrin whenever you are around or feel up to it, give me a ping. I have a question about an elementary matrix you used to show all symmetric matrices can be written as $EDE^t$

@D.C.theIII I'm here now, leaving soon for lunch. What's up?

I'm trying to see how you got the matrix $E$ in this case

I got all the mechanical work and got the quadratic form, but I'm trying to generalize the idea

This was the last question in the chapter on 2nd derivative test

Right. The point was that you don't get $LDL^\top$ when the matrix $A$ is not nice.

I'm also curious as to why you did what you did to get the matrix $B$ he...

But we can turn it into a nice matrix with the correct row and column opertations right?

You need to do $PA$, where $P$ is an appropriate permutation matrix.

Yea i figured as much....I've been trying to find said matrix

I thought it could possibly be a combination of $E_1 L$ where $L$ came from doing the $LDL^t$ to my matrix $B$

it was "close-ish" but that just may be because the numbers are small

I haven't thought about this in forever, but the point is that $B$ can be written as $LDL^\top$, and so you get $A$ by using that.

Yeah, my original $P$ comment is wrong, because we need to preserve symmetry. So perhaps do $PAP^\top$, or perhaps not.

what's a quick and dirty lower bound for an nth degree complex polynomial

What are you talking about?

i meant, a quick and dirty lower bound for an nth degree complex polynomial's modulus

0?

i want to show i can make it bigger than the modulus of its constant term

That contradicts the Fundamental Theorem of Algebra.

@DC My permutation comment is not right.

why would it contradict the fundamental theorem of algebra?

Because the correct lower bound is $0$.

that's the greatest lower bound, no?

@TedShifrin Ok...that was actually how I was approaching things. I had envisioned something of the sort $ B = LDL^t = E_1AE^t$ and then bringing over $E_1$ it would result in: $E_1^{-1}LDL^t(E_1^t)^{-1}$

negative reals are also bounds for moduli because moduli are absolute values

no?

Yes, but $0$ is the tight lower bound.

What is the point of this?

But what's throwing me off is I still had to do a row operation on $B$ to get it into the form where I could rewrite it as $LDL^t$.....I'll go fiddle around. If I don't hit the discovery here I know the idea is going to come back around.

@DC To get $LDL^\top$, you definitely need a pivot in the $(1,1)$ slot. No, $B$ is already $LDL^\top$.

it's to prove the fundamental theorem of algebra. if i get a lower bound, i can show the polynomial's modulus becomes bigger than its constant's modulus, so i can make a neighborhood that has a complex root, and then use the min-max theorem for real numbers to show that the modulus function attains its minimum

You're proving the FTA by contradiction. So you assume by contradiction that the inf is $>0$, in fact the minimum attained (by a compactness argument) at some point $z_0$. Then you show that you can make it smaller yet on a neighborhood of $z_0$.

hm ok, thanks

Ah ok...actually one other matter, I see how it worked out nicely, but what was the reason for creating the matrix $B$ from the original matrix $A$? Was that $1/2$ operation equivalent to what is done with dividing by 2 when using the completing the square algorithm?

You can do lots of different versions of $B$, but you need a pivot there.

Try something different.

Ok will do

had my global pass interview this morning. once again they pointed out that i have little by way of fingerprints

can't find them on a timer if you don't have em to begin with

i always wipe the timers afterwards

You really need to stop filing your fingertips!

@copper.hat 😯😯😯...... really are a 00-agent

"99, I asked you not to tell me that!"

i remember a hawaii 50 episode in the 70's i think, where the culprit worked at a pinapple factory and as a sonsequence the fingerprinted were indistingusihable.

i was on the no fly list once

Pineapple erodes skin?

I was about to ask the same

according to that episode, yet

creative licesne perhaps

@TedShifrin I’m still here. Enjoying the rain. Last time it rained here I missed it.

I know it does soften meat tissues when you use it in a marinade.

the acid. i imagine any acidic fruit would do so...

that sounds like something the actual hawaii police would ask to be written into an episode so they could catch people better.

@robjohn Glad to hear it!!

but i use a special mix of hydrochloric and...

it also makes you invisible to surveillance cameras.

oops.

OK, lunch break for me.

enjoy!

what's on the menu?

@D.C.theIII my interactions with authorities are invariably edgy

you're here doing consulting work and you may right well have yourself a good novel lurking in your mind. ...the Irish Tom Clancy

@copper.hat boundary interactions

my novel would be titled "What can go wrong?"

@robjohn dunno. everytime i get my hands completely swabbed and my knees, crotch & tailend patted down. no, i do not enjoy it :-)

Comedic Spy Thriller starring the MSE crew........I think we can guess who will be the Evil Mastermind at the end...

i have had a few second level exams

while those that know me realise the internal peculiarites, i am not in any way an outlier by way of appearance.

i expected & understood when passing through heathrow during the troubles, but now i have no explanation...

beuracracy takes time to catch up, especially with how the UK is now.........

so the systems are still using the commodore data and processing

commodore? as in pet or 64?

64

black screen with green text.........

was a beauty

@copper.hat I have my PET in storage. I added a 48K RAM/prom programmer/ADC converter board from MTU. That gave me 32K of usable RAM and 8K of PROM storage for 3D graphics software I wrote. I did some nice 3D (monochrome) graphics with it.

The ADC converter had 5 channels. I transcribed some songs, including "Shine On You Crazy Diamond"

i liked the pet, it was simple enough that you could figure out how things worked. i tried to make an adc, but my resistors were not of high enough tolerance, so it did not work well well.

lol...I didn't know pet was a computer. I thought when he mentioned pet that there was some obscure animal that I had not heard of that was called a commodore..... isn't there a bird that is a commodore?

I ask all these things and I can just go search it up instead of exposing my stream of conciousness..........

@D.C.theIII Personal Electronic Transactor

sounds like the type of computer that would be created in the 70's....

it was aawesome. plus the basic.

I wrote a 6502 assembler for the PET in BASIC

cool!

That was how I wrote the graphics and audio routines.

nice

@ILikeMathematics Not so. You would need to have $f'''(x)\gt0$ for $a\lt x\lt b$

Just having those conditions at those points is not enough

@robjohn If we restrict $a$, $x_0$ and $b$ to be the only inflection points of $f$ on the interval [a, b], then it should hold though, right?

@ILikeMathematics By inflection points, you mean $f''(x)=0$, then I think so.

Thank you

@AlessandroCodenotti I think that this is false. Take a convergent sequence of points.

@ILikeMathematics i don't think so. suppose $f''(x) = x^2-1$, $a=-1,b=1$

pick $x_0$ so that $f'''(x_0)>0$.

i'm not exactly sure what you mean by concave up and down

i won't. i could but i won't. right over the plate, that one is.

i am missing something...

i am become woke

@copper.hat $x_0$ would have to be $-1$ or $1$, since $f''(x_0) = 0$, and since $a < x_0 < b$, there would be no $x_0$ in that case

Concave up means the derivative of $f$ is strictly monotonic increasing, concave down means the derivative of $f$ is strictly monotonic decreasing

The assumptions were $f’’(x)=0$ at $3$ points.

How's life