hey guys. I thought that you meant spectra by MO & MSE but now I see that it's not the case

yeah, it was MSU, not MSE. I really should've studied more instead of procrastinating

not that I do know the difference tho

the sad truth is that I wanted to ask something about irreducible polynomials

what's the generalized method for finding minimum distance between a point and a surface? like $P(x_0,x_1,..,x_n)$ to something of the form $\sum_{i=0}^{k}a_ix^{i}$

I don't understand the notation GF(p^m) for finite fields. but as far as I remember there's only one finite field for every prime power. if so then I understand

might have iterated that wrong but I just wanted something really general and abstract that would encompass a point to a point, point to a line, point to a plane, etc..

@Obliv project a point onto a surface, then take the coordinates of a point minus coordinates of a projection, and take a module of that vector

if you have no problem with any of that then it should work

@PetyaBalabanov what is a module? I thought that was the method as well, but in my class we did something else. For example, we had $z=\sqrt{1-2x-2y}$ with $P(-4,1,0)$ and we did $d = \sqrt{(x-x_0)+(y-y_0)+(z-z_0)}$ and did the discriminant test to find a relative minimum

oh i see why now lol..

@PetyaBalabanov G for Galois.

because it's the same form wow

Module you mean multiple

I was thinking it had to do with setting the inner product of the gradient and some line you make with the surface equal to 0 or something

@Obliv yep. the distance is just a length of a perpendicular to smth

oh yes @TedShifrin that should say multiple I see

norm?

Oh, modulus … yeah, length or norm.

Hard to decipher

agreed

@TedShifrin thanks, I was having trouble with understanding what the lecturer tried to say in his notes, but it was probably just an algorithm for constructing a finite field. some sophisticated stuff with polynomials

Yes, you construct them as quotient rings.

I was thinking about method of finding all irreducible polynomials over Z_p, p prime, of degree \leq n. for small p and n there's not so much polynomials, so I think if we'd be fine if we just take all products of smaller degrees and cross them out of all polys of degree n

yeah, that works

if you think about it, irreducibles are those who are not in that form. yeah.

Sieve of Eratosthenes, upgraded.

I'm so tired of trying to remember stuff for exams, it'd be so easier to finally understand smth)

exactly

Strive to understand and work examples/counterexamples

I forgot that. good analogy

yeah. doing smth with your hands is a good way to understand things

not only in math I guess

there's a formula that I tried to understand, regarding the number of irreducibles over Z_p

finite fields are cool but i know no more about them than what you get in a basic algebra sequence

And that was decade(s) ago.

so i guess i thought they were cool, but not cool enough

@leslietownes yeah, I have similar attitude towards them. just have to take the discrete math exam, and there they like finite fields for some reason, I dunno why. probably because of encryption or smth

you can do actual linear algebra (not just fake stuff with modules and crap), and also things like counting and induction arguments that can't work in the infinite setting, what's not to like

cryptography is certainly an interesting application, and probably good for using in grant applications even if you aren't serious about it

Strange

How do you get 2 upvotes in 46 seconds...

i grew up in Ireland where all fields are finite.

@leslietownes Also round-robin tournament scheduling!

@copper.hat And brown.

@TedShifrin If it rains too much, yes, muddy fields too...

Grassing at straws.

I will be passing through Dublin for 4h in less than a week.

Exciting!

i have two cousins in Ireland. never contacted them yet tho

I have a cousin I have never met. Long & complicated story.

i have 3 cousins i have never met! i win!

I haven’t seen cousins in 50 years. There are certainly children and children of children, so I’m missing uncountably many.

Why For compact surfaces $E$ does, of course, contain all the exact differentials?

What the hell is this notation?

It's from farkas kra riemann surfaces

That is not helpful.

You have to make a question self-contained.

So what are $E$ and $E^*$?

Can anyone please help me plot the graph y=logx -x+1, x>0 analytically?

I could conclude that f(x) is increasing for x\in (0,1) and decreasing when x>1

I need to sketch this function roughly

But I can't seem to do it!

Why not? What about concavity?

yeah, you've got a limits as x goes to 0+ and as x goes to +oo to compute, and the value of the function at 1, and some concavity, that should be enough for an OK picture.

these pictures are generally only as accurate as what you specifically intend them to reflect, but thats OK.

If you’re thinking you have a linear asymptote, that will lead to a problem.

that's a good point. it wouldn't have occurred to me, but sometimes stuff people are expected to do with e.g. rational functions might suggest that it's generally possible to "draw" finer asymptotics for general functions more easily than you can

@TedShifrin I'll upload the question on MSE later on.

@TedShifrin , @leslietownes well f''(x) is always negative i.e 1/(-x^2) due to which it is everywhere concave downwards

Do you know what $E$ is?

Right, Franklin. So why can’t you graph it?

@TedShifrin I dont have an answer to that 😔. But, I just dont know where the function is positive and negative. Isn't that necessary to graph it as well ?

You know $x=1$ is its global max.

you should be able to infer that from what you've done and what f(1) is.

admittedly, if you haven't computed f(1) and the derivative stuff, it might not be clear why you wouldn't also maybe do a separate analysis of x-intercepts and signs and stuff.

but this does not explain the inertia. you can move forward

Let’s have just a moment of inertia!

@TedShifrin It's the $L^2(M)$ closure of $\{df:f\in C^\infty_c(M)\}$ where $M$ is a Rieamm surface and $L^2(M)$ is the hilbert space of measurable square integrable 1-forms.

this is kind of a fortuitous problem. if it was log(x) - x + 2, you'd maybe like to do more, and maybe struggle algebraically with finding x-intercepts and intervals of positive/negative.

but the problem you have doesn't present that difficulty.

So what is true if $M$ is compact? Isn’t everybody $L^2$?

everybody in L^2? sounds like communism to me.

@leslietownes Or give up on said struggle.

i've mentioned this before, but i ran into this all the time TAing at berkeley. less so at iowa. the question of, 'what if the problem was slightly different in a way that made it more complicated to approach, what then,' where the answer was just, 'but it isn't different in that way, and if it somehow were, which it isn't, you just wouldn't be able to do more.'

easily one question a week like that, every time in TAing calculus.

Sorta like DC trying to invent his own modifications of my exercises.

at iowa maybe i was just better at presenting the material in a way that didn't make people worry about that stuff. or they trusted me more when i said i wouldn't ask questions like that.

Not a bad thing in principle …

In linear algebra, a minor trivial alteration usually causes computational chaos.

i'd usually say something at the beginning of the semester like, in this class, polynomials of degree 3 or more are going to have at least one rational root, and it's probably going to be an integer. and we're not going to detour from basic exercises in whatever we're trying to do, to having fractions with denominators like 59 and 137 in them.

and sometimes that got through and sometimes it didn't. maybe saying it only made people think about it, and thinking about it made it worse.

@TedShifrin Yes, 1 is a point of maximum

What is $f(1)$?

@TedShifrin f(1)=0

So doesn’t this answer what you said you were stuck on?

Ohh, so the complete function isnegative

@TedShifrin just asec...

I think I got it...

well it's not negative at 1, but the sign analysis on either side of 1 is as you describe. :)

And there are no inflection points.

Um …

log is "ln" here. has to be.

this is a software thing.

@TedShifrin Ah ok.

But still log (1)=0

@leslietownes you literally saved my day!

Typing ln fixes it

Yes, I have been able to complete it.

If you use s graphing calc, you don’t need any thinking!

if you plot the function on a device that things "log" is base-10 log, yes, you will see another mystery root between 0 and 1 - and 1 is no longer a max, which is a red flag conflicting with your analysis that it was.

your computer is a witch and you should probably burn it

@TedShifrin no no...that was just to validate my rough sketch

Ofc using graphing calculator, no analysis has to be there!

One thing, that still bothers me that how can we conclude, x=1 is the only point of extremum.

You proved it.

@TedShifrin ohh, it's because, f"(x) is always negative

how many points have $f'(x) = 0$?

@Franklin um ?

@copper.hat Only 1 i.e x=1

that answers your last question then

@TedShifrin that's the reason that no other points of extremum exists....

Oh, that does do it. But look at what you told us at the very beginning.

What?!?

well, not completely, you also have $f(x) \to -\infty$ as $x \downarrow 0$ and $x \to \infty$.

hence there is at least one extremum.

You said increasing to 1, decreasing after 1.

This snap consists of the definitions, I use

haha that typesetting. who set that type, gutenberg?

i'm surprised there's a screenshot of that book, it should be chained to a bookcase in a monastery somewhere

i love the typem

No cavemen?

@TedShifrin yes , certainly that's true. That 's determined by 1st derivative

caveman jokes are reserved for your education, ted (which predated books and maybe paper)

well, personally i use ogham stones

So can there be any more critical points?

@TedShifrin No. And that's just the reason to be more specific, right?

(Reason for only one point of extremum)

f' f' f' f' f' f' set to some europop

The domain is x>0

@copper.hat what's that supposed to mean? 😕

You asked why no other extrema.

@TedShifrin Yes! And since there are no critical points except x=1, so ....

That's the complete reason

If I were to say, more accurately !!!

at an extremum you must have $f'(x) = 0$. Since there is exactly one such point there can be at most one extremum. Then, since $f$ 'goes' to $-\infty$ at the

@TedShifrin Thanks a lot! I do get it now!

edges' of the domain, it muct have a $\max$ somewhere. Hence there is at least one extremum. Now mix together and add some whiskEy

@copper.hat just a little addition f'(x) might not exist as well

Since a point of extremum is a critical point.

it is smooth on the domain. unless you are getting into nodifferentiable analysis...

its past my bedtime, i am getting punchy

@copper.hat that's true! Can't agree with you more !

@copper.hat good night!

@Franklin Good night & good luck!

Have sweet dreams !😋

@leslietownes this gave me a good laugh 😂😂😂

That's really an old book by IA Maron , a Russian

Better say a Soviet citizen

The book is Problems in Calculus of One Variable

But The point is: I got this graph now with a fair understanding or rather a pretty good understanding!

Thanks!

I was looking at this nice problem :math.stackexchange.com/questions/132621/…, there the answerer uses an inequality $f(t)-f(1)=\int_1^t {1\over x^2+f^2(x)}\,dx\le \int_1^\infty{1\over 1+x^2}$. I have no idea about how is this inequality true ? Can anyone please help me with this ?

the expression for f'(x) shows that f'(x) is positive for all x. since f(1) = 1 and f'(x) is positive for all x, f(x)^2 is at least 1^2 for all x >= 1, so 1/(x^2 + f(x)^2) is at most 1/(x^2 + 1) for such x

Like, of course $\frac{1}{x^2+f^2(x)}\leq \frac {1}{x^2+1} $, but what bothers me is how did they write

the upper and lower limits

Of the integration

It was 1 to t and then they changed it to 1 to infty

would it help if you mentally inserted a "<= integral from 1 to t 1/(x^2 + 1) dx <" before the integral from 1 to infinity

the integrand 1/(1 + x^2) is positive too, so taking the integral out to infinity only makes it larger

int 1...t g(x) dx < int 1..infty g(x) dx holds for any positive g(x)

they're presumably going out to infinity just so they get an upper bound that would hold for all t without any further detailed analysis

@leslietownes can we do that always? I mean to say, this is the first time I have encountered integration and inequalities on same page ! Let's say, f(x)\leq g(x) and they are both differentiable in a common domain D=[a,b], then can we always say $\int_a ^b f(x)\leq \int_a^b g(x)$ ? Also can we write, that $\int_a ^b f(x)\leq \int_a^t g(x)$ , if $g(x)$ is also differentiable in D'=[a,t] such that t\geq b(considering the validity of my previous statement)?

Are these statements that I asserted valid in general ? 🤔

you want \leq in both of those second integral displays, but yes

@leslietownes Yes, I have fixed the typo. Is this valid in all sense ?

in the second one you need to assume that g(x) >= 0 or something on [b,t] (try to find the precise statement)

you can "apply the same definite integral to an inequality of functions", that's one always true thing

the second thing you're asking about is under what circumstances adding integral from b to t of g(x) dx will change the intequality you get of the integrals from a to b of both functions

@leslietownes Actually, I haven't learnt all these things in a real analysis course, or more precisely, in a rigorous manner. So the point is: if we know in the second case, g(x)\geq 0 in D', we can assert it , right?

if g(x) >= 0 on [b,t] then it's definitely not going to change it

you've learned them, but maybe they aren't emphasized as such

early in riemann integration, yoou might learn the definition in terms of riemann sums, which makes clear that if h(x) >= 0 on [a,b] then the integral from a to b of h(x) dx is >= 0

@leslietownes Maybe

(roughly speaking, because adding nonnegative numbers together gives you a nonnegative number)

@leslietownes that's the intuition played in my mind

so if h(x) = g(x) - f(x) this tells you that if f(x) <= g(x) on [a,b] then integral from a to b of [g(x) - f(x)] dx >= 0

and if you also learned the additivity of the integral [i.e. that inegral (h-k) = integral h - integral k)

then it's just what we were talking about above

@leslietownes Yes, exactly, if $h(x)=g(x)-f(x)\geq 0$ in a domain D, then, if, g(x),f(x) are differentiable on D as well, then, we can say, $\int_a^b g(x)\geq \int _a^b f(x)$

I think this is the general statement

Now, if we consider a domain $D'=[a,t]$ such that g(x) is differentiable upon D' as well , then we must check whether $\int _a^b f(x)\geq \int _a^b g(x) +\int _b^t g(x)$. If this is true, then, $\int _a^b f(x)\geq \int _a^t g(x)$ is obviously true.

@leslietownes I think this is the general idea

yes

hey guys. how's it goin

i guess this stuff is maybe so close to the definitions involved in integration that maybe some books wouldn't separate it out and state it as a theorem

or just bits and pieces of it (roughly, additivity and positivity of the integral)

Is it true that Galois theory has the following fact? x^(p^n) - 1 is a product of all irreducible polynomials of degrees that divide n. I'm not sure if I recall details correctly

I was told that it's somewhere in the beggining of Galois theory

i think x^{p^n} - x? the polynomial having all elements of GF(p^n) as roots.

@leslietownes thanks a lot!

something like that is indeed true and probably in at least most books on galois theory that cover finite fields. some treatments (e.g. chapters in 'abstract algebra' books) might not.

@leslietownes that's interesting, thanks, I'll think about that

i don't find it expressly stated in an answer on MSE, but the comments in math.stackexchange.com/questions/288120/… cover some of it

@Obliv Please! I will be obliged :)

(sorry for replying so late)

Determine $d$ in such way that $f(x) = -\sin(x) + dx - 1$ has saddle points.

We can find all inflection points and pick the ones that have a slope of zero out of them.

So $f''(x) = 0$, and then check the $x$ we obtain with $f'''$, if $f'''$ is $0$ at some $x$, then check if $f''$ changes signs at that $x$.

The solution to this seems to omit the step where you check your obtained solutions from $f''$, it right away assumes that all $x$ obtained from $f''(x) = 0$ are inflection points.

That's "cheating" a bit, right?

@ILikeMathematics What is your definition of a "saddle point"? In my world, it doesn't make sense to talk about the "saddle points" of a function of only one variable.

probably it's an inflection point

@SineoftheTime I can make all kinds of assumptions about what @ILikeMathematics means, but until they answer... *shrugs*

@Silent , like I said, I don't know how to find that solution I just used wolfram alpha to compute its solution to be $y(x) = i c_2\sinh(x^2) + c_1 \cosh(x^2) - \sin(x^2)$

Yes, it's an inflection point, with slope $0$

but I think if you ask again today someone could help you.

@ILikeMathematics What is your definition of an inflection point? Presumably, it is a point $c$ where the second derivative changes sign, e.g. there are $a < c < b$ such that $f''(x) < 0$ for $x \in (a,c)$ and $f''(x) > 0$ for $x \in (c, b)$?

With the additional assumption that $f'(c) = 0$?

(removed)

@Obliv oh! alright :) thanks

@XanderHenderson Yeah, a point $x$ where the second derivative changes signs and $f'(x) = 0$.

The second derivative is $\sin(x)$. Every zero of $\sin(x)$ corresponds to a change in sign, so the inflection points are those zeros, i.e. $\{k\pi : k \in \mathbb{Z}\}$. The first derivative is $-\cos(x) + d$, which is going to be $\pm 1 + d$ at $k\pi$, depending on the parity of $k$. Thus it is possible to get inflection points when $d=1$ (these will occur at odd multiplies of $\pi$), and if $d=-1$ (these will occur at even multiplies of $\pi$).

I don't see any reason to look at the third derivative.

@XanderHenderson Oh, if you say that the zeros of sine right away imply a change of the sign, then yeah, there shouldn't be a reason to look at the third derivative. Thanks

@ILikeMathematics I mean, you know what the sine function looks like, right?

Yes, generally, you would have to check the third derivative, but since we are dealing with trigonometric functions, we can use that to our advantage

I suppose that you could look at the third derivative, but it is much easier to know that $\sin(x) < 0$ for $x \in (-\pi, 0)$, that $\sin(x) > 0$ for $x \in (0,\pi)$, and that it is $2\pi$-periodic.

This is all stuff which can be "easily" derived directly from the definition of $\sin$ with respect to the unit circle, hence there is no need to use heavier hammers.

Yes, thank you

quick question on diff geometry

regularity of a curve depends on the parametrization of the curve? for example $(cost,sint)$ has always non zero derivative so is regular. but $(cos(t^2),sin(t^2))$ is not regular and both curves are a circle.

On the other hand if a parametrization has non zero derivative then every reparametrization of that curve has also non zero derivative

@Jam You just gave a counterexample to that!

ok wasnt sure cause ive seen a wrong answer

that regularity does not depend on the parametrization

it could be that some source effectively made that statement true by adopting it into the definition of 'parametrization.' i could imagine some source doing that, and also some readers not noticing that difference between sources.

LESLIE

@leslie makes a good point. Check the definitions in your source(s).

Some texts may require reparametrization by a diffeomorphic change of coordinates, rather than just a smooth one.

hi anak

@TedShifrin are we supposed to be able to compute all integrals by hand without the assistance of technology? Reason being I was computing $\int_{0}^{1} \int_{x^2}^{x} \frac{x}{1+y^2}dydx = \int_{0}^{1}x [tan^{-1}(x) - tan^{-1}(x^2)] dx$.......only way I figured out the integral was through mathematica..

the real question is do you want technology to domineer over you, to hold you in the palm of its hands and do whatever it wants of you, like a silly little marionette

I am VERY averse to that actually Shin and purposely go out of my way to do things that would be easier with tech and do it without using it.

Sometimes to my detriment

@D.C.theIII You’re obviously missing the main point.

switch the limits?

of integration

Although this can be done by straightforward int by parts. Yes. Fubini.

Switch the order…

had a feeling I should've resorted to that once I got stuck

actually this question asked me to integrate it both ways

So you purposefully had us do the hard way first, to "appreciate" Fubini.

Oh no! Someone got some Toneli in my Fubini!

dc3 i think generally no, nobody is supposed to be able to compute all integrals by hand, but if that is specific to a textbook and specifically ted's textbook, the definite rule is that if you set up a textbook exercise one way and it seems impossible, the textbook might be trying to tell you something.

I usually follow that logic myself, but this exercise explicitly said compute the integral both directions...

or to put it another way, while various theorems will guarantee that any number of different setups of a multivariable integral as an iterated integral will be equal to one another, it's definitely not true that all of the various setups will be equal or even approximately equal in terms of difficulty of computation by hand.

ted's just banging you over the head with his point, like he always does.

@D.C.theIII If you are asked to integrate a random function, it is nigh certain that you are not going to find an antiderivative in terms of elementary functions, and that you will not be able to do anything more than come up with a good numerical approximation scheme.

But that isn't what textbook problems are. Textbook problems are, if done right, made to emphasize the concepts taught in the preceding section or sections.

And those problems are meant to be done by hand.

Oh when I got to my sticking point Xander it was made clear...hence the frustration

where students get the idea that their profs are little tricksters trying to fool and punish them at every turn, i'll never know.

I'm just saying that there is a psychological / pedagogical approach to answering questions. :D

Hi 🥱

it's a bit like how sometimes a multiple-choice question will be premised in part on awareness of the multiple-choice format. "which of the following is a root of [4th degree polynomial with 2 as a root]. A) 1 B) 2 C) -1 D) 4" is not an exercise that requires the quartic formula

@leslietownes the answer is B :)

@D.C.theIII Perhaps you should reread.

@TedShifrin got any ideas for how to address this question?

not that it's not posed very well atm (too vague)

@leslietownes reminds me of problems on the Physics GRE. you -could- compute the electric field of a line charge by hand...or you could realize that every other option given had no hope of being right. (wrong units, incorrect limiting behavior, etc)

@TedShifrin "Evaluate each of the folowing iterated integrals. In addition, interpret each as a double integral $\int_{\Omega}f dA$, sketch the region $\Omega$, change the order of integration, and evaluate the alternative iterated integral."

Oh, yeah, I missed the beginning. Anyhow, yes, integration by parts is expected.

semi: heh, i remember a problem or two on the math GRE like that. some counting problem where, yes in principle you could set it up and find the answer, or you could compare the asymptotic behavior or even behavior for n = 2, 3, 4 in the given options.

@leslietownes Oh, man! The math subject GRE! I decided that I was going to apply to a PhD program, realized that I needed to take the math subject GRE, and took it a week later. I was super prepared!

Weirdly enough, I did really well on every type of problem except for the stupid integrals. It had been a decade since I had looked at integration techniques.

when i took it the consensus was that the exam format was busted. something close to 20% (no exaggeration) of takers got the top score. and yeah, i think even then they used multivar integrals to distinguish between people.

they later rewrote or rebalanced it somehow, but what are you going to do if the syllabus for advanced material is basically "uh, everything that somehow makes it into algebra and analysis classes at every university" so... chapter 1 of three or four books?

A group is a set with a ____ operation on it making [definition of a group]. A) binary B) unary C) trivalent D) quadruped

TRIVALENT!

You see, a "quadruped" is a monster with four legs, "binary" means either male or female, and "unary" is what a baby wears. By elimination, it must be (C).

there probably is some formalization of the group concept using a single ternary operation. someone without english as a native language may have called it trivalent. file an appeal, citing that line of important work, in translation if necessary.

Unary:

OMG...found my mistake.....when I multiplied the square roots I forgot to remove the square root....very much slap deserved.

Would you say "-3^2" is 9 or -9?

It is $-9$.

Cheers