Mathematics - 2022-08-21

What about though: $\{ U_a : a \in \Bbb{Z}\}, U_a = \{ an^k - 1: n \in \Bbb{Z} \}$? It should form a topo basis, and $f(n) = an^k - 1$ should be continuous (or something).

Ted Shifrin

We are not listening.

Prime gap homology stoner mon.

Do you mean reading?

Ted Shifrin

ears covered

copper.hat

3:05 AM

preparation g

Prime gap homology stoner mon.

I just proved $\Bbb{P}_{\text{odd}} - \Bbb{P}_{\text{odd}} = 2\Bbb{Z}$. where $\Bbb{P}$ are the prime numbers and $-$ is elementwise.

Iow, the set of differences of odd primes covers all even integers.

It's an immedate consequence of Helfgott's finiteness on Ternary, but could be had at even larger than ternary sums of primes, just any finite sums $q_1 + \dots+ q_n$ covering (almost all of) $2\Bbb{N}$ or $2 \Bbb{N} + 1$

will do

copper.hat

3:38 AM

I'm searching for the second even prime.

Jai Sri Krishna

4:00 AM

Is calculus by James Stewart good?

Jai Sri Krishna

4:13 AM

@leslietownes

copper.hat

i think leslie is out swimming at the moment.

Ted Shifrin

5:09 AM

@JaiSriKrishna It’s standard. Good for what? Good for whom?

user4539917

Not a "gold" standard, Dr. @TedShifrin?

/jk

robjohn

@copper.hat $-\frac1{12}$

Positive integers are closed under addition and that is $1+2+3+4+\cdots$. It is not divisible by another integer.

user4539917

5:27 AM

Hi pal.

leslie townes

i used stewart. it is a fine, standard-issue calculus book.

@copper.hat today was my day off swimming.

first day in many days.

grandad did not get the day off.

he's swimming still.

copper.hat

6:12 AM

@robjohn At some point I did come across an April's fool joke about the even primes.

@leslietownes he's just sampling the water quality

Jai Sri Krishna

6:38 AM

@leslietownes Thank you It's good

Koro

I want to eat samosa but it's not available here.

Jai Sri Krishna

@TedShifrin Thank you I wanna learn calculus so I asked!

@Koro Even I wanna eat it 😥😥

Koro

yeah, it tastes so good with a good sauce while it is medium hot. Add some green chilly and some ginger tea, it tastes even better.

Jai Sri Krishna

😋😋

Find an expression for the area under the graph of f as a limit

f(x)=x^1/4 1<=x<=16

Is the answer Lim n-->oo sigma x=1 to 16 (15/n)x^1/4

Koro

6:58 AM

do you know how the graph of f looks like?

Jai Sri Krishna

Yeah

user4539917

Making a sketch may help

Jai Sri Krishna

Yeah

So isn't my answer right?

Since in the integral a to b f(x)dx, we could replace x with any variable we call it the dummy variable?

Koro

@JaiSriKrishna it doesn't look right. What is sigma x=1 to 16?

Jai Sri Krishna

oh

user4539917

7:12 AM

Where did you get the 15/n factor?

Koro

@JaiSriKrishna you have to be careful in definite integrals. Upper/lower limits may change when you substitute something for x.

user4539917

Label everything you do on your sketch.

Koro

@JaiSriKrishna so try approximating the area using 'small' rectangles.

Jai Sri Krishna

@Koro Yeah I am trying it

user4539917

Take your time :-)

Jai Sri Krishna

7:16 AM

Btw what if the point ( left end point ) is not in the domain of the function?

For computing area using left end point

Koro

@JaiSriKrishna you mean computing area on (a,b], a<b? In such a case, you try to find integral from a+h to b, h>0 and then take limit $h\to 0$.

In such a case, if the function f is bounded on [a,b] and continuous on (a,b], then the limiting value of the integral $\int_{a+h}^b f (x) dx $ as h tends to 0 gives you $\int_a^b f(x) dx$

Jai Sri Krishna

7:31 AM

Oh

So I should take the limit of f(a+h) as h tends to 0?

Koro

no. can you see math formulas here?

If not, then get the bookmark from LATEX in chat link available in this room's description.

Jai Sri Krishna

No I can't see the formulas correctly It's in mathjax format

Oh yeah I can see the formulas clearly now

user4539917

coolio

Koro

:)

user4539917

but, for now, rely more on your sketch and the labels

Jai Sri Krishna

7:44 AM

What if after the computation of integral it is F(x) what if 'a' is not present in the domain of F(x)

:)

yeah sure

user4539917

Arranging the given information in a chart may be helpful.

Jai Sri Krishna

Integral $(\int_{a}^{b}f(x)\,dx)$ = F(x) what if a isn't defined in the domain of F(x)

user4539917

Separate the different cases in your chart.

Each with a sketch

copper.hat

why do authors write 'homotopic to zero' when they really mean 'homotopic to a point'?

Jai Sri Krishna

Is David Burton's Number theory a standard book?

robjohn

8:09 AM

@copper.hat zero is a point, is it not?

if you make an algebra of curves, what would be the zero curve?

stabbing in the dark, here.

Koro

you mean sigma algebra?

or algebra in measure theory?

:)

Astyx

@robjohn You've got a point there

Curious Mind

8:27 AM

N

robjohn

N?

Curious Mind

9:39 AM

Math is evil. You have to spend so much time feeling stupid.

When I spend time cooking food, I get the food to eat, but when I spend time learning math, I get nothing.

user4539917

You're supposed to get food for thought, ie
something to think about.

Feynman_00

But when you finally understand the true beauty of what you are learning, you are overwhelmed with wonder and feel at peace with yourself

I mean, studying Math and struggling through it is surely frustrating but eventually it gives you so much it's worth it

user4539917

Indeed, the mental nourishment helps you grow!

Feynman_00

9:55 AM

Regarding the "feeling stupid" I guess we can't help it, I feel that all the time too. Nonetheless, remember that the mathematicians who proved the theorems you read on your books didn't just wake up and write down the proof

Jai Sri Krishna

Is Indefinite Integration "defined as" the inverse process of differentiation

Feynman_00

Let $f:I\rightarrow\mathbb{R}$ The indefinite integral of $f$ is defined as the set of all functions $F:I\rightarrow\mathbb{R}$ such that $F'(t)=f(t)\quad\forall t\in I$.

The indefinite integral is a family of functions

Jai Sri Krishna

Yeah

So is the function defined as inverse of derivatives?

Also the author mentions $\int f(x) dx $ is really infinite valued

what does that mean?

Feynman_00

10:22 AM

@JaiSriKrishna If you have a differentiable function $g$, then $g\in\int g'(x)dx$ i.e. $g$ belongs to the set of the antiderivatives of $g'$, but there are infinitely many. I guess "infinite value" is a crude way to say this

It turns out that all the elements of such set are of the form $g(x)+c\qquad c\in\mathbb{R}$

Jai Sri Krishna

Oh Yes @Feynman_00

I get it

porridgemathematics

can someone help me understand why $D$ needs to be integral over $A$ in this proof?

i dont see where that assumption is being used

'Lemma 7.14' here refers to 'If $A$ is a subring of $B$, and $I$ is an ideal of $A$, then the integral closure of $I$ in $B$ is $\overline{I} = \sqrt{I \overline{A}}$

Thorgott

11:45 AM

what's the bar notation? wouldn't make sense to mean integral closure

porridgemathematics

the way its stated verbatim is 'the integral closure of $I$ in $B$ is $\sqrt{ I \overline{A}}$

where $\overline{A}$ is the integral closure of $A$ in $B$

Thorgott

oh, I misread a part

yeah, you don't need to assume $D$ to be integral over $A$

it's redundant since you assume $x\in D$ to be integral anyway

and the argument does not change if you replace $D$ by $A[x]$ (by which I mean the subring generated by $A$ and $x$ in $D$, not the polynomial ring over $A$), which is integral over $A$ if $x$ is

porridgemathematics

12:03 PM

I see, thanks!

also in the proof, am I right in thinking they tacitly work in some field extension of $A[x]$ where $x$'s minimal poly splits?

(that may be larger than $D$)

Thorgott

oh yeah, for sure

didn't even register for me cause I always implicitly work inside an algebraic closure

porridgemathematics

yes of course, I just paused a bit because up till this point they've been pretty painstaking in stating every step explicitly (its an introductory commalg course)

come to think of it though this is par for the course at least in my experience, 12 week short courses tend to go really slow for the first 75% of the course, and then at least 3x faster in the last 25

Thorgott

12:45 PM

I don't really like integrality as a topic in a comm alg course, but perhaps that's just cause I'm not super fond of the topic to begin with

it's pretty useless unless you get into algebraic number theory

porridgemathematics

oh i see, im going through these notes because im taking an algebraic geometry course when semester starts

so maybe I should skip some of this stuff in interest of time,

ive been spending a while on these 'going up/down' results

are they needed for AG?

(the course i will be taking tries to cover hartshorne ch1 and 2.1-2.5)

and the notes im using roughly follow eisenbuds book

HelpMeToUnderstandContours

1:55 PM

Hii, my friend is preparing for JEE and he showed me 3 short-cut formulas for finding second order derivatives.

Here are the formulas.
$$\dfrac{d^2y}{dx^2} = \dfrac{\left(\frac{dx}{dt}\right)(\frac{d^2y}{dt^2}) - \left(\frac{dy}{dt}\right)(\frac{d^2x}{dt^2})}{\left(\frac{dx}{dt}\right)^3}$$

$$\dfrac{d^2x}{dy^2} = \dfrac{-\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^3}$$

$$\dfrac{d^2y}{dx^2} = \dfrac{\frac{dx}{dt}\cdot \frac{d^2y}{dt^2} - \frac{dy}{dt}\cdot \frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^3}$$

Some application questions are,

1.) If $y = t^2 + t$ and $x = t + \frac1t$, then find $\dfrac{d^2y}{dx^2}\bigg|_{t=3}$.

2.) If $\cos^{-1}(\frac y2) = \ln(\frac x5)^5, |y| < 2$ then which of the following equation is true?

(i) $x^2y'' + xy' - 25y = 0$
(ii) $x^2y'' - xy' - 25y = 0$
(iii) $x^2y'' - xy' + 25y = 0$
(iv) $x^2y'' + xy' + 25y = 0$

3.) If $y = \frac{2x^2 + 3}{5x + 4}$, then find $\frac{d^2y}{dx^2}$.

Koro

2:28 PM

@HelpMeToUnderstandContours Let dy/dx=p. Then, $\frac{d^2 y}{dx^2}=\frac{dp}{dx}=\frac {d}{dx}(\frac 1{\frac 1p})$

Please ignore that later half of previous message. $\frac {dp}{dx}=\frac 1{dx/dp}$

Koro

2:42 PM

Using $p$ is obscuring what's going on. $\frac {d^2y}{dx^2}=\frac d{dx}(\frac 1{dx/dy})=-(dx/dy)^{-2}\frac d{dx}(dx/dy)$

Use Chain rule again to write: $\frac d{dx} ()=\frac d{dy}() dy/dx$.

Thorgott

3:17 PM

all of this reads like borderline gibberish to me

stuff like $dy/dx$ is nonsense in general, let alone thinking about analogous higher derivatives

HelpMeToUnderstandContours

@Thorgott Yes, same here :(

monoidaltransform

3:37 PM

closed and bounded subsets of the isometry group of $\mathbb{R}^n$ are compact?

leslie townes

probably, but what do you mean by 'bounded'? not saying that this can't be given a meaning. just that it might not come with one.

porridgemathematics

@monoidaltransform yes, if you mean w.r.t the compact-open topology

the proof is basically use arzela-ascoli on a compact exhaustion

and then extract a subsequence by taking the diagonal, which gives you sequential compactness

since the compact-open topology is metrized by the usual metric related to local uniform convergence, a subset is compact iff its sequentially compact

monoidaltransform

@porridgemathematics whats the metric on the isometry group of a metric space $X$ ?

porridgemathematics

i thought we already discussed what it is before?

monoidaltransform

yes yes, but I can't seem to find it

porridgemathematics

3:48 PM

(assuming $X$ is hemicompact)

monoidaltransform

assuming $X$ hemicompact

leslie townes

you can think of isometries on R^n as pairs (a,b) where a is an element of R^n and b is an element of O(n), so these things sit inside R^(n + n^2), which you can metric-ify. that's one family of ways to get a metric on there.

i guess that doesn't work for a more general metric space.

Thorgott

leslie is absolutely right

porridgemathematics

yeah but you will need to use something about the structure of the isometry group to get equicontinuity anyway

so your method is more straightforward

Thorgott

the isometry group of $\mathbb{R}^n$ is $\mathbb{R}^n\rtimes O(n)$

so this property is an immediate consequence of good old Heine-Borel

porridgemathematics

3:52 PM

@monoidaltransform anyway, assuming $X$ is compact, just set $\rho(f,g) = \sup_{x \in X} d(f(x),g(x))$, for hemicompact take an increasing exhausting sequence, and consider $\rho_n = \frac{1}{2^n} \frac{\rho(f,g)}{1 + \rho(f,g)}$ on each $X_n \uparrow X$

then take the sum

of course this coincides with the straightforward approach to the question, just identify the group with something that is clearly a proper metric space

(you get the same topology)

uh but actually this is true for a general hemicompact metric space (and its isometry group)

because equicontinuity is immediate $d(f(x),f(y)) = d(x,y)$

that plus uniform boundedness on compact subsets will give you sequential compactness

(you get uniform boundedness on compact subsets if your set is bounded in this metric)

copper.hat

@robjohn it certainly is an identity in the right space.

one potato two potato

5:33 PM

recent years our math department has moved very toward the applied side

leslie townes

boooo

porridgemathematics

7:01 PM

in the last sentence of this, shouldnt it say $A$ is complete if every cauchy sequence has a unique limit?

going by their definition of complete, where they define it to mean the map $A \rightarrow \hat{A}$ is a (ring) isomorphism

since it seems like every cauchy sequence having a limit would only give surjectivity

and unique would give injectivity

Thorgott

what does Cauchy even mean here?

porridgemathematics

i took it to mean how its defined for a topological vector space, so $\{x_i \}$ is cauchy if for any nhood of $0$ of $V$, there is some $N$ s.t. $n,m \geq N$ implies $x_n - x_m \in V$

(the note itself doesnt elaborate)

Thorgott

ok, that makes sense

I agree with you

injectivity is equivalent to $\bigcap\mathfrak{a}^n=0$, which is a nontrivial condition

porridgemathematics

7:18 PM

ah right, that makes sense

so in fact injectivity is equivalent to $A$ being hausdorff too

Thorgott

yeah

one of my favorite examples of a case where this condition is not met is the ring $A=C^{\infty}(\mathbb{R})$ with the ideal $\mathfrak{a}=(x)$

if you only take germs instead, this is an example of a non-noetherian local ring with f.g. maximal ideal

porridgemathematics

ah yeah, thats neat

is non-noetherian here being seen via krull intersection theorem?

(and the $\exp(-\frac{1}{x^2})$ thing)

Thorgott

you can quote Krull intersection to get it immediately, but there also is a fun explicit argument

the functions in $O(\exp(-1/nx^2))$ form a proper infinite increasing chain of ideals

a funny related fact is that if you consider germs of continuous functions at $0$, they still form a local ring, but the maximal ideal is not f.g. anymore

anak

Hey Thorgott

and mr. porridge

porridgemathematics

nice, I think I get the argument

hi :)

robjohn

7:41 PM

@austere1993 I've added quite a bit of explanation to my answer.

Ted Shifrin

8:07 PM

howdy @robjohn

anak

@TedShifrin this is a calculus of variations computation to compute geodesic equations. The integral over M is just part of the metric. I can tell the second line is just the product rule, but I can't figure out what is going on to get the last line. It doesn't seem to be the adjoint connection afaik

Ted Shifrin

It doesn't make much sense to me. I assume there's an $\varepsilon$ parameter in $q(t,x)$?

anak

Yeah they are viewing $q$ as a 1-parameter variation of a smooth path.

Ted Shifrin

We need to know how $\partial_\varepsilon$ and $\partial_t$ are related, clearly.

anak

The $\varepsilon$ is the variation, and the $t$ is the time of the path.

Ted Shifrin

8:13 PM

Oh, this looks like some of the standard stuff with first and second variation that's in DoCarmo, etc.

anak

Yeah, I just don't know what could possibly happening to get the very last line.

Just purely to change the integrand. I am effectively ignoring everything outside it.

Ted Shifrin

We're swapping $t$ and $\varepsilon$ derivatives.

anak

Is that a routine thing to do? I am not sure how one would swap them in practice.

Or do you mean it's like equivalence of second order partials?

That would actually make sense with the adjoint of the covariant derivative.

Because if you could get $g(\nabla_{\partial_t}\partial_\varepsilon q(t,x), \partial_tq(t,x))$, then taking the adjoint, you get something with a - sign, and then the divergence. But I think the divergence would probably be zero...

Ted Shifrin

This kind of stuff appears, as I said, in DoCarmo and other standard treatments. See, for example, p. 195.

anak

Do Carmo's surfaces book?

Ted Shifrin

8:19 PM

No, the grad text.

It's in the surfaces book, too, I guess.

They use the notation $D/dt$, $D/ds$, etc., for the covariant derivatives.

anak

"Riemannian Geometry" is the grad one? I actually have never looked at his grad book.

Ted Shifrin

Yes.

anak

I will have a look, thanks!

Ted Shifrin

It's one of the standard resources, along with various French ones.

anak

Yeah, so he does some similar swaps on page 195

anak

8:49 PM

So yeah, it's actually just that $\nabla_\varepsilon \partial_t - \nabla_t \partial_\varepsilon = [\partial_\varepsilon, \partial_t] = 0$

So you can swap them, as you said.

robjohn

Hey, @Ted! How are things?

Xander Henderson

So, I went to a sporting event last night. Based on the final score (16-7), you might think it was a football game.

Sadly, it was not.

robjohn

Either a very active hockey game, or a boring basketball game.

Xander Henderson

It was 8 innings of baseball, followed by half and inning of amateur hour, little league BS.

robjohn

Ah

Xander Henderson

8:54 PM

At the bottom of the 8th, the D-Backs were losing 8-7. It had been a tense game, in which Bumgarner managed to hold Pujoles to only two home runs and a base hit.

But in the top of the 9th, the wheels fell off.

Even the worst batter on the Cardinals team (season average: 0.190) managed to score a grand slam. Without the ball leaving the park.

Technically, it was a triple, plus a one base error.

But four runs scored.

Stupid Diamondbacks and their terrible team this year. And Oakland is doing even worse. Why can't any of the teams I care about ever win any games. :(

robjohn

9:14 PM

I can stop a store or restaurant from carrying an item by saying I like it.

Xander Henderson

@robjohn Ouch. That sucks.

robjohn

It does, even though one might think it gives you a kind of control. It is a Midas touch.

porridgemathematics

@anak how do you go from that commutation formula to the thing in the picture?

copper.hat

9:40 PM

@robjohn why would they do that?