Mathematics - 2018-03-07 (page 1 of 4)

Martin Svanberg

00:04

@LeylaAlkan

Hit enter too fast

$x^2+y^2$ is a hint that you should substitute polar coordinates

But on the lhs you're integrating over the entire real plane, so to integrate over a disk you need to take the limit as the radius goes to infinity

Leyla Alkan

How do we do the substitution?

Martin Svanberg

$x=r\cos(\theta), y=r\sin(\theta)$. If you draw a circle of radius $r$ you can draw a triangle with the radius of the circle as the hypotenuse to find that this is how you specify coordinates on the circle

Not sure how advanced your course is but the added $r$ in the integration is known as the Jacobian and is added when going from cartesian to polar coordinates to scale the infinitesimal areas from small squares to something more circular

The thing to google for is "polar coordinates double integrals"

Leyla Alkan

Okay, let me check, thank you!

Leaky Nun

is it true that $m \otimes n = 0$ iff ($m = xy$ and $xn = 0$ for some $x$ and $y$, or $n=xy$ and $xm=0$ for some $x$ and $y$)?

(@MatheinBoulomenos)

Ted Shifrin

What module over what ring, @Leaky?

Leaky Nun

00:17

any module over any commutative ring

Ted Shifrin

Let me be more specific. You're doing $A\otimes_R B$ for what $A$ and $B$?

Leaky Nun

modules over R

in fact I'm trying to prove that A is absolutely flat => S^-1 A is absolutely flat. I might be in the wrong direction (just in case this becomes an xy problem, which is highly likely I reckon)

Ted Shifrin

I see how to prove $\impliedby$ but I'm suspicious about $\implies$.

Leaky Nun

is this easy?

Ted Shifrin

I haven't thought about this in way long.

Leaky Nun

00:22

ok then

Akiva Weinberger

@LeakyNun R meaning any ring or R meaning $\Bbb R$

I guess the former

Leaky Nun

any ring

Semiclassical

hi chat

Akiva Weinberger

Is there always a map from $A\otimes_RB\to Rm\otimes Rn$?

I guess you need maps $A\to Rm$ and $B\to Rn$

I guess not... there's no map $\Bbb Q\to\Bbb Z$

Ted Shifrin

00:43

Hi DogAteMy

Akiva Weinberger

What's the field of fractions of $R$ called?

Leaky Nun

@AkivaWeinberger you just named it

Ted Shifrin

The field of fractions

MatheinBoulomenos

Ohh

I was thinking about faithfully flat

Ted Shifrin

No one is faithful here.

Akiva Weinberger

00:47

No I mean what's the notation

Leaky Nun

$\Bbb Q[R]$ I think

Ted Shifrin

I've never seen a notation.

People say "let $K$ be the field of fractions of $R$"

Akiva Weinberger

That makes no sense @LeakyNun

The Testosterone Fanatic

is anyone here using OpenID to log in to their stackexchange account?

then I've got some bad news for you

Leaky Nun

I think I've also seen Quot(R)

The Testosterone Fanatic

00:48

68

Q: Support for OpenID ends on July 1, 2018

Stack Overflow was an early and strong supporter of OpenID. We built our sign up/log in flow around it. We were idealistic and had high hopes, but these hopes weren't realized. Over the years people have wondered if OpenID is dead. We've had to remove support as OpenID providers pulled support or...

Ted Shifrin

Oh, lovely.

Leaky Nun

@AkivaWeinberger no it's just a notation it isn't like Q-algebra

Akiva Weinberger

Right so call it $K$. Is there always a map from $A$ to $Km$? We can assume $A$ is finitely generated I guess

MatheinBoulomenos

@LeakyNun one possible proof is proving that absolutely flat is equivalent to von-Neumann regular, i.e. $\forall a \in A \exists x \in A: a=axa$

Leaky Nun

@AkivaWeinberger what is Km?

@MatheinBoulomenos right, I already have that in an earlier chapter

MatheinBoulomenos

00:49

that obviously passes to localizations

The Testosterone Fanatic

if anyone here is using OpenID account, please associate with google or facebook

just a PSA

Leaky Nun

so a=axa and s=sys, and then a/s = (a/s)(?)(a/s) I can't solve this

Akiva Weinberger

I guess I mean $K\otimes Rm$?

Ted Shifrin

Thanks, @TheTestosteroneFanatic. I think I have an alternative login, but perhaps it's lapsed.

Leaky Nun

@MatheinBoulomenos nice to know that it passes to the non-commutative case

@AkivaWeinberger and what is Rm?

$R_m$ you mean? R localized away from m?

Akiva Weinberger

00:50

What do you call the ring generated by $m$

MatheinBoulomenos

@LeakyNun what about x/y?

Leaky Nun

you can't generate a ring...

Akiva Weinberger

Oh I mean module

Leaky Nun

@MatheinBoulomenos y isn't in s

Akiva Weinberger

Smallest thing in $A$ with $m$ in it

Leaky Nun

00:51

ideal :)

I write $(m)$ or $mR$ for that

MatheinBoulomenos

@LeakyNun replace $S$ by its saturation, then

user537069

@AkivaWeinberger But why though?

Leaky Nun

@MatheinBoulomenos oh god what is wrong with me

that was stupid

user537069

My whole difficulty right now is proving/seeing how this result even comes about

Leaky Nun

I just proved the god*** result

(that I can replace with its saturation)

Ted Shifrin

00:52

doesn't remember what saturation is

Leaky Nun

@TedShifrin let $S$ be a multiplicatively closed subset of a ring $R$

the saturation $\overline S$ of $S$ is where you add in every element $x$ and $y$ where $xy \in S$

MatheinBoulomenos

that's the largest multiplicatively closed subset containing $S$ with the same localization

Leaky Nun

$\overline S = \{x \mid \exists y, xy \in S\}$

Akiva Weinberger

The point is if there's always a map from $A$ to $mK:=mR\otimes K$ which maps $m$ to $m$ then we can have a map from $A\otimes B$ to $mK\otimes nK$ which means $m\otimes n$ will be $0$ in the former only if it's $0$ in the latter

MatheinBoulomenos

@LeakyNun okay here's a direct approach: Any $S^{-1}R$-module is of the form $S^{-1}M=S^{-1}R \otimes_R M$ for a $R$-module $M$. Let $N$ be any other $S^{-1}M$-module. We have $S^{-1}M \otimes_{S^{-1}R} N \cong M \otimes_R S^{-1}R \otimes_{S^{-1}R} N \cong M \otimes_R N$. One checks that this is a natural isomorphism, so we have an isomorphism of functors $S^{-1}M \otimes_{S^{-1}R} - \cong M \otimes_R -$. This implies in general that if $M$ is flat over, then $S^{-1}M$ is flat over $S^{-1}R$

Leaky Nun

01:01

why is any $S^{-1}R$ module of the form $S^{-1}M$?

MatheinBoulomenos

If you have an $S^{-1}R$-module $M$, then let $M'$ be the same abelian group, but with restriction of scalars to $M$ (I don't remember the notation from AM for that), then one can show that $M \cong S^{-1}M'$. This is not difficult. Because $M'$ was originally an $S^{-1}R$-module, for any $s \in S$, the map $M' \to M', m \mapsto sm$ is an isomorphism.

It's very similar to the proof that every ideal of $S^{-1}R$ is an extended ideal from $R$

(I know that's not a full proof, but I don't want to spell out all details right now)

Basically any element in $S^{-1}M'$ can be written uniquely in the form $\frac{m}{1}$, because of what I said above about multiplications with elements in $S$ being isomorphisms. This implies that the obvious map $M \to S^{-1}M'$ is an isomorphism

Jake Rose

can somebody give me a hand with this?

\$\int_{x=-a}^{x=a} \int_{y=x^2}^{y=(1-x^2)^(1/2)} dydx\$

nitsua60

Which part of it?

Ted Shifrin

@JakeRose: That doesn't look right. Is $a$ supposed to be something in particular? What was the original question?

Jake Rose

Yeah sorry $(\root{5} -1)/2$

poop

Ted Shifrin

01:10

Oh, ok.

Jake Rose

$(5^(1/2)-1)/2$

Ted Shifrin

So what part are you having trouble with?

you want \sqrt5

Jake Rose

Thank you

$(\sqrt5 -1)/2$

Okay so I initially integrated with respect to y

Ted Shifrin

OK.

nitsua60

What'd you get?

Jake Rose

01:12

Leaving me with $\int_{-a}^{a} (\sqrt(1-x^2) -x^2 dx$

Sorry it takes me a while to type latex

Ted Shifrin

OK.

nitsua60

(no worries)

Jake Rose

But I cant figure out how to integrate the first term

nitsua60

What level of calculus are you familiar with? Have you encountered "trigonometric substitution"?

Secret

@AkivaWeinberger this is what happens when you type * too fast

nitsua60

01:15

(That's not quite the way to approach it, but it informed my thinking.)

Jake Rose

Yep

nitsua60

How 'bout integration by parts?

Jake Rose

Somewhat decent level calculus Id say

Yup

Ted Shifrin

You really don't want integration by parts for that.

nitsua60

It gets that sqrt into a denominator, right where I like it =)

Leaky Nun

01:16

@MatheinBoulomenos hmm...

Secret

Philosophical question: How do mathematical identities form?

Ted Shifrin

But then you end up with $x^2/\sqrt{1-x^2}$ to integrate. You need a trig sub for that.

Secret

Sure, we can prove them easily, but is there anything in common that turn them into identities?. I guess, what I mean is, what is the structural feature for a bunch of mathematical object to have a mathematical identity

Ted Shifrin

Just do the damn trig sub to start with.

nitsua60

Alright, good night, everyone.

Ted Shifrin

01:19

@JakeRose: So let $x=\sin\theta$ and make sure you don't forget to change the limits of integration, too.

Jake Rose

What trig sub?

And how did you spot it was a trig sub?

Ted Shifrin

Because I have been doing these for 50 years. :)

Jake Rose

Ahhhhh

Is there like a general form or anything?

Ted Shifrin

You recognize trig identities when you see $\sqrt{a^2\pm x^2}$ without an $x$ to go with them.

Jake Rose

How would a humble fresher natsci spot it

Ted Shifrin

01:21

If there's an $x$ you let $u=a^2\pm x^2$. If there's not, you let $x=a\sin\theta$ or $x=a\tan\theta$ depending on the sign.

You should have seen this in your integral calculus class.

Jake Rose

We go at quite a fast pace. Dont get shown many examples

Ted Shifrin

But now you're at double integrals?

Jake Rose

Yup

Ted Shifrin

Curious. Where are you a student?

Jake Rose

Cambridge

Natural sciences though not mathematics

Ted Shifrin

01:22

Ah ... so the lectures are all theory ...

Jake Rose

Yeah quite so

It depends on the lecturer tbh

Some lecturers do more examples and some do less

Ted Shifrin

Although there's a lot of theory in my lectures, there are also lots of computational examples (and even physics applications), so you might want to check out my multivariable math lectures on YouTube (linked in my profile).

Just skip all the fancy theory that you're not doing.

Jake Rose

You know when you say 'an x to go with them' do you mean the x in the root?

Secret

Substitution in a nutshell
$$\int f(x) dx$$
Let $x = g(u) \implies dx = \frac{dg}{du}du$
$$\int f(g(u))\frac{dg}{du} du$$
The idea is that with a suitable $g$, $f(g(u))\frac{dg}{du}$ is easier to integrate by exploiting mathematical identities of $g$

Jake Rose

Yeah man will do ! THnaks

Always love a new resource

Ted Shifrin

01:24

No, I meant something like $\int x\sqrt{a^2\pm x^2}\,dx$ or $\int x/\sqrt{a^2\pm x^2}\,dx$.

Jake Rose

Ah okay

And where do you go from there?

Ted Shifrin

As I said, you let $u=a^2\pm x^2$. Note that $du = 2x\,dx$, which works perfectly.

But you don't have the $x$. So make the direct trig substitution.

I need to leave, but hang in there. :)

Jake Rose

Thanks a lot

Gosh how long does it take for you to prepare for a lecture???

Ted Shifrin

I'm sure I'll see you in the future ...

Not long for those — I taught the course 13 times and wrote the textbook :)

Jake Rose

That wasnt sarcastic btw I hope it didnt come across like that

Ted Shifrin

01:27

I didn't hear any sarcasm.

Jake Rose

I couldnt even imagine doing that from memory

Ted Shifrin

It's not from memory. It's off the cuff. :)

Hi/bye DogAteMy.

Jake Rose

Adios

Ted Shifrin

Adios.

Akiva Weinberger

Hlello

Secret

01:29

$$\int \sqrt{h(x)} dx$$
Let $h(x) = g(u) \implies \frac{dh}{dx}dx = \frac{dg}{du}du$
$$\int \sqrt{g(u)}\frac{\frac{dg}{du}}{\frac{dh}{dx}} du$$

bleh, I really need a good way to handle $\frac{dh}{dx}$ in terms of u if $h$ is not invertible

Leyla Alkan

Does anyone possibly have solution manual for Gerald B. Folland's Advanced Calculus?

Jake Rose

@Secret thanks

Secret

$h(x) = g(u) \implies h^{\leftarrow} (g(u)) = x$
$\frac{d(h^{\leftarrow} (g(u)))}{dg}\frac{dg}{du} du = dx$

$$\int \sqrt{g(u)}\frac{d(h^{\leftarrow} (g(u)))}{dg}\frac{dg}{du} du$$

agh, this is too messy

if $g(u) = a^2±x^2$ then $x = a^2 \pm g(u)^2$

$\pm x \mp a^2 = g(u)^2 \implies \pm\sqrt{\pm x \mp a^2} = g(u)$

nope, that's wrong

Martin Svanberg

02:13

Wikipedia says that GL(n, F) and GL(V) are isomorphic when V has finite dimension. What about when it doesn't? Does it make sense to talk about $GL(\infty, F)$?

anon

02:44

@BalarkaSen I almost didn't watch the show since the characters and first few episodes were off-putting, but it's one of my favs now.

Secret

04:15

$g(u) = a^2 \pm x^2 \implies g(u) - a^2 = \pm x^2 \implies \pm \sqrt{\pm g(u)\mp a^2} = x$

$h(x) = a^2 \pm x^2 = a^2 \pm (\pm (g(u)-a^2)) = a^2 + g(u) - a^2 = g(u)$

It works

$\pm \sqrt{\pm g(u)\mp a^2} = x \implies \frac{g'(u)}{2\sqrt{\pm (g(u) - a^2)}} du = dx$

$$\therefore \int \sqrt{h(x)} dx = \int \frac{\sqrt{g(u)} g'(u)}{2\sqrt{\pm (g(u) - a^2)}} du$$

$$= \int \frac{\sqrt{k}}{2\sqrt{\pm (k - a^2)}}dk$$

yh016

Hello all, does normality of data set refer to seeing whether the empirical distribution of the data set is approximately normal, or refer to the likelihood that the data set is drawn from a population with normal distribution?

Secret

$ = \int \frac{1}{2\sqrt{\pm (1 - \frac{a^2}{k})}}dk$

Let $v = \pm (1-\frac{a^2}{k}) \implies \frac{a^2}{1\mp v} = k$

$\frac{\pm a^2}{(1\mp v)^2} dv = dk$

$= \frac{\pm a^2}{2}\int \frac{1}{(1\mp v)^2\sqrt{v}} dv$

Let $(1 \mp v)^2 \sqrt{v} = w \implies :::::::::::$

Error: Does not have closed form inverse relation

bleh, terribly hard to work on this without a piece of papefr

0

Q: Normality test with large data set

I think I am dealing with this issue http://www.r-bloggers.com/normality-tests-don%E2%80%99t-do-what-you-think-they-do/ I have large data set(10k data points) that slightly diverges from normal, and I get p-value of 0. I am interested in having perhaps a more crude test that tells me if data is ...

hypothesis-testing normality kolmogorov-smirnov

Perturbative

05:27

If $a, b, c \in \mathbb{R}$ and $a \leq b + c$ does it follow that $\min \{1, a\} \leq \min \{1, b\} + \min \{1, c\}$?

Perturbative

06:36

Yeah it does, I was just lazy to work out the cases

Hey @TedShifrin :)

Mesentery

07:39

Hello! Math.SE. In, 1100/10.2 by using significant figure rules, the answer is 110. But in, 1100 m/s divided by 10.2 m/s the answer is 108. The actual answer is 107.8431373. Can anybody please explain that why 110 turned 108 just by using units? Thanks!

108 is also by using significant figures rules

Sam

Hello all

user685252

Heya

Sam

Can someone help me - I'm working through a linear algebra playlist currently encoding linear transformations as matrices. And the guy has now confused the hell out of me

user685252

in The h Bar, 8 mins ago, by user685252

@Mesentery what significant figure rules are you using?

Sam

khanacademy.org/math/linear-algebra/matrix-transformations/… In practice problem 1 (of that link).. the right answer (C) seems to have inverted the column values to what I assumed the matrix should look like :S Could you sanity check it for me and tell me where I'm going wrong?

I had the matrix:
0.5 1
1 0.5

Ah wait. I was confusing myself. Durp.

$\begin{bmatrix}1 & 0.5 \\ 0.5 & 1 \end{bmatrix}$

Tobias Kildetoft

08:17

Let's say I have been given complex numbers $a$ and $b$. Can I always find complex numbers $c$ and $d$ with $d\neq 0$ such that both $2c + db$ and $d(a - bc) - c^2$ are non-negative reals?

Vrouvrou

Hello, someone have an idea about that: math.stackexchange.com/questions/2680408/…

Mesentery

The rule is "Suppose in the measured values to be multiplied or divided , the least number of significant digits be n, then the product or quotient, the number of significant digits should also be n"

@user685252

Mesentery

08:43

See the last part in this image

Secret

10:31

E--Eb----Eb--E----F#----G--Eb----Eb--E----

madcrazydrumma

10:41

Hey guys! I'm currently in the process of writing a research proposal in the field of Mathematics and Computing. The desired area of study is the detachments of directed graphs, and since studies have been extensively done on detachments regarding undirected graphs, little has been done on directed graphs.

I was wondering if someone can help direct me to areas of interest in this topic so that I can hopefully find a topic title for my proposal.

Leyla Alkan

Hi, could someone help me with the part I marked blue in this example:

Particularly, how are the first and the second inequalities obtained?

@TedShifrin @AkivaWeinberger Are you guys around? :)

Lembik

why does $(p-1)^2 \bmod p ==1 $ when $p$ is prime?

ÍgjøgnumMeg

10:57

@Lembik Expand $(p-1)^2$ and you'll see, it's actually true for any $m$.

Lembik

ah! thanks

can there be other $x,y \in \{0,\dots,p-1\}$ so that $x^2 = y^2 \bmod p$?

Isomorphism

Hello, I am trying understand the proof of Steiner's problem from wiki. https://en.wikipedia.org/wiki/Steiner%27s_conic_problem

I have not studied any algebraic geometry. Is there any simple exposition of this problem.

Niing

11:15

Hello, I forget how to explain the $e^x$, which $x$ is real number instead of natural number...

ÍgjøgnumMeg

@Lembik Look at how $\bmod p$ is defined; if $x \equiv y \bmod p$ then $x = y + kp$ for some $k \in \Bbb Z$. Then $x^2 = y^2 + 2kp + (kp)^2 \equiv y^2 \bmod p$.

Lembik

thanks

ÍgjøgnumMeg

@Lembik An more interesting question is when a number is a square modulo $p$, i.e. when there is a solution to the equation $x^2 \equiv a \bmod p$.

Lembik

good point

ÍgjøgnumMeg

@Lembik I'm sure you'll meet this question soon if you're doing a course in elementary number theory :)

Lembik

11:24

:) thanks

Mary Star

Hello!!

We have 10 numbered balls in a box. We pick three balls with replacement.
The probability that the second ball has the number "1" is equal to $\frac{1}{10}\cdot 1\cdot \frac{1}{10}$, or not?
Is the probability that we get twice the same number the same, i.e. $\frac{1}{10}\cdot 1\cdot \frac{1}{10}$ ?

Mary Star

11:46

Hello @LeakyNun !! Do you have an idea about my question?

Leaky Nun

@MaryStar $1 \cdot \frac 1 {10} \cdot 1$

twice the same number: $1 - \frac{10 P 3}{10^3}$

Mary Star

probability that the second ball has the number "1" : Ah yes!

probability that we get twice the same number: Does it hold that P(twice the same number) = 1 - P(all 3 different OR all 3 the same) ?

@LeakyNun

Leaky Nun

depends on your interpretation

i.e. whether (2,2,2) is allowed

I allowed it

Mary Star

The statement is "exactly two of the three picked balls have the same number", so I think (2,2,2) is not allowed.

Then it holds that P(twice the same number) = 1 - P(all 3 different OR all 3 the same) = 1 - [ P(all 3 different) + P(all 3 the same) ], or not? @LeakyNun

Leaky Nun

yes

Mary Star

11:56

Does it hold that P(all 3 different) = 10*9*8/10^3 and P(all 3 the same) = 1/10 * 1 * 1 ? @LeakyNun

Leaky Nun

yes

Mary Star

Great! Thank you!! :-) @LeakyNun

Alessandro Codenotti

12:49

Hi @Mathei

MatheinBoulomenos

hi @Alessandro

Alessandro Codenotti

How is it going with the logic course?

MatheinBoulomenos

There's no logic course, I'm just reading a bit on my own

Alessandro Codenotti

Ah, I see, so how is it going with your logic self study?

MatheinBoulomenos

I've been a bit lazy the last few days, but I enjoy it

So far I did completeness and compactness for propositional logic, I'm currently working on the basics of first-order logic, in particular the proof of completeness for first order logic (not neccesarily of a countable language)

Leyla Alkan

12:57

Hi guys, do you know how is this part that I pointed done?

Alessandro Codenotti

@MatheinBoulomenos Via the Henkin property stuff?

MatheinBoulomenos

yeah

Alessandro Codenotti

Once you have completeness you can easily do compactness and the two Löwenhein-Skolem, which is the essential toolbox

MatheinBoulomenos

yeah, after that I'm going to skip the optional chapter on logic programming, then the next chapter is some model theory

Alessandro Codenotti

Cool, I'm reading some model theory notes too now

I looked at the proof of the compactness theorem via ultraproducts instead of going the long route through completeness, you'd probably like it as it's much more algebraic in flavour. I'm looking at quantifier elimination now

MatheinBoulomenos

13:05

oh, that sounds cool

Leyla Alkan

No one knows ? :(

MatheinBoulomenos

the book I'm reading right now only has 3 pages on ultraproducts

Alessandro Codenotti

The notes I'm reading don't even mention them, they prove compactness in a third way, via Hintikka sets, I had to find another reference for ultraproducta

Semiclassical

@LeylaAlkan they do a simple u-sub to do the antiderivative

MatheinBoulomenos

I once read somewhere that some logician proved compatness differently each time he taught it

Semiclassical

13:11

Note that, for each such integral, x is being held fixed

MatheinBoulomenos

@AlessandroCodenotti where did you find the ultraproducts proof of compacntess?

Semiclassical

(That’s how you’d do the integrate formally, anyways. Once you have enough experience, you’ll be able to do that integral by inspection ie you can infer what functuon you’d differentiate to get the integrand)

Alessandro Codenotti

@MatheinBoulomenos In some notes by a professor from Turin, they're written in English, I can link them to you later if you want (I'm on my phone now)

MatheinBoulomenos

@AlessandroCodenotti please do!

Leyla Alkan

@Semiclassical But when I do u-sub I get $\frac 1 x \log (1+xe^y)$ instead of $\frac {e^y} x \log (1+xe^y)$ And also How is this done : $F'(x)=\int_0^1 \frac {e^y}{1+xe^y}$ ?

Shobhit

13:25

Hello all. Can someone provide a hint for this question : If $Q$ is a polynomial with distinct roots $a_{1},..., a_{k}$ and if $P$ is a
polynomial of degree $<n$, how to show that $$\frac{P(z)}{Q(z)}=\sum_{k=1}^n \frac{P(a_{k})}{Q'(a_{k})(z-a_{k})}$$, i checked it using examples it checks out, all i see is if i take L.C.M of the denominators in the summation i will get $Q(z)$ in the denominator, and an ugly numerator.

Semiclassical

@LeylaAlkan hmm. I actually agree with your answer over the one written there

The check is that you should be able to differentiate the antiderivative with respect to y and recover your integrand

But if the answer were of the form e^y*log(stuff) then differentiation won’t get rid of that log

Whereas your answer doesn’t have that issue and seems correct

As for the integral itself, they’re arguing that you can justify moving the differentiation with respect to x inside the integral

Alessandro Codenotti

@Mathei domenicozambella.altervista.org/creche

He introduces products and quotients of structures first to treat ultraproducts as a special kind of them, some authors prefer to just introduce ultraproducts directly

MatheinBoulomenos

@AlessandroCodenotti thanks!

Leyla Alkan

Okay, thank you @Semiclassical

Semiclassical

Np

0celo7

13:47

@Adeek One of my professors suggested I make the font size 12. Now it's 126 pages

Secret

14:00

Last night dream chemistry seen in a 2 page (including reference) journal article in a dream

I suspect once I start readng journal articles in my research intensely again, my brain should get used to the state enough to finally simulate the scene where I can read those journal articles in the dream in full

...provided that Mechanism 1 is not triggered first

Secret

$$\int \sqrt{f(x)} dx$$

$f(x)=u^2 \implies x = f^{\leftarrow}(u^2) \implies dx = \frac{df^{\leftarrow}(u^2)}{d(u^2)}2u du$

$\int \pm 2u^2 \frac{df^{\leftarrow}(u^2)}{d(u^2)}du$

mike239x

14:18

Hey ppl

what is the most correct way to deal with an edit which I assume was done wrong?

like, the post itself was a bit of a mess, and the editor didn't quite get the point of OP (I think)

Slereah

@mike239x did you try Zorn's lemma

mike239x

@Slereah eeehm... what?

Slereah

(it's a joke)

mike239x

yeah, I sadly didn't get it

sorry

user537069

14:41

in an epsilon delta proof is it valid to have something like |f(x)-L| < non-epsilon-related-constant?

Astyx

Do you have a more concrete example ?

user193319

Problem: $\Bbb{R}^\omega$ in the box topology is completely Regular. Attempt: Since translations are homeomorphisms of $\Bbb{R}^\omega$, it suffices to consider $0=(0,0,...)$ and some closed set $C \subseteq \Bbb{R}^\omega$ not containing. The goal is to find a continuous function $f : \Bbb{R}^\omega \to [0,1]$ such that $f(0)=0$ and $f(C) = \{1\}$. My first thought was to define $f$ in terms of the uniform metric $\rho(x,y) = \sup_{i \in \Bbb{N}} \overline{d}(x_i,y_i)$ some way...

but I couldn't figure it out...I tried $f(y) = \rho(y,C)$, $f(y) = 1 - \rho(y,C)$, and many other variants...I could use a hint.

user537069

15:01

@Astyx I was googling around and found en.wikibooks.org/wiki/Calculus/Proofs_of_Some_Basic_Limit_Rules

for product rule, their third assumption uses |g(x)-M| < 1 (i.e. not a function of epsilon)

this seemed odd to me because i always assumed the limit definition was meant to work for all epsilon > 0

whereas |g(x)-M| < 1 might be false if epsilon is too large

Astyx

You are confusing some things I think

$\lim_{x\to a} f(x) = l$ means $$\forall \epsilon >0, \exists \delta>0, \forall x, |x-a|<\delta \implies |f(x)-l|<\epsilon$$

In particular, taking $\epsilon =1$, there exists a $\delta$ such that if $|x-a|<\delta$, $|f(x)-l|<1$

user537069

yes but then it's not going to be true for all epsilon > 0 anymore

like imagine some huge epsilon, |g(x)-M| < 1 might be false

Abcd

15:19

$\tag{mathematical physics}$

After differentiating it I obtained:

$v= a\omega \cos(\omega t)\hat{i}+ a\sin(\omega t)\hat{j} $

Kasmir Khaan

@anon hey anon :D

@anon text me when you see this :D i kinna need some help :D

Abcd

Then I used $\int dx= \int vdt$

But it didn't help.

GFauxPas

hmm, doing practice problems so this isn't homework,

Abcd

How do I solve it then?

GFauxPas

True or False? If $\displaystyle \int_K f \, \mathrm d\lambda = 0$ on every compact set in $K \subseteq\mathbb R^n$, and $f$ is $L^1$ measurable, then $f = 0 \text{ a.e.}$

it feels true :P

if $f$ were given to be a limit of a suitably well behaved sequence of functions $f_n$ then I can prove it's true, I think

can I always do that?

@user537069 I think we only need it to be true for $\epsilon = 1$, for the proof

user537069

15:40

but then doesn't this contradict the requirement that it be true for all epsilon > 0?

GFauxPas

We're saying, it's true for epsilon >0, therefore, we can, in particular, analyse the case of it being one

If i told you i saw a movie every day last week, then you can ask me, what movie did i see on Sunday? And know there's an answer

user537069

Yes, but we're trying to derive an eventual truth that is true for all epsilon > 0

how are we showing this to be true if we're only picking a specific epsilon = 1?

GFauxPas

Let me look at the proof again

Well we're saying, take the min of those three deltas

Anyway, if epsilon is more than 1,

You'll have |...|<1<epsilon

Since < is transitive

Niing

15:56

I'm trying to thinking about exponential function $e^x$, with $x$ is irrational number... any advices? Sorry for interrupt your discussion...

user537069

For example is it saying that we are making these assumptions:

$$\forall \epsilon>0, \exists \delta_1 > 0 : (0 < |x-a| < \delta_1 \implies |f(x) - L| < \frac{\epsilon}{2(1+|M|)})$$

$$\forall \epsilon>0, \exists \delta_2 > 0 : (0 < |x-a| < \delta_2 \implies |g(x) - M| < \frac{\epsilon}{2(1+|L|)})$$

$$\forall \epsilon>0, \exists \delta_3 > 0 : (0 < |x-a| < \delta_3 \implies |g(x) - M| < 1)$$

Or is it saying

$$\exists \delta_1 > 0 : (0 < |x-a| < \delta_1 \implies |f(x) - L| < \frac{\epsilon}{2(1+|M|)})$$

$$\exists \delta_2 > 0 : (0 < |x-a| < \delta_2 \implies |g(x) - M| < \frac{\epsilon}{2(1+|L|)})$$

$$\exists \delta_3 > 0 : (0 < |x-a| < \delta_3 \implies |g(x) - M| < 1)$$

Alessandro Codenotti

@GFauxPas that's true, $\int_A f=0$ implies that $f$ is zero a.e. on $A$

GFauxPas

Niing, the easiest way to think of it, though it's not the easiest definition to use in practice, is as a limit of e^x_n, where x_n is a Cauchy seq of rationals

But i only have its true for K compact, I can't cover all of Rn that way

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$Mathematics$ Mathematics