Mathematics - 2018-03-20 (page 2 of 4)

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2:00 PM

because $T$ contains all star domains, and open curves, and also some open sets with very complicated shape (example shown in next diagram) such that it becomes disconnected whenever we remove a point. So that means, at maximum, every set can be described as a at most continuum of union of open curves hence $\bigcup_{k\in\mathfrak{c}}\gamma_{k}$ where $\gamma : [a,b] \to \Bbb{R}^2$

@AkivaWeinberger I know what you mean but that statement is still amusing

Akiva Weinberger

Can you show that, if $X$ has measure zero, then $\int_X\varphi$ equals $0$ (regardless of what $\varphi$ is)?

Hley Semi

(Probably one should insert “compared to the reals” in there)

Hi

@AkivaWeinberger Yes. I already proved that.

user image

2:02 PM

@Secret : hence you got, even, the open disc (which is not in T), no ?

Akiva Weinberger

@user193319 Hint: Break the integral into two parts.

This strange set is basically like a star domain except it is made of unaccountably many open curves stitched together so that it will become disconnected whenever we remove a point (which my drawing does not do any justice)

Akiva Weinberger

(Remember that $\int_{A\cup B}=\int_A+\int_B$ where $A$ and $B$ are disjoint)

“Starred ensemble” sounds like a description of the starboard on bad days

@Dattier A more rigorous construction is for example take the union of all straight lines with irrational angles centered on the origin. That is not homeomorphic to the open disk

2:04 PM

@AkivaWeinberger Hmm...I was thinking about that myself, but $m(E) = \infty$, and I don't have a way of dealing with that case yet.

@Secret : bravo

Akiva Weinberger

Consider $A=\{x\in E:\varphi(x)=0\}$, that is, the set of points on which $\varphi$ is zero

Then $\int_A\varphi$ equals $\int_A0$, since $\varphi=0$ everywhere on $A$.

That's one reason why I spend a lot of time on infinities, because I want to be able to illustrates these weirdos

@AkivaWeinberger, how could you find those normalizer examples so quickly?

@AkivaWeinberger I understand that, but the problem is I can't use that "splitting" theorem because $m(E)$ might be infinite.

2:06 PM

Here’s a linear algebra question that’s bugging me

Akiva Weinberger

I thought $\int_{A\cup B}=\int_A+\int_B$ (disjoint $A,B$) was always true, even if they have infinite measure

@Silent Conjugations on $S_n$ have a fairly simple form

ok

@AkivaWeinberger It may, but my book has only dealt with the finite measure case.

Suppose I’ve got a square matrix U and an orthogonal non-identity matrix S such that $U=US$.

@user193319 what are you trying to prove?

2:08 PM

@0celo7 I am trying to prove that if $\varphi$ is a simple function that is $0$ almost everywhere on $E$, which may have infinite Leb. measure, then $\int_E \varphi = 0$.

@user193319 $\sigma$-finite measure space?

Akiva Weinberger

@Silent Like, for example, say we have $h=(1,2,3)$ and $g=(1,10)(2,20)(3,30)$, then $ghg^{-1}=(10,20,30)$

@0celo7 Umm...$E$ is a subset of $\Bbb{R}$ with the Lebesgue measure. Not sure if that answers your question.

Anyway, the point is any set in $T$ is going to be at most a continuum union of open curves in $\Bbb{R}^2$, thus its cardinality can only be $\text{card}(T) = \text{card}(\bigcup_{k\in \mathfrak{c}} (2\mathfrak{c})^{\mathfrak{c}}) = \text{card}(\bigcup_{k\in \mathfrak{c}} \mathfrak{c}^{\mathfrak{c}}) = \text{card}(\bigcup_{k\in \mathfrak{c}} 2^{\mathfrak{c}}) = 2^{\mathfrak{c}}$

@user193319 It does.

Prove it for $\int_{E\cap (-n,n)}\varphi$ and let $n\to\infty$.

Akiva Weinberger

2:10 PM

Like, conjugating with $g$ applies the permutation $g$ to the numbers in the cycle-notation of $h$

if that makes sense

If U were nonsingular, I could multiply both sides by U inverse and get I=S in contradiction with the assumption

@AkivaWeinberger omg! I will come to this after i will cover conjugation thoroughly. Thanks for mentioning this.

So U must be singular.

What I’m trying to figure out is what happens if I allow U to be rectangular.

Akiva Weinberger

$U=US,\quad SS^\top=I$?

Right.

In particular, I want U to have more columns than rows

2:12 PM

@0celo7 Okay. I'll try that. Thanks!.

Akiva Weinberger

$U_{m>n}$ and $S_{n\times n}$

@user193319 Now, you do have to justify that the limit is still $\int_E\varphi$. But you probably have some theorem that says that.

Other way around, I should think

Idk, do the $\varphi=\varphi^+-\varphi^-$ trick and use monotone convergence or something.

Dimension of S is same as number of columns

Akiva Weinberger

2:14 PM

I thought like $A_{a\times b}B_{b\times c}=C_{a\times c}$ (columns x rows)

So we need $U_{m\times n}=U_{m\times n}S_{n\times n}$

Yes, but m is the number of rows

And I want more columns than rows

Akiva Weinberger

Isn't it "columns x rows" and not the other way around?

Oh wait

You're right sorry

Kk

yesterday I was offered a problem that I still do not know how to solve, show that for Legendre's polynomials Bonnet's formula characterizes the same polynomials as for Rodrigues' formula (the strangest thing is, I did not manage to find a proof, on French wiki, whereas it is supposed to be a classic.)

Akiva Weinberger

It's "height times width", but "height" is the number of rows, not columns

Point is this is a fat/wide rather than a skinny/tall matrix

2:17 PM

Right

@Dattier There might be a combinatorial argument to be made

Which means that I know right off the bat that the rank of U is at most n<m

@Lozansky : I don't understant why there are not this proof in french wiki

And the question is whether U=US further constrains the rank of U.

Akiva Weinberger

$\begin{bmatrix}1&0&0\end{bmatrix} \begin{bmatrix}1&0&0 \\ a&b&c \\ d&e&f\end{bmatrix} = \begin{bmatrix}1&0&0\end{bmatrix}$

as an example

2:19 PM

@Lozansky In english wiki too

Guys, Zorn lemma question: Is it possible for the hamel basis of $\Bbb{R}(\Bbb{Q})$ to be the vector space itself?. Also do zorn lemma allow us to find a maximum, besides a maximal element under some conditions?

Yeah. That’s automatically rank 1, though

So you can’t make rank any smaller in that case

Alessandro Codenotti

@Secret No, if $v$ is in a basis $qv$ isn't, for $q$ rational

Akiva Weinberger

$\begin{bmatrix}1&0&0 \\ 0&1&0 \end{bmatrix} \begin{bmatrix}1&0&0 \\ 0&1&0 \\ d&e&f\end{bmatrix} = \begin{bmatrix}1&0&0 \\ 0&1&0 \end{bmatrix}$

@Lozansky I think Rodrigues keep this justification secret

2:21 PM

Beat me to it

Akiva Weinberger

I think you just take any solution of $US=S$ for square matrices and delete some rows of $U$

and randomize the corresponding rows of $S$

S isn’t necessarily orthogonal in that example

Akiva Weinberger

What I just said makes no sense actually

@Dattier I think there is a clever way to show it

But I’m not sure it not being orthogonal is a big deal

2:24 PM

@Lozansky : maybe, but I don't know it

Rodrigues is a strange personnality, he want build with an other french people new christianism

Actually, is that S only orthogonal if d=e=0 and f=1?

@AlessandroCodenotti got it

Alessandro Codenotti

Regarding the second question usually if there is a maximum you don't even need to use Zorn's lemma

with Saint-Simon : fr.wikipedia.org/wiki/Olinde_Rodrigues

@Lozansky

I am actually not sure where we will expect a maximum in an infinite set that is not an ordinal though in non set theory topics?

2:27 PM

Hey guys say we have $ a \cdot b = c \cdot b$

Can i just remove the b?

And have a=b?

so I guess I am a bit puzzled

I can cant I

@JakeRose What's your underlying algebraic system?

Oh no I cant

More than one vecotr could give that answer

Vectors sorry

Depends what you’re doing

Yeah, definitely not

2:29 PM

I forget thats used in other things too

dot products are not cancellative in general

so no

Cool thanks guys

Note that that’s the same as asking whether $(a-b)\cdot c=0$ implies a=b or c=0

@Lozansky are you Freemason ?

A more interesting thought will be. if I have $a \cdot W^{\perp} = b \cdot W^{\perp}$ where the W thing is the orthgonal complement of $a$ and $b$, can it cancellate?

2:30 PM

@Dattier Not to my knowledge :P

So it’s really about whether $x\cdot y=0$ can be true for nonzero x,y

And it clearly can since orthogonal vectors are a thing

@Lozansky where did you find this problem?

Only if a=b? @secret

@Dattier "Fourier Analysis and its Applications" by Vretblad

ah right, cause any vector parallel to $a,b$ will dot to zero with the orthogonal complement of them

2:33 PM

@Lozansky : does he give the solution ?

Ohhhhhhhh

I didnt spot that

Good one

@Dattier Nope

That's one reason why I like abstract algebra, it "parallelise" a lot of operations by operating things at the set level

@Lozansky some hint ?

None

2:34 PM

This question seems to be quite popular now on Philosophy SE" Why is the complex number an integral part of physical reality?

@AkivaWeinberger I think this S is only orthogonal when d=e=0 and f=1?

And if you get bored of the Rodrigues'/Bonnet question, try proving $\int_{\mathbb{R}} (-1)^{n+m} e^{x^2} D^{n} (e^{-x^2}) D^{m} (e^{-x^2}) dx = \delta_{nm} n!2^n \sqrt{\pi}$

complex numbers really show itself in the physics by having the amplitudes to be able to interfere without the total probability changing

it's the simplest way to do that

@Lozansky which is pretty much a physics problem in disguise

@Semiclassical Hmm?

2:36 PM

@Lozansky : the gnostic often act like that he gives a result for which they have no proof, and by pretending that he knows one, it motivates the researchers, who think that there is in one, and when someone finds it we act as if it's been known for a long time ...

Lol, I am certain there is a proof @Dattier

Yep. It shows that the eigenfunctions of a quantum harmonic oscillator form an orthogonal set

@Lozansky It's probably not the same question, but the same recurrenct appears in this question: Recurrence relation for Legendre polynomials. (Perhaps the link given in the answer posted there might be useful to you. I found it by searching for the recurrence in Approach0 - maybe you can find some other related posts in this way.)

@Lozansky : why don't I find it, on french wiki ?

or anywhere

Ahh, they use the generating function for $P_n$ as well

2:38 PM

Oh, wait. Hrm

No, they don't give a proof for Rodrigues formula

@Lozansky

I’m being silly. It’s still possible to formulate it in terms of harmonic oscillator stuff, I think, but my above statement was too hasty

With harmonic oscillator eigenstates, you start with exp(-x^2/2) and act on the left with (x-D) some number of times

And you get nice stuff

@Lozansky well give me a proof of Rodrigues formula (even a link), that's closed that

This is just powers of D, so it’s not quite the same

Hello. How Can I write in wolfram alpha function : $\begin{cases} 0 &\text{for } |x| < \pi/2 \\1 &\text{for } \pi/2 \le |x| \le \pi \end{cases}$.

2:45 PM

@PabloZ392 I think WA should recognize the UnitBox function from Mathematica. So that’d be “UnitBox[x/Pi]”

Yep, just tested it

...oh, wait. I misread your question

Bleh

If you only plot from -Pi to Pi, though, doing “1-UnitBox[x/Pi]” should work

kind of a stupid question but if I take the derivative of a constant function f(x) = c, then lim h->0 (f(x+h) + f(x)) / h = lim h->0 (c-c)/h = lim h->0 0/h

if I plug in 0 for h this becomes an indeterminate form which usually requires l'Hopital

but most proofs online I see jump straight to 0

what am I missing here?

Hello all

Can anyone help me understand this proof?

screencast.com/t/zg1sm8199

script P (S) means the space of probability measures of S

@Semiclassical Thank you! :)

Np

@Lozansky : I find it (the proof of Rodriguez formula) it basis on the Sobolev trick

$\int_0^1 f'(x)g(x)\text{d}x=-\int_0^1 f(x)g'(x) \text{d}x$

when $f(1)=f(0)=g(0)=g(1)=0$

2:59 PM

@MartinSleziak hmm. I’m sorta of two minds there. On the one hand, the ways in which we define and write complex numbers is clearly dependent on some human-chosen conventions. In that respect, complex numbers are a human invention and so are a contingent not necessary part of reality

@Dattier So what does the proof look like?

On the other hand, there is a certain gut-level Platonism in me which resists the suggestion that the complex numbers wouldn’t exist if not for humans

@Lozansky it basis on the calculus of $<R_n,R_m>$

And to the extent that the complex numbers represent a certain kind of pattern/structure then it doesn’t seem absurd that it could be part of our physical reality

@Dattier Mkay, I'm guessing repeated integration by parts?

3:05 PM

I think complex numbers naturally encoded rotation in its multiplication structure, in that it mixes the horizontal and vertical components

@Lozansky yes

@Semiclassical You might recognize that integral as the inner product of the Hermite polynomials of degree $n,m$

Does anyone have experience with the 3rd and 4th edition of Royden's Real Analysis (the 4th edition is entirely the work of a later author, Fitzpatrick). I've heard that Fitzpatrick ruined Royden's text and that it is better to work with the third edition? Is this true?

@Dattier Do you believe in its validity now?

@Lozansky : now, yes, but true one day is not true forever

I am not intuistionist, but empirist

for recall, the reasoning empirist is :

3:12 PM

My professor mentioned that historically there might have been philosophical objections to the idea that a matrix with real entries has (only) complex eigenvalues. Does anyone know if that was the case? What's older, matrix algebra like $\det(\lambda I - A)=0$, or the theory of complex roots of polynomials with real coefficients?

A is exact if with ten examples, and without know counterexample

For example, do you know in France, in the past, 1 was a number prime

and now, it's not true

1 is not prime (in France)

well Dattier that's more a matter of convention, we wanted the prime decomposition of a number t0 be unique

no?

do you speak french ?

@GFauxPas

Alessandro Codenotti

Hi @Akiva

nope

3:18 PM

so in this case, 0 is not number for decomposition in basis 2 to be unique, no ?

sorry, I meant

it the same case

a number $\ge 1$

(with the empty product convention)

6=2*3*1 or 6=2*3 yes ?

Alessandro Codenotti

Hi @Dami

3:21 PM

It the same case here : 3=2+1 or 3=2+1+0

Alessandro Codenotti

Did you actually take that model theory course you mentioned as a possible course a while back?

if want the unicity (of décomposition in basis 2) you must choose 0 is not a number

interesting Datt

well, why 1 is not prime and 0 is a number ?

it's inconsistent

@Lozansky I was forgetting that you could obtain the Hermite polynomials in a few different ways

So yeah, I buy that now

3:23 PM

Decomposing a number into a sum of primes is not unique anyway..

@GFauxPas ok ?

This sounds perilously close to “is zero a natural number?”

hi Semi

Which is pretty tiresome

Hi

so 1 is a prime number

3:24 PM

my number theory professor told me

logicians call 0 natural, number theorists start at 1, but it doesn't matter as long its not ambiguous in context

My rule of thumb is that 0 is a natural number when I only care about addition

But it’s not when I care about multiplication properties

0

Q: 3rd and 4th Edition of Royden's Real Analysis Book

Does anyone have experience with the 3rd and 4th edition of Royden's Real Analysis (from my understanding, the 4th edition is due entirely the work of a later author Fitzpatrick). From the reviews on Amazon, the 'consensus' seems to be that Fitzpatrick ruined Royden's text and that it is better t...

reference-request book-recommendation

@GFauxPas here it's a porblem, 1 is not prime for unicity reasoning so for the same reasoning 0 is not a number, no ?

what is unicity

@Dattier What uniqueness are you trying to preserve by using the convention that $0$ is not a number?

3:27 PM

Hence a power series starts at n=0 but a Dirichlet series starts at n=1

unicity of the décomposition

what decomposition?

oh, right, American standard word is "uniqueness" in convention Dattier even though they're exact synonyms

anyway

in product of number prime and addition of power of 2

"every number $\ge 1$ is uniquely decomposed as a product of primes"

(convention that $1$ is the product of no primes)

3:29 PM

why 1 is not prime ?

because then numbers will not be uniquely decomposed as a product of primes

Because it makes things annoying

we show the unicity is not a good argument, so why ?

Every number is divisible by 1 arbitrarily many times

and we'd pick a convention that allows us to define uniqueness (= unicity) in a convenient way

rather than saying "and when we say unique, we mean in the sense that ... "

3:30 PM

And having to distinguish between how many times we divide by 1 is just annoying

right, we could do it, but it would be annoying, so we don't

It makes everything more complicated without providing any insight

@Semiclassical : even every number can be added 0

without change it

so why 1 is not prime ?

recalling in France in the past 1 was prime

because it's inconvenient

why ?

3:33 PM

why is the product of no numbers 1, $\displaystyle \prod_{i \in \varnothing} x_i = 1$?

How many times does 1 divide 10?

because it's convenient

no it the same think with the addition and 0, but 0 stay a number

You’re drawing an irrelevant analogy

because then you don't have to think about questions "how many times does 1 divide 10 " because that question doesn't give you insight into how numbers work

3:33 PM

@Dattier your argument that $3 = 2+1 = 2+1+0$ is not a good argument; $3 = 1+1+1$ too. The point of uniqueness of prime factorisation is that it really IS unique, unless you include units like $\pm 1$. Then we have the concept of "associatedness" which says that associates have essentially the same divisibility properties, and are the same up to multiplication by a unit

With addition there’s no notion of unique decomposition even if you exclude zero

@ÍgjøgnumMeg : no I talk about the unicity of written in basis 2

3=2^1+2^0+0=2^1+2^0

Why only positive coefficients?

does it matter what representation you use?

If you allow negative ones, you could just as well write 3 =2^2-2^0

3:36 PM

Recalling I want to know why the french mathematician change thier opinion for the german's opinion

It’s only unique because you made some assumptions as to how you construct the base 2 representatiob

If 0 is number, it's concerned

And the point is that if you want to do the same thing but using prime decomposition then it only works if you exclude 1 as a prime

It's the same thing for uncity of decomposition in basis 2

you must exclude 0 of number

I dunno if you're listening or not

3:40 PM

another exemple : the R[x] polynome of degree 1

P(X)=X or P(X)=X+0

so if 0 is a polynomial is not unique

you are understanding ?

You can go ahead and define $1$ to be prime if you want to

yes, so now for me 1 is prime

just.. don't complain when your wrist is sore from writing $6 = 2 \times 3 = 2 \times 3 \times 1 = 2 \times 3 \times 1 \times 1 = \dots$

$$\mathop{\huge{\times}}_{i \in \Bbb{N}} 1$$

and when proofs that invoke uniqueness of prime decomposition suddenly have to be completely rewritted

lol Secret

with the convention that $0 \in \mathbb N$ :D

or just use $i \in \mathbb Z$

whynot

3:46 PM

lol

[Random]

I think one consequence of having additive absorbers in a ring like structure is now you have something called "additive roots"

(I have not investigate the valuation analogues suggested by mathein yet)

To illustrate, let $u$ be an additive absorber. Then we have for all $a$, $a + u= u + a =u$

well we had that anyway

we just write it as $a \circ e = e \circ a = a$

in a group for example

well, addition and multiplication are not different if you only have one binary operator such as in groups

What makes multiplication stands out in all ring like structures is it distributes over addition

4:00 PM

Blarg

Dieing how's everyone else?

(cont.) One can then wrote something analogous to the zero product lemma, e.g. $ac+bc=u \implies ac=u \lor bc=u$.

So, it is possible things get a bit more interesting if we have:

Guys, it holds that $\int_a^b(b-x)(x-a)dx=1/6(b-a)^3$. I was wondering if there is a smart way of going about this, without brute-force solving the intergral (I mean, it not that tedious, but if there is a quick way about it, I would like to know it)

$(x^2+bx)(x+1)=0$ and $x^2+bx=u$

Then we have $x=0 \lor x=-b \lor x=-1$ and $x^2=u \lor bx=u$

Hello

Then we can see that $x=0,-1$ will lead to a contradiction (gives $0=u$ and $-1=u$ respectively), leaving behind $x=-b$. Thus $b^2=u \lor -b^2=u$ for example

not sure what that will be useful for, will needed to study about additive analogues of polynomials to find out

4:07 PM

Is there a computational science chat room?

Computer Science

General discussion for cs.stackexchange.com

je suis de retour

I'm back

R[x] P(x)=X and P(X)=X+0 so the decomposition is not unique if 0 is a polynimal

ok ?

P(X)=X+0=X+0+0+0=X+0+0+0+0+0=...

well, I don't understand the polynomial rings stuff you and ÍgjøgnumMeg are discussing. Btw I am heading to h bar for the night before quickly go to sleep

I'll see you guys tmr

oh, the debat is closed

I have forgotten

Conlusion : now for me 1 is prime

So, it's an illustration of this sentence : true one day, is not true forever

Yestarday : 1 is not prime

And Today : 1 is prime

lol

4:26 PM

Is anyone here familiar with elliptic integrals?

@ShaVuklia Parabolic area? You have width $b-a$ and height $(b-a)^2/4$ so the area is $\dfrac{2}{3}(b-a)(b-a)^2/4 = (b-a)^3/6$ as desired

Suppose that $f_n \to f$ pointwise, that $h(x) \le f(x)$, and that $h_n = \min \{f_n,h\}$. Why must $h_n \to h$ pointwise?

Also, all the functions are nonnegative.

It's obvious that $\lim_{n \to \infty} h_n(x) \le h(x)$, but I don't see how to get the other inequality.

$h_n(x)=\min\{f_n(x),h(x)\}$ $\min$ is continuous so $\lim h_n(x)=\min\{f(x),h(x)\}=h(x)$

@Dattier None of the functions are continuous, just measurable.

No, it's only the continuity of $\min$, I use

ok ?

4:42 PM

@Dattier Oh! Wow! Of course. Thanks!

Akiva Weinberger

4:53 PM

How are people

@AkivaWeinberger I don't understand

@Dattier Referring to the same $h,h_n,f$ and $f_n$ as above, why is it true that $\int_E h = \lim_{n \to \infty} \int_E h_n \le \lim \inf \int_E f_n$? I am trying to understand the proof of Fatou's lemma.

I understand why $\int_E h = \lim_{n \to \infty} \int_E h_n$ holds, but I don't understand why the inequality holds.

what's the definition of $f_n$ ?

@user193319

@Dattier $\{f_n\}$ is just some sequence of measurable functions converging pointwise to $f$.

ok

5:06 PM

Can two bases generate the same topology where one of the bases is countable and other is uncountable? I am trying to prove something, and truth or falsification of this will help.

@Shobhit Yes. $\{(a,b) \mid a,b \in \Bbb{Q}\}$ and $\{(a,b) \mid a,b \in \Bbb{R}\}$ both generate the standard topology on $\Bbb{R}$.

because $h_n(x) \leq f_n(x)$

ok. got it. ty @user193319

@Dattier Okay. That's true, but I am having trouble seeing how that shows that $\lim_{n \to \infty} \int_E h_n \le \lim \inf \int_E f_n$. I am probably being too dense at the moment to see it.

@Shobhit No problem.

$\lim \int h_n=\liminf \int h_n(x) \leq \liminf \int f_n$

5:10 PM

@Dattier Yup. I was being dense! Thanks again for the help!

with pleasure

@Dattier Want to look at the problem I posted before?

@Lozansky can you give a link ?

$\int_{\mathbb{R}} e^{x^2} D^n(e^{-x^2})D^m(e^{-x^2}) dx = \delta_{mn} n! 2^n \sqrt{\pi}$

Problem: prove it

1/show $\forall n\in \mathbb N, \lim \limits_{x\rightarrow +-\infty } D^n[e^{-x^2}]=0$

2/ use an intergration by parts, and maybe conclued.

@Lozansky can you answer to the first step ?

5:21 PM

@Dattier Interesting

Give me a second

I need to prove that $T = (X, \tau)$, indiscrete topological space is not second countable. My try: let $\beta$ be a countable bases for $T$, then for every open set $U$ in $T$, and for every $x \in U$ there exist $B \in \beta$ such that $x \in B \subseteq U$. Since on indiscrete topology every set is open, since $\tau = P(X)$, therefore $\{x\}$ is open for every $x\in X$.

Therefore for every $x \in X$, there is $B_{x} \in \beta$ such that $x \in B_{x} \subseteq \{x\}$, therefore $B_{x}=\{x\}$. Now since $X$ is uncountable, $|\beta|=|\cup_{x \in X}B_{x}|$ is uncountable. Will this work?

@AkivaWeinberger, Rudin says that 'if $f$ defined on a segment $(a,b)$ and if $a<x<b$, then $f'(a)$ and $f'(b)$ are not defined in this case.' I wonder why $f'(a)$ and $f'(b)$ not defined even if limit of quotient at $a$ is defined?

Akiva Weinberger

Confucius say, he who stand on toilet, high on pot

Is that for me?

I don't understand that

Akiva Weinberger

Nah it was just a joke

5:26 PM

@Silent since $f$ is defined only for $(a,b)$, you dont know anything about $f(a)$ or $f(b)$, so dont know about their derivatives

Akiva Weinberger

Yeah, think about the definition of the derivative

But derivative is a limit, you don't need knowledge of function at limit point.

Akiva Weinberger

It's $\lim\dfrac{f(x+h)-\color{Red}{f(x)}}h$

You don't need knowledge of the function at the limit point $h=0$. But if $f(x)$ isn't defined, you don't have knowledge of the function for $h={}$anything.

@AkivaWeinberger that was really helpful.

@Dattier I can see how 1) makes sense, since $e^{-x^2}$ dominates $x^k$, $\forall k \in \mathbb{N}$ in the limit

5:32 PM

@Lozansky yes

what about the step 2/

@AlessandroCodenotti can you tell me if my proof is correct? its just above, i don't know how to link it to this message.

@Lozansky 2-1/Show $D^n[e^{-x^2}]=P_n(x)\times e^{-x^2}$ with $degree(P_n(x))=n$

Nice, Mister Rogers marathon on Twitch for his birthday

@Lozansky 2-2/Determinate $a_n$ the dominant coefficient of $P_n(x)$.

2-3/Find a hint and conclued

Akiva Weinberger

If they were to show hours footage of marathon races from around the world and over the years, would that be a marathon marathon?

5:52 PM

"Eat as long as the pasta is hot" Confucius's wife

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Mar '1820

$Mathematics$ Mathematics

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