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12:20 AM
Are you getting ready to be banned again?
 
1:10 AM
livvy reached 500 followers on instagram. i am going to be able to monetize this sooner or later.
 
@copper.hat PERMA-BANNED!
 
your SE-branded orange jumpsuit will arrive via fedex in, let's say, august.
 
1:25 AM
as long as the jumpsuit has the se logo, i'm good
i send my se socks to my daughter in london so she can impress her friends
i wonder if there are many father daughter combos on mse?
 
i'll get my daughter on this. no problem.
if you see someone replying to a random post and yelling "I DON'T WANT THIS, I WANT A CHEESE STICK," that's her.
 
Cheese isn't vegan. I call foul.
Heya @robjohn
 
1:42 AM
we aren't raising her vegan. breast milk wasn't vegan.
 
Thank goodness. I think that's something an adult should decide.
 
even my wife cheats, she ate sushi yesterday.
 
Well, my sister became vegetarian as a teen-ager. Close enough.
That's serious cheating.
I don't do instagram, but nor have I put Screech on FB.
 
my daughter eats red meat with about the same frequency that i do, maybe one meal per month. but freely eats fish, poultry, and the aforementioned cheese sticks.
you should put screech on instagram. cat instagram is a huge thing.
an enormous number of my cat's followers are in japan. i don't know why this is. #2 nation is turkey.
black cat instagram may be different from general housecat instagram.
 
We’ll pass ;)
 
1:53 AM
what's screech again? orange tabby? my mental image is orange tabby.
 
working hard on my psqs to get the jump suit
i mean, really, that would be serious motivation. an mse jumpsuit.
well, my daughter did call me a nerd yesterday because i said like convex stuff.
 
which you are
 
@TedShifrin Hey there! Just got back from walking the dog, had to write a note to my boss, and now we are off to get dinner. Be back in a while.
 
may all your paths be rectifyable.
 
2:19 AM
you and your boss are off to dinner? weird.
i'm reminiscing with a now-former coworker who left. she claims that on her first day i spent the whole day in her office talking about inane distracting BS instead of letting her work.
i disagree, it was at most half of the day. i have the parking lot records to prove it.
 
More than the whole day sounds like you!
 
i mean i did waste at least 4 hours of her time
but that's not a day
 
That’s only your claim. The evidence shows otherwise.
 
2:34 AM
I'm back again
 
Oh yeah, “at least.” You are culpable, indeed.
Hi Under
 
i only arrived at noon, i at most wasted noon-6 hours of her day
 
I saw someone mentioned a biography about Alan Turing. Has anyone read Erdös's Biography: The Man Who Only Loved Numbers? I'm reading it right now.
 
3:00 AM
@copper.hat mine today seemed to be space filling.
 
3:24 AM
:-)
I read some of Erdös's Biography. I went through a maths biography phase.
Norbert Wiener's left the biggest mark.
 
I'm going through that phase right now. The last one I read was Halmos'. I did a bit of a deep dive on Ramanujan and Hardy for a school assignment, so I didn't end up reading the actual biography.
@copper.hat Norbert Weiner? I'm not sure if I've heard of him.
 
Many I could not finish because I lost interest.
 
It's about the journey, not the destination.
 
I had a similar phase with some poets/playwrights
I stopped when I hit politicians & government figures. The Dulles Brothers left me so depressed I stopped that phase.
 
Oh ☹
 
4:11 AM
@robjohn thank you! I could solve the problem about the series with Cauchy condensation test.
 
since $\frac{2^n}{\log(n)^n}$ is summable
 
I wish you had a good day :-)
 
It’s easier to do direct comparison!
 
not sure if you've heard of norbert wiener. oh, boy.
 
Hi @TedShifrin
 
4:17 AM
Hi Alex
 
one of my advisor's prized possessions was a book, authored by NW, and signed by NW after some event somewhere. the pure joy he had in having it was palpable.
his proof of the PNT is the only one that makes sense to me.
 
PNT=Prime Number Theorem?
 
I think there are several proofs about that, inclusive a topological proof , I think
Why is the other doesn't make sense for you?
 
there are a lot of proofs out there. a lot of what wiener did with the fourier transform has just so become part of what people use the fourier transform to do, that it's second nature now.
 
4:30 AM
Interesting, well I know that $\displaystyle \pi(x)\underset{+\infty}{\sim}\frac{x}{\log x}\iff p_{n}\underset{+\infty}{\sim}n \log(n)$. I read a bit about this a while back in a number theory book. However, my memory is not good.
It is an interesting problem about NT.
The series $\displaystyle \sum_{n=1}^{+\infty} \frac{\mu(n)}{n^{s}}$ converges for all $s>\frac{1}{2}$. Show that this statement implies the Riemann Hypothesis.
 
@UnderMathUate Yes. I enjoyed the Erdős bio. I also read the Ramanujan one, The Man Who Knew Infinity. It's good too.
 
5:16 AM
what is the integrating factor for: $\frac{\mathrm{d}y}{\mathrm{d}y}= \frac{2xy^2+y}{x-2y^3}$?
I can solve this ODE and here's my solution: the simplification leads to $2xy^2\mathrm{d}x+y^2 \mathrm{d}(\frac xy)=0$ and integrating both sides gives an implicit solution.
I think that the given ODE can't be converted to an exact differential equation.
 
5:30 AM
@Koro I suppose there is a typo and the correct problem is $dy/dx=\frac{2xy^{2}+y}{x-2y^{3}}$ that is not exact equation as you said. But the integrate factor is $\mu(y)=\frac{1}{y^{2}}$ then we have a exact equation. Finally we get the solution: $x^{2}+y^{2}+\frac{x}{y}=C$ with $C$ a constant.
 
5:59 AM
@Alex yes, I meant $\frac{\mathrm{d}y}{\mathrm{d}x}= \frac{2xy^2+y}{x-2y^3}$
and the solution is $x^2+y^2+\frac xy=c$
But I was wondering why the method of finding IF fails here.
Someone downvoted my question and apparently Ted's answer as well to that question.
 
which of your questions got downvoted. a quick perusal did not show an obvious downvote
 
Ted's answer was not downvoted, I added a compensatory upvote :-)
 
I thought the downvoter had a change of heart and came back to upvote.
Thanks copper. :)
 
unexplained downvotes really bother me
unexplained upvotes are just fine :-)
 
6:16 AM
I got an unjustified downvote elsewhere earlier. I think in some cases people have no good reason.
 
Why am I not getting integrating factor using formula?
for this ODE: $\frac{\mathrm{d}y}{\mathrm{d}x}= \frac{2xy^2+y}{x-2y^3}$
Integrating factor $\mu$ is given by $\mu_y=-\color{red}{\frac {M_y-N_x}{M}}\mu$ if the red colored expression is free of $x$.
and by $\mu_x=\color{purple}{\frac{M_y-N_x}{N}}\mu$ if the purple colored expression is free of $y$.
But still, the IF is $\frac 1{y^2}$.
 
6:35 AM
Ah, I got it.
I made calculation mistake, again!
$\frac{M_y-N_x}M$ is free from $x$ actually.
 
@TedShifrin If your downvotes are like mine, they are usually without reason.
 
Operator method of finding particular solution of ODE is really amazing.
 
@Koro which method is that?
 
Given $p(D)y=e^{at}V(t)\implies y_p=\frac{e^{at}}{p(D+a)}V(t)$.
 
How do you get $\frac{e^{at}}{p(D+a)}$?
 
6:51 AM
blackberry used to make one
 
@copper.hat Palm did, too
 
i had one a long time ago
 
So did I (Palm, not Blackberry)
 
$p(D)$ is a polynomial operator with constant coefficients. For example if we have $y''+ay'+by=g(x)$ then $p(\frac {\mathrm{d}}{\mathrm{d}x})y=g(x)$, where $p(z)=z^2+az+b$. :)
 
I wanted to play with the 'handwriting' recognition.
 
6:53 AM
@Koro Yes, but how do you get the inverse operator?
 
I think you need to use the Laplace transform or similar if you want to formalise that?
 
@copper.hat you had to write in just the right way
 
There was that funny way of writing.
 
@copper.hat nah, you can use integrating factors to invert $D-a$. Then just factor $p$
it is messy and not easy
 
Ok. If $(D-a)y=g(x)$ then $y_p=\frac 1{D-a} g(x):=e^{ax}\int g(x) e^{-ax}\,dx$
Using this, there is a proof using which I wrote the earlier expression.
Using this, if we take $g(x)=e^{bx}$ then $p(D)y=e^{bx}$ gives $y_p=\frac 1 {p(D)} e^{bx}=\frac 1{p(b)} e^{bx}$
if $p(b)\ne 0$
 
7:00 AM
Good night folks!
 
what if $p(b)=0$? Then, it can be shown that $y_p= \frac x{p'(b)} e^{bx}$, if $p'(b)\ne 0$
Good night, copper!
And if $p'(b)$ is also zero, then $y_p=\frac{x^2}{p''(b)}e^{bx}$, if $fp''(b)\ne 0$.
Usually, we find particular solution using variation of parameters or indetermined coefficients method. But operator method gives a very short solution. :)
 
 
1 hour later…
8:06 AM
@Koro btw, the result you told me last night, has a name?
 
8:20 AM
I think I told that. I had added a note in my comment.
 
Oh let me see
Complementary subsequences?
 
@Odestheory12 yes
 
Do you know where can I find a proof of that
 
You can prove it using contradiction.
Suppose that the sequence $(x_n)$ does not converge to L. This means that there is a $d>0$ and a subsequence $(x_{n_k})$ such that $|x_{n_k}-L|\ge d$ for every $k\in \mathbb N$.
 
Yeah and that means $A\cup B \neq \Bbb N $.
 
8:34 AM
Now, note that either $\{n_k:k\in \mathbb N\}\cap A$ or $\{n_k:k\in \mathbb N\}\cap B$ is an infinite set.
WLOG. let the former be an infinite set. So we can make a subsequence of $(x_{n_k})$ out of those indices. Such subsequence is also a subsequence of $(x_a), a\in A$. :)
Hence the subsequence must converge to $L$, which contradicts out assumption for some $k$.
 
Thanks :P
 
I hope you got it.
Also, A and B are ordered (both have elements in increasing order) subsets of $\mathbb N$. I think I skipped the word ordered earlier.
 
 
2 hours later…
10:31 AM
@Koro other demo I made if you are interested :P For $\epsilon > 0$ by hyphotesis there are $m_1,m_2 \in \Bbb N$ such that $|x_a - L| <\epsilon$ for all $n\ge m_1$ and $|x_b - L| <\epsilon$ for all $n \ge m_2$. Now, every element of $\{x_n\}$ is either an element of $\{x_a\}$ or $\{x_b\}$. So, given $n\ge \max(m_1,m_2)$ we have $|x_n-L|<\epsilon$
I wonder if the assumption $A\cap B = \emptyset$ is needed.
I think it suffices that $A\cap B = C$ where $C$ is finite.
 
10:54 AM
Consider 2 eigenvectors v1 and v2 of a matrix M. v1 generates the eigenspace A and v2 generates the eigenspace B. Are v1 and v2 orthogonal? Or just if A is symmetrical?
 
Since we can make a matrix which takes $(1,0)$ to $(2,0)$ and $(1,1)$ to $(3,3)$, eigenvectors do not need to be orthogonal, so symmetry is needed.
$\begin{bmatrix}2&1\\0&3\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}2\\0\end{bmatrix}$
$\begin{bmatrix}2&1\\0&3\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\3\end{bmatrix}$
 
11:11 AM
@Odestheory12 The assumption that $A\cap B=\emptyset$ is not needed. Such is the beauty of taking limits, ‘eventuality’ or should I say ‘ultimate behaviour of the sequence’ matters and you can ignore first 10 billion or 20 trillion etc. terms of a sequence and still the limit of the sequence will remain the same.
The assumption that $A\cap B$ is a finite set works just fine.
 
@Koro I don't see why $A\cap B$ is of any consequence. It seems to me that as long as $A\cup B=\mathbb{N}$, and that $\lim\limits_{j\in A}x_j=\lim\limits_{j\in B}x_j=L$, we have $\lim\limits_{j\in\mathbb{N}}x_j=L$
 
11:27 AM
@Koro Actually, I didn't notice this.
If a sequence does not converge to $L$ does it has infinitely many subsequences that won't converge to $L$?
Or Am I missunderstanding the claim?
 
@robjohn $A\cap B=\emptyset$ was placed just for the sake of defining ‘complementary sequences’.
*complementary subsequences.
@Odestheory12 yes.
 
It makes sense, the contrapositive is more intuitive.
I feel like the notes I am following are leaving a lot of gaps in my understanding of sequences theory.
 
I only negated limit definition and reached a contradiction.
 
Yeah, your proof is pretty. Its just.. I am trying to prove everything I can using the definition of convergence.
(just to practice)
 
 
2 hours later…
1:27 PM
@robjohn Thanks!
 
2:08 PM
@Koro I had never seen complementary to mean disjoint, only that their union makes a whole.
 
2:23 PM
Hi there, more a question on methods of proving I guess:
I have a conjecture about the general form of some derivative, is it enough to check /match coefficients, then prove by induction?
My personal view is that a proof from elementary means is to be preferred, but sometimes it's just so much work...
 
VLC
$\bigcup_{i=1} ^{\infty}A_i = \{x| \exists i \in \mathbb{N};x\in A_i\} = A_1$

Can somebody explain why does the upper equal $A_1$
 
2:51 PM
@VLC This seems true if $A_{i+1}\subset A_i$, but not in general.
@1010011010 it is unclear what you are asking, but it seems that this would be something you would need to discuss with your instructor.
 
dtn
Hello everyone. I want to know from specialists in polynomials - is there any progress in solving a system of equations consisting of Vieta formulas? For instance:
Solution for x1, x2 and x3.
A very interesting topic, but as far as I understand, there is not much progress in this area yet?
 
3:12 PM
@robjohn There's no instructor :D
 
@1010011010 Then there is no way to tell without more information. It may depend on the intended audience.
 
It's appendix material for a math phys / integrability project but it's got a lot of basic calculus in it like cesaro summation or generalisations of partial fractions
 
3:33 PM
anyone who's involved in any kind of math research?
 
4:22 PM
dtn: that seems equivalent to finding the three roots of a cubic with coefficients given by d/a, c/a, b/a, 1. lots of work done on polynomial root finding in general. maybe some of it is useful, depending on your desired interpretation of what it means to 'solve' a polynomial.
 
VLC
What about: $\bigcap A_i = \{x | \forall i \in \mathbb{N} ; x \in A_i\} = \{0\}$
Why is only zero in that set ?
 
what are the sets A_i?
 
VLC
Sorry: $A_i = \{x| x \text{even natural numbers} \wedge x\geq i \}\union \{0\}$
 
ok. so no odd number is going to be in the intersection.
and no even positive number is going to be in the intersection, because any even positive t is not going to be in A_{t+1}.
but 0's in all of those sets by definition, so it will be in the intersection.
 
VLC
4:39 PM
I get it now thanks
 
my daughter's very confused right now. our phones blew up with a tsunami warning, i guess because of that thing in the south pacific. no evacuations are ordered, so it's basically a warning for people who might literally be on the beach to move their sand castles.
our daughter heard some mild discussion of this and saw a video clip of what was going on in the south pacific and now thinks that it's going to get very wet here.
it started raining, which didn't help. "see? i told you"
 
VLC
Does anybody get what does the following mean:

if $f:A\rightarrow B$ is injective and $g:A\rightarrow B$ is surjective, then there exists $h:AA\rightarrow B$ that is injective.
I don't really get it, does it mean that there is another function that can be found that is injective or ?
 
i agree that as stated, it is confusing. what is AA?
this seems to be setting up some kind of thing where only the cardinalities of A and B are relevant, but the desired conclusion about h is confusing as stated.
i was expecting the conclusion to be that there is a bijection between A and B. may involve the axiom of choice if A and B are infinite sets.
 
VLC
5:06 PM
@leslietownes I am sorry, I did a Typo: $g:A\rightarrow B$ is surjective, then there exists $h:A\rightarrow B$ that is injective.
It's only A not AA
 
Do we need $f$, $g$, and $h$ to be distinct? If not, then maybe $f = h$?
 
i like soupless's idea.
this may be a typo or something in the exercise. it seems broken somehow.
 
Maybe $h$ needs to be bijective.
 
VLC
They are distinct yes
It says here that it should be injective
It's a theorem that we haven't named
 
if A and B have only one element then it is not possible for f, g, and h to be distinct. the exercise is broken.
 
VLC
5:09 PM
So: f injective and g surjective means that there exists h that is injective
hmm, ok maybe destinction isn't necessary
but why name them different letters then
 
@VLC Maybe $g \circ f$?
Wait, no.
 
oh, well that's a general thing. things named with different letters should not be assumed to be different (or required to be different) unless explicitly stated.
different naming usually just reflects potentially different roles or properties in an argument, but not actual distinction.
 
@leslietownes Isn't this a part of existence and uniqueness proofs? Something like, assume that $x$ and $x'$ are both this, then $x = x'$?
 
VLC
Well but why would a theorem exist then, that's like saying that if x =2 then there exists y = 2 but why would it exist in such a stated way
 
Not that it is just that, but is seen more often there
 
5:12 PM
sometimes textbooks do mess this up, saying stuff like "let a, b, and c be integers" and from context it's clear that they intend a, b, and c to be distinct integers. but without context and in generality, labels are just labels.
VLC: this gets back to my suggestion that the exercise is broken. if f and h aren't required to be different, soupless's suggestion works. take h = f.
i don't know why an exercise would ask "if A and B, then A." that led me down the path of guessing what the exercise might have been intended to do.
 
@leslietownes Is there a function $f: A \to B$ such that $A$ and $B$ are null sets or this is just nonsense?
 
VLC
I think that the professor did a typo and met that h is bijective
 
the empty set would satisfy that definition somewhat vacuously. i don't think there is a nonempty example.
 
VLC
so if f is injective and g is surjective, then there exists h that is bijective
does that make sense ?
 
VLC: it does. are A and B finite? in the infinite setting this will involve some version of the axiom of choice, so if that is unfamiliar to you, it may be a signal that something else was intended.
 
VLC
5:16 PM
Yes it says that the proof for infinite sets will be skipped because its too hard, but it holds for both
 
i think. all of this has the air of something where i post a comment on main and then asaf karagila responds with three comments curb-stomping my comment into the ground.
 
soupless: I think that the set of all such functions is defined to have cardinality $1$. I think that's done so that cardinal arithmetic $|B|^{|A|}$ holds with the definition that $0^0=1$.
 
Schroeder-Bernstein variant
 
yeah, S-B or even the pigeonhole principle, in the finite context.
 
wait, what is that cardinal arithmetic? i can't keep up. never mind, asking questions only that i can understand for now. thanks, Koro and leslie townes
 
5:23 PM
you could fiddle with f and g and argue that if f is not a bijeciton then B is in bijection with a proper subset of itself.
 
@TedShifrin is that a new Corona mutation?
 
You never know!
 
my daughter cracked open a package of old masks that we don't use anymore because they aren't the N95/KN95/etc standard. she applied four masks to her stuffed animal and two masks to me.
that's how my morning is going.
 
My KN95 shipment is weeks delayed. Sigh.
 
she tried to mask the cat but olivia's head is not large enough to provide the elastic with any purchase.
 
5:26 PM
at least she cares more about you, or perhaps she thinks that might shut you up?
 
olivia was surprisingly tolerant of this.
robjohn: i think the latter.
i took the masks off after she left the room and then she saw i had done that and told me to put them back on.
 
bad daddy
 
Reflecting what’s been said to her …
 
children are but mirrors...
 
How to do I solve this IVP: $(x\sin\frac yx-y\cos\frac yx)+x\cos \frac yx\frac{\mathrm{d}y}{\mathrm{d}x}=0, y(0)=0$?
I think that this question makes no sense.
At $(0,0)$, functions on LHS aren't even defined.
 
5:29 PM
why do children reverse wrong and right but not up and down?
 
like a mirror? (wrong =left)
 
no... right=wrong
 
koro: yes. if you're feeling sinister.
 
I ended up with $|x\sin \frac yx|=c_1$. I'm not sure how to apply initial value condition here. 🥲
 
@leslietownes not dexterous
 
5:31 PM
ambi-dextrous?
 
No, robjohn, dextrose!
 
time for my oral dose
 
dextrose implies the existence of sinistrose
 
@leslietownes wasn't he a villain on the Green Lantern?
 
robjohn: right+right=left :P
 
5:34 PM
i think so.
 
@Koro no, right+right+right=left
 
Ah, yes!!
 
There is a No Right Turn sign going into a parking structure at UCLA, positioned so that you see it right in front of you as you are walking into the structure. Unfortunately, you need to turn right to walk into the structure. When I had my son with me, I would make 3 left turns to walk in.
 
EM4
@robjohn thank you for the help on my question :).
 
@EM4 glad to help
 
EM4
5:39 PM
can I show you the way I solved it?
 
sure
 
EM4
oops I can't uploaded it.
 
I just noticed, since the set of rationals is a field, and it is a subset of the set of reals, this forces the set of reals to be a field. Is this correct?
 
No, definitely not.
I can put the rationals inside lots of rings that are not fields.
 
Hmm. Hasty generalization. Thanks.
 
5:48 PM
if $y=x\cos x$ is a solution of the ODE $y_n+a_1y_{n-1}+...+a_n y=0$ where $a_i$'s are real constants and $y_j$ denotes jth derivative of $y$ w.r.t. x. Then what is the minimum possible value of $n$?
 
Horrid notation.
 
I agree with you professor Ted.
 
So how does $xe^{ax}$ show up as a solution of a linear ODE?
 
We solve such ODEs by substituting $y=e^{rx}$. This substitution gives us a polynomial equation in $r$. Repetition of roots is same as multiplication to $e^{rx}$ by $x$.
 
Right. It shows up when $a$ is a double root of the indicial equation.
 
5:51 PM
So to get $x\cos x$, we should have $r=i$ as root of $r^n+a_1r^{n-1}+...+a_{n-1}r+a_n=0$.
 
Careful.
 
But since coeff. are real $-i$ is also a root.
If i repeats, so should -i.
 
Ah, now you're repeating. Cool. So ...
 
$e^{ix}, e^{-ix}, xe^{ix}, xe^{-ix}$ so $4$ is the minimum value?
 
depending on your ODE background theorems i wonder if you can get only a candidate for a minimum value, and then maybe have to check by hand that the smaller ones don't work.
 
5:57 PM
Thinking more on this: the general solution is $y=c_1e^{ix}+c_2e^{-ix}+c_3 xe^{ix}+c_4 e^{-ix}=a_1\cos x+a_2\sin x+a_3 (x\cos x)+a_4x\sin x$
But I'm not sure if it is correct.
 
the delight of your problems, koro, is that it's never clear exactly what the background is. :)
 
You left off an $x$, but why not sure?
 

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