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00:01
@AkivaWeinberger Nice, thank you!!!
DogAteMy: What's the question with that mapping?
Which points get sent to $0$ @TedShifrin
And does the boundary of the set that gets sent to zero have a nice polynomial description
Huh, really? Are $x,y\in\Bbb C$? Are you sure you don't have a typo?
$x,y\in\Bbb R$, I think
Scroll up, @Semiclassical talked about it not too long ago
The question is trivial unless you mistyped your map.
00:06
Well, $x^2 + y^2 = 0$ means $x = y = 0$.
Not true unless you're in an ordered field, @Balarka.
Ohh, yeah, typo
Well, @Akiva said $\Bbb R$. That's ordered to me.
Which points get sent to $0$ when you iterate the map.
Oh, I didn't see any $\Bbb R$.
00:08
Sorry. It's about iterating the map.
Well, surely we need $|y|<1$ for starters.
Isn't that just the interior of the unit disk?
@PVAL Pretty sure it's smaller
GGG
GGG
@TedShifrin I've rectified that. However, I think I still have another mistake somewhere.
$\displaystyle \mathbf {(AB)}_{ik} = \sum_{1 \le j \le n}a_{ij}b_{jk}$, call this $(1)$. Let $\mathbf{A} \mapsto \mathbf{AB}$ and $\mathbf{B} \mapsto \mathbf{C}$ and reindexing we've $\displaystyle (\mathbf{(AB)C})_{i\ell} = \sum_{1 \le k \le n}(\mathbf{AB})_{ik}c_{k \ell} = \sum_{1 \le k \le n}\bigg(\sum_{1 \le j \le n}a_{ij}{b_{j k}}\bigg) c_{k \ell} =\sum_{1 \le k \le n}\sum_{1 \le j \le n}a_{ij}{b_{j k}} c_{k \ell}$.

Now, this time letting $\mathbf{B} \mapsto \mathbf{BC}$ in $(1)$ reindexing again similarly we end up with
Correct, @GGG. So switch the names of $j$ and $k$ and remember that you can do a finite sum in any order you want (so you can switch the order of summation).
GGG
GGG
00:11
Ah, thank you! @TedShifrin xD
@AkivaWeinberger What do you mean by sent to $0$? Which points converge to $0$ under successive iteration?
Most welcome, @GGG :)
@TedShifrin the question is which points are eventually sent to zero, and which to infinity
right, that's what I thought.
i can get a mathematica graph of it, hang on
or something that should resemble it, at any rate
00:14
Yeah, so, for example, a point that starts on the unit circle other than $(0,\pm1)$ will get sent to infinity, methinks.
first, ignore the stuff at the top and lower left; that's just visual artifacts
(And it's symmetric about the axes, of course.)
@Semiclassic. Do AspectRatio->Automatic, please.
@TedShifrin Seems true to me.
Oops, I typed the wrong thing. I meant other than $(\pm 1,0)$.
00:16
better?
Hmm, so I wonder if it's a true ellipse.
it's not
i tested it last night
Didn't think so.
to be precise about what that's a plot of
Calling the map $S$, perhaps there's a map $F$ such that $FSF^{-1}$ has a simpler form.
00:19
that's $|f^9(x,y)|$, with the blue and pale yellow (tope?) regions being below $e^{-5}$ and above $e^{5}$ respectively
it appears to go to zero or infinity with doubly-exponential speed, so it's pretty sharp; i had to go that far in tolerance to get a nice-looking boundary
tope = taupe :)
i bow to your color expertise
no, just spelling expertise
unless, of course, I'm wrong
no, you're right. though it looks like taupe isn't even close to that color
amusingly, 'tope' is a word albeit archaic
"drink alcohol to excess, especially on a regular basis."
Oh, I tope on a regular basis. For Balarka, I might even say on a normal basis. But I digress.
00:25
I tope on an orthonomal basis.
hah. in any case, i'm curious what can be said about that boundary
@TedShifrin I am confused where you infer that fact from.
I wasn't suggesting you toped, @Balarka. I was doing regular->normal for you.
I'm wondering (while reading) if the boundary is smooth (or how you can show it).
00:27
Guys, what happened to I'm an artist?
I clicked on hide posts
I have had her hidden for months, @Michael, so I have no idea. And Jasper has disappeared again. Come to think of it, robjohn has too.
@TedShifrin Perhaps you should make it "I tope on a Galois basis".
Something fishy is going
It's probably more fun to tope transcendentally, @Balarka.
00:28
I was gonna ask her for one of her challenging integrals
Someone's killing all the integrators.
but not one that's years above my league
I can give you one from the Putnam exam a few years ago, @Michael.
@Balarka can think about it as he does chapter 7 in my book, too :P
@TedShifrin My body is ready. Lemme see it :)
@PVAL well, the choice of map was from a question on chat. and there it was to show that the points which are eventually mapped to zero is a domain
just for fun, here's the boundary plus an ellipse
00:29
@Michael: Evaluate $\int_0^{\pi/2} \frac1{1+(\tan\theta)^\pi}d\theta$.
oh cool
@Balarka: Less jokes; more work or sleep. Your choice which.
@Michael: Warning — you can't do it by any conventional means.
Not even substition?
any type of substituion?
00:30
@MikeM: You mean fewer, not less. puts red pen away
substitution*****
or 'less joking'
Not that I know how to do, @Michael.
Howdy everyone!
hi @Julian :)
00:31
but, ew, transcendental power
Hi @Ted! :)
@Semiclassic: I'll change the power to $\sqrt2$ if you prefer.
Do you make some questions for those exams @TedShifrin ?
No, @Michael, I never have.
hah, fair enough. i'm guessing it works for some range of exponents?
00:32
I do know people who have, however.
@MikeMiller I'll sleep. I promise I'll work on your problems tomorrow.
my taste in things is such that i immediately generalize that to: what are the Fourier coefficients of $(1+\tan(\theta)^p)^{-1}$?
Yikes, good luck, @Semiclassic.
@TedShifrin Ok I haven't been taught anything similiar to solve this. GImme a hint pleaseee
00:35
Think I'd just guess zero, and then pretend like I knew it if I was right and I was joking if I was wrong.
Chris'ssis probably won't like my hint. My hint is to consider the function $F(x) = \int_0^{\pi/2} \frac1{1+(\tan\theta)^x}d\theta$.
heh..... blankly stares at wall
OK, sorry, @Michael. Pretend you're an engineer. What would you do?
@TedShifrin discuss with fellow engineers
00:36
Fine, then :)
plug in a few values?
I meant to do math without worrying about rigor :P
If I was a physicist I'd pretend that I came up with this technique as a freshman undergrad (am I close?).
What exactly is rigor?
@PVAL: I doubt you came up with it as a freshman undergrad, but perhaps you were shown it in E&M ? :)
@Michael: Rigor is having a proof for everything you do.
00:38
hi
SSSShhhhhhhhh @Semiclassic.
is this considered a function f(x) = do nothing
@Balarka will soon be proving that he can do that, however.
ohhhhhhhhhhhhhhhhhhhhhhhh
OHHHHHHHHHHHHHHHHHHHHHHHHHHH
sigh, i spoiled it
00:38
Someone very famous claimed to have come up with it as a freshman undergrad.
smacks Semiclassic
I should've said "Pretend I'm Feynman?"
Feynman, I'm sure, @PVAL.
I think it was well known before Feynman was an undergraduate.
And lots of us just do it without trying to prove that it works :)
Hence the pretend.
Is this considered a function: f(x) = do nothing (map x to no point y)?
00:39
i don't think he claimed to have come up with it. the story i remember is that the book he learned out of taught him different tools than the standard ones for integration
no, @Puzzled, there has to be something to send $x$ to.
It's a one line application of the multivariate chain rule (I'm sure you knew that @Ted)
No, the Leibniz rule for differentiating $\int_a^{g(x)} f(x,t)dt$ is an application of that, @PVAL. But I have a proof using Fubini in my exercises in my book.
I was gonna buy it, but cash stack is limited by these bio textbooks...
So, now that we gave you the hint, @Michael, go ahead and play with it. Differentiate with respect to $x$.
00:41
and so he was able to do problems which people couldn't do by standard complex contour integration, and thus picked up a reputation for integrals
Maybe Chris'ssis would have respected Feynman.
back later
ok
No partial differntiation right?
Well, officially, yes, you're partially differentiating with respect to $x$ (cuz the integrand is a function of $x$ and $t$).
oh, my partial differentiation isn't at putnam level
00:50
You don't know how to differentiate $a^x$?
$x(a)^{x-1}$
Um, what?
How do you differentiate $e^x$?
keep it the same
oh
facepalm
nods at the facepalm :)
oh, we were never taught this
its actually cool
00:53
@TedShifrin I at least did a case of this as an answer a year ago. math.stackexchange.com/questions/1091544/…
Yes, you're doing precisely what I referred to. But it assumes you know how to differentiate under the integral sign.
Don't worry Mr. Shifrin, I am learning that right now!
@Michael: You just need to write $a^x = e^{x\ln a}$.
Yah I know how to solve it now
it is always given by
$a^xln(a)$
Yup. Of course, we're assuming $a>0$.
00:58
^
Im learning leibniz integral rule as we speak
Oh, ok.
Wow this partial derivative business is pretty cool
What is a hessian matrix?
Second-order partial derivatives as the entries of the matrix.
oh ok
When does one start learning about convex functions and convex sets?
01:04
Dunno, @Michael.
actually, i think there's an elementary solution to that integral
just by symmetry, in fact
GGG
GGG
What's the easiest proof that $\det(AB) = \det(A)\det(B)$?
Good, @Semiclassic. That's another good approach.
not sure i'd have spotted it if i hadn't plotted it, though
Depends on what you know, @GGG. Thinking about writing nonsingular matrices as a product of elementary matrices. And then doing the singular case separately.
01:09
@TedShifrin: still planning a visit to LA?
Oh, @robjohn is still alive. Yes, arriving Thursday.
GGG
GGG
You can do the integral by Weierstrass substitution as well I think.
Planning lunch at Google in Venice on Friday, dinner in WeHo Friday night, but the rest of Friday is currently open.
What's Weierstrass substitution?
tangent of half-angle?
The $\tan(u/2)$ substitution?
That only works for rational functions.
01:11
yeah. though it might lead to some interesting versions for the case of noninteger exponenets
@Semiclassic: You are too easily distracted from your research.
distracted isn't the right word
GGG
GGG
I think it works in this case as well because the integral is symmetric. We write it over two interval after applying the substitution and then add them together again.
discouraged?
That's the word I'd usually use.
yeah
not the research itself, mind. i still love that
01:13
@PVAL, I can't count how many times I gave up in grad school.
but writing continues to be a brick wall, and just triggers my anxiety to the extent that i can lose days when it gets bad
though the time i lose isn't so much due to anxiety this week as the resulting exhaustion
Is this difficulty with writing the technical bits or the expository bits?
it's with writing, period. or at least writing in this particular context of a paper for an oral exam
This is a research paper?
Or something else?
summary of research, but yeah
which at some level i can tell myself is silly, since the point of the oral exam here is to show that you're ready to do research
and i've been doing that for three years now
but my brain really isn't interested in logical arguments.
01:18
That's why you hang around us math geeks so much. :)
Have you tried writing something that is not meant for public consumption and then fixing it up into something that's eventually ok.
well, this isn't something that gets published
only myself and the people on my committee will see it
They're not your enemy, @Semiclassic.
didn't say they were
That is what I do when I make talk notes. I just start writing the easiest things there are to explain, and then eventually I can fill in the gaps.
01:19
I know ... but your subconscious thinks of them that way.
if there's anything that's an enemy here, it's my own brain.
honestly, it's not them i find myself anxious about
I agree. You need to be hypnotized :)
part of the problem i have with this line of conversation, though, is that while i can talk about the anxiety, i really can't explain it. it's not something whose dynamics happen at the level of arguments and logic.
I know. I was suggesting an irrational solution :)
Anyhow, just know that we care about you and wish you the best and don't wish to make it worse.
thanks, i appreciate it.
it doesn't help that, while i know my adviser tries to be supportive, i don't feel like he gets how utterly overwhelming it gets for me
01:24
I would suggest talking to a counselor, but I don't want to get personal in public here.
yeah, i understand.
and i know that advisor-advisee relations being difficult isn't exactly an unusual thing
Some advisers are schmucks, but usually they want nothing but the best for their protégés.
oh, i don't doubt it
OK, I have cooking to do. See you later.
01:32
What is your research gonna be on?
GGG
GGG
Here's what I was referring to: let $\theta = \arctan{t}$

$\displaystyle \begin{aligned}I & = \int_0^{\frac{\pi}{2}} \frac{1}{1+\tan^x (\theta)} d\theta \\& = \int_{0}^{\infty}\frac{1}{(t^2+1 )( t^x+1)}\;{dt} \\& = \int_{0}^{1}\frac{1}{(t^2+1)( t^x+1)}\;{dt}+ \int_{1}^{\infty}\frac{1}{(t^2+1 )( t^x+1)}\;{dt}\end{aligned}$

Let $t \mapsto 1/t$ for the second integral then:

$\displaystyle \begin{aligned} I & = \int_{0}^{1}\frac{1}{(t^2+1)( {t^x}+1)}\;{dt}+\int_{0} ^{1}\frac{1}{(t^2+1)(1/t^x+1)}\;{d t} \\& = \int_{0}^{1}\frac{t^x+1}{(t^2+1) (t^x+1)}\;{dt} \\&= \int_{0}^{1} \frac{1}{t^2
what did the upper bound become infinity in the second step?
$\tan\pi/2=\infty$
Ok where did the second integral come in in the third step?
what i did was just $\theta\to\pi/2-\theta$ and recognized that the integrand was mapped to one minus itself
01:38
OH they are the same, nvm my bad
averaging those two gives $\int 1/2\,d\theta=\pi/4$
GGG
GGG
@Michael write the integral from splitting the integration from $[0,1]$ and $[1, \infty]$.
@GGG yah ty
GGG
GGG
@Semiclassical I think these two approaches are equivalent, though I can't explain it. :p
What is t↦1/t
01:39
oh, i agree
it's because $1/t = \cot\theta =\tan(\pi/2-\theta)$
GGG
GGG
@Michael sub $u = 1/t$, then $u$ becomes dummy variable. So you change it back to $t$. That's why I wrote $t \mapsto 1/t$.
@Semiclassical Ah, thanks.
oh
whats a dummy variable
GGG
GGG
@Michael $\displaystyle \int_{0}^{1} \frac{1}{(u^2+1)(1/u^x+1)}\;{du} = \int_{0}^{1} \frac{1}{(t^2+1)(1/t^x+1)}\;{dt}$. It doesn't matter whether you denote it $u$ or $t$ or $x$ or whatever you still get the same value. That's what a dummy variable is. So we change it back to $t$ so that we can add the two integrals conveniently.
oh kk ty
I get it
hi
how is inf(R_+) = 0 if 0 is not in R_+?
01:52
inf does not equal min.
but i thought infimum must be in the set
You should read your textbooks definition of inf and sup.
I am reading wikipedias definition
it is saying a subset S of a partially ordered set T
what does that mean?
Perhaps this is why you should use the definition given to you in the same source that the exercise was.
Wikipedia is being more general than you need, there.
Just know that $\Bbb R$ is an example of a partially ordered set.
(If you're curious, a partially ordered set is a set together with an order $\preceq$ satisfying several axioms that we expect an ordering to have. The "partial" bit means that it's possible for us to have neither $x\preceq y$ nor $y\preceq x$.)
01:58
I am wondering how can the infimum not be an element of the set then as in how is inf(R_+) = 0 if 0 is not in R_+?
The infimum of a set is its greatest lower bound. Do you remember what a lower bound is?
(Also, out of curiosity, do you have LaTeX installed on this chat room? Like, do you see dollar signs everywhere?)
Ah, how do i install?
The relevant bit is the "start ChatJax" bit
In any case, that's not terribly important. A lower bound of a set is a number that's lower than anything in the set.
For example, 0 is a lower bound of R_+. So is -1. And so is -100.
In fact, for your set, every non-positive number is a lower bound for R_+ !
The infimum is defined to be the greatest lower bound of the set.
0 is the greatest lower bound of R_+. All other lower bounds are smaller than it.
Thus, 0 is the infimum.
Note that nowhere do we require 0 to be in R_+.
I just thought it might since it is in a "subset S of a partially ordered set T"
Here S is R_+ and T is R.
02:05
then how is 0 in S
0 isn't in S.
It's in T, though.
ah
i see
yes i misunderstood
If T is a partially ordered set (basically, if it has an ordering relation on it, like "<"), and S is a subset of T,
then the infimum of S (if it exists) should lie in T.
we could also have S = T
right?
Right. Then, the infimum of S would be the lowest element of T (if it exists).
For example, if T=[1,2], the set of numbers from 1 to 2 inclusive,
and S=T,
then inf(S)=1.
02:08
how do you now what S and T are if they aren't stated
Generally, T is the set of real numbers (written "R").
You probably won't need anything more general.
for example in this question i ask math.stackexchange.com/questions/1667916/…
user AJY says $\sup \{ f(x) : x \in \mathbb{R} \} = \max ( \sup \{ f(x) : - N \leq x \leq N \}$
That's not the full equation s/he wrote
It was:$$\sup \{ f(x) : x \in \mathbb{R} \} = \max ( \sup \{ f(x) : - N \leq x \leq N \} , \sup \{ f(x) : |x| > N \})$$
I.E. The left hand side is equal to the larger of $\sup \{ f(x) : - N \leq x \leq N \}$ and $\sup \{ f(x) : |x| > N \}$
02:12
Right, but by sup I believe AJY meant max because if not then it wouldn't be in f(x)
Sup can be in the set. It isn't always in the set, but it can be.
he said he used suprememum instead of max in case the function didn't have a bound
Weren't you trying to prove that the function attained its maximum?
yep that's righ
At that point in the proof, you hadn't yet proven that it attained it's maximum
02:13
yes
So AJY wrote sup instead of max, because s/he hadn't proven it had a max yet.
I'm saying let's say supremum doesn't lie in f(x), then what have we shown
By f(x), you mean the set $\{f(x):x\in\Bbb R\}$?
Do you know the Extreme Value Theorem?
He uses it in his proof
02:16
yes, states that on a compact set a continuous function obtains a min and max
0
Q: Proof of existence and uniqueness of the exponential function using ODEs

Jessy CatIn our lecture notes for our complex analysis class, we were given the following theorem: Theorem: There exists a unique complex function $f$ such that $f(z)$ is a single valued function $f(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ and $f(1)=e$. $\forall z_{1},z_{2} \in \mathb...

So that's how he knows $\{f(x):-1\le x\le1\}$ and $\{f(x):-N\le x\le N\}$ have maximums. Because the sets $[-1,1]$ and $[-N,N]$ are compact.
If anybody in here can help, please do. I only need help with part (a).
He defines $M$ to be the maximum value of $f$ on $[-N,N]$.
And, because of the way he chose $N$, he knows that $f$ never gets a value bigger than $M$ on the set $(-\infty,-N)\cup(N,\infty)$ (i.e. the rest of the reals).
So, what do we know?
We know that $M$ is an upper bound (because it's an upper bound both on $[-N,N]$ and $(-\infty,-N)\cup(N,\infty)$).
And we know that nothing smaller than $M$ is an upper bound, because $f$ attains the value $M$.
Thus, $M$ is the least upper bound; that is, it's the supremum.
And since we already know $f$ attains the value $M$ (for some $x$ in $[-N,N]$), we get that $M$ is the maximum, as well.
So, $f$ attains its maximum.
02:35
@JessyCat If $f$ is complex differentiable , what can you say about lim f(z_n)-f(z)/(z_n-z) for any sequence where $z_n$ approaches $z$.
 
1 hour later…
03:40
hi
can anything in math be traced back to the axioms?
03:55
Assuming you have a good enough set of axioms, sure
I need help again:
Just please no proofs using either taylor sums or logarithms.
 
1 hour later…
05:08
Hi @robjohn
3
Q: A problem on real valued functions in $\mathbb{R}^2$ with least variation

Rajesh DachirajuLet $\alpha(s) = (x(s),y(s))$ be the arc length parametrization of a plane, smooth, closed, convex curve, of length $L$. Let $J:(0,L)\to\mathbb{R}$ be a smooth and Bounded variation (BV introduced after comment from Dirk of possibility of blow up at 0 or L) function. Let $\Omega$ be the set of al...

05:36
Is there a continuously differentiable function $f:\mathbb{R}\to\mathbb{R}$ such that $\exists x,f'(x)=0$ and $f$ is either increasing or decreasing on both sides of $x$, yet there is no inflection at $x$?
GGG
GGG
In trying to find the inverse of $[1066]_{2009}$ I've found that $1 = 315 \cdot 2009-346\cdot1829$. What do I do next?
[cont.] Suspicion is no, but I do feel like I'm brainfarting.
05:49
@brody not sure it helps but: defining $g(x)=f'(x)$, that'd amount to finding $g(x)$ such that $g(0)=0$ (picking $x=0$ WLOG) and $g(x)>0$ for $0<|x|<\epsilon$ but for which $g'(0)\neq 0$
Morning
morning, mike
@Semiclassical Thanks for the response. Trying to conjure such an example, if there is one. :)
i think i've an example, but i'm always paranoid myself that i'm forgetting a detail in these counterexamples
GGG
GGG
Oh actually there's nothing left to do at all. $$[1]_{2009} = [315\cdot2009-346\cdot1829]_{2009} = [-346 \cdot 1829+2009(1829)]_{2009} = [1663 \cdot 1829]_{2009}$$
05:55
@Brody No use Taylor's theorem (assuming $f''(x) \ne 0$) to approximate $f$.
GGG
GGG
But there MUST be a way to find modular inverse without using Euclidean algorithm. I'd appreciate if someone could link me to it, 'cause I can't find it.
I'm having a bad case of brain flatulence. It just blew a wet one: try $g(x)=|x|$?
nvm
I'd assume you meant $f$ was at least $C^2$.
If you are talking about points of inflection.
$|x|$ was my thought too. integrating, it gives $f(x)=x^2\,\text{sgn}(x)$
I believe it meets your criteria @Semiclassical but its primitive (which I believe is $f(x)=\sgn(x)x^2$) changes concavity at $x=0$, but I want otherwise.
05:59
(aside from a factor of 1/2 which is irrelevant here)
depends on how you mean inflection, i guess

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