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00:57
@XanderHenderson I've been using $$f:\begin{cases}\mathbb R\to \mathbb R\\x\mapsto x^2\end{cases}$$ but I don't know how widespread this notation is
@Astyx I have seen $$\begin{align} f :{}& \mathbb{R} \to \mathbb{R} \\ &x \mapsto x^2,\end{align}$$ which seems to be basically the same idea.
Yeah that's also what I've seen, which motivated my question
I just got notified that starting in an hour, the site will be read only for 3 hours.
This includes chat
I like/use Xander's notation.
01:27
@robjohn Oh n...o and i'm just starting my Lin Alg block ......😱
@D.C.theIII Hopefully, you can postpone until 5 UTC (midnight EST, 9 PM PST)
I'll manage to survive.........
somehow, someway.....
@robjohn it was announced in banners. Rejoice! You can watch a good movie!
02:21
managed to drop my phone and break the screen.
Bad @copper!
@copper.hat :(
i am receiving admonistments from a broad variety of sources....
however, i did want to replace my phone, just not tonght
i replaced my phone the other week. first time in 7+ years.
02:43
wow!! that's cool.
I'll try to keep mine for minimum 10 years.
I am surprised that chat is still working.
Suppose that f and g are both bijective functions (domain, ranges are subsets of R). Suppose that $\lim_{x\to a} f(x)/g(x)=1$. What can be said about $\lim_{x\to a}\frac{f^{-1}(x)}{g^{-1}(x)}$?
Do we know anything about $f(a)$ and $g(a)$?
let's take a to be infinity so that this question does not appear.
This is to understand the following step: Proving the prime no. theorem i.e. $\pi(x)\sim \frac{x}{\log x}$ is equivalent to proving $p_n\sim n \log n$.
,where $p_n$ is the nth prime. $h\sim g$ means that $\lim_{x\to \infty}\frac{h(x)}{g(x)}=1$.
(this seems to suggest that the answer to my question is 1. This is so because $\pi(p_n)=n$ so $p_n$ (as a function of n) can be regarded as 'the inverse' of $\pi(x)$(as the function of x). )
 
1 hour later…
04:01
@robjohn We must be hallucinating.
We can't be talking since the site is down.
I’m not talking; are you?
No, why didn't you ask?
looks skyward and inquires
why does my hallucinatory experience have ted shifrin in it? is this a "bad trip"?
04:16
@leslietownes Just don't try to fly, even though you know you can.
hi, i'm not here
πŸ‘€
3
Leslie will do his Yellow Submarine impersonation.
Is the approximation $\pi(x)\gt c\frac{x}{\log x}$ true for all $x\ge 2$ or only for large x?
The proof seems to be proving it for large x and then saying that it is true for all $x\ge 2$.
This is proposition 2.4.4 in Ireland and Rosen's.
I found the answer here: math.stackexchange.com/a/1890792
This indeed proves for $n$ but not for any x (I think): Substituting $n=[x/2]$, we get $\pi(x)\ge \pi(2n)\ge \log 2\frac{n}{\log x}\ge \frac{\log 2}2\frac{x-2}{\log x}$
= $c\frac x{\log x}-\frac{\log 2}{\log x}$, where $c=\frac {\log 2}2$
04:50
If you can prove this for large enough $x$, cannot you then obtain that it is valid for $x\ge2$ simply by modifying the constant $c$?
on RHS we can modify the constant, but inside $\pi(x)$?
At least I understand the claim as: there exists a constant $c$ such that...
@Koro I don't really follow what you mean by "inside $\pi(x)$"
yes. c should be valid for all x>1
All I' saying is this.
Claim 1. There exist $x_0$ and $c_1$ such that $\pi(x)>c_1\frac{x}{\ln x}$ for each $x\ge x_0$.
Claim 2. There exist a constant $c_2$ such that $\pi(x)>c_2\frac{x}{\ln x}$ for each $x\ge 2$.
To me it seems that Claim 1 and Claim 2 are equivalent.
In fact, I would add another formulation which seems equivalent.
Claim 3. There exist $n_0$ and $c_3$ such that $\pi(n)>c_3\frac{n}{\ln n}$ for each integer $n\ge n_0$.
I mean that suppose we have $\pi(x)>cx/{\ln x}$ for all $x\ge K$. K>0 is some number. Then set $x=yK/2$ so that we have $\pi(yK/2)>c' x/{\ln x}$
Claim 3 is true for all $n\ge 2$ as the linked answer shows.
04:56
@Koro assuming the behavior of $\pi(x) $ the behavior of $p_n$ is deduced easily using limit laws and $(\log x) /x\to 0$ as $x\to\infty$. I think the reverse implication should also be not too difficult
so in my last message, I want to simplify $\pi(yK/2)$.
Well, naturally all these claims are true - they are consequences of PNT.
I'm not at PNT yet. The said bounds will be used later in the proof of PNT.
But I think that with some work each of them can be shown to imply the other ones - without relying on PNT.
In any case, it's time for me to take the shower and then commute to work.
See you later - and have a nice day!
But I think $pi(yK/2)$ should be easily related to $\pi(y)$ by a multiplicative constant.
See you later. Have a nice day! Thanks. @MartinSleziak
I was also thinking about 'scaling' down x to prove it for all $x\ge 2$ but this $\pi(yK/2)$ part confused me. But it seems doable now. I'll try to solve this part.
@ParamanandSingh But for this to be true, shouldn't we also need a definite answer to this:
2 hours ago, by Koro
Suppose that f and g are both bijective functions (domain, ranges are subsets of R). Suppose that $\lim_{x\to a} f(x)/g(x)=1$. What can be said about $\lim_{x\to a}\frac{f^{-1}(x)}{g^{-1}(x)}$?
for definiteness, we suppose $a=\infty$ to answer the above question.
05:25
Not necessarily, I understand how $p_n$ looks like an inverse to $\pi(x) $ but the prime counting function is not a bijection
Is there a name for a function that has linearity-like behavior when some or all of its inputs are modified together? So perhaps $f(ax,\ ay,\ z)\ =\ af(x,\ y,\ z)$ and similar for superposition.
Bilinear, multilinear, semilinear don't seem to describe it.
Homogeneous of degree 1 in certain of the variables.
That wouldn't incorporate the superposition though, right?
I have no idea what you’re talking about.
Linearity = Superposition + Homogeneity of degree 1? Let me think about how to write it in functional form for the several variables example.
05:50
Okay I guess the overall statement including both superposition and homogeneity would be $f(ax + bv, ay + bw, z) = af(x, y, z) + bf(v, w, z)$...perhaps.
Bit harder to think about than regular linearity.
Write vectors as $(x,z)$, with $x\in \Bbb R^2$. It’s linear in $x$.
Okay, I was thinking that wasn't sure if there was a fancier term.
I’ve never seen this precise notion before. Sesquilinear, sure.
06:17
No. of primes is infinite: Consider the numbers $f(n):= 2^{2^n}+1$. $(f(n),f(m))=1$ for $m\ne n$. This gives us infinitely many pairs of relatively prime nos. It follows that no. of primes is infinite.
Is this the proof by G.Polya?
@ParamanandSingh true, $\pi(x)$ is not a bijection. But then how do we get the equivalence? Given $\pi(x)\sim x/{\ln x}$, it follows that $n\ln p_n\sim p_n$.
ahh we do have: $n\ln p_n \sim n\ln n$
hmm, it makes sense now. @ParamanandSingh
thanks a lot :).
 
3 hours later…
09:29
Let's say I have f(t,y) = tsin(y). I know that y=3t. Then the partial derivative of f respect to y is tcos(y) or 3tcos(y)?
 
2 hours later…
11:10
Is $e^x$ quasi concave?
yes it is
11:45
How to find an example of a group which has 2 elements of order 2? I am new to group theory and could not find any example :( I modified Z_n, U(n) but didn't work
12:00
when does equality hold in the following version of Holder's inequality?
$|\sum_n x_ny_n|\le \|(x_n)\|_1 \|(y_n)\|_\infty$.
12:39
When $y_n$ is constant and $x_n$ all have the same sign.
13:05
thanks :).
13:15
Is there an example of a Banach space that is not isomorphic (with norm preservation) to its dual?
13:47
@Koro $\ell^1$?
$\ell^\infty$?
Indeed, most Banach spaces are not self-dual.
Being isometric to your dual is a pretty special property.
Off the top of my head, I can't think of any example which is not a Hilbert space.
Or am I misunderstanding your question?
Whoa... you can do this: math.stackexchange.com/questions/65609
On the main site: Isometric to Dual implies Hilbertable? Well, you were faster.
@MartinSleziak I just linked to that. :D
Ninja'd!
:P
*smoke bomb*
that'll do. Thanks a lot both :-).
dual of $l_1$ is l infinity.
BTW if some moderator is bored, this comment could be edited so that $X^{**}$ is rendered properly:
@Theo: well, you took the time to give more detail, so you deserve the points! Tangentially related: I believe it is an open problem whether $X$ and $X^{\ast\ast}$ are necessarily isomorphic whenever $X$ and $X^{\ast\ast}$ are isomorphic to complemented subspaces of one another. Meanwhile, it is known that $X$ and $X^{\ast\ast}$ needn't be isomorphic if $X^{\ast\ast}$ is isomorphic to a complemented subspace of $X$. — Philip Brooker Sep 18, 2011 at 22:39
13:59
and dual of l_3 is $l_p, p=3/2$.
@Koro You fixed it.
:-)
@MartinSleziak I don't know what you're talking about.
Also, do these gas lamps look dim to you?
@XanderHenderson those red marks look like they need fixing.
@Koro Again, I don't know what you are talking about...
14:02
thanks :-)
 
1 hour later…
15:10
So to define linear functionals on a finite dimensional spaces, I just have to define a map only on a basis and extend linearly.
But what if the space is infinite dimensional having a Schauder basis?
I think it still is true that a linear functional is completely determined by its values on a Schauder basis.
Let me try to prove this: Let $e_n$'s be a Schauder basis of some space X. Let $f$ be a functional on X. Suppose there is another functional $g$ such that $g(e_i)=f(e_i)$ for all i. I want to now show that $f=g$. To that effect, let $x\in X$ be arbitrary. There exists a sequence of scalars $x_n$ such that $x= \lim_{n\to \infty}(\sum_{j=1}^n x_n e_n)$.
Let's suppose that $f,g\in X'$. It follows that f and g are continuous. So $f(x)=\lim_{n\to \infty}(\sum_{j=1}^n x_n f(e_n))= \lim_{n\to \infty}(\sum_{j=1}^n x_n g(e_n))=g(x)$.
This is true for every x in X. This proves the statement for all linear functionals in the dual of X.
Hence the following theorem: For every normed space X with a Schauder basis, every element of the dual of X i.e., X' is completely determined by its values on a Schauder basis of X.
15:40
how to show that $l_p$ and $l_q$ where q=(q-1)p are not isomorphic?
or that $l_1$ is not isomorphic to $l_\infty$.
@Koro Why are you making that restriction? And do you mean isometric?
In general, if $p \ne q$, then $\ell^p$ is not isometric to $\ell^q$.
yes, isometric.
I looked it up and the proof seems complicated.
 
1 hour later…
17:05
Is it true that for a complex line bundle $L$, its first chern class $c_1(L)$ vanishes iff $L$ is trivial?
smoothly, yes
holomorphically, no
alright, I'm only interested in the smooth case
do you happen to know the argument?
or a reference
I think there's multiple ways to argue depending on where you come from. Here's one: complex line bundles on, say, a paracompact space $X$ are classified by homotopy classes maps $X\rightarrow B\mathbb{C}^{\ast}\simeq BU(1)=BS^1$, but $\mathbb{CP}^{\infty}$ is a model for $BU(1)$ and also a model for $K(\mathbb{Z},2)$, so these homotopy classes are naturally identified with $H^2(X;\mathbb{Z})$. Thus, such a line bundle is trivial iff the corresponding element in $H^2$ is trivial.
It remains to know that the corresponding element in $H^2$ is the first Chern class. As far as I'm concerned, that can be taken to be the definition, but you might want to use another one.
17:21
oh hm, for me the first chern class is (the cohomology class of) the trace of the (local) curvature matrix-valued forms corresponding to a connection
don't know if you're familiar with Tu's diffgeo book, but basically all I know about characteristic classes is the very brief treatment in Tu's book
If $X$ is nice enough (paracompact and locally contractible, perhaps), you can also use the exponential sequence $0\rightarrow\underline{\mathbb{Z}}\rightarrow\underline{\mathbb{C}}\rightarrow\underline{\mathbb{C}^{\ast}}\rightarrow0$ (SES of constant sheaves), which yields a LES in sheaf cohomology $H^1(X,\underline{\mathbb{C}})\rightarrow H^1(X,\underline{\mathbb{C}^{\ast}})\rightarrow H^2(X,\underline{\mathbb{Z}})$. The middle group is identified with isomorphism classes of complex line bundles by Cech theory, the latter group is identified is identified with $H^1(X,\mathbb{Z})$ by gener
@ShaVuklia yeah, that's a terrible definition from an algebro-topological standpoint or for any classification theory
I think you'll have to first invoke Chern-Weil theory to translate this into reasonable alg top
I mean, there's a uniqueness theorem for characteristic classes in the appropriate sense (following from classification theory), so I guess you can skirt around it
Hi everyone,
Is there any conclusion I can draw about this function $\dot{V}$? Unfortunately, I'm not getting anywhere with checking if the function is positive definite or at least positive semi-definite.
B is positive definite for sure.
but what about A?
I should specify that all I've said was in the continuous setting. But, if $X$ is a smooth manifold, topologically trivial and smoothly trivial are the same by some genericity arguments.
(this is different from holomorphic triviality if $X$ is a complex manifold, which is much stronger than topological triviality)
17:37
yea I have to admit that little you wrote really rang a bell (except that I recognised some algtopo notation). I guess I'll just have to have look at the more algtopo-oriented definitions at some point, but I'll do that after my diffgeo exam
my question originated from a) in the exercise above, but looking at b), it doesn't seem like (according to the definitions we use) it holds that $c_1(L)=0$ iff $L$ is trivial
otherwise that b) would have been phrased differently
oh, I see one potential point of friction
you're doing differential forms, so you're only defining the image of the Chern class in $H^2(X,\mathbb{R})$, not the "true" Chern class that lives in $H^2(X,\mathbb{Z})$
the latter can be non-trivial even when the former is trivial
perhaps the phrasing in b) is to address this torsion phenomenon, but I don't really know how this stuff works on the differential geometry end
will have to wait for Ted to come around
hm, we do get an integral differential form (according to Tu)
since Tu adds a factor $i/2\pi$
I wonder if being integral (a $k$-form is integral is whenever you integrate over a compact oriented $k$-submanifold you get an integer) has to do with this $\mathbb Z$-cohomology group, I assume it does
17:53
it's probably equivalent to lying in the image of $H^k(X,\mathbb{Z})\rightarrow H^k(X,\mathbb{R})$
but the issue is that this homomorphism is not injective
this characterization of being in the image is not quite trivial either, tho
the de Rham theorem tells you it's in the image if you get an integer when integrating over all smooth $k$-cycles
@Thorgott Good to know that these notions are not equivalent. I felt quite confused in the past few hours as I was trying to find the statement on stack
but I don't see immediately why it suffices to integrate over compact oriented $k$-submanifolds
5
Q: A complex line bundle is trivial if and only if the first Chern class is zero

ShiquanLet $\xi$ be a complex line bundle over a CW-complex $B$. I want to prove that $\xi$ is trivial if and only if $c_1(\xi)=0$. My attempt: Suppose $c_1(\xi)=0$. Then the Euler class $e(\xi)=0$. Since $e(\xi)=o_2(\xi)$, there exists a nonvanishing cross section of $\xi|_{sk^2 (B)}$, denoted as $X\...

this is similar to what I said earlier (slightly less general)
speak of the devil, hi @Ted
xD hi Ted
18:05
@CroCo Have you tried ChatJax?
@robjohn not really. What's it?
In the room description, there is a link after "$\LaTeX$ in chat". That leads to an installation page for ChatJax. It allows your browser to render the MathJax in chat.
It is far preferable to post MathJax rather than images in chat
Images have their place here too, just not in a formal symbolic mathematical sense.
@robjohn It's always a mystery to me how you guys manage to read the latex format in the chat. I've installed it and I can see now. :)
There was a place where I thought it had been copied and pasted.
18:21
@CroCo Glad to help. I still think it would be better if SE incorporated something like this without needing outside helpers like ChatJax.
totally agree.
@robjohn The counter-argument has long been that most SE sites don't use MathJax, and that loading MathJax increases page load times and local resource use. So they don't want to implement, since it will annoy most users, while only benefiting a small number of us.
There is always trouble with mathematicians. :)
@XanderHenderson they don't need to load MathJax, They could provide a link to something like ChatJax or better. That way, the onus is on the browser and it does not affect the SE page load time.
It could even be a room option so that there is not necessarily a link that is never used in a room that never uses MathJax.
or it could be provided at all times as a room option.
Many variations to deal with that problem.
@robjohn You don't have to convince me.
18:30
Preaching to the choir :P
I have tried talking to the devs, but to no avail. Perhaps the set of devs has changed and I should try again.
This chat seems very basic compared to other websites. Uploading a picture and typing at the same time is not possible for me.
I think there's been a change in the devs.
At least in stack overflow.
Hi :) I have an idea for a question. There's a bunch of different, equivalent definitions of linear algebraic groups out there. It'd be helpful to have them in one place, hence the question title would be something like, "What are the definitions of linear algebraic groups? Let's list them!" I want to mention it here, though, in case I have missed an obvious place to find them all.
Does someone have time to look at my question?
0
Q: Proof that if $(X,Z)\stackrel{d}{=}(Y,Z)$ then $X\stackrel{d}{=}Y$

user1294729 Let $X,Y,Z$ be random variables on $\Bbb{R}$ such that $(X,Z)\stackrel{d}{=}(Y,Z)$. I want to show that $X\stackrel{d}{=}Y$. My idea was the following. Let me take any $A\subset \Bbb{R}$ then $$\Bbb{P}(X\in A)=\Bbb{P}((X,Z)\in A\times \Bbb{R})\stackrel{*}{=}\Bbb{P}((Y,Z)\in A\times \Bbb{R})=\Bb...

18:36
@robjohn I would appreciate it if you could take a look at my question. Please let me know if you would like me to type it here using latex format.
@Shaun I would suggest a thorough googling before doing that.
2
@Thorgott What did I do this time?
we need a differential geometer to resolve Sha's question
Warning: When you talk about triviality above, you're talking as a continuous complex line bundle, not as a holomorphic line bundle. $c_1$ detects only topological. But, yeah, in non-simply connected spaces, a flat bundle needn't be globally trivial because of monodromy.
is the monodromy what detects whether $c_1$ is trivial in the kernel of $H^2(X,\mathbb{Z})\rightarrow H^2(X,\mathbb{R})$, then?
Yeah. Such a bundle will be torsion.
18:42
@CroCo I cannot really tell what is going on in that question. You mention $V(\cdot)$, but what is the argument?
very nice, so everything does fit together in the end
So, yes, if $c_1(L) = 0$ and we're on a surface, then Euler class tells us that there's a nowhere zero smooth section. The line bundle may still be holomorphically nontrivial. Hence the Picard variety.
so, to summarize:
the real Chern class is 0 iff there is a flat connection
the integral Chern class is 0 iff the bundle is trivial (topologically)
if the real Chern class is 0, the integral Chern class is torsion and it is 0 iff there is a flat connection with trivial monodromy
@TedShifrin right, I brought up how to see this difference in the continuous and holomorphic exponential sequence above
Hey everyone
Heya Demonark
18:46
How's everything going?
@robjohn My hope is to show $\dot{V}(\cdot)$ is negative definite to prove the stability of the control signal but I was stuck at the term $(JJ^T-I)$. Is there any conclusion I can draw from this in terms of definiteness?
this is what I've done
\begin{align*}
y^TAy &\geq 0, \\
y^T(JJ^T-I)y &\geq 0, \\
y^T JJ^T y - y^Ty&\geq 0, \\
(J^Ty)^T (J^T y) - y^Ty&\geq 0, \\
\|J^Ty\|^2 - \|y\|^2 &\geq 0, \\
\|J^Ty\|^2 &\geq \|y\|^2.
\end{align*}
This is all I can see.
@XanderHenderson What this time?
> It may seem pedantic, but this is mathematics.
Them are fightin' words :P
@TedShifrin I'll redouble my efforts.
When does something pedagogic become pedantic?
19:00
Generally, IMHO, pedantry rarely makes good pedagogy .
Nit picking over the details rarely helps learning.
off to another appointment. Is there life between appointments?
Cya pal.
Hello!
19:11
I have arrived to whine about trig and play Wordle variants, and I’m never out of Wordle variants.
@TedShifrin Someone doesn't know how to read your book.
I assume that it is their problem, not yours. But it is more fun to poke you.
But then again, removing too much chocolate from the broccoli isn't that appetizing either @TedShifrin
@robjohn Good luck!
… what hath I missed?
@XanderHenderson Not mine, unless I don’t know what you’re talking about.
I got today’s Wordle in 2. French Wordle not for several hours.
19:15
got it in five
I for one am a big pedantry proponent
would’ve been four if it weren’t for ANAGRAMS (annoyed parz noises)
1
Q: Is Ted Shifrin's definition (8-2.3) missing the verbalization of the pullback of the product of a function with a basis?

Steven Thomas HattonMy question pertains to Ted Shifrin's Multivariable Mathematics Section 8-2.3 Pullbacks. We are to use the following pieces, given in the definition, to build the pullback $\mathbf{g}^{*}\omega\in\mathcal{A}^{k}\left(U\right)$ of $\omega$ by $\mathbf{g}.$ The symbol $\mathbb{I}$ is a multi-index....

nice of him to link to the publisher's site, and not the free copy available at greatbooks dot library dot ebiz dot ru
I think this is a fair question and it has an even better answer.
19:32
Yeah, I didn’t comment explicitly that pullback is an algebra map.
It’s fair.
19:42
Question about interpretation of the question: SHould I be only concerned with the $x$ that are in the basis only or should I be thinking about all $x$ in the vector space?
Whoops
I totally forgot about wordle
Gaussian elimination is a pain
dc: in the hypothesis? assume only for x in the basis
and the question is indeed, does knowing that allow you to get the same thing for all x
@Gokuγ‚«γ‚«γƒ­γƒƒγƒˆ Oh?
@leslietownes Thanks Leslie. and the answer is no. Counterexample: $U(1,0) = (1,0), U(0,1) = (1,0)$ Both basis vectors satisfy the hypothesis, but the matrix will not be unitary
19:56
In order to prove that an equation only has one unique solution, all I need to do is prove its monotonicity right?
@D.C.theIII What if we say $U$ is injective?
dc: good example.
ooh, ted is making it more interesting.
@TedShifrin Just going off my counterexample, though that may not be enough. I would say that it is unitary.
BUt at least to get the ball rolling in the right direction
calling a vector space $V$ and an operator on it $U$ should be criminalized
why THor?
20:02
Well, $T$ is only one letter over.
@TedShifrin thanks. This appt is for my wife. We decided to have cancer surgeries on top of each other. Hers in December and mine in January.
The family that surgers together recovers together?
I hope you're both doing much better.
I remember my appendix removal surgery from last year. Wasn't as bad as I thought
thorgott: how about $V: U \to U$...
But $T\colon U\to U$ is fine :D
20:07
I'm late to hearing the news, but may my well wishes to and your wife be well received @robjohn
and somehow in differential geometry, lowercase x gets to be a map. what a world.
What is this inside math joke about $U$ and $V$ ?
@leslietownes $U : v \to u$.
dc: no inside joke, just visceral reaction to notation choices. it is not too uncommon for uppercase U and V to be used at the same time for things of the same 'type'. thorgott doesn't want U to be a map on something called V.
imagine reversing the roles of f and z in cauchy's integral formula for f(z). and the games you could play with the variable of integration.
I've seen the most common use of u and v together for integration
20:11
One of my favorite colleagues at UGA used to talk about a faculty member (before my time) who gave his calculus class test questions to find $dx/df$.
@TedShifrin Heh.
But that's mean. :D
makes sense....some authors just enjoy being disruptors to thetraditions of mathematical academia
A girl in my calculus class throws all of that out the window by using hearts as "substitution" for calculus.....
it's bad enough in calc 3 when you have regions in the xy plane and there's sometimes actually a good reason to think of x as a function of y
Well, I hate people who write $df/dg$ in the chain rule. (For excellent reason.)
20:12
@TedShifrin That's a good idea for maybe one question, just to bang home the idea of not getting accustomed to notation in a certain way.
@Gokuγ‚«γ‚«γƒ­γƒƒγƒˆ I often teach substitution / change of variables using smiley faces.
Sure, it's ok to have $y=f(x)$ and $x=g(y)$.
Because I get really tired of students telling me "Oh, just do a $u$-sub!"
@leslietownes Step one: relabel the axes. :P
that comes up a lot with integration by parts. sometimes you've done a u-sub already and don't get to choose what "u" is.
@XanderHenderson you should've been her teacher then
20:15
It's funny how certain things teachers do stick in students' minds. My students from 15+ years ago still refer to "script Q" (which I used to denote quadratic forms) with affection.
@leslietownes that's when I do $U_2$ or something like that
 
1 hour later…
22:07
Hey, @Ted!
@XanderHenderson Joshua Tree is my wife's and my favorite album to play on our way to Mammoth, but we wait until we hit the altitude where you can see Joshua trees.
@robjohn Is there an easy way to get Mathematica to shade between two graphs?
The Filling setting might be what you want.
Sorta involved, though. Thanks.
@TedShifrin It can be, but depending on what you want, it might not be too bad.
I miss my Illustrator. πŸ™πŸ™„
22:31
good news everyone
im back
lol
i just wanted to sound like that prof from futurama :)
anyway here is a new question from me.
0
Q: Roots and analytic continuation of $T(s)=\sum_{n>0} (n^s + n^{-s})^{-1} $?

mickLet $s$ be a complex number. $$T(s)=\sum_{n>0} (n^s + n^{-s})^{-1} $$ This is well defined for $Re(s) > 1$. It seems $T(s) = T(-s)$ but then again we have it only defined for $Re(s)>1$ for now. We try analytic continuation by $$\sum_{n>0} (n^s + n^{-s})^{-1} - n^{-s} = - \sum_{n>0} (n^{3s} + n^{...

Hi Rob. hope you are feeling well.
22:57
and another one
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Q: $f(x) = \sum_{n=1}^{\infty}(-1)^n \frac{x^{2n-1}}{(2n)! \ln 2n}$ is totally monotone?

mickI think $f(x) = \sum_{n=1}^{\infty}(-1)^n \frac{x^{2n-1}}{(2n)! \ln 2n}$ is totally monotone. Bernstein's theorem and his integral transform convinced me. Am I correct ? Probably related or useful is a function like $f(x,k) = \sum_{n=1}^{\infty}(-1)^n \frac{x^{2n-1}}{(2n)! \ln (2n+k)}$ which foll...

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Q: Suppose $A$ and $B$ are diagonalizable matrices. Prove or disprove that $A$ is similar to $B$ iff $A$ and $B$ are unitarily equivalent. Alt-Solution?

D.C. the IIISuppose $A$ and $B$ are diagonalizable matrices. Prove or disprove that $A$ is similar to $B$ iff $A$ and $B$ are unitarily equivalent. Is there a way to disprove this without using the idea of showing: $\Sigma_{i,j}^n|A_{i,j}|^2 = \Sigma_{i,j}^n|B_{i,j}|^2$? I ask because the question that esta...

mm, any invertible diagonal matrix is similar to I, but only one diagonal matrix is unitarily equivalent to I.
you could write out specific matrices and a similarity transformation if you wanted to. the nonexistence of a unitary implementing an equivalence between I and something that isn't I should be pretty clear.
@leslietownes HUH?
The only matrix similar to $I$ is $I$.
okay, maybe not I, but the eigenvalues are unitary invariants, the nonzeroness isn't.
Think about similarity in terms of change if basis, DC.
23:10
That's what I was fiddling with right now.
What does unitary equivalence imply about the bases?
That's also the question I'm asking myself and getting stuck on ...lol
the eigenspaces of a diagonal matrix are orthogonal to one another.
well what I recall is about this but I don't think it directly pertains to this is that the bases is orthonormal
@leslietownes But not those of a general diagonalizable one!
I think in this case it’s actually less confusing to think about linear maps than about matrices.
23:16
@TedShifrin sure.
ok, well unitary equivalence in that sense means that if $\beta$ is an orthonormal basis for $V$ then $T(\beta)$ is also an orthonormal basis.
can you find a 2x2 with distinct nonzero eigenvalues whose eigenspaces are not orthogonal to one another?
ted's right, thinking in terms of matrices isn't much of a help
23:45
So for $A$ and $B$ to be unitarily equivalent there must exist some unitary matrix $P$ such that $A = P^*BP$. In what I read again the discussion talked about this in terms of unitarily equivalent to a diagonal matrix $D$. In that case the columns of this change of basis matrix would be columns of eignevectors...but I can't make anything from this.
What does diagonalizability tell you without mentioning matrices?
ordered basis, eigenvalues, all distinct eigenvalues implies diagonalizability
You don’t know all the eigenvalues are distinct.
You left out the most important word.
characteristic polynomial of $T$ splits........but that's probably not what you want
Nope
Basis consisting of ….

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