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12:04 AM
Thistles notwithstanding
12:43 AM
I always associate thistles with Scotland, their national flower.
1:12 AM
And I think of Winnie the Pooh stories. Eeyore and poor, poor Tigger.
2 hours later…
3:06 AM
math.stackexchange.com/questions/3811426/… can anyone explain me why it put (ab+bc+ca)=k(a^2+b^2+c^2) if a^2+b^2+c^2=ab+bc+ca?
3:24 AM
ted: we saw the coyote again today while checking our mailbox.
Too close for comfort!
3:58 AM
he ran away.. this time.
4:26 AM
Q: If math is so deductive, why is it so hard to discover new math?

XeonSome considerations: The conclusions of much latter/new math may be said to be already existent within the premises of current math The importance of deduction changes depending on if math is said to be invented or discovered Some parts of math are inductive and not necessarily deductive Spe...

asked in Philosophy SE
1 hour later…
5:50 AM
Hello respected mathematicians. I am here with a text-reference request, for singular value decomposition applied to curve fitting problems. I can find plenty of online resources and lecture notes for this, but can this be traced to some comprehensive treatment in some linear algebra textbook. Gilbert Strang has the rudiments put together in the appendix, but that hardly qualifies as a detailed treatment.
Any other book/source that you may be aware of, for this?
2 hours later…
8:02 AM
@299792458 This doesn't directly answer your question, but this question math.stackexchange.com/q/99299/207316 has some nice examples of using SVD to fit a plane to a collection of points. And it links to math.stackexchange.com/q/3869/207316 which has more general info, & links to tutorials.
Here's a crude Sage / Python demo, with interactive 3D output:
8:26 AM
@PM2Ring Thanks, I have followed the first two before. They are popular brilliant documentations. I was just wondering if there would also be some introductory textual reference to back these up.
i.e. starting from the very root of the svd problem
going on to discuss curve fitting.
Thanks nevertheless for our response. The third one was new to me.
9:20 AM
@299792458 Several of the regulars know about SVD. Hopefully, they'll have some useful suggestions for you. I only partly understand it myself, and it seems semi-magical to me. Fortunately, I can write code without fully understanding the algorithms I'm using. ;)
9:33 AM
@PM2Ring Haha. Thanks.
9:58 AM
I have got some response to the question above
But yes it seems a bit difficult to understand the answer..
I am wondering why the Schur complement of the whole matrix removing the first and last row/ column has to be taken?
Would any GR expert be willing to help me understand this? web.archive.org/web/20030912200701id_/http://…
More precisely equation 2.2
I'm stuck on the beginning of the paper
10:20 AM
suppose $P(z) = \prod_{i=1}^n (z-\alpha_i)$ is a reciprocal (irreducible) polynomial, with distinct roots satisfying $\alpha_i = \frac{1}{\overline{\alpha_{n-i+1}}}$ for all $i=1...n$, and no roots lying on $S^1$, note this means $\prod_{i=1}^n \alpha_i = 1$ and $n$ is even,define $h_i = \sqrt{(z-\alpha_i)(z - \alpha_{n-i+1})}$ to be asymptotic to $z$ for large $z$, and set $h=\prod_{i=1}^{\frac{n}{2}} h_i = \sqrt{P}$, why is it obvious that $h(0) =1$?
here $h_i$ is defined outside the segment $[\alpha_i,\alpha_{n-i+1}]$
also $n > 1$, so zero is not a root of $P$
ideally using $P$ is reciprocal, i want to establish something like $z^{\frac{n}{2}}h(\frac{1}{z}) = h(z)$ everywhere $h$ is defined, which would make $h(0)=1$ clear
not totally sure how to prove that though
the domain $h$ is defined is mapped conformally onto itself by $z \rightarrow \frac{1}{z}$ because $P$ is reciprocal, so that part isn't an issue
10:44 AM
I summarized my question here in case anyone is interested in answering: math.stackexchange.com/questions/4510154/…
2 hours later…
12:25 PM
Hi. Is my understanding correct that in order to prove a limit exists using the epsilon definition, all we need to prove is that there exists an inequality x > f(epsilon), where f can be any function that is defined (at least) on an interval (0, a], where a can be any number greater than 0?
@Cosinux What situation do you have in mind?
@Jakobian Proving limits using the epsilon definition in general. My thinking is that as long as X is on the "left" side of the inequality, the result will be within epsilon of the limit for all x above X, and as long as the "right" side is defined immediately above 0, an infinitesimal epsilon can be chosen.
Limit of a function $f$?
LaTeX in chat if you don't know how to enable it
1:29 PM
Socialize me, chat room.
@Jakobian No, not of $f$. I'm asking about the required conditions to prove an infinite limit using the epsilon condition. I identified two, and I'm wondering whether they are correct and/or complete.
$$\lim_{x \to \infty } g(x) = L \Leftrightarrow \exists X, \left | g \left ( X \right ) - L \right | < \varepsilon, \varepsilon > 0, \left | g \left ( X \right ) - L \right | < \varepsilon \equiv X > f \left ( \varepsilon \right ), (0, a] \subseteq def\left ( f \right ), a > 0$$
This is what I came up with, assuming my notation is correct.
1:47 PM
How to evaluate the definite integral as limit of a sum
3 hours later…
4:49 PM
> $$
\text { If } f(x)=\left\{\begin{array}{cl}
\left(\sin ^{-1} x\right)^{2} \cos ^{-1}\left(\frac{1}{x}\right) & x \neq 0 \\
0 & x=0
\end{array} \text { then } \mathrm{f}(\mathrm{x})\right. \text { is }
(A) Continuous nowhere in $[-1,1]$
(B) Continuous everywhere in $[-1,1]$
(C) differentiable nowhere in $[-1,1]$
(D) differentiable everywhere in except $x=0$
Isn't the question wrong? As $\cos^-(t)$ is defined only for $t\in [-1,1]$
5:07 PM
maybe they mean reciprocal, not arccos?
I think this is a typo. Remove that $-1$.
i think sin^{p}(x) usually means to the pth power, like sin^2(x) isnt sin(sin(x))
No. It’s arcsin, very clearly.
Remove the “exponent” on cos. What’s the answer?
ah okay, i see [-1,1] is emphasized
And they have another exponent outside.
5:11 PM
right.. spoke too soon
@Cosinux I'm not completely sure what you mean by this statement. $$\lim_{x\to\infty} g(x) = L \iff \forall_{\varepsilon > 0}\exists_N \forall_{x\geq N} |g(x)-L| < \varepsilon$$
Who writes subscripts for quantifiers?
@TedShifrin Ah! yes, then it matches with the given answers B and D. Thanks.
@TedShifrin I do
5:24 PM
Why? Totally non-syntactical and hard to read. Seriously.
Of course, writing math in symbols is unprofessional unless you’re a logician.
It never got to the point of being unreadable
You don’t have old eyes.
$\text{Do }_\text{you }{}^\text{really }\text{think }_\text{so?}$
Readability is my standard as well :)
I stand by my comments … all of them. It makes no sense and, besides, real mathematicians don’t write sentences in symbols.
5:27 PM
I must have picked it up somewhere along the way when I was learning math in the beginning
I'm sticking to it out of habit
Munkres made a huge point of that day 1 of the point set class I took from him.
Well, please avoid sentences in symbols in here. They’re bad for the newbies.
And very annoying to me.
It was written using quantifiers in the first place
Besides, how they're going to learn how to write things using symbols if they don't see examples of people doing so
Yeah, cosinux’s post was unintelligible.
hm, that statement does make sense to me
Is the isometry group of a metric space with the compact open topology metrizable?
5:32 PM
Most of us do not ever do it. I never taught my students to write that way. As I said, I was trained early by a real mathematician.
probably better if the LHS said $x \rightarrow +\infty$ though
Any string of symbols with only a comma or two is unreadable.
I think part of the job of the “more advanced” members of this chat is to model/teach good mathematical style.
@monoidaltransform in general, its metrizable if the metric space is hemicompacy
i.e. there is some increasing sequence of compacts, s.t. every compact is contained in one of the sequence members
In some cases, it's easier to use symbols than to explain it in words, I think uniform continuity and continuity are a good example of that.
So if the space admits an exhaustion by compact sets?
5:35 PM
the metric is just given by defining metrics on each of the compacts in the sequence, given by sup-norm, and then summing them all with weights $2^{-n}$ so the thing converges
I totally disagree, Jakobian. Totally.
I’ve never heard of hemicompact before.
uh, not sup-norm sorry, because you dont get a norm
Yeah, never heard of hemicompact before as well
5:36 PM
just means there is an exhausting sequence of compacts
I did hear about it. It's not something I remember though
thats all
its useful for topology of compact convergence
(to see when its metrizable)
its actually a necessary condition too..
Okay, so if $X$ is a proper metric space then Iso(X) is metrisable.
or local uniform convergence if you like
whats a proper metric space?
closed balls are compact
5:38 PM
also i wasnt trying to metrize Iso(X), i was trying to metrize all of C(X,X)
of course Iso(X) gets a metric by inheritance
sure, Iso(X) contained in C(X)
i dont think closed balls are compact will be enough though, unless you also have something like second countability
cant you have a metric space where closed balls are compact but the space doesnt have a compact exhaustion?
i.e. you dont have a sequence of compacts exhausting your space, because the space is too large
*the topology is too large
Can't you just fix $x\in X$ and then take $B^c(x,1)\subseteq B^c(x,2) \subseteq B^c(x,3)\subseteq...$
If you browse Dugundji's chapter about compactness then it introduces you to all kinds of compactness-like properties. I think hemicompactness is one of them
why is that necessarily an exhaustion?
it unions to the whole space, but why does every compact have to be a subset of one of those balls
5:41 PM
Why isn’t it?
its a hemicompacy
oh okay, never mind, indeed it is an exhaustion
my bad :)
yeah, proper should work then
Seems right. The distance from $x$ to any compact set is bounded.
yeah I brain farted, if it unions to the whole space, a compact is covered by the union -> extract finite subcover
@TedShifrin More like the Hausdorff distance
5:44 PM
uh, although you need to take interiors
one might want $K_n\subset K_{n+1}^{\circ}$ for exhaustions in general
Just use the distance function and the max value theorem. Way easier.
@KarlKroningfeld Agreed.
yeah, although in this case we have that too
Just to clarify, so proper metric spaces admit exhaustions which is a hemicompacy?
yeah, you just proved proper metric space -> hemicompact
5:49 PM
okay cool, just never heard of hemicompact before.
a more commonly heard of one is sigma-compact
that just requires your space to be a countable union of compacts
but its weaker than hemicompact
Yes, that term I know.
Yeah I know sigma compact
has anyone faced the issue: macbook not connecting to wifi?
Why is it weaker? I can turn it into an increasing union …
5:51 PM
im sure more than one person here has
(faced your issue, lol)
@TedShifrin doesnt guarantee compacts will be contained in some compact in that increasing union
or connecting with lot of difficulties but disconnecting again?
no, it seems like Dugudnji doesn't introduce hemicompactness
classic example is $\mathbb{Q}$
Your finite subcover argument proves that, doesn’t it?
it relied on interior containment though
5:53 PM
So it’s a Baire issue.
Baire-ly an issue
Soon we’ll have polar decomposition.
6:04 PM
It seems that my wifi will not connect tonight, laptops are so annoying. I want to ask: is the following way of creating a
sequence of disjoint sets out of an infinite $sigma$- algebra valid? Suppose that M is an infinite sigma algebra on set S. Clearly, S is an infinite set here. Take E from M such that E is non empty and not equal to S, and call it $F_1$. Now choose a non empty and not equal to $E_1$ from M-{$F_1$} and call it $E_2$ and then define $F_2=E_2\setminus E_1$
Similarly, I define {F_n}
Surprisingly, the issues that I face are unprecedented in history, like no one has faced them before.
@Jakobian I'd like to know if your statement is logically equivalent to mine. The reason I wrote it that way is because I'd like to understand whether rearranging $| g(x) - L | < \varepsilon$ in such a way that $x$ is isolated and checking the two conditions is enough to prove that $L$ is the limit.
6:24 PM
Your post makes absolutely no sense to me, Cosinux. Can you explain how you would prove a concrete example function limit?
6:37 PM
Hello Guys!
I am trying to show that f: C → C defined by f(z) = z for all z ∈ C is not
differentiable at any point in C. But, I get differentiable at all points in C...
You are correct.
Presumably you're missing a conjugate.
oh yes f(z)=z conjugate in question
But, I took z conjugate only while proving.
I did this by taking z=x+iy and using the definition of differentiability
What kind of differentiability are we talking about?
f (z0 + h) − f (z0)
You need to learn to write in MathJax.
So you're talking about complex differentiability. Have you learned the Cauchy Riemann equations? Presumably not yet.
6:46 PM
Yes, what?
I have learned Cauchy Reimann Equations
So use that. What are $u$ and $v$?
oh yes $u=x$ and $v=-y$
OK. Cauchy-Riemann equations will fail in an obvious way.
6:49 PM
yep got it I was trying by using defination.
You can see this directly by looking at your limit when $h$ is real and when $h$ is imaginary.
When you do the limit with $h\to 0$ in $\Bbb C$, you have to get the same limit no matter how $h$ approaches $0$.
I'd prove that $\lim_{N\rightarrow \infty }\left | \frac{\left ( -3 \right )^{N+1}}{4^{N-1}} \right |=0$
by simplifying and rearranging $\left | \frac{\left ( -3 \right )^{N+1}}{4^{N-1}}-0 \right |< \varepsilon$
until I isolated $N$: $N> \log _{\frac{3}{4}}\frac{\varepsilon }{12}$.
Then I'd check that $N$ is on the left hand side of the $>$ symbol, and that the right hand side is defined arbitrarily close to $0$ in $\mathbb{R}^{+}$.
Is that an adequate proof?
Using $N$ for your variable is very unfortunate.
@TedShifrin okay.. so lim h->0 h conjugate / h will be not same
Right @Rover.
6:51 PM
@TedShifrin Thanks!
@Cosinux What do you need to know about the right-hand side as $\epsilon\to 0^+$? You need to know something here.
And your algebra needs to be if and only if steps at every stage.
I also don't get where your $\epsilon/12$ came from?
7:18 PM
Is this any better?
$\lim_{n\rightarrow \infty }\left | \frac{\left ( -3 \right )^{n+1}}{4^{n-1}} \right |=0, n\in \mathbb{R}
\Leftrightarrow \exists N \in \mathbb{R}, \left | \frac{\left ( -3 \right )^{N+1}}{4^{N-1}}-0 \right |< \varepsilon , \varepsilon \in \mathbb{R}^{+} \Leftrightarrow$
$\Leftrightarrow \frac{3^{n+1}}{4^{n-1}}< \varepsilon
\Leftrightarrow \frac{3\cdot 3^{n}}{\frac{1}{4}\cdot 4^{n}}< \varepsilon
\Leftrightarrow 12\cdot \left ( \frac{3}{4} \right )^{n}< \varepsilon
\Leftrightarrow \left ( \frac{3}{4} \right )^{n}< \frac{\varepsilon }{12} \Leftrightarrow$
It has to be defined. Are there other requirements I'm not seeing?
Also, I apologize. I accidentally used n instead of N in the 2nd line. You mentioned my choice of N was unfortunate. Why is that? I saw it used in the epsilon proof of a limit at infinity.
8:08 PM
Yes, we usually use $N$ for $n\ge N$ or $x\ge N$. So you need to know that $(3/4)^n\to 0$. You’ve shoved that into the log function.
So at some point (perhaps your course has proved this) we need some fundamental facts to be proved.
8:20 PM
in our analysis class we proved that for $x$ with $|x|<1$, we have that $x^n \to 0$ way before we covered logarithms. If I remember correctly, we used Bernoulli's inequality (which can be proved by induction)
|x|^n is monotone and bounded, limit g has to satisfy g = |x|g, so g = 0
easier than Bernoulli
8:43 PM
Excellent point.
1 hour later…
10:03 PM
I Bonds are compounded semi-annually. What's the equation for an I Bond?
How does one derive that the symmetric difference of two sets A and B is (A\B) \cup (B\A) or (A \cup B) \ (A \cap B) if you're not given one of these two definitions to start, just the verbal 'in either A or B, but not both' definition?
Overall Rate = [Fixed interest rate + (2 x bi-annual inflation rate) + (Fixed interest rate x bi-annual inflation rate)]

So if the first 6 months is 9% and the next six months are 6%... what is the overall rate?
(Fixed interest rate = 0.1% for both half years.)
Is it the overall rate done twice, added and then divided by two?
10:56 PM
Your fixed interest rate comment makes no sense.
@TedShifrin It makes sense. They use it here: wallstreetmojo.com/series-i-bond
If I had to guess, it's just to make the rate come out non-negative.
The effective annual interest rate $r$ satisfies $1+r=(1.09j(1.06j=1.1554$.
I figured it out:

$1000*9%*0.5 = afterSix = 45
(1000+afterSix)*6%*0.5 = 31

= $1076
Maybe your 9%, 6% were annualized rates, so should be halved in my computation. Shrug.
@upanddownintegrate in either A or B, not in both gives literally $A\cup B \,\backslash\, A\cap B$ by definitions.

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