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Huy
Huy
15:02
@MikeMiller: sorry, what exactly do you mean by "positive dimensional Lie group"? also, doesn't compactness already imply having a 1-parameter subgroup?
Mike Miller
Mike Miller
@Huy: A Lie group having positive dimension, i.e. dimension > 0?
Recall the definition of 1-parameter subgroup: it's a Lie group homomorphism $\Bbb R \to G$. If $G$ is 0-dimensional the only such thing is the trivial homomorphism. I want a nontrivial one.
(I don't see how compactness shows up there. I only used it to go from infinite to positive-dimensional.)
Huy
Huy
@MikeMiller: because on compact Lie groups the exponential maps in the Lie and Riemannian sense coincide and therefore the latter is onto
Mike Miller
Mike Miller
Compact connected Lie groups.
Huy
Huy
oh
Mike Miller
Mike Miller
It can't possibly surjective when it's not connected!
Huy
Huy
15:06
ok, so your Lie group dimension is the manifold dimension?
Mike Miller
Mike Miller
Sure... I agree that I don't know what dimension means for an arbitrary group
Whenever someone says "a Lie group of dimension blah" they mean the underlying manifold
Huy
Huy
ok, now why do you get a 1-parameter subgroup if not by my argument which only holds for compact connected?
Mike Miller
Mike Miller
@Huy: Like you said, the exponential map gives you one. All you need is for the exponential map to be nontrivial (so that the Lie group is positive dimensional).
Your argument shows that there's a 1-parameter subgroup containing a given element of the Lie group. I've dropped that last clause entirely. I just want a 1-parameter subgroup at all.
Huy
Huy
ah, so you take the exponential map at the identity and this is a 1-dimensional subgroup but just not onto
Mike Miller
Mike Miller
@Huy: I'm a little bit confused. The 1-dimensional subgroup won't generally be onto; in fact it will if and only if the Lie group is 1-dimensional.
or I guess 0-dimensional.
Huy
Huy
15:10
yes, of course. the exponential map is what is onto sometimes
ok, I get what you mean though, I think
Mike Miller
Mike Miller
In fact, this is all unnecessary. You just needed to construct elements of the isometry group arbitrarily close to the identity. As long as you knew that the isometry group was a positive-dimensional manifold, this follows because... you can pick a chart around the identity
and then pick points arbitrarily close to it.
Happy Festivus, by the way.
Huy
Huy
likewise
I think I get the argument as you stated it now, but don't understand what you were doing earlier with the 1-parameter subgroup.
Mike Miller
Mike Miller
@Huy: So the idea was then just that $\varphi(1/n)$ would be a sequence of elements arbitrarily close to the identity.
(Which is all you're looking for!)
Huy
Huy
$\varphi$ being the 1-param subgroup, I suppose
Mike Miller
Mike Miller
Right
To recap: We assumed infinite, but because Isom is a compact Lie group, infinite implies positive dimensional (as a manifold). Which we used to get elements arbitrarily close to the identity
(how does the argument proceed from here? I don't know how to do the thing you're trying to prove in the end)
Balarka Sen
Balarka Sen
15:18
@Huy if Mike hasn't helped already, you can ask me.
Huy
Huy
I'm still thinking about your argument btw.
Balarka Sen
Balarka Sen
(although I can't guarantee I can help)
Mike Miller
Mike Miller
@Balarka: Progress on CP2 # CP2?
Huy
Huy
I don't remember all the details, but isn't it the case for compact manifolds that all smooth vector fields are complete and hence induce a 1-parameter subgroup?
Balarka Sen
Balarka Sen
@MikeMiller That was problem 2, which I have not thought about. I am thinking about the $2\alpha$ problem. Not much progress.
Mike Miller
Mike Miller
15:20
@Huy: Of Diff(M).
Huy
Huy
oh
Mike Miller
Mike Miller
@Balarka: That was problem 1. 2\alpha was part a.
@Huy: The appropriate statement for Isometries is that the flow of a Killing field is a 1-parameter subgroup of Isom(M).
Balarka Sen
Balarka Sen
yeah, it was. I though I had to do the problems in the order given?
(sorry for oh?-ing by the way. it has become a bit of a bad habit, I was about to edit the minute I posted it)
Mike Miller
Mike Miller
@Balarka: I never said anything about order above... I was just clarifying that CP2 # CP2 refers to problem 1
since there's ne'er a CP2 in problem 2
Balarka Sen
Balarka Sen
ok, I see. I don't see CP^2#CP^2 in here though.
Mike Miller
Mike Miller
15:24
Ayup, you're right, my mistake
was confused about something I was thinking about the other day
Huy
Huy
@MikeMiller: the next step is to show that there exists some $\epsilon > 0$ such that any homeomorphism satisfying $\sup d(x, \varphi(x)) < \epsilon$ is isotopic to the identity. intuitively with isometries this makes sense, is there a way to "approximate" homeomorphisms by isometries?
Mike Miller
Mike Miller
@Huy: Probably not, no.
Balarka Sen
Balarka Sen
ok, no problem.
Mike Miller
Mike Miller
I only say probably because you haven't specified what you mean by "a way to approximate"; if you did, I could probably disprove it.
Actually, here's some specificity: There's a hyperbolic genus 3 surface with trivial isometry group. You can't possibly mean "every homeomorphism is homotopic to an isometry", because that's not true here; and you can't mean "every homeomorphism is arbitrarily close to an isometry", because that's definitely not trhe.
Balarka Sen
Balarka Sen
off-topic, but makes me wonder if I could approximate homeomorphisms by diffeomorphisms. here, by approximate I mean given a $f$ in $\text{Homeo}(X)$, I can get a sequence of diffeoms $f_n$ converging to $f$, given the usual topology on Homeo(X).
X is a smooth manifold. bad notation.
Mike Miller
Mike Miller
15:29
No. First non-example I can give is in dimension 4. Should be true in dimension 3.
Huy
Huy
@MikeMiller: ok. I think the only trick we've seen to show isotopy of two homeomorphisms (in particular isotopic to identity) was using Alexander's trick. this usually involves cutting the surface into disks. can this be applied to $Isom(S)$?
Balarka Sen
Balarka Sen
@MikeMiller interesting, what's the counterexample?
Huy
Huy
don't answer it if the answer is yes, I have something in mind
Mike Miller
Mike Miller
@Huy: I mean, here's an argument. Isom(S) is a compact Lie group. Let $i=1,\dots, n$ enumerate the xonnected components. Let $\varepsilon$ be less than the distance from the identity to any other xonnected component.
Then if f is an isometry within epsilon of the identity, it's in the identity component, and so, by definition of component, there's a path from the identity to it as desired
Huy
Huy
@MikeMiller: yes, but that's an isometry
Mike Miller
Mike Miller
15:33
Ah, misread what you were asking, sorry
Balarka Sen
Balarka Sen
@Huy If you still have that Heegaard decomposition question, you should ask it away because I'm about to leave. If you don't want to have multiple discussions at once, ask Mike later on.
Mike Miller
Mike Miller
I don't really see how this will help us in the end
Huy
Huy
@BalarkaSen: some other time, I'll try to finish this first
Balarka Sen
Balarka Sen
sure.
see you later.
Mike Miller
Mike Miller
@Balarka: Homeo is locally contractible so if this were true any homeomorphism would be homotopic to a diffeomorphism. But there are examples of 4-manifolds X with a specified cohomology class $K_X \in H^2(X)$ such that any diffeomorphism preserves $K_X$
Have a little patience if you want answers, man, I'm answering two questions at once while on my phone :/
Anyway results of Wall imply yo can do most anything you want to $H^2$ with homeomorphisms; there's a homeomorphism inducing any automorphism of the intersextion form
ok he's gone so I guess I won't bother continuing to write that?
Huy
Huy
15:39
if I take $\epsilon$ less than the smallest distance of the identity component to a different one, all homeomorphisms satisfying $\sup < \epsilon$ map the identity component to the identity component. now I was going to argue that such homeomorphisms must be the identity on the boundary and then apply Alexander's trick, but I don't think that's true.
then again I feel like this intuition is false, since only Isom is a Lie group and Homeo isn't
dorothy
dorothy
does anyone here understand the multivariate central limit theorem by any chance?
specifically, when you can apply it
Mike Miller
Mike Miller
@Huy: They don't have to be the identity on the boundary but they so have to be isotopic to the identity
Huy
Huy
why though?
also, I keep thinking of connected components and distance and the like but is Homeo even a topological group?
Mike Miller
Mike Miller
you're just going a dimension down, yeah? "Every homeomorphism of the circle with sup dist small is isotopic to the identity"
This is something provable by hand; in fact every orientation-preserving homeomorphism of the circle is isotopic to the identity
Huy
Huy
sorry, I'm not as comfortable with isotopy yet as I wish, I find it much harder to imagine isotopy for homeomorphism as opposed to homotopy for paths.
Mike Miller
Mike Miller
15:44
No need to apologize
Huy
Huy
that was Huy-lingo for "I don't get it yet"
:P
why are they isotopic to the identity already when they map the connected component of the identity to itself?
Mike Miller
Mike Miller
I cannot parse that sentence
Huy
Huy
:(
Mike Miller
Mike Miller
The statement was that if f was in the identity component of Isom(M) it's isotopic to the identity
Nothing about mapping components
Huy
Huy
ok
I just don't really see how to work with what we know about the isometry group already to prove this statement for general homeomorphisms
Mike Miller
Mike Miller
15:52
They appear to me to be completely unrelated facts.
Huy
Huy
but then I have even less of a clue how to do this with homeos
Mike Miller
Mike Miller
You already started an argument!
First you isotope it to one that's the identity on the boundary circles.
Huy
Huy
but the picture I'm using for it is the one I have of the Isometry group
Mike Miller
Mike Miller
That doesn't make sense to me, man. I don't even know how you could have a picture of doing this with the isometry group when you're trying to prove the isometry group is finite.
Huy
Huy
I know right?
Mike Miller
Mike Miller
15:53
And Alexander's trick - which you wanted to use! - has nothing to do with Isometries.
Huy
Huy
that's what I'm saying: I don't have a clue of how to do this
ok
so back to Alexander's trick
I know that there's a pants decomposition for $S$ of genus $\geq 2$. can I use this somehow?
(trying to cut up into disks)
Mike Miller
Mike Miller
Yes.
morphic
morphic
Wow Mike turned into Sun Wukong
Huy
Huy
@MikeMiller: say I have a pair of pants lying on the floor. now I cut it horizontally and obtain two hexagons. by scaling, I get two disks. I'm worried about the boundaries here.
Mike Miller
Mike Miller
@Huy: Ok, I'm running out of time. The point is that you can isotope your homeomorphism to fix the line you're cutting along.
(And the boundary circles.)
Huy
Huy
16:04
@MikeMiller: and then can I choose epsilon smaller than the radius of the "smallest hexagon" and then I get isotopy to the identity?
Mike Miller
Mike Miller
Then when you Alexander trick you don't ever modify what's going on on the boundary.
@Huy: That does not make sense to me, no.
Huy
Huy
hm.
how do I get epsilon then?
it must have something to do with the decomposition here?
Mike Miller
Mike Miller
I think we're around the point where you're just asking me to write the proof of the theorem...
Huy
Huy
not really, but I don't see why that epsilon isn't what I think it is
also, doesn't Alexander's trick require the homeomorphism to be the identity on the boundary? or is this what you mean by "you can isotope your homeo to fix the line you're cutting along"?
Mike Miller
Mike Miller
Yeah, @Huy. I'm going to have to tag out now.
Good luck with this.
Huy
Huy
16:12
ok, thanks
OFFSHARING
OFFSHARING
16:45
@Huy What was the command to measure the calculation time in Mathematica?
Ah, got it
// Timing
Balarka Sen
Balarka Sen
17:14
@MikeMiller I didn't mean to seem impatient. I have a buckload of schoolwork, so only taking short breaks in between studies.
I see what you mean, I should have figured that out. If a homeom is arbitrarily close to a diffeom, it can be perturbed a bit by homotopy to get a diffeom since Homeo(M) is locally ctrctible. But certainly there are manifolds admitting homeomorphisms not homotopic to a diffeomorphism (we have discussed this before, and one of Milnor's exotic sphere works.)
@MikeMiller I'd be glad if you write that up and ping me with it (if you have the time and care about it, of course). I don't know how one can construct such a 4-manifold.
Mike Miller
Mike Miller
17:30
@Balarka: 1) I can construct an example with exotic spheres but don't otherwise see the relevance. (In particular I don't think my construction is what you have in mind.) 2) I do not consider local contractibility of Homeo(X) at all obvious. I don't even know how to prove it.
evinda
evinda
17:52
Hello!!!
@DanielFischer According to my notes the following set (where I drawed the lines) is closed and unbounded. Could you explain me why it is closed?
user image
Jake1234
Jake1234
18:04
I'm not DanielFischer, but if you take the functions (of 2 variables) that define those two long lines - $kx + ly = m$ -then if you consider the function $f=kx +ly >m$ you get one open set, as clearly $f$ is continuous, and preimage of an open set, which $\{a>m\}$ is, you get an open set.
evinda
evinda
@Jake1234 So you mean that it is not closed? We call the set F and it's the set of feasible solutions of a linear programming in caconical form.
Jake1234
Jake1234
I'm sorry, it depends on what the set you painted is.. does it contain the lines, or is everything above the lines (but not including the lines) ?
DanielSank
DanielSank
@DanielFischer I realized after our discussion yesterday that I really want to understand how to handle $\delta(f(x))$ in the case $f: \mathbb{R}^N \rightarrow \mathbb{R}$. The case $f:\mathbb{R}^N \rightarrow \mathbb{R}^N$ is straightforward as you pointed out.
Posted on main site in case you're interested: math.stackexchange.com/questions/1586969/…
I think the answer is $\int_{p|f(p)=0} \left[ \prod_i \delta(x_i - p_i) \right] / \left \lvert \nabla f (p) \right \rvert$.
evinda
evinda
@Jake1234 I don't know... It is not written something about it. I wanted to understand why the set of feasible solution can be unbounded and closed, but now looking at the proof that the set is closed I saw that we don't use the boundness or not of F, but only that A (Ax=b) is continuous.
@Jake1234 Anyway... thanks for answering...
Mike Miller
Mike Miller
@Balarka: 3) The simplest example to describe might be a degree 6 hypersurface in $\Bbb{CP}^3$ but the intersection form and $K$ would be a pain to write down (this manifold has $b_2 = 106$). Easier to write down $K$ are the so-called Dolgachev surfaces, which I won't define, but will tell you are homeomorphic (but not diffeomorphic!) to $\Bbb{CP}^2 \# 9\overline{\Bbb{CP}^2}$. This has diagonal intersection form $(1,-1,\dots,-1)$.
DanielSank
DanielSank
18:12
However, I mostly think that because of "argument by symbolic similarity and squinting at the details".
Puzzled417
Puzzled417
Hi
Mike Miller
Mike Miller
I think, based on some brief reading, that the Dolgachev surface I'm thinking of has $K = (18,-6,\dots,-6)$. Then because of Wall's result and the intersection form there is a homeomorphism that negates the second and third copies of $\Bbb Z$ in $H^2$ and only those. But this does not preserve $K$ or send $K$ to $-K$, so there is no diffeomorphism doing this.
Puzzled417
Puzzled417
Is it possible to inscribe a sphere inside a square pyramid that isn't equilateral?
Mike Miller
Mike Miller
But I am not used to these computations so I may have written down the wrong $K$.
I should really read Griffiths and Harris eventually.
Jake1234
Jake1234
@evinda well I think in linear programming you generally use greater or equal, not striclty greater - the solutions to $kx+ly>a$ is open, but if you change the > to greater or equal, the solutions are closed.
Behzat
Behzat
18:16
lel
evinda
evinda
@Jake1234 Why is it like that? Could you explain it to me?
Balarka Sen
Balarka Sen
18:42
@MikeMiller 1) There is a orientation reversing homeomorphism S^7 --> S^7. If $M$ is an exotic sphere, then we get a orientation reversing homeomorphism of $M$ too. Since this has degree -1, it can be homotopic to a diffeomorphism only if there is a diffeomorphism of degree -1, i.e., which reverses orientation.
In our previous discussion, I asked you whether there exists exotic spheres with no orientation reversing diffeom, and you mentioned such a thing must be a 2-torsion in $\Theta_7$. Thus, pick an element which is not a 2-torsion.
Mike Miller
Mike Miller
Ok, yup, that's what I was thinking of.
Balarka Sen
Balarka Sen
2) Isn't every smooth manifold a CW-complex? Then Homeo(M) has the homotopy type of a CW-complex by the theorem you mentioned yesterday, hence must be locally contractible, yeah?
Mike Miller
Mike Miller
Think about what you just said.
Balarka Sen
Balarka Sen
Er, sorry, local contractibility need not be homotopy equivalence-invariant I think.
Yeah, not true. Sorry.
Mike Miller
Mike Miller
Not the problem I'm pointing out.
Though I guess that is a problem, huh, never noticed before.
Balarka Sen
Balarka Sen
18:53
Ok, I see the problem. Map(M, M) has the homotopy type of a CW-complex and hence is locally contractible, but that doesn't mean Homeo(M) (subspace) is too.
Mike Miller
Mike Miller
Correct. It says nothin at all about Homeo.
Balarka Sen
Balarka Sen
@MikeMiller Fun, thanks.
Daniel Fischer
Daniel Fischer
@DanielSank The idea is that if I have a sufficiently well-behaved function $f \colon U \to V$, and a sufficiently well-behaved function $g$ on $V$, then I get a sufficiently well-behaved function $g\circ f$ on $U$. This defines a distribution on $U$ in the usual manner, for a test function $\varphi$ we set $$T_{g\circ f}[\varphi] = \int_U g(f(x))\cdot \varphi(x)\,dx.$$ Now we would like to express that as an integral over $V$. We can't plain make a change of variables $y = f(x)$,
since we don't assume that $f$ is that nice. But we can write the integral as an iterated/a double integral, $$T_{g\circ f}[\varphi] = \int_V g(y)\biggl(\int_{f^{-1}(\{y\})} \varphi(x)\,d\sigma(x)\biggr)\,dy,$$ where $d\sigma$ denotes the ($\dim U - \dim V$-dimensional) surface measure on the level set $f^{-1}(\{y\})$ of $f$.
You're probably very familiar with that for polar/spherical coordinates as a physicist. If $f$ is nice enough, the inner integral $$I_{\varphi}(y) = \int_{f^{-1}(\{y\})} \varphi(x)\,d\sigma(x)$$ is well enough behaved that you can't just integrate it against nice functions like $g$,
but you can apply more general distributions to it (whether you can apply all distributions or only some, e.g. distributions with compact support, or with order $\leqslant k$, or with other constraints depends on $f$). For distributions where it is possible, we thus define $T\circ f \colon \varphi \mapsto T[I_{\varphi}]$. In particular for $\delta$ we have $$(\delta\circ f)[\varphi] = \int_{f^{-1}(\{0\})} \varphi(x)\,d\sigma(x).$$
I have never learned to interpret the physics notation with $\delta$s inside integrals, so I can't say whether what you have is correct or not, but it looks like it may be. [Also, I have forgotten how $d\sigma$ is expressed in terms of the derivatives of $f$, and I don't want to figure that out now.]
Mike Miller
Mike Miller
19:08
@DanielF: Wow, an essayist.
Balarka Sen
Balarka Sen
That's bigger than anything I have ever written.
Mike Miller
Mike Miller
19:23
@Balarka: So I take it I shouldn't have written up the 4-fold bit? :P
Balarka Sen
Balarka Sen
I read it, and thanked you for it, but I can't make much of it yet :(
Khallil
Khallil
I'm going to screen-cap that and make it my wallpaper, @DanielFischer. What an effort! :-b
Daniel Fischer
Daniel Fischer
@Khallil The biggest effort was cutting it into pieces short enough to be posted in chat. The message length restriction should not apply to moderators.
Khallil
Khallil
Start the motion for no length restrictions, @DanielFischer!
DanielSank
DanielSank
@DanielFischer Any particular reason you didn't write it as an answer to the question I posted/linked?
Balarka Sen
Balarka Sen
19:28
@MikeMiller I am curious why so many 4-manifolds emerge from algebraic geometry though.
Mike Miller
Mike Miller
I don't really know what that means.
Balarka Sen
Balarka Sen
Vague question, nevermind.
Ignore that.
Daniel Fischer
Daniel Fischer
@DanielSank Because it's just an informal "here's the general idea", for an answer I would have to dive into the technical details, and that - ugh.
Balarka Sen
Balarka Sen
I don't have anything mathematical to say about your 3rd part, unfortunately.
Jake1234
Jake1234
@evinda it's a consequece of the fact that for a continuous function $f$, $f^{-1}(A)$ is an open set, for $A$ an open set. $kx + ly$ is a continuous function, and $A = \{x: x>0\}$ is an open set, so $f^{-1}(A)$ is an open set.
Mike Miller
Mike Miller
19:30
ok
Semiclassical
Semiclassical
hi
DanielSank
DanielSank
@DanielFischer I'm sure that I and many other users find not-completely-rigorous answers like the one you've written here helpful. On that basis I don't see a good reason to avoid writing them on the main site.
Mathematical rigor and just plain understanding what's going on are both important.
Anyway, thanks much!
Jake1234
Jake1234
@evinda but for a set like that, you really should just prove it directly... if you take a point under that is at under at least one of the long lines you drew ( the complement of where you drew the short lines, let's call it $A$), you can always find a small neightborhood of that point that is fully contained in $A$, so $A$ is open, and thus your set is closed (complement of an open set is a closed set).
Daniel Fischer
Daniel Fischer
@DanielSank You're welcome. If nobody writes a good answer on main, feel free to paraphrase the above and post it there.
idonutunderstand
idonutunderstand
Does anyone else find the chalk-on-chalkboard sound in some online lecture videos irritating?
Mike Miller
Mike Miller
19:39
There are lots of non-rigorous answers on main. Some are intentionally so. Others...
 
1 hour later…
robjohn
robjohn
20:46
@OFFSHARING Congratulations. A new computer is always fun.
@OFFSHARING Not yet. I have been very busy this week.
@OFFSHARING Sounds interesting. If you care to share, I would be interested.
OFFSHARING
OFFSHARING
@robjohn hehe, yeap. It's only I3, but I could have bought an I5. However I keep some money for the desktop version.
Huy
Huy
@OFFSHARING: note that i3, i5, i7 for laptops aren't the same as the usual i3, i5 and i7s in desktop PCs
OFFSHARING
OFFSHARING
@Huy I know less about it. It moves pretty fast.
Huy
Huy
as long as you're satisfied with it, everything's fine. :)
OFFSHARING
OFFSHARING
@robjohn that was a point! :-)
:26399552 Precisely! :-) There is somethign amazing around that I prepared for a magazine (as a proposed problem). :-)
:26399552 Have you ever seen anything like that?
@Huy You suggested is a huge difference between i3 and i5 (for notebooks)?
Huy
Huy
20:59
@OFFSHARING: depends on what you do and which i3 and which i5. to be honest, I'm not that well-informed anymore about notebook CPUs, more about desktop CPUs. as far as I remember, on notebook CPUs, the difference is not as big as on desktop CPUs
OFFSHARING
OFFSHARING
@Huy I intend to buy the best I7 for the desktop (when my pocket is prepared for that).
8 cores
Huy
Huy
I suppose it's mostly for Mathematica?
OFFSHARING
OFFSHARING
@Huy Yes. :-)
Huy
Huy
if so, inform yourself if that actually makes a big difference. most programs don't take advantage of more cores (yet).
OFFSHARING
OFFSHARING
@Huy I have a lot of stuff for a need verifications, but for some it takes hours ... :-(
Huy
Huy
21:01
I think one of the few programs that actually take advantage of an i7 over an i5 are AutoCAD and the like
OFFSHARING
OFFSHARING
@Huy Hmmm, you might be right though ...
Huy
Huy
just google or ask some people who really know this stuff before you spend another $200 for something that makes your computations maybe 1% faster
(I don't know about Mathematica, sorry)
OFFSHARING
OFFSHARING
Intel Core i7-5960X
cpubenchmark.net/high_end_cpus.html
@Huy see above how poweful that CPU is. Unfortunately it is pretty expensive now ... (to get a used one? hmmm, not sure if this would be a good idea)
Huy
Huy
@OFFSHARING: Yes, it is very powerful, but you have to check whether Mathematica actually takes advantage of that power. Some programs do, some programs don't. If the programs you use don't, it would be like buying a Ferrari but never driving faster than 100 km/h.
OFFSHARING
OFFSHARING
In the past I was thinking that I lose my value as a mathematician if I make use of computational systems. Later I realized that they are actually great for checking your work, especially when your work is truly crazy.
@Huy Sure, but compared to what I have as a desktop, yes.
Huy
Huy
21:12
Sure it will be an improvement, but you surely want to save money where you can (without loss of computation power in Mathematica)? :)
OFFSHARING
OFFSHARING
@Huy Definitely. I only said I wish I had that and I intend to buy it but not at that price. Never at that price.
However, I wish I have it ... :-)
Karl Kronenfeld
Karl Kronenfeld
@MikeMiller I don't understand your tone. Just provide a sufficiently rigorous answer or break the given answer if it is bad.
Mike Miller
Mike Miller
What tone did I have?
Karl Kronenfeld
Karl Kronenfeld
You're better at answering that question than I, but it came across as a form of dissatisfaction.
OFFSHARING
OFFSHARING
@robjohn that series is the devil himself, but I slayed it (I would dare to say I did it in an exceptionally nice way).
Daniel Fischer
Daniel Fischer
21:22
@KarlKronenfeld Context, Mike was just saying that being non-rigorous was not necessarily a reason to not post it as an answer on main.
Karl Kronenfeld
Karl Kronenfeld
Hah!
Mike Miller
Mike Miller
@KarlKronenfeld Fair enough; as Daniel said I was being silly and suggesting that lack of rigor, if appropriate, is perfectly fine for the site. In about a quarter the words that necessitates. My excuse for not using the quote-arrow is that I was on my phone :)
Huy
Huy
@OFFSHARING: Apparently, Mathematica comes with benchmarking tools. Try

Needs["Benchmarking`"]
BenchmarkReport[]

Note that while the i7 is leading in benchmarks, it is hardly any better than a Xeon (which are well known for being a good substitute for an i7 but at a cheaper price). It is also not *that much* better than a good i5. What you must know is that the dependency between price (to buy) and computational power is unfortunately not linear but rather exponential, so most of the times to get a **noticable** difference, you'll have to pay **a lot** more.
EwokNightmares
EwokNightmares
Is functional data analysis getting curve fits to histogram data?
Mike Miller
Mike Miller
@KarlKronenfeld: As recompense for my sins, enjoy this article about Neil deGrasse Tyson.
OFFSHARING
OFFSHARING
21:26
@Huy That's sure. I also have another possibility, that is to see the offer from AMD, they plan to launch Zen CPUs which, at least from rumours, seem to be endowed with much computational power (and far cheaper).
That will happen during the next year.
Karl Kronenfeld
Karl Kronenfeld
@MikeMiller OK that's pretty hilarious.
Huy
Huy
@OFFSHARING: I don't know a lot about AMD CPUs, just that almost everyone who is serious about their work gets an Intel. The former are far cheaper though, but I have no experience with them at all.
OFFSHARING
OFFSHARING
@Huy Yeap, that's true also. However, I cannot say AMD didn't give me satisfaction, and for a long period of time I could do without problems all kind of stuff. I only wish they offer something much more powerful and to be in the same range with I7.
Huy
Huy
Yes, I don't think it's bad. Just maybe Intel is better. I have an AMD graphics card in my PC and am perfectly satisfied with it and an equivalent NVIDIA would have costed around 30% more.
OFFSHARING
OFFSHARING
@Huy Well, that's the point.
Huy
Huy
21:30
:P
OFFSHARING
OFFSHARING
:D
Huy
Huy
do you intend to build your PC yourself or buy a pre-built one?
OFFSHARING
OFFSHARING
@Huy I intend to build it myself, I did it many times in the past too.
Huy
Huy
@OFFSHARING: consider asking on r/buildapc on reddit for some advice if you're unsure. sometimes there are some people around who can tell you a lot that you didn't know yet.
I built 4 PCs so far and still go there for advice sometimes because there are many people who have just a lot more experience and know a lot more about the components. :P
OFFSHARING
OFFSHARING
@Huy It's very hard to figure out what the reality is. The benchmarks presented by great magazines contradict each other.
The reality is something you fell better than any other magazine that present to you benchmarks. That means that you need to have some experience with hardware stuff.
@Huy I have a friend which has an IT store and he told me: DON'T BUY EXPENSIVE STUFF FROM INTEL!
And he could tell me to buy his expensive stuff from his store, but he didn't. I was always told to try AMD.
EwokNightmares
EwokNightmares
21:40
do math professors and experts get emails from random readers about their books ever?
Karl Kronenfeld
Karl Kronenfeld
I've done it ("randomly" sent an email to an author).
EwokNightmares
EwokNightmares
@KarlKronenfeld Is it normal or something frowned on?
Karl Kronenfeld
Karl Kronenfeld
@EwokNightmares He responded promptly and kindly.
No idea if it is normal and whatnot.
Mike Miller
Mike Miller
If you're emailing someone to say "I enjoyed your thing!" it's hard to believe that could be received poorly.
They might not respond, but most people would probably be happy to know someone is reading their work, much less enjoying it.
EwokNightmares
EwokNightmares
It is much more selfish reason. I want to get laymen translation if I should spend the effort on their topic for my own uses.
I think the general thought will be figure out yourself
Mike Miller
Mike Miller
21:56
Without knowing more specificity, probably most people won't respond. They've got prefaces, introductions, table of contents for that sort of thing
EwokNightmares
EwokNightmares
I want to know when I am taking data samples of a measurement how many measurements are enough to estimate the average within a limit of error.
so I can stop taking samples once I reach that error limit
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