*sigh* looks like I won't be getting any sleep tonight with all these fireworks, speaking of Canada Day. Not that I sleep at a healthy time on any night anyway. But still, annoying :/
king tsin was not that good but it was very reasonably priced. my parents loved how they computed your tab on an abacus, which felt a little theatrical to me.
Line 7 is needed to handle cases where $\overline{\mathbf{H}}$ has zero or negative eigenvalues. The flooring described in Line 7 may be done as follows: do the Singular Value Decomposition $\overline{\mathbf{H}}=\mathbf{U D V}^{T}$, then let $\hat{\mathbf{D}}$ be a floored version of $\mathbf{D}...
Let $X:=\{X_t\}_{t\in[0,T]}$ be the unique strong solution to the Itô diffusion
$$
dX_t = a(t,X_t)dt + b(t,X_t)dW_t,
$$
where $a,b$ are such that the conditions for the existence of the invariant measure are satisfied. How would one go about showing that for any $f\in\mathcal{C}(\mathbb{R})$:
$$
...
Line 7 is needed to handle cases where $\overline{\mathbf{H}}$ has zero or negative eigenvalues. The flooring described in Line 7 may be done as follows: do the Singular Value Decomposition $\overline{\mathbf{H}}=\mathbf{U D V}^{T}$, then let $\hat{\mathbf{D}}$ be a floored version of $\mathbf{D}$ with diagonal elements $\hat{d}_{i}=\max \left(d_{i}, \epsilon \max _{i} d_{i}\right) ;$ then let $\hat{\mathbf{H}}=\mathbf{U D} \mathbf{U}^{T}$ (note: the use of $\mathbf{U}$ on both sides is not a typo). This has the effect of flipping the sign of negative eigenvalues, and then imposing a floor …
Can someone explain me that last line This has the effect of flipping the sign of negative eigenvalues, and then imposing a floor of $\epsilon$ times the largest eigenvalue. and how V changes into U is best approximation for Hbar
If $\gamma : [0,1] \rightarrow O \subset \mathbb{C}$ is a closed (continuous) curve, and $O$ is open, is it always possible to find a smooth curve $\gamma' : [0,1] \rightarrow O$, such that $\gamma$ and $\gamma'$ are (freely - no endpoints need to be fixed) homotopic in $O$?
intuitively it seems like if I choose $\gamma'$ close enough to $\gamma$, I should be able to take the straight line homotopy
This is an old question, but here are some visual examples that help to suggest that indeed the Jordan curve theorem is actually not so intuitively obvious.
Both are not rectifiable. We have relatively good intuition about rectifiable curves but not non-rectifiable ones. These examples may s...
The linked post gives examples of continuous curves that can be closed up to form your γ, making it impossible to just use a straight-line homotopy to a "close enough" γ'.
user21820 assuming for any $\epsilon > 0$, there is a smooth curve $\gamma' : [0,1] \rightarrow O \subset \mathbb{C}$ for which $sup_{t \in [0,1]} |\gamma'(t) - \gamma(t)| < \epsilon$, and given $\epsilon > 0$ is a lebesgue number for $\gamma([0,1]) \subset O$, at every point $t \in [0,1]$, all line segments of length $< \epsilon$ starting at $\gamma(t)$ are contained in $O$, now consider the homotopy $u_{t}(s) = F(t,s) = (1-t)\gamma(s) + \gamma'(s) t$ is a homotopy in $O$
im not sure why the jordan curve theorem is involved here, perhaps what you are saying makes it easy to point out a flaw in my argument?
@porridgemathematics Look, as I said, just imagine a straight-line homotopy from my first example to a smooth curve. It will utterly fail regardless of the smooth curve, because a smooth curve is asymptotically linear around the point that maps from the centre of my example.
@user21820 math.stackexchange.com/questions/1441749/… it seems the answer to this question is the same proof I just gave in essence, it remarks we do not need $\gamma$ to be rectificable, just continuous is enough. I do not think I was being stubborn
@porridgemathematics This doesn't quite make sense yet (you're taking a Lebesgue number for a cover by one open set?), but I think your idea is correct and the same as in the post you just linked. The same idea, with perhaps a slight bit more work, shows that any continuous curve in a smooth manifold is homotopic to a smooth curve (even with endpoints fixed). In fact, the same holds for any continuous map between smooth manifolds (the idea for this isn't more complicated either, it's just more technical).
@porridgemathematics You're wrong. Eric's answer does not justify that the straight-line homotopy is actually a homotopy. You cannot halfway have self-intersection.
Hmm. Maybe I'm using a different meaning of "homotopy".
@user21820 These pictures look like curves in the plane. Any two closed curves in the plane are freely homotopic. If you mean to view them as curves in some other open subset of C, you have failed to make your intent clear.
The recent kerfuffle about the Riemann hypothesis inspired me to write a little Sage / Python script to explore the graph of the cumulative Liouville function, A002819. There are a couple of graphs on Wikipedia, but they're pretty crude, and you can't zoom in.
My script takes ~15 seconds to factorize the numbers <500,000 and plot the whole graph, but then you can zoom into any sub-domain. You can change the domain by adjusting the m parameter in the script. I've tried it with 2,000,000. If you go too high, the server will just kick you off. ;)
But anyway, looking at that graph, I find it strange that someone would claim that the Liouville function "obviously" has a mean value of zero. Maybe it does, eventually...
@Thorgott yeah my argument wasn't exactly up to par I agree, I meant to say a lebesgue number for some cover using open sets contained in $O$, also the statement of the problem should definitely have had endpoints fixed, otherwise the resolution is trivial
but the general theme was that eventually the straight line homotopy works (after some fiddling to get a smooth approximation)
@Thorgott I do not know how viable what I am saying is, but maybe one can modify your proof which was in the case of Riemann surfaces to general complex manifolds, to give a cohomological criteria for ampleness. What you did was working because $\mathcal{O}_X(-p-q)$ were invertible sheaves. This won't happen in general. Instead what I think we can do is blowup $X$ at $p,q$ and then consider the exceptional divisors and proceed with your argument. I have to iron out some details but it can work
What is the differences between a singularity and a discontinuity as it pertains to functions? It seems similar, but I think there are some subtle differences that I'm unsure about
@LearningCHelpMe Fine :). Well, plot the graph of y=arctan(1/x). It doesn't have a singularity at x=0. but it is discontinuous there so a discontinuity and singularity aren't always the same thing.
@LearningCHelpMe I've just finished A Levels, cool! Which textbook did you find that in? I wasn't aware that the syllabus covered continuity.
Unless it's from the chapter about numerical methods?
I understand that there are several types of discontinuities. Such as, removable, jump, infinite discontinuities. For the function 1/x at x= 0, could the function at x = 0 be classed as an infinite discontinuity aswell as a singularity?
Are they just two terms that can be used interchangeably in most cases?
@LearningCHelpMe in that article, the section about real analysis only names the fourth discontinuity type as a "singularity", and "essential" one in fact.
if I understand correctly, an example of such singularities would be $x=0$ for $f(x)=\sin(1/x)$
For removable discontinuity, I'd think of $y=x^2/x$ at $x=0$.
Jump discontinuity, the floor function at all positive integers.
Looking at the floor function, I think we can see that according to your textbook's definition of a singularity, a singularity and discontinuity are not necessarily equivalent.
The lyrics of a song by The Chameleons made me wonder:
It's just coincidence
Well you can talk that way
But I have to say
I don't believe in it
After considering these lyrics, I have the following questions:
Is everything just coincidence? Doesn't fate exist? Are all the people meeting ju...
so then how does 5x0=0 when what ever you are multiplying does not disappear. as in, i have five apples. i want to multiply them by 5 so i have 25 now. so then if i multiply them by 0 were do they go? if 5 things dont exist, then its 0x0.
@toothlessgrinn When we say we have 0 of something, that means we have none of that particular thing. Now, when we say axb, we mean that we are finding how much we are left with of we have a lots of b, or equivalently b lots of a. In this case, suppose that a=0 and b=5. ...
...(continued)Then 0x5 means 0 lots of 5 which means we have no lots of 5, ie nothing at all. Similarly, if we were to write 5x0 this means 5 lots of 0, but 0 is nothing so 5 lots of nothing is still nothing. @toothlessgrinn Is that any clearer?
saying 5 lots of nothing is not multiplying. then its 0x0. 5 represents somthing and it doesnt matter what. if you multiply it by 0 it doesnt disappear. and if 0=1 then in turn its 5x1 yet 5x0 and they both have the same answer.
you act as if multiplication is a physical process in which you take two physically represented quantities and somehow multiply them. this isn't the case. the "real world interpretation" of multiplication is that it describes something being iterated, like A-Level Student already explained.
you cant make up your own rules just cuz you dont like the facts. ask a rabbit how physical multiplying is and get back to me. 5 STILL represents SOMETHING and it does not matter what. it represents something. so in saying if you multiply it by 0 you still have the same number of whatever. SHOW me how im wrong. is it so hard? dont slather me with propaganda. i want proof. if you cant prove it then dont answer.
@BalarkaSen Deschele seems to be going through a bit of a manic phase at the moment... He's an interesting character but he has a tendency to get caught up in pretty wild ideas, and that seems to have intensified in the last few weeks.
FWIW, I made my Liouville cumulative sum calculation much more efficient than the version I posted here. It can now go upto 5,000,000 in around 1 minute. I'll post the new version, if anyone's interested.
Hint: Prove the result when $A$ is an interval, then prove it when $A$ is a disjoint union of intervals. Then, use that any measurable set can be approximated by open sets from outside.
I'm trying to make sense of the hint, i.e. assuming for open interval and then showing it holds for a disjoint union. Here's what I got:
Suppose $$\lim_{n\to\infty}\int_A \cos nx\ dx=\lim_{n\to\infty}\int_A \sin nx \ dx=0$$ for an open interval $A$. Let $B$ be an open set in $[0,2\pi]$. Then, $B = \bigcup_{i=1}^\infty A_i$ where $A_i$ are open. We have $$\lim_{n\to\infty}\int_B \cos nx\ dx = \lim_{n\to\infty}\lim_{m\to\infty}\sum_{i=1}^m \int_{A_i} \cos nx\ dx$$ $$\lim_{n\to\infty}\int_B \sin nx\ dx = \lim_{n\to\infty}\lim_{m\to\infty}\sum_{i=1}^m \int_{A_i} \sin nx\ dx$$
Now I'm not sure if we can interchange the limits.
If a function is symmetric about the y-axis, does that necessarily mean that it is an even function? If you defined some piecewise function which is symmetric about that y-axis, but doesn't obey the $f(-x) = f(x)$ could you still call it even?
$$\int_a^b\sin(nx)\,\mathrm{d}x = \frac{1}{n}(\cos nb - \cos na) \le \frac{2}{n} \xrightarrow{n\to\infty} 0$$ There is no lower bound here though. I mean, the lower bound is not zero.
@epsilon-emperor Cannot directly exchange, but you can increase n until integral over the first interval stays close enough, then increase n some more until over second interval stays close enough, then ... Make sure all the close enough adds up to ε.
@robjohn Actually I cheated, because "be after" is a phrasal verb, as is "give up", so it doesn't obey the prescriptivist rule that prepositions should not be left dangling.
@epsilon-emperor Now there are the next two hints.
You were trying to do too much at once. The idea in most investigations is to try simple cases and once you see how those work, go on to more complicated cases.
@robjohn And it's often the case in measure theory that you first prove the result for elementary sets (disjoint union of boxes) and then build from there.
So the intuition is to just keep increasing n to satisfy more and more intervals. Now if you have countably many disjoint intervals that are all stuck within a bounded region, you can ignore the sufficiently tiny ones, leaving finitely many intervals.
Of course, there may be other slicker proofs, but this concrete intuition via approximation is easier to find.
@user21820 I don't think I've seen a slicker proof of RL. At least at the level where it is being proven, pounding with the available hammers is the way to go.
I would worry that a slicker proof might be circular.
Since $\int_{A_i} \cos nx\ dx \to 0$ as $n\to\infty$, there exists $N_i$ such that for $n\ge N_i$, $-\epsilon < \int_{A_i} \cos nx\ dx < \epsilon$. For $n\ge \max\{N_1, ..., N_m\}$,
@robjohn Sometimes, on the real line, you can get away with weird tricks. For example, I didn't check but how about letting c = sup { x : x∈[0,2π] ∧ claim holds for A ⋂ [0,x] }, and deriving contradiction unless c = 2π?
Thanks rob was looking at the graph incorrectly, silly me. So there is no cases where a function can be made symmetric about the y-axis unless it obeys $f(x) = f(-x)$?
$$\int_B \cos nx\ dx = \lim_{m\to\infty}\sum_{i=1}^m \int_{A_i} \cos nx\ dx$$ We can choose $M$ such that for all $m\ge M$, we have $$\left|\int_B \cos nx\ dx - \sum_{i=1}^m \int_{A_i} \cos nx\ dx \right| < \epsilon/2$$ Is this what you're saying?
Also are you suggesting the following now? Taking limit $n\to\infty$, we have for $m\ge M$, $$\left|\lim_{n\to\infty}\int_B \cos nx\ dx - \lim_{n\to\infty}\sum_{i=1}^m \int_{A_i} \cos nx\ dx \right| < \epsilon/2$$
This seems weird, it is as good as swapping limits
yes, and knowing that $B$ is measurable implies that there is a collection of open sets so that $B\subset\bigcup\limits_kA_k$ and $\sum\limits_km(A_k)\le M(B)+\epsilon/2$
@epsilon-emperor Why do we need a countable disjoint union? $\sum\limits_{k=1}^\infty \mu(A_k)\lt\infty$ implies that there is an $m$ so that $\sum\limits_{k=m}^\infty\mu(A_k)\lt\epsilon/2$
@epsilon-emperor Yes, that is a countable collection of open intervals, but we only need a finite collection to come within $\epsilon/2$ of the measure of $B$
The measure of the rest of the $A_k$ is less than $\epsilon/2$
Let $B$ be a measurable set. Then, there exists an open set $V = \biguplus_{k=1}^\infty A_k$ such that $B \subset V$ and $m(V\setminus B) < \epsilon/2$. Since $m(V) < \infty$, there exists $m$ such that $\sum_{i=m+1}^{\infty} m(A_i) < \epsilon/2$. $$\int_B \cos nx\ dx \le \sum_{i=1}^m \int_{A_i} \cos nx\ dx + \sum_{i=m+1}^\infty \int_{A_i} \cos nx\ dx < \sum_{i=1}^m \int_{A_i} \cos nx\ dx + \epsilon/2$$ Taking $n\to\infty$, $$\lim_{n\to\infty} \int_B \cos nx\ dx \le \epsilon/2 < \epsilon$$
Since $\epsilon > 0$ is arbitrary, I think we're done.
In fact, $m(V\setminus B) < \epsilon/2$ seems unnecessary
All we need is that some open set $V \supset B$ exists so that $m(V) < \infty$. Right?
hi, im not sure why this justification imgur.com/a/nls9tDq makes sense, for example if $f(z) = -1$, then surely we don't have $log(-1 ) + log(\overline{-1}) = log(|-1|^2)$ for any branch of the logarithm defined in a neighbourhood of $-1$?
Oh, I see your point. Well, OK, those differ at most by a constant if you change branches, anyhow. Just compute away from branch issues and use continuity.
It's not a matter of neater. It's a matter of valid. I think you have to listen to what robjohn was telling you. I was just trying to give you more understanding of what you did wrong. I'm not thinking about the question.
ah ok I see, so in other words besides at most some half line, we can ensure $log(f(z)) + log(\overline{f(z)}) = log_{P}(|f|^2 (z))$ holds , i.e. $\Delta^c(log_{P}(|f|^2(z))) = 0$ everywhere but that half line, and by continuity, also on that half line
is that the idea?
here $log_P$ is principal branch (i.e. outputs wholly real values on positive real line)
Dror Varolin has a book on Riemann surfaces and complex analytic geometry which discusses a lot of several variables. Though my favorite is Gunning and Rossi's text
I use it as a reference whenever there is some analysis I do not understand, and it is one of my favorite texts on analysis. It's chapters on Stein Spaces and Sheaf cohomology are excellent
i never owned it, i just borrowed it from the library. after a certain time in grad school you could borrow things in perpetuity unless they were recalled back. so i had it on my shelf
Same here. A friend and I are doing this seminar series on Fourier Mukai transforms, so we have been giving an extended period on borrowing all algebraic geometry texts and related areas. So we just keep cycling all these textbooks
Typically, advanced graduate-level texts have few or no exercises. None of these has exercises. Griffiths/Harris has no exercises. My favorite SCV book, Hörmander, has no exercises.
one time as a prof i needed a text and the library said it was checked out for 12 months and i said, just tell me their name and i will bother them directly or take it out of their office when they aren't there. they gave me the name.
That's why I favor Huybrechts. Lots and lots of crazy good exercises. Read stuff from Griffith-Harris, solve exercises from Huybrechts
Though his text on Fourier Mukai transforms has very boring exercises. I guess you cannot blame him there, most of that stuff is out in the papers, so except routine checks I do not think much can be given in terms of exercises
I have a typewriter. I like to take it out time to time, clean it and then keep it back. I used to write a lot on it when I was kid but I am afraid I might damage it now
@TedShifrin Regarding our conversation a day or two ago about $o(xt)\neq xo(t)$, I have made a post on the main site about the problem related to the question. It's about statistics, but the calculations are very basic calculus.
This is an old question, but here are some visual examples that help to suggest that indeed the Jordan curve theorem is actually not so intuitively obvious.
Both are not rectifiable. We have relatively good intuition about rectifiable curves but not non-rectifiable ones. These examples may s...
The lyrics of a song by The Chameleons made me wonder:
It's just coincidence
Well you can talk that way
But I have to say
I don't believe in it
After considering these lyrics, I have the following questions:
Is everything just coincidence? Doesn't fate exist? Are all the people meeting ju...
