Why is the zeroth homotopy set the same as all path components of the space? Since it is defined with a basepoint, wouldn't it be the set of all points that can be path-connected to the basepoint? Thanks!
@TedShifrin My definition is $\pi_0(X,x_0)=[S^0,1;X,x_0]$ which is the set of homotopy classes of based maps between the sets, so since $1$ always maps to $x_0$, doesn't it mean what I said above?
@robjohn thanks a lot professor Rob :) I never thought about that recursive sequence problem in the way you have considered. Fantastic solution and generalization :)
semiclassical, one time i had a paper rejected, citing a paper whose result was false, if you followed the reasoning or considered even a single example in my paper.
we had to write a letter to the editor saying, uh, this reason they cited for rejecting our paper is wrong, see example 1.2, and this is even more reason to publish the paper.
it happened at a very inconvenient time in my life, i was trying to run up my publication count and maybe get an academic career going. months matter in this, in terms of what you put on your CV.
it was, pardon my french, excremental to have someone stomping on my result when they didn't have a leg to stand on.
so i said f- it and lived off my girlfriend for a year. and now i am able to support the two of us and our child doing something else.
i don't mean to lure people out of academia, there is a lot of unfair stuff that happens in my business. it's just different stuff and not so much tied up with people's personal identities.
with all the junk that is on Google, they get rid of the content of decent newsgroups. The largest post I made on sci.math is probably dwarfed by one image posted here.
I understand what the question wants me to do. However, I would like to find out if there a clever way or a pattern that can be observed to find the sum the of $\sum \alpha^n$? I brute forced the $\sum \alpha^2$ and $\sum \alpha^3$. But I think the question wants me to observe a pattern or something, but I don't see it. By $\sum \alpha^3$, I mean $\alpha^3 + \beta^3 + \gamma^3 = (\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\alpha\gamma)+3\alpha\beta\gamma $
Nevermind, figured it out. Seems instead of expanding out $(\alpha + \beta + \gamma)^n$ to get an identity, you were supposed to make use of the cubic equation. I.e. $\alpha^3 = 2\alpha^2 - k$ and from there it's easy to find $\alpha^3+\beta^3+ \gamma^3$, $\alpha^4+\beta^4+ \gamma^4$ and $\alpha^5+\beta^5+ \gamma^5$
The following is an Example 6.7.8 in Spanier AT
If $C^*$ is the relative Alexander presheaf of $(X,A)$ (with some coefficient module $G$), the kernel of $\alpha:C^*\to\hat{C}^*$ is $C^*_0$ (the locally zero functions). To show that $\alpha$ satisfies condition (b) (which is gluing property), let...
in ZF, is $\{A\} = \{x: x = A \lor x = \varnothing \}$ an exhaustive definition of $\{A\}$? I'm trying to figure out if $x \in \{A\} \iff x = A$ is a tautology
de morgan wrote a very funny book, 'a budget of paradoxes,' compiled from his reviews of the work of what we might call 'cranks,' and correspondence with same. it's on the project gutenberg website and in a dover hardcover from 1910 or so. he doesn't just poop on them. when he sees original ideas he praises them for what they are worth.
his father in law was obsessed with the idea that negative numbers did not exist. that comes up a few times.
i used to have a habit of putting bills of large denomination that would be hard to pass at normal stores in my copy of that book. then i forgot about the habit. a few years went by and i opened up the book and $400 fell out.
i remember when the soviet union began to disintegrate. we were camping in the middle of nowhere and my dad gave me a quarter to go down and buy a newspaper. it was the failed coup attempt in the USSR.
i took one look at it as i brought it back and said "dad, you might want to read this."
Hello, everyone. I'm convinced I've finally found something that I can use for computing quotients in general, however, I just have one question now: what is the formal name of the algorithm or process by which you find the largest element common among a number of discrete sets?
Took me long enough to realize that you can compute $\frac{x}{y}$ in binary by computing the quotient of the individual powers of two, then finding the largest integer that is less than or equal to all the quotients.
err I suppose in this case, less than or equal to the smallest quotient.
Eh, something like that, actually.
A simple example: $\frac{7}{5}$. $5 = 2^2 + 2^0$. $\frac{7}{4} = 1$; $\frac{7}{1} = 7$. The largest value common to the quotients 1 and 7 is 1, therefore the quotient of 7 and 5 is 1.
If $n = a\cdot 2^b$ ($a$ is odd) is given natural number and $f:\Bbb N\to\Bbb Z$ is a function given by $f(n) = b$ (i.e. counts the number of $2$ in the unique factorization in to a prime number). Then $\sum_{k=1}^n f(k)\leq n-1$.
I want to prove this by induction. Base case is clear and I consider the case $n$ (and suppose the statement is true up to $n-1$, $n\geq 2$).
$\sum_{k=1}^n f(k) = \sum_{k=1}^{n-1}f(k) + f(n)\leq n-2+f(n)$ by induction.
I need to conclude $f(n)\leq 1$ which is false in general.
I need to approach in different way. Could you help?
@AMDG however, since these are defined only on integers, it does not need to be any logarithm. We could even define $f(p)=1$ for all primes $p$ and get a function that is not a logarithm.
i'm surprised at the level of seasonal variation around here. it is less than what i grew up with but people who say LA doesn't have seasons are overlooking a lot of obvious signs.
it doesn't have months of crummy weather, is what they're saying.
Ok, so assuming what I have here is a valid algorithm for computing quotients, then the theoretical optimum will be O(n) to compute the power of two quotients, O(lg(n)) to compute the comparisons. On Zen 2, that should translate to 1n cycles for n concurrent shifts, and 6n cycles for n concurrent compares.
In theory that means a theoretical optimum of 7c for a single integer divide compared to 41c for div/idiv.
So it would be only $2 \frac{2}{3}$ times slower than multiply.
We can use an n-bit LUT to get us closer to this theoretical optimum, accounting for cache misses, data location (icache is optimal), cache line, and fetch size (32 bytes iirc).
To remove the cache misses, and in consideration of the 32 bytes that are pulled from main memory every time there is a cache miss, we can make a 5-bit unconditional jump table to keep us in icache and our data to operate on at $\lceil \frac{64}{5}\rceil = 13$ as the practical software optimum implementation.
Then our value $n$ here can be computed from $13$ as $\lceil\frac{13}{8}\rceil = 2$ assuming you can operate on eight bytes at a time per 64-bit SIMD register of your choice.
@robjohn Since you asked for me to mention it, I figure now is a good time to mention it "officially" and ask what your thoughts are. I'll have to respond later, though, as we'll be going to a farm for groceries right now.
@AMDG most of what I've seen you mentioning are divisions with small quotients. How does it work with things with larger quotients, say $\frac{2^{63}-1}{11}$?
> @schn well, I mean that for different $u$, the convergence of $\frac{f(h)}{h^m}$ to $0$ might be altered. The limit may be $0$, but the rate of convergence might change. It depends on the context how and if this affects the result. That is why it is difficult or impossible to create a calculus for little-o.
Take for instance the Taylor expansion. Are you saying that one needs to know the rate of convergence of $\frac{f(u,h)}{{(hu)}^m}$ to $0$ for every $u$?
@schn it all depends on the application to which it is being applied. There is no way to make a generalization that will be true in every case. Some cases require a uniform convergence and some don't.
As I said before, look at actual cases where little-o is being used before trying to generalize things.
it looks simple, and in definition it is, but it can be tricky to use in some situations.
Use the definition. work from there. $f(x)=o(g(x))$ near $0$ if $\lim\limits_{x\to0}\frac{f(x)}{g(x)}=0$
that is the extent of the definition. That there is only one variable, and often there are more in a given problem, leaves a bit to be handled by the author.
it is a tool, and one often needs experience or training to use a tool
Question: Let $P$ be the stationary transition probability matrix of the Markov Chain $\{X_n:n\ge0\}$ which is irreducible and every state has period $2$. Further, suppose that Markov Chain $\{Y_n:n\ge 0\}$ on the same state space has transition probability matrix $P^2$. Both the chains are assum...
Hi, I was trying to figure what’s the dimension of 2 by 2 hermitian matrices consider as a vector space over C. If these 2 by 2 hermitian matrices are considered a vector space over the reals I know the dimension is 4 when considered as a vector space I think their dimension should be 3 because we will have one constraint that off diagonal terms are complex conjugates of each other. But when I tried to find a basis for these hermitian matrices considered as a vector space over C, I couldn’t
Can anyone help me figure out a basis for hermitian matrices considered as a vector space over complex numbers.
Hello, any idea why $for ~ i \leftarrow 1~ to~ (n-1)$ takes $3n-1$ operations? I guess assignment to 1 is 1 op, comparing to $i<(n-1)$ takes another 1 op and finally $i++$, but $i++$ is supposed to take 2, so the total should be $4n-1$. Is not it please?
@TedShifrin Ok I see what you mean, H_n is hermitian but iH_n isn’t hermitian. So it won’t be a vector space in the first place
@robjohn I suppose you meant “think of (1+i, 1+i)”. Well that is just equal to (1+i)(1,1)+0(i,-i). Sorry what did you want show there, I couldn’t get. Can you give a further hint or something
@robjohn Also I am confused how do you have as a basis vector (i,-i) when you are considering a real vector space. The thing inside the “(“ and “)” should be a real number but “i” is an imaginary number which is not in R. So how is it a basis vector in the first place….
@Shashaank The two dimensional real vector space with basis $\{(1,1),(i,-i)\}$ has the element $(1,1)$. This cannot be realized as a complex vector space because if we multiply this element by $1+i$, we get $(1+i,1+i)$ which is not in the space.
@robjohn $11 = 2^3 + 2^1 + 2^0$. $\frac{2^63 - 1}{8} = 2^60 - 1$. $\frac{2^63 - 1}{2} = 2^62 - 1$. $\frac{2^63 - 1}{1} = 2^63 - 1$. $2^60 - 1$ is the largest integer common to all of these quotients, but we can analytically determine that it is not the answer because this value multiplied by 11 is greater than the dividend, hence the issue. The brute force approach would be to decrement this coefficient and test to see if it is less than or equal to the dividend.
This is primarily why I asked what the formal name of the process or algorithm of finding the largest common integer (which I will define as best I can) is, as that is effectively what I'm doing here.
@Shashaank The complex numbers can be considered as a real vector space with the basis $\{1,i\}$. The basis vectors do not need to be in the base field.
I'll begin by assigning a mnemonic for brevity: $\text{greatest common integer} \sim \operatorname{gci}(A, B)$. Let $\operatorname{gci}(A, B) := \operatorname{gcd}(A, B) \text{ for } A[i],B[j] \in \mathbb{N} \text{ where gcd is greatest common divisor.}$
Someone posted an easy question about geodesics on the $n$-sphere and deleted the question after I'd made some obvious comments that he/she had not previously understood. I think that was more a matter of embarrassment, but we never know now.
I'm not much of a believer in justice these days. Despite having been vaccinated months ago and having taken great care for a year and a half, I believe I'm now the proud owner of a breakthrough case of COVID. Just went for a test. I am so pissed off at the selfish, stupid assholes.
Yes, math. One was a line integral over an implicitly-defined curve. The other was a differential systems (differential forms + geometry) question.
I'm sure there are others I've spent nontrivial time on, but those stand out in my mind (because in both cases I got a little help from a colleague or another).
@TedShifrin Hm, well, case in point: if you received a vaccine, then it should protect against what it vaccinates against. If it does not protect against what it vaccinates against, then it is not a vaccine, but a farce.
