Let's take some random sampling of topics. (1) Synthetic geometry. (2) Analysis. (3) Algebra (though I don't see how it's related here). (4) Measure theory (for fun).
In this, I somehow find that synthetic geometry alone somehow presupposing that we are going to study R^2, may be R^3.
For Euclid it might have been ok because the world is after all 3-dimensional. But surely we would want a theory that can generalise to any $n$, no?
I forgot topology in the list. The list doesn't matter.
I'm fine with assuming that synthetic geometry studies R^2 or R^3. May be you can write axioms defining R^4, but that's alright. In fact, I don't even care that much.
Here's what I do care about.
So what about non-Euclidean geometry? I know of the traditional account that you get it by taking Euclid's axioms minus the parallel postulate, adding some other replacement (I'm not sure about this).
But does it mean that non-Euclidean geometry is somehow "two-dimensional"?
Ok, I'm done with my talking. If you need some clarifications about what I said, then I will elaborate. Otherwise, you may take the stage now. =)
You could leave a comment. He seems to check the site still.
Lots of clever answers there.
I really enjoy Bill Dubuque's answers but sometimes he seems to rag on other responses. Maybe we should all just have thicker skin and be able to take criticism, for our betterment. I don't know.
I forgot why you do what you do when localizing a graded ring at a homogeneous prime and it isn't really in Atiyah-Macdonald, Eisenbud, Matsumura, what have you.
Maybe this isn't very deep but I always find it confusing.
A-M talk about it for a bit when they do completions.
> The development team, at this time, included Jeff Golden (language, compiler, etc), Bill Gosper (special functions, summation), Howard Cannon (user interface, optimization), and consultant Bill Dubuque (integration, equation solving, database, optimization). Other developers made major contributions in numerical analysis, graphics, and help systems.
@tb Say you write an answer but don't post it; then it's saved, right? Won't it disappear if I clear the browser cache? I thought everything is stored locally, but QED says otherwise.
@Gigili Yes, I mean a unique identification of the book. I'm pretty sure that in every written language there exist at least a dozen books with that title. I honestly don't understand the point of being secretive about it.
@DylanMoreland It is not circular but it doesn't answer anything. It just hides the difficulty in step 2 without mentioning how it is done. A blatant non-answer, nothing smart-ass about it.
The real question is this: If a square matrix A has a left inverse, then it has a right inverse and they coincide. There's an irreducible core of difficulty here; and I don't find it addressed in that post.
@QED You might get a correct proof; I'm not disagreeing with that. What I am thinking is that the all the group-theory stuff mentioned there will turn out to be extra stuff that is not relevant to this problem.
For example, in a solved example, finding the roots of $x- \sin x=0$ by NR method .. it says $\frac{x_{n+1}-x_n}{x_n - x_{n-1}}$ .. putting $n=6$then $ x_7=0.02909 x_6=0.04364 x_5=0.06547$ , from $\frac{x_7-x_6}{x_6 - x_5}$ which is equal to $\frac{m-1}{m}$ m would be $m \approx 2.9985 $ (4D)
@tb Nothing secretive, I just think the context wouldn't help in this case ... I know there are a dozen books with that title but it's not translation, it's originally in Persian .. "Numerical analysis , Ismail babelyan " I just don't know how to type the author's name in English .
I added this comment "It's not known that A and B are invertible - so you have to prove that before using group theory. Proving A and B invertible solves the problem without any group theory"
Maybe I should read the answers again: did anyone say that $AB = I$ implies $A$ surjective and $B$ injective, hence $A$ and $B$ are bijective, inverses are unique, blah blah?
@Gigili I see. Since I don't read Persian I can't help, of course... Doesn't the author say anywhere (maybe at the beginning of the solutions) what those suffixed parentheses are supposed to mean? Maybe (4S) could mean four significant digits, or something like that.
@DylanMoreland If A is surjective and B injective, it is not at all clear that A and B are bijective. In fact, again that is what the question asks basically.
@DylanMoreland On the contrary, that will only confuse a beginner in this thread. Note that there's the subtlety of left-inverse, right-inverse and honest to goodness inverse.
@QED That's why I didn't say the identification of the book cause I thought it wouldn't be any of help. No, there's nothing about it but I'm wondering if it's relevant to the first chapter about errors and ways they can be arisen , something like that .. Or maybe round to n significant digits? So you think the question on the main site would be vague or off-topic?
@Gigili I haven't read that many numerical analysis texts, but as I said, I've never seen such a thing. The problem is that this may well be a notational peculiarity of the author and as such I think it's not a very good question for the main site. But that's just my opinion, I might be completely mistaken.
@Gigili I'm looking at those. The missing information in there is that the seed was not given. On the other hand, the slow convergence is expected since $x=0$ is a triple root for the function $x-\sin\,x$ (note the Maclaurin series).
@Gigili: I'm guessing it's respectively "significant figures" and "decimal places". I'm not sure why the notation is needed; I'd think one'd already have the habit of displaying only the figures you can justify...
e.g. if your starting data is only good to six or so figures, then there are rules on how many figures should you be retaining in the final result. That being said, one should always do the rounding at the end, and not in intermediate stages.
Someone (elsewhere) just asked about an exercise I couldn't do back when I did Linear Analysis. The question was, given a unitary operator $U$, how do we deduce that its spectrum is non-empty by considering the operator below? $$i (U + \textrm{id})(U - \textrm{id})^{-1}$$
It seems to be hinting at a reduction to the spectral theorem for self-adjoint operators, but we only learned the spectral theorem for compact self-adjoint operators in that course.
@ZhenLin It's the inverse Cayley transform. However, this is an unbounded self-adjoint operator in general (note that $1$ might be in the spectrum of $U$, so the inverse isn't defined everywhere.
(On the other hand, that does look like something that somebody too lazy to write many Pochhammer symbols would do... it would probably pop up if you're dealing with those generalizations of hypergeometric functions.)
@tb Don't worry about it. If I can't remember the solution then that probably means it was non-trivial / involved mysterious tricks / couldn't be solved.
So, if $1 \notin \sigma(U)$ then $(1-U)$ is invertible and it's not hard to check that $S$ is actually self-adjoint. However I don't think it's compact in general (the spectrum of $U$ is mapped bijectively to the spectrum of $S$ via $\lambda \mapsto i(\lambda+1)(\lambda-1)^{-1}$ and there's no reason for the image to be a sequence converging to $0$).
@ZhenLin In fact, the Cayley transform $\kappa(S) = (S-i)(S+i)^{-1}$ is a bijection from the densely defined symmetric operators and the isometries $U$ on $H$ with the property that $(U-I)$ has dense range. Moreover, $U$ is unitary if and only if $S$ is self-adjoint.
@robjohn: hi. I think I'm happy now with your Stirling answer. I still need to do some scribbling (I still would like to figure out how your approximation relates to the error function I'm used to: $$\mu(z) = \log \Gamma(z) - (z - \frac{1}{2})\log{z} + z - \frac{1}{2} \log{(2\pi)} = -\int_{0}^\infty \frac{w - [w]-\frac{1}{2}}{z+w}dw,$$ but I'll manage to do this myself.)
@tb But I am still working on the easiest way to finish off the answer. I had tried to give the general idea without cluttering with what I think are usual details, but I guess they are not so usual to everyone.
@robjohn I mainly managed to confuse myself a lot by staring at the graph of $u^2/2$ without actually thinking... It's certainly not your fault. I really like the recursion for the approximating coefficients that you can give.
What is the age of the people and is there any discount?
@Ilya You can check out the railroads page of Belgium: b-rail.be/main/E You can also go to the ticket boots and ask what would be the cheapest since for groups it can be cheaper if you buy a "10x ticket".
@JonasTeuwen thanks a lot. could you tell me, where is the city center? From central station I can go either in the Zoo direction or Stadspark direction
This question reminded me of an exercise I forgot how to do: Given a function which is analytic and non-injective in some ball $\{|z| \lt \rho\}$ then there are two points $z_1 \neq z_2$ with $|z_1| = |z_2|$ such that $f(z_1) = f(z_2)$.
As I said, I don't remember how to do it at all. However, the assumptions do not prevent the function to be injective on some small ball, so I don't see how the local picture helps.
@Ilya Ah, you are talking about the evening before. We spent that with some sparkling cider with cheese and crackers before midnight. We packed it in soon after midnight.
@Matt I don't understand what you say: you try to reduce to $f \geq 0$. So assume it's already solved and let $f$ be arbitrary. Write $f = f^+ - f^-$ with $f^{\pm} \geq 0$ and approximate them with simple functions $s^{\pm} \geq 0$. Then approximate $f$ by $s = s^+ - s^-$ and use the triangle inequality.
@Matt I think the point that is being made is just that you weren't very explicit about the reduction and commenter spells it out and tells you that you should do so, too.
@tb Got it. Thank you. You don't want $s$ here to be a simple function because simple functions are positive so writing them as positive and negative parts is unnecessary. If you assume that it's solved then $s \in L^1$ with finite support.
