@schn little-o is really only a one variable family. When we have two variables involved the estimates may or may not be uniform in the second variable. These need to be handled on a case-by-case basis, which is why the calculus of little-o that you seem to want may not always be what you want.
does rudin have something on the existence theorem for first order differential equations that states that approximations to the solution of an initial value problem converge to the solution?
it seems like the way which we use for solving homogeneous linear ordinary differential equations with constant coefficients.
I had never seen it before. I wanted to understand how to choose the particular/trial solution in case the recursion is not homogeneous. My understanding is that it is chosen by hit and trial
For example if I have a recursion: $x_n= a x_{n-1}+b_{n-2}+\sin n$, then what can we choose as particular trial solution. In case of differential equation, if I remember correctly we choose A cos n +B sin n. But here?
@robjohn I meant that the method on that link looks similar to the one used in case of linear ordinary homogeneous differential equations such as $y"+ay'+b=0$ where we assume the solution to be $y=e^{\lambda x}$ and solve for $\lambda$.
I was using the fact $|f|=-|f|$ implies that $|f| = f$. Don't we have that in the above case? (I realize you're right but I'm trying to figure out where I went wrong)
For example if I have a recursion: $x_n= a x_{n-1}+b_{n-2}+\sin n$, then what can we choose as particular trial solution. In case of differential equation, if I remember correctly we choose A cos n +B sin n. But here?
Though, I have never faced such question before so my guess is $A\cos n+B\sin n$
I thought I could do the inverse operation on the absolute value and integral on both sides, but I'm realizing I didn't think of the inverse operation of $\int_{x_0}^x$
@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.
basing estimates on something tending to $0$ is a lot less precise than basing them on something bounded by a constant.
big-O can fall prey to the same problems if the constant involved might change with some parameter in the equations.
but it is usually easier to account for these changes in the constant than it is to account for the change in convergence to $0$
Is there any good basic FOL textbook which covers the following things: (1)FOL syntax (2)A logical deductive system (Natural Deduction fitch style) (3)Conventional and self contained (4)Only requires pen and paper.
So I was teaching some divisors and somebody asked me this and I have never thought about this before (which is embarrassing but okay). So suppose we have a cartier divisor $D$ which is effective. Corresponding to this there is a line bundle $[D]$. In terms of locally free sheaves, this line bundle is $\mathscr{O}(D)$, and also the dual $\mathscr{O}(-D)$.
The sheaf $\mathscr{O}(-D)$ is what I thought were sections of the dual line bundle $[-D]$. But in the effective case, the definition of $\mathscr{O}(-D)$ implies we are looking for holomorphic sections. But for instance in something like a RS ($\Bbb{P}^1$), $D = p$, $[D] = \mathscr{O}(1)$ and $[-D] = \mathscr{O}(-1)$. But then the notion that $\mathscr{O}(-D)$ is the sheaf of sections of the line bundle $[-D]$ doesn't work. Am I missing something trivial?
Here by $[-D]$ I mean the dual line bundle to $[D]$, where $[D]$ comes from using the cartier divisor definition
I have $e^{-Px}\int Q'(x)\frac{e^{Px}}{P} \ dx$, with constant $P$ and $Q'(x) \to 0$ when $x \to \infty$. Clearly, expanding the integral with integration by parts will result in the integrated $e$ term being canceled by the $e^{-Px}$ in front of the integral so the whole thing will converge to $0$
but... how do I head in that direction more rigorously?
we agree. there are questions i haven't asked lately because i couldn't come up with anything resembling an 'attempt.' they are cooler about this on math overflow but i forgot my account name there.
MO also seems a little more focused on the status or situation of the questioner as opposed to the quality of the question. maybe that's unfair but it's the impression that i get.
MO is cool but i see even less activity there ? less questions and answers ? ofcourse they have a higher level so that is a good excuse ... but is it growing or shrinking ?
i dont know ... i never wanted to call those ppl by name , bc that would escalate and be ungrateful ( they answer alot often too ) but after all those years , perhaps that is the only way ...
i feel the context argument should be removed entirely. math is math , no explaination needed. imo homework is ok too , but should be tagged and filtered
@Leslie, further to yesterday’s discussion: proof for the sequence $(x_n)$ defined by $x_n=\cos n$ does not have a limit : since $\{\cos n: n\in \mathbb N\}$ is dense in [-1,1], there must exist subsequences of $x_n$ converging to 1 and -1 respectively that is $\limsup x_n=1\ne -1=\liminf x_n$ whence it follows that $\lim x_n$ does not exist.
on my site we sometimes close older questions as duplicates in favor of pointing people to newer and better specified ones, though that's because (barring some exceptions) questions on our site can't actually be answered, as in there is no concept of "OP has gotten the answer they wanted, no more is to be added"
here it may be similar though where the duplicate closure favored the one that wasn't a PSQ, but I don't know what Math.SE's policies are on actually retroactively dupe-closing questions especially if they're that old.
koro that is an interesting idea. it is possible to prove that lim cos(n) does not exist without relying on the density result. if the limit exists it has to satisfy a ton of trig identities.
yeah we can without many trig identities. we can argue like this: suppose on the contrary that $\lim \cos n=l$, it follows from identity $\cos (n+2)-\cos n=-2\sin (n+1)\sin 1$ that $\lim \sin (n+1)=0$, and trig. identity $\sin (n+2)-\sin n=2\cos (n+1)\sin 1$ whence we get $\lim cos (n+1)=0$ which is a contradiction as it’s in violation of the identity $\sin^2 t+\cos ^2t=1$
Hi professor Ted. I think that density of $\{\cos n:n\in \mathbb N\}$ makes life easy :)
Leslie, I can’t stop thinking about your yesterday’s suggestion of using “continuous functions are completely determined by their values on dense sets” for proving density of some sets. I think that’s a very powerful application which I never thought of. Thank you so much for that :)
@hyper-neutrino This is, more or less, correct. The goal of marking a post as duplicate is to provide a roadsign for future askers. The "best" version of the question should be at the root of the tree of duplicates. This can mean that an older question may be closed as a duplicate of a newer question.
@XanderHenderson Thanks for the explanation. Yeah, that makes sense, and I agree with it too; I've never considered duplicates really as "this question has been asked before" but rather "this question is the same or a subset of this other question, and we would like to point you to a place with more resources / that you should answer instead".
Granted, that usually is the same as "this has been asked before", but like in this case, not always.
Problem:
Assume we have the following points:
$(x_0,y_0), (x_1,y_1), (x_2,y_2), (x_3,y_3)$ where $x_0 = -3$, $x_1 = -2$,
$x_2 = -1$ and $x_3 = 0$.
Given the function $f(x) = Ax^2 + Bx + C$
find the constants $A$,$B$ and $C$ such that
$f(0) = y_3$ and
$$d = \sum_{i = 0}^{2} (f(x_i) - y_{i})^2$$
is...
@copper.hat. For example, given high dimensional data of size 60 (60 dimensions), how I can model that with density function and then find out low density areas of this fuction by projecting it to 2d manifold for example.
Hello, if we have to prove "A(n) iff B(n) for all n>1" , and we previously showed that B(n) is false for all n>1; does it suffice to prove that A(n) is false for all n>1?
How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y>0$ ?
($x,y$ are integers.)
