Say I have a section of some fiber bundle, the total space of which is a complex manifold. Suppose the image is a holomorphic submanifold (with the induced complex structure). Then why is it absurd if we reach the conclusion that the image is horizontal?
@MikeMiller So if I call the total space $S$, $TS= V\oplus H$ splits into vertical and horizontal fibers, the splitting is induced by the Levi-Civita connection on the base $M$ in this case (I didn't think the details were relevant to this question---maybe they are?)
Hey! Don't know if this is the right place but: I have a small question: First: There is not enough (for example) C Programs for all of the subsets of the natural numbers. So for example let M be a subset of natural numbers and P be a C Program which gives 1 with input n if n is an element of M. We cannot find for all subsets such a corresponding C Program. The reason is that there exists uncountably many subsets but only countably many C Programs. I am trying to find another proof/counterexample for this question. We can encode M as an 0-1 sequence (n is in M iff the nth element of the seq…
@PVAL-inactive The argument is: Suppose the total space is (not just almost) complex and the image of the section is a holomorphic submanifold. The conclusion is that the image is horizontal---this is supposed to be a contradiction.
@TedShifrin This question is from a mathematics lecture and I think it is a discrete mathematics question and somewhat interdisciplinary. I am just trying my chances.
@Konformist: If one set is countable (although I guess I have no idea why the set of C programs must be countable) and the other is uncountable (that I know), they can't be in correspondence. I don't know what you mean by a counterexample. One particular subset doesn't establish uncountability.
@TedShifrin No, for example all even natural numbers can. See my first message again, where I explain the input and output of the corresponding program.
@Danu If the total space is an actual complex manifold, so is the base, and the fiber bundle projection is holomorphic then yes, you're literally solving CR.
@TedShifrin We have a subset of natural numbers M. We want to create a program which gives 1 with input n if n is an element of M and 0 if not. And we can see this subset M as a 0-1 sequence. So for example 0,1,0,1,0,1,0,1,0,1,0,1,0,1... is a 0-1 sequence and the corresponding subset is even numbers (don't include 0 to natural numbers) and we can write a code which gives 1 if the input is even and 0 otherwise easily.
@TedShifrin And the claim was that there are some 0-1 sequences, for which there is no corresponding program and one can prove this using the uncountable/countable argument. I want to find a direct counterexample which uses the idea of non-periodic (or a similar definition) seqeunces.
I understand the abstract claim, of course. But you're pretending that you have a recipe that can tell me if a particular sequence cannot be programmed. So explain that to me.
I mean there are probably simple languages where you can describe definitely everything you could possibly do in these languages (even if they allowed for infinite programs)
,but I doubt thats possible to do in any functional one.
@TedShifrin 'I' don't have such a recipe. If non-existence of a recipe indicates the impossibility of finding a counterexample, than I would like to understand why there is no such recipe. And also I don't follow why this indication is true.
Pi in diadic should be (theoretically) very easy. There are series approximations for pi and one has plenty of error upper bounds for these series estimates to tell you when you have calculated a correct digit.
My intuition (really wild guess) is that as long as a language can encode any finite Turing machine, there shouldn't be a way to tell if a given sequence can be encoded.
I understand that if we have 2 arbitrary sets, one is countable one is uncountable and a fixed element from the uncountable set, than we can choose a (non-surjective) injection from the countable set to uncountable set which cover this fixed element. So in this sense, we cannot find a counterexample. But these sets are not arbirary, so I am not sure about this argument.
Let $f(n)$ be the largest real solution of
$$x^n - x^{n-1} = 1 $$
As $n$ grows to positive infinity we get the asymptotic :
$$ f(n) = 1 + \frac{\exp(2)}{n} + ...$$
Where the value $\exp(2)$ is optimal !
( and $...$ means smaller term(s) )
Notice $f(2)$ is the golden mean.
How to show this ...
> Just curious, for which rational $c_i$ can the algebraic number $\sqrt{c_1+c_2\sqrt{c_3}}$ be "flattened," i.e., written such that there are no nested radicals? It's possible for some $c_i$, such as if $c_1=d_1^2+d_2^2d_3,c_2=2d_1d_2,c_3=d_3$ for some $d_i$ Then, $\sqrt{c_1+c_2\sqrt{c_3}}=d_1+d_2\sqrt{d_3}$ (excepting negative things, etc., they can be dealt with separately) But yeah, are these the only solutions?
@Legion: That's a classic question that's usually addressed in abstract algebra courses (Galois theory). $\sqrt{a+\sqrt b}$ can be written as a sum of square roots of rationals if and only if $\sqrt{a^2-b}$ is rational.
You can derive it intuitively, but probably not prove what I said, without more, @Legion.
But that is the answer.
If $\alpha=\sqrt{a+\sqrt b}=\sqrt c+\sqrt d$ then it'll turn out that $\beta = \sqrt{a-\sqrt b}=\sqrt c-\sqrt d$ (if you make the right choice). Now play with it.
Let $f(n)$ be the largest real solution of
$$x^n - x^{n-1} = 1 $$
As $n$ grows to positive infinity we get the asymptotic :
$$ f(n) = 1 + \frac{\exp(2)}{n} + ...$$
Where the value $\exp(2)$ is optimal !
( and $...$ means smaller term(s) )
Notice $f(2)$ is the golden mean.
How to show this ...
