C.r.u.d.e chat room is intended to help with various "janitorial" task such as closing, reopening, (un)deleting, editing and improving post. I am posting here this information since a few users are currently trying to resurrect this room - it was rather inactive for some time.
U(P,f)U(P,f) is non-increasing. Also, the ultimate goal is not to approximate upper/lower sum. The goal is to make the gap U(P,f)−L(P,f)U(P,f)−L(P,f) smaller and smaller so that whats in between, namely, the definite integral, gets squeezed in.
@AkivaWeinberger Suppose x^2 + ax + b = 0 for integers a and b and real number irrational x. It then follows that there exists a unit y in Z[x] such that 1 < y and there do not exist any units w such that 1 < w < y.
that statement is sufficient to then prove that all units in Z[x] are of the form $\pm(y^n)$
This question arises after the realisation that countable sets and uncountable sets does not necessary differ in the way where the former has gaps while the latter has no gaps. The best topological notion that capture the property of gaps is total disconnectedness. Quoting Wikipedia:
A totally d...
The recurrence I had in mind is a variant of the second recurrence in this table: https://math.stackexchange.com/questions/873899/successive-ratios-of-a-sequence-is-this-limit-true The recurrence for the rationals goes outside the borders to the top of the matrix of rationals.
Proof: Take the set of elements in the Cantor set that have a successor. This set is dense (in the order-theoretic sense) and countable, and thus it is order-isomorphic to $\Bbb Q$.
Someone told me a pretty neat problem today: Assume you have a rectangle of size $A\times B$ which is covered by smaller rectangles of sizes $a_i\times b_i$ for $i\in \{1,\dots, n\}$ for some $n\in \mathbb{N}$. And assume further that for each $i$ at least one of $a_i$ or $b_i$ is an integer. Show that at least one of $A$ or $B$ is an integer.
if they are at least aligned, then we can at least do some a divides b type argument to figure out why integers, but this just throws that out of the window
@Secret You can't cover (with disjoint interiors) a rectangle with a finite number of rectangles unless those smaller rectangles have sides parallel to the large sides
It was sort of an "on a lighter note" after we had spent some time talking about whether a certain thing was really a knot invariant and if so whether it could distinguish the unknot
@Ted for what we were talking about yesterday, if we know that $f$ is holomorphic and nothing else, could we try to integrate it to get a function which we know is $C^1$, then have that it's analytic, and backtrack to say that $f$ is analytic? It feels cheap but it could work...
If both sides are not of integer length, you know there's a disparity, which you can tell by first placing it such that it aligns with the corner and checking it there
But then shifting it around shouldn't change anything
That or just place the large rectangle at a corner, if it didn't have integer side length it'd cover different white and black, but then you partition it into rectangles which do. Contradiction.
Anyway gtg, and @Alessandro you'd need to prove that holomorphic functions have harmonic real/imaginary parts, and to know that you need to be able to take second derivatives of real and imaginary parts, which is why I was hesitant on that note
@Tobias this solution is secretly looking at signed measures for what it's worth. Anyway, see you guys!
@Daminark you can weaken your notion of solving so that you don't have to do any of that
i think it's true for example that $f \in L^{1}_{loc}$ and only satisfies C-R in the sense of distributions then it's analytic a.e. and this is kind of an instance of more general kinds of regularity results you see
@AkivaWeinberger Here is a sketch of the proof. For a quadratic real integer a + bx another element c + dx is greater in real magnitude if and only if a > c and b > d. Therefore, a unit greater than one must be of the form a + bx where a > 1 or b > 0. However if 1^2 = 1, then a^2 > 1 = 1. Therefore a > 1 and b > 0. Because the integers are discrete this set is discrete for all elements greater than 1. Therefore, we can select the unit succeeding 1 in the real numbers.
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Have to prove a few properties first but that is actually how I proved it for Z[sqrt(3)] with the unit 2 + sqrt(3)
@TobiasKildetoft in Z[x] where x is a real irrational such that x^2 + ex + f = 0 for integers e and f there exists a unit y such that y > 1 and no units are between y and 1.
its a rough sketch
It builds up to showing that all units are integer powers of y or negatives of powers of y.
Assuming you mean in the ordering of the reals, it fails because your characterization is not even a total order
If you mean in terms of the norm on the quadratic extension, you don't get an ordering and anyway, the characterization is still wrong since increasing the second component decreases the norm.
@TobiasKildetoft if you refer to when I state that a > 1 and b > 0 the reasoning there is that the norm of a is a^2 > 1 = 0. Therefore we need b to be nonzero to decrease the norm. Therefore, b is greater than 0.
@TobiasKildetoft "For a quadratic real integer a + bx another element c + dx is greater in real magnitude if and only if a < c and b < d" <=== This? Alright then. I'll prove it. Give me a moment to prove it. It's just a matter of working from memory. :-) I know for a fact that it is true when x is real and irrational. Perhaps you are thinking of rational solutions or complex solutions?
In the text "An Introduction to Measure and Integration by Rana" i'm having trouble gaining intuition behind the following Proposition in $(1)$
$(1)$
$$\text{Proposition}$$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, ...
