to prove that J(x) - J(x-1) is about x/2 We investigate J^-1 ( J(x) - J(x-1) ) ... and then a contour integral seems usefull. YOU see now why its intresting ? @BalarkaSen
I am wondering how to calculate surface of the church tower in the picture, for painting purposes. Especially, I am interested in the two 'onion-shaped' parts.
I am thinking, that it is not really round, it consists of 6 equal parts, that are twisted, but would look like this on plane: like this:...
I did calculus this year(Differential and integral calculus of functions of one variable, differential equations, partial derivatives, vector geometry, matrix algebra, complex numbers, Taylor series) @JasperLoy
@robjohn I think it can be done in terms of factorials and lots of sums of geometric series. Something along the lines of summing them over an order n! times
I am wondering how to calculate surface of the church tower in the picture, for painting purposes. Especially, I am interested in the two 'onion-shaped' parts.
I am thinking, that it is not really round, it consists of 6 equal parts, that are twisted, but would look like this on plane: like this:...
@Sabಠ_ಠ I would say that the 2 Apostol books and the 3 Marsden books cover about the same material. Apostol is slightly more rigorous in some parts and has some more topics, but that rigour can be made up for in Rudin later on and those topics can be better studied elsewhere later on.
Let $x>0$.
Let $J(x) = \sum_{n=0}^{oo} \dfrac{x^n}{2^{n(n-1)/2} n!}$
Let $J^{-1}(x)$ be the functional inverse of $J(x)$.
How to show that for all $x$ there exists a fixed positive real constant $C$ such that
$(J^{-1}(J(x)-J(x-1))-\dfrac{x}{2})^2 < C $
@Sabಠ_ಠ C, AC, RA can mean many different things. One has to look at the content of the course. Even looking at your syllabus it is hard to figure out exactly what level the instructor will treat the topics.
@robjohn by the way, that generalisation I showed you is one of the greatest achievements in the last months. Absolutely amazing. (the difficulty level is pretty high, a Ramanujan difficulty level)
@BalarkaSen I object to the notation $S_r[x]$ (not only because the $S$ should be a $B$, but the brackets are prone to confusion), but yes, that's right.
Consider a different situation : Let $(X, d)$ be a metric space. $y$ be a limit point of $S_r(x)$. Then for every $r'$, $S_{r'}(y) \cap S_r(x)$ is nonempty.
takes no notices of @DanielF's notational complaints
@BalarkaSen welcome back !! I recently (10 min ago or so ) asked a question that was the inspiration for the question we discussed. see my last link. Maybe you can answer ?
@DanielFischer Now, let $y$ be outside of $S_r(x)$. Then for every $r'$, there is an element $z$ in $S_{r'}(y) \cap S_r(x)$. $d(x, y) \leq d(x, z) + d(z, y)$ by triangle inequality.
quick question here:
In my proofs class we had a problem that after a little work we end up with:
$x(x-y)=(x+y)(x-y)$ where $ x = y $. Now, I know this is pretty basic, but my teacher said that for the next step, one cannot cancel out $(x-y)$ from both sides as $(x-y) = 0 $.
Can someone explain...
@UserX I mean that once you develop some thinking skills to work through problems, you can apply them to all problems instead of depend too much on others.
I came across the use of Descartes' theorem while solving a question.I searched it but I could only find the theorem but not any proof.Even Wikipedia also, just states the theorem!!I want to know the procedure to find the radius of the Soddy Circle??
I apologize if its duplicate and to mention ...
:18263548 It says that the square of the sum of the curvatures of $n+2$ mutually tangent spheres in $\mathbb{R}^n$ equals $n$ times the sum of the squares of the curvatures.
In any case, my corollary, which I did a lot of research to conclude it was not known in this form, is that the average of the centers of these spheres weighted by their curvatures equals the average of the centers weighted by the square of their curvatures.
Given the centers and curvatures of $n+1$ of the spheres, these two theorems allow you to get not only the radius of the $n+2^{\text{nd}}$ sphere, but the location of its center.
There are actually two possible $n+2^{\text{nd}}$ spheres, but they come as a pair of solutions to the quadratic equation for the radius of the last sphere
@Chris'ssis I wish I had seen that question... I should post my proof... It is much simpler.
Hello, One question please : Let $K$ be a non-empty compact, connected. If the path connected composant of $K$ are reduced to a point, does $K$ is reduced to point too?
@DanielFischer quasicompact is about Borel-Lebesgue and compact spaces is related to hausdorff : this course will start the next week but it said a few words of it yesterday.
@MarcGato So compact means quasicompact and Hausdorff? That's good. On the one hand. On the other, I don't know of a Hausdorff example of a connected quasicompact space whose path-components are all single points. As far as quasicompact spaces are concerned, $\mathbb{N}$ with the cofinite topology gives an example.
@DanielFischer Ok. it works, nice. another question (completely not related to the first one), I have seen an exercise where the author started with : Let $T$ be a triangle with integer coordinates, show that the are is 'something with the interior'. How can I define a triangle in term of topology?
I have to prove that the area is $|\mathring{\mathcal{T}} \cap \mathbb{Z}^2| + \frac{1}{2} | \partial \mathcal{T} \cap \mathbb{Z}^2 | - 1$ (just to be complete)
@MarcGato It requires more than just topology. The "integer coordinates" strongly indicate that the ambient space is an $\mathbb{R}^n$ (most likely $\mathbb{R}^2$), and there you can define a triangle as the convex hull of three points. You can do it more generally, on nice enough Riemannian manifolds, you can define triangles with arbitrary vertices, but you need more than just topology.
@MarcGato Pick's formula, $\mathbb{R}^2$ hence.
You have a naive notion of the interior and exterior of a triangle there. That happens to be correct.
@huy---Noticed this on Wikipedia (see here), and it looks like a more general version of the result you're interested in (with a rescaling of the integration variable) $$\dfrac{d}{dt}e^{X(t)}=\int_0^1 e^{\alpha X(t)}\dfrac{dX}{dt}e^{(1-\alpha)X(t)}$$
@huy: it's without proof, alas, but there may be some useful references on that page
@MattN. If I understand it correctly, Eric wanted to see the additivity of the derivative of a differentiable function by manipulating difference quotients for directional derivatives. Ted's answer told him that that doesn't work.
Let $x>0$.
Let $J(x) = \sum_{n=0}^{oo} \dfrac{x^n}{2^{n(n-1)/2} n!}$
Let $J^{-1}(x)$ be the functional inverse of $J(x)$.
How to show that for all $x$ there exists a fixed positive real constant $C$ such that
$(J^{-1}(J(x)-J(x-1))-\dfrac{x}{2})^2 < C $
I came across the use of Descartes' theorem while solving a question.I searched it but I could only find the theorem but not any proof.Even Wikipedia also, just states the theorem!!I want to know the procedure to find the radius of the Soddy Circle??
I apologize if its duplicate and to mention ...
