@Sophie Thank you Sophie for your suggestion, and yes inlatexbot is working but the problem is that it's not working in a group. It's just working on it's own chat.
Here it is okay:
But here not okay:
@robjohn My aim is to get latex rendered on a chat app called Telegram. People say that they have developed bots (I really don't know if they mean a kind of program by that) which will convert our latex code to the images (like the first image above). But unfortunately it's not working.
@Knight $\Large\textbf{Description and general usage}$
This bot converts LaTeX expressions to .png images when called in inline mode (@inlatexbot <your expression>) and then shows you the image it has generated from your code. Everything happens live, so you can just gradually type and continuously monitor the result. When expression contains errors, they will be displayed instead of the image. Finally, when you are ready, click or tap the suggested picture to send it.
@Knight that was not at all my issue with what you showed in the screenshot this message replies to. In that screenshot, you have already sent the messages, whereas in the screenshot above, the message is not yet sent
The above quoted description mentions "Everything happens live", which does not seem to be possible for sent messages, so together with the screenshot, i guessed you were doing it wrong
Hello all! I'm looking for a book that contains the following results:
Given a bijection between the points of two lines that conserves the double ratio, the geometric place of the lines that connect each point of one of the lines with its image in the other is a hyperboloid of a leaf. If the application also retains the simple ratio, then the site is a hyperbolic paraboloid.
Do you know books about this and its proof? Thanks!!
Let $V$ be a $\Bbb{R}$-vector space. In the paper I am reading, the author wants to "equip $V$ with a locally convex topology, called the 'algebraic topology' (also known as the finest locally convex topology), by declaring that any convex set $C$ such that $int(C) = C$ is open." Why is this a topology? The union of two convex sets is not necessarily convex, so the collection might not be closed under arbitrary unions. By the way, $int(C)$ denotes the set of 'algebraic' interior points.
That is, $c \in C$ is an algebraic interior point if for every $v \in V$, there is a $t \in (0,1]$ such that $(1-t)c + tv \in C$.
i didn't expect that "algebraic" can modify not only "topology" the subfield of mathematics, but also "topology" the collection of sets...
so algebraic topology does not study algebraic topologies...
@user193319 From Encyclopedia of Mathematics https://encyclopediaofmath.org/wiki/Locally_convex_topology : $\textbf{Locally convex Topology}$ A (not necessarily Hausdorff) topology $ \tau $ on a real or complex topological vector space $ E $ that has a basis consisting of convex sets and is such that the linear operations in $ E $ are continuous with respect to $ \tau $. A locally convex topology $ \tau $ on a vector space $ E $ is defined analytically by a family of semi-norms (cf. Semi-norm) $ \{ {p _ \alpha } : {\alpha \in A } \} $ as the topology with basis of neighbourhoods of zero con…
In Communism they say the capital will be distributed equally, so what I think do they pay equal salary to everyone? For example equal salary of General manager and an employee?
Or even if we consider the 19th century world, did Communist countries use to pay equal salaries to Lieutenant and an artillery officer?
@LeakyNun Oh sorry! I meant to write “Sociology” but wrote “socialism”.
There were revolutionary ideas brewing in the late 19th century in Russia of course. And of course communism as an ideology was established in the 19th century by Marx, and the Paris commune happened during that time
There was no such thing as a communist country in 19th century
@BalarkaSen certainly not, you had people such as Proudhon and Fourier and Owen long before, and their ideas still have sway amongst various people who call themselves the c-word (often more than Marx I'd guess)
Off-topic. Let $f : M_1 \to M_0$ be a local diffeo and $g : M_2 \to M_0$ be a smooth map. Then the projection $M_1 \times_{M_0} M_2 \to M_2$ is a local diffeo.?
man... I don't even know what are the charts on the fiber product. we proved its manifoldness using transversality so I never bothered to construct an explicit atlas.
What is the dimension of the fiber product? The fiber product is just the inverse image of the diagonal in M_0 x M_0 under the map f x g : M_1 x M_2 -> M_0 x M_0. The diagonal has codim m_0. So the fiber product should have dim m_1 + m_2 - m_0.
The tangent space to a point $(p_1, p_2) \in M_1 \times_{M_0} M_2$ is going to be $T_{p_1} M_1 \times_{T_{f(p_1)} M_0} T_{p_2} M_2$, and because $T_{p_1} M_1 \to T_{f(p_1)} M_0$ is an isomorphism this fiber product of vector spaces projects isomorphically to the second factor
@TedShifrin IIRC, I had the pleasure of teaching @Khallil what a Fréchet derivative was for a map between Banach spaces a full year before the supposed first functional analysis class
Well, better than Univ Chicago's crazy honors course where they learn functional analysis and never learn a decent amount of multivariable calculus/analysis at all. But I digress.
@Calvin: A proper multivariable calculus course (such as mine), even though it's just in $\Bbb R^n$, should give you all the tools for stuff in Banach spaces without a problem.
provided we have some random variable $X$ following e.g. a lognormal distribution, what properties must a function $f$ then have to ensure that the random variable $f(X)$ still has a valid density function?
If the question is not well-formulated please let me know
presumably f is 0 outside of [a,b], as a function on [a,b], its $C^1$ and increasing, so $P(f(X) \in [c,d] ) = P(X \in [f^{-1}(c) , f^{-1}(d)])$ which you can write as some integral which should eventually lead to what you want @CharlieShuffler
By definition:
Let $\mathbb Z^+_n = \{[0],[1],[2],\ldots,[n−1]\}$
$\mathbb Z^+_4 = \{[0],[1],[2],[3]\},$
but how $\mathbb Z^*_{12}$ is $\{[1],[5],[7],[11]\}$ ?
Well, turns out I need it for a meromorphic function
(@Thorgott if it was holomorphic then the difference between two such functions would have a maximum in the closed disk which must be on the boundary, so it is everywhere zero)
Oh yea I remember seeing that maximum principle proof
Another way to do it for a holomorphic $f \colon U \to \mathbb{C}$, where $D \subset U$ is a closed disk containing $a \in U$, is to take an arbitrarily small $D$ and bound above the difference between $$ \left| f(a) - \dfrac{1}{2\pi i} \int_{\partial D} \dfrac{f(z)}{z-a} \text{ d}z \right|. $$
Let $f$ be a meromorphic function in the unit disk that is zero on the boundary and continuous in the closed disk (as a function to the Riemann sphere). Does it follow that $f$ is identically zero?
Let $f: \mathbb D \to \widehat {\mathbb{C}}$ be a meromorphic function inside the unit disk.
Assume that $f$ is zero on the boundary and continuous in the closed disk (as a function into $\widehat {\mathbb{C}}$).
Is $f$ necessarily identically zero?
If $f$ is not surjective then we can take $a\...
Oh, maybe I didn't even need to bother with that. I didn't want to mess up the boundary value condition, but multiplying by $\prod (z-a_j)^{\mu_j}$ won't mess it up.
it's continuous on the closed disk, so if the poles were to accumulate, they would have a limit point, which cannot itself be a pole, but a sequence of poles accumulating to a finite value would contradict continuity (arranged poles as sequence and for each pole pick a point with distance <1/n from the pole and absolute value >n)
It has finitely many poles because they don't accumulate on the interior (definition of meromorphic) and not to the boundary (it's zero on the boundary)
In the original question, there's a polynomial $p(z)$ so that $p(z)f$ is holomorphic on the unit disc, continuous up to the boundary, zero on the boundary. So $|p(z)f|^2$ is harmonic, zero on the boundary, hence zero.
The function can certainly be continuous on the boundary and be zero at a limit point of zeroes. So our function could have poles accumulating at a point on the boundary which maps to $\infty\in\Bbb P^1$.
For the first time in my MSE experience, there are two people asking a plethora elementary differentiable manifolds questions without, apparently, the basic background. I'm just avoiding commenting or answering these two people now.
I've experienced this sort of thing with other fields, of course, before.
By definition:
Let $\mathbb Z^+_n = \{[0],[1],[2],\ldots,[n−1]\}$
$\mathbb Z^+_4 = \{[0],[1],[2],[3]\},$
but how $\mathbb Z^*_{12}$ is $\{[1],[5],[7],[11]\}$ ?
Let $f: \mathbb D \to \widehat {\mathbb{C}}$ be a meromorphic function inside the unit disk.
Assume that $f$ is zero on the boundary and continuous in the closed disk (as a function into $\widehat {\mathbb{C}}$).
Is $f$ necessarily identically zero?
If $f$ is not surjective then we can take $a\...
