@TedShifrin Yep, that's what I was looking for. Thank you!
I had an idea for a proof by contradiction, but I'm not sure it would actually work.
Suppose you do have an oval with only one minimum and only one maximum. Then, by using the intermediate value theorem, we can find that there are two points on the oval whose tangent lines are parallel and which have the same curvature. Consider the two portions of the curve delimited by those two points.
In one portion, the curvature is everywhere greater than the curvature at those points; in the other portion, it's everywhere less than the curvature at those points.
Yet it seems like the portion with greater curvature must have endpoints which are closer together than the portion with less curvature, which is a contradiction.
@TerranSwett the best looking egg I've seen has been made of four circular arcs, two with greater, unequal, curvatures and two with lesser, but equal curvature on the sides.
@Huy I don't think that is known, but alternative expressions using elliptic integrals are known.
@robjohn Recently found out about Chandrupatla's algorithm for root-finding, which may interest you. Essentially it chooses root-finding methods based on how close the previous midpoint was to the previous secant line between endpoints.
x.x I feel like I was born a few decades late! Really nice simple 1D root-finding methods are often amazingly recent discoveries...
Let $G \leqslant K^{\times}$ be a multiplicative subgroup of the group of units of a field $K$. Then $\sum_{g \in G} g = 0$ necessarily.
**Proof.** If $\sum_{g \in G} g = u \neq 0$ then $u \in K^{\times}$ so that $0 = u^{-1}\cdot 0 = u^{-1} \sum_{g \in G} g = \sum_{g \in G} u^{-1}g = 1$.
@robjohn Sir, should I spend some more time in learning manipulative techniques ? My current level of manipulation is of Higher Algebra (Hall and Knight) level (I.e. I don’t get much problem in solving). Should I need to be more prepared?
(please don’t say “I don’t know that depends on you” like few other MSE users, you know what I’m asking and you know that your judgment matters)
*To find the minimum value of $\frac{(a+x)(b+x)}{(c+x)}$ we do something like this* $$ \text{Let} (c+x) = y \\ a+x \rightarrow a-c+y ~~~~~~~b+x \rightarrow b-c+y \\ \frac{(a+c)(b+c)}{(c+x)} = \frac{(a-c+y)(b-c+y)}{y} \\ \frac{(a+c)(b+c)}{(c+x)}= \frac{(a-c)(b-c)}{y} +(b-c)+( a-c) + y $$
Now: $$ \frac{(a+c)(b+c)}{(c+x)}= \frac{(a-c)(b-c) }{y} + (b-c)+(a-c) +y \color{blue}{+2\sqrt{(a-c)(b-c)} - 2\sqrt{(a-c)(b-c)}} \\ \frac{(a+c)(b+c)}{(c+x)}= \frac{(a-c)(b-c) }{y} + y \color{blue}{-2\sqrt{(a-c)(b-c)}} +(b-c)+(a-c) \color{blue}{+2\sqrt{(a-c)(b-c)}} \\ \frac{(a+c)(b+c)}{(c+x)}= \left( \frac{ \sqrt{(a-c)(b-c) } } {\sqrt{y} } - \sqrt{y} \right)^2 + (b-c)+(a-c)+ \color{blue}{+2\sqrt{(a-c)(b-c)}} $$ Hence, the minimum value of $\frac{(a+c)(b+c)}{(c+x)}$ occurs when $\left( \frac{\sqrt{ (a-c)(b-c) }}{\sqrt{y}} - \sqrt{y} \right)^2 $ is least and therefore when it is zero.
Hence, the minimum value of $\frac{(a+c)(b+c)}{(c+x)}$ is $$ \color{red}{ (a-c) +(b-c) + 2\sqrt{(a-c)(b-c)} } $$
${\large \text{By mistake, at every LHS I have written c instead of x in numerators} }$
@SpecterProphet Although I haven't studied analysis yet, this question has dumbfounded me. Just like the asker, I'm of the view that such a function does not exist, but all sorts of weird functions exist in higher math so I'm curious to know what the answer actually is.
enumerate the rationals as $r_1, r_2, \cdots$, look at $\sum_{k = 1}^n (x - r_1) \cdots (x - r_n)/100^{100^n}$. this should converge as $n \to \infty$ to some function like that
i guess the choice of the denominator really depends on your enumeration
u want to divide by something which grows much faster than $r_1 \cdots r_n$
Suppose that the function $p(x)$ is defined on an open subset $U$ of $\mathbb{R}$ by a power series with real coefficients. Suppose, further, that $p$ maps rationals to rationals. Must $p$ be defined on $U$ by a rational function?
Take a genus $2$ surface $S$, pick a meridian $\alpha$ around one of the holes. You can pick a hyperbolic metric on $S$ such that $\alpha$ is as long as you want, yeah? (In fact a lot more, this $\alpha$ is part of Fenchel-Nielsen coordinates for the Teichmuller space of $S$)
So fix a basepoint for $\alpha$, travel sometime for more than half of the circle so that you have an arc of $\alpha$ which is a local geodesic but not a minimal geodesic.
Like you do to get non minimal local geodesics given any closed geodesic really
Ok so call this $\gamma$, and it's locality constant can be as large as I want it to be since I have control on the metric of $S$. It's like half of the length of $\alpha$ or something
I thought it would maybe be because when you jiggle the metric to get your example to have appropriate length for $\alpha$, it would jiggle the constants in this theorem so as that your "$2\delta$-neighborhood " is huge or something
You can't lift that back up canonically to the actual subgroup of $PSL_2 \Bbb R$; every element can lie at a different point in the circle-fibers (so long as they multiply correctly)
So you don't know just from the image what the action is
Say you're acting on the point $p$ with stabilizer $\Gamma_p$. If you have two representations $\pi_1 S \to PSL_2 \Bbb R$ which differ by a rep to $\Gamma_p$ then they should have the same image
But $\Gamma_p$ is not the stabilizer of every point, just that point. If you act somewhere else then $\Gamma_p$ will actually contribute to the action
I have a coalgebra $C$ (with comultiplication $\Delta$ and counit $\varepsilon$) and a vector subspace $V$. I define $C(V)$ as $\{x\in C\mid ((\Delta\otimes I)\circ\Delta)(x)\in C\otimes V\otimes C\}$, and I want to show that $C(V)$ is the biggest subcoalgebra of $C$ contained in $V$
I think it should boil down to the fact that if $x\in C\setminus V$, then $\Delta(x)\in (C\setminus V)\otimes (C\setminus V)$, does that seem sensible?
Look at this nice ring: $\mathcal{O}_F^{\sharp} := \widetilde{\Theta}_{\mathbb{C}_p}(W(\mathcal{O}_F)_L \otimes_{W(\mathcal{O}_{\hat{L}_\infty^\flat})_L} \mathcal{O_{\hat{L}_\infty^\flat}})$
@AlessandroCodenotti Hm ok. Anyway the idea for this should be to use that $L\circ(\varepsilon\otimes I)\circ\Delta=I$, where $L:K\otimes C\to C$ sends $k\otimes c$ to $kc$
I'm a bit confused now though because this seems to show that if $x\in V$,then $x\in C(V)$
Because $(\epsilon \otimes I \otimes \epsilon) \circ [(\Delta \otimes I) \circ \Delta](x)=x\in V$, but for $\epsilon\otimes I\otimes\epsilon$ of an element of $C\otimes C\otimes C$ to be in $V$ the central part should be in $V$?
(there are a bunch of multiplication maps $K\otimes C\to C$ and $C\otimes K\to C$ left implicit here and there)
For instance, suppose $x$ has $\epsilon(x) = 0$. Choose $y \not\in V$ and $z$ so that $\epsilon(z) = 1$. Then $x \otimes y \otimes z $ is not in $C \otimes V \otimes C$, but $$(\epsilon \otimes I \otimes \epsilon)(x \otimes y \otimes z) = \epsilon(x) y \epsilon(z) = 0 \in V$$
No need for the sum to appear here. Just the fact that $\text{ker}(\epsilon)$ is nonzero, which it had better be if your coalgebra is interesting!
Hi! I'm currently a computer/electrical engineer, highest degree conferred MS. I'm considering a change to mathematics, with a goal of teaching at the college/community college level. I'm 40 years old, is this just a pipe dream at my age?
I’m looking for a classic introductory Real Analysis book. Something of early 20th century or late 19th century.
Just like for Algebra the classic one in Hall and Knight, for Geometry we have SL Loney, for Calculus we have Joseph Edwards’. So, in a similar fashion what would you suggest for Real Analysis?
@AlessandroCodenotti You were probably imagining everything was simple tensors and that you could express the result as a simple tensor between things not in V
I think maybe try doing this with a subspace of a group algebra to see what happens in practice
Hiya @TedShifrin , I asked a question earlier about some terminology. How would you interpret an "$f$-linear endomorphism $\varphi : M \to M$" where $f : R \to R$ is a ring hom and $M$ an $R$-module?
Hmm, I was not aware that I had coauthored a paper titled "On good Formula Not Shown-filtrations for rational G-modules". But if it is in Google Scholar, it must be the correct title :)
Scholar also seems fairly confused with a couple of my papers, having several of them split up as if they were not the same paper
Really not sure what happened. The correct title is further up on the list, with an equivalent reference, so Scholar ought to be able to tell that they are the same paper, except $(p,r)$ has been replaced by Formula Not Shown.
Can someone decode the question: There is 1-to-1 correspondence between maximal ideals in $R$ and homomorphisms into $\Bbb Z_2$ which are not identically zero. Where R is an ideal in Boolean ring $(R,\cdot, +)$.
Is it asking for a bijective function on $R$ to homomorphism $\varphi : X \to \Bbb Z_2$ ?
@AbstractAlgebraLearner Take $K = \Bbb R$. Each pair of real numbers $x$ and $1 - x$ cancels to $1$, so your sum won't converge. Or maybe actually take the pair $x$ and $2 - x$
@flowian it is asking for a bijective function between the set of maximal ideals in $R$ and the set of non-zero homomorphisms from $R$ into $\mathbb{Z}_2$
@Thorgott pairing $A = \{ x : x \in \Bbb{R}\}$ and $B = \{1 - x : x \in \Bbb{R}\}$ would yield $\sum A + \sum B = 2 \sum_{r \in \Bbb{R}} r$ since $A$ covers $\Bbb{R}$ exactly once but so does $B$, and so by symmetry we have $2\cdot 0 = 0 = \sum_{r \in \Bbb{R}} r$
So as you can see your counter argument has failed
If: By removing 1 point from the set of points that make up a sphere does not change the cardinality Then: Why is the punctured sphere homeomorphic to the plane, but not also the sphere? Can it be bijective at least?
I.e. is it the 'continuous' restriction that falls apart? Or the bijection altogether?
@Threnody $[0,1]$ is in bijection to $\Bbb R$, as well as $\Bbb R^n$ for any $n$, or even if you like to the space of sequences $(a_n)$ with $\sum |a_n| < \infty$, etc ...
It's not very hard for things to be in bijection
Convince yourself that most functions that exist are not continuous
On the other hand, there is the following curious fact: If $f\colon\mathbb{R}\rightarrow\mathbb{R}$ is any function, there exists a dense subset $D\subseteq\mathbb{R}$ such that $f\vert_D$ is continuous.
Yeah, $f\vert_D$ is a function with domain $D$, $f$ is a function with domain $\mathbb{R}$. To say that $f$ is continuous at a point $x\in D$ means that for all sequences $(x_n)_n$ of real numbers with $x_n\rightarrow x$, we have $f(x_n)\rightarrow f(x)$. To say that $f\vert_D$ is continuous at $x\in D$ means that for all sequences $(x_n)_n$ of numbers in $D$ with $x_n\rightarrow x$, we have $f(x_n)\rightarrow f(x)$.
So in the second case, we are considering possibly way less sequences.
An instructive example is the indicator function of the rational numbers. $\chi_{\mathbb{Q}}\colon\mathbb{R}\rightarrow\mathbb{R},\,x\mapsto\begin{cases}1,&x\text{ is rational},\\0,&x\text{ is irrational}\end{cases}$. This function is nowhere continuous. But $\chi_{\mathbb{Q}}\vert_{\mathbb{Q}}\colon\mathbb{Q}\rightarrow\mathbb{R}$ is a constant function (with constant value $1$), so is very much everywhere continuous.
To say it is continuous on $D$ just means that it is continuous at every point in $D$. The point is that the domain of the function determines which sequences we have to take into account.
there's a model of the hyperbolic plane in the first quadrant? I feel like nobody uses this and just uses the half plane model or poincare model or Klein model
Let $K$ be a field of characteristic $\gt 2$ that is not necessarily finite.
Start to define the notion of arbitrary summation of certain subsets $X \subset K$ via:
If $-X \subset X$, then $\sum\limits_{x \in X} x := 0$.
Call any such set $X$ arbitrarily summable to zero. In a field of charact...
Suppose that the function $p(x)$ is defined on an open subset $U$ of $\mathbb{R}$ by a power series with real coefficients. Suppose, further, that $p$ maps rationals to rationals. Must $p$ be defined on $U$ by a rational function?
Let $K$ be a field of characteristic $\gt 2$ that is not necessarily finite.
Start to define the notion of arbitrary summation of certain subsets $X \subset K$ via:
If $-X \subset X$, then $\sum\limits_{x \in X} x := 0$.
Call any such set $X$ arbitrarily summable to zero. In a field of charact...
