alright, is the statement "$y = \pm \sqrt{1-x^2}$ is not a function of $x$" actually, if we want to speak proper formalese, "the solution set of $y = \pm \sqrt{1-x^2}$ is not a function of $x$"?
@Thorgott. I would like to cite an answer from mathexchange in my paper, is this allowed or not please? Or I can only cite papers and articles in a paper like springer for example?
once you've made the statement precise, it is of course pretty easy to demonstrate
the more remarkable thing is that, in a certain sense, the solution set can be locally expressed as the graph of a function, but you will hear about that if you learn about submanifolds some day
I'm not trying to prove Borsuk-Ulam. It's a different problem because I have one fewer dimension so I think I can get one more dimension in my output object, i.e. a line instead of a point
so I imagined that if you look at all the meridians connecting the poles and each of them has a root of g, so there must be a ring going around the planet where the temperature is equal to the temperature of the antipode
i have achieved the summit of formalization. no one can stop me now.
i am pure abstractness
"There is no function $f(x)$ such that, for $S:= \{(x,y) \in \mathbb{R}^2 : y = \pm \sqrt{1-x^2} \}$, $S=F$ where $F := \{(x,y) \in \mathbb{R}^2 : y = f(x)\}$"
As part of the reciprocal algorithm I'm working on, that limit there at the bottom makes it even more convincing. I just hope I'm not speaking too soon, so I'm not getting my hopes up too much.
Oh no, I just looked up modular arithmetic. This is the incorrect term. I'm not sure what you would call this, then, but it does have a modulus of 16.
For computing the reciprocal, what I have as a preliminary sketch is as follows: given a divisor $m$, compute $a + 16n$ for integers a, n such that $\frac{1}{a}$ is in a table with precomputed reciprocals, and $a + 16n \equiv m$. Then $\operatorname{rcp}(m) = \frac{a}{a+16n} (\frac{1}{a})$. Please excuse any potential mistakes.
There are in fact redundant entries in a 4-bit integer reciprocal table. I am not competent to prove this, but having a reciprocal with a table in range $[\frac{1}{2}, \frac{1}{10}]$ suffices, and not all values in that range are needed if one can can cheaply compute those reciprocals (e.g. 1/2, 1/4, and 1/8 on binary computers).
Namely, I am not competent to prove that this small range of values is sufficient to reproduce every other integer reciprocal with this particular algorithm(?).
And lastly, in case it needed to be said at all, I make no guarantees about $m$ less than $1$.
If this works out, then I can't wait to make a fixed-point and floating-point implementation!
Well as exhilarating as this has been, I'm quite exhausted now, so I shall retire for today. Good night!
now that I have become Abstractness, Master of Formalese, I can say $\{ (x,y) \in \mathbb{R}^2 : y = \pm \sqrt{1-x^2} \} = \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \}$
the power of logic compels you to agree they are the same
(but more seriously I wrote it with the $\pm$ because it was exactly what was confusing me, so if I could solve the $\pm$ case I could understand the rest)
the argument is actually a proof by contradiction by supposing $y = f(x)$ and $y = \sqrt{1-x^2} \lor y=-\sqrt{1-x^2}$ have the same truth value
Wouldn't the solution set be all the points on the line $y = x$? Clearly it isn't, but why? Surely you can just inverse sine both sides and arrive at $y = x$ ?
I took two lines of the general form $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ and then found their intersection and used slope - intercept formula but...
Let $\displaystyle f$ be differentiable on $\displaystyle [ a,b]$ such that $\displaystyle f( a) =0$ and there exists a real number $\displaystyle A$ such that $\displaystyle |f'( x) |\leq A|f( x) |$ for all $\displaystyle x\in [ a,b] .$Then prove that $\displaystyle f( x) \equiv 0$.
Let S = {1, 2, 3} be a sample space. Let P be a probability measure defined on 2S (the collection of all subsets of S) such that
P(i) = xi , for i = 1, 2, 3.
Select the correct statement from the following and complete it. 1. (x1, x2, x3) can be any point in a sphere centered at the origin, having radius R = . 2. (x1, x2, x3) can be any point in a triangle whose vertices are v1 = , v2 = , and v3 = . 3. (x1, x2, x3) can be any point in a square whose vertices are v1 = , v2 = , v3 = , and v4 = .
Can someone tell how this question should be approached?
@Koro here is a sketch of another approach. note that $|f(x)| \le A \int_a^x |f(t)|dt$. let $M=\max |f|$. so $|f(x)| \le M$ to start with. Now use this estimate to get $|f(x)| \le A M(x-a)$. Keep repeating to get $|f(x)| \le A^k M {(x-a)^k \over k!}$. conclude that $f=0$.
Integration starts at a later chapter but since I know it, i'll consider this
the first assertion that is, $|f(x)|\le A\int_a^x |f(t)|dt$, why is it true? I mean I agree with $f(x)\le A\int_a^x |f(t)|dt$ but with mod on left side?
Assuming your assertion to be true, the last expression is bounded by M for large k. That is we'll have $|f(x)|\lt M$ for large k giving us a contradiction as f will attain supremum on [a,x]
I am assuming that your first assertion is true (true or not, let's get back to that later). Given the truthfulness of first assertion, I think that's how we'll get contradiction.
@Koro here is one approach: We have $|f(x)| \le A |f(\xi)|(x-a)$ for some $\xi \in (a,x)$. Since $f$ is uniformly continuous, pick $\delta'$ such that if $|x-y|\le \delta'$ then $|f(x)-f(y)| \le 1$. Now choose $\delta = \min(\delta', {1 \over 2A})$. Consider $x \in I=[a,a+\delta]$. Note that $|f(x)| \le 1$ on $I$. Hence the formula above gives $|f(x)| \le {1 \over 2}$ for $x \in I$. Repeating gives $|f(x)| \le {1 \over 2^k}$ for $x \in I$. Hence $f(x) = 0$ for $x \in I$. Now repeat on the interval $[a+\delta,a+2 \delta]$, etc.
being away from work due to pandemic caused me to forget a lot about the appearances of people i work with. some people project as tall on zoom but are not in real life. and vice versa. people say i look shorter on zoom, i don't know what that means.
some of it is distribution of height in legs vs upper body. if it's mostly in your legs it goes away when you sit down. this is me to a t. my wife is the other way around. i think she looks taller than me when we're both sitting at a table.
we both have difficulty finding clothes that fit, for opposite reasons.
i had a wheel stolen off of my last bike, and i took it to a shop for a replacement and the owner said my bike was sized all wrong for my body. which i think was technically correct, but i don't do races or even long rides, i really just wanted the wheel. no sale on that italian thing.
chemistry is a fascinating subject. a few of my cases involve it right now. lots of deep stuff there. sometimes i get to have 2 hour zoom calls with chemists and just pummel them with questions. 95% are relevant to some issue and 5% is just me being curious.
I say, let $t\in [a,x_n]$ and then $|f(t)|\le A|f(\xi)|(t-a)|\le A\sup |\{f(t): t\in [a,x_n]||(x_n-a)|\lt s$, where $s=\sup |\{f(t): t\in [a,x_n]|$ and the claim is true because $x_n\to a$ so there exists some $N$ such that for all... etc. Since $|f|$ is continuous on compact set $[a,x_n]$ it must attain its supermum, hence a contradiction so $f\equiv 0$ on $[a,x_0]$, which is a contradiction since $f(x_0)\ne 0$ and hence proved.
First I chose $x_0$ such that $f(x_0)\ne 0$ and then created a decreasing sequence $(x_n)$ such that $x_n \to a$ as is clear from above linked message.
a guy described small scale optical detection as astronomy in the daytime. the stars are there, you just can't see them easily because there is so much other stuff lighting up. and if you add more detectors, depending on the modality, each one has their own noise and you're adding that too, mucking it all up. everything has to be tuned just so.
@copper.hat $(x_n)$ is a sequence such that $x_0$ is well defined as per my choice above and $x_{n}\lt x_{n-1}$ for all $n\in \mathbb N$ and also $f(x_n)f'(x_n)\ne 0$
yeah. the point is this: I want to say that there is a decreasing sequence $(x_n)$ such that $x_n\to a$ (so i tried to just construct one to show its existence). $f'(x_n)f(x_n)\ne 0$ helps me conclude that $x_n\to a$.
I say, let $t\in [a,x_n]$ and then $|f(t)|\le A|f(\xi)|(t-a)|\le A\sup |\{f(t): t\in [a,x_n]||(x_n-a)|\lt s$, where $s=\sup |\{f(t): t\in [a,x_n]|$ and the claim is true because $x_n\to a$ so there exists some $N$ such that for all... etc. Since $|f|$ is continuous on compact set $[a,x_n]$ it must attain its supermum, hence a contradiction so $f\equiv 0$ on $[a,x_0]$, which is a contradiction since $f(x_0)\ne 0$ and hence proved.
Here, the first step follows from MVT.
Copper, is my idea still not making sense? or still you feel it's missing details, please let me know
@copper.hat all conditions are satisfied, nothing is going beyond $[a,b]$ and added advantage (in my opinion) with this approach is that I don't have to consider $[a,a+\delta], [a+\delta, a+2\delta]...$ etc.
it helps to delineate the goal, formally, and then a series of steps via which you achieve the goal. with words that signal where and what you are actually achieving before doing it. substeps are OK but should tie in with steps and not to some uncited theorem from two chapters ago. rudin is particularly bad at this.
I say, let $t\in [a,x_n]$ and then $|f(t)|\le A|f(\xi)|(t-a)|\le A\sup |\{f(t): t\in [a,x_n]||(x_n-a)|\lt s$, where $s=\sup |\{f(t): t\in [a,x_n]|$ and the claim is true because $x_n\to a$ so there exists some $N$ such that for all... etc. Since $|f|$ is continuous on compact set $[a,x_n]$ it must attain its supermum, hence a contradiction so $f\equiv 0$ on $[a,x_0]$, which is a contradiction since $f(x_0)\ne 0$ and hence proved.
the conclusion is flawed. :'( so not posting it on main for now. @copper
Is finding a lyapunov functional for a dynamical system like comparing it to a gradient flow? : Given a dynamical system say in $\mathbb{R}^d$, $$\dot{x}=G(x)$$, a strict Lyapunov function for the system is an $F:\mathbb{R}^d\to\mathbb{R}$ (which has continuous first derivatives ) such that $$\langle G , -\nabla F \rangle \geq 0$$ and $\nabla F(x)=0 \implies G(x)=0$.
Assume now that the dynamical system is a gradient flow, i.e $G=-\nabla g$, then clearly we can choose $F=g$ as the Lyapunov functional
Looks like the laurent series for $\frac{x}{x+n}$ converges rather quickly. What methods for making series converge faster exist in general? Wolfram keeps mentioning in related queries at the bottom something called "Pade approximation", but is there anything better? Can I use argument reduction with this somehow?
re lyapunov functionals i have never used them from but from my vague readings i think that at a high level that understanding sounds reasonable to me.
@orangeisthenewf Correct, the Lyapunov condition is the same as saying $F$ strictly decreases along the flowlines of the dyanmical system given by $x' = G(x)$
I would add a note regarding the earlier question though, indicating that modern literature calls the Lyapunov condition as the "pseudo-gradient" condition
the only grammar i ever formally learned was spanish grammar in a language class for 9th graders who were native speakers of spanish but did not know the grammar. i thought, they might have offered one of these for native english speakers too.
i wish that the kadison-singer conjecture wasn't true.
andre bloch of bloch's constant fame murdered several members of his family and did a lot of his work in an asylum. we can probably still call it bloch's constant because he was probably very ill and traumatized from WWI.
Hello. Quick question here please, why we did not choose Chemistry as the root node (based on alphabetical order) please? The question is:, "Form a binary search tree for the words mathematics, physics, geography, zoology, meteorology, geology, psychology, and chemistry (using alphabetical order)."
avra it appears that the order of the words in the statement of the problem is relevant. i presume that the first word is supposed to be the root and you create successive nodes based upon whether there is something in the tree other than the root that precedes it alphabetically (in which case you add on to that branch) or not (in which case it's a new branch. not 100% sure.
avra i think we agree that the implicit rules of the question are not fully known until you see the answer. unless this is some routine thing in data structures that i am simply ignorant of, it wouldn't occur to me to regard the listed order of inputs as relevant in the first instance.
although i guess you could imagine that making sense in practice, if you've got data coming in from some pipeline and you have no idea what's next, you're forced to use the order that they come in.
Look at it to make sense what I am about to say. In my textbook, it was written $\frac{\bigtriangleup ABC}{\bigtriangleup ABD}= \frac{\frac{1}{2} AB \times CN}{\frac{1}{2} AB \times DM}=\frac{m_1}{m_2} because we considered $\frac{CN}{DM}=\frac{CE}{DE}=\frac{m_1}{m_2}$.
@TedShifrin I mean the formula is applicable and I need to use it in many maths, the problem is in derivation.
Most people do these sorts of calculations with no signs.
Unless your book defined $\times$ as something with a sign, it's just not possibly correct. And, as I already said, the sign has to be the negative of what you expect.
Anyhow, I need to leave in a few minutes. We're not going to settle this.
Theorem: Let $\displaystyle f:[ a,b]\rightarrow \mathbb{R}$ be a function differentiable on $\displaystyle [ a,b]$ such that $\displaystyle f( a) =0$ and that there exists $\displaystyle A\in \mathbb{R}$ such that $\displaystyle |f'( x) |\leq A|f( x) |$ for all $\displaystyle x\in [ a,b]$, then $...
