I know that the volume for a pyramid is 1/3*b*h, and the height is the radius of the sphere, and b is area of the base. But why does it state that 4rA/3=hA/3? Where does that relationship come from?
@LeakyNun So, an example of polynomial which does not have extensionality would be $x^2+2x+1=x^2+1$ over $\Bbb Z/2\Bbb Z$? But I think this should not be considered as violation of extensionality as 2=0 in $\Bbb Z/2\Bbb Z$.
Let ${a_n} \to a$ and ${b_n} \to b$ in $\Bbb R$. If $a_n\le b$ and $b_n \le a$ for infinitely many $n \in \Bbb N$ then $a = b$.
I think it is correct, because $a_n-b_n\le b-a$, but for 'those' infinitely many $n$s, we see $a_n-b_n\to a-b$, hence $a-b\le b-a$. Similarly, $b-a\le a-b$.
Will this hold even after some modification: Let ${a_n} \to a$ and ${b_n} \to b$ in $\Bbb R$. If $a_m\le b$ and $b_n \le a$ for infinitely many $m\in \Bbb N$ and $n \in \Bbb N$ respectively, then $a = b$. ??
Hi, I have some function $f: \Bbb R^2 \to \Bbb R$ which I want to maximize on a compact set $K$. How would I do this in general? If I can show that $K$ is a submanifold, then I can use lagrange multipliers, right?
With lagrange multipliers I would determine all candidates for local extrema and then compute the hessian of the lagrange function to determine if the candidates are a local max/min/saddle point or neither
I have a concrete example where $f$ attains a local maximum at a point according to Mathematica and the plot, but if I compute the hessian for it I get that it his indefinite...
The hessian should be positive definite, no?
I'm trying to find the mistake in my calculations but can't find it. Does anybody want to help?
@philmcole I'm confused. If $M$ is a submanifold of $\Bbb R^2$ locally given at that point $\mathbf{a}$ by $g(x, y) = c$ say, then according to what you say, $f$ attains a local minimum at $\mathbf{a}$ with respect to the constraint $g(x, y) = c$. $\mathbf{a}$ not a local minimum of $f$ as a function of $\Bbb R^2$ on the whole.
We have a theorem in the text which says approximately "if the Hessian of the Lagrange function at a candidate point obtained through Lagrange multipliers is positive definite then this is a local maximum, negative definite -> local minimum, indefinite -> saddle point"
I know its not the Hessian of the function $f$ I want to maximize like if $K$ was an open set
The Lagrange function is $F(x, y, \lambda) = f(x, y) + \lambda(g(x, y) - c)$. The Hessian of this guy at $\mathbf{a}$ is the block matrix $[A, B; C, 0]$ where $A$ is the $2\times 2$ matrix $Hf(\mathbf{a}) - \lambda Hg(\mathbf{a})$, $B$ is the $2 \times 1$ vector $\nabla g(\mathbf{a})$ and $C$ is the $1 \times 2$ vector $\nabla f(\mathbf{a})$.
That's the bordered Hessian.
@philmcole I don't see how this is true. The matrix has a $0$ at the lower-right corner, which forces the Hessian matrix to have kernel (as a quadratic form)
The theorem I refer to talks about the quadratic form associated with the lagrange function being positive/negative/in-definite. I assumed that I can just check the Hessian of the lagrange function like we did when we worked with open sets, where we just checked the Hessian of $f$. Maybe this is wrong?
Wiki says that I need to check the determinant of the bordered hessian now.
For my example this would fit (the bordered hessian has positive determinant and the point is a local maximum)
In the theorems we always had a statement about the quadratic form being +ve/-ve definite.
In practice we just checked if the hessian matrix was +ve/-ve definite
This seems to not work here anymore then
I wonder why it's suddenly different though...
Anonymous
08:22
Hi. I was facing some problem with notation. What does $C[a,b]$ mean? Does it refer to functions which are continuous in $[a,b]$? But does that mean it is not defined outside of $[a,b]$?
Are there noncommutative rings in which the set of nilpotent elements forms an additive subgroup?
Well, I just realized that division rings like $\Bbb H$ satisfy my question the way I've asked it. What I'm really looking for is a case where $a,b$ are nilpotent, $a + b$ is nilpotent, but $ab \neq ba$.
@AlessandroCodenotti Yeah. I work from home on Tuesdays, so I take the opportunity to make it since otherwise I would not be able to let it simmer for long enough.
Reality is that what is exist does not exist, for it dies, not exist and not it does, exists
The unreality is a construct, a replica and not even the Original
Consider the following:
Given an object $X$ and a noninjective map $Z$ we have $ZX$ but we cannot recover $X$. What if $Z$ maps probablistically to different values of its image. For example:
I am reading a paper on privacy that says we can model something as an arbitrary probabilistic function from $X^M \to X^M$. I'm trying to figure out what exactly that means. I saw another paper that defined a probabilistic function as a function $\mathcal{F}:X \times Y \to [0, 1]$ that satisfie...
message from the future: That is posted repeatedly not because it is interesting, but because it is unanswered, but that is not even a question, because otherwise it will be obvious
NB: That message that this future message referred to has not appeared yet
As always, math chat is a shadow of what should be happening on this instability known as the h bar
messages are delocalised in space and time, just to add to the obfusication
It is perhaps the easiest question, for to not be and end your life, you don't stop the evil. But to be, there is a chance on a near measure zero set that you will stop them
The practicalities then resides on how to integrate this infintesimal nothingness such that they join together into a unity and hence bring the stop to all despair
The dimension of the vector subspace $W$ of $M_2(\Bbb C)$ given by $W = \left\{\begin{pmatrix}a&b\\c&d\\ \end{pmatrix}: a, b, c, d\in\Bbb C, a + b = c, b + c = d, c + a = d\right\}$ is equal to ....
I took this approach: we see that $a=b$, and $c=2a$ and $d=3a$, so, $W$ consists of matrices $\begin{pmatrix}a&a\\2a&3a\\ \end{pmatrix}$, hence dimension is 1. Am I right?
Here is a corollary that appears in my Algebra book: An $A$-module is finitely generated if and only if it is a quotient module $A^n$ for some integer $n$...What exactly is this saying? The statement seems incomplete. Is it saying that $M$ is a finitely generated module if and only if $M$ is isomorphic to $N/A^n$ for some module $N$ and some natural number $n$?
Let $X$ be some metric space and $A$ some subset. Does $$\mu(A) = \inf \{\sum_{k=1}^\infty diam(U_k) \mid \{U_k\}_{k=1}^\infty \mbox{ is an open cover of } A\}$$ define a measure on $X$?
all the proofs im finding for $|\mathcal P (\mathbb N)| = |\mathbb R|$ are using the existence of a $n$-ary representation of a number. Is there a way to prove that without invoking such a thing?
the existence of an $n$ary representation is a very strong theorem and I'd prefer a proof that doesn't need to rely on such heavy machinery. Unless there is no such proof otherwise
You don't really need the unique representations. You know that $\mathcal{P}(\mathbb{N}) = 2^\mathbb{N}$, and this is just sequences of 0's and ones. You can probably adapt the proof from there
or maybe not
I don't actually know set stuff
I guess part of the problem is how you actually define $\mathbb{R}$
Sam I'm just going through the prologue of Folland's Analysis book and he just says "here's review or whatever of the concepts and notation you need for this book, I'm gonna go fastish because it's not the point of the book"
I have this question: The dimension of the vector subspace $W$ of $M_2(\Bbb C)$ given by $W = \left\{\begin{pmatrix}a&b\\c&d\\ \end{pmatrix}: a, b, c, d\in\Bbb C, a + b = c, b + c = d, c + a = d\right\}$ is equal to .... I took this approach: we see that $a=b$, and $c=2a$ and $d=3a$, so, $W$ consists of matrices $\begin{pmatrix}a&a\\2a&3a\\ \end{pmatrix}$, hence dimension is 1. Am I right?
Also, can you please help me with this: Consider the real sequences $a_n$ and $b_n$ such that $\sum a_nb_n$ converges. If $a_n$ is unbounded then $b_n$ is bounded. Is this statement true?
yes I think you can make an example where you space out zeros in $a_n$ and $b_n$ so that by themselves theyre unbounded but when you multiply them you only get 0s
tbh im not 100% right now so my thinking maybe off
In the complex plane, If I want to transform the disc $|z-i| \leq 2$ into the unit disc... if I have to move it downwards which would be make $z \to z-i$ right?
That is the step to center it at the origin, right?
hi, im trying to find an injection from $P(\Bbb N) $ to $P(\Bbb N) - P_{fin}(\Bbb N)$ where $P_{fin}(\Bbb N)$ are the finite subset of $\Bbb N$, someone can help?
hey @Semi, do you happen to have a source where they explain why the spin of a 2-particle system is $s_1+s_2,\dots,\vert s_1-s_2\vert$, where $s_1$ and $s_2$ are the spins of the separate particles? I'm able to work it out for simple numbers, but Griffiths refers to a text for the general proof that I cannot find online. I would just like to see somewhat of a more abstract argument (I don't necessarily need a proof)
If $M$ and $N$ are modules, and $f : M \to N$ is a module homomorphism, what exactly is the canonical projection of $M$ onto the cokernel? What is it defined?
Surely it isn't $\pi : M \to N/f(M)$ defined by $\pi(m) = \overline{f(m)}$, as that would be the $0$ map.
@LeakyNun Okay. Referring to the same book and section, it says "Given a six term exact sequence $$0 \rightarrow K' \rightarrow M \stackrel{f}{\rightarrow} N \rightarrow C' \rightarrow 0$$ $f$ is injective iff $K'=0$ and surjective iff $C'=0$. My question is, how is that sequence well-defined if only one function is involved in the sequence?
I am reading a paper on privacy that says we can model something as an arbitrary probabilistic function from $X^M \to X^M$. I'm trying to figure out what exactly that means. I saw another paper that defined a probabilistic function as a function $\mathcal{F}:X \times Y \to [0, 1]$ that satisfie...
