Suppose there is a street of houses lined up horizontally, numbered $1$ through $N$.
My family lives in house $k$. My neighbor's family gets along with mine. Without loss of generality, let's assume the neighbor's family lives in house $k + 1$.
One morning, every family wakes up at a different ho...
@BigSocks , if I create all possible labellings of vertices from my computer , i.e., make all adjacency matrix and if atleast 1 matches, can I say it's an isomorphism?
I want to calculate the difference a-b so that the polynomial $f(x)=x^5+ax+b$ is divisible by $(x-4)^2$ do we have to apply Euclidean algorithm and take the remainder equal to 0 ? Or is there also an other/easier way?
I guess you're feeling historical, what with the impeachment trial quoting the Constitution.
I was expecting a partial credit score, instead. :P
Speaking of history, I suppose I should figure out how many years I've been in chat. I do remember that Peter (before I talked him into becoming Pedro) dragged me in here for some multivariable analysis, which, as I recall, we never figured out.
Had a quick question @TedShifrin, with regards to a question in the text and your favorite aspect....formalism. It was Sec 2.2 - 1)c: The question asks to determine if the set $C = \{(x,y): y > 0 \}$ is open, closed, or neither. So I've chosen to show it is open.
Using the idea of the open rectangle, I've worked out to let $\delta = |b|$, where I'm taking $\mathbf{a} = (a,b)$ to be the middle of my open rectangle.(I should probably use $\frac{|b|}{2}$ to be safe, but anyways...)
So if I let $\mathbf{x} \in R(\mathbf{a}, |b|)$ (where $R$ is for rectangle), then I only concern myself with…
At any rate, I don't know what your notation even means, @dc3rd. What is $R(\mathbf a,r)$?
And I certainly don't follow your sentence. Supposing I knew what this meant, you say ... "let $\mathbf x \in S$. I'm having trouble actually showing $\mathbf x\in S$." Huh?
Ok. well if I use the open ball based off of my rectangle idea, $\delta = \frac{|b|}{2 \sqrt{2}}$.............$R(\mathbf{a}, r)$ was just my notation for the open rectangle.
You don't need absolute values, since we know the $y$-coordinates of points in the set are all positive. So, yes, show that the ball $B(\mathbf a,b)$ is contained in the set.
You need to show that for every $\mathbf x=(x,y)\in B(\mathbf a,b)$, we have $y>0$.
So your triangle inequality is the right idea, but you definitely do NOT want absolute values of $y$. If $\|\mathbf x-\mathbf a\|<b$, then don't we know directly that $|y-b|<b$? So what does this say?
As for my sentence of "let $\mathbf x\in S$. I'm having trouble actually showing $\mathbf x\in S$."..................I was asking this based off of the process of if we want to show a set is open, I have to choose a point in the neighbourhood of my ball, but I also have to verify that said point is in the neighbourhood based on the characteristics of my over arching set.
Kind of.....see what you mean with regards to the rectangle. But to be sure I got the idea of it. You're implying that the distances of the open rectangle may not work.
Yes, and all you had to show was that $y>0$ for the points in that ball!! Done.
Of course, you're right that if you choose any rectangle with height $b$, the same thing will work, but you haven't specified one yet. And, according to the definition, I'd still have to fit a ball inside. So, yes, a square fits, and a ball fits inside the square easily.
So I'm a little uneasy with part of my solution. If I'm letting $ r = |b|$, then I would have:
$||\mathbf{x} - \mathbf{a}|| = ||(x,y) - (a,b)|| \leq |x-a| + |y - b| < 2b$........this is why we have to use the idea of $\frac{r}{\sqrt{n}}$? or specifically in this case $\frac{b}{\sqrt{2}}$.
Since we know component wise $|y-b|\le \|\mathbf x-\mathbf a\|$ (for the $x$ component as well), using the idea of the open ball the distance of each individual component will not be larger than the distance of all the components. And since we stated that the distance of $\|\mathbf{x} - \mathbf{a}\| < b$ then after the manipulation we establish the condition of the set $y > 0$.
Yes I agree, this is why I'm verifying these ideas with you, and thnk you by the way. Because I can do the "algebraic manipulation", but I've been getting to the point that I was applying it as an algorithm without thinking about what is truly happening
You wrote the notation as if to suggest that I knew exactly what rectangle this defined. If you're only going to specify one real number, all I can think of is a square with that (half-)side centered at $\mathbf a$.
You derive higher-dimensional intuition from the lower dimensions. Sometimes it's misleading. For example, the volume of the $n$-dimensional ball of fixed radius goes to $0$ as $n\to\infty$.
Ok...I just briefly thought about this volume idea you mentioned and it is perplexing....lol........I won't play around with it right now and wait until the tools are sharpened.........
Not closed and bounded. That's only for subsets of $\Bbb R^n$ with the usual metric. (There's a lie in that sentence.) But every sequence having a convergent subsequence will do, @dc3rd.
sanity check: if $V,W$ are complex vector spaces, then $\overline{V\otimes W}=\overline{V}\otimes\overline{W}$, yes? (the bar is conjugate structure and the tensor product is over $\mathbb{C}$)
I'm looking for the Fourier transform of $$\tilde{\chi}(\omega)=\frac{Nq^2}{Vm\epsilon_0}\frac{1}{\omega_0^2-\omega^2-\mathrm{i}\omega\gamma}=\frac{Nq^2}{Vm\epsilon_0}\frac{1}{2\sqrt{\omega_0^2-\frac{\gamma^2}{4}}}\left(\frac{1}{\omega+\mathrm{i}\frac{\gamma}{2}+\sqrt{\omega_0^2-\frac{\gamma^2}{4}}}-\frac{1}{\omega+\mathrm{i}\frac{\gamma}{2}-\sqrt{\omega_0^2-\frac{\gamma^2}{4}}}\right) \ .$$
So I'm looking for $\chi (t) = \int_{-\infty}^{\infty}\tilde{\chi}(\omega) e^{i\omega t} \mathrm{d}\omega$. Are there any tricks to simplify this integral? I have not taken any course in complex analysis yet.
@Thorgott yes, the conjugate vector space has elements v' with iv' = (-iv)'. Then the simple tensors in Vbar o Wbar are v' o w' and i(v' o w') = (iv') o w = (-iv)' o w = v' o (iw') = v' o (-iw)', both of which are equal to (-i(v o w))'
@Mike ok, that's what I was thinking as well, thanks for confirming
@schn I don't think there is an easy way to compute this by hand without complex analysis. It's possible to "guess" the inverse and appeal to Fourier inversion, though.
@Thorgott thanks for the reply. Is there an easy way to compute it actually using complex analysis? It's a problem related to electrodynamics, where $\chi$ is the real (time dependent) or complex (frequency dependent) electric susceptibility.
Hello. When solving differential equations, when can we assume things about... heat? For example, if we model something about how coffee cools over time, then this problem states that it's never going to be lower than room temperature, but when can we assume that? It's not true for all things in the world that they're never going to be lower than room temperature
So if $V,W$ are complex vector spaces, $\overline{V}\otimes_{\mathbb{C}}W$ should be isomorphic to $V\otimes_{\mathbb{C}}W$ as real vector spaces (since the $\mathbb{R}$-action on both is the same) and the orientations should be the same/reverse depending on whether the product of the complex dimensions of $V,W$ is even/odd?
It's intuitive from thermodynamics, but it's just stated as a fact in the math textbooks. The substance cools at a rate proportional to the difference between its temperature adn the ambient temperature. This is even somewhere in the exercises of Chapter 18 of Spivak, probably.
Consider $x \in \ZZ(p^{\infty})$, i.e., $$x=\overline{\frac{a}{p^m}}\, \text{ where } 0 \leq a < p^m \text{ and } \gcd{(a,p^m)}=1\, .$$ Then, note that if $x \in \text{ker }\phi$, then $$\overline{\frac{a}{p^{m-t}}}=\phi\left(\overline{\frac{a}{p^m}}\right)=\phi(x)=\overline{0}\, , \text{ or equivalently } \frac{a}{p^{m-t}} \in \ZZ$$ from which it follows that $a=0$ or $t \geq m$.
I want to be sure. $a=0$ implies $x=overline{0}$. When $t \geq m$, then $\displaystyle x=\overline{\frac{a}{p^m}}=\overline{\frac{a\, p^{t-m}}{p^t}} \in \ZZ(p^{t})$.
uh oh....so I'm here again, @TedShifrin. I know I can just use the idea that the set $D$ is just the negative of the set I proved previously, but I was just trying to do it explicitly to get more comfortable with the right algebraic manipulation.
Right, you want the radius to be the distance from $b$ to $0$, which is $|b|=-b$.
There's a reason Spivak spends a lot of time on absolute value and manipulating the triangle inequality in his first chapter :) This stuff does show up.
OK, so you have $|y-b|<-b$. How do you conclude that $y<0$?
@Thorgott regarding my earlier posted problem about the complex integral with the two poles $i\gamma \pm \sqrt{...}$. If $\chi(t)=0$ for $t<0$, does this possibly simplify the integral $\chi (t) =\frac{1}{2\pi} \int_{-\infty}^{\infty}\tilde{\chi}(\omega) e^{-i\omega t} \mathrm{d}\omega$?
Oh. I am not knowledgeable about the esoteric sorts of elliptical integrals. If you're talking about meromorphic functions on a curve of genus $1$, then maybe.
In general, I typically had a lot of basic analysis, algebra, topology, geometry, complex analysis, and even some applied stuff at my fingertips throughout most of my career. Most mathematicians probably do not.
Well I had a function $f = \wp' - a\wp - b$, and I was asked to show that I can choose a,b so that it has zeros at $z_1,z_2,-z_1-z_2$ (where all are non-zero).
I found a,b so that it works for z_1 and z_2
But the -z_1-z_2 part is confusing me and I think it might be under my nose
My favorite interpretation is very much not what you'll be seeing, but I learned it from Griffiths and it's beautiful. Riemann Roch is about osculating spaces at certain points on the canonical embedding of the curve.
But once you have Serre duality, it's about $H^0$ and so the cohomology disappears and you can interpret just in terms of dimensions of natural vector spaces.
For the set $E = \{(x,y): y > x\}$ I think I have a $\delta$ that works, but I'm not too comfortable with it.
Let $\mathbf{x} = (x,y) \in B(\mathbf{a}, \delta)$, where $\mathbf{a} = (a,b)$ and $\delta = b - x$.
Then $|y-b| < b-x \Rightarrow x < y < 2b - x$
The reason I'm not comfortable with this is because it is depending on $x$, now I know my $\delta$ would be dependent on the position of my point, but even though it "looks" like it could work, I don't feel it is valid. I did try to build off the idea of the previous questions though
