my wife came back from a week away and my daughter was really happy until she realized that my wife wants to do things like enforce "rules" and monitor her "bad behavior."
i was running camp dowhatchawanna.
the running theme in my life is people not recognizing how cool i am.
we got an incident report from day care the other day. she and her bff were playing at being dinosaurs and my daughter scratched him with her fingernails, like, for real.
she was not a fan of the scratching situation. there's a ton of stuff that happens at day care that embarrasses her because she takes it personally.
i mean, i don't like it either, but i don't read too much into the details. our daughter has been punched and kicked before, and yeah it sounds crazy until you realize, these are three year olds, they don't really appreciate what they're doing.
so the idea of documented their stated reasons for doing what they did strikes me all as a little silly and i don't see it as reflecting negatively on me as a parent.
she just got too into impersonating a dinosaur. her bff is still her bff. i know this because we have to go to his stupid birthday party.
there's no way i'm tearing up an incident report, it's too funny of a record of how she behaved. i'm photoing it and showing it to everybody at her high school graduation.
the report, despite giving the names of three witnesses to the dino incident, does not specify what the 'scratching' actually involved. could have been just unwanted touch, or something that drew blood, or something in between.
like i said, silly to report this stuff. there may be some legal requirement that they do so.
@leslietownes Yes, I believe so. When my mom would get into skirmishes in the nursing home, they had to write it up and report to the caregivers of all involved.
there's weird stuff where the image of the embedding isn't closed. weird spirally stuff. i see a lot of those hypotheses as just confirming 'this is what you'd imagine if you weren't having nightmares.'
ok. what do i know, i never paid too much attention to this stuff. i did not see those hypotheses as adding more than "this is what you need to be able to draw pictures and not be lying about it" or "this is what you need so it isn't 'point-set topology'"
Find all continuous functions $f$ satisfying $$F(x) = \int_0^x f = (f(x))^2 + C \quad\text{for } C \neq 0,$$ assuming $f$ has at most one zero.
Since $f$ is continuous on $\mathbb{R}$, $F$ must be differentiable on $\mathbb{R}$ and $F'(x) = 2f'(x)f(x) = f(x)$. But $f$ isn't differentiable at $x$ where $f(x) = 0$, even though $F$ is. What's going on?
$f$ is differentiable when $f$ is nonzero because we are assuming $F(x) = (f(x))^2 + C$ and $F$ is differentiable, so $f(x)^2$ is as well which also has $f$ in its derivative
Then we assume that the integral, which is also equal to $F(x)$, is equal to $(f(x))^2 + C$ for some nonzero constant $C$. Since both sides are equal, we get to take the derivative to find that $F'(x) = 2f(x)f'(x) = f(x)$
To be honest I would have to look them up in the book, but if I recall correctly, then $f(g(x))$ is differentiable at $x$ if $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$
I am so stressed. I don't understand what else we can conclude.......
By assumption, $f^2$ is diff everywhere, so it's diff at $f(x)$, and we also require that $f$ be diff at $x$. If that is satisfied, then $f = 0$ or $f' = 1/2$
So when $f \neq 0$, we have $f' = 1/2$, which should mean $f'$ is differentiable at $x$ for which $f(x) \neq 0$
Okay I've got it. Oh my god. I haven't been that frustrated in ages.
The trick is to notice that $|f| = \pm f$ still implies $f$ is differentiable when $f^2 \neq 0 \Rightarrow f \neq 0$
Since $f$ is zero at no more than one point, so $f$ will not change sign more than once, if at all.
Then even if $|f| = -f$ for $x < r$ but $|f| = f$ for $x > r$ where $f(r) = 0$, we would have that $f$ is differentiable since $-f$ being diff implies $f$ is diff
@BAYMAX: you want to solve the equations simultaneously, so go with the first suggestion, take the difference of the equations. That gets rid of the $xy$ term and leaves you with a linear relation between $x$ and $y$.
Solve that linear relation for $y$ and substitute back into one of the original equations. You will get a quadratic in $x$. After solving that, use the formula for $y$ in terms of $x$.
Take the local model of an elliptic Lefschetz fibration over $D^2$ with a single singularity at the origin, monodromy being a Dehn twist about the meridian $a$ of the fiber torus $T^2$. In some sense, the total space of this fibration is $T^2 \times D^2$ with $a \times \{1\}$ killed.
$1$ is a boundary point of $D^2$. What I mean is that the total space is naively $T^2 \times D^2$ with a disk $D$ attached to $a \times \{1\}$, which would be the Lefschetz thimble.
yes thorr i am actually trying to prove the first statement you write
Also, why do we say for an upper triangular matrix we can write it in the form D(I +N) where as D is diagonal and N is nillpotent, i thought more that we can write D+N
Another method is as follows. An invertible upper triangular matrix has the form
$A=D(I+N)$ where $D$ is diagonal (with the same diagonal entries as $A$)
and $N$ is upper triangular with zero diagonal. Then $N^n=0$ where $A$
is $n$ by $n$. Both $D$ and $I+N$ have upper triangular inverses:
$D^{-1...
here is the refrence to where he uses this
@Thorgott Also , do you have a good link for a proof of the first part (Flag = trig )
@Astyx This is part of what I am trying to visualize. Let me be clearer as to what I believe the total space is. $T^2 \times D^2 \cup_{a \times \{1\}} D$ is not a manifold, so to correct for that one has to attach a thickened disk (or what we call a 2-handle) $D \times D^2(\varepsilon)$, instead.
To describe this, it's easiest to draw some Kirby diagrams. Let me demonstrate.
@Astyx: Here is some setup. Take an unknot $S^1 \subset S^3$, and let $D^2 \subset D^4$ be a disk bounded by this unknot $S^1 = \partial D^2$ inside the 4-ball bounding $S^3 = \partial D^2$. Then I claim $D^4 \setminus \nu(D^2) \cong S^1 \times D^3$, where $\nu(D^2)$ denotes an open neighborhood of $D^2$ in $D^4$.
Proof: Interior of $D^4 \setminus D^2$ is $\Bbb C^2 \setminus \Bbb C \cong (\Bbb C \times \Bbb C) \setminus (\Bbb C \times \{0\}) \cong \Bbb C \times (\Bbb C \setminus \{0\}) \cong \Bbb C \times \Bbb R \times S^1 \cong \Bbb R^3 \times S^1$
Check that these homeo(diffeo!)morphisms extend all the way to the boundary as well.
Take this link in $S^3$; I'll explain what the markings mean.
Carve two disks, each bounding the dotted circles $a$ and $b$ respectively, out of $D^4$ bounding $S^3$.
Then attach a $D^2 \times D^2$ to the undotted circle $C$ (the one marked by a $0$) by a homeomorphism $\partial D^2 \times D^2 \to \nu_{S^3}(C)$ where $\nu_{S^3}(C)$ is a closed toral neighborhood of $C$ inside $S^3$.
There are many homeomorphisms $S^1 \times D^2 \to S^1 \times D^2$, though. The marking $0$ specifies what this homeomorphism is. Any homeomorphism $S^1 \times D^2 \to S^1 \times D^2$ is isotopic to one which is $(\theta, z) \mapsto (\theta, e^{i k \theta} z)$. Here, $0$ means $k = 0$.
Let me know if the construction makes sense. I will then explain why it is $T^2 \times D^2$.
To see why this is $T^2 \times D^2$: First, ignore the undotted circle $C$. What you have is a pair of dotted circles in $S^3 = \partial D^4$. Chop $D^4$ along a $D^3$ which separates $a$ and $b$, i.e., $a$ and $b$ are contained in the the two separate components of $D^4 \setminus D^3 \cong D^4 \sqcup D^4$.
can anyone explain this result, if $K$ is sequentially compact then isnt the first statement of (3.3.4) (the convergence of a sub-sequence) immediate??
Carving along each of these then gives rise to $S^1 \times D^3 \sqcup S^1 \times D^3$ by this discussion. Glue back along the $D^3$'s. So the result is the so-called "boundary connect sum" $S^1 \times D^3 \sharp S^1 \times D^3$
Finally, the circle $C$ has the homotopy class of $aba^{-1} b^{-1}$ in the complement of the pair $a \sqcup b$ in $S^3$. This implies it has the same homotopy class in $S^1 \times D^3 \sharp S^1 \times D^3$ where $a$ and $b$ denotes the obvious generators of the fundamental group of this thing.
So now you see that it is just a thickened version of the torus $T^2$, which is $S^1 \vee S^1$ (of which $S^1 \times D^3 \sharp S^1 \times D^3$ is a thickening) with a 2-cell (of which a 2-handle $D^2 \times D^2$ is a thickening) glued along $aba^{-1} b^{-1}$.
I am slow as well, so don't worry about that. So, for one, note that the new link component with $+1$ on it is the vanishing cycle. It gets killed along the disk (the core of the 2-handle that is attached along that link component). The singular fiber is hidden somewhere; in fact the projection map itself is not clear
This is like the total space in abstract. But I think it is possible to draw the projection map.
The homotopy class of the vanishing cycle, at least, is correct. It is indeed the meridian of the torus $T^2 \times \{pt.\}$ in $T^2 \times D^2$
I have a thought on how I would draw the projection map, based off @AkivaWeinberger's pictures on a related question earlier.
@BalarkaSen Akiva, IIRC, was trying to understand the trivial fibration $T^2 \times D^2 \to D^2$ from this picture.
And I think this is what he did. Try to understand the "boundary fibration" $T^2 \times S^1 \to S^1$ instead.
As preparation, try to see what carving or adding a 2-handle does to the boundary $S^3$
In either case, the red circle or the parallel curve to the unknot becomes nullhomotopic and the blue or the meridian curve to the unknot survives. The effect on the boundary is the same: cut out a solid torus neighborhood $S^1 \times D^2$ of the unknot and paste a new solid torus back so that the parallel curve dies. ($D^2 \times S^1$)
(If $k \neq 0$ during handle attachment it gets more complicated, you'd end up doing a so-called surgery on $S^3$ along the unknot)
@BalarkaSen With this in mind, forget the $b$ circle here and arrange things symmetrically.
The complement of a solid torus nbhd of $b$ is a solid torus (because $S^3 \setminus S^1 \cong S^1 \times D^2$). $a \sqcup c$ is sitting inside this solid torus.
One can foliate the exterior of this link in the solid torus by torii which are as drawn above.
It's a little complicated to describe but I can say more, or you can see Akiva's cool pictures here, if interested
if i have a matrix $\begin{bmatrix} a&&b&&c\\a^2&&b^2&&c^2 \end{bmatrix}$ How can i simplify it by removing a b c? Sorry for dumb questions i am not good in linear algebra
I know you can multiply scalar with the whole matrix.
I have a Markov chain with values in $\mathbb{Z}^3$ which can move from point $x$ to $x+(\pm 1, \pm 1, \pm 1)$ (eight different possibilities) with equal probability
I am to decide if this Markov chain is recurrent or transient
The way the book does it, is that $$p_{xx}(2n) = \sum_{n_1+n_2+n_3+n_4 = n} \binom{2n}{n_1, n_1, n_2, n_2, n_3, n_3, n_4, n_4} \frac{1}{8^{2n}}$$
It's just $(\mathbf{X}_n, \mathbf{Y}_n, \mathbf{Z}_n)$ where $\mathbf{X}_n, \mathbf{Y}_n, \mathbf{Z}_n$ are iid 1D SRW. Product of Markov chains is Markov
The transition matrix is the tensor product of the matrices
I just told you. Your process is given by $(\mathbf{X}_n, \mathbf{Y}_n, \mathbf{Z}_n) = (\mathbf{X}_{n-1}, \mathbf{Y}_{n-1}, \mathbf{Z}_{n-1}) + \mathbf{V}$ where $\mathbf{V} \sim \mathrm{Unif}(\pm 1, \pm 1, \pm 1)$. That's the same as in distribution as the vector $(\mathbf{E}_1, \mathbf{E}_2, \mathbf{E}_3)$ where $\mathbf{E}_1, \mathbf{E}_2, \mathbf{E}_3 \sim \mathrm{Unif}(1, -1)$, iid.
I also don't like the way that intro Markov chains courses just want you to keep counting and estimate sums of return probabilities to prove transience/recurrence
So if we let $a_n = \sum_{n_1+n_2+n_3+n_4 = n} \binom{n}{n_1, n_2, n_3, n_4}^2$ then this formula says that $$2\sum_{k=0}^{n/2-1} \binom{4k}{2k}a_{2k} + \binom{2n}{n}a_n = \binom{2n}{n}^3$$ which allows us to compute values of $a_n$ recursively for even $n$.
Combinatorics goes out of the window if I give you the nearest neighbor lattice $\Bbb Z^3$ and chuck a sparse set of edges. No combinatorist can tell me if its recurrent or transient with a million years of computations
@politeproofs This is still garbled. If $f^2$ is differentiable at $a$ and $f(a)>0$, then $\sqrt{f^2}$ is differentiable at $a$ by the chain rule. By continuity, $f=\sqrt{f^2}$ on an interval around $a$, so $f$ is differentiable at $a$. Similarly, if $f(a)<0$, then $-\sqrt{f^2}$ is differentiable at $a$. We know nothing whatsoever if $f(a)=0$.
Indeed, you should think about the example $$f(x) = \begin{cases} x, & x<0 \\ 0, & 0\le x\le 5 \\ x-5, & x>5 \end{cases}.$$ Note that $f^2$ is everywhere differentiable.
The way you should understand the basic example is this: If you take the wedge product $\omega$ of $r$ orthonormal $1$-forms in $\Bbb R^n$, then $\star\omega$ satisfies $\omega\wedge\star\omega = dx^1\wedge\dots\wedge dx^n$.
Then what is what? What you wrote originally makes no sense.
What they wrote is what I said if you don't worry about upper and lower indices.
