Can someone help me with this? https://math.stackexchange.com/questions/3235882/a-simple-complex-function-spivak-27-12a/3235904?noredirect=1#comment6656318_3235904
I think I'm experiencing a language barrier with the person who answered.
Most of the errors in earlier chapters got caught by those of us who taught out of the book numerous times (14 times for me), but I never covered that chapter.
The problem itself has the square root in it (at least, in the last edition), but it's a sloppy error in the answer book.
You're most welcome, @Simone. There are some excellent problems in that book, so enjoy them.
If the problem is correctly stated, then it becomes obvious that the answer book is just silly. But there have been occasional mis-stated problems over the years.
I am totally not careless, but nevertheless I found typesetting errors/printing errors in my first book that drove me crazy. Several books later, I came to accept the notion that even perfect authors are human and make errors.
>The following matrix over $\mathbb{Z_5}$ is given: $$ \begin{bmatrix} 2 & 2& 2& 3\\ 1&3&1&3\\ 3&0&2&2\\ 4&1&0&4\\ 1&2&2&0 \end{bmatrix} $$ Specify the dimension of the row space and the column space. Specify a base BS for the row space and a base BS for the column space.
I probably shouldn't do this until (unless?) I get tenure but at some point later on I'd like to write books on random topics I have nothing to do with to learn about them. Sounds like it could be fun
Now my final try hopefully: $B=\{(2,1,3,4,1),(2,3,0,1,2),(2,1,2,0,2)\}$ I took the first, second and third vector, because these are the non zero rows in the final row echelon form.
Yeah I guess that's fair. Tomorrow's is gonna assume people know about automorphic forms and it conflicts with a class so I won't be going. I have epsilon hope for today. At least it'll be fun even if I don't get it
With the way my brain works, I find it easier to take the column space of the transpose than the row space of the original. That's because I tend to think of vectors as columns rather than rows.
Basically the speaker was talking about how shapes aren't actually a good criterion for fair districting, and that a possibly better idea is to use Markov chains
Demonark, Tufts Univ. ran a course for mathematicians last summer on fairness in making districts. I actually don't know the content, but I know at least one person who went to be trained.
@TedShifrin i always thought this was weird cuz it should be well known that people actually in practice have drawn districts to expressly make them less fair
For some positive integer n, generate a 3-row array such that 1) each row contains each number from -n to n exactly once, and 2) each column sums to 0.
hey here is a great charity to donate to www.drugpolicy.org isn't it nice to see logic and common sense at work as opposed to some open ended statement that has no real impact and a bunch of guys in blue shiny costumes that only the elderly look up to bragging about stealing a small amount of cash and drugs, being hail by the media as heroes literally every week on the news
to be honest its a risk asking any question on any social media network but this is more expert that facebook messenger in that sense you are in the right place good sir
you should probably hire a Stack Exchange team for that no problem getting hedge funds to pile in for the return in investment look they are literally paying people for mathematics in this room
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@AlessandroCodenotti If $f : X \to Y$ is a map, $E$ is a vector bundle over $Y$, then essentially $f^* E$ is the vector bundle over $X$ with fiber over $x$ being the fiber of $E$ over $f(x)$, so to speak.
You can make this idea somewhat precise. Do you know how to build a vector bundle from the data of transition functions?
Ah, I think I'm seeing it, if $f$ is $z\mapsto z^n$ pulling back the trivial $1$-dimensional $S^1$ bundle along $f$ looks like rolling a cylinder until it overlaps with itself $n$ times
People care about them because Grothendieck's relative point of view says you should consider Sch/S, schemes over a fixed scheme S and then products in Sch/S are fibered products in Sch
@BalarkaSen Yes, for non affine stuff you need to glue together a bunch of those
@Alessandro Given a section $s : Y \to E$, take $sf : X \to E$, and upgrade that to $s' : X \to f^* E$ by defining $s'(x) = (sf(x), x)$. Does that work?
@BalarkaSen Alright, looks like this works then. If $\pi:E\to Y$ then $\pi(sf(x))=f(x)$ since $s$ is a section, so $s'(x)$ lands in the right place, and it is also a section of $f^\ast E$, just because it does nothing to the second coordinate
So now you can argue as follows: $E$ is trivial over $Y$ iff there are sections $s_1, \cdots, s_n$ such that for every $y \in Y$, $s_1(y), \cdots, s_n(y)$ spans $E_y$. Pull these sections back to $X$ to obtain a basis of sections for $f^* E$ as well.
That proves pullback of trivial bundle is trivial no problem
Given any vector bundle $E$ over $Y$, choose a trivializing atlas $(U_i, \varphi_i)$ for this. Then for every pair $i, j$ of indices, denote $U_{ij} = U_i \cap U_j$. You have the various maps $\varphi_j \varphi_i^{-1} : U_{ij} \times \Bbb R^k \to U_{ij} \times \Bbb R^k$ which is of the form $x \mapsto (x, g_{ij}(x))$ where $g_{ij}(x) \in \text{GL}_k(\Bbb R)$
$g_{ij} : U_{ij} \to \text{GL}_k(\Bbb R)$ are called the transition functions of this atlas
The data of the vector bundle $E$ over $Y$ and the data $(U_i, \varphi_i, g_{ij})$ of the atlas and the transition functions are equivalent in the following sense; $E = \bigsqcup U_i \times \Bbb R^k/\sim$ where $(x, v) \sim (x, g_{ij}(v))$ whenever $x \in U_i$ and $j$ is such that $U_i \cap U_j \neq \emptyset$, and the projection $E \to Y$ is given by forgetting about the $\Bbb R^k$ factor
So I claim that for any map $f : X \to Y$, a vector bundle $E$ over $Y$ with a trivializing atlas $(U_i, \varphi_i)$ and transition functions $g_{ij}$, $f^* E$ has a natural trivializing atlas and transition functions
if $f_n:\mathbb{R}\to\mathbb{R}$ are Lebesgue integrable functions, the integrals are uniformly bounded, i.e. there is such $M$ that $|\int f_n|\leq M$ for all $n$, and the pointwise limit $f$ of the sequence $f_n$ exists and is finite almost everywhere, then is $f$ necessarily integrable?
yeah the pointwise limit $f$ will be measurable, that i know
fatou's lemma says that lim inf (which is lim in this case) is integrable if $f_n$ are nonnegative, let's see if i can decompose $f_n$ into positive and negative part and use fatou's lemma twice
or maybe look at the sequence $|f_n|$ instead... anyway i'm off for now, thanks
actually it doesn't hold with those assumptions, counterexample: $\chi_{[-n,0)}+\chi_{[0,n)}$, i think it needs $\int |f_n|\leq M$ instead... ok now i'm off for real :)
Depends what you’re asking. As an equation which involves derivatives, it’s perfectly sound. However, the typical context in which you’d see that DE is where g(x) is known and f(x) is unknown
(You can solve that ODE to get g(x) in terms of f(x) easily. It’s harder if you want to get f(x) in terms of g(x), which is the more interesting case)
An infinite set A is $\textbf{countable}$ if $\mathbb{N} \sim A$. (There exists a bijective function from the naturals to A).
WTS:
Let A be an infinite set. A is countable $\iff$ $A=$ $\{$ $a_1,a_2......$ $\}$, where $a_i \neq a_j$ for $i \neq j$.
proof
Assume A is countable. So $\mathbb{N} \si...
An infinite set A is $\textbf{countable}$ if $\mathbb{N} \sim A$. (There exists a bijective function from the naturals to A).
WTS:
Let A be an infinite set. A is countable $\iff$ $A=$ $\{$ $a_1,a_2......$ $\}$, where $a_i \neq a_j$ for $i \neq j$.
proof
Assume A is countable. So $\mathbb{N} \si...
