yeah. for some of them you could wonder if they "believed" it, or just saw which way the wind was blowing (i mean who cares, but you could at least wonder). not that guy.
there's a scholar of languages, victor klemperer, who kept a set of diaries during that time and chronicles the experience of being an academic persecuted by other academics. makes it seem like quite a mix of ardent believers, and people who just stood there, and then, like, standard academic jerks who maybe just wanted to get his office.
they named it after him anyway. i remember reading a series of blog posts, some more interesting than others, on whether people should continue to refer to it that way.
in my field (functional analysis) there tended not to be huge spans of theory all named after one people, but a lot of theorems did have names, and i was always horrible at remembering whose name was associated with what. i also don't remember faces very well.
i liked it because it had analysis, but wasn't PDE, and it didn't require me to visualize things, or alternatively, it required me to find the 'right' way of visualizing something finite dimensional (usually the wrong way of looking at it for other purposes) that would generalize immediately. just enough algebra to keep it interesting, but not algebra.
i like matrices but not numerical linear algebra or, like, matrices over weird fields.
I want to show that in $\Bbb C$, if we define $\phi:\Bbb C-\{0\}\to \Bbb C$ by \phi(z)= Re(z)/z, it's impossible to extend $\phi$ to $\Bbb C$ so that the resulting map is continuous. My thought is that the limit does not exists at 0 so the conclusion follows. Am I right?
The following question is from Kutner's Applied Linear Statistical Models - Ch 2 - 2.12
To answer the question a few pieces of information are needed, provided below:
What I gather the question is asking me is that if I take the limit as $n \to \infty$ then what happens to the variance of...
stat is its own set of notational conventions and notation. it's the kind of thing where a lot of math.SE people could unravel it after ten, twenty, x minutes of thinking, and maybe deriving theorems 2.6 and 2.7 and 2.8 without realizing it, only then to think about the problem. stat people live and breathe that stuff.
my wife got an MS in stat to buff up her resume. i could do her homework but it took me half an hour of flipping through her book just to figure out which variables were being suppressed in stat notation that i would have expected to see written out.
whereas if you immerse yourself in the world long enough, you know exactly what symbols you can play with, and maybe even what some of them mean.
it's a bit like the leibniz notation for derivatives in multivariable calculus. lots of sort of hidden conventions going on there, not at the surface of the notation, that you need to be either clever or experienced or just comfortable with being wrong about some of the time, or it doesn't work. it's not intuitive.
once my wife was crunching some numbers for a problem that involved a lot of data. she got some greek letter was 0.00000463. something very suspiciously small. i asked, "does that mean your model is a really good fit? or is that actually zero and we just got a lot of roundoff errors somewhere in the computation?" she didn't know. i didn't know. one of these mythical stat people would know.
they frighten me.
it turned out, not to leave everybody hanging, that it was roundoff error. it was going to be zero no matter what the data was, because the model focused on zeroing out that thing as the measure of what the result of the model was. we all lived happily ever after.
@dc3rd I've answered one question on CV. It used too much math, and although I tried to explain the calculus of variations in one paragraph, it only got a sympathy vote from the mod who asked me for more explanation.
@dc3rd It's interesting that that question actually bears on astrophotography (one of my hobbies). If you collect more time on an image, the random action of the atmosphere (the variance) can be reduced, giving smaller (sharper) star images. However, the physical limitation of the aperture size (Dawe's Limit) gives a minimum to the star image size.
To show $\phi:\Bbb C\to \Bbb C$ given by $\phi(z) =z/Re(z)$ is not continuous on $\{z:Re(z) =0\}$ if I assign some value $\phi(ib) \in\Bbb C$, then $lim_{x\to 0, y=b} (x+iy)/x$ is infinity so the limit does not exists. So we can't define $\phi(ib)$ making the map continous.
"The Brauer group was generalized from fields to commutative rings by Auslander and Goldman. Grothendieck went further by defining the Brauer group of any scheme."
I kept saying formal Laurent series in $\pi$ with coefficients in $O_v^\times \cup \{0\}$, but you can just identify them upto $\pi$. That's the same thing as $O_v/\pi$
if $X_k$ are random variables and $\sum_{k=1}^{\infty} X_k$ exists almost surely, then why is its characteristic function the infinite product of $X_k$'s characteristic function ?
there is no assumption on mutual independence of $X_k$
@Astyx Given a number field $K$ and a CSA $A$ over $K$, if $A_\mathfrak{p}/K_\mathfrak{p}$ splits for all primes (including infinite) then $A/K$ splits.
If $A = (L/K, \sigma, a)$ is even a cyclic algebra then splitting of $A$ is equivalent to $a \in K^\times$ being a norm from $L$, so for cyclic algebras the local-global principle for CSAs/K is equivalent to the local-global principle "if $a$ is a norm in $L_\mathfrak{p}/K_\mathfrak{p}$ for all $\mathfrak{p}$ then $a$ is a norm in $L/K$."
Which is the Hasse Norm Theorem
it's not CFT, it's just a local-global principle
altho I don't yet know why this norm condition is equivalent to splitting of the algebra
The utility obtained when a person works for $x$ hours at job $A,$ and for $y$ hours at job $B$ is given by $$ f(x, y)=2 \sqrt{x}+\sqrt{y}, x \geq 0, y \geq 0 $$ How many hours should the person work on each job to maximise the utility, if the person works for a total of 10 hours ?
So it should be solved by Lagrange Multiplier with constraint g(x,y) = x+y-10 right ?
Given a set $S$ of size $n$ and $A \subseteq S$ of size $k$, how many subsets of $S$ contain $A$?
I think it should be $2^{n} - 2^{n-k}$ ? Please tell is it correct or not ?
@mathsstudent isn't this set in bijection with the set of subsets of $S \setminus A$? Where a subset $U$ of $S \setminus A$ corresponds to $A \cup U$? I see an $n-k$ in your guess, which suggests you're thinking about this.
for example if $S = \{1,2,3,4\}$ and $A = \{1,2\}$ the subsets $S$ containing $A$ correspond to subsets of $\{3,4\}$ via the correspondence I mentioned. I don't think your formula works there, it seems to be overcounting.
in fact it seems to be counting the sets that don't contain $A$.
i'll agree to disagree with you there. it's almost always simplest to write down a bijection and prove it's a bijection, than to say "to choose a subset, you do blah blah, and this is that many choices, so it's blah blah." counting "choices" never feels as real to me because it's never clear in English whether the choices are independent of one another, and (if the problem is complicated) whether the narrative is fully exploring the universe of choices.
if you give S an ordering in which A comprises the first k elements, then subsets of S are in correspondence with bit strings of length n (the jth element of the string is 1 if the corresponding element of S is in the subset, and 0 otherwise). subsets of S containing A are in correspondence with bit strings of length n, the first k elements of which are 1. so you're being asked to count bit strings of length n, the first k elements of which are 1.
or the number of ways of constructing a bit string of length n-k. n-k independently made choices from two possibilities.
the 'in correspondence with' stuff is hiding the word bijection. i dunno.
a guy on math.SE has just made my observation above in even less detail. he didn't even give you a bijection!
i realize i am in the minority on this. when i had to teach discrete math type stuff, i was always telling people to look for bijections, and they wanted to make choices, and think of every counting problem as a iterative decision process. it seems more natural to most people. i do like using recursive relations established between sizes of different sets (established via bijections) to count stuff. i just want to write down those bijections before i do it.
@user586228 it looks like they're graphing profit percent a.k.a. percent profit for you. i don't know how to back out revenue or investment or profit from that. if the data on that was given elsewhere and the graph is the output, it's obtained by applying those formulas to that data. i realize this doesn't help very much. do you have more information about what is given and what is to be solved for?
it is atrocious notation. i think it is trying to solve the problem that $\bigoplus$ sometimes coincides with other things in simple examples, but not in more complicated ones. it's saying, yes, this really a direct sum. this is actually a coproduct in our category where we refer to it as a sum, and not any other thing that's isomorphic to it a lot of the time but not always.
@user586228 I would come up with symbols for revenue and investment for each company (even though not all of these are explicitly asked for in the problem), write down the relations you know from the text of problem 1, and then also look at the graph (and the definitions of profit percent) to write down more relations from the numbers on the graph. it seems like a large number of variables, but it will reduce a bit.
and you're only trying to solve for a ratio, not the investments themselves.
If $R$ is commutative ring with unity and $M,N$ are $R$-modules, then how can I prove that $Tor_n^R(M,N)$ is independent of a choice of resolution $F_\bullet$ of M? My definition of $Tor_n^R$ is first from the free resolution of $M$, and tensor with $N$. Then $n$th homology of that sequence if $Tor_n^R(M,N)$
Need to show it directly using chain homotopy (Not using free then projective. In fact I didn't learn projective module)
In fact I also didn't learn chain homotopy but professor said us to search by ourselves
but more like module theory? and homological algebra? I don't know if it's also called righ theory
I first thought if (F_\bullet, f_\bullet) and (G_\bullet,g_\bullet) are free resolutions of $M$, then it induces complexes $(F_\bullet\otimes N,f_\bullet\otimes 1_N)$ and $(G_\bullet\otimes N,g_\bullet\otimes 1_N)$. Then construct two chain maps $h_n:F_n\otimes N\to G_n\otimes N$ and $h'_n:G_n\otimes N\to F_n\otimes N$ and show $hh'\simeq 1$ and $h'h\simeq 1$ so $h$ is an quazi-isomorphism
don't say "yet." that implies that you might take them. be proud about what you don't know. i certainly am.
what's a spectral sequence? i don't know, some guy tried to tell me once, and i told him to buzz off. he said, make me. i said, i'm certainly about to. then he punched me in the chest and face and took my cell phone and wallet. in retrospect, he may not have been trying to tell me what spectral sequences were. i may have misheard.
@astyx i very much like that image. i am going to meme-ify it across my socials in ways that nobody will understand because i do not have math friends in real life anymore.
when i was in grad school i had a little latex alias that renamed \qedbox into a crude pixellated drawing of an extended middle finger. i can resurrect it off of an old laptop if there is interest.
specifically they say earned more profit. i agree it's ambiguous if they just said earned more. that would sound more like revenue.
but it wouldn't be clear.
in real life, if the companies are publicly traded, profits and investment are independently reported and there's none of this smoke and mirrors about solving for x. and if the companies aren't publicly traded, sometimes they announce something resembling profits (usually with a weird definition to make it sound better), and they don't announce investment unless they're thinking about scaring competitors.
that's not part of the question. i'm just throwing that in there.
publicly traded on a US exchange, i should say. some markets have minimal reporting requirements.
@Thorgott what is that mean? sorry I'm not familiar with those words. Does that mean a chain map between those two complexes is unique up to chain homotpy?
precisely, if you have a hom $M\rightarrow M^{\prime}$ and free resolutions $F_{\bullet}$ of $M$ and $F_{\bullet}^{\prime}$ of $M^{\prime}$, then there is a chain map between these two complexes that is the given map in least degree and this chain map is unique up to chain homotopy
the proof is an inductive construction repeatedly using freeness
the one where $M$ and $M^{\prime}$ live, respectively
I mean, a free resolution looks like $\dots\rightarrow F_1\rightarrow F_0\rightarrow M$, where all the $F_i$ are free. So a chain map between the two resolutions contains a homomorphism $M\rightarrow M^{\prime}$ at the rightmost part (in lowest degree). What I'm saying is that homomorphism on the right inductively determines the rest of the chain map up to chain homotopy.
If I have a matrix of the form $X = \begin{bmatrix}D&A\\0&-D\end{bmatrix}$ where $D$ is diagonal and $A$ is symmetric, is there an easy way to find $\exp(X)$? Unfortunately, it's got the -D in there instead of D so it's not as easy as the usual upper-triangular cases.
@Thorgott Once I show two chain maps from $F_\bullet$ to $F'_\bullet$ is chain homotopic, then also two chain maps from $F_\bullet\otimes N$ to $F'_\bullet\otimes N$ is chain homotopic. What happens next? By the way, is my basic plan I wrote before right?
I'm convinced this must be false but does anyone see an easy counterexample? Let $X$ be a topological space and $x\in X$. Suppose that there is a nbhd basis $U_i$ around $x$ such that $X\setminus U_i$ has exactly $n$ connected components for all $i$. Must $X\setminus x$ have exactly $n$ connected components?
Hello everyone! here is a link to a new discord server dedicated to the discussion in combinatorial topics discord.gg/aDpQdu7VBA People from any background with an interest in combinatorics are welcome! The server is full of resource material in different combinatorial topics!
@TedShifrin if it was not a negative D, but instead D again, then you'd have the commutativity of the block diagonal D matrix and the 0 matrix with A in the corner.
Let $R$ be a commutative ring with unity and $M,N$ be an $R$-module.
Let $(F_{\bullet},f_\bullet)$ be a free resolution of $M$. Then there is an induced complex $(F_\bullet\otimes_R N,f_\bullet\otimes 1_N)$. Define $\text{Tor}_n^R(M,N)$ be an $n$th homology of $(F_\bullet\otimes_R N,f_\bullet\ot...
@anakhro yes i am having a mid life crisis. i turned 40. everything is frightening now. at 39 i was breakdancing in the streets, or whatever people do that might be suggestive of heedless abandon. now i'm only thinking about my mortgage, the adequacy of my life insurance, and something that looks like mold on the roof of my garage.
axiom 1 is, like, i dunno, you've got a set with a two-sided identity element, or something. you pick.
my daughter's very interested in optics. at this time of year the sun hits a light fixture on our staircase that causes rainbow patterns to be displayed on the walls. when i woke up this morning, she was on the landing, and said "hi dad, i'm looking at rainbows." and she was. maybe a physicist someday.
although i think optical methods have gone about as far as they can go. we'll have to get her into other forms of detection.
There are probably still radioactive mineral deposits around.
Maybe check the government archives for list of prospecting claims in the area and you can take a Geiger counter to a field and show her that rocks make it go beep beep beep
LA county is not as bad as santa barbara. i don't think any of it is comparable to iowa.
i had to sign stuff related to radon just to rent a house in iowa. and there was asbestos all over the place in the roofing (undisturbed, which is mostly fine, but i had to acknowledge it). our house here is probably OK.
that was one of my coolest experiences in high school, actually. my chemistry teacher taking something the size of a golf ball out of a container, and putting it next to a geiger counter, and have it go crazy. if i ever die, i've told my estate to sue him for causing it.
we do have some americium in our house on account of that.
>Unknown to officials, his mother, fearful that she would lose her house if the full extent of the radiation were known, had already collected the majority of the radioactive material and thrown it away in the conventional garbage.
i like the expression "conventional garbage." it implies the existence of nonconventional garbage. who's picking that up? i only have two receptables, [conventional] garbage and recycling.
i kind of liked it as a kid. we didn't have a compost infrastructure, we'd just bury organic waste in a part of our yard. and the next year - roma tomatoes! without even paying for a seed packet.
that's a shame. i wanted to have a discussion about ghosts.
my mother claims, non ironically, to have seen ghosts. not in the context of spectral sequences. has anybody here seen a ghost? my daughter has. she says that the ghost says to give her a cookie. i think i understand the psychology there.
I see why I was struggling with coming up with the concept after you pretty much shoved me over the line. I was expecting the area of the parallelogram to be used in a more explicit way. Not stating that we are restricting the side to be $||\mathbf{c}||$. I was anticipating having to set something equal to that area the same way the dot product was used and end up solving for $\mathbf{x}$. Which I'm sure I'll be getting to eventually in your book, just not at this point.
average in terms of age/number of users? i see so many 1 and 10 point reputation people that the average is close to 20.
@dc3rd yes.
i think about 20 people who are about my age or older or much older basically handle the back end. and then goofballs like me wander around the site. but the front end is extremely young people who need their homework done yesterday.
It is a good problem.....multiple times the algebraic solution with a formula was tossed out to me, but I purposely ignored it because it wasn't telling me anything. I'm glad I did that geometry review and also glad I have been forcing myself to look at it geometrically
mathoverflow was started when i was in grad school, or shortly afterward, by a bunch of people i knew in grad school. if there was a math.SE then i was unaware of it. so i think because of that, everyone with ten zillion reputation is older than me, or needs a social life.
@dc3rd i will only say this because i think that ted might not be around. it is helpful to look at things geometrically. you can wade through formulas all day but unless you think about what they 'mean' (= thinking geometrically) it's just symbol pushing.
if anyone asks you later, someone put a gun to my head and told me to type that.
if you really wanna do symbol pushing, functional analysis is a start. although the topologies suck, so you have to be very careful in terms of the symbols that you decide to push.
i think i was about four years into my thesis when i realized the pictures i was drawing for "the two dimensional case" weren't even two-dimensional pictures because the dimension was measured in C instead of R. it didn't stop me from understanding things.
but again, some guy has a machine gun pointed at my forehead and is requiring me to say that.
the guy might even be ted, for all i know. he could have acquired weapons and driven up here.
I've been too obsessed with symbol pushing over my young math journey and it is what got me in this fix in the first place....I alwyas did ask "why this means what it means?", but I never went and ventured for the answer and just plugged along
@vitamind in the USA i think people would normally say 'graduate student' after a bachelors degree. in math in the USA, many departments do not give terminal masters degrees so you can assume phd student. in europe i understand it is very different and maybe the language needs to be more precise.
ted i've looked at your differential geometry book. it's really good and i wish i had learned out of it. we learned out of something - it was out of print, wu liked it, got the rights to a limited republication, i bought it at a copy shop - that did the same thing worse in about twice as many pages.
the book spent a huge amount of time on curves, which i liked. but it did surfaces up to the invariance of the gaussian curvature, and then it had a bunch of crap sections on "here's what riemannian geometry might look like," "here's a manifold," it was all nonsense to me. i could have used more exercises.
all i remember about the book is that it had two authors. which is weird because i normally remember everything. which is a sign of how suboptimal the book was.
it's driving me crazy for two reasons. one, defect in memory, two, the postal service lost about half of my books when i moved across the country, and that book was one of them. so i can't go down to the space above my garage and just find out what it was. it might well have been milman and parker.
this is what is so mind boggling about it. i also lost a book that paul chernoff gave me, and now he's dead. buying a new edition wouldn't be the same.
if congress wanted to abolish the USPS i would be ok with it, as long as there was a provision to fully compensate me for the loss of my books in 2012.
which means i guess that i wouldn't be ok with it.
interesting. i'm also annoyed because there was a guy in my class who was sick for half the semester, and i gave him my notes, and he never gave them back. i have a very incomplete record of math 140, is what i'm saying.
i was there, though. wu's wife proctored the final and brought lemon cupcakes.
bringing food near the end of the semester appears to be a tradition in a lot of departments. it is not the norm at UCs or CSUs. it is appreciated when people try to import it.
I didn't usually do that. For many years I did a huge party in the spring to which I invited students, friends, and some colleagues. I usually cooked two weeks for it.
i had a professor in law school who prepared a paella about the size of a bathtub for an end-of-year party. i told his wife, "i wouldn't invite people over for something like this, it would be repugnant to me to have this many people in my house." the way she laughed, i could tell she didn't want me to be there.
i say the quiet part out loud. it hasn't been that much of a leg up, if i'm being honest with you
i was told that if i just got a phd in math i could find a secure career designing weapons or surveillance systems for an overpaid contractor in the DC area. these promises did not materialize. the rest of my life has been a very internalized, misdirected, sublimated form of vengeance.
the universe. and several advertisements in notices AMS. "join us! your cup will overflow!" i think they used that language. they might not have. in retrospect it would be weird to advertise a job like that.
people slime the law but i've never done anything immoral. the closest i got was withdrawing our representation from a case where the client had probably done something immoral. and yet you see these category theory people just commuting diagrams left and right for anybody who asks them to. who's the real villain?
i just learned that my stepsister got married over zoom. i'm torn between congratulating her and interrogating her as to who this guy is. (i know who the guy is, but why is he on zoom?)
ok. What you expressed was the vector normal to $a$ and $a \times c$, which would be the vector $c$, because $a$ is orthogonal to $a\times c$ as I have stated it.
The way I stated it is under the assumption that $||a \times c|| = ||c||/||a||$, to get the same expression with only $||c||$ I have to somehow make sure my $\mathbf{x}$ has a norm of $||a||$. Is that what you meant?
@dc3rd i went through this last night while going to bed. what would $a \times c$ be in magnitude without any restriction on where $a$ and $c$ came from. take that and build from that.
or look at it in some even smarter way, i don't know.
this is the sign of a really good exercise. let me just say that. this is like a two weeks in the gym level of exercise.
i'm hypothesizing because i've never spent an hour in a gym
The following question is from Kutner's Applied Linear Statistical Models - Ch 2 - 2.12
To answer the question a few pieces of information are needed, provided below:
What I gather the question is asking me is that if I take the limit as $n \to \infty$ then what happens to the variance of...
Given a set $S$ of size $n$ and $A \subseteq S$ of size $k$, how many subsets of $S$ contain $A$?
I think it should be $2^{n} - 2^{n-k}$ ? Please tell is it correct or not ?
