you see stuff like that on the internet billed as interview questions for the prestige employer of the day (used to be google, maybe now it's somewhere else). i would walk out of the room, close the zoom, whatever, if someone popped one of those questions on me.
and i say this as someone who was asked to evaluate a limit once during a job interview as an attorney.
i did one of those in an interview once and got it . i was asked to talk through my process and the interviewer was disappointed that my estimation process drew mainly on stuff i saw in my own life and not something 'smarter.'
One of my very smart students interviewed for a summer internship on Wall Street. The only question he bungled was: You bicycle 5 miles, stopping for a coffee, whatever, in NYC in precisely an hour. Was there a 30-minute interval in which you covered precisely half the distance?
i was supposed to go to the whitey bulger trial one day because a friend of mine worked in the courthouse somewhere and could get me a seat, but i didn't go because i was tired. i should have gone, he flipped out in the courtroom and there was high drama that day. woops.
The only way I see to approach it is by the following argument: $f_{}:H_n(X,A)\rightarrow H_n(Y,B)$ is an isomorphism for each $n$ then $f_{}$ is a chain equivalence. Hence $f^{*}$ is a cochain equivalence hence an isomorphism on cohomology groups
@anak: Assuming $a_{t-1}\le \frac1t$, is it true $a_{t-1}\ge \frac1{t+1}$? I do not know how to show that. Nor do I have any idea currently of other approaches.
Hi all. I'm doing my online homework and I'm really bamboozled by the answer. Even though $P_y=Q_x$, the homework system is telling me the field $F$ is not conservative. Is there an error in the system?
@TedShifrin We haven't learned about curl yet, but I'm simply testing if the force field F is conservative by checking if the mixed partials are equal $P_y=Q_x$
i used to live a few blocks from a restaurant that was basically a window next to the sidewalk. it was called PRIME. they did not offer math based discounts.
just to add to my last comment: prime ideal definition as I know it is: a proper ideal I of ring R such that if for any a,b in R, ab is in I then either a or b is in I.
it had a nice atmosphere. they had four or five tables in what resembled a driveway. we'd take visitors there if only to see the look on their faces when we showed up.
i miss living in a truly urban area. no good restaurants within walking distance now.
Is it okay to spend 10-15 years total on a research question? Is that too long? Say 10 years is learning the material to be able to do the actual research and then 5 is reading, writing and then publishing a paper
consider $\Bbb R^3$ and it's image under a coordinate change - is it possible for a surface $S$ to be a minimal surface of both $\Bbb R^3$ and it's image?
If I have the function $f(z) = \frac{1+z}{z}$, I want to use the argument principle to find $\frac{1}{2\pi} \Delta \textrm{arg}f(z)$, to do so I need to find the number of poles and the number of zeros.
Hello, I have a question in topology, I have this definition that says the following: If B is a basis for a topology on X, the topology T generated by B is described as follows: U \in T is said to be open if for all x \in U, there exists B* \in B s. t. x \in B* and B \subseteq U.
I took an example: Let X = {a, b, c} and let B be the basis which is: B = { {a}, {b}, {c}}. Now, I just applied the definition to get a topology T, so I got T = { {a}, {b}, {c}} but T is not a topology here since neither X nor phi are subsets of T. Do I have something wrong
By the way the definition came from James Munkers, Topology A First Course
@AlessandroCodenotti because B is a basis for a topology on X, the topology T generated by B is described as follows: U \in T is said to be open (i.e. to be an element of T) if \all x \in U, there exists B* \in B s. t. x \in B* and B* \in U. So I applied this definition to get a topology and I got T = { {a}, {b}, {c}} only and not included {a,b}
there are a few things going on here. a basis for a topology need not itself be a topology. that's one thing.
another is, you can consider the topology generated by any family of sets. if you have some topology on your space already, then you might or might not get that topology from some family of sets. you might get a different one.
if you consider a set X, and consider the set S of all one-element subsets of X, then the topology generated by S turns out to be the discrete topology. i.e. the set of all subsets of X.
so for example if R is the real numbers with the usual topology, S = {{x}: x in R} is a family of subsets of R that generates the discrete topology on R. S happens to be basis for the discrete topology. S is not a basis for the usual topology on R (e.g. because its members are not open in the usual topology of R).
whether a given collection of subsets is or is not a basis can depend on what topology you are considering.
@JoeShmo Did you read past the first page? My recollection is that, starting three or four years ago, they started including an "undergraduate" section on the exams.
I think in general past grad school quals are a good source if the exercises in undergrad books or the exams for undergrad courses are too easy for you. I've personally only studied the algebra ones, though
@anak It is maybe not that I dislike it as I don't think that it is a very good book to teach / learn out of. It is an encyclopedia, which seems more interested in providing the most terse, elegant proof of every statement, and provides very little by way of intuition or explanation.
It is a good reference if you already know the topic, but is not great if you are learning it.
@anak That is how 5.8 is phrased in my edition, but it is really, secretly, a theorem about the correspondence between a space and its dual, and establishes nice embeddings. The argument is completed in Theorem 5.25.
Having opened up the book again, my real objection to Theorem 5.8 is that only the last statement in the laundry list of statements is terribly interesting---the other statements are really just lemmata that Folland uses to prove the important one.
@anak No, it is the Riesz Representation Theorem: if $f \in \mathcal{H}^*$, there is a unique $y\in\mathcal{H}$ such that $f(x) = \langle x,y\rangle$ for all $x \in \mathcal{X}$.
Does Riemann's hypothesis help in knowing the distribution of prime numbers, or is it to improve prediction of where numbers are located or something else?
There are many hypotheses about prime numbers. Why does the Riemann hypothesis have so much momentum compared to other hypotheses?
i don't know what other hypotheses you have in mind, or what you mean by 'momentum', but RH and its generalizations have a lot of consequences. there are lots of things in number theory that people would learn more about if they knew more about RH.
