The plane spanned by r' and r'' is the osculating plane that a curve is "closest to lying in" at that point. This generalizes to higher dimensional hyperplanes for higher dimensional curves, right? Like, (by thinking about Taylor polynomials?) the 3d hyperplane spanned by the first three derivatives is the best fitting one.
@N.Maneesh (1) says the integral of it is exactly $2\pi$, so the integral of the real part is $2\pi$. But (2) says the most the real part could be is 1, so it needs to be 1 almost everywhere on the interval. Since f is a polynomial, it must be identically 1.
analysis-2 eg was half a semester of metric spaces and half a semester of multivariable calculus done badly with riemann integration (which is one of the major things one should learn in a first analysis course imo) was sandwiched half-assedly in between
it was terribad
one learns more analysis by going through rudin than sitting in these courses
@SubhasisBiswas i mean mmath is fundamentally a bad course and most of the students dont have proper mathematical foundation which is bad because ur a grad student who'll presumably persue a phd, so
@SubhasisBiswas its one thing to not know something and another thing to be paid to learn something but still not doing it, which is my criticism of the students in isi really
I don't think ability really matters at the end of the day, what matters is sticking to it, whatever you are doing. You spent a whole day and solved a problem that you wanted to solve; that's how one does math
@BalarkaSen very basic actually. I don't even have much grasp on $D_8$.
I know small things about cyclic groups, even lesser amount of knowledge on permutation groups, then comes normal and quotient groups.
very little actually. And yes, no idea about sylow theorem and conjugacy classes. I won't say that's because of lack of cleverness. That's because of lack of hard work and discipline
@BalarkaSen you will see that most people in most math colleges in India are "sub-par". Everyone just wants to solve Olympiad problems. It's frustrating after I point.
its a serious problem because on one hand you have thousands of students cramming for engineering colleges and the scarce few who do get in science just want to enjoy college life or whatever
we have broken up our education cycle beyond repair
@BalarkaSen well, same applies to me too, kind of. I wasted so many countless hours gaming and on social media (and chasing girls), that nothing can fix it
it's not for everyone.
@BalarkaSen will, but what do I do with this problem? (generalisation)?
@BalarkaSen True. I mean I don't really see a problem with enjoying college life. You need to know your priorities if you truly want to learn something.
I believe that if you truly want to develop academically you should develop as a person as well and that implies developing yourself socially and academically. That will happen I guess if you interact outside our comfort zones such as academics etc.
@Albas the notion of enjoying college life is synonymous to lowkey adolescent hedonism to be honest so i dont think those two are compatible. i mean the process of learning doesn't pay off in the short term, it's painful and unpleasant because you realize how pathetic you are and there's so little you actually understand and even though you understand it how slow you are to internalize it
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its sort of depressing if you linger on it really
but ofc you can learn and socialize at the same time its just that the postmodern culture equates socialization with partying
Yup that's the issue isn't it, conforming people to certain norms which are mostly hollow. But again I don't have much of a problem with partying even though I don't really prefer it. If people are partying, drinking etc it's their thing. If they know what their priorities are then they know how to get themselves back. I know many quite distinguished people in science (mathematics specifically) who have done "partying" in their undergraduate/graduate studies but are no less in learning
I think that it's more of a problem of how mathematics is put forward as a subject. It's said to be a subject of prodigies and geniuses. A subject if you don't have anything unique in yourself you cannot succeed. This kind of an outlook forces people to build deep insecurities in themselves about their own knowledge and they start shying away from learning anything new, experimenting anything new.
Whereas it's actually a lot of hard work, a lot of failures. There aren't any moments. There are culminations.
sure, but i dont think its a result of the social view of math. anyone with minimum working knowledge in math can experience this learning mechanism, where you see some technique, try to apply it to solve problems, fail innumerably many times, and then culminate in to a proper understanding of the technique
the idea is that's how learning always ever works, this should not be a point of demotivation
Can someone give me the definition of a real subspace of a complex vector space? Is it the one which is closed under the scalar multiplication by real numbers?
@ThomasShelby That sounds correct to my ears. Any $\Bbb C$-vector space $V$ can be thought, by restriction of scalars, as a $\Bbb R$-vector space. A subspace of that real vector space is what you want
groups of order $p^r$ has a non trivial centre (had to look it up, I forgot). So, in case of $p^2$, there are only a few possibilities. Centre is of order $1$ (impossibility); of order $p$ and of order $p^2$.
suppose, $o(Z) \neq p^2$
then $Z$ is cyclic. But, $G/Z$ cannot be non trivial cyclic.
@BalarkaSen Okay, thank you! Do you happen to know any theorem which states that a real subspace of $M_n(\Bbb C)$ is (topologically) closed? $M_n(\Bbb C)$ is the space of all n by n matrices.
I know. The problem is that here I get a scholarship which kind of forces me to do a "summer project" each summer where I have to write up a report and give a seminar in the end. It's mostly learning stuff from a book but that's what they call it.
The questions though that you gave @SubhasisBiswas is a very nice exercise. I remember it troubling me a little in my group theory course.
Yea it's something that atleast makes people learn something new.
Though we have to kind of make sure that these projects are somewhat interdisciplinary. The institute kind of pushes the entire notion that interdisciplinary work is very important.
I am just studying the two at the moment and let's see where that takes me. There was this person here who did his project on the topic of the atiyah singer index theorem and it's relations to the heat equations. That was kinda cool.
I only learnt the Toeplitz index theorem and some basic things about Fredholm operators because I had to write a report on the works of an Indian mathematician for a non-credit course on intro to mathematics in my first sem, and I chose Patodi
I didn't actually write anything about the super general Atiyah-Patod-Singer index theorem, but just mentioned Toeplitz and Gauss-Bonnet as examples of index theorems
I just understand Gauss Bonnet that too only for surfaces. I just have heard this but isn't atiyah singer index some generalisation of the Riemann Roch theorem. The guy who did his project was talking about it in his presentation.
I am taking this course on algebraic curves next semester(will Riemann Roch pop up there?)I know like algebra till rings and modules. Is that going to be enough? Do I need to do some more stuff? I think they will follow Fulton or Shafarevich
@BalarkaSen: Did you receive my message above where I mentioned my email? I just checked my email and saw that there has been no email from you. In any case, if you send me those images via email, please give me a ping.
What does it mean to quotient by the same subgroup when the two groups are different? Do you mean if $G/H \cong G'/H'$, $H \cong H'$, then $G \cong G'$?
@SubhasisBiswas Well, at this level of generality, there's no hope of having this. Take $G = G' = S_n \times A_n$, $H = A_n \times \{e\}$ and $H' = \{e\} \times A_n$.
I know this because I have thought about this before.
Ok, a counterexample for you is easier. Take $G = G' = \Bbb Z_4 \times \Bbb Z_2$. Quotient by $H = \Bbb Z_2 \times \Bbb Z_2$ and $H' = \Bbb Z_4 \times \{0\}$.
@SubhasisBiswas Everything's abelian here, and the quotient $G/H \cong G'/H' \cong \Bbb Z_2$, so... :)
Here's something which is true. If there is a homomorphism $\varphi : G \to G'$ such that $\varphi(H) \subset H'$ so that $\varphi|_H : H \to H'$ is an isomorphism and the induced homomorphism $G/H \to G'/H'$ from $\varphi$ is an isomorphism, then $\varphi$ is an isomorphism between $G$ and $G'$.
This is the "corrected" statement for the statement I was giving a counterexample of at first
For yours, the correct statement is simply that if there is an isomorphism $\varphi : G \to G'$ such that $\varphi(H) \subset H'$ such that $\varphi|_H : H \to H'$ is also an isomorphism, then the induced homomorphism $G/H \to G'/H'$ is an isomorphism
@SubhasisBiswas I just realized I again gave a counterexample to something different: $G \cong G'$, $G/H \cong G'/H'$ but $H \not \cong H'$. Sigh. You are asking $G/H \cong G'/H'$, $H \cong H'$ but $G \not \cong G'$. Take $G = \Bbb Z_6$, $G' = S_3$ and $H = H' = \Bbb Z_3$. Both quotients $G/H$ and $G'/H'$ are $\Bbb Z_2$.
Tons of example, eg with $\Bbb Z_2 \times \Bbb Z_2$ and $\Bbb Z_4$, with the diagonally embedded $\Bbb Z_2$ in the first and the cyclic subgroup $\Bbb Z_2$ in the latter (this is even abelian)
This is known as the "extension problem": Given $N$ a normal subgroup of $G$, and the quotient $G/N$, if you know the isomorphism type of $N$ and $G/N$, how well do you understand the isomorphism type of $G$?
Sorry for the confusion. Basically there are tons of counterexamples to all possible variations to the problem, is the takeaway
Guys, can you answer my question? Median is 50th percentile. But there are two kinds of definition of percentiles, exclusive and inclusive. Which definition of a percentile do we use when calculating median? Or can be just choose arbitrary definition and use it to calculate the median?
This is what I have done so far (the prev question).
@BalarkaSen, although I have missed a short part of the proof, here is what I have done far. I am trying to fully complete it (the missed part).
Let $a \in G$, with $o(G)=6$. Then either $o(a)=3$, or $o(a)=2$. The number of elements of order $2$ in $G$ must be odd. Therefore, the number of elements of order $2$ is either $3$ or $1$.
It cannot be the case where $|\{a \in G: o(a)=2\}|=1$ (I missed the proof here, will come back to it). So, let the elements be $ p, q, r , x, y, e$, with $ o(p)=o(q)=o(r)=2$ and $o(x)=o(y)=3$. Now, we ass…
Suppose that $V,W$ are normed linear spaces such that $\mathcal{L}(V,W)$, the space of bounded linear operators $V\rightarrow W$ with the operator norm, is complete. Is $W$ necessarily complete? The converse of this is well-known to be true.
