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"Meanwhile, I was doing well in mathematics. It was fun to solve mathematical problems, but in a deeper sense mathematics was boring and empty because for me it had no purpose. If I had worked on applied mathematics I would have contributed to the development of the technological society that I hated, so I worked only on pure mathematics. But pure mathematics was only a game. I did not understand then, and I still do not understand, why mathematicians are content to fritter away their whole lives in a mere game."


Sep
8
comment Weierstrass $\wp$ function doubly periodic
@glebovg Indeed. I'm sure not sure what I was thinking here.
Aug
29
awarded  Good Question
Aug
20
answered What is the determinant value of $J-I$ if $I$ is identity matrix and $J=(1)_{101\times 101}$?
Aug
13
awarded  Nice Answer
Aug
6
comment $L_H(X)$ is real vector space,
Do you know what a vector space is?
Aug
4
comment Group of automorphisms of a compact hyperbolic Riemann surface is finite
Thanks! $\textbf{}$
Aug
4
accepted Group of automorphisms of a compact hyperbolic Riemann surface is finite
Aug
4
comment Group of automorphisms of a compact hyperbolic Riemann surface is finite
Great, this is exactly what I was looking for. Do you know of a reference for the facts you mention, or a place where this proof is given in detail?
Aug
3
comment Group of automorphisms of a compact hyperbolic Riemann surface is finite
@QiaochuYuan I should probably clarify. I know about Hurwitz's bounded. I was wondering if there is a simpler way to obtain the weaker statement that the automorphism group is finite.
Aug
3
asked Group of automorphisms of a compact hyperbolic Riemann surface is finite
Aug
3
comment Improper integral $\int_{B}\frac {1}{|x|^\alpha}dV$
@LunaSage It happens to simplify nicely. You can find the final formula on Wikipedia, I believe.
Aug
3
comment Improper integral $\int_{B}\frac {1}{|x|^\alpha}dV$
A dyadic decomposition would also work in a pinch.
Jul
29
comment Show that there is sequence of homeomorphism polynomials on [0,1] that converge uniformly to homeomorphism
@jibounet It won't. There's no way to guarantee those polynomials are homeomorphisms from $[0,1]$ to itself.
Jul
22
comment How do you self-study Functional Analysis?
I never liked Rudin's book on functional analysis. Folland's Real Analysis is the standard graduate real analysis book today and covers all the topics you mentioned. Real and Complex Analysis by Rudin is also good. I don't understand your remarks about being "hesitant to commit" -- there's no commitment involved. If you don't like the book, stop reading it. Both books are going to be terse. Graduate level mathematics is hard, so you better get used to it.
Jul
21
awarded  Popular Question
Jul
20
revised invertibility of self adjoint operators
added 95 characters in body
Jul
20
comment invertibility of self adjoint operators
Good, detailed answer. Bravo.
Jul
20
comment Olympiad number theory question
By the way, I think you can piece a solution together from the comments here: nrich.maths.org/discus/messages/153904/149403.html?1281107322
Jul
20
comment Olympiad number theory question
I don't think your first line is correct. Suppose they are all greater than two. Then indeed all of the products of the form $pq-1$ are even. But that isn't a problem (even numbers can have odd prime divisors), and it doesn't immediately imply anything interesting.
Jul
20
comment Interview riddle
This is not mathematics. This is numerology.