Reputation
6,428
Next privilege 10,000 Rep.
Access moderator tools
Badges
2 11 31
Impact
~61k people reached

Aug
24
comment May Algebraic Geometry be appropriate for me?
@PeteL.Clark The differences in role between letters and concepts I mentioned were supposed to show that hating certain letters is fundamentally und utterly different to hating certain concepts.
Aug
23
comment May Algebraic Geometry be appropriate for me?
By the way, Majin-Bu, out of curiosity: What happens if you think of the polynomial ring $R[X]$ of a commutative ring $R$ as the free commutative $R$-algebra over $X$? Like viewing polynomials as the left adjoint $R[•]\colon \mathrm{Set} → R\mathrm{-CAlg}$ to the forgetful functor?
Aug
23
comment May Algebraic Geometry be appropriate for me?
… 3) letters all share the same functionality, 4) concepts do come with a way to think, and 5) disliking concepts (like zero, imaginary numbers, set theory, general nonsense) has been done by professional mathematicians. Therefore, to me it doesn’t seem to be that much immature to hate a certain mathematical concept as a mathematician. Is it yet?
Aug
23
comment May Algebraic Geometry be appropriate for me?
@PeteL.Clark Saying “sounds like a would-be writer averring that he we try to avoud isng the letter “g” as much as possible” – isn’t that a bit overstating it a bit? This is meant as a real question, because I’m not a professional mathematician, but I don’t see the analogy “letters – concepts” at all: 1) Words are used by writers as material, concepts are used by mathematicians as tools, and letters are “submaterial”, writers actually don’t use letters. 2) you cannot introduce new letters to better express yourself as a writer, letters are sort of determined once and for all, and …
Aug
23
comment Existence of holomorphic function from unit disc to itself.
I’m not sure, it might help to check out Schwarz Lemma and the characterization of automorphisms of the unit disc (which are certain Möbius transformations).
Aug
22
comment Existence of holomorphic function from unit disc to itself.
Well, yes (if $a ≠ 0$), since then for $\lvert z \rvert < 1$, you’d have $\lvert f(z) \rvert ≤ \lvert az \rvert + \lvert b\rvert < \lvert a \rvert + \lvert b \rvert ≤ 1$. But this doesn’t help you here, because such an $f$ has $f’ = a$, so condition “$f’(3/4) = 1$” forces $a = 1$, and $\lvert a \rvert + \lvert b \rvert ≤ 1$ then forces $b = 0$, so $f = \mathrm{id}$, but that doesn’t do the trick with ”$f(1/2) = -1/2$”.
Aug
21
comment Existence of holomorphic function from unit disc to itself.
Well, $0 ∈ \mathbf D$, but $f(0) = 0 - 1 ∈ \mathbf D$?
Aug
17
comment let $f$ be continuous on $[a,b]$. Prove that $f$ is integrable on $[a,b]$
A function $f$ is integrable if …?
Aug
15
comment Why doesn't Zorn's lemma apply to $[0,1)$?
To stress the actual point that the questioner was missing, you should add “in $[0,1)$” to the end of “… states that if each chain in [0,1) has an upper bound”.
Aug
10
comment Are Lang's books reliable?
@martycohen Or safety goggles for what it’s worth!
Aug
7
revised Question concerning proving a function is not analytic?
added null sequence requirement
Aug
7
revised Question concerning proving a function is not analytic?
edited tags
Aug
7
answered Question concerning proving a function is not analytic?
Jul
23
comment Is there a way to show the following equality without induction
Nah, apparently it is what the questioner wants, as she or he accepted xpaul’s answer. Not sure how the site’s policy is on duplicate answers, I’d just leave it until someone complains, possibly.
Jul
23
comment Is there a way to show the following equality without induction
@SimonS Ok, that seems like a legit interpretation. One should nevertheless state that induction is still needed to prove the result when collapsing the telescope sum.
Jul
23
comment Is there a way to show the following equality without induction
You’re like the fifth person to answer that.
Jul
23
comment Is there a way to show the following equality without induction
By the use of “$\cdots$”, you are implicitly carrying out induction.
Jul
23
comment Is there a way to show the following equality without induction
I’m actually not even sure if it is possible to do that without induction, because finite sums are defined inductively. Maybe if you can give a non-inductive definition of finite sums …?
Jul
23
comment Is there a way to show the following equality without induction
But “$\ldots$” is hiding induction.
Jul
23
comment Is there a way to show the following equality without induction
To all the hints: Please note that using telescope sums just postpones the use of induction to showing that $\sum_{k=1}^n (a_{k+1} - a_k) = a_{n+1} - a_1$ (unless, of course, you have a non-induction way of showing that). I think it would be nice to have an answer which doesn’t suggest hiding the induction, but really has an idea of avoiding it.