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visits member for 3 years, 9 months
seen May 3 at 19:12

Oct
17
awarded  Yearling
May
28
comment Surprisingly elementary and direct proofs
Can you please explain a bit about the part "the computation can be done directly without appeal to the general minimum modulus principle"? This interests me very much! (I didn't do that proof in my lecture because I didn't have the general minimum modulus principle available.)
May
26
comment How do you respond to “I was always bad at math”?
@Steve: Maybe it depends on what you mean by "hard" and "struggle". The first thing I remember for which I'd use those words is closure operators, and this was in my first year at university. But that doesn't mean that I understood everything before immediately at the first read. So I come to the same conclusion as Brian, and would only say that math is hard for most people.
May
25
comment Infinite series and its upper and lower limit.
@Brian: Ah, thanks, now I got it! I interpreted the fact as the positive outcome of the ratio test for the modified series, which induced me to post my first comment. Thanks again!
May
25
comment Infinite series and its upper and lower limit.
@Brian: Ah, I think now I get your point. The result I use is simple enough, but it has to be proved, and one has to say that one uses it. Agreed! However, I maintain that your formulation "you can’t make use of the fact to show that it converges" is not true. Or maybe I don't understand "can’t make use" well enough, I'm not a native speaker? I'd rather say something like "this fact alone doesn't yet show that it converges".
May
25
comment Infinite series and its upper and lower limit.
@Brian: OK, let me formulate it differently. The series has positive terms, so the sequence of its partial sums is increasing. And an increasing sequence is convergent if any of its subsequences is convergent. The OP proved convergence of one subsequence, so we're done.
May
25
comment Infinite series and its upper and lower limit.
@Brian: "you can’t make use of the fact to show that it converges" is not true. You should make use of that fact, and indeed you should do it properly. The series in question has positive terms, so that fact proves that the series is absolutely convergent.
May
13
comment How to represent the floor function using mathematical notation?
@Nathaniel: We're talking about the $(-\frac\pi2,\frac\pi2)$ branch (the tangent is $\pi$-periodic) :-) Of course it's only convention, but it's the standard mathematical convention to use this branch and not another one.
May
13
comment How to represent the floor function using mathematical notation?
@Clayton and agksmehx: Using this as an implementation for the floor function would be ludicrous (pardon my French). I'm pretty sure that any implementation of $\tan$ involves the floor function in some way of another.
May
13
comment How to represent the floor function using mathematical notation?
@agksmehx: Still, the factorial should be defined as a (finite) product, not via the Gamma function!
May
13
comment How to represent the floor function using mathematical notation?
@Nathaniel: Why do you say that $\arctan$ is multi-valued? Isn't it widely accepted that it's the inverse function of $\tan\colon(-\frac\pi2,\frac\pi2)\to\mathbb R$? (For me as a mathematician this is the standard definition, and I don't know any programming language that would use another definition.)
May
12
comment How to represent the floor function using mathematical notation?
@agksmehx: Zev's answer should be the accepted one - no chickening out there, just the plain professional definition. (Of course this answer is a nice idea, too.)
May
10
comment How far can one get in analysis without leaving $\mathbb{Q}$?
Yes - compactness is at the heart of analysis!
Jan
29
awarded  Revival
Jan
12
comment What's wrong with $\sum_{i=0}^{\infty}x^i = \frac{1}{1-x}$
A small TeXnical tip: if you use \lvert, then you should also use \rvert. Thus, to denote the absolute value of x, the best usage is \lvert x \rvert. Note that |x| also works (but |\sin x| has spacing issues that \lvert \sin x \rvert does not have).
Dec
28
comment counterexample for the fixed point theorem
COncerning "$f$ has no fixed point", there's no need to assume the contrary. I'd say it's by its very definition that $f$ has no fixed point.
Nov
18
comment Is every connected metric space with at least two points uncountable?
@Mathematics: Yes, it avoids a singleton, and also the more pathological case of the empty space containing no points at all (which is connected by definition!).
Nov
18
comment Is every connected metric space with at least two points uncountable?
@Mathematics: Yes, that true for any metric space.
Nov
18
comment Is every connected metric space with at least two points uncountable?
@Mathematics: No, at least not if $X$ is a metric (or more generally a Hausdorff) space. And that's the point! The intermediate value theorem is not valid, and $X$ is not connected. (Take $f(x_0)=0$ and $f(x_1)=1$; then $f$ is continuous(!) and doesn't take the intermediate values in the open interval $(0,1)$.)
Nov
18
comment Is every connected metric space with at least two points uncountable?
For me, the main point is that a metric (or topological) space $X$ is connected if and only if the intermediate value theorem is valid on $X$; see user49521's answer.