Reputation
12,131
Top tag
Next privilege 15,000 Rep.
Protect questions
Badges
2 23 50
Impact
~241k people reached

Aug
15
comment Generalized modulo arithmetic
@MichaelBurr: Interesting thought. Since all elements of the sequence are already included in this algebraic structure, adding all the $p$-adics would be sort of a completion. However as far as I can see, there should also be other weak limits, for example sequences $\mod k!$, so I guess the completion would probably contain more than the $p$-adics.
Aug
15
comment Generalized modulo arithmetic
@IvoTerek: Thanks for the hint at the sandbox; unfortunately that would require that I anticipate that MathJax would get unbearably slow in my post. At the time when I notice, it's already too late.
Aug
15
revised Generalized modulo arithmetic
added 856 characters in body
Aug
15
asked Generalized modulo arithmetic
Aug
15
comment The logarithm is non-linear! Or isn't it?
Actually any invertible function $f: V\to X$ from a vector space $V$ to an arbitrary set $X$ can be made linear by giving $X$ the vector space structure defined by $x+y := f(f^{-1}(x) + f^{-1}(y))$ and $\lambda x := f(\lambda\,f^{-1}(x))$. Of course, due to invertibility, it doesn't matter on which side the original vector space sits: for $f: X\to V$ just use the same construction on the inverse function.
Aug
13
comment Proving conditionals
Err … in your proof table, the last two columns are equal. You want to have a T in the last row/last column.
Aug
13
revised How to prove or disprove $n^{28} = O(2^n)$
added 4 characters in body
Aug
9
comment If $f(\alpha x, \alpha y) = f(x,y)$ is $f$ some special function?
Being homogeneous of degree $d$ does not require that the origin is excluded. If $d>0$ the function can even be continuous at the origin (e.g. the function $f(x)=\|x\|$ is homogeneous of degree 1, and well defined and continuous at the origin).
Aug
9
comment convergent sequence $\{a_n\},\{b_n\}$, does it convergent $\{ {a_n}^{b_n} \}$?
But good example for another convergent value than $1$ (and of course that can easily be made into an example of non-convergence by interleaving two such sequences).
Aug
9
comment convergent sequence $\{a_n\},\{b_n\}$, does it convergent $\{ {a_n}^{b_n} \}$?
But the question was only about the sequence being convergent, not about the value it converges to.
Aug
9
answered Determinant of matrix with unit length rows
Aug
9
comment convergent sequence $\{a_n\},\{b_n\}$, does it convergent $\{ {a_n}^{b_n} \}$?
There are also other examples with convergence, despite only using positive vaues. For example with $a_n=b_n=1/n$, $a_n^{b_n}$ converges to $1$. Indeed, I've not been able to find any two strictly positive series converging to $0$ where the power does not converge to $1$ (which doesn't necessarily mean they don't exist, only that you probably have to think harder to find one).
Aug
9
comment False proof: $\pi = 4$, but why?
@biffletsbq: Ah, with that I can agree. :-)
Aug
9
comment False proof: $\pi = 4$, but why?
Wait, why is $2\pi\frac{\pi r^2}{2} = \pi^2r^2$ blatantly false? I would consider it obviously true, as you get the right hand side by cancelling the $2$ on the left hand side. I certainly cannot derive any value of $\pi$ from that equation.
Aug
9
comment convergent sequence $\{a_n\},\{b_n\}$, does it convergent $\{ {a_n}^{b_n} \}$?
Intuitively I expect that bad things can happen if both $a_n$ and $b_n$ converge to $0$.
Aug
9
answered What is the difference between dense and closed sets?
Jul
18
awarded  Nice Question
Jul
13
comment Is there a simple sufficient condition for a function to depend “only on $r$”?
A necessary (but not sufficient) condition for dependence only on $r$ is that you must get the same function if you exchange $x$ and $y$. That simple test would already have identified the student's answer as wrong. More generally, if $g$ has any symmetry, that symmetry must also show up in $f$.
Jul
13
comment Typesetting the mathematical expression “const.”
The reason could of course just be that if you just type const. without any further commands in $\rm\LaTeX$ formulas, you get italics.
Jul
13
comment When a function is a dimension?
@JonathanHebert: So what would you then call a criterion in the everyday-language sense that is not necessary (for example, a sufficient criterion)? The "being seven" criterion sounds ridiculous because it's ridiculously narrow. More similar to my dimension sentence would be "an obvious criterion for the term "numbers" making sense for elements of an algebraic structure is if that structure includes the integers."