Reputation
32,125
Next tag badge:
315/100 score
19/20 answers
Badges
6 50 101
Newest
 Enlightened
Impact
~648k people reached

16h
comment Isomorphism between $\mathbb R^3$ and the the Heisenberg group.
What do you mean by 'as a group'? There's no group isomorphism between $\mathbb{R}^3$ and the Heisenberg group because the former (under the usual group operation, at least) is abelian while the latter isn't. If you mean a three-dimensional parametrization of the Heisenberg group, there are several, with the easiest coming just by writing $z\in\mathbb{C}$ as $a+bi : a,b\in\mathbb{R}$...
1d
comment Prove that for $k>1$ $a_n$ is a perfect square
Have you tried explicitly computing $P(3)$, $P(4)$, etc. to see if you can find a pattern?
1d
comment $n \equiv 7 \pmod{8}$, prove $\sigma(n) \equiv 0 \pmod{8}$
This is a much better answer than the hint I had in comments - really nice!
1d
comment $n \equiv 7 \pmod{8}$, prove $\sigma(n) \equiv 0 \pmod{8}$
A broad hint: $\sigma(n)$ is a multiplicative function, and therefore so is $\sigma(n)\bmod 8$ (why?). If $n\equiv 7\pmod 8$, what are the possible factorizations of $n$?
2d
awarded  Enlightened
2d
awarded  Nice Answer
Apr
24
comment How do I prove the following equation involving $\int_{0}^{+\infty}\left(\frac{sin(x)}{x}\right)^ndx$
It's easy to make a case that the integrals on the left are manifestly positive: since $\sin x\geq 0$ for $0\leq x\leq \pi$ and $\sin x\leq 0$ for $\pi\leq x\leq 2\pi$ and $\sin(x+\pi)=-\sin x$, then $\left|\frac{\sin(x+\pi)}{x+\pi}\right|\lt \left|\frac{\sin x}{x}\right|$ and so the 'negative' piece of each $2\pi$ period is smaller than the positive piece just before it - i.e., break the integral into integrals from $0$ to $\pi$, $\pi$ to $2\pi$, etc. and make the case that each integral is smaller in absolute value than the one before and they're alternating in sign with the first positive.
Apr
19
comment If $G/Z(G)$ is cyclic, why is $G$ only abelian and not also cyclic?
I'll go a step further - start with the simplest case, $C_2\times C_2$, and see what goes wrong.
Apr
18
comment Is there an explicit irrational number which is not known to be either algebraic or transcendental?
@lulu I believe that number is known to be transcendental. It's the sum of a rational number ($\frac19$) and a theta-value at a rational argument that I'm pretty sure is known not to be algebraic.
Apr
15
comment A sum of irrational numbers is an algebraic integer
If you need to find a suitable polynomial 'by hand' then you can peel off: $x-\sqrt{2}+\sqrt{17}(\frac12(7-\sqrt{13}))=\sqrt[3]{5}$, so you can cube both sides of this, etc. This gets painful in a hurry, though. Instead, you might consider what operations algebraic integers are closed under...
Apr
15
comment Bijection between $[\mathbb N\to \{0,1\}]$ and $[\mathcal P(\mathbb N) \to \{0,1\}]$
@user3286435 If you work in a system where all relations etc. must be computable, then you're right that $\mathcal{P}_c(\mathbb{N})$ is countable in the 'base model' - that is externally - but there's no computable bijection between $\mathcal{P}_c(\mathbb{N})$ and $\mathbb{N}$ - that is, $\mathcal{P}_c(\mathbb{N})$ is 'internally' uncountable.
Apr
15
answered Bijection between $[\mathbb N\to \{0,1\}]$ and $[\mathcal P(\mathbb N) \to \{0,1\}]$
Apr
13
comment Can you comb the hair on a 4-dimensional coconut?
Spoiler from that Wikipedia article: the short answer is that it alternates, so that $S^3$ has a tangent field, $S^4$ doesn't, $S^5$ does, etc.
Apr
12
comment How to integrate $x^2\sin(x^2)$?
Whoops - yes, it is. Fixed, though of course that doesn't materially affect the result here.
Apr
12
revised How to integrate $x^2\sin(x^2)$?
Fixed a matho
Apr
12
answered How to integrate $x^2\sin(x^2)$?
Apr
12
comment What is the Order (Big O) of this polynomial?
Two pieces of relevant nitpickery: (1) I presume you mean as $n\to\infty$, since $n$ is usually used for integer variables, but note that $p(n)$ is $\in O(1)$ as $n\to 0$, for instance - knowing what your asymptote is matters! (2) You (or your instructor) probably pedantically mean $\Theta()$ - it's also completely true to say that $p(n)\in O(n^{14})$ as $n\to\infty$, because $O()$ only talks about upper bounds, not precise orders. (Also, (3) strictly speaking $p(n)$ isn't a polynomial, but that's another matter entirely.)
Apr
5
comment Rewriting ∃! using predicate logic expressions ( “=” excluded)
It's a little strange for = to not be a valid symbol - in the language of set theory there are some reasonable definitions for equality, but most of those require at least $\in$. You might have a look at en.wikipedia.org/wiki/First-order_logic#Equality_and_its_axioms , but I would speak to your professor first.
Apr
5
comment Rewriting ∃! using predicate logic expressions ( “=” excluded)
@P.Lance No - in fact, what you've written is exactly equivalent to $\exists x A(x)$; can you see why?
Apr
5
comment Finding the limit $\lim_{n\to\infty}{\sum_{i=0}^n \frac{(-1)^i}{i!}f^{(i)}(1)}$.
What have you tried so far? Do you know any formulas that involve sums over the derivatives of a function, suitably scaled?