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11h
answered Name and role of a particular finite group?
2d
comment How can the ball reach the wall when it always has to travel halfway?
Zeno? Is that you?
2d
comment How do i evaluate this integral $ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $?
Have you tried the usual Weirstrass substitution? $t=\tan\frac x2$ followed by $u=\sqrt{t}$ seems like it would lead towards something very like a traditional elliptic integral...
Jul
4
comment Is there always a square between two consecutive cubes?
I suspect you've misunderstood the question, though this is an interesting answer to the question of 'when can the differences between two successive cubes be a square?' - that isn't what OP was asking, though...
Jul
3
comment Example of subgroup of $\mathbb Q$ which is not finitely generated
@YoTengoUnLCD Actually, it would be $\{\frac{j}{2^k}:j,k\in\mathbb{Z}\}$. These are commonly known as the dyadic rationals and you can find more information on them searching under that name.
Jul
3
comment Finding the convergence radius of $\sum_{n=1}^\infty n! x^{n!}$
Two hints: for $x\gt 0$, your series is bounded above by $\sum_n nx^n$ (why?); for $x\gt 1$, it's bounded from below by $\sum_nx^n$ (again, why?).
Jul
3
comment Is anybody researching “ternary” groups?
That generalization looks like it's shading towards combinators (and combinatory logic); I wonder if there's a clean translation there.
Jul
3
answered Newton-Raphson For Integer Factorization
Jul
1
comment In the equation $2n^3 + 3n^2 = 500,000$, what does $n$ equal?
Note that the 'nice' solution to this problem can be found without going through the general cubic equation by using some guided knowledge with the Rational Root Theorem, but I see no point in fleshing out more details since OP has provided no additional context and has already accepted an answer.
Jul
1
comment Multiplying two ordinals where one has been raised to power of $\omega$. Term order matters?
This is an awkward argument - it's also the case that $\delta\times\omega\lt\omega^\omega$ for all $\delta\lt\omega^\omega$, so you can't really talk about limits in the way you're trying to.
Jun
30
comment How to find $I=\int_{-4}^4\int_{-3}^3 \int_{-2}^2 \int_{-1}^1 \frac{x_1-x_2+x_3{-}x_4}{x_1+x_2+x_3+x_4} \, dx_1 \, dx_2 \, dx_3 \, dx_4$
Is there any reason to believe that it converges? You have non-isolated singularities in the domain (the entirety of the plane $x_1=-x_2, x_3=-x_4$) so at best you can look at PVs.
Jun
30
comment any example of non-zero ordinals $a,b,c$ for which $a < b$, but $a^c \ge b^c$?
@ChrisBedford Note that it's not really dependent on ordinal exponentiation, eiither; the same choices work for cardinal exponentiation. The real lesson here is that $\lt$ isn't continuous through limits, which you already know in a different disguise: $1\lt\frac{n+1}{n}$ for all $n$, but $1\not\lt\lim_{n\to\infty}\frac{n+1}{n}$.
Jun
29
comment any example of non-zero ordinals $a,b,c$ for which $a < b$, but $a^c \ge b^c$?
A big hint: under ordinal exponentiation, what is $2^\omega$?
Jun
29
comment Does $\zeta(s)^2 \pm \zeta(1-s)^2$ have roots at the $\rho$s?
I wouldn't necessarily put a lot of faith in Newton-based methods to converge on the complex plane. Do you get any convergence when you just ask it to find the zeroes of $\zeta(s)$?
Jun
29
comment Does $\zeta(s)^2 \pm \zeta(1-s)^2$ have roots at the $\rho$s?
What algorithm are you using for rootfinding? If $s=\frac12+it$ is a zero of the zeta function, then $1-s=\frac12-it$ will also be a zero, and so your expression will have a zero there - but it may have some 'spurious' zeroes too.
Jun
29
comment Evaluating $\int{ \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}}dx$ using Pascal inversion
I can't say specifically where you went wrong, but I think notation is dragging you down - for instance, you write $I_n(m)$, but there's a dependence on $x$ there that you don't capture at all.
Jun
29
revised Evaluating $\int{ \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}}dx$ using Pascal inversion
Tweaked some language and cleaned up formulas
Jun
28
comment Find whether $f(n) = o(g(n))$ or $g(n) = o(f(n))$
Your formula for $f$ is slightly wrong - you probably want $n^{\log\log n}$ there. You may find it easier still to write both as functions of the form $2^{a(n)}$ and then compare the corresponding functions $a()$ - but be careful; this slightly modifies your conditions for the $f\in O(g)$ relationship and you need to understand how it changes them.
Jun
26
comment prove that $p(n) := n^2 + n + c$ is not prime
@TheoBendit Ahhh, good point - I'd missed the obvious there; thank you.
Jun
26
comment prove that $p(n) := n^2 + n + c$ is not prime
This answer falls down in the case where $c=\pm 1$ and a slightly more extended argument is needed to cover that.