1,665 reputation
416
bio website reddit.com/r/…
location meatspace
age
visits member for 2 years, 6 months
seen Sep 11 at 23:52

when someone smiles at me, all I see is an ape bearing its teethe


Aug
2
asked If $N$ is a normal subgroup of $G$ with $N$ and $G/N$ solvable, prove that $G$ is solvable
Jul
27
comment Proving $\frac{-\theta + \theta^2}{2}$ is an algebraic integer in $K = \mathbb{Q}(\theta)$, given that $\theta^3 + 11\theta - 4 = 0$
Thank you for the help Robert. However I would like to check my understanding: The reason its min poly couldn't be a quartic or higher is because $[\mathbb{Q}(\theta):\mathbb{Q}]=3$, and since $\theta$ and $\frac{\theta^2 -\theta}{2}$ can each be constructed from the other in $\mathbb{Q}(\theta)$, this means it couldn't be quadratic or lower. Thus its min poly had to be a cubic. Thanks.
Jul
27
accepted Proving $\frac{-\theta + \theta^2}{2}$ is an algebraic integer in $K = \mathbb{Q}(\theta)$, given that $\theta^3 + 11\theta - 4 = 0$
Jul
27
comment Proving $\frac{-\theta + \theta^2}{2}$ is an algebraic integer in $K = \mathbb{Q}(\theta)$, given that $\theta^3 + 11\theta - 4 = 0$
Thanks for the elaboration.
Jul
27
asked Proving $\frac{-\theta + \theta^2}{2}$ is an algebraic integer in $K = \mathbb{Q}(\theta)$, given that $\theta^3 + 11\theta - 4 = 0$
Jul
25
accepted How to prove that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer
Jul
25
comment How to prove that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer
I see that it does lie in said splitting field, but could you explain more what you mean by the rest of your comment?
Jul
25
revised How to prove that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer
added 82 characters in body
Jul
25
asked How to prove that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer
Jul
23
accepted My book states that $\sum_{n=1}^{\infty}r^{-n} = \frac{1}{r-1}$ for $r > 1$
Jul
22
accepted Strange application of Cauchy's Integral Theorem
Jul
22
comment Strange application of Cauchy's Integral Theorem
Ohhh I see it! Thank you for this.
Jul
22
asked Strange application of Cauchy's Integral Theorem
Jul
19
comment My book states that $\sum_{n=1}^{\infty}r^{-n} = \frac{1}{r-1}$ for $r > 1$
thankssssssssss
Jul
19
comment My book states that $\sum_{n=1}^{\infty}r^{-n} = \frac{1}{r-1}$ for $r > 1$
Ohhh gosh ok that clears things up, thanks.
Jul
19
asked My book states that $\sum_{n=1}^{\infty}r^{-n} = \frac{1}{r-1}$ for $r > 1$
Jul
17
comment Show that $(x+1+O(x^{-1}))^x = ex^x + O(x^{x-1})$ for $x\rightarrow \infty$
Yes but in this case we are ok right?
Jul
16
answered Show that $(x+1+O(x^{-1}))^x = ex^x + O(x^{x-1})$ for $x\rightarrow \infty$
Jul
16
comment Show that $(x+1+O(x^{-1}))^x = ex^x + O(x^{x-1})$ for $x\rightarrow \infty$
Ok so past some point $x_0$ or within some neighborhood of a point, that makes sense; thanks! I think I'm going to spend some time proving those identities you listed to help my understanding.
Jul
16
comment Show that $(x+1+O(x^{-1}))^x = ex^x + O(x^{x-1})$ for $x\rightarrow \infty$
So when we say $g=O(f)$ what we're saying is g is some element of the set of all functions whose absolute value is bounded by a constant multiple of $f$, is that a correct understanding?