$X^2-\beta X+1$ is the minimal polynomial. To show it is irreducible, it suffices to show that $\alpha$ and $\beta-\alpha$ are not in $\mathbb{Q}(\beta)$
If $f\in F[x]$ with degree $n$, then it has at atmost $n$ roots in $F$. Suppose it has all n roots. Can we have a field extension $E$ of $F$ containing other roots?
@TedShifrin because what i'm trying to say is if I let $P=X^2-X\beta +1\in \mathbb{Q}(\beta)$ , with root $\alpha$, then the only other possible root is $\beta - \alpha$ and we can't have $\alpha \in \mathbb{Q}(\beta)$ because $|\mathbb{Q}(\alpha):\mathbb{Q}|=12$ and this implies that $\beta - \alpha \notin \mathbb{Q}(\beta)$
He deduced it by saying that $\mathbb{Q}(\alpha)$ has a unique subfield $L$ such that $|L:\mathbb{Q}|=6$. Then he used an argument to show that $L= \mathbb{Q}(\beta)$
Yeah, he used the following proposition: If $H\leq Gal(\mathbb{Q}(\xi_{p})/\mathbb{Q})$ then setting $\alpha = \sum_{\sigma \in H}\sigma(\xi_{p})$ we have $\mathbb{Q}(\alpha)=\mathbb{Q}(\xi_{p})^{H}$
yes, that's what concerns me. the unanswered list.
a person asked a kind of interesting question in operator theory a while ago. it was obviously wrong, but it might have been a garbled version of a good question. about what happens at the boundary of the spectrum when you pass to a subalgebra.
i gave an answer that suggested, please say more about what motivated this. dead silence.
my daughter successfully negotiated ice cream as her afternoon snack.
i love la jolla, it's a beautiful place. i was last down there for work. i sat in a poorly air conditioned office building and deposed somebody for six hours.
when it was done we were all talking with the stenographer and videographer about our plans for the rest of the day. i said i was going to legoland. the witness said, oh, is your daughter with you? i said no. i don't think anybody understood that i was joking.
@TedShifrin My mom's brother and family lived near La Jolla for a while. He was in the Navy, so San Diego was their home except when he was stationed in Japan.
oh, cool. one of my friends' husbands used to command a group, squadron, something or other at pendleton. i'd pop in and say hi to him when i was down there.
ted: i just beat rush hour. i think i made it in 2.5 hours.
my grandfather was a major for a while in WW2 but they were tossing out promotions like souvenir t-shirts for a while. everyone else in my family was 1LT or lower.
we had a substitute teacher in high school that we made fun of for being old and out of it, and when i googled him later he had a silver star for doing something nearly suicidal in combat and somehow surviving.
The only thing I can find online is that he commanded VAW-114 in the 70's. My mother had said that he was in command of a carrier, but I cannot find online evidence of this.
in 1966, as a Lt Cmdr, he made the first Greyhound carrier onboard delivery (COD) in the combat zone on board Kitty Hawk, from NAS Cubi Point, Philippines, in the Gulf of Tonkin.
@robjohn: hello! Did you see the outline of the proof? In the centenary edition of the book, (. ) open interval is used in place of [] closed interval as symbols. I think the set theory didn’t exist that time.
> Celebrating 100 years in print with Cambridge, this edition includes a Foreword by T. W. K?rner, describing the huge influence the book has had on the teaching and development of mathematics worldwide. Hardys presentation of mathematical analysis is as valid today as when first written: 1908
Review comment: It's a 5 star book with a great knowledge of calculus and complex numbers also with additional theorems in calculus which can be found no where in books available in India
> Between the years 1874 and 1897, the German mathematician and logician Georg Cantor created a theory of abstract sets of entities and made it into a mathematical discipline...
i do think of axiomatic set theory as an early 20th century thing but certainly they had the 'theory' of intervals worked out. the notation was just less standardized.
i think that cantor got into set theory because of fourier series. the sets on which they converge and diverge can get ugly fast. but not a lot of that in textbooks, then or even now.
students got multiple tries on this exam. i think you could fail it twice before things got serious. people who failed it usually failed because they didn't take preparation seriously and assumed it would be 'easy undergraduate stuff,' which it is, but it covers a lot of easy undergraduate stuff, and was a timed exam.
they're slightly more advanced than GRE questions (and not multiple choice), but yeah, a lot of similar subject matter. on the GRE i found it very helpful that questions were given with candidate answers.
when i took the subject exam, it did cover, i want to say maybe chapters 1 and 2 of each introductory book in abstract algebra, real analysis, etc. it's helpful to study the specification of the exam to realize what won't be on it.
i think the hardest problems when i took it were multivariable integrals. just calculus homework.
when i took it there were definitely experimental questions on the subject exam. i don't know if there were experimental questions on the math portion of the general exam.
'There exists a continuous function from{f(x; y) belongs to $R^2 : 1 < x^2 + y^2 < 2}$ onto {x belongs to R : |x| > 1}' what does it mean? |x|>1 is codomain?
But I often plot stuff using plain Python, outputting directly in SVG.
SVG knows how to do quadratic & cubic Bézier curves, which is rather handy, if you have the Bézier control points.
In fact, vector graphics systems like SVG, PDF, and Postscript use Bézier curves to draw circles and ellipses, since that avoids square root extraction
One problem is something like this (problem 6B): in a square matrix of order n by n, if diagonal elements dominate sum of other elements in each row then the matrix is invertible. It looks familiar as it's stated in Strang's book (without proof, iirc).
I just noticed that Paul de Casteljau is still alive. I bet he never imagined back in 1959 that the algorithm he developed for plotting Bézier curves would be used so extensively today. Imagine if he & Bézier got royalties for that stuff. ;)
@leslietownes Square roots are fast on modern hardware, but they're still slower than division, which is slower than multiplication. When you're doing a lot of that stuff in a loop, it pays to optimize it.
Modern fonts use Bézier curves. So to render this Web page, your browser is rendering a lot of Bézier curves.
@leslietownes Yes, it's real time. That's why you can zoom in & out without pixellization.
The renderer doesn't recompute the curve at every step of the zoom, though. It uses simple linear interpolation to speed things up, but the exact curve is computed when you stop zooming, so you can see some pixellization while the zoom is happening.
Kangaroos are pretty chilled-out, too. Wild ones don't like it if you get too close, though. They hop away if you get within 3 metres or so. But tame ones love interacting with humans. I used to hand-feed some kangaroos when I was a kid. There was a place near my aunite's house that had a small mob of tame roos.
From 2010 to 2017, I lived near Woolgoolga. We'd sometimes get roos in our yard. It's not unusual to see roos in the Woolgoolga town centre. They often lie around on the football field. They sometimes even cross the main road, so you have to watch out for them while driving. Kangaroos don't have much road sense. ;)
Woolgoolga has a big Indian population (and the 1st Indian temple in Australia), so I expect there might be some captive peacocks in the district, although I didn't see any. We did get the occasional bush turkey in our yard, though.
And lots of parrots, who loved our bird bath. Mostly rainbow lorrikeets, but I counted 6 different parrot species, including the rare black parrots, who never actually landed in our yard, just flew overhead.
our peacock population is due to some deluded people in the 19th century who thought it would be fun to keep them as pets. we also have feral parrots, conures mostly, for similar reasons.
at our old house we had a magnolia tree and a dozen of the birds would land in it and eat all the seeds and make huge messes on the sidewalk.
When our umbrella tree was in season, a pair of king parrots would visit to eat the fruit. The female has nondescript green plumage, but the male has a vivid red breast.
Birds are not known for their table manners. This has generally worked out well for the plants that the birds feed on.
Suppose $f(z) = \sum_{n=0}^\infty a_nz^n$ has a radius of convergence $R>1$. If $f(z)\in\Bbb R$ for all $|z| = 1$ then $f$ is constant on $D(0,R)$ i.e, open ball of radius $R$ centered at $0$.
We had a few different corvid visitors, too. Mostly a couple of species of magpie, but one day we had some amazing-looking ravens at the bird bath, with deep chocolate plumage. They didn't bathe, though.
The proof I know is letting $a_n = \alpha_n + i\beta_n$ for each $n$ then $\sum_{n=0}^\infty a_n z^n = \sum_{n=0}^\infty (\alpha_n\cos n\theta-\beta_n\sin n\theta)+ i\sum_{n=0}^\infty (\alpha_n\sin n\theta+\beta_n\cos n\theta)$. So the first summand $\sum_{n=0}^\infty (\alpha_n\cos n\theta-\beta_n\sin n\theta) = 0$ for all $\theta\in [0,2\pi]$. And this series is in fact uniformly convergent on $[0,2\pi]$. Then integrate this series with $\cos m\theta$ and $\sin m\theta$, $\alpha_m =\beta_m =0$.
Here's an Australian Coastal Carpet Python sunbathing in my back yard a few years ago. The snake spent the winter in that rock wall, but disappeared in the early spring. Sorry I don't have a better photo. It was hard to get even that close without scaring the snake.
all. I can't see how existence of degree 24 polynomial over ~~ is equivalent to $S_5$ has a subgroup of index 24. Doesn't look like a correspondence theorem.
@robjohn Not sure, what my question here is exactly. But when you solve the individual separable DE's in the factorization above. For the parts with $i$, you get $-i\sqrt{y^2-1}=x+A$ and $i\sqrt{y^2-1}=x+B$. But when you solve $\frac{dy}{dx} = \pm \frac{\sqrt{1-y^2}}{y}$. You get real solutions. I haven't learned complex calculus yet, so I am not even sure if the above solutions are right, but wolfram says so. So I guess my question is are they equal?
If you don't understand the complex numbers, don't factor over the complex numbers. Just use that if $a^4=\left(b^2-1\right)^2$, then $a^2=\pm\left(b^2-1\right)$, so $a=\pm\sqrt{\pm\left(b^2-1\right)}$
@LearningCHelpMeV2 furthermore, those are real if $|y|\le1$
@dc3rd I use Mathematica. here is the output of ContourPlot[y^2-x^3-x^2==0, {x, -5/4, 1/2}, {y, -1/2, 1/2}, AspectRatio->Automatic, PlotPoints->50, ContourStyle->Thickness[1/300], GridLines->Automatic, ImageSize->600]
The extra options are to match PM2Ring's image :-)
@robjohn The other day I asked if $f(x)$ and $g(t) \sin(xt)$ being in $L^{2}(\mathbb{R})$ implies that $$\int_{\mathbb{R}^{2}} f(x) g(t) \sin(xt) \, \mathrm dt \, \mathrm dx =\int_{\mathbb{R}^{2}} f(x) g(t) \sin(xt) \, \mathrm dx \, \mathrm dt. $$ But isn't this just Plancherel's theorem for the sine Fourier transform? Could this somehow be false if $g(t)$ itself is not in $L^{2}(\mathbb{R})$?
how to prove it? OK, if there's an element of order 4 its cyclic. otherwise all non identity elements have order 2. math.stackexchange.com/questions/238171/… might be the result to use in that case.
yeah. it's worth keeping a list of ways of classifying groups of small order. with three or four theorems you can get into the 20s before things get complicated.
when you get to order 32, you're basically dealing with all of the complexity of groups.
This is stronger than the corollary of Lagrange' theorem that says that an element of order $n$ appear as a multiple of $\phi(n)$, where $\phi$ is Euler's function.
Hi, I'm studying the uniform convergence of $\sum_{n\geq 1} \frac{(-1)^{n}}{\sqrt{n}}x^{n}$ on $E=[0,1]$. But M-Test seems does not work, because: with $f_{n}(x)=\frac{(-1)^{n}}{\sqrt{n}}x^{n}$ then $|f_{n}(x)|\leqslant \frac{1}{\sqrt{n}}$ on $E$ but $\sum \frac{1}{\sqrt{n}}$ diverges.
Well, setting $s_{N}(x)=\sum_{k=1}^{N}\frac{(-1)^{k}}{\sqrt{k}}x^{k}$, then we are looking for $(s_{N}(x))\underset{N\to +\infty}{\longrightarrow} f(x)$ for the pointwise convergence. I do not know how calculate this limit.
Well, so we can use partial summation to get an estimate of Cauchy sum and then use Cauchy criterio. With $a_{n}=\frac{(-1)^{n}}{\sqrt{n}}$, then $\sum_{n\geq 1}a_{n}$ converge, then $\sum_{n\geq 1}a_{n}x^{n}$ converge uniformly. But can we find the limit function $f$ for the pointwise convergence?
Than koro and leslie, I had completely forgotten that the convergence for the coefficients $a_{n}$ of $x^n$ were enough to ensure the convergence of the series.
