Students in my area can take precalculus at the college as juniors, then calculus as seniors.
I am sort-of kind-of toying with the idea of building a four semester sequence of courses which assumes high school algebra and geometry as a prerequisite, and gets to multivariable calculus in semester four (basically, a semester of precalc, with an emphasis on polynomials and a deemphasis of transcendental functions; followed by the standard 3 semester calc sequence, more or less).
The idea being that students who are successful in that sequence finish high school with an AS or AA and math prereqs for the sciences done.
@XanderHenderson yep, it was the fact that I really love mathematics (and good at it) that kept me relevant and even top of the list in mathematics classes back there, I suspect a "normal" american student who doesn't exactly enjoy math would have a much harder time there. Especially since the competition is VERY real. Everyone is studying.
Also, what kind of anime did you happen to come across?
Also, on the subject of a university bound student not taking many courses outside their major, that is true, but it is somewhat compensated for by the fact that extracurricular and club activities are actually really important in Japan, especially if you want to maintain a good track record in High school. That's why I had to be part of at least one club otherwise I wouldn't be able to graduate
@XanderHenderson I'm assuming it was some "high school" anime? Because anime like Dragon Ball isn't exactly a great indicator of life in Japan, or life in general.
A fair bit of trig can probably stay in the precalc portion of the class---most of the elementary stuff is pretty geometrical, and doesn't require appeals to continuity or completeness.
@TedShifrin I'm not really proposing that any material be left out entirely---I am only suggesting that topics get moved around.
And I know that such a program can be successful, because I've taught that sequence at UCR---except that sequence is done in four quarters, not four semesters.
@Gokuカカロット In any event, I can't really think of any of the trash anime that led to my understanding of Japanese high school. The things that I remember are the things that are, in my opinion, not trash.
I just finished binge watching a bunch of old classic anime over the holidays: Fist of the North Star, Street Fighter, Ninja Scroll, Sakigake Otokojuku, Tekken,
You have to understand that, in the 90s and early 2000s, the popular conception of "anime" in the US was that it was all hentai, watched by lonely nerds. If you wanted to watch anime, you had to get 8th generation fansubbed bootleg VHS copies of things, shipped in from Hong Kong.
@XanderHenderson you're right about that. I didn't watch all the one's I mentioned when I was a kid and they had just hit the scene because of how difficult it was to find them. But I had a trip down memory lane that sparked a light in the recesses of my mind to go back and watch
@TedShifrin its from a 1984 comic, and Goku in question is an 11 year old clueless boy who's only human interaction was his adoptive grandfather who he accidentally killed, cut him some slack
@XanderHenderson hey, to be fair Toei animation didn't show everything just that the clothes were off, that's all. If they wanted they could've gone further like many anime did back then
@D.C.theIII it was more of a gag trope than anything else. Toriyama thinks stuff like this is funny and he's far from the only mangaka who thinks that even ignoring those that took clear inspiration from him
Even Saint Seiya wasn't immune to such tropes. Its literally known for Yaoi....
Until I started my graduate studies, I had a vivid memory. It is very difficult for me to understand how stress can have a detrimental effect on memory.
You are truly an inspiration to me for developing your skills without the aid of the internet when you were a student.
oh, i had the internet, it just wasn't as useful. google indexing the contents of PDFs was big. that happened, i forget when, but feels like the middle of grad school.
Point $A, B$ and $C$ are points on the circle with centre $O$. $CD$ is tangent to the circle $O$ at point $C$ and the points $A, O$ and $C$ are collinear (lie in a straight line). Given that the equation $\frac{a\pi - b\sqrt{3} - c}{d}$ represents the area of the shaded region in its simplest form, what is $a + b + c + d$?
Is it a necessary to have the condition that a, b, c, d are relatively prime to each other in the quesiton body?
It is to prove that $\sum_{d|n} \phi(d)=n$, where $\phi$ is the Euler phi function.
Proof: Reduce $1/n, 2/n,...,\frac{n-1}n, n/n$ to smallest form. Denominators will all be divisors of $n$. If $d|n$, then there are $\phi(d)$ many $d$ in the denominator. QED.
I guess there are various ways to formulate basically the same proof.
You're asking when the denominator of $\frac kn$ is equal to $d$ (after simplification).
You can use that $\gcd(k,n)=t \Leftrightarrow \gcd(\frac kt,\frac nt)=1$.
To get the denominator equal to $d$, you need $\gcd(k,n)=\frac nd$.
I.e. $\gcd(k,n)=\frac nd \Leftrightarrow \gcd(\frac {kd}n,d)=1$.
So $k\mapsto\frac{kd}n$ is a bijection between the set all numerators $k$ in fractions where the simplified fraction has the denumerator equal to $d$ and the set $\{j\in\mathbb N; 1\le j\le d; \gcd(j,d)=1\}$.
I have no doubts that there are better ways how to formulate this.
Clearly $n$ counts the number of elements in the set $ \{1,\ldots,n\}$. This suggests that to get a combinatorial proof we should count the number of elements in this set in a different way and get $\sum_{k \mid n} \varphi(k)$.
For $k \mid n$, let $S(k)$ be the set of $m \in \{1,\ldots,n\}$ suc...
@MartinSleziak of course. I don't know what I was thinking.
I understand it now: the idea is that the every denominator in reduced form of 1/n,..., n/n divides $n$.
so we have after reduction: a/b, (a,b)=1. Suppose a/b is not 1/n. Then how many such a's are there? $\phi(b)$. So there are $\phi(b)$ many a/b's such that (a,b)=1.
We do this for every $b$; since total no. of fractions is n, we have $\sum_{d|n}\phi(d)=n$.
At the same time, I would consider something like this perfectly fine: "In the following, whenever an index in a sum or in a product is denoted by $p$, this means that we're summing (multiplying) only primes from the given range."
In the other words, to have a convention that $\sum_p \frac1p$ means some only over primes.
The Wikipedia article has this formulation in one place: "In the following, a sum or product taken over p always represents a sum or product taken over a specified set of primes."
BTW I think I have seen $\mathbb P$ for the set of all primes. (But I do not think that this notation is used too often.)
> “Math-aware” means you can add math expression(s) as some of your keywords to have search engine help you find similar expressions and return those documents/topics that you may find relevant to your query. In short, a typical search engine plus math search.
If $\{a_n\}$ is a sequence in $\Bbb C\setminus\{0\}$ then $\prod_{n}(1+a_n)$ converges iff $\sum\log(1+a_n)$ converges. Suppose $a_n\to 0$ as $n\to\infty$, then the sum converges iff $\sum\log(1-a_n)$ converges?
The following example was left in exercise of my topology class. I think I would need help on how to prove the asserstion asked.
Let $D^n$ denote the unit ball in the n-th dimensional euclidean space and let $S^{n-1} $be the unit sphere.
Show that $D^2 $ is not homeomorphic to $D^n /S^{n-1}$.
T...
If I take an interval [0,1]. I keep wrapping it around and around and I think that it does give me a disk. If I connect 0 to 1 and just one winding, I get a 1-sphere.
@TedShifrin I don't know much beyond the fact that dyadic rational multiples of $\pi$ are constructible, as well as dyadic multiples of $\pi / 3$ and $\pi / 5$. However, according to Wikipedia, dyadic multiples of $\pi / n$, where $n$ is the product of distinct Fermat primes.
@Koro While $S^2$ is simply connected ($\pi_1(S^2)=1$) and it's compact and connected, there are other tools we can use to show that it's not contractible and therefore not homeomorphic to $D^n$ for any $n$.
For one thing, $\pi_2(S^2)$ (the second homotopy group) is $\Bbb Z$. However, though $\pi_2$ is relatively easy to define, it is hard to calculate in general. Showing that $\pi_2(S^2)=\Bbb Z$ isn't trivial.
For another thing, $H_2(S^2)$ (the second homology group) and $H^2(S^2)$ (the second cohomology group) are both $\Bbb Z$. There are good algorithms for calculating homology and cohomology in many cases, but they take some machinery to set up and define.
@Koro Here's a simple proof that $S^2$ is not homeomorphic to $D^n$ for any $n$: if you delete two points, $S^2\setminus\{p,q\}$ is not simply connected while $D^n\setminus\{p,q\}$ is for $n\ge3$. And $S^2\setminus\{p\}$ is simply connected while $D^2\setminus\{p\}$ isn't if $p$ is an interior point.
(and of course $D^1\setminus\{p\}$ isn't even connected if $p$ is an interior point of $D^1=[-1,1]$.)
This doesn't show you that $S^2$ isn't homotopy equivalent to $D^n$, however, and it doesn't deal with $S^m$ for $m>2$.
@XanderHenderson I think you don't need the Fundamental Theorem of Galois Theory for this, though you can use earlier Galois theory topics
It means a semidirect product. In general, we can write
$$G = N:Q$$
for $$G = N\rtimes Q$$
(or $G=Q\ltimes N$) for a semidirect product, where $N$ is normal in $G$.
@Shaun Are you looking for feedback because you don't know if your answer is any good? or are you looking for feedback because you want to understand the downvote?
If the latter, I would suggest that your request is futile. Only the downvoter knows why they voted as they did.
I would guess, however, that the person who downvoted your answer is the same as the person who downvoted the question, and that the downvote is meant to send the message "please don't answer low-quality or unclear questions".
Or, frankly, after reading the question in more detail, I wonder if it is even an appropriate question for Math SE? It seems like the GAP documentation is the right place to look---the question isn't really about mathematics, but about how a particular software package notates things.
Let me assume that $f\in L^p(\Bbb{R})$, then let me consider $\int_{-\infty}^{-n} |f(x)|^p ~d\lambda$ where $\lambda$ is the Lebesque measure on $\Bbb{R}$. Now I want to consider the limit as $n\rightarrow \infty$. I know that $\int_{-\infty}^{-n} |f(x)|^p ~d\lambda=\int_\Bbb{R} \Bbb{1}_{(-\infty,-n]}(x)|f(x)|^p~ d\lambda$ now using dominated convergence we can take the limit inside ad get $\int_{-\infty}^{-n} |f(x)|^p ~d\lambda\rightarrow 0$.
My question is, since I know that $\int_{-\infty}^{-n} |f(x)|^p ~d\lambda\leq \int_\Bbb{R} |f(x)|^p~d\lambda<\infty$ can I also immediately say that…
I have been working on a new quadrature technique for oscillatory integrals that pop up in my work with a British numerical analyst but they seem to be a near miss for my original cases. i was daydreaming a little bit while i was doing that and it gave me a fun little idea to explore.
right otherwise it can also be infinity, it can not be another constant since then it would mean that the function is somehow bounded which would then mean that it goes to $0$ as $n$ tends to infinity right?
My first approach would be the dominated convergence which is clearly true. But when we work with series, we also said if the serie is finite then $\lim_{n\rightarrow \infty} \sum_{k=n}^\infty ...=0$. Without using DCT. Or mabye it is in the background and I don't see it. Therefore I wanted to argue as follows:
Since I know that $\int_{-\infty}^{-n} |f(x)|^p ~d\lambda\leq \int_\Bbb{R} |f(x)|^p~d\lambda<\infty$ can I also immediately say that the "tail-integral" goes to $0$ as $n\rightarrow \infty$, since the integral itself is bounded.
I found something interesting related to simple conectivity: If $M\subset\Bbb R^n$ is a smooth submanifold of dimension $m<n-2$, then $\Bbb R^n\setminus M$ is simply connected.
Let me assume that $f\in L^p(\Bbb{R})$, then let me consider $\int_{-\infty}^{-n} |f(x)|^p ~d\lambda$ where $\lambda$ is the Lebesque measure on $\Bbb{R}$. Now I want to consider the limit as $n\rightarrow \infty$. I know that $\int_{-\infty}^{-n} |f(x)|^p ~d\lambda=\int_\Bbb{R} \Bbb{1}_{(-\infty,-n]}(x)|f(x)|^p~ d\lambda$ now using dominated convergence we can take the limit inside ad get $\int_{-\infty}^{-n} |f(x)|^p ~d\lambda\rightarrow 0$.
here's a somewhat non-trivial generalization: if $M$ is a smooth manifold and $N\subseteq M$ is a smooth submanifold, then the inclusion $M\setminus N\rightarrow M$ is $(\mathrm{codim}(N)-2)$-connected
@user123234 I guess I wanted the comment that $\int_0^\infty - \int_0^n = \int_n^\infty$. Are we assuming $\int_0^\infty = \lim\int_0^n$ by definition or what?
That's certainly how we define it with the Riemann integral. In the case of the Lebesgue integral, I think you're trying to prove it (that's what your dominated convergence theorem argument does).
If $V,W$ are Banach spaces over a $\Bbb{K}$ and $A:V\rightarrow W$, $B:W^*\rightarrow V^*$ are linear operators such that $\langle l,Av\rangle=\langle Bl,v\rangle $ for all $v\in V$ and for all $l\in W^*$ how can I show that if $A$ is bounded also $B$ is bounded?
I first thought about cauchy-schwarz inequality but this does not help me
@leslietownes but there they talk about the adjoint operator why do I know that $B$ is the adjoint of $A$? Is this always when one operator goes from $A\rightarrow B$ and the other from $B^*\rightarrow A^*$
no, i think they are using langle rangle to denote the dual pairings between V, W, and their duals. i.e. $\langle l, Av \rangle$ is shorthand for what you might more traditionally write as $l(Av)$ because $l$, an element of $W^*$, is a function on $W$, and $Av$ is an element of $W$.
I am studying the series with terms $a_k=\frac{1}{\sqrt{k}\ln{k}}$ and it is claimed that, for large enough $k$, $a_k\geq \frac{1}{k}$. I would be grateful if someone could shed some light on why that inequality holds, that is $$\frac{1}{\sqrt{k}\ln{k}}\geq \frac{1}{k}$$.
The following example was left in exercise of my topology class. I think I would need help on how to prove the asserstion asked.
Let $D^n$ denote the unit ball in the n-th dimensional euclidean space and let $S^{n-1} $be the unit sphere.
Show that $D^2 $ is not homeomorphic to $D^n /S^{n-1}$.
T...
