That's crazy. The concept of the most immediately used and well known uncountably infinite set is full of vastly more unused numbers than used numbers.
There are countably many algebraic numbers and uncountably many reals. There are no numbers that is neither algebraic nor transcendentals. It follows the rest are transcendentals
@Secret I dunno, man. I challenge anyone <cough>not in this room<cough> to name more than five of them without having to look some up. Whereas my six year-old already knows thousands of algebraic numbers =)
The Bernoulli-Bernoulli number, and (of course) the Bernoulli-Bernoulli-Bernoulli number.
(headcanon: mathematical historians centuries hence will be scratching their heads trying to "rediscover" the Bernoulli-Bernoulli number. The Bernoulli-Bernoulli-Bernoulli number, of course, is known to all as the breakthrough that got us teleportation.)
Liuoville theorem meanwhile had nothing to do with Liuoville numbers
Liuoville numbers are the first class of numbers proved to be transcendental. They are also the class of numbers which are "indefinitely" close to any nearby rationals
This is captured by the theorem in diophataine approximation:
@Abcd $P(A|B)$ will be the fraction of $B$ that contains $A$. This is different from $P(A \cap B)$ as the whole space is used when the same segment is referred
Consider the real projective space $\Bbb P^4$ and let $\alpha,\beta$ be planes such that $\alpha+\beta=\Bbb P^4$. What are the possible configurations of $\alpha, \beta$ and a generic line $\ell$?
Here's my reasoning:
$\alpha+\beta=\Bbb P^4$ implies that $\dim(\alpha+\beta)=\dim\alpha +\dim\bet...
and since you already argued that it's not a rotation, it'd better be a reflection
If you want to make that more precise, you should work out the eigenvectors/eigenvalues of A
If it's a reflection, then you'll have a +1 eigenvalue (eigenvector points along the axis of reflection) and a -1 eigenvalue (eigenvector perpendicular to the axis of reflection)
If you always win, then you're always getting X=+1 as your outcomes. Hence the mean value is just +1, and therefore $X-\mu=0$. So in that case you'd get zero variance
Similarly, you'll also get zero variance if you always lose or always tie
@chandx Yep. And that's directly related to that |F(X)-F(Y)|=|X-Y| definition
(continuing) On the other hand, suppose you win/tie/lose at equal frequencies. Then the probability of each outcome is 1/3, so $\mu=\sum_i p_i x_i = 1/3(+1)+1/3(0)+1/3(-1)=0$
so the mean value is zero in that case, just as it would be if you always tied.
Have a lot of practice teaching laymen about standard deviation and variance.
TL;DR; It is something like average of distances from the average. (which is a bit confusing and misleading in such concise version. So read the full article)
I assume layman knows about average. I give a talk of Impo...
@Semiclassical I found the example here super-helpful^^^
in that case, we'd need to be careful when writing out the mean value as $\overline{x}=\sum_i x_i$
It's still true, but that one value of $X$ shows up more than once
more precisely, you'd have it as $\overline{x}=\dfrac{\sum_i n_i x_i}{\sum_i n_i}$ where $n_i$ is the number of times each outcome shows up
If $n_i=1$, that collapses to $\overline{x}=\dfrac{\sum_i x_i}{\sum_i 1}=\dfrac{1}{N}\sum_i x_i$
The point, I guess, is that there's two ways to regard $\overline{x}$: write down all N outcomes, ignoring whether any are the same, and compute $\sum_i x_i$
Suppose you flip a coin 100 times, winning 2 dollars when you win and losing 1 dollar when you lose. If it's a fair coin, you'd expect to see heads as often as tails
So the expected value of flipping 100 times would be 50*(+2)+50*(-1) = 50
and therefore the average value of a single flip is +1, i.e. if you played 1000 games the expected value of those 1000 games is +1000
That doesn't mean you'd actually see that for sure, of course. But if your gain after 1000 games is a lot different than that, then you'd question whether the coin is actually fair in the first place.
(The use of the variance is, as the word might suggest, to get a sense of how much variation after 1000 flips is typical. e.g. is being up +950 after 1000 games a sign that it's not a fair coin?)
(If I did want random reals, I'd do RandomReal[1,10^6]. random reals in the range 0 to 1, sampled 10^6 times)
@Abcd I mean, you're basically asking about the concept of "expectation value" as a whole
where you have $E[f(X)]=\sum_i p_i f(x_i)$
empirically, those $p_i$ are obtained as $p_i = \frac{n_i}{\sum_i n_i}$ i.e. the frequency $n_i$ with which $X=x_i$ occurs, relative to the total number of outcomes $\sum_i n_i$
so empirically you'd have $\sum_i \frac{n_i f(x_i)}{n_i}$
which, if $n_i=1$, is just $\frac{1}{N}\sum_i f(x_i)$
So you gather your samples, compute $f(x_i)$ for each of them, and take the average value over all samples
Inspired by a question on the main site: Suppose I've got $X,Y,Z$ as quartic polynomials in $t$. It turns out that there exists a fourth-order homogeneous relation between these polynomials, i.e. there's a fourth-order homogeneous polynomial $f(x,y,z)$ such that $f(X,Y,Z)$ vanishes identically
What kind of math would that go under?
Seems like something out of commutative algebra
(To put things in more gory detail, the polynomials are $(X,Y,Z)=(t^4,3t^3+4t,t^2+2)$ and the fourth-order polynomial is annoying)
I can see some things. for instance, X^4 is the only degree-4 monomial in X,Y,Z with degree 16 in t, so x^4 can't show up at all in f(x,y,z)
but I don't know whether there's a systematic approach
@Semiclassical (frenkel is quite a character!) did not know too much about langlands program until reading this book, didnt realize it related to FLT. very impressive that he just won abel prize (at ~80). find it a pity that many elite prizes are awarded decades later in late age when the respondents dont have as much chance to enjoy the fruits of their labor.
Ok just wait its kind a large sentence and sorry for my english btw
Let a>0 and let J:= [-a,a]. Let f: J--->R be a bounded function and let P* the set of all partitions P of J that contain 0 and are symetric (that is x€P iff-x€P). Show that L(f)= sup{L(P;f) : P€P*}
I have a friend who is taking a masters real analysis course, and it looks like they're going to fail it. Unfortunately, all the masters courses are year-long, so they won't be invited into the second semester of analysis and they also will have a hard time swapping into anything else at that level. I tried looking for professors who might be amenable to them transferring, and I'm not optimistic. Any idea what they might do?
Consider the real projective space $\Bbb P^4$ and let $\alpha,\beta$ be planes such that $\alpha+\beta=\Bbb P^4$. What are the possible configurations of $\alpha, \beta$ and a generic line $\ell$?
Here's my reasoning:
$\alpha+\beta=\Bbb P^4$ implies that $\dim(\alpha+\beta)=\dim\alpha +\dim\bet...
Have a lot of practice teaching laymen about standard deviation and variance.
TL;DR; It is something like average of distances from the average. (which is a bit confusing and misleading in such concise version. So read the full article)
I assume layman knows about average. I give a talk of Impo...
