And if you're not over $\mathbb{F}_2$, then that should be equivalent to the fact that swapping pulls up a negative sign, so right now I'm reasonably well-convinced that this is what you do
Yeah I remember when Ronno told us about the dimension of exterior power, it showed how being alternating, multilinear, and mapping the identity to 1 characterized it completely
Oh actually, m-vectors are cool because there's a Gauss-Green type formula that replaces the exterior unit normal with an m-vector field corresponding to the tangent planes
So I didn't know that street, it's just that my dad had something to do in Houston, I was like 5 or something, and I was like oh, where's that on the map?
And after he showed me that this was how it was pronounced I was like "You know what? I don't care, I'm saying Houston anyway because the other pronunciation is stupid"
Maybe I am being stupid, but I had a question concerning the usual proof in R^n that every derivation of $C^\infty(R^n)$ is a tangent vector (say in Lee's smooth manifolds book, pg. 53).
The problem I have is showing the surjectivity of the map $T_pR^n \to Der$
If d is any derivation, you can plainly find the corresponding tangent vector by evaluating d at the coordinate functions, call them e^i. So v^i = d(e^i), for each n.
Now the problem I have is showing that d = D_{v}, the directional derivative in the direction of v, evaluated at p.
Lee and Tu and most books use Taylor's theorem.
But I thought that since they agree on a basis (e^i) of C^\infty(R^n), then they have to be equal by R-linearity.
@anakhronizein If $f=\sum f_i e^i$ then $\Bbb R$-linearity implies $d(f)=\sum d(f_i e^i)$, but how do you intend to go from $d(f_i e^i)$ to $(\partial f_i/\partial x_i)d(e^i)$?
So I've got my workshop over the next two days, I'm gonna pack now and if I will be substituting sleep with caffeine I will let you know and we will continue
If you want to pick up some measure theory, look up Rudin Real and Functional Analysis, or for perhaps a faster treatment look up Bass Real Analysis for Graduate Students
For a way quicker one check the notes I sent
If you do Kolmogorov, make sure you know what's up in the chapter on topology but you'll prob have that down
A measure space is a trio $(X,\Sigma,\mu)$ where $X$ is a set, $\Sigma$ is a $\sigma$-algebra, and $\mu$ is a countably additive map from $\Sigma$ to $[0,\infty]$ such that $\mu(\emptyset) = 0$
Now, in $\mathbb{R}^n$, you want to start with finite unions of boxes and the volume you expect, but that doesn't form a $\sigma$-algebra. There's this neat little theorem called Caratheodory extension which allows you to build a measure out of this
Nifty thing to note about it is that it's a Radon measure. Measure is the sup over compact subsets, inf over open supersets, and every Borel set is measurable (converse does not hold, there are only $\aleph_1$ Borel sets and $\aleph_2$ measurable sets).
In particular, every measurable set is a $G_{\delta}$ set minus a null set
Hey @Alessandro , is the following the correct in the nutshell understanding of bayanesian vs frequentist?
Frequentist knows the probability distributions of all events they are interested in, and that repeated experiments should corresponds to said probability distributions. Meanwhile Bayesianist don't know or don't have the probability distribution of all events they are interested in. They first assign a prior probaility to some given question, and then each trial of an experiment or other incoming evidences will serve to update the probability distribution,
I mean the basic idea is still like that isn't it? for general $f$ you're going to define upper/lower Riemann sum over a partition using what you just said right now. just you're using $\chi$ of really strange subsets than just intervals
But yeah so, a function is measurable iff it's the pointwise limit of simple functions (this is the idea of Lebesgue sums, you're kinda partitioning the range instead of the domain of a function)
So for a non-negative function $f$, you have $\int_X fd\mu = \sup_{g\le f} \int_X gd\mu$ where $g$ is simple
I'm not convinced, take as $f$ the indicator function of the irrationals on $[0,1]$, there is no positive simple function with $g\le f$ apart from the one which is always $0$
If $A\subset X$ has finite measure, then for any $\epsilon$, there's some set $E$ such that $\mu(A\setminus E) < \epsilon$ and $f\restriction_E$ is continuous
So if you know this, let $f$ be non-zero on $A$ with positive measure. I can find a subset $E$ with positive measure such that $f$ is continuous on it and positive, take a compact subset $F$ of positive measure, find the minimum of $f$ on it, your integral is at least $\mu(F)\min_{x\in F} f(x)$
Yesterday my sir asked us a question:"How can you find the value of tan2° without using the calculator? " I asked, whether he is asking the formula of tan 2A or something, but he said no its tan 2°. I tried my head out in every possible way even tried out the approximation method of differentiati...
(I actually just came up with this on the spot since our professor didn't really deal with this point whatsoever, she just said "take the set of functions whose absolute pth power has finite integral")
Yeah, I was thinking about limits, since they are so good at dealing with indeterminate forms
but I suspect it might be tricky here since $f^+$ and $f^-$ are part of the same measurable function $f$ so their relative growth or something might be less obvious
I see, hmm, so it seems these cases, the integration becomes "path dependent" since how you reach infinity for the positive and negative portions of the function will affect the answer
Hmm, I wonder if I can apply the monotone convergence theorem for any measurable function $f$ by slicing up $f$ such that each segment is monotone and bounded, i.e.:
@Daminark that statement strongly reminds of the delta distribution in physics, because as $k \to \infty$ the thing get infinitely high to maintain the same area
I also don't know anything serious about distributions, delta stuff, or hyperreals, so really that was just a mashup of words that is either insightful or just cool
So I am guessing, lesbegue integral might tend to give unexpected behaviour if some of the sets are measure zero sets, but that's a really overgeneralised wild guess
That'd follow from linearity and the fact that if $g=0$ ae, then $\int g = 0$, so you just subtract it off
Now, I also don't know shit about Fourier analysis
But like, so apparently you usually think about Fourier in $L^2$ theory
If you sorta think about periodic functions, then renormalize and look at $\mathbb{R}/\mathbb{Z} = S^1$, thereby you can kinda think about it as a function on the circle (use exponential I guess to do it smoothly)
And that $e^{inx}$ stuff in Fourier series is basically a statement that those functions form an orthonormal Schauder basis for $L^2(S^1)$
Now, the problem with it is that you don't think of the limit as taking on definite values at all
But apparently pointwise convergence of Fourier stuff is like, really hard
Anyway the more I rattle off the higher the chance I'll say something really stupid so I'll hold back for now
If $\tan\theta$ = $n\tan\phi$,then what is the maximum value of $tan^2(\theta-\phi)$?
My attempt:
First I tried to reach $tan^2(\theta-\phi)$ using the given condition, but couldn't.
Next, I used the property that $AM>=GM$ to reach:
$\dfrac{\sin^4(\theta-\phi)+\sec^4(\theta-\phi)}2 $ >= $\...
We all know that if a ring with unity is of characteristic p then it will contain Z_p as a subring but if the ring is without unity then how can we show it?
If $\tan\theta$ = $n\tan\phi$,then what is the maximum value of $tan^2(\theta-\phi)$?
My attempt:
First I tried to reach $tan^2(\theta-\phi)$ using the given condition, but couldn't.
Next, I used the property that $AM>=GM$ to reach:
$\dfrac{\sin^4(\theta-\phi)+\sec^4(\theta-\phi)}2 $ >= $\...
