particularly I like the vocational training. Not every kid is university material, and there's absolutely no justification to going to college to study pottery at a price tag of $60K/year
youre very difficult to talk to, I have to be frank
Hey. I have the following theorem: Prove the following version of the Mean-Value Theorem: If $f(x, y)$ has first partial derivatives continuous near every point of the straight line segment joining the points $(a, b)$ and $(a+h, b+k)$, then there exists a number $\theta$ satisfying $0<\theta<1$ such that $$ \begin{aligned} f(a+h, b+k)= & f(a, b)+h f_1(a+\theta h, b+\theta k) \\ & +k f_2(a+\theta h, b+\theta k) \end{aligned} $$
This part: the first partial derivatives are continuous near every point of the straight line segment joining the points $(a, b)$ and $(a+h, b+k)$
Does it mean the partial derivatives are continuous in $(a,b)$?
I am using a similar theorem to prove continuous partial derivatives implies differentiability in $\mathbb{R}^2$ But using the theorem i just wrote above shouldn't work and I don't see clearly why.
Suppose that there is a field F. Let's say that it is totally ordered (i.e., partially ordered + Trichotomy condition). Can we say that i) $a>b\implies a+c>b+c$ and ii) $a>b, c>0\implies ac>bc$?
The field that is totally ordered and also satisfies i) and ii) is called ordered field. $\mathbb C$ is not ordered.
According to Calculus: A complete course by Adams & Essex, a version of the mean value theorem in several variables is given by
If $f_1(a,b)$ and $f_2(a,b)$ [notation for partial derivatives] are continuous in a neighbourhood of the point $(a,b)$, and if the absolute values of $h$ and $k$ are...
@Goku This was not a general question, but to understand the context that they raise the question about the ubiquity of complex numbers, which cannot be answered without specific motivation.
@CroCo Usually Q is symmetric, but it is not necessary, since you can replace any non-symmetric Q by (Q+QˆT)/2.
@Goku Sure. Doubling the cube is one of the classic unconstructible problems. "the nonexistence of a compass-and-straightedge solution was finally proven by Pierre Wantzel in 1837. [...] the degree of the field extension generated by a constructible point must be a power of 2"
> but even in ancient times solutions were known that employed other tools.
The search for proofs that cube duplication and angle trisection are unconstructible eventually led to more general questions about polynomial roots, which led to Galois theory.
Here's a little diagram that illustrates that $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$ and $\tan^{-1}(1)+\tan^{-1}(1/2)+\tan^{-1}(1/3)=\pi/2$.
@Goku Ok. Of course, 15° is easy to construct. Annoyingly, 1° requires trisection, but 1.5° is constructible. Generally, if the trig ratio is nice, the angle is ugly, and vice versa, except for the handful of well-known cases.
OTOH, we can easily find nice relations of the form $\tan^{-1}(1/c)=\tan^{-1}(1/a)+\tan^{-1}(1/b)$, where $a,b,c$ are integers. These are useful to compute pi from the Taylor series for arctan.
In the old days, when people did this stuff by hand, or using primitive calculators, it was less work to compute arctan of integer reciprocals because all the numerators in the series are 1. It's still kinda handy on computers if you want to compute lots of digits of pi, although there are other ways to do arctan, especially if you have cheap sqrt.
I guess some of those constructible polygons aren't so well-known. Gauss did some impressive work on the Fermat prime polygons when he was still quite young, which established him as an outstanding geometer & mathematician. en.wikipedia.org/wiki/Constructible_polygon
But getting back to arctans, let $c^2+1=df$. That is, $d,f$ are a factor pair of $c^2+1$. Let $a=c+d, b=c+f$. Then $\tan^{-1}(1/c)=\tan^{-1}(1/a)+\tan^{-1}(1/b)$
Using the sum of tan identity, $c=(ab-1)/(a+b)$
machination.eclipse.co.uk has a huge collection of such identities. It writes $\tan^{-1}(1/x)$ as $[x]$. In that notation, my diagram illustrates $[1]=[2]+[3]$.
@PM2Ring Years ago, I designed a worksheet for a precalc/trig class which walked students through finding trig ratios related to $\pi/5$. It was fun. I like that construction / argument.
A moderately interesting question which I have never really had time to think about in any detail: if you can only approximate trig ratios using dyadic multiples of $\pi$, $\pi/3$, and $\pi/5$, how quickly can you get approximations to converge?
@PM2Ring I've spent time thinking about trig rations related to the pentagram! HAIL SATIN!
@XanderHenderson Well, you can easily get 3°=30°+45°-72°. If you want finer granularity you need bisection. I suppose there will be optimal strategies, but I haven't spent much time looking for them, compared to the time I've spent messing around with Machin-like identities.
2 weeks back I did a factorisation for a problem, and had noted down the result of it (which at that time felt obvious to me) and today, when I re check the results and try to re factorise it, i wasnt able to do it, and the idea clicked a few 10 mins later
@nickbros123 Not really. When you looked at it today, your brain didn't have the same context that it had when you worked on the factorisation two weeks ago. Writing notes can be helpful to remind you of the context. Similarly, programmers write comments to provide context for tricky pieces of code.
Can anyone help me see why $\int \psi(\mathbf x)dF(\mathbf x)=E\psi(\mathbf X)$?
$\mathbf X$ is a random vector and $dF(\mathbf x)$ Steljies notation where $F$ is a CDF and $\psi(\mathbf x)$ is a measurable function from positive reals to $[0,1]$
here's what i know: $\int x f_X(x)~dx=E[f]$
and $\int \psi(\mathbf x) f_{\mathbf X}(\mathbf x)~d\mathbf x=E\psi(\mathbf X)$
in ted's day, only one person could hold the clay tablets at a time, so these issues of the OP deleting the question while it was being answered didn't come up.
@TedShifrin You won't like the real reason. I'm too lazy. I have "saved" about 50 geometry questions that I know for sure I can answer easily. Why haven't I? Too lazy.
a while ago olivia accidentally trapped herself in my dresser by jumping into an open drawer with such momentum that it closed by itself. i didn't find her for about 20 minutes.
@JohnRennie There's a fairly new Greg Egan short story, available free online. https://www.gregegan.net/BIBLIOGRAPHY/Online.html https://clarkesworldmagazine.com/egan_04_22 (although financial contributions to Clarkesworld are welcome). Dream Factory is set in the very near future. It involves YouTube, Bluetooth, and cats.
@Goku Unless they say exactly the same thing you can just post you answer. What about writing 3 paragraphs and see "this question has been deleted" pop up? :P
So this is a probability question which I keep confusing myself with. I start in one of two states 0, 1 with equal probability. If I start in state 1, I then change to state 0 with probability p and otherwise remain in state 1. What is the probability that, with certainty, I can deduce which state I started in?
But I’ll try again. I have two coins, one which only gives heads and the other which is unfair: heads with probability p and otherwise tails. I put one coin in each hand and have someone pick between them. I then flip that coin and reveal the result. What’s the probability they can deduce which coin they picked?
"deduce" is a weird word to me here. aren't they always just guessing (maybe according to some rule hopefully maximizing the probability of being right), and you're asking the probability they guessed right?
No, deduction. If they get tails, they definitely picked the unfair coin. But if they get heads, they may have picked either coin. So that would lead me to (1-p)/2. (I think I mixed up p vs 1-p in my original formulation.)
Locally this is always true. If you have an embedded $k$ dimensional submanifold of $\mathbb{R}^n$ then you can write the manifold locally as a graph over it's tangent plane, i.e.
$$M\cap U = \{(x_{k+1} , \dots, x_n) = (f_{k+1}(x_1, \dots,x_k), \dots, f_{n}(x_1, \dots,x_k)\}$$
where the coordinat...
I believe your professor does want $0$ to be a regular value because otherwise there's nothing special about a submanifold. Next, orientability is required to glue together local functions and make sure you still have $0$ as a regular value for the resulting sum.
I thought if I had a function $f:A\to R$ and $g:B\to R$ I could use the function $f(x)\phi_1(x)+g(x)+\phi_2(x)$ or something like that to get a function on the union of $A,B$ but that doesn't use the orientability so it must be wrong
According to Calculus: A complete course by Adams & Essex, a version of the mean value theorem in several variables is given by
If $f_1(a,b)$ and $f_2(a,b)$ [notation for partial derivatives] are continuous in a neighbourhood of the point $(a,b)$, and if the absolute values of $h$ and $k$ are...
