@Astyx Well this is just lovely I must say. Quadratic instead of exponential decay since the partial sum is $$(i,j)\mapsto \sum_{k=1}^{i+j-1}k + j = \frac{1}{2}(i+j-1)(i+j) + (i+j-1)j$$ if I did everything correctly, right?
Define "infinite binary tails" to be the equivalence classes of infinite binary sequences where two sequences are equivalent if they disagree at finitely many points
Then without the axiom of choice, you cannot prove that the set of infinite binary tails can be given a total order
Hi. I am solving an exercise that's telling me to estimate the value of $\sqrt{4.1}\cdot \sin\left ( 0.8 \right )$ using the partial derivatives of $f\left ( x,y \right )=\sqrt{x}\sin y$, but I'm unsure what value to use as an approximation for $y$, or rather what value the exercise expects me to use. Choosing $y=\frac{\pi }{4}$ gives a very good estimate, but the result (assuming my solution is correct) contains both $\pi$ and $\sqrt{2}$, both of which are irrational. Choosing $y=1$ instead yields a rational number, but the error is substantial (almost 40% of the true value).
@XanderHenderson I remember seeing the Weaire-Phelan structure years ago, but that question has inspired me to work on an interactive 3D diagram. So far, I've only done the pyritohedron.
You can parametrize pyritohedra with a parameter $0<h<1$. h=0 gives a cube, h=1 gives a rhombic dodecahedron. h=1/2 gives the pyritohedron used in the Weaire-Phelan structure, and h=1/phi gives a regular dodecahedron. Here's an anim from Wikipedia which illustrates the family.
Here's my Sage / Python script which lets you view any member of the family in 3D.
The following is a sketch of the proof and notations are abused frequently.
Let $p\in\partial M$ and choose a smooth boundary chart $(U,\varphi)$ with local coordinate $(x^i)$. Since $f$ is a smooth function, $f:U\to [0,\infty)$ is a smooth function. For $v\in T_pM$, write $v = v^i{\partial\over\...
I answered my own question. Can someone please check if this argument is correct?
for m; i selected $4$ girls and let them be a group, they can be arranged in $4!$ ways; so total ways= $5_{C_4}\times4!\times7!$ but this would include the cases in which 5 girls are arranged together as well so $m=5_{C_4}\times4!\times7!-5!\times6!$ but in the soln they have subtracted $2\times5!\times6!$ why an extra factor of 2? I understand other solutions but wanted to know my mistake in mine
@TedShifrin If say $Y$ is $G_1G_2G_3G_4$ and $X$ is $G_5$ then $XY$ will be $G_5G_1G_2G_3G_4$ and $YX$ will be $G_1G_2G_3G_4G_5$ won't both these cases be already included in $6!\times 5!$?
ok, so in the first one the mapping function g is C^1 on the closure of a 'figure', and in the second the mapping function (now called f) is only continuous? is that right? i guess 14.6 is more general.
oh, there might also be a gap between measure zero and 'content zero' but maybe not. i think under the usual definition, not.
i think it's the same theorem, up to whether it's clear that 'measure zero' and 'content zero' are the same thing. they probably are, but they might not be defined that way.
Both are weaker than needed. You don’t need the one-to-one and invertible derivative everywhere. The second one seems weaker as it requires those conditions on a larger set than that on which you will integrate.
I thought I had a (hand-wavy) argument for 1=>2 (assuming additionally that in 2 it is assumed that the open containing the figure has boundary with content zero)
In my book it's said that the set of all probability measures on a compact Hausdorff space $X$ is identified, using Riesz representation theorem, with a $w^*$-compact set $B_{C(X)^*}$.
Here's a famous example of the counter-intuitive behaviour that you want.
The Dieudonné measure $\mu$ is a Borel probability measure on the compact space $X=[0,\omega_1]$,
where $\omega_1$ is the first uncountable ordinal.
It has the property that $\mu(K)=0$ for every compact subset of $X\setm...
Apostol in his Advanced Calculus book has a very general proof of Green's Theorem. I don't see the point of it.
I have never needed to apply it more generally than what I proved in my book (any region that can be covered by a finite union of parametrizations by rectangles). Of course, we could prove it with triangles near the boundary, with the edges of the triangles converging to the boundary curve, but ... meh.
I'm actually AGAINST proving things in their greatest possible generality. Pedagogy is more important than that unless you're a pedantic graduate student.
Here's a famous example of the counter-intuitive behaviour that you want.
The Dieudonné measure $\mu$ is a Borel probability measure on the compact space $X=[0,\omega_1]$,
where $\omega_1$ is the first uncountable ordinal.
It has the property that $\mu(K)=0$ for every compact subset of $X\setm...
