@TedShifrin Well, note that the indicator function of $A$ is discontinuous precisely at $\partial A$. Thus $$\int_R \chi_A$$ exists $\iff$ $\mu(\partial A)=0$.
@TedShifrin I proved what I told you the other day, that if $f$ is diff. with diff. inverse then every closed form on $U$ is exact $\iff$ every closed form on $f(U)$ is exact.
When Spivak says differentiable he means $C^\infty$.
He cleared that out at the start of the book, or something.
Can that fail if $f$ is merely differentiable, or say $C^1$?
We shall tacitly assume that forms and vector fields are differentiable, and "differentiable" will henceforth mean $C^\infty$; this is a simplifying assumption that eliminates the need for counting how many times a function is differentiated in a proof.
It's probably ok, but I'm not used to keeping careful counts of how many derivatives are used where. @skull ... peter's comment seems to be gone, but I added one.
Ok, @Peter, I say smooth where I don't to worry. I warn my students that diff does nor imply $C^1$.
Suppose $a_1=1, a_{k+1}=\sqrt{a_1+a_2+\cdots +a_k}, k \in \mathbb{N}$. Find the limits
$$i)\space \lim_{n\to\infty}\displaystyle \frac{\sum_{k=1}^{n} a_{k}}{n\sqrt{n}}$$
$$ii)\space \lim_{n\to\infty}\displaystyle \frac{\sum_{k=1}^{n} a_{k}}{n^2}$$
I'm puzzled with it. What to do?
I take it that under the "flags" tab, I see flags made by other users
There have been a couple of flags which I disagree with, and so I chose put it as an Invalid flag
I did this because the answers were flagged as "not an answer" and I think that they were indeed answers, however the quality of them could be questioned
Nevertheless, all my "invalid flags" has been disputed even though the answers are still here
So it seems to me that the "not an answer" flags was declined/disputed otherwise the answer would have been deleted (right?), but at the same time my "invalid flag" was also disputed
also I cant edit it now but here is a fixed version of above: $\newcommand{\CapPsi}{\mathop{\vphantom{\sum}\mathchoice{\vcenter{\Huge\Psi}}{\vcenter{\Large \Psi}}{\Psi}{\Psi}}}$ $$\Large\sum_{k_1+k_2\dots k_m=n}=\sum\CapPsi^{m-1}_{j=1}\;\;\;\sum^{n-\sum^j_{i=0}k_{m-i}}_{k_{m-j}=0}$$
Let $$\sum_{p=1}^mk_p=n$$ Then the multinomial theorem is as follows: $$\Large\sum_{k_1+k_2\dots k_m=n}=\sum\CapPsi^{m-1}_{j=1}\;\;\;\sum^{n-\sum^j_{i=0}k_{m-i}}_{k_{m-j}=0}\prod^m_{p=1}\binom{\sum^p_{q=1}k_q}{k_p}x_p^{k_p}=(x_1+x_2\dots x_m)^n$$
(Sorry to spam just want to test a comment: $\lvert\frac{\partial{\boldsymbol{r}}}{\partial{\theta}}\times\frac{\partial{\boldsymbol{r}}}{\partial{\psi}}\rvert d\theta\ d\psi$ )
Is that rendering properly for you guys, my client doesn't like the the second partial's numerator?
@PeterTamaroff I think you went into way too much detail in the construction of $n$, and went into no detail in the most important part--the verification that the rationals really intersect every interval.
you can associate a topological space to such aperiodic tilings, and the invariants of this space tells you something about the tiling you started with.
My previous statement would follow if the residues of all primes $>p_n$ modulo $p_n\#$ are just the union of $\pm 1$ with $\pm$ the primes between $p_n+1$ and $\frac{p_n\#}2-1$ for $n>2$. Is this true?
Suppose $a_1=1, a_{k+1}=\sqrt{a_1+a_2+\cdots +a_k}, k \in \mathbb{N}$. Find the limits
$$i)\space \lim_{n\to\infty}\displaystyle \frac{\sum_{k=1}^{n} a_{k}}{n\sqrt{n}}$$
$$ii)\space \lim_{n\to\infty}\displaystyle \frac{\sum_{k=1}^{n} a_{k}}{n^2}$$
I'm puzzled with it. What to do?
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