when i had an appendectomy i was offered (and refused) morphine.
my dr came by and said that he understood, but some might help me relax and heal quicker, so i relented and took a half dose.
for the next few hours i would drop off, have intensely vivid dreams, lurch awake sending my stomach into spasms. so that was the last time i took morphine.
part of me was hoping the percocet would bring back the dreams but no dice
During the last semester of my masters program, my ex-wife woke me up at about 2 in the morning, abut 8 hours before my very last final. Her gall bladder had... failed, and she was writhing around on the bathroom floor in tremendous pain. I took her to the hospital, they gave her morphine, and I got a C on that final, and a B+ in the class. It is the only non-A I got in the masters program.
@TedShifrin I don't know... in retrospect, it might be the kind of thing she would do...
(Honestly, I don't actually care all that much about the grades---after two years as an undergraduate, my GPA was 1.95; I managed to finish a bachelors with a 3.5 overall average; and I was kind of over "classes" by the time I started my phd---I still managed a good GPA overall, but didn't give no sh*ts. It is just ever so slightly annoying to me that almost managed a 4.0 over two and a half years, only to not quite cross the line).
Is there some efficient way to count number of partitions of $k$ into $n$ parts where each part $\le n$ where $k >n$? More specifically I'm interested in $k=193$ and $n=60$.
none of that seems exactly on point but its all pretty close
there are some recursion relations satisfied by that as a function of k and n (for recognizing the recursion it maybe even helps to let the bound on each part's size vary separately, independently of n)
if $G_1,G_2$ are open (bounded) simply connected disjoint neighbourhoods of $z_1,z_2 \in \mathbb{C}$, then is it necessarily the case that the sets $A(G_i - z_i) + \frac{1}{A} z_i ,i=1,2$ are also disjoint? Here $A$ is some non-zero complex number.
Let $I=(-a,a)$ be an interval (finite or infinite). Suppose that $(K_n)_{n=1}^\infty$ is a sequence of real-valued, Riemann-integrable functions defined on $I$, with the following properties: 1) $K_n(s)\geq 0$, 2)$\int_{-a}^a K_n(s)ds=1$, and 3) if $\delta>0$, then $\lim\limits_{n\to\infty}\int_{\delta<|s|<a} K_n(s)ds=0.$
$(K_n)_{n=1}^\infty$ is called a positive summability kernel. I wonder the following; are these kind of kernels even functions? It looks to me as if $K_n(s)$ is a kernel, then $K_n(-s)$ is a kernel too, but I'm not sure if this defines it as an even function.
ok. I meant probably a whole sequence, i.e. if $K_n(s)$ is a kernel for all $n$, then so is $K_n(-s)$ for all $n$, right? From the definition, I agree, they don't have to be even.
@Jakobian ok, the reason for the question is that, in Stein and Shakarchi, they claim that $$\lim_{n\to\infty}\int_{-\pi}^\pi K_n(s)f(s_0-s)ds=f(s_0),$$ for $f$ being continuous at $s_0$. In my textbook, they claim that $$\lim_{n\to\infty}\int_{-a}^a K_n(s)f(s)ds=f(0),$$ if $f$ is continuous at $0$.
I'm not sure how I obtain the former limit from the latter, or vice versa? Making the change of variables $u=-s$ in the former integral, I get $$\int_{-\pi}^\pi K_n(-u)f(s_0+u)du.$$ Setting $s_0=0$, we should obtain the latter limit of the integral provided the entry $K_n(-u)$ is a kernel.
I sent an e-mail to Erwann Aubry (see Georges Elencwajg's answer) and asked for a scan of this proof from the book "Calcul Différentiel". He replied and send me the proof translated into English. Here is his original paper. And this is how the proof goes:
Theorem. Every open star-shaped set $...
Non-empty open star-shaped subset is diffeomorphic to $\mathbb{R}^n$
Take a disk centered at the origin and remove the (relatively closed) outer portion of one ray. So the distance goes from $1$ for all other radial directions to, say, $1/2$ for that particular one.
Okay, I am thinking of something different; the distance function to each $e\in E$ is a continuous function and the distance to $E$ is often considered to be $\inf\limits_{e\in E} d(x,e)$
and the infimum of continuous functions is continuous. But that is not what you are thinking about.
perhaps that is not correct...
Okay, the infimum of infinitely many continuous functions is not necessarily continuous.
Sorry, I'm in the middle of cleaning today. Not thinking clearly.
@TedShifrin This is a problem which is near and dear to my heart. What do the neighborhoods of the boundary look like? How messed up can they be? How does that change diffusion on the set near the boundary? etc.
I'm trying to plot the function $f(t)=t^4-2\pi t^2, \ |t|<\pi, f(t+2\pi)=f(t)$ in python, to verify to myself it is continuous at $t=\pi$. The reason; I'm trying to compute $\zeta(4)$ and for this you require continuity at $t=\pi$, for the Fourier series to equal the function. I get a pretty continuous graph, but I'm wondering, why did they specify it like this...$|t|<\pi$? Why not $|t|\leq\pi$?
I can't think of any reason to assume off the bat that either $\sum \log(n)/n^2$ or $\sum \log(n)/n!$ would have a nice analytic closed form. I'd be pretty surprised if either did.
You'd be surprised at the craziness/f****-upedness in many families. Dealing with my mom's Alzheimers and the stress that put me under was no walk on the beach. That was a 15-year ordeal or so.
@TedShifrin I am not sure the limit has a nice closed form. The log is computable as $\sum\limits_{n=2}^\infty\frac{\log(n)}{n!}$. This came from extending a duplicate question that I saw yesterday. Following up on Approach0, the best upper bound is $2$.
I'm not too sure why, but I can't focus at all when reading a book, I continue to re-read the same page over and over, only beginning to understand something after a while
@TedShifrin It was really disappointing when I figured that out.
First, you tell me that I can't divide by zero, which is fine. Then you tell me that $i$ is a thing... okay. But then you tell me that you can't actually integrate or sum anything in a nice way? *flips desk*
I sent an e-mail to Erwann Aubry (see Georges Elencwajg's answer) and asked for a scan of this proof from the book "Calcul Différentiel". He replied and send me the proof translated into English. Here is his original paper. And this is how the proof goes:
Theorem. Every open star-shaped set $...
