I'm gonna take this opportunity to remind everyone that the words overmorrow and ereyesterday exist and that it's a tragedy that they're no longer in use
They were used before weekdays were invented. To keep track of days people used to say "I will be there in overoverover...overmorrow." Then when they meet the person that day, they will "See here I am, as I promised ereereere...ereyesterday." Thank god for the names of the days.
I just calculated the pdf of the sum of 4 independent, Unif[0,1]-distributed random variables by hand using convolution, because a friend dared me to. It was a bad decision
Does a morphism of schemes induce a morphism on sheaf/cech cohomology? Something like $f:X\to Y$ induces $H^i(X,\mathcal{F})\to H^i(Y,f_*\mathcal{F})$, or maybe for $\mathcal{F}$ on $Y$ we have $H^i(Y,\mathcal{F})\to H^i(X,f^*\mathcal{F})$
and the Mathematica program I wrote does not give as beautiful a plot as the New Kind Science pyramid looking plot. It was easier to format it as a triangle instead.
I just calculated the pdf of the sum of 4 independent, Unif[0,1]-distributed random variables by hand using convolution, because a friend dared me to. It was a bad decision
I think this is coming from a problem in Apostol that I have assigned numerous times over the years (and even put into Spivak's book when I wrote extra problems for it).
You can give a pretty direct argument using two-dimensional normal distributions, I think, but convolution was the only tool available when we were given that task
@Thorgott It's not true. Take something like $\sum \frac1{n^2}\chi_{(1/(n+1),1/n]}$ (and $0$ at $0$, of course), so it'll be Riemann integrable, continuous at $0$ but on no neighborhood of $0$.
We defined continuity on arbitrary subsets of $\mathbb{R}$ in the first analysis lecture I listened to, but what you're saying is probably what was intended.
What I was thinking of is: Let $D$ be the set of points where $A$ is differentiable. Let $f$ be differentiable at $c$. Then $A$ is differentiable at $c$ with $A'(c)=f(c)$. By the Lebesgue Differentiation Theorem, $D$ is of full measure, so, in particular, $c$ is a limit point of $D$ and then the continuity of $A'$ at $c\in D$ is a well-posed and non-trivial question.
Yeah, you're at the advanced undergraduate or — in the US — graduate level, first of all. I still say that unless you're in an advanced context, this is not the right interpretation of continuity.
I would never say the function $f(x) = \begin{cases} 1, & x=1/n, \\ 0, &\text{otherwise}\end{cases}$ is continuous at $0$. Nor would I if the function forgets to be defined at $1/n$.
Unless it's clear I'm working in the measure theory world where functions are only unique a.e.
But this is why it's so important to get posters to make their context and efforts clear. Often people post answers that are just in a different universe from the questioner.
My issue was that simply asking whether $A'$ is continuous requires $A'$ to be defined somewhere in the first place. So choosing the domain as the set of points where $A'$ exists seemed to me to be the natural choice.
The actual question ends up being "Is A necessarily differentiable in a neighborhood of $c$?", but this seems somewhat backwards cause the original question is not exactly well-posed
But yeah, I think you are definitely right with regards to the intent.
To me, that's the whole point of the question. The context in the books is that there's a whole sequence of questions, which makes the global intent clearer.
I often post on questions: Show your effort and where you got stuck, because we can't help if we don't know what you know and what you are allowed to use.
This is particularly true for me with differential geometry questions, but certainly true with most subjects.
I just complained to someone who posted a differential topology question with no work. I had to ask him what his definition of degree of a map is.
Hmm, so because $f$ is Riemann integrable, it has to be continuous except on a set of measure 0. So if you throw out that set of measure 0, won't it follow, then, just from the usual FTC?
I'm concerned about the points at which $A$ is differentiable, but $f$ is not continuous.
If, say, the set of continuity points of $f$ is $C$, then $C\subseteq D$, $A'=f$ on $C$ and hence $A'$ is continuous on $C$, but I'm not too sure about $D\setminus C$
So those are the removable singularities of $f$. It's the jump singularities that cause issues with $A$.,
Let's see, if $f$ has a singularity of the remaining sort, it can be very subtle. For example, for the function $f(x) = \begin{cases} \cos 1/x, &x\ne 0 \\ 0, & x=0\end{cases}$, I believe $A'(0)$ exists.
But then $A'$ certainly won't be continuous at $0$.
