What $1-$ and $n+$ seem to signify is that we have an interval $[a,b]$ that contains $[1,n]$ such that $a\in (0,1)$ with the assumption that $1/x$ is $0$ at x=0. :(
That's why I wanted to write the version free from 1- and n+.
@Koro Since $\mathrm{d}\lfloor x\rfloor$ has a mass of $1$ at each integer, you need to approach from one side or the other to assure that the point is or is not included in the integral.
After an integration by parts, you can drop the direction of approach.
@Koro when the mean is $0$, the integral can be written as a continuous function and that is what allows one to drop the approach direction.
@robjohn professor Rob, how? Can you please elaborate that a little more?
I'm not familiar with it. I think I have never seen mean of a function being used anywhere before except for the exercises that say: "find the mean of this function over [a,b]$, a<b."
Since $\{x\}$ is not continuous, if we simply integrate $\{x\}$, we get a pretty ugly piecewise quadratic function.
However, if we integrate $\{x\}-\frac12$, then we get $\frac12\{x\}^2-\frac12\{x\}$, which is good for all $x$. Note that it is the same at $0$ and $1$
but after integration by parts you have $\{x\}$ on the inside of the integral. Then you integrate it and get an ugly antiderivative, unless you integrate $\{x\}-\frac12$, which gives $\frac{\{x\}^2-\{x\}}2$ (which is good everywhere, not just $[0,1]$).
Then you want to integrate that, and to avoid an ugly antiderivative, you need to add $\frac1{12}$
Yes, I know RS-integration and I also know that $\int_a^b f(t) d(u(x))= \sum f(x_k)\times $ jump at $x_k$, where $a=x_1<x_2<...<x_n=b$ is a partition and $u$ is a unit function. having jump $1$ at $x_i$'s.
Since $\{x\}=x$ on $[0,1)$, we know that $\int_0^x\left(\{t\}-\frac12\right)\mathrm{d}t=\int_0^x\left(t-\frac12\right)\mathrm{d}t=\frac{x^2-x}2=\frac{\{x\}^2-\{x\}}2$ for $0\le x\lt1$. However, the integral of $\{x\}-\frac12$ over $[0,1]$ is $0$, we get that the antiderivative is also periodic: $\frac{\{x\}^2-\{x\}}2$.
I meant that the difficulty arises (as you said, that it gets complicated) if I simplify further the expression that I got without using any periodicity.
I'd go about it the other way, $\int_0^x\{t\}\,\mathrm{d}t=\int_0^x\left(\left(\{t\}-\frac12\right)+\frac12\right)\,\mathrm{d}t=\frac12\{x\}^2-\frac12\{x\}+\frac12x$
Sometimes my equations in Latex become so lengthy in a straight line and don't load fully so I use \\ to add new lines or try to write the equations the other way (so that the equations fit in one one line or so).
Hi! I'm trying to prove the convergence of the series $\displaystyle \sum_{n=n_{0}}^{+\infty} \frac{1}{(\log \log n)^{\log n}}$ for all $n_{0}\geq 2$. I know as we can solve the problem using the comparastion test with $\sum 1/n^{2}$, for that reason the series converge. But can we use the Cauchy condensation's test?
I know the Cauchy condensation's test says that $$\sum a_{n}<+\infty \iff \sum 2^{n}a_{2^{n}}<+\infty$$ only if the sequence $(a_n)$ is a non-increasing sequence.
For that reason we can apply the test for the series $\sum \frac{1}{(n\log n)^{2}}$ since the the sequence $a_{n}$ is non increasing for $n$ sufficiently large.
But what's about the sequence $n\mapsto \frac{1}{(\log \log n)^{\log n}}$ can we apply the Cauchy condensation's test?
Hello! I am trying to show that $\int_{\mathbb{R}^d} f(|\nabla f|^2+ \nabla f \cdot \nabla g) dx\geq 0 $ where $f,g: \mathbb{R}^d \to \mathbb{R}$, $f$ is a probability density, and $g$ I am free to choose some assumptions ( obviously choosing $g$ constant gives the result immediately) .
Also can someone help me get mathjax to render in this chat forum ? is that possible?
your asking if a sequence is convergent then is every subsequence also convergent ?
I am trying to show that $\int_{\mathbb{R}^d} f(|\nabla f|^2+ \nabla f \cdot \nabla g) dx\geq 0 $ where $f,g: \mathbb{R}^d \to \mathbb{R}$, $f$ is a probability density, and $g$ I am free to choose some assumptions ( obviously choosing $g$ constant gives the result immediately) .
Question: is there an analogue concept of a pseudovector (axial vector) but for matrices?
The motivation is something like this: A vector field F(r) transforms under a "rotation" g via gF(r) = (gF)(g^{-1}r); a pseudovector does the same, but incorporates an additional factor det(g). A "matrix field" M(r) transforms similarly like gM(r)g^{-1} = (RM(g^{-1}r)R^{-1}), with R denoting a matrix-representation of g. Now, if I have a matrix field that has an additional factor of det(g) that then seems very analogous to the pseudovector concept.
I'm watching a movie about S. Ramanujan. I'm at an emotional scene where Ramanujan decides to go back to India and tells prof. Hardy that he's received the tickets to go back to India. Hardy says that the vehicle he arrived in had a boring number on it 1729. To that, Ramanujan replies that it's the smallest positive integer that can be written as sum of two cubes of positive integers in two different ways.
Does the triangle inequality hold for negative values? E.g if $a_n, b_n > 0$ and $\lim a_n = L$ can I conclude that $|a_n-L+(-b_n) \le |a_n-L| + |b_n|$?
The opening scene of the movie was so funny. Teacher: If we divide a number by itself, we get 1 just as when we divide 20 apples to 20 people each gets one. Kid Ramanujan says: if we divide 0 by 0, we should get 1 just like when we give 0 apples to 0 people each should get one apple. haha Teacher then says: 0 is nothing. Kid Ramanujan: How can that be nothing? If you put 0 to the side of 1, you get 10 and append one more 0, you get 100 etc. The teacher gets shocked. :)
@robjohn I was thinking about how to prove that it was decreasing using the definition. Is decreasing for an $n$ sufficiently large, correct? For some reason I thought it wasn't decreasing and so the test didn't work.
I have to say I'm not a fan of sensationalized biographies in film format. I think it's fair to attempt to recreate the character and play back a story but being overly dramatic is retarded.
ted were you ever the chair of your department? my wife now is, and is doing meetings all day. i understand the purpose of about 20% of them. the rest is just like, i don't remember me or anybody i know needing to bother people about this.
No, I was associate department head for 8 years. Close enough. There are meetings with the dean and other things that come up, but I suspect that these sorts of things may be worse the lower the level of the institution.
that makes sense. a lot of it seems to be one-on-one, or near one-on-one, meetings with people with grievances or wanting reasons for something that is not really anything that a chair can fix.
her department is 'interdisciplinary' which increases the likelihood of unpredictable and unproductive crosstalk.
Oh, you meant meetings with people in the department? I assumed you mean external such. ... I handled all sorts of student complaints and dealt with all course scheduling issues, but ultimately the department head was the recourse of those who were dissatisfied.
The least fun part of my job was course transfer credit issues. I had some pretty violent parents with that ... oy ...
it's mostly unhappy faculty, but there's also stuff with deans and provosts. i haven't seen student focused stuff. i don't know what a provost does. the one she's working with isn't raising money. that has come up. i hope that isn't part of their job.
my view from 30,000 feet is that tiny liberal arts schools hire people with less professional experience than i have for administrative positions to do [?????? huge blank here] and then they underperform and are thrown out.
this may be reflective of the level of institution my wife is working with
i think the smaller the school, the more something can evolve in such a way that it is incomparable to other things, and you have to be inside it to understand why it is the way it is.
i do think munchkin could run a better meeting. her default criticism when she's done with something is "I DON'T WANT THIS" and slapping the screen full of the things she doesn't want.
ODE: i would post whatever you want to post. for context, this is nominally a chat about math, and the comments immediately preceding your question are about my daughter hitting a laptop.
Assuming $S_n = \displaystyle \sum_{k=1}^{n} a_k$. If $\lim \{S_{3n}\} = L$ and $\lim \{S_{3n} - S_{3n-1}\} = 0$ and $\lim \{S_{3n} - S_{3n-2}\} = 0$ I want to prove that $\lim \{S_n\} = L$ Can I proceed as follows? We note that $S_{3n} - S_{3n-2} = a_{3n-1} + a_{3n} $ and $S_{3n} - S_{3n-1} = a_{3n} $. Given $\epsilon > 0$ and an $m\in \Bbb N$ we have that $|S_{3n}-L| < \epsilon/3$, $|a_{3n}|<\epsilon/3$ and $|a_{3n-1}|<\epsilon/3$. Hence $|S_{3n-1}-L| =|S_{3n}-L- a_{3n}| \le |S_{3n}-L|+ |a_{3n}| < 2\epsilon/3 $ and $|S_{3n-2}-L|=|S_{3n}-L- a_{3n}-a_{3n-1}| \le |S_{3n-2}-L|+ |a_{3n}| + |a_…
koro: thanks for asking. i think my daughter has the newest covid variant, she has a very ugly-sounding cough and her school is closed because a classmate and some staff had it. but she is not in any distress. and we haven't bothered to find tests to make sure.
ODE: Let me state a version of what you're trying to prove: If A and B are infinite subsets of $\mathbb N$ such that $A\cup B=\mathbb N$ and $A\cap B=\emptyset$ and it is given that the subsequences $(x_{a})$ and $(x_{b})$ both converge to the same limit L, then the sequence $(x_n)$ converges to L. Note: $(x_a)$ and $(x_b)$ are often called complementary subsequences of the sequence $(x_n)$.
Professor Ted: sorry for the interruption. I stated one result that ODE can take a look at, if they want after your discussion. I didn't mean to interfere in the discussion.
My skin is a mess throughout, @leslie, nothing to do with covid, I believe. It's a combination of lichen planus, from which I suffer, and bad reactions to cat bites and scratches.
@Odes: Here's what you're trying to do (with easier notation). If you know $\lim a_n=\ell$ and $\lim(a_n+b_n) = m$, then how do know $\lim b_n$ exists and how do you find it?
No, that's my whole point. It is perhaps a trick the first time you see it, but this is an important technique. How do you get $b_n$ from $a_n$ and $a_n+b_n$?
By the way, @Koro, your general statement doesn't apply here. You still have to see the other subsequences have the same limit. That's the whole point I'm working on here.
