@robjohn to avoid further complications, we can simply consider $$\int_0^{1} \left\{\dfrac1{x+1}\right\} \left\{\dfrac1{x}\right\} \left\{\dfrac1{1-x}\right\}\ dx$$
@robjohn it's a very nice series you shouldn't miss. :-)
@AndrewG There are 2 sensible interpretations. First, by choosing your base field to be C(z) and define a function field of logarithm over C(z), i.e., take the closure by including transcendentals log(z), log(z+1), log(z)+z, etc. This is called C(z, log). [For C, however, you cannot extend (C is it's own logarithmic closure) but "de-extend" by taking the subfields which are logarithmically closed, i.e., log(k) \in F whenever k \in F and take the intersection of those fields]
Second, by equipping your base field with a differential operator D(u+v) = Du + Dv and D(uv) = uDv + vDu, u and v \in F. Take F to be C(z) and D the usual differentiation. You can have a simple transcendental extension K over C(z) generated by some x such that Dx = Dz/z. This is the "so-called" differential logarithmic extension.
Let $H^0 = (H^0)^*$ and $H^1 = (H^1)^*$ be linear, self-adjoint operators and let $E^0$ be an isolated eigenvalue of $H^0$. Let $E_k^1$ be the eigenvalues of $H^1$. How can I immediately see, that $$H^0+ \varepsilon H^1$$ has exactly the eigenvalues $E_k = E^0 + \varepsilon E_k^1$ for $\varepsilon > 0$?
@BalarkaSen no. I think you can turn it into a simple continued fraction by moving the denominators down (or perhaps more care is needed), but with rational numbers (and the rational implies periodic fact I only know for integer entries)
the question is whether a rational can have a nonperiodic nonsimple continued fraction, not whether an irrational can have a periodic nonsimple continued fraction
actually among simple continued fractions, rationals have finite, not periodic, expansions
If we look at $\hat {\Bbb C}$ as the set of all (mostly) holomorphic functions. We could look at it as the fullest extension. So we need to strip it down.
Let $H^0 = (H^0)^*$ and $H^1 = (H^1)^*$ be linear, self-adjoint operators and let $E^0$ be an isolated eigenvalue of $H^0$. Let $E_k^1$ be the eigenvalues of $H^1$. How can I immediately see, that $$H^0+ \varepsilon H^1$$ has exactly the eigenvalues $E_k = E^0 + \varepsilon E_k^1$ for $\varepsilon > 0$? I also know that the eigenspace corresponding to $E^0$ is finitely dimensional, but I'm not sure if that is necessary. @DanielFischer: Can you maybe help me, please?
@Chris'ssis I came across a nice inequality: for distinct positive integers $\{a_i\}_{i=1}^n$, such that the $2^n$ sums $\displaystyle \sum\limits_{i} \epsilon_i a_i$, ($\epsilon_i = 0 \textrm{ or } 1$) are distinct .. show that $\displaystyle\sum\limits_{i=1}^n \dfrac{1}{a_i} < 2$ :)
@Huy That's specifically for the trivial example $H^0 = E^0\cdot \mathbf{1}$. It's meant to illustrate that generally the single eigenvalue $E^0$ splits under perturbances. In the special case where $H^0$ is a multiple of the identity, you have the eigenvalues of $H^1$, multiplied by $\varepsilon$, and shifted by $E_0$. Generally, it's not as easy, and the following treats the general case.
@DanielFischer: I see. I am not quite sure how to justify the trivial example though. Thinking of finite dimensions, I like to diagonalise matrices and directly see the eigenvalues. The two matrices $H^0$ and $H^1$ are simultaneously diagonalisable and thus we can add the eigenvalues like stated. How can I justify this for operators on infinitely dimensional spaces?
@Huy For a multiple of the identity, nothing changes when looking at the infinite-dimensional case. Generally, even in the finite-dimensional case, $H^0$ and $H^1$ are only simultaneously diagonalisable if they commute. If they commute, you're back in the trivial case, since the eigenspaces of $H^0$ are $H^1$-invariant, and you look at the restriction to an eigenspace of $H^0$. If they don't commute, and that is the general situation, it is simply far more complicated.
Then the eigenspace of $H^0$ is not $H^0 + \varepsilon H^1$-invariant (for $\varepsilon\neq 0$). But, the eigenvectors of $H^0$ are, for small $\lvert\varepsilon\rvert$, almost eigenvectors of $H^0+\varepsilon H^1$.
So one "expects" that the true eigenvectors are close to those of $H^0$, and the eigenvalues are close to $E^0$.
One more question: Right above equation (6.3), there is an expansion $$(H^0+\varepsilon H^1 - E^0 - \varepsilon E_k^1 - \varepsilon^2 E_k^2 - \dots) ( \psi_k^0 + \varepsilon \psi_k^1 + \varepsilon^2 \psi_k^2 + \dots) = 0.$$ This notion slightly confuses me: $H^0, H^1$ are the operators and $E^0, E_k^1$ the respective eigenvalues. Now what exactly is $E_k^2$? Is this supposed to mean $(E_k^1)^2$? Also, what about the $\psi$s? What is $\psi_k^0$, what is $\psi_k^1$ and $\psi_k^2$?
Are these the eigenvectors of $\varepsilon H^1$ to the power of $0,1,2, \dots$?
Hello. Off topic: What does a polydisc look like? For example, if $D(0,r_i)$ is an open disk in the plane centered at $0$ with radius $r_i$, then how do I visualize the Cartesian product $D(0,r_1)\times D(0,r_2)$?
@Huy The idea is that the eigenvalues depend on $\varepsilon$ (they evidently do), and that they have expansions $$E_k(\varepsilon) = E^0 + \varepsilon\cdot E^1_k + \varepsilon^2E^2_k + \varepsilon^3 E^3_k + \dotsc\,.$$ The superscripts on $E^m_k$ are just indices, but the first order coefficient $E^1_k$ is the same as the eigenvalue of $H^1$ (I don't know for sure whether that's necessarily true, but seems reasonable enough).
Then in $H^0 + \varepsilon H^1 - E^0 - \varepsilon E^1_k - \varepsilon^2 E^2_k - \dotsc$, the $E^m_k$ stand for $E^m_k\cdot \mathbb{1}$.
Analogously for the $\psi_k$, it is assumed that the eigenvectors $\psi_k(\varepsilon)$ to the eigenvalue $E_k(\varepsilon)$ have expansions as power series in $\varepsilon$, and the $\psi^m_k$ are the coefficients of that series.
@BalarkaSen Hmm yes diff galois theory. I fear though, if I try and being to learn it I will neglect other mathematics. So I will try when I am confident and ready.
Why does studying Mathematics have to hurt so much? I Love it very passionately but I can't remember the last day I studied it without feeling like a f**king idiot for at least 20 minutes :/
@Shaun Don't feel like an idiot. An idiot is somebody less smarter than everybody else. Everyone feels the same way in reality. Therefore you are not an idiot.
For a bit of context I spent most of the last week making progress with some computer programme to answer some question in Semigroup Theory, only to find today that I made a stupid mistake early on that ruined it all. I'm back to square one.
So yeah, that sucks . . .
I just wanted to express myself. I feel better now :)
Yes that one @DanielFischer You gave a definition of the dyadic squares $A = \prod (c_i,d_i)$ such that $o^m_{1,2} \cap A$ takes up exactly half of $A$ or nothing if $A$ doesn't meet $O$.
Don't know if you remember the post I am referring to.
I am referring to this post. To show the last part of the proof in the post where you show that $\int gf_{m} = \int \frac{1}{2}\int g$ I used the fact that for large $m$ $o^m_{1,2} \cap A$ takes up exactly half of $A$.
@DanielFischer I showed this idea, you said it was fine I think...kind of.
@DanielFischer Do you know how to show how the linear span of these characteristic functions on dyadic squares $A$ is norm dense in $O$?
@LucioD In $L^1(O)$, not in $O$. The best method depends on what you already know. If you already know that $C_c(O)$ is dense in $L^1(O)$, the simplest way is to use that fact.
Anybody know how to fix the comments in this question - I made a typo in a comment, and they are all displaying horribly ever since, with "edit" and "delete" buttons "underneath" the links on the right. math.stackexchange.com/questions/880637/…
Good enough, @LucioD. You can obtain the denseness of $C_c(O)$ in $L^1(O)$ from that, but we don't need that. You know that you have a canonical embedding $L^1(O) \hookrightarrow L^1(\mathbb{R}^n)$ by extending the function by setting it to $0$ outside $O$. So fix $f\in L^1(O)$, and $\varepsilon > 0$. Let $\tilde{f}$ be the trivial extension of $f$ to all of $\mathbb{R}^n$. Pick a $g\in C_c(\mathbb{R}^n)$ with $\lVert \tilde{f} - g\rVert_{L^1} < \varepsilon/2$.
Now, @Lucio, $g$ is uniformly continuous, so there is a $\delta > 0$ such that $\lvert x-y\rvert < \delta \implies \lvert g(x) - g(y)\rvert < \frac{\varepsilon}{2\cdot m(O)}$. Then choose your dyadic cubes so small that their diameter is smaller than $\delta$ (sidelength $< \delta/\sqrt{n}$), and set $ = \sum g(x_i) \chi_{A_i}$, where $\{A_i\}$ is a partition of $O$ into disjoint small dyadic cubes.
Then you have $$\sup_{x\in O} \lvert g(x) - h(x)\rvert \leqslant \frac{\varepsilon}{2\cdot m(O)},$$ and that implies $\lVert g-h\rVert_{L^1(O)} \leqslant \frac{\varepsilon}{2}$.
Will I be ready for an introductory course in manifolds with good grades in the following subjects?: Calc 1, 2, 3, Lin. Alg, Abs. Alg, Disc. Math, Topology
@DanielFischer Are you still taking the dyadic cubes to be defined as $A= \prod (c_{i}, d_{i})$ where $d_{i}-c_{i} = 2^{-k}\cdot R$ where $c_{i} = x_{i} + n \cdot 2^{-k }R$?
@LucioD Yes. (But that $x_i$ is not the same as in the definition of $h$)
The point is that a uniformly continuous function does not change much on a small cube, and therefore it can be well approximated by functions that are constant on small enough cubes.
@LucioD Then you can in general not exhaust all of $O$ with finitely many cubes. That makes things a bit more complicated. It's easier if you take $2^{-k}$ times the side length of the big cube.
@LucioD For general $L^1$ functions, that is only true "except on sets of arbitrarily small measure".
If we take $n = 1$, and a function with a jump discontinuity, that changes much on every small interval containing the jump in its interior. But you can make that interval as small as you like.
let $f$ is derivative on $[0,+\infty)$, if such $\lim_{x\to+\infty}
let $f$ is derivative on $[0,+\infty)$, if such $\lim_{x\to+\infty}\left([f'(x)]^2+f^3(x)\right)=0, show that $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$
i.e., if $\sigma$ is in Gal(K/F) then $\sigma(x + y) = \sigma(x) + \sigma(y)$, $\sigma(xy) = \sigma(x)\sigma(y)$, $\sigma(1) = 1$ and $\sigma(f') = (\sigma(f))'$
@DanielFischer The proof looks good so far, everything follows. One question, the dyadic cubes as was previously defined as $A = \prod (c_i,d_i)$, where $d_i-c_i = 2^{-k} \cdot R$ for some $k$ and all $i$ and $c_i = x_i + n\cdot 2^{-k}R$. Everything seems fixed there, what exactly changes to produce many cubes. $x_i$ is the center of the cube, $R$ is the length of $O$, $n$ is the dimension of $\mathbb{R}$ and $k$ is chosen to be fixed. Seems like that just defines how to produce one cube.
let $f$ is derivative on $[0,+\infty)$, if such $$\lim_{x\to+\infty}\left([f'(x)]^2+f^3(x)\right)=0$$, show that $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$$ can you someone see this problem?
@LucioD $k$ varies through the non-negative integers. $n$ varies through integers so that altogether the entire cube $O$ is covered by the $2^{n\cdot k}$ cubes with sidelength $2^{-k}\cdot R$. Look at the unit square in $\mathbb{R}^2$ to get the picture. You can cover it with one square with sidelength $1$, 4 squares with sidelength $\frac{1}{2}$, sixteen squares with sidelength $\frac{1}{4}$ etc.
@LucioD Oh, wait, I used $n$ as both, a parameter and the dimension of the space above, that's not good. Let's use $m$ for the parameter, better, $m_i$, since we need one parameter for each dimension. If $O = \prod_{k=1}^n [x_i, x_i+R)$, then we consider for each $k \in \mathbb{N}$ the cubes $$A(k,m) = \prod_{i=1}^n \left[x_i + m_i\cdot 2^{-k}R, x_i + (m_i+1)2^{-k}R \right)$$ where $0 \leqslant m_i < 2^k$ for $i \in \{1,2,\dotsc,n\}$ and $m = (m_1,m_2,\dotsc,m_n)$. Then the system
$\mathcal{A}(k) = \{ A(k,m) : 0 \leqslant m_i < 2^k\}$ is a partition of $O$ into dyadic cubes of sidelength $2^{-k}R$.
We look at the system $$\mathcal{A} = \bigcup_{k\in\mathbb{N}} \mathcal{A}(k).$$
Then $\{ \chi_A : A \in \mathcal{A}\}$ is dense in $L^1(O)$.
@LucioD No, $O$ is a subset of $\mathbb{R}^n$. $\mathcal{A}$ is a family of subsets of $\mathbb{R}^n$. But yes, for every $k$ we have disjoint cubes partitioning $O$.
@LucioD Each $\mathcal{A}(k)$ is a family of subsets of $\mathbb{R}^n$ (a partition of $O$). $\mathcal{A}$ is the union of the collection of these families, $$\mathcal{A} = \left\{ A \subset \mathbb{R}^n : \bigl(\exists k\bigr)\bigl( A \in \mathcal{A}(k)\bigr)\right\}.$$
