TITLE: If calculus can be applied to non-numbers.
QUESTION [3 upvotes]: Trying to understand differential structure.
So (from Wikipedia) two spaces with a homeomorphism are considered the same. A manifold is a topological space that locally resembles the Euclidean space. A Euclidean space is a space over the real numbers. That is the first question (1) below. A differentiable manifold is a manifold locally similar enough to a linear space to allow one to do calculus. That leads to my next question (2) below. Finally, a differentiable structure on a set makes it into a differentiable manifold, which is a topological manifold with some additional structure to allow doing differential calculus on the manifold.

Can a manifold be a topological space outside of the numbers / matrices / complex numbers that has some sort of mapping/relation ("homeomorphism") to the Euclidean space, and in that way be considered "locally resembling the Euclidean space". And in this way calculus can be applied to a topology (non-numbers).
Or is calculus only done over numbers / matrices / complex numbers, or can calculus be done on arbitrary spaces such as groups, rings, fields, etc. (all generated by non-numbers).

Basically I'm wondering if you can construct a manifold by constructing a topological space without numbers, and what it would look like. If not, then if you can construct a topological space $X$ without numbers, then construct a space $Y$ with numbers, construct a mapping $f : X \to Y$, and then say the topological space is a manifold. Wondering if an example could be outlined showing what this would look like. Wondering if calculus can then essentially be applied on non-numbers in this fashion.

REPLY [1 votes]: As @Yucoub Kureh said, formal derivatives can be defined for the polynomial ring $R[X]$ or over the ring of formal power series $R[[X]]$. But obviously $R$ itself may have no topological structure and so I am not really so sure if you can call this calculus. For example, one defines formal derivatives over polynomials in $\mathbb{Q}[X]$ in Galois theory but the reason we do this is to show that given any irreducible polynomial, it is a separable polynomial as $f$ and its formal derivative $f'$ are coprime polynomials. The notion of "continuity" or "topology" is not really relevant one defines the formal derivative in this context.
I guess the inner product over the space of continuous functions is an example where we have a topological structure.  (For example, $C[0,1]=\{f:[0,1]\to\mathbb{R}\text{ such that }f\text{ is continuous}\}$ and we have $<f,g>=\int_{0}^{1}fgdx$. This inner product defines a metric space structure an subsequently you can have a topological structure.) However, again not sure if you can call this calculus because I don't have an equivalent thing to "differentiation".
A manifold is a topological space $X$ with a collection of charts which is a homeomorphism $\phi_{\alpha}$ from an open set in $X$ to an open subset of $\mathbb{R}^n$ or $\mathbb{C}$.
However, given two charts $\phi_{\alpha}:U_{\alpha}\to\mathbb{R}^n$ and $\phi_{\beta}:U_{\beta}\to\mathbb{R}^n$, they need not agree on $U_\alpha\cap U_\beta$. However, $\phi_\beta\circ\phi_\alpha^{-1}:\phi_\alpha(U_\alpha\cap U_\beta)\to\phi_\beta(U_\alpha\cap U_\beta)$ needs to be a suitably well-behaved function. (In the context of Riemann surfaces, that means it needs to be analytic as we are dealing with complex analysis, in other contexts this may mean that it is a continuous or a smooth map from $D\subset\mathbb{R}^n\to\mathbb{R}^n$).


TITLE: Splitting fields over $\mathbb{Q}$
QUESTION [0 upvotes]: Find a splitting fields over $\mathbb{Q}$ for:
i)$x^4+4=(x^2-2x+2)(x^2+2x+2)$ (both factors are irreducible). The roots: $x_1=1+i,\ x_2=-(1+i)$. So the splitting field is $\mathbb{Q}(i)$, which has deg 2 over Q.
ii)$x^6-8=(x^2-2)(x^4+2x^2+2)$. The roots: $x_1=1, \ x_2=-1$. 4 left so it will be $e^{\frac{2 \pi i}{6}}$. So the splitting field is $\mathbb{Q}(e^{\frac{2\pi i}{6}})$ And deg is also 2?
iii)$x^4+5x^2+6=(x-i)(x+i)(x-i\sqrt5)(x+i\sqrt5)$. So the splitting field will be $\mathbb{Q}(\sqrt5,i)$ and deg 4?
Is that correct?

REPLY [2 votes]: For i) The roots are the powers of $\sqrt{2}e^{\frac{i \pi}{4}}$ and a splitting field is $\mathbb Q(\sqrt{2}e^{\frac{i \pi}{4}})$.
For ii) The roots are the powers of $\sqrt{2}e^{\frac{i \pi}{3}}$ and a splitting field is $\mathbb Q(\sqrt{2}e^{\frac{i \pi}{3}})$.
For iii)
$$x^4+5x^2+6=(x^2+2)(x^2+3)=(x-i\sqrt{2})(x+i\sqrt{2})(x-i\sqrt{3})(x+i\sqrt{3})$$ and a splitting field is $\mathbb Q(i\sqrt{2},i\sqrt{3})$


TITLE: Triangular Numbers Modulo $k$ - Hit All Values?
QUESTION [2 upvotes]: The sequence of triangular numbers is well known. Consider now the triangular numbers modulo some positive natural number $k$. That is, let $T_{n,k}$ be the remainder of dividing $T_n$ by $k$. For the regular sequence of triangular numbers, we can take as convention $T_n = T_{n, \infty}$.
Question: For all values of $k, i < k$, does there exists some $n$ such that $T_{n, k} = i$

REPLY [1 votes]: With Lulu's argument, and permission (thanks Lulu) I will answer this. The answer to the question is no. The counterexample is $k=3, i=2$.
We want to prove that for all $n$, $T_{n, 3} \neq 2$, noting that $T_{n+1} = T_n + n + 1$. Rather than prove this, we prove a theorem that implies this:

$T_{n, 3} = 1$ whenever $n \equiv 1$ (mod $3$)
$T_{n, 3} = 0$ otherwise.

We proceed by induction on $n$, noting that $T_{n,k} = r$ is the same thing as saying $T_n \equiv r$ (mod $k$).
Base case: $n = 0$, $T_n = 0$, $T_{n, 3}=0$ and we're done.
Inductive case: There are three sub-cases to consider:

If $n \equiv 1$ (mod $3$), then $n+1 \equiv 2$ (mod $3$), and by I.H. $T_n\equiv 1$ (mod $3$). By triangle number definition, $T_{n+1} = T_n + n + 1 \equiv 0$ (mod $3$), as required.
If $n \equiv 0$ (mod $3$), then $n+1 \equiv 1$ (mod $3$), and by I.H. $T_n\equiv 0$ (mod $3$). By triangle number definition, $T_{n+1} = T_n + n + 1 \equiv 1$ (mod $3$), as required.
If $n \equiv 2$ (mod $3$), then $n+1 \equiv 0$ (mod $3$), and by I.H. $T_n\equiv 0$ (mod $3$). By triangle number definition, $T_{n+1} = T_n + n + 1 \equiv 0$ (mod $3$), as required.


TITLE: Integral of $\frac{1}{t^{2}-1}$
QUESTION [1 upvotes]: When I solve the indefinite integral $\int \frac{1}{t^{2}-1}$  with partial fraction decomposition I obtain
$$\int \frac{1}{t^{2}-1} dt=\int \frac{1}{2(t-1)} - \frac{1}{2(t+1)} dt=\frac{1}{2}(\ln(t-1)-\ln(t+1))+C$$
But when I compare it with Wolfram Alpha, they have
$$\frac{1}{2}(\ln(1-t)-\ln(t+1))+C$$
$\ln(t-1) \neq \ln(1-t)$ , did I calculate something wrong or?

REPLY [3 votes]: They're both wrong. It should be$$\frac12\left(\ln(|t-1|-\ln|t+1|\right)+C.$$


TITLE: Confidence interval without standard deviation
QUESTION [1 upvotes]: From a list of stock mutual funds, 52 funds were selected at
  random.
       Of the funds chosen, it was found that 19 required a minimum
      initial investment under \$1,000.
(a) Construct a 90 percent confidence interval for the true proportion
  requiring an initial investment under \$1,000.
(b) May normality be assumed? Explain.

I am not sure how to go about solving this. I am used to answering these kinds of questions when I know the mean and standard deviation.
How do I determine these?
Thanks

REPLY [1 votes]: Say the true proportion of funds that require an initial investment of less than
\$1000 is $p$. If we select $n$ funds at random with replacement, then the probability that of the $n$ funds we sampled, $k$ require an initial investment of less than \$1000 follows a binomial distribution with probability parameter $p$. Under certain conditions, you can use a normal approximation to this binomial distribution to calculate a confidence interval. You do not know $p$, but you can estimate it with $\hat{p}$, the sample proportion. See here for more information. The method I've outlined is not the best way to solve this problem, but I believe it to be the simplest. Better methods are detailed in the article I linked to.


text:
\section{Related Work and the Dual Formulation}
A standard approach to adress problem \eqref{eq:global} is to express it as a {constrained} optimization problem
\begin{equation}\label{primal}
\min_{\X \in \R^{nd}}
F(\X):= \frac{1}{n} \sum_{i=1}^n f_i(x_i) \quad \mbox{such that} \quad x_1=x_2= \cdots = x_n \in \mathbb{R}^d\,, \tag{P}
\end{equation}
where $\X = [x_1;x_2; \cdots x_n] \in \R^{nd}$ is a concatenation of the vectors.
To lighten the notation, we introduce the global mixing matrix $\W = W \otimes I_d \in \R^{nd \times nd}$, {where $\otimes$ denotes the Kronecker product}, and let $\| \cdot \|_{\W}$  denote the semi-norm induced by $\W$, i.e. $\|\X \|^2_{\W} = \X^T \W \X $. With this notation in hand, we briefly review existing literature on decentralized algorithms.

\paragraph{Decentralized Gradient Descent} The decentralized gradient method \cite{nedic2009distributed,yuan2016convergence} has the update rule
\begin{equation}
  \X_{k+1} = \W \X_k - \eta \nabla F(\X_k).  \tag{DGD}
\end{equation}
{However, with constant stepsize, the algorithm does not converge to a global minimum of \eqref{primal}, but rather to a neighborhood of the solution~\citep{yuan2016convergence}.} A decreasing stepsize schedule may be used to ensure convergence, but this yields a sublinear convergence rate, even in the strongly convex case.

\begin{algorithm}[tb]
	\caption{Decentralized Augmented Lagrangian framework} \label{algo:AL}
	$\,\,$ {\bf Input:} mixing matrix $W$, regularization parameter $\rho$, stepsize $\eta$. \\
	\vspace{-0.4cm}
	\begin{algorithmic}[1]
		\FOR{$k = 1, 2, ..., K$}
		\STATE $\X_{k} = \argmin  \left \{ P_k(\X) := F(\X) + \mathbf{\Lambda}_k^T \X +  \frac{\rho}{2}  \| \X \|^2_{\W}   \right \} $.
		  \STATE $\mathbf{\Lambda}_{k+1}  = \mathbf{\Lambda}_k + \eta \, \W \X_{k}$.
		\ENDFOR
	\end{algorithmic}
\end{algorithm}

\paragraph{Linearly convergent primal algorithms}

By and large,  recent methods that achieve linear convergence in the strongly convex case \cite{shi2014linear, jakovetic2014linear, shi2015extra, qu2017harnessing, nedic2017achieving, nedic2017geometrically,sun2019convergence} can be shown to follow a general framework based on the augmented Lagrangian method, see Algorithm~\ref{algo:AL}; The main difference lies in how subproblems $P_k$ are solved.   \citet{shi2014linear} apply an alternating directions method; in  \cite{shi2015extra}, the EXTRA algorithm takes a single gradient descent step to solve~$P_k$, see Appendix~B for details. \citet{jakovetic2014linear} use multi-step algorithms such as Jacobi/Gauss-Seidel methods. To the best of our knowledge, the complexity of these  algorithms is not better than  $O \left ((1+\tau) {\kappa_f \kappa_W}  \log (\frac{1}{\epsilon}) \right )$, in other words, they are non-accelerated.

The recently proposed algorithm APM-C  \cite{li2018sharp} enjoys a square root dependence on $\kappa_f$ and $\kappa_W$, but incurs an additional $\log(1/\epsilon)$ factor compared to the optimal attainable rate.

\paragraph{Optimal method based on the dual formulation}
By Assumption \ref{ass:W}.\ref{ass:W_spec}, the constraint $x_1=x_2= \cdots = x_n$ is equivalent to the identity $\W \cdot \X = 0 $, which is again equivalent to $\sqrt{\W} \cdot \X = 0$. Hence, the dual formulation of  \eqref{primal} is given by
\begin{equation}\label{dual}
    \max_{\mathbf{\Lambda} \in \mathbb{R}^{dn}} -F^*(- \sqrt{\W} \mathbf{\Lambda}).  \tag{D}
\end{equation}
Since the primal function is convex and the constraints are linear, we can use strong duality and address the dual problem instead of the primal one. Using this approach, ~\cite{scaman2017optimal} proposed a dual method with optimal accelerated rates, using Nesterov's accelerated gradient method for the dual problem (\ref{dual}). As mentioned earlier, the main drawback of this method is that it requires access to the gradient of the dual function which, unless the primal function has a relatively simple structure, is not available.

One may apply a first-order method to approximate the dual gradients inexactly at the expense of an additional $\sqrt{\kappa_f}$ factor in the computation cost~\cite{uribe2020dual}, but this woul make the algorithm no longer optimal. This indicates that achieving optimal rates when using primal updates is a rather challenging task in the decentralized setting. In the following sections, we provide a generic framework which allows us to derive a primal decentralized method with optimal complexity guarantees.

config: arxiv


TITLE: Product of $n$ consecutive positive integer is not a $n$th power?
QUESTION [3 upvotes]: If $n>2$ and $k$ is positive integer, then there is no positive integer $m$ satisfy that $$k(k+1)\cdots (k+n-1)=m^n\, ?$$
I tried to prove this problem, but I don't know how to prove it. I know that every product of consecutive integers is not a power, but I want to simple proof of my question. Thanks for any help.

REPLY [7 votes]: I think I found a simple solution, will need someone to verify.
suppose $m$ exists such that $k(k+1)⋯(k+n−1)=m^n$
then $k < m < k + n - 1$
$gcd(m,m+1) = 1$
$m+1$ divides the LHS but doesn't divide the RHS


TITLE: what is the probability distribution for the angle of an approximate laserbeam
QUESTION [5 upvotes]: I'm trying to simulate the light distribution characteristics from a Gaussian laser beam, but having difficulty with the angular distribution.
I need to generate a large number of points on an x/y plane, along with pointing vectors down the z-axis, such that their aggregate approximates the power distribution of a laser beam. The distribution on the x/y plane is a Gaussian, while the pointing vector down the z-axis approximates the laser divergence.
For example, a beam with a 1 mm beam waist with a 1.5 mrad divergence has the following beam irradiance, E(r):
$E(r) = exp(-r^2/b^2)/(\pi b^2)$
where $r$ is the radius from the beam center, $b$ is the beam waist (1/e radius). Therefore, I can sample from this distribution by the following equation:
$r = b\sqrt(-ln(1-\mathbb{R}))$
where $\mathbb{R}$ is a random number uniformly distribution on [0,1], to get the radius from the origin on the beams starting point on the x/y plane.
OK, so the question I'm having trouble with, is how do I randomly choose the polar angle $\theta$ of divergence such that it approximates the laser divergence of 1.5 mradians? Do I just choose the polar angle to be uniform on $[0,1.5e^{-3}]$ radians?
I guess I'm getting confused by whether the distribution should be uniform over polar angle, or uniform irradiance over the solid angle, and how to sample from that. Below is an illustration to help sort things out. I'm trying to determine how to distribute the polar angle in order to approximate the laser divergence.

REPLY [1 votes]: This process is treated in this paper: http://www.springerlink.com/content/2ww1gtp5cvbrerhm/fulltext.pdf
The relevant equations are:
$x_s=\frac{w(-d)}{\sqrt{2}}\text{Erf}^{-1}(2r_1-1)$
$y_s=\frac{w(-d)}{\sqrt{2}}\text{Erf}^{-1}(2r_2-1)$
$x_f=\frac{w_0}{\sqrt{2}}\text{Erf}^{-1}(2r_3-1)$
$y_f=\frac{w_0}{\sqrt{2}}\text{Erf}^{-1}(2r_4-1)$
These give the x and y positions of rays at the beam waist ($x_f$) and at a point far away ($x_s$), with d much greater than the Rayleigh range.
$w(z)$ is the standard gaussian beam waist formula, and the $r_i$ are uniform random variates on (0,1).
EDIT:
Realized I didn't really address your divergence issue. The above is valid for a collimated diffraction limited beam, so to add excess divergence you could apply the ABCD matrix for a thin lens to the rays in order to create a diverging beam.


TITLE: $\lim\limits_{h\to0} \frac{(x+h)^3 -x^3}{h}$ Compute the following limit
QUESTION [1 upvotes]: I can not seem to figure out the direction to go with this problem. I'm not quite sure how to break it up.

REPLY [1 votes]: Others have pointed out here that $(x+h)^3 = x^3+3x^2h+3xh^2+h^3$.
Here's different approach.  Recall that $a^3-b^3=(a-b)(a^2+ab+b^2)$.  Consequently
\begin{align}
(x+h)^3-x^3 & = \Big((x+h)-x\Big)\Big( (x+h)^2 +(x+h)x + x^2 \Big) \\[12pt]
& = h\Big( (x+h)^2 +(x+h)x + x^2 \Big).
\end{align}
Now we might be tempted to do the routine simplifications of the expression in the large parentheses, but it's quicker to do this:
$$
\frac{(x+h)^3-x^3} h = \frac{h\Big( (x+h)^2 +(x+h)x + x^2 \Big)} h = (x+h)^2 +(x+h)x + x^2.
$$
We want the limit of that as $h\to0$, and now we can just plug in $h=0$ and get $x^2+x^2+x^2$.


TITLE: why is this formula not valid for n=0 and i=0?
QUESTION [0 upvotes]: $i$ and $n$ are integers.
$\exists (i)\Big(\big((n\geq 0) \wedge (0\leq i)\wedge (i\leq n)\big)\implies (3\leq n)\Big) $
It should be not valid for n=0 and i=0.

REPLY [1 votes]: $i$ is a bound variable, so if you want to analyze it for a given value of $i$ you are ignoring the leading $\exists$.  For $i=0, n=0$ all of the terms before the $\implies$ are true and the term after the $\implies$ is false, so the whole implication is false.  If you consider the leading $\exists i$ the predicate is true for $n=0$ as $i=1$ is a witness.  Both the antecedent and the consequent of the implication are false, which makes the implication true.


TITLE: Strictly positive element in unital $C^*$ algebra is invertible
QUESTION [3 upvotes]: Let A be a unital $C^∗$ algebra. $a\in A_+$ is called strictly positive if for all positive linear functional $\phi \in A_+^*$ and $\phi \neq 0$, we have $\phi(a)>0$. Prove that $a\in A_+$ is strictly positive iff $a$ is invertible.
My attempt: if $a\in A_+$ is invertible, then there exists $x\in A$ such that $a=x^*x$ and $a^{-1}=x^{-1}(x^{-1})^*$, Using Cauchy-Schwarz inequality: $\forall \phi \in A_+^*$
$$\phi(a)\phi(a^{-1})=\phi(x^*x)\phi((x^{-1})^{**}(x^{-1})^{*})\ge |\phi(x^*(x^{-1})^*)|^2=|\phi(1)|^2=\|\phi\|^2 > 0 $$
then we have $\phi(a)\neq 0$, then $\phi(a)>0$.
But for other direction, i have no idea. I knew that $a\in A_+$ is called strictly positive also can be definted that $\overline{aAa}=A$, and from this definition i can prove that if $a$ is strictly positive, then $axa$ is invertible, then $a$ is invertible. but i want to prove $a$ is invertible from defintion which i give firstly.
I guess if $\phi(a)>0$, then $1-\phi(a) <1$, and let $\phi$ be some specific function ,then apply functional calculus, but i can't prove that.
Any idea will be appreciated!

REPLY [2 votes]: EDITED Since I originally just reproved what you already included in your statement.
Suppose $a\in A$ is strictly positive.  Since $A$ is unital, the state space of $A$ is weak$^*$-compact, and it follows that $\varepsilon:=\inf\{\phi(a):\phi\text{ is a state on }A\}>0$. It then follows that $a-\varepsilon\cdot1$ is positive, from which it follows that the spectrum of $a$ is contained in $[\varepsilon,\infty)$, i.e., $0$ is not in the spectrum of $a$, i.e., $a$ is invertible.
Here's another proof of the reverse direction:
If $a\in A$ is positive and invertible, its spectrum is a compact subset of $(0,\infty)$, and thus $a-\varepsilon\cdot 1$ is positive for some $\varepsilon>0$.  Thus if $\phi$ is a non-zero positive linear functional on $A$, we have
$$\phi(a)=\varepsilon\phi(1)+\phi(a-\varepsilon\cdot1)\geq \varepsilon\|\phi\|+0>0.$$

REPLY [2 votes]: For the other direction, let $a$ be a positive non-invertible element. Then $0\in \text{Spec}(a)$ and there is a unital *-homomorphism $\Phi: C^*(a) \rightarrow C(\text{Spec}(a))$ mapping $a$ to the identity function on $\text{Spec}(a)$. Here $C^*(a)$ is the $C^*$-subalgebra generated by $a$ and $C(\text{Spec}(a))$ the continuous functions on $\text{Spec}(a)$. Define a state $\phi$ on $C^*(a)$ by $\phi(x)=\Phi(x)(0)$.
$\phi$ vanishes on $a$ and we can extend it to all of $A$.