Number News

I was thinking whether algorithms in general are predictable or not, ie. not knowing anything about the algorithm but knowing its input, can we tell something about the output? Imagine a Turing machine with an arbitrary program with an arbitrary number of states. Let its alphabet contain 2 symbols (call it 0 and 1). Let the machine start at a tape with 0 in every cell (or any other given starting sequence) and run its program. If the program halts (or rather halts within some given period of time), we look at all the cells visited by the machine and count how many of them contain 0 and how many contain 1. The question is, if we repeat this process infinitely and count the output for all the programs, on average what fraction of all symbols would be 0 or 1? Would it be equal, ie. half of all symbols are 0 and half are 1, or would some symbol appear more often? If it is impossible to know, how do you prove it? Are there any other ways to think of how predictable algorithms are?

$ABC$ is an equilateral triangle. The path $ABC$ is twice as long as $AC$. Similarly the path $ADEFC$ is also twice as long as $AC$, as is the path $AGHIEJKLC$, and so on. Breaking down the jagged path into smaller and smaller jags, the deviation of the jaged path from the straight line $AC$ goes to zero, so, in a sense, the line $AC$ is the "limit" of the sequence of jagged paths. This seems to suggest that the length of $AC$ is twice the length of itself! This is obviously wrong, but why$?$

I am trying to prove that $f(x) = x + 2e^{-a^2/2}(x\cosh(ax) - a\sinh(ax)) \ge 0$ for $0 \le x \le \frac{a}{2}$. From the plots that I have done this seems to be true. However, I have not been able to show this. The first thing that I have tried is finding a lower bounding function $g$ for $f$ on $[0, \frac{a}{2}]$ for which it is clear that $g(x) \ge 0$ on $[0, \frac{a}{2}]$. Since $f(0) = 0$ and $f(\frac{a}{2}) = \frac{3}{2}ae^{-\frac{a^2}{2}} > 0$, I have also tried analyzing the derivative $f'(x) = 1 + 2e^{-\frac{a^2}{2}}(ax\sinh(ax)-(a^2 - 1)\cosh(ax))$ to see whether I would be able to find something. If $0 \le a \le 1$ then $-(a^2 - 1) \ge 0$ and consequently $f'(x) \ge 0$, which implies that $f(x) \ge 0$. However, for $a > 1$ in general, I cannot seem to find the way to show this. Which approach could I take for this proof?

In a book I am reading, the author is stating that the following equation holds for any Lie operator (or any linear operator, more generally): $\exp(A\cdot\nabla)\exp(F\cdot\nabla)\exp(-A\cdot\nabla)=\exp(e^{[A\cdot\nabla,\,\circ]}\,F\cdot\nabla)$, where $[X,\circ]$ is the adjoint of $X$, such that $[X,\circ]\; Y = [X,\,Y] = XY - YX$. One step of this proof is the question in the title, from which we can apply a similarity transform to get to the final expression with $e^{[A\cdot\nabla,\,\circ]}$ in the exponent. I can prove the latter the similarity transform $e^X Y e^{-X} = e^{[X,\circ]}Y$ by constructing functions $U(t)=e^{tX}Ye^{-tX}$ and $V(t)=e^{t[X,\circ]}Y$ and showing their initial values and derivatives agree.

A large section of the book "Complex Analysis with Applications" by Asmar and Grafakos is dedicated to the evaluation of real improper integrals using complex analysis, namely contour integrals. One exercise asks to establish the identity: $$\int_{0}^{+\infty} \frac{x^p}{x(1-x)}dx =\pi \cot(p\pi),$$ for $0<p<1.$ We consider the following complex integral $$\int_{\gamma} \frac{z^p}{z(1-z)}dz$$ where $\gamma$ is the contour suggested by the authors (I hope everything is fine with the copyrights): The idea is to let $R \to +\infty$ and $r \to 0^+$, where $r$ denotes the radius of the small circumferences around the singularities. Let me point out that the dashed line is a branch cut of the logarithm, that is the ray through the origin along which a branch of the logarithm is discontinuous. Thus, I guess the authors here are choosing $$x^p = \exp(p \log_0(x)),$$ where $\log_0(z) := \ln|z| + i \arg_0 z$ and $0 < \arg_0(z) \le 2 \pi$ (not a multivalued function); in general the complex power is a multivalued function, since it is defined as $$z^a := \exp(a \log z),$$ where $\log z = \ln|z| + i \arg z$ and $\arg z = \{\text{Arg}z + 2k \pi: k \in \mathbb{Z}\}$ with $-\pi < \text{Arg} z \le \pi.$ Let us focus on the two semicircles around $1$. Let $\gamma_1$ be the circle around $0$ and follow the direction of the arrow, so that the semicircle around $1$ which is above the $x-$axis will be denoted by $\gamma_3$ and the semicircle around $1$ which is below the $x-$axis will be denoted by $\gamma_7.$ Are the integrals over $\gamma_3$ and $\gamma_7$ zero or not? For example, let me consider $\gamma_3$ only. Reason why I believe the integral over $\gamma_3$ is null: $$\left | \int_{\gamma_3} \frac{z^p}{z(1-z)} dz \right | = \left | \int_{\gamma_3} \frac{1}{z^{1-p}(1-z)} dz \right | \le \int_{\gamma_3}\frac{1}{|z^{1-p}||1-z|} dz = \int_{\gamma_3}\frac{1}{|z|^{1-p}|1-z|} dz \le \int_{\gamma_3}\frac{1}{r^{1-p}(1-r)} dz \le \frac{\pi r}{r^{1-p}(1-r)} = \frac{\pi r^p}{1-r} \xrightarrow{r \to 0^+}0$$ Note: here the tricky steps are $\frac{z}{z^p} = z^{1-p}$ and $|z^{1-p}| = |z|^{1-p}$ for $1-p \not\in \mathbb{Z}$, but they seem to work even as multifunctions, because of the properties of the exponential function. Reason why I believe the integral over $\gamma_3$ is not null: we need a lemma, which is called the "Shrinking Path Lemma" in the book. Suppose that $f$ is a continuous complex-valued function on a closed disk $\overline {B_{r_0}(z_0)}$ with center at $z_0$ and radius $r_0$. For $0<r\le r_0$, let $\sigma_r$ denote the positively oriented circular arc at angle $\alpha$, consisting of all $z=z_0 + r e^{i\theta}$, where $\theta_0 \le \theta \le \theta_0 + \alpha$, $\theta_0$ and $\alpha$ are fixed, and $\alpha \neq 0.$ Then $$\lim_{r \to 0^+} \frac{1}{i\alpha} \int_{\sigma_r} \frac{f(z)}{z-z_0} = f(z_0)$$ We would like to apply this lemma to $f(z):=\frac{z^p}{z}$. Observe that $\log_0(z) = \ln|z| + i \arg_0 z$, $0 < \arg_0(z) \le 2 \pi$ is discontinuous on the positive real axis. Therefore we use the following trick: in the first quadrant $\arg_0(z) = \text{Arg z}$, so $\log_0(z) = \text{Log z}$, which is continuous. Thus $$ \int_{\gamma_3} \frac{z^p}{z(1-z)} dz = -\int_{\gamma_3} \frac{z^p}{z(z-1)} dz = -\int_{\gamma_3} \frac{\exp(p \cdot \text{Log z})}{z(z-1)} dz \xrightarrow{r \to 0^+} i \pi \frac{\exp(p \cdot \text{Log 1})}{1} = i \pi$$ (note the change of sign due to the negative orientation of $\gamma_3$). I know that the substitution $x = e^t$ in the first integral helps overcome most of this trouble, but I would like to understand what is going on. I suspect the trouble lies in the complex power being a multifunction. If this is the case, please explain how to work with it properly.