# Back Door Man

This cover by The Doors can be terrifying. I think it has at least two interpretations – who are the little girls? Morrison, Hendrix & Cobain all died at 27. Alcohol was involved with Morrison & Hendrix: http://www.youtube.com/watch?v=tKpOWdA1h9Y

If a human brain is like a computer, judged partly by speed and reliability, then where does the code come from? Don’t listen to these songs after using too much meth! I could have punched someone to get a beer.

# (iϕ)²=1=(ϕi)²

Let

$\phi = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$

be the golden matrix, which satisfies $\phi^2=\phi+1.$ It maps $\begin{pmatrix} a \\ b \end{pmatrix}$ to $\begin{pmatrix} b \\ a+b \end{pmatrix}.$ Then let $i=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},$ which satisfies $i^2=-1.$ Then

$\begin{array}{rcl} i\phi &=& \frac{1}{i\phi} \\ \phi i &=& \frac{1}{\phi i} \\ \{i\phi,\phi i\} &=& i \end{array}$

where $\{,\}$ is the anticommutator. In fact, all 2×2 matrices must satisfy some quadratic equation, the characteristic equation. It is $\mu^2=1$ for $\mu=i\phi$ & $\nu = \phi i.$ Also, in general

$(a+b)^2 = a^2+b^2+\{a,b\}$

From this, the complex plane $(1,i)$ can be square rooted into the $(\mu,\nu)$ plane:

$\begin{array}{rcl} \sqrt{x+yi} &=& \frac 1 2 (\sqrt{x+2y}+\sqrt{x-2y})\mu+\frac 1 2 (\sqrt{x+2y}-\sqrt{x-2y})\nu \\ &=& \sqrt{\frac{x+\sqrt{x^2-4y^2}}{2}}\mu+\sqrt{\frac{x-\sqrt{x^2-4y^2}}{2}}\nu \end{array}$

for the region $x\ge 2y\ge 0.$

Also, if $\varphi$ is the golden ratio, then $\frac{1}{\sqrt 5}(i\varphi)^n-(i\varphi)^{-n})$ is the $n$‘th term of the golden spiral, $i^nF_n,$ where $F_n$ is the $n$‘th Fibonacci number given by $F_1=F_2=1$ & $F_{n+2}=F_{n+1}+F_n.$ The spiral starts with $0,$ $i$ & $-1.$

# Cut out the middle-names

I haven’t been able to prove I’ve ever had middle names, so I’m officially “Konstantinos Nitsopoulos”. It seems lawyers need proofs, too!

Also, it turns out you can encrypt text simultaneously under different PGP keys, using gpg4usb! Furthermore, that program allows copy-pasting of the keys, in clipboard. Last year I posted to the contrary, due to a lack of experience. Most of my advice was alright, though. PPGP is apparently not secure, and PGP4Win is outdated. I recommend PWGen to create long passwords which are saved into a text document which is encrypted using a memorised password into a .zip archive using 7-Zip, which is then backed up! AVG can be used to shred files, and Recuva to recover files otherwise emptied from a Recycle Bin. The FBI still owns the most valuable bitcoin address in existence (144 thousand coins), from the Silk Road seizure. The owner Ross Ulbricht (Dread Pirate Roberts) was caught in a library while using Linux, and in November was charged with paying more than $700 thousand in conspiracies to murder six people. The new Silk Road is now controlled by Defcon, who claims to have enforced encryption, the lack of which brought down the original marketplace. Buyers are still required to deposit coins before making a purchase, which is dodgy. I still believe PGP will be the future of most digital communication, maybe as a new email standard. The current lack of email privacy we endure is shameful, but acceptable. I’ve also updated my post from a couple years ago regarding the resistance between adjacent vertices of an octahedron, which is only $\frac{5}{12}\Omega$ when each edge has resistance $1\Omega.$ This result can be deduced intuitively. # xCFD profit xCFD is a New-Zealand (NZ) based CFD (Contract-for-Difference) trading company. A CFD is a financial derivative, not the commodity itself. So xCFD does not allow users to purchase commodities such as gold and bitcoin, but only to take long and short positions on them. Unfortunately the company is often listed as being a currency exchange, like here on YouTube, and here on the main bitcoin wiki. During December 2013 I deposited into the xCFD platform thinking it was a currency exchange, not just an options exchange. Ever since 26 Dec I’d been trying to withdraw. The tech guy is Ivan Kuznetsov (ike@xcfd.com). I didn’t have to prove my identity and address to deposit, but only to withdraw. My ID was approved on 3 Jan 2014, and withdrawal approved on 4 Jan with a deadline of one week. I’m guessing legal action would be a nuisance because they’re based in NZ. They also have hidden fees and misleading advertising. I was charged a$67 currency conversion fee, which overdrew a bank account without warning. They also promised a bonus trading amount for deposits made last year, without making clear that money can never be withdrawn.

Incorporated Aug 2013 in Napier, NZ, but the only shareholder and director is Alexey Kirienko of Moscow, Russia. My money was in Russia.

Update (16 Jan): Withdrawal came through yesterday. I can’t really call it a scam when I made a \$16 profit overall! I got lucky trading a few CFDs while I was waiting for my identity to be verified.

# Γ from ζ

$\Gamma(s)$ is the unique log-convex analytic continuation of $(s-1)!,$ which maintains the recursive relation $\Gamma(s)=\frac{\Gamma(s+1)}{s}.$ $n!$ is the number of permutations of $n$ crystals, $n(n-1)\cdots 1.$ As of 2014 MathWorks, the owners of Matlab, have not implemented the unreal $\Gamma$ function. Nor has the open source clone Octave. It has been implemented in Mathematica which I don’t have.

Aside: Octave is not friendly on Windows, but doesn’t take long to load on Linux. By loading into RAM, Octave allows many Matlab scripts to be incorporated into open source software. Programmers need a cheap and easy way to incorporate advanced maths into their software.

Strangely, Matlab does have the complex-valued Riemann $\zeta(s).$ Hence my code uses the reflection formula

$\Gamma(s) = \frac{(2\pi)^s}{2\cos(\frac{\pi}{2} s)}\frac{\zeta(1-s)}{\zeta(s)}$

This formula is problematic at integers, so for those I use the standard “gamma”. The only other problems are near non-trivial $\zeta$ zeros, on the critical strip between 0 & 1. For those, I escape the strip using recursion.

I (Tinos Nitsopoulos) copyright zetaGamma.m under GNU.

One application is to investigate $\Gamma$ along the critical line, $\Gamma\left(\frac 1 2+iy\right).$ This algorithm is always accurate (the selling point), but slow. For example,

>> zetaGamma([0 3 1/2 1/2+i])
ans =
Inf         2.0000         1.7725        0.3007 - 0.4250i
>> abs(ans(end))
ans =
0.5206
>> sqrt(pi/cosh(pi))
ans =
0.5206

From the reflection formula for $\Gamma,$

$\begin{array}{rcl} |\Gamma(\frac 1 2+iy)| &=& \sqrt{\frac{\pi}{\cosh(\pi y)}} \\ |\Gamma(iy)| &=& \sqrt{\frac{\pi}{y\sinh(\pi y)}} \\ y\tanh(\pi y) &=& \left|\frac{\Gamma(\frac 1 2 + i\pi y}{\Gamma(iy)}\right|^2 \\ \frac{e^{\pi y}}{\pi} &=& \frac{1}{|\Gamma(\frac 1 2+iy)|^2} + \frac{1}{y|\Gamma(iy)|^2} \end{array}$

That is, $\Gamma$ produces hyperbolic functions. Hence the appearance of $\sqrt{\frac{\pi}{\cosh(\pi)}}$ in the numerical example. $\log|\Gamma(s)|$ is the real part of the analytic continuation of $\left(\log\sqrt{\frac{\pi}{\cosh(\pi y)}}+i\text{Phase}\right)$ from the critical line. The solution to Laplace’s equation is not unique, given only $\log|\Gamma(\frac 1 2+iy)|.$

Aside: As Feynman points out in his lectures (Vol. II Ch. 7), analytic $f(x+iy)$ may solve a physics problem. What problem does $\zeta(s)$ solve?

# Eight consecutive tails

$2^{-8}\approx .4\%$ is the probability of eight consecutive tails with a fair coin. An example of the gambler’s fallacy is the bet that the eighth coin will land heads after seven tails, because eight tails are very unlikely. On the contrary, I think the coin is likely to be biased towards tails! If you’re already in trouble, why not take the eighth bet? Apparently the woman’s sister couldn’t have children, so she was married to her sister’s husband.

Suicide in China is more common among women than men, by 3:1. In Greenland more than 0.1% of all men commit suicide! Australia ranks #50, although the statistics in developing countries may be unreliable.

# Band-aids vs super-glue

A band-aid restricts blood-flow to a wound, covers it and protects it from infection. However, after the wound has clotted blood-flow may continue to be restricted. In my experience, leaving it for several hours can worsen the wound. Vietnam veteran Dave Herbert uses super-glue.