# Eikonal Blog

## 2010.02.26

### Moore-Penrose inverse for light-cone vectors

Filed under: mathematics — Tags: — sandokan65 @ 16:54

This is continuation of previous post https://eikonal.wordpress.com/2010/02/17/some-examples-of-the-moore-penrose-inverse/

Example (added 2010.02.25 Fri): $D=1+1$ light-cone vector (i.e. $R_\mu R^\mu \equiv 0$) $R_\mu$ ($\mu=0,1$) can be cast via $D=1+1$ Dirac matrices $\left\{\gamma^0 = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}, \gamma^1 = \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}\right\}$ into the nilpotent (i.e. $\not{R}^2 \equiv 0$) matrix $\not{R}:\equiv R_\mu \gamma^\mu = R_0 \begin{pmatrix}1 & \epsilon \\ -\epsilon & -1 \end{pmatrix}$, where $R_1 = \epsilon R_0$ (and $\epsilon=\pm 1$).

To find the MPI of that matrix, start with the transposed matrix $\not{R}^T = \begin{pmatrix}R_0 & -R_1 \\ -R_1 & -R_0 \end{pmatrix}$. It is also nilpotent (i.e. $(\not{R}^T)^2 = 0$) but products following (singular) product are non-zero:

• $\not{R}^T \cdot \not{R}= 2R_0^2 \begin{pmatrix}1 & \epsilon \\ \epsilon & 1 \end{pmatrix}$,
• $\not{R} \cdot \not{R}^T = 2R_0^2 \begin{pmatrix}1 & -\epsilon \\ -\epsilon & 1 \end{pmatrix}$.

Now calculate $\not{R}^+$ via following limiting process:

$\not{R}^+ = \lim_{\delta\rightarrow 0} (\not{R}^T \not{R} + \delta {\bf 1})^{-1} \not{R}^T$
$= \frac1{2 R_0^2} \lim_{\delta\rightarrow 0}\begin{pmatrix}1+\delta & \epsilon \\ \epsilon & 1+\delta\end{pmatrix}^{-1} \not{R}^T$
$= \frac1{2 R_0^2} \lim_{\delta\rightarrow 0} \frac1{2\delta+\delta^2} \begin{pmatrix}1+\delta & -\epsilon \\ -\epsilon & 1+\delta\end{pmatrix} \not{R}^T$
$= \frac1{2 R_0} \lim_{\delta\rightarrow 0} \frac1{2\delta(1+\frac12\delta)} \begin{pmatrix}\delta & -\epsilon\delta \\ \epsilon\delta & -\delta\end{pmatrix}$
$= \frac1{4 R_0} \begin{pmatrix}1 & -\epsilon \\ \epsilon & -1\end{pmatrix} = \frac1{4R_0^2} \not{R}^T$.

Direct check verifies that all four defining properties of MPI are satisfied.

Note that $\not{R}^+$ corresponds to the $D=1+1$ vector $(R^+)_\mu = \left\{\frac1{4R_0}, -\frac{\epsilon}{4R_0}\right\} = \frac1{4R_0^2} \left\{R_0, -R_1\right\} = \frac1{4R_0^2} \hat{P}R_\mu\hat{P}$, where $\hat{P}$ is the space-parity operator.

The two products $\Pi_1:\equiv \not{R}\cdot \not{R}^+ = \frac12 \begin{pmatrix}1 & -\epsilon \\ -\epsilon & 1\end{pmatrix}$ and $\Pi_2:\equiv \not{R}^+ \cdot \not{R} = \frac12 \begin{pmatrix}1 & +\epsilon \\ +\epsilon & 1\end{pmatrix}$ are projectors: $\Pi_a \Pi_b = \delta_{a,b} \Pi_b$ ($a, b \in \{1,2\}$), $\Pi_1 + \Pi_2 = {\bf 1}$. They project on two light cone directions defined by vectors $R_\mu$ and $R^+_\mu$.

More at this blog:

## 2010.02.25

### Information disclosure sites

Related here: WikiLeaks 2010 – https://eikonal.wordpress.com/2010/12/29/wikileaks-2010/ | ACTA – https://eikonal.wordpress.com/2010/07/16/acta/ | Law vs Technology – https://eikonal.wordpress.com/2012/05/16/law-vs-technology/

.

### With a little help of the government

Filed under: opression, surveillance — Tags: — sandokan65 @ 11:07

Universal surveillance:

• The Snitch in Your Pocket (Law enforcement is tracking Americans’ cell phones in real time—without the benefit of a warrant), by Michael Isikoff (NEWSWEEK; Feb 19, 2010.02.19; from the magazine issue dated 2010.03.01): http://www.newsweek.com/id/233916

Dirty deeds:

• “The Chemist’s War” (The little-told story of how the U.S. government poisoned alcohol during Prohibition with deadly consequences.), by Deborah Blum (Friday, Feb. 19, 2010, at 10:00 AM ET) – http://www.slate.com/id/2245188/

## 2010.02.22

### Public network gateway sites

Filed under: networking — Tags: , , , — sandokan65 @ 14:31

## 2010.02.20

### Leonard Cohen’s Suzanne

Filed under: music — Tags: — sandokan65 @ 19:46

Suzanne:

## 2010.02.19

### Weighted Moore-Penrose inverse

Filed under: mathematics — Tags: — sandokan65 @ 15:13

This is a generalization of the original concept of Moore-Penrose inverse (MPI). The weighted MPI $A^{+(N,M)}$ of a matrix $A \in {\Bbb F}^{n\times m}$ is defined by the following four properties:

• (A): $A \cdot A^{+(N,M)} \cdot A = A$,
• (B): $A^{+(N,M)} \cdot A \cdot A^{+(N,M)} = A^{+(N,M)}$,
• (C)_N: $(M \cdot A \cdot A^{+(N,M)})^c = M \cdot A \cdot A^{+(N,M)}$,
• (D)_M: $(A^{+(N,M)}\cdot A \cdot N)^c = A^{+(N,M)}\cdot A \cdot N$.

where the weighting matrices $M$ and $N$ are of the orders $n\times n$ and $m\times m$.

When weighting matrices are equal to the corresponding identities, the above definition reduces to ordinary MPI $A^c$.

Source: R. B. Bapat, S. K. Jain and S. Pati “Weighted Moore-Penrose Inverse of a Boolean Matrix”; Linear Algebra and Its Applications 225:267-279 (1997); North-Holland; pg. 692-704. http://www.math.ohiou.edu/~jain/077.pdf.

More at this blog:

## 2010.02.18

Filed under: unix — Tags: , , — sandokan65 @ 14:23

## Unix humor

### Danger of low-frequency electromagnetic radiation

Filed under: health — Tags: , — sandokan65 @ 14:02

## 2010.02.17

### Some examples of the Moore-Penrose inverse

Filed under: mathematics — Tags: — sandokan65 @ 16:49

Source: T2009.02.12

Definition: For rectilinear matrix $A \in {\Bbb F}^{n\times m}$ there exist a unique matrix $A^+ \in {\Bbb F}^{m\times n}$ (called the Moore-Penrose inverse [MPI]) s/t:

• (A): $A A^+ A = A$,
• (B): $A^+ A A^+ = A^+$,
• (C): $(A A^+)^c = A A^+$,
• (D): $(A^+ A)^c = A^+ A$.

where $M^c$ is the appropriate conjugation defined on the field ${\Bbb F}$, i.e. $(M^c)^c = M$ ($\forall M$).

In particular, for ${\Bbb F} = {\Bbb R}$:

• (A): $A A^+ A = A$,
• (B): $A^+ A A^+ = A^+$,
• (C): $(A A^+)^T = A A^+$,
• (D): $(A^+ A)^T = A^+ A$,

where $A^T$ is the transposition of the matrix $A$.

and for ${\Bbb F} = {\Bbb C}$:

• (A): $A A^+ A = A$,
• (B): $A^+ A A^+ = A^+$,
• (C): $(A A^+)^\dagger = A A^+$,
• (D): $(A^+ A)^\dagger = A^+ A$,

where $A^\dagger :\equiv (A^*)^T$ is the Hermitian conjugation of a matrix $A$.

## Properties

• The matrix equation $A \cdot \underline{y} = \underline{x}$ has infinitely many solutions
$\underline{y} = A^{+} \underline{x} + ({\bf 1}_n - A^{+} A) \underline{q}$
parametrized by an arbitrary $\underline{q}, provided that the consistency condition $A A^{+} \underline{x} = \underline{x}$ is satisfied. • If $B$ is a square positive semidefinite matrix, then for each $\underline{z}$ the following inequality holds: $(A\underline{z}-\underline{x})^T \cdot B \cdot (A\underline{z}-\underline{x}) \ge \underline{x}^T C \underline{x}$. Here $C:\equiv B - B A (A^T B A)^{+} A^T B^T$. That inequality gets to be an equality for $\underline{z}=(A^T B A)^{+} A^T B \underline{x} + [{\bf 1}_n - (A^T B A)^{+}(A^T B A)]\underline{q}$. • $\Pi_1:\equiv A^+ A$ and $\Pi_2:\equiv A A^{+}$ are projectors: $\Pi_{1,2}^2 = \Pi_{1,2}$. ## The LU method • 1) Separate $A$ to an LU-product $A=L_0 \cdot U_0$. • 2) Then trim $U_0$ by dropping the empty rows (getting $U$), and trim $L_0$ by trimming the corresponding columns (getting $L$). Note that it is still valid that $A=L \cdot U$. • 3) Finally calculate $\Phi:\equiv U^T (U U^T)^{-1}$ and $\Psi:\equiv (L^T L)^{-1} L^T$, to get $A^{+} = \Phi \cdot\Psi$. ## Examples Example 1: • $A=\begin{pmatrix} 1&1&0\\ 0&1&1 \end{pmatrix}$, • $A^{+}=\begin{pmatrix} \frac23&-\frac13\\ \frac13&\frac13 \\ -\frac13&\frac23 \end{pmatrix}$, • $A\cdot A^{+} = {\bf 1}_2$, • $A^{+} \cdot A = \begin{pmatrix} \frac23&\frac13&-\frac13 \\ \frac13&\frac23&\frac13 \\ -\frac13&\frac13&\frac23 \end{pmatrix}.$. Example 2: • $A=\begin{pmatrix} a\\ b \end{pmatrix}$, • $A^{+}=\begin{pmatrix} \frac{a}{a^2+b^2}& \frac{b}{a^2+b^2} \end{pmatrix}$, • $A^{+}\cdot A = {\bf 1}_2$, • $A\cdot A^{+} =\begin{pmatrix} \frac{a^2}{a^2+b^2}& \frac{ab}{a^2+b^2 } \\ \frac{ab}{a^2+b^2}& \frac{b^2}{a^2+b^2} \end{pmatrix}.$. Example 3: • $A=\begin{pmatrix} \underline{a} \end{pmatrix}$, • $A^{+}=\begin{pmatrix} \frac1{\underline{a}^T\cdot\underline{a}}\underline{a}^T \end{pmatrix} = \frac1{\underline{a}^T\cdot\underline{a}} A^T$, • $A^{+}\cdot A = 1$, • $A\cdot A^{+} = \underline{a} \underline{a}^T$. Example 4: • $A = \left(U | \underline{v}\right)$, • $A^{+} = \begin{pmatrix} U^+ - \frac1{\underline{r}^T\cdot\underline{r}} U^{+}\underline{v}\underline{v}^T (1-U^{+T}U^T) \\ \frac1{\underline{r}^T\cdot\underline{r}}\underline{v}^T (1-U^{+T}U^T) \end{pmatrix}$, where $\underline{r} :\equiv (1- U U^{+}) \underline{v}$, • $A^{+}\cdot A = \begin{pmatrix} U^{+} U & 0 \\ 0 & 1 \end{pmatrix}$, • $A\cdot A^{+} = U U^{+} + \frac1{\underline{r}^T\cdot\underline{r}}\underline{r}\underline{r}^T.$. More at this blog: More: ### PanoptiClick Filed under: surveillance — sandokan65 @ 15:46 Browser fingeprinting: http://panopticlick.eff.org/ ## 2010.02.16 ### A su(1,1) realization Filed under: mathematics — Tags: — sandokan65 @ 14:13 For assembly of bosonic particle creation $\hat{a}_{\underline{p}}^\dagger$ and annihilation $\hat{a}_{\underline{p}}$ operators, the following bilinear combinations • $\hat{{\cal L}}_1 + i \hat{{\cal L}}_2 :\equiv \hat{a}_{\underline{p}}^\dagger \hat{a}_{-\underline{p}}^\dagger$, • $\hat{{\cal L}}_1 - i \hat{{\cal L}}_2 :\equiv \hat{a}_{\underline{p}} \hat{a}_{-\underline{p}}$, • $\hat{{\cal L}}_3 :\equiv \frac12 \left(\hat{a}_{\underline{p}}^\dagger \hat{a}_{\underline{p}} + \hat{a}_{-\underline{p}}^\dagger \hat{a}_{-\underline{p}} + 1 \right)$. form a representation of the $su(1,1)$ algebra: • $[\hat{{\cal L}}_3 , \hat{{\cal L}}_1] = + i \hat{{\cal L}}_2$, • $[\hat{{\cal L}}_2 , \hat{{\cal L}}_3] = + i \hat{{\cal L}}_1$, • $[\hat{{\cal L}}_1 , \hat{{\cal L}}_2] = - i \hat{{\cal L}}_3$. Source: G.E. Volovik “Vacuum in Quantum Liquids and in General Relativity”; arXiv: gr-qc/0104046v1, 2001.04.15; http://arxiv.org/abs/gr-qc/0104046. ### Casimir pressure Filed under: physics — Tags: , — sandokan65 @ 13:06 Casimir pressure for spherical shell of radius $R$: $P_c = -\frac{dE_c}{dV} = \frac{K}{8\pi} \sqrt{-g} \left(\frac{\hbar c}{R}\right)^4$ where • $K=-0.443,9\cdots$ for the Neuman b.cs (boundary conditions), • $K=+0.005,639,\cdots$ for the Dirichlet b.cs (boundary conditions), Source: G.E. Volovik “Vacuum in Quantum Liquids and in General Relativity”; arXiv: gr-qc/0104046v1, 2001.04.15; http://arxiv.org/abs/gr-qc/0104046. ## 2010.02.15 ### Magnus’ Bracket Operator Filed under: mathematics — Tags: — sandokan65 @ 22:46 Definition: Define $z(x,y)$ via product of exponentials $e^z :\equiv e^x e^y$. Properties: • $z = x+y+\frac12 [x,y] + \frac1{12} (\{xy^2\} + \{yx^2\}) + \frac1{24}\{xy^2x\} + \cdots$ • $e^{-y} e^x e^y = e^{x+\frac1{1!}[x,y] + \frac1{24}\{xy^2\}+\cdots} = e^{\{x e^y\}_M}$ Definition: Here “the bracket operator” $\{\cdots\}_M$ is defined by following properties: 1. $\{G\}_M = G$ if $G$ is homogeneous of the first degree, 2. $\{r G\}_M = r \{G\}_M$ if $r$ is a rational number, 3. $\{G_1 + G_2\}_M = \{G_1\}_M + \{G_2\}_M$, 4. $\{G x\}_M = [\{G\}_M , x]$, $\{G y\}_M = [\{G\}_M , y]$ So: • $\{x^2 G\}_M = \{y^2 G\}_M = 0$, • $\{x y^n\}_M = x y^n - \binom{n}{1} y x y^{n-1} + \cdots + (-)^k \binom{n}{k}^k y^k x y^{n-k} + \cdots + (-)^n y^n x$. • $\{\omega_n\}_M = n \omega_n$ for $\omega_n$ homogeneous in $x$ and $y$ of degree $n$, i.e. $\{\omega\}_M = \left(x D_x + y D_y\right) \omega$. Source: T1272 = W. Magnus “A Connection between the Baker-Hausdorff formula and a Problem of Burnside”; Annals of Mathematics; Vol 52, No 1, July 1950. Related here: Magnus on matrix exponentials – https://eikonal.wordpress.com/2010/01/26/magnus-on-matrix-exponentials/ ### Lutzky’s expansion Filed under: mathematics — Tags: — sandokan65 @ 22:16 $\frac{d e^{z}}{dt} = e^z \left\{\frac{dz}{dt},\frac{e^z-1}{z}\right\}$ $e^{z(t)} :\equiv e^{x}e^{t y}$ • $z(t) = \sum_{l=0}^\infty z_l t^l$ • $z_0 = x$ • $z_1 = \left\{ y , \frac{x}{e^x-1}\right\} = y - \frac1 [y,x] + \frac1{12} \{y,x^2\} - \frac1{720} \{y,x^4\} + \cdots$ Here the expansion $\frac{x}{e^x-1} = \sum_{k=0}^\infty \frac{B_k}{k!} x^k$ was used. ## Calculating $z_2$ Definition: • $\frac{x e^{\xi x}}{e^x-1} = \sum_{k=0}^\infty \varphi_k(\xi) x^k$ • $\frac{e^{\xi x} -1}{e^x-1} = \sum_{k=0}^\infty \Psi_k(\xi) x^k$ • $\varphi_k(\xi) = \frac1{k!} \sum_{m=0}^\infty \binom{k}{m} B_m \xi^{k-m}$ • $\partial_\xi \varphi_{k+1}(\xi) = \varphi_k(\xi)$ • $\varphi_k(0)= \frac1{k!}B_k$, • $\varphi_k(1)= \varphi_k(0)$ (for $k\ne 1$), • $\varphi_0(\xi) \equiv 1$ • $\int_0^1 \varphi_k(\xi) d\xi = \delta_{k,0}$, • $\Psi_{k}(\xi) = \varphi_{k+1}(\xi) - \varphi_{k+1}(0)$, • $\int_0^1 \varphi_k(\xi) \Psi_{l}(\xi) d\xi = \delta_{k,0}\varphi_{l+1}(0) \theta(l) + \delta_{l,0}\varphi_{k+1}(0) \theta(k)$, Now: $z_2 = - \sum_{l=1}^\infty \sum_{k=0}^\infty \{y,x^l,y,x^k\} \frac{B_{l+1}B_k}{(l+1)!k!} -\frac12 \sum_{k=1}^\infty \sum_{l=1}^\infty \sum_{j=0}^\infty I_{kl} \{\Delta_{kl}, x^j\} \frac{B_j}{j!}$ where: • $\{y,x^k,y\} :\equiv [\{y,x^k\},y]$, • $\{y,x^l,y,x^k\} :\equiv \{\{y,x^l,y\},x^k\}$, • $\Delta_{kl} :\equiv [\{y,x^l,\{y,x^k\}] = - \Delta_{lk}$, • $I_{kl} :\equiv \int_0^1 \varphi_k(\xi) \varphi_{l+1}(\xi)$, ($I_{l+2,l}=0$) The more compact expression is: $z_2 = -\frac12 \left\{ J, \frac{x}{e^x-1}\right\}$ where $J :\equiv \int_0^1 d\xi \left[\{z_1,e^{x\xi}\}, \left\{z_1,\frac{e^{x\xi}-1}{x}\right\}\right] = \int_0^1 d\xi \left[\{y,\frac{xe^{x\xi}}{e^x-1}\}, \left\{y,\frac{e^{x\xi}-1}{e^{x}-1}\right\}\right].$ The Hausdorff recursion formula: $z_n = \frac1{n} \left(\left\{y, \frac{x}{e^x-1}\right\} D_x \right) z_{n-1}$. Source: T1265 = M. Lutzky “Parameter Differentiation of Exponential Operators and the Baker-Campbell-Hausdorff Formula”; Journal of Mathematics Physics, Vol 9, N 7, July 1968. ### Perl online ## Hashes ## Files ## Chomp() ## Control structures ## Tidbits ### Rename files Alex Batko says (at http://www.cs.mcgill.ca/~abatko/computers/programming/perl/):  Here is a brilliant program for renaming one or more files according to a specified Perl expression. I found it on page 706 of Programming Perl (3rd edition). #!/usr/bin/perl$op = shift; for( @ARGV ) { $was =$_; eval $op; die if$@; rename( $was,$_ ) unless $was eq$_; }  In the code above, the second last line calls the built-in function “rename”, not the program itself (which is named “rename.pl”). Below are a few examples of use. % rename.pl 's/\.htm/\.html/' *.htm # append an 'l' % rename.pl '$_ .= ".old"' *.html # append '.old' % rename.pl 'tr/A-Z/a-z/' *.HTML # lowercase % rename.pl 'y/A-Z/a-z/ unless /^Make/' * # lowercase  ### Printing hashes Starting with an input file with data in two columns separated by coma (,):  #/bin/perl -t my %TempHash = (); my$InputFile = shift; print "Input file = ",$InputFile,"\n"; my ($line,$column1,$column2,); #reading input file to generate hash open (INPUTSTREAM, '<', $InputFile) || die ("Could not open$InputFile"); while ( $line = ) { chomp; #print$line; ($column1,$column2) = split ',', $line;$TempHash{$column1}=$column2; #print $column1," ==> ",$TempHash{$column1}; } close (INPUTSTREAM); ## printing hash - way #1 print "The following are in the DB: ",join(', ',values %TempHash),"\n"; ## printing hash - way #2 while (($key, $value) = each %TempHash) { print "$key ==> $value"; } ## printing hash - way #3 foreach$key (sort keys %TempHash){ print "$key ==>$TempHash{$key}"; }  ## Removing white spaces Sources:  # Declare the subroutines sub trim($); sub ltrim($); sub rtrim($); # Perl trim function to remove whitespace from the start and end of the string sub trim($) { my$string = shift; $string =~ s/^\s+//;$string =~ s/\s+$//; return$string; } # Left trim function to remove leading whitespace sub ltrim($) { my$string = shift; $string =~ s/^\s+//; return$string; } # Right trim function to remove trailing whitespace sub rtrim($) { my$string = shift; $string =~ s/\s+$//; return $string; } # Here is how to output the trimmed text "Hello world!" print trim($string)."\n"; print ltrim($string)."\n"; print rtrim($string)."\n"; 

Related: Regular Expressions – https://eikonal.wordpress.com/2010/04/02/regular-expressions/ | Command line based text replace – https://eikonal.wordpress.com/2010/07/13/command-line-based-text-replace/

## 2010.02.13

### Relatedness and Hamilton’s rule

Filed under: evolution, population genetics — Tags: , — sandokan65 @ 01:27

Definition: Coefficient of relatedness (Sewall Wright 1889-1988) $r$:

• $r=0.5$ for first degree relatives,
• $r=0.25$ for 2nd degree relatives,
• $r=0.125$ for f3rd degree relatives, etc.

Hamilton’s rule (William Donald Hamilton (1936–2000)): $r > \frac{c}{b}$ where $c$ is cost of gift to donor, and $b$ is benefit of gift to recepient.

Source: Dylan Evans & Oscar Zarate “Introducing Evolutionary Psychology”

More:

### Twistors

Filed under: mathematics, physics — Tags: — sandokan65 @ 01:09

Twistor variables:

• direct $\{{\cal Z}^\alpha|\alpha=1,2\}$,
• conjugated: $\{\bar{{\cal Z}}_\beta|\beta=1,2\}$.

Quantization: $[{\cal Z}^\alpha, \bar{{\cal Z}}_\beta] = \hbar \delta^\alpha_\beta$.

Canonical representation: $\bar{{\cal Z}}_\alpha \rightarrow -\hbar \frac{\partial}{\partial {\cal Z}^\alpha}$.

Helicity: $s :\equiv \frac12 \left\{{\cal Z}^\alpha,\bar{{\cal Z}}_\alpha\right\} = -\frac12\hbar \left(2+ {\cal Y} \right)$,
where ${\cal Y} :\equiv {\cal Z}^\alpha\frac{\partial}{\partial {\cal Z}^\alpha}$ is the Euler’s homogeneity operator.

Source: Roger Penrose “The Road to Reality”

### TeX sites

Filed under: chemistry — Tags: , — sandokan65 @ 00:33

## Physics wikis

• SklogWiki – sklogwiki.org – dedicated to the thermodynamics of simple liquids, complex fluids, and soft condensed matter
• Qwiki – qwiki.stanford.edu – a quantum physics wiki devoted to the collective creation of technical content for practicing scientists
• Quantiki – quantiki.org – dedicated to the quantum information science

Related here:

## Cubic equation

The roots of 3rd order algebraic equation $x^3 + bx^2+cx+d=0$ are given by the Tartaglia-Cardano formulas:

• $x_0 = X_0 + X_{+} + X_{-}$,
• $x_1 = X_0 + \omega X_{+} + \omega^2 X_{-}$,
• $x_2 = X_0 + \omega^2 X_{+} + \omega X_{-}$,

where:

• $\omega:\equiv e^{i\frac{2\pi}{3}} = \frac{-1+i\sqrt{3}}2$,
• $X_0 :\equiv -\frac{b}3$,
• $X_{\pm} :\equiv \left\{ - \left(\frac{27 d -9bc+2b^3}{54}\right) \pm \sqrt{\left(\frac{27 d -9bc+2b^3}{54}\right)^2 - \left( \frac{b^2-3c}{9}\right)^3} \right\}^{\frac13}$,
• $X_{+} X_{-} = \frac{b^2-3c}{9}$,
• $X_{+}^2 + X_{-}^2 = - \frac{27 d -9bc+2b^3}{54}$.

More:

### Special case: reduced cubic equation (Cardano’s formula)

Change of variables $x = y -\frac{b}3$ reduces the general third-order algebraic equation $x^3 + bx^2+cx+d=0$ to the reduced form $y^3+p y + q = 0$. This one can be solved using the Cardan’s ansatz

• $y = \sqrt[3]{A+B} + \sqrt[3]{A-B}$ for some not yet determined $A$ and $B$.
• This gives: $y^3 = 2A+ 3y \sqrt[3]{A^2-B^2}$
• Comparing coefficients with the $y^3 = -p y - q$ gives $2A=q$ and $3 \sqrt[3]{A^2-B^2} = p$
• Their solution is: $A=\frac{q}2$, $B = \sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}3\right)^3}$.

So, finally, one root of the reduced equation $y^3+p y + q = 0$ is (the Cardano’s formula):

$y_1 = \sqrt[3]{\frac{q}2+\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}3\right)^3}} + \sqrt[3]{\frac{q}2-\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}3\right)^3}}$

The remaining two roots are solutions of the quadratic equation $y^2 + y_1 y + y_1^2+p=0$:

$y_{2,3} = \frac12(-y_1 \pm \sqrt{-3 y^2_1 -4 p})$.

## Quartic equation

### Special case: $y^4+y=x$

The solution is $y = \frac12 \epsilon \sqrt{A} + \frac12 \epsilon' \sqrt{B}$ for $\epsilon, \epsilon' = \pm 1$, where $B = -A - \frac{2\epsilon}{\sqrt{A}}$, and $A$ is any of three roots of the cubic equation $A^3+4xA-1=0$. For example, we may take the Cardan’s solution: $A_1 = \sqrt[3]{a+b} + \sqrt[3]{a-b}$ where $a=\frac12$ and $b = \frac1{2\sqrt{3}}\sqrt{27+256 x^3}$. Putting all elements together,

$y = \frac12 \epsilon \sqrt{\sqrt[3]{\frac12+\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} + \sqrt[3]{\frac12-\frac1{2\sqrt{3}}\sqrt{27+256 x^3}}} + \frac12 \epsilon' \sqrt{- \sqrt[3]{\frac12+\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} - \sqrt[3]{\frac12-\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} -\frac{2\epsilon}{\sqrt{\sqrt[3]{\frac12+\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} + \sqrt[3]{\frac12-\frac1{2\sqrt{3}}\sqrt{27+256 x^3}}}}}$

Derivation: Start from the ansatz $y = \frac12 \epsilon \sqrt{A} + \frac12 \epsilon' \sqrt{B}$ for yet unknown A & B (Q: How? Why? ;-). Then

• $y^2 = \frac14 (A+B) + \frac{\epsilon\epsilon'}2 \sqrt{AB}$,
• $y^4 = \frac1{16} (A+B)^2 + \frac{\epsilon\epsilon'}4 (A+B) \sqrt{AB} + \frac14 AB$,
• Require that $A+B = \frac{\alpha}{\sqrt{A}}$ (Again: How? Why?)
• Then $y^4 = \frac{\alpha^2}{16 A} -\frac{A^2}4 + \frac{\epsilon\epsilon'\alpha}4 \sqrt{B} + \frac{\alpha}4 \sqrt{A}$,
• and $y^4 + y = \frac{\alpha^2}{16 A} -\frac{A^2}4 + \left(\frac{\epsilon\epsilon'\alpha}4+\frac{\epsilon'}2\right) \sqrt{B} + \left(\frac{\alpha}4 +\frac{\epsilon}2\right)\sqrt{A}$.
• Setting $\alpha=-2\epsilon$ simplifies that equation by eliminating roots: $y^4 + y = \frac{\alpha^2}{16 A} -\frac{A^2}4 = x$
• which is equivalent to the cubic equation $A^3+4xA-1=0$ solvable via Cardano’s procedure.

Example: A simple (and trivial) example is for $x=0$, when $y^4+y=0$ has roots $\{y_1=0\} \cup \{y_{n}=e^{i\frac{\pi}3(n-1)} | n=2,3,4\}$. Here $b=a=\frac12$, $A_1=1$, $B_1=-1-2\epsilon$ and $y_{\epsilon,\epsilon'}(A_1) = \frac{\epsilon+\epsilon' \sqrt{-1-2\epsilon}}2$, i.e.

• $y_{+,+} = \frac{1+i\sqrt{3}}2 \equiv y_2$
• $y_{+,-} = \frac{1-i\sqrt{3}}2 \equiv y_3$
• $y_{-,+} = 0 \equiv y_1$
• $y_{-,-} = -1 \equiv y_4$

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