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.


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2010.02.25

Information disclosure sites

Gaining access to the supressed information:


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

Public gateways

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.


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2010.02.18

Unix links

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..

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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

2010.02.12

Physics sites

Physics seminars

Collections of historic papers


Physics blogs

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

Astronomy and Astrophysics

Related here:

Complexity (Integrability, Exact Solvability, Chaos, Small Worlds, Networking, etc)

Various

Related information on this blog

Roots of algebraic equations

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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