Mathematics

Algebra

(reference 2.1)

Laws


Commutative	$a+b = b+a ,\, ab = ba$
Associative	$a+(b+c) = (a+b)+c$
Distributive	$a(b+c) = ab+ac$

Identities

Exponents	Logarithms
$a^x a^y = a^{x+y}$	$\log_b{b} = 1$
$\left( ab \right) ^x = a^x b^x$	$\log_b{1} = 0$
$\left( a^x \right) y = a^xy$	$\log_b \left( MN \right) = log_b{M} + log_b{N}$
$a^{mn} = \left( a^m \right) ^n$	$\log_b \left( \frac{M}{N} \right) = \log_b{M} - \log_b{N}$
$a^0 = 1$ If $a \neq 0$	$\log_b \left( M^p \right) = p \log_b{M}$
$a^{-x} = \frac{1}{a^x}$	$\log_b \left( \frac{1}{M} \right) = -\log_b{M}$
$\frac{a^x}{a^y} = a^{x-y}$	$\log_b{ \sqrt [q]{M}} = \frac{1}{q} \log_b{M}$
$\sqrt[x]{ab} = \left( \sqrt[x]{a} \right) \left( \sqrt[x]{b} \right)$	$\log_b{M} = \left(\log_c{M} \right) \left(\log_b{c} \right)= \frac{\log_c{M}}{\log_c{b}}$
$a^{\frac{x}{y}} = \sqrt[y]{a^x} = \left( \sqrt[y]{a} \right)^x$
$a^{\frac{1}{y}} = \sqrt[y]{a}$
$\left( \sqrt[x]{a} \right) \left( \sqrt[y]{a} \right) = a^{\left( \frac{1}{x} + \frac{1}{y} \right)} = \sqrt[xy]{a^{x+y}}$
$\sqrt{a} + \sqrt{b} = \sqrt{a + b + 2\sqrt{ab}}$

Equations

Quadratic Equation

$ax^2 + bx + c =0$

Two roots, both real or both complex

$x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Cubic Equation

$y^3 + py^2 + qy + r = 0$

Three roots, all real or one real & two complex

Let $y = x - \frac{p}{3}$ to rewrite equation in form of $x^3 + ax + b = 0$

where $a = \frac{3q - p^2}{3}$ and $b = \frac{2p^3 - 9pq - 27r}{27}$

let

$A = \sqrt[3]{-\frac{b}{2} + \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}}$

and

$B =\sqrt[3]{-\frac{b}{2} - \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}}$

then

\begin{align} x_1 &= A + B\\ x_2 &= \frac{-(A + B)}{2} + \frac{\sqrt{-3}}{2} (A - B)\\ x_3 &= \frac{-(A + B)}{2} - \frac{\sqrt{-3}}{2} (A - B) \end{align}

Special cases:

If $\frac{b^2}{4} + \frac{a^3}{27} < 0$ , then the real roots are

$x_{1,2,3} = 2 \sqrt{\frac{-a}{3}} cos \left( \frac{\phi}{3} + 120° k \right)$

where $k = 0,1,2$

and

$cos\phi = + \sqrt{\frac{ \frac{b^2}{4} }{ \frac{-a^3}{27} }} \;\;\;\text{if}\; b < 0$

$cos\phi = - \sqrt{\frac{ \frac{b^2}{4} }{ \frac{-a^3}{27} }} \;\;\;\text{if}\; b > 0$

If $\frac{b^2}{4} + \frac{a^3}{27} > 0$ and $a > 0$ , the single real root is

$x = 2 \sqrt{\frac{a}{3}} cot \left( 2\phi \right)$

where $tan\phi = \sqrt[3]{tan\psi}$

and

$cot \left( 2\psi \right) = + \sqrt{\frac{ \frac{b^2}{4} }{ \frac{-a^3}{27} }} \;\;\;\text{if}\; b < 0$

$cot \left( 2\psi \right) = - \sqrt{\frac{ \frac{b^2}{4} }{ \frac{-a^3}{27} }} \;\;\;\text{if}\; b < 0$

If $\frac{b^2}{4} + \frac{a^3}{27} = 0$ , the three real roots are

$x_{1} = -2 \sqrt{\frac{-a}{3}}, \; x_{2,3} = +\sqrt{\frac{-a}{3}} \;\;\; \text{if} \; b > 0$

$x_{1} = +2 \sqrt{\frac{-a}{3}}, \; x_{2,3} = -\sqrt{\frac{-a}{3}} \;\;\; \text{if} \; b < 0$

Quartic (biquadratic) Equation

For

$y^4 + py^3 + qy^2 + ry + s = 0$

let $y = x - \frac{p}{4}$ to rewrite equation as

$x^4 + ax^2 + bx + c = 0$

let $l$ , $m$ , $n$ denote roots of the following resolvent cubic:

$t^3 + \frac{1}{2} at^2 + \frac{1}{16} \left( a^2 - 4c \right) t - \frac{1}{64}b^2 = 0$

The roots of the quartic are

$x_{1} = + \sqrt{l} + \sqrt{m} + \sqrt{n}$ $x_{2} = + \sqrt{l} - \sqrt{m} - \sqrt{n}$ $x_{3} = - \sqrt{l} + \sqrt{m} - \sqrt{n}$ $x_{4} = - \sqrt{l} - \sqrt{m} + \sqrt{n}$

Symbols Derivative, Integral & Laplace Tables