Difference between revisions of "User:Temperal/The Problem Solver's Resource4"

(Third-degree and Quartic Formulas: cubic euqations)
(Third-degree and Quartic Formulas: spacing)
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The three roots <math>r, s, t</math> of a cubic polynomial equation <math>x^3 + ax^2 + bx + c = 0</math> are given implicitly by
 
The three roots <math>r, s, t</math> of a cubic polynomial equation <math>x^3 + ax^2 + bx + c = 0</math> are given implicitly by
<math>r</math> <math>=</math> <math>- \frac {a}{3} + \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}</math>
+
<math>r=- \frac {a}{3} + \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}+\left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}</math>  
    <math>{} + \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}</math>  
+
<math>s=- \frac {a}{3} - \frac {1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}</math>+ \frac { - 1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}<math>  
<math>s</math> <math>=</math> <math>- \frac {a}{3} - \frac {1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}</math>
+
</math>t=- \frac {a}{3} + \frac { - 1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3} - \frac {1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}$
    <math>{} + \frac { - 1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}</math>  
 
<math>t</math> <math>=</math> <math>- \frac {a}{3} + \frac { - 1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}</math>
 
    <math>{} - \frac {1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}</math>
 
  
 
Quartic formulas are listed [http://www.josechu.com/ecuaciones_polinomicas/cuartica_solucion.htm here].
 
Quartic formulas are listed [http://www.josechu.com/ecuaciones_polinomicas/cuartica_solucion.htm here].

Revision as of 16:25, 8 December 2007



The Problem Solver's Resource
Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing page 4.

Algebra

This is a collection of algebra laws and definitions. Obviously, there is WAY too much to cover here, but we'll try to give a good overview.

Elementary Algebra

Definitions

  • A polynomial is an expression that has exponents that are positive integer constants, and has no higher-level operations or functions.
  • A polynomial has degree $c$ if the highest exponent of a variable is $c$.
  • A quadratic equation is a polynomial of degree $2$. A quartic is of degree $4$. A quintic is of degree $5$.

Factor Theorem

Iff a polynomial $P(x)$ has roots $a,b,c,d,e,\ldots,z$, then $(x-a)(x-b)\ldots (x-z)=0$, and $(x-a),(x-b)\ldots (x-z)$ are all factors of $P(x)$.

Quadratic Formula

For a quadratic of form $ax^2+bx+c=0$, where $a,b,c$ are constants, the equation has roots $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Fundamental Theorems of Algebra

  • A polynomial of degree $n$ has at least one root, real or complex.
  • A polynomial of degree $n$ has exactly $n$ roots, real or complex.

Third-degree and Quartic Formulas

If third-degree polynomial $Q(x)=ax^3+bx^2+cx+d$ has roots $r,s,t$, then:

  • $rst=\frac{-b}{a}$
  • $r+s+t=\frac{-c}{a}$
  • $rs+st+rt=\frac{-c}{b}$

The three roots $r, s, t$ of a cubic polynomial equation $x^3 + ax^2 + bx + c = 0$ are given implicitly by $r=- \frac {a}{3} + \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}+\left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}$ $s=- \frac {a}{3} - \frac {1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}$+ \frac { - 1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}$$ (Error compiling LaTeX. Unknown error_msg)t=- \frac {a}{3} + \frac { - 1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3} - \frac {1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}$

Quartic formulas are listed here.

The general quintic equation (or an equation of even higher degree) does not have a formula.

Determinants

The determinant of a $2$ by $2$ (said to have order $2$) matrix $\left |\begin{matrix}a&b \\ c&d\end {matrix}\right|$ is $ad-bc$.

General Formula for the Determinant

Let $A$ be a square matrix of order $n$. Write $A = a_{ij}$, where $a_{ij}$ is the entry on the row $i$ and the column $j$, for $i=1,\cdots,n$ and $j=1,\cdots,n$. For any $i$ and $j$, set $A_{ij}$ (called the cofactors) to be the determinant of the square matrix of order $n-1)$ obtained from A by removing the row number i and the column number j multiplied by (-1)^{i+j}. Thus:

$\det(A) = \sum_{j=1}^{j=n} a_{ij} A_{ij}$

Cramer's Law

Consider a set of three linear equations (i.e. polynomials of degree one)

  • $ax+by+cz=d$
  • $ex+fy+gz=h$
  • $ix+jy+kz=l$

Let $D=\left|\begin{matrix}a&e&i\\b&f&j\\c&g&k\end{matrix}\right|$, $D_x=\left|\begin{matrix}d&h&1\\b&f&j\\c&g&k\end{matrix}\right|$, $D_y=\left|\begin{matrix}a&e&i\\d&h&l\\c&g&k\end{matrix}\right|$, $D_x=\left|\begin{matrix}a&e&i\\b&f&j\\d&h&l\end{matrix}\right|$ $x = \frac{D_x}{D}$, $y = \frac{D_y}{D}$, and $z = \frac{D_z}{D}$. This can be generalized to any number of linear equations.

Abstract Algebra

Incomplete.

Diophantine Equations

Incomplete.

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