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

Revision as of 12:56, 23 November 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:

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

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. Back to page 3 | Continue to page 5

@@ Line 27: / Line 27: @@
 If third-degree polynomial <math>Q(x)=ax^3+bx^2+cx+d</math> has roots <math>r,s,t</math>, then:
-<math>rst=\frac{-b}{a}</math>
+*<math>rst=\frac{-b}{a}</math>
-<math>r+s+t=\frac{-c}{a}</math>
+*<math>r+s+t=\frac{-c}{a}</math>
-<math>rs+st+rt=\frac{-c}{b}</math>
+*<math>rs+st+rt=\frac{-c}{b}</math>
 <!-- actually, I'm not sure if this is right. Could someone check this? -->

Page

Toolbox

Search