{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Polynomials__Finite_Fields__and_Secret_S

# Polynomials__Finite_Fields__and_Secret_S - CS 70 Fall 2006...

This preview shows pages 1–3. Sign up to view the full content.

CS 70 Discrete Mathematics for CS Fall 2006 Lecture 11 Polynomials Recall from your high school math that a polynomial in a single variable is of the form p ( x ) = a d x d + a d - 1 x d - 1 + ... + a 0 . Here the variable x and the coefficients a i are usually real numbers. For example, p ( x ) = 5 x 3 + 2 x + 1, is a polynomial of degree d = 3. Its coefficients are a 3 = 5, a 2 = 0, a 1 = 2, and a 0 = 1. Polynomials have some remarkably simple, elegant and powerful properties, which we will explore in this lecture. First, a definition: we say that a is a root of the polynomial p ( x ) if p ( a ) = 0. For example, the degree 2 polynomial p ( x ) = x 2 - 4 has two roots, namely 2 and - 2, since p ( 2 ) = p ( - 2 ) = 0. If we plot the polynomial p ( x ) in the x - y plane, then the roots of the polynomial are just the places where the curve crosses the x axis: Property 1: A non-zero polynomial of degree d has at most d roots. Property 2: Given d + 1 pairs ( x 1 , y 1 ) ,..., ( x d + 1 , y d + 1 ) , there is a unique polynomial p ( x ) of degree (at most) d such that p ( x i ) = y i for 1 i d + 1. Let us consider what these two properties say in the case that d = 1. A graph of a linear (degree 1) polynomial y = a 1 x + a 0 is a line. Property 1 says that if a line is not the x -axis (i.e. if the polynomial is not y = 0), then it can intersect the x -axis in at most one point. CS 70, Fall 2006, Lecture 11 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Property 2 says that two points uniquely determine a line. Polynomial Interpolation Property 2 says that two points uniquely determine a degree 1 polynomial (a line), three points uniquely determine a degree 2 polynomial, four points uniquely determine a degree 3 polynomial, . .. Given d + 1 pairs ( x 1 , y 1 ) ,..., ( x d + 1 , y d + 1 ) , how do we determine a polynomial p ( x ) = a d x d + ... + a 1 x + a 0 such that p ( x i )= y i for i = 1 to d + 1. We will give two different efficient algorithms for reconstructing the coefficients a 0 ,..., a d and therefore the polynomial p ( x ) . In the first method, we write a system of d + 1 linear equations in d + 1 variables: the coefficients of the polynomial a 0 ,..., a d . The i - th equation is: a d x d i + a d - 1 x d - 1 i + ... + a 0 = y i . Since x i and y i are constants, this is a linear equation in the d + 1 unknowns a 0 ,..., a d . Now solving these equations gives the coefficients of the polynomial p ( x ) . For example, given the 3 pairs ( - 1 , 2 ) , ( 0 , 1 ) , and ( 2 , 5 ) , we will construct the degree 2 polynomial p ( x ) which goes through these points. The first equation says a 2 ( - 1 ) 2 + a 1 ( - 1 )+ a 0 = 2. Simplifying, we get a 2 - a 1 + a 0 = 2. Applying the same technique to the second and third equations, we get the following system of equations: a 2 - a 1 + a 0 = 2 a 0 = 1 4 a 2 + 2 a 1 + a 0 = 5 Substituting for a 0 and multiplying the first equation by 2 we get: 2 a 2 - 2 a 1 = 2 4 a 2 + 2 a 1 = 4 Then, adding down we find that 6 a 2 = 6, so a 2 = 1, and plugging back in we find that
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

Polynomials__Finite_Fields__and_Secret_S - CS 70 Fall 2006...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online