5. Polynomial Functions and Equations

Transcription

1 5. Polynomial Functions and Equations 1. Polynomial equations and roots. Solving polynomial equations in the chemical context 3. Solving equations of multiple unknowns

2 5.1. Polynomial equations and roots Consider the polynomial function y = f(x). o For every value of x within the definition domain of f, there exists one (and only one) associated value of y. o Conversely, for any given value y 0 R, there might be 0, 1 or more associated values of x. If such an x exists, it is the solution (or root) of the equation y 0 = f(x). o Note that: (x) y f(x) y 0 g(x) 0. f 0 0 Solving the polynomial equation f(x) = 0 corresponds to finding the values of x that cross the x-axis (y = 0) in a graphical representation of the function y = f(x). Consider a polynomial function of degree n: f (x) a i x. n i0 o If x is a root of the polynomial equation, then f( ) 0, and ( x ) is a factor of the polynomial equation: f(x) (x ) o The expression c i0 n-1 with other coefficients! n 1 i i x n 1 i0 c i x i is another polynomial of degree i

3 Polynomial of 1st degree: f(x) ax b o Graphically, this represents a straight line of slope a and ordinate at the origin b. o Solution of the equation f(x) = 0: b b ax b ax 0 x with a 0 a a Polynomial of nd degree: f(x) ax bx c 0 o From graph, a polynomial of degree can have two, one, or zero roots. o By trial and error one might be able to factorize: roots 1 and : (x) (x )(x ) 0 f 1 1 root : f(x) (x )(x ) (x ) 0 0 roots: can t factorize. o If factorization is difficult, one uses the formula for the roots of a quadratic equation ax bx c 0: x b b a 4ac

4 o b 4ac is known as the discriminant: If D > 0, two different roots exist, If D = 0, one root exists (= two identical roots), If D < 0, no real roots (roots are complex numbers). o Example: f(x) x 4x 3: Factorizing yields: f(x) (x 3)(x 1), hence the roots are: x = 3 and x = 1. Using the formula: x 1. n1 Polynomial of degree n: f(x) anx an1x... a0 have 0, 1,,, or n roots at most. n can 5.. Solving polynomial equations in a chemical context In practice, the solution (roots) of polynomial equations is problematic: o Try to factorize by trial and error, and establish the new coefficients of the polynomial of degree n-1: n1 n f(x) (x 1)(cn1x cnx... c0) o Reiterate the procedure for the new polynomial of degree n-1, and so on o If not possible, then use a graphical representation of the remaining polynomial (and deduce the values of f(x) crossing the x-axis (that is, f(x) = 0). o Alternatively, one can use computer algebra software (Maple, Mathematica).

5 Polynomial equations of higher degree arise in many areas of physical chemistry: o Electronic structures, through the determination of molecular orbitals, constructed as linear combinations of atomic orbitals (LCAO): the simplest -type molecular orbital for HCN (in its ground state) involves seven atomic orbitals 1s H, 1s C, 1s N, s C, s N, p C and p N leading to a polynomial of degree 7 for the molecular orbital energies. o Characteristic frequencies of molecular vibrations: in the case of HCN, e.g., there are four vibration frequencies that may be calculated from a polynomial of degree 4 (from appropriate assumptions) Solving equations of multiple unknowns Consider a simple 1 st order polynomial equation f(x) = 0. This equation can always be solved, because the function f(x) is function of only one variable. The equation is said to have one unknown, x. Now, consider a simple 1 st order polynomial equation f(x,y) = 0. This equation might never be solved, because the function f(x,y) is function of more than one variable. The equation is said to have two unknowns, x and y.

6 In general, solving for n unknowns requires at least a set of n equations. Consider the following set of two equations related to the two unknowns x and y (a, b, c, m, n, and p are coefficients): ax by c 0 (1) mx ny p 0 () o We can solve for the two unknowns by substitution: From (1): by c x a (3) From (): nx c y p m (4) Inserting (3) in (): by c m ny p 0 a Solving for y: mby mc any ap 0 cm ap y an bm (5) Inserting (5) in (1): bp cn x an mb Hence, x and y are determined. o If coefficients are alike in both equations, one can solve by adding or subtracting (1) and (). E.g. if b = n: ax by c 0 (1) mx by p 0 () From (1)(): p c (a m)x p c x, y =... a m