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Numerical Solutions of Equations

Overview of the Programme

  • Focus on analytic and numerical methods for solving the equation in a single variable:
    f(x)=0f(x) = 0.
  • Understanding functional relationships in ordered pairs instead of algebraic expressions.
  • Introduction to interpolation methods for estimating values of f(x)f(x) at intermediate points.

Fundamental Theorem of Algebra

Statement of the Theorem
  • The Fundamental Theorem of Algebra states that:
    • Every polynomial expression of degree nn can be factored into nn linear factors:
      f(x)=anxn+an1xn1+ext+a1x+a0f(x) = a_n x^n + a_{n−1} x^{n−1} + ext{…} + a_1 x + a_0
      can be expressed as:
      f(x)=an(xr1)(xr2)(xrn)f(x) = a_n (x - r_1)(x - r_2)…(x - r_n)
      where r1,r2,,rnr_1, r_2, …, r_n are the roots of the polynomial.
    • There are nn values of xx such that f(x)=0f(x) = 0, which are termed the roots.
    • Roots can be real, imaginary, or complex, and they may or may not be distinct.
Examples of the Theorem
  1. For the quadratic equation:
    x2+5x+6=0x^2 + 5x + 6 = 0

    • Factored as:
      (x+2)(x+3)=0(x + 2)(x + 3) = 0
    • Distinct roots found: x=2,3x = -2, -3.
  2. For the quadratic:
    x24x+4=0x^2 - 4x + 4 = 0

    • Factored as:
      (x2)(x2)=0(x - 2)(x - 2) = 0
    • Coincident roots found: x=2x = 2.
  3. For the quadratic:
    x2+x+1=0x^2 + x + 1 = 0

    • Factored as:
      (x+a)(x+b)=0(x + a)(x + b) = 0
    • Roots are found as: x=a,bx = -a, -b.
Formula for Quadratic Equation Solutions
  • The solution to the quadratic equation:
    ax2+bx+c=0ax^2 + bx + c = 0 is given by:
    x = rac{-b ext{±} rac{ ext{√}(b^2 - 4ac)}{2a}}.

Relationship Between Coefficients and Roots

General Form
  • Consider a cubic polynomial:
    x3+px2+qx+r=0x^3 + px^2 + qx + r = 0
    Let a,b,ca, b, c be the roots:
    x3+px2+qx+r=(xa)(xb)(xc)x^3 + px^2 + qx + r = (x - a)(x - b)(x - c)
    Expanding gives:

    x3(a+b+c)x2+(ab+ac+bc)xabcx^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc.
    This leads to the following relationships:

    • Sum of roots: a+b+c=pa + b + c = -p
    • Sum of products of roots (two at a time): ab+ac+bc=qab + ac + bc = q
    • Product of roots: abc=rabc = -r.
Example Relationship Calculation
  1. Given roots a,b,ca, b, c of the polynomial:
    • For:
      f(x)=3x3+2x2+5x+7=0f(x) = 3x^3 + 2x^2 + 5x + 7 = 0
    • Roots: a+b+c=23a + b + c = -\frac{2}{3}
    • Products: ab+ac+bcab + ac + bc
    • Product: abc=73abc = -\frac{7}{3}.

Example Problems

Example 1 - Solving a Cubic Equation
  • Equation: x3+8x2+9x+18=0x^3 + 8x^2 + 9x + 18 = 0 with a+b=5a + b = 5
    • Derived:
      (a) a+b+c=8a + b + c = 8
      (b) ab+ac+bc=9ab + ac + bc = 9
      (c) abc=18abc = -18
  • Solving shows roots as: x=1,3,6x = -1, 3, 6.
Example 2 - Roots in Arithmetic Sequence
  • Equation: 2x3+3x211x6=02x^3 + 3x^2 - 11x - 6 = 0.
    • Roots represented as: (ak),a,(a+k)(a - k), a, (a + k).
    • Sum = 3a3a, and product of roots = a(ak)(a+k)a(a - k)(a + k) equals to the constant term.
Example 3 - Roots in Geometric Sequence
  • Equation: x3+3x26x8=0x^3 + 3x^2 - 6x - 8 = 0
    • Roots represented as: a,ak,ak2a, ak, ak^2
    • Solving gives roots as: 4,2,1−4, 2, −1.

Cubic Equations

General Form Description
  • Each cubic polynomial can be expressed as
    f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d
    and factored into three linear factors.
  • Cubic equations may have real or complex roots, and complex roots appear in conjugate pairs.
Finding Real Roots
  • Utilize Tartaglia's method for solving cubic equations defined as:
    x3+ax+b=0x^3 + ax + b = 0
    with the formula:
    x=bext±ext(b2+27a3)3ax = \frac{-b ext{±} ext{√}(b^2 + 27a^3)}{3a}.
Example of Tartaglia's Method
  1. For the cubic: x3+2x+5=0x^3 + 2x + 5 = 0.
    • Solve for the real root using Tartaglia's method and predict the convergence of roots by substituting the values iteratively.

Numerical Methods

Bisection Method
  • A numerical method for finding f(x)=0f(x) = 0 involves:
    1. Identifying x=ax = a such that f(a) < 0.
    2. Identifying x=bx = b such that f(b) > 0.
    • The root lies between aa and bb.
Example Problem Bisection
  • Find positive xx that satisfies x22=0x^2 - 2 = 0 by evaluating midpoints iteratively.
Iteration Methods in Excel
  • Set up iterative processes using Excel to determine roots more efficiently than manual calculations.

Newton-Raphson Iterative Method

Overview
  • The Newton-Raphson method provides a means of improving approximations iteratively to find a root of f(x)=0f(x) = 0, formulated as:
    xn+1=xnf(xn)f(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.

Summary of Study Problems

  1. Identify quadratic equations with complex roots and determine corresponding quadratic formulas.
  2. Solve cubic equations based on known root properties.
  3. Find exact values of roots that establish polynomial divisibility.
  4. Utilize the methods of bisection and Newton-Raphson to solve polynomial equations, ensuring high accuracy.