Unit 4: Theory of Equations and Inequalities

Relation between Roots and Coefficients

For a polynomial equation, the roots (α, β, γ, ...) are directly related to the coefficients (a, b, c, ...).

Quadratic Equation: ax² + bx + c = 0 (Roots: α, β)

• Sum of roots: α + β = -b/a
• Product of roots: αβ = c/a

Cubic Equation: ax³ + bx² + cx + d = 0 (Roots: α, β, γ)

• Sum of roots (one at a time): α + β + γ = -b/a
• Sum of roots (two at a time): αβ + βγ + αγ = c/a
• Product of roots: αβγ = -d/a

Biquadratic Equation: ax⁴ + bx³ + cx² + dx + e = 0 (Roots: α, β, γ, δ)

• Sum (one at a time): Σα = α + β + γ + δ = -b/a
• Sum (two at a time): Σαβ = αβ + αγ + ... = c/a
• Sum (three at a time): Σαβγ = αβγ + αβδ + ... = -d/a
• Product: αβγδ = e/a

Transformation of Equations

This involves finding a new equation whose roots are related to the roots of a given equation, without first solving for the roots.

Let the original equation be `f(x) = 0` with roots `α, β, ...`. We want a new equation in `y` with roots `g(α), g(β), ...`.

Method:

1. Set `y = g(x)`, where `x` is a root of the old equation.
2. Solve this relation for `x`. You will get `x = h(y)`.
3. Substitute this expression for `x` back into the original equation: `f(h(y)) = 0`.
4. Simplify the resulting equation in `y`. This is the new equation.

Examples of Transformations

• Roots with opposite sign (y = -α):
  Set `y = -x`, so `x = -y`. Substitute `f(-y) = 0`.
• Roots multiplied by k (y = kα):
  Set `y = kx`, so `x = y/k`. Substitute `f(y/k) = 0`.
• Roots are reciprocals (y = 1/α):
  Set `y = 1/x`, so `x = 1/y`. Substitute `f(1/y) = 0`.

Descartes' Rule of Signs

This rule gives an upper bound on the number of positive and negative real roots of a polynomial `f(x)` with real coefficients.

1. Positive Real Roots: The number of positive real roots of `f(x)` is either equal to the number of sign changes in the sequence of its coefficients, or is less than this number by an even integer (0, 2, 4, ...).
2. Negative Real Roots: The number of negative real roots of `f(x)` is either equal to the number of sign changes in the coefficients of `f(-x)`, or is less than this number by an even integer.

Example: `f(x) = x³ + x² - 17x + 15`

1. Coefficients of f(x): `+1, +1, -17, +15`
   Sign changes: `+` to `+` (no), `+` to `-` (yes, 1), `-` to `+` (yes, 2).
   Total sign changes = 2.
   Number of positive roots is either 2 or (2-2) = 0.
2. Find f(-x): `f(-x) = (-x)³ + (-x)² - 17(-x) + 15 = -x³ + x² + 17x + 15`
   Coefficients of f(-x): `-1, +1, +17, +15`
   Sign changes: `-` to `+` (yes, 1), `+` to `+` (no), `+` to `+` (no).
   Total sign changes = 1.
   Number of negative roots is exactly 1.

Conclusion: The equation has 1 negative root, and either 2 positive roots or 0 positive roots. (Since it's a cubic, the total roots must be 3. The possibilities are: 1 neg, 2 pos, 0 complex OR 1 neg, 0 pos, 2 complex).

Bisection Method

The Bisection Method is a simple numerical technique to find a real root of an equation `f(x) = 0`.

Pre-condition

You must have a continuous function `f(x)` and an interval `[a, b]` such that `f(a)` and `f(b)` have opposite signs. This guarantees (by the Intermediate Value Theorem) that at least one root lies between `a` and `b`.

Algorithm

1. Find the midpoint of the interval: `c = (a + b) / 2`.
2. Evaluate `f(c)`.
3. • If `f(c) = 0`, then `c` is the root. (You're done!)
   • If `f(a)` and `f(c)` have opposite signs, the root is in `[a, c]`. Set `b = c` and go to step 1.
   • If `f(c)` and `f(b)` have opposite signs, the root is in `[c, b]`. Set `a = c` and go to step 1.
4. Repeat until the interval `[a, b]` is sufficiently small (i.e., the desired accuracy is reached).

Binomial Equations

A binomial equation is an equation of the form xⁿ - A = 0 or xⁿ = A.

To solve this, we use the polar form of complex numbers and De Moivre's Theorem. This is the same method used to find the n-th roots of unity, but for a general complex number A.

Method

1. Write A in its general polar form: `A = r(cos(θ + 2kπ) + i sin(θ + 2kπ))`.
2. Take the n-th root of both sides: `x = A¹/n`.
3. `x = [r(cos(θ + 2kπ) + i sin(θ + 2kπ))]¹/n`
4. Apply De Moivre's Theorem: x = r¹/n [ cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n) ]
5. Find the `n` distinct roots by using `k = 0, 1, 2, ..., n-1`.

Inequalities

Arithmetic Mean - Geometric Mean (AM-GM)

For any set of `n` non-negative real numbers `a₁, a₂, ..., aₙ`:

AM-GM Inequality: (a₁ + a₂ + ... + aₙ) / n ≥ (a₁a₂...aₙ)¹/n
(The Arithmetic Mean is always greater than or equal to the Geometric Mean).

Equality holds if and only if all the numbers are equal (`a₁ = a₂ = ... = aₙ`).

Cauchy-Schwarz Inequality

For any two sequences of real numbers `(a₁, ..., aₙ)` and `(b₁, ..., bₙ)`:

(a₁b₁ + a₂b₂ + ... + aₙbₙ)² ≤ (a₁² + a₂² + ... + aₙ²)(b₁² + b₂² + ... + bₙ²)
Or, in summation notation: (Σ aᵢbᵢ)² ≤ (Σ aᵢ²)(Σ bᵢ²)

Equality holds if the two sequences are proportional (i.e., `aᵢ = k * bᵢ` for all i).

Chebyshev's Inequality (not explicitly on syllabus, but related)

This relates to the ordering of sequences. If `a₁ ≤ a₂ ≤ ... ≤ aₙ` and `b₁ ≤ b₂ ≤ ... ≤ bₙ` (both sequences are similarly ordered), then:

n(Σ aᵢbᵢ) ≥ (Σ aᵢ)(Σ bᵢ)

If one is increasing and one is decreasing (oppositely ordered), the inequality sign flips.