Polynomials | Electrical Energy Technology

1. Introduction to Polynomials

Polynomials are expressions consisting of variables and coefficients combined using only addition, subtraction, multiplication, and non-negative integer exponents. They serve as the basis for approximating complex functions and modeling diverse engineering phenomena.

In engineering applications, polynomials are used to:

Describe system responses and transfer functions
Model physical behaviors and relationships
Approximate more complex functions (as in Taylor polynomials)
Analyze circuit characteristics in the s-domain
Describe control system stability and performance

Graph of a cubic polynomial — Graph of a third-degree (cubic) polynomial

2. Definition & Basics

2.1 Formal Definition

A polynomial in a variable \(z\) is an expression of the form:

\[ p(z) = a_n z^n + a_{n-1} z^{n-1} + \ldots + a_1 z + a_0 \]

Where:

\(a_n, a_{n-1}, \ldots, a_1, a_0\) are constants called coefficients
\(n\) is a non-negative integer representing the degree of the polynomial
\(a_n \neq 0\) for a polynomial of degree \(n\)

Example: Polynomial Identification

Which of the following expressions are polynomials?

\(z^2 + 1\) ✓ (Polynomial of degree 2)

\(z^3 + 1 + z^{-1}\) ✕ (Not a polynomial due to the negative exponent \(z^{-1}\))

\(i\) ✓ (Constant polynomial of degree 0)

\(z^{2.5} + z + 1\) ✕ (Not a polynomial due to the non-integer exponent)

\((z+1)^2\) ✓ (Polynomial of degree 2 when expanded)

2.2 Expanding Polynomials

Polynomials are often presented in factored form or with expressions that need to be expanded. Here's how to expand some common expressions:

Example: Expanding Polynomials

\((3z + 5)^2 = (3z)^2 + 2(3z)(5) + 5^2 = 9z^2 + 30z + 25\)

\((3z + 2)(3z - 2) = (3z)^2 - 2^2 = 9z^2 - 4\)

\((z + 5)(z + 6) = z(z + 6) + 5(z + 6) = z^2 + 6z + 5z + 30 = z^2 + 11z + 30\)

\((z - 1)(z^2 + z + 1) = z(z^2 + z + 1) - (z^2 + z + 1) = z^3 + z^2 + z - z^2 - z - 1 = z^3 - 1\)

2.3 Polynomial Operations

Basic operations with polynomials:

Addition & Subtraction

Combine like terms by adding or subtracting coefficients of terms with the same power.

Example:

\((3z^2 + 2z - 5) + (z^2 - 3z + 4) = 4z^2 - z - 1\)

\((2z^3 - 4z + 7) - (z^3 + 2z^2 - 5) = z^3 - 2z^2 - 4z + 12\)

Multiplication

Multiply each term of the first polynomial with each term of the second, then combine like terms.

Example:

\((2z + 3)(z - 4) = 2z(z - 4) + 3(z - 4)\)

\(= 2z^2 - 8z + 3z - 12 = 2z^2 - 5z - 12\)

Polynomial Division

Division of polynomials works similarly to long division with numbers. The division algorithm can be expressed as:

\[ p(z) = d(z)q(z) + r(z) \]

Where \(p(z)\) is the dividend, \(d(z)\) is the divisor, \(q(z)\) is the quotient, and \(r(z)\) is the remainder. The degree of \(r(z)\) is less than the degree of \(d(z)\).

Example: Long Division

Divide \(3z^2 + 2z - 5\) by \(z - 1\)

        3z + 5
    ________
z-1 ) 3z² + 2z - 5
      3z² - 3z
      ________
         5z - 5
         5z - 5
         ________
           0

Therefore: \(3z^2 + 2z - 5 = (z - 1)(3z + 5) + 0\)

Since the remainder is 0, \(z - 1\) is a factor of \(3z^2 + 2z - 5\)

3. Polynomial Roots

3.1 Definition of a Root

A root (or zero) of a polynomial \(p(z)\) is a value of \(z\) that makes the polynomial equal to zero.

\[ \text{If } p(\rho) = 0 \text{, then } \rho \text{ is a root of } p(z) \]

3.2 Fundamental Theorem of Algebra

Fundamental Theorem of Algebra:

Every non-constant polynomial with complex coefficients has at least one complex root.

This means that a polynomial of degree \(n\) has exactly \(n\) roots when counted with their multiplicities.

3.3 Connection Between Roots and Factors

A fundamental relationship exists between the roots of a polynomial and its factors:

\[ \rho \text{ is a root of } p(z) \text{ if and only if } (z - \rho) \text{ is a factor of } p(z) \]

Example:

Let's verify that \(z = 2\) is a root of \(p(z) = z^3 - 6z^2 + 12z - 8\):

\(p(2) = 2^3 - 6 \cdot 2^2 + 12 \cdot 2 - 8 = 8 - 24 + 24 - 8 = 0\)

Using polynomial division, we find:

\(p(z) = (z - 2)(z^2 - 4z + 4) = (z - 2)^3\)

This means \(z = 2\) is a root with multiplicity 3.

3.4 Multiplicity of Roots

If \((z - \rho)^m\) is a factor of \(p(z)\), but \((z - \rho)^{m+1}\) is not, then \(\rho\) is a root of \(p(z)\) with multiplicity \(m\).

Example:

For the polynomial \(p(z) = (z - 3)^2(z + 1)(z - 5)\):

\(z = 3\) is a root with multiplicity 2
\(z = -1\) is a root with multiplicity 1
\(z = 5\) is a root with multiplicity 1

A polynomial of degree \(n\) has exactly \(n\) roots when counted with their multiplicities. Here, the polynomial is of degree 4, and we have 4 roots counting multiplicities: 3 (twice), -1, and 5.

4. Second-Degree Polynomials

4.1 Quadratic Formula

For a second-degree (quadratic) polynomial \(p(z) = az^2 + bz + c\) with \(a \neq 0\), the roots are given by:

\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

4.2 The Discriminant

The expression \(D = b^2 - 4ac\) is called the discriminant of the quadratic polynomial. It determines the nature of the roots:

If \(D > 0\): Two different real roots
If \(D = 0\): One real root (a double root)
If \(D < 0\): Two complex conjugate roots

Example: Solving Quadratic Equations

Find the roots of \(2z^2 - 4z + 10\)

Calculate the discriminant: \(D = (-4)^2 - 4 \cdot 2 \cdot 10 = 16 - 80 = -64\)

Since \(D < 0\), there are two complex conjugate roots:

\(z = \frac{4 \pm \sqrt{-64}}{2 \cdot 2} = \frac{4 \pm 8i}{4} = 1 \pm 2i\)

The roots are \(z_1 = 1 + 2i\) and \(z_2 = 1 - 2i\)

5. Factorization of Polynomials

5.1 Complete Factorization

Every polynomial of degree \(n\) can be factored as:

\[ p(z) = a_n(z - \rho_1)(z - \rho_2)\cdots(z - \rho_n) \]

Where \(\rho_1, \rho_2, \ldots, \rho_n\) are the roots of the polynomial, with repetitions allowed for multiple roots, and \(a_n\) is the leading coefficient.

5.2 The Division Algorithm

The polynomial division algorithm is a systematic way to determine if a polynomial divides another polynomial with no remainder. If \(d(z)\) is a divisor of \(p(z)\), then:

\[ p(z) = d(z)q(z) \]

This is essential for finding factors of polynomials.

Example: Using the Division Algorithm

Check if \(z + 3\) is a factor of \(2z^2 + 3z - 9\)

Performing the division:

        2z - 3
    ________
z+3 ) 2z² + 3z - 9
      2z² + 6z
      ________
        -3z - 9
        -3z - 9
        ________
           0

Since the remainder is 0, \(z + 3\) is a factor of \(2z^2 + 3z - 9\)

Thus, \(2z^2 + 3z - 9 = (z + 3)(2z - 3)\)

5.3 Synthetic Division

Synthetic division is a shorthand method for dividing a polynomial by a linear factor of the form \(z - r\):

Example: Synthetic Division

Divide \(z^3 - 2z^2 - 4z + 8\) by \(z - 2\) using synthetic division

  2 | 1  -2  -4   8
    |    2   0  -8
    ----------------
      1   0  -4   0

The last row gives the coefficients of the quotient (1, 0, -4) and the remainder (0)

Thus, \(z^3 - 2z^2 - 4z + 8 = (z - 2)(z^2 + 0z - 4) + 0\)

Simplifying: \(z^3 - 2z^2 - 4z + 8 = (z - 2)(z^2 - 4)\)

Further factoring: \(z^3 - 2z^2 - 4z + 8 = (z - 2)(z - 2)(z + 2) = (z - 2)^2(z + 2)\)

6. Polynomials with Real Coefficients

6.1 Complex Conjugate Pairs

For polynomials with real coefficients, an important property holds:

Key Property:

If \(\rho\) is a complex root of a polynomial with real coefficients, then its complex conjugate \(\overline{\rho}\) is also a root.

This means that non-real roots always come in complex conjugate pairs for polynomials with real coefficients.

Example:

The polynomial \(p(z) = z^2 - 4z + 13\) has roots \(z_1 = 2 + 3i\) and \(z_2 = 2 - 3i\)

We can verify these are complex conjugates, and the polynomial can be factored as:

\(p(z) = (z - (2 + 3i))(z - (2 - 3i)) = (z - 2 - 3i)(z - 2 + 3i)\)

6.2 Factorization with Real Coefficients

Every polynomial with real coefficients can be factored into a product of:

Linear factors \((z - a)\) where \(a\) is a real root
Quadratic factors \((z^2 + bz + c)\) where \(b^2 - 4c < 0\), corresponding to pairs of complex conjugate roots

Example: Factoring with Real Coefficients

Let \(p(z) = z^4 + 5z^2 + 4\)

This is a biquadratic polynomial (only even powers). Substituting \(u = z^2\):

\(p(u) = u^2 + 5u + 4 = (u + 1)(u + 4)\)

Substituting back \(z^2 = u\):

\(p(z) = (z^2 + 1)(z^2 + 4)\)

The quadratic factor \(z^2 + 1\) corresponds to the complex conjugate roots \(±i\)

The quadratic factor \(z^2 + 4\) corresponds to the complex conjugate roots \(±2i\)

7. Third-Degree Polynomials

7.1 Existence of a Real Root

A key property of third-degree polynomials with real coefficients:

Key Property:

Every cubic polynomial with real coefficients has at least one real root.

7.2 Finding Roots of Cubic Polynomials

A general approach to solving cubic polynomials:

Graph the polynomial to identify potential real roots
Test integer values that might be roots using the Rational Root Theorem
Once a real root \(\rho\) is found, use polynomial division to obtain \(p(z) = (z - \rho)q(z)\)
Solve the resulting quadratic polynomial \(q(z)\) using the quadratic formula

Example: Solving a Cubic Equation

Find all roots of \(p(z) = z^3 - 3z^2 - 7z + 33\)

By graphing or testing, we find that \(z = -3\) is a root:

\(p(-3) = (-3)^3 - 3(-3)^2 - 7(-3) + 33 = -27 - 3(9) + 21 + 33 = -27 - 27 + 21 + 33 = 0\)

Using polynomial division:

\(z^3 - 3z^2 - 7z + 33 = (z + 3)(z^2 - 6z + 11)\)

Solving the quadratic factor using the quadratic formula:

\(z = \frac{6 ± \sqrt{36 - 44}}{2} = \frac{6 ± \sqrt{-8}}{2} = \frac{6 ± 2\sqrt{2}i}{2} = 3 ± \sqrt{2}i\)

All roots are: \(z = -3\), \(z = 3 + \sqrt{2}i\), and \(z = 3 - \sqrt{2}i\)

8. Taylor Polynomials

8.1 Approximation by Linearization

The simplest form of approximating a function with a polynomial is linearization, which uses the tangent line at a point:

\[ \tilde{f}_1(t) = f(t_0) + f'(t_0)(t - t_0) \]

This is a first-order approximation valid near \(t_0\).

Example: Linearization

Find the linearization of \(f(t) = e^t\) at \(t_0 = 0\)

\(f(0) = e^0 = 1\) and \(f'(t) = e^t\) gives \(f'(0) = 1\)

The linearization is \(\tilde{f}_1(t) = 1 + 1(t - 0) = 1 + t\)

8.2 Taylor Polynomials

Taylor polynomials extend the idea of linearization to higher orders:

\[ \tilde{f}_n(t) = f(t_0) + f'(t_0)(t - t_0) + \frac{f''(t_0)}{2!}(t - t_0)^2 + \cdots + \frac{f^{(n)}(t_0)}{n!}(t - t_0)^n \]

Where \(f^{(n)}(t_0)\) is the nth derivative of \(f\) evaluated at \(t_0\).

Example: Taylor Polynomial

Find the 3rd order Taylor polynomial for \(f(t) = \ln(t)\) at \(t_0 = 1\)

Computing the derivatives:

\(f(t) = \ln(t) \Rightarrow f(1) = 0\)

\(f'(t) = \frac{1}{t} \Rightarrow f'(1) = 1\)

\(f''(t) = -\frac{1}{t^2} \Rightarrow f''(1) = -1\)

\(f'''(t) = \frac{2}{t^3} \Rightarrow f'''(1) = 2\)

The 3rd order Taylor polynomial is:

\(\tilde{f}_3(t) = 0 + 1(t-1) - \frac{1}{2}(t-1)^2 + \frac{2}{6}(t-1)^3 = (t-1) - \frac{(t-1)^2}{2} + \frac{(t-1)^3}{3}\)

8.3 Common Taylor Series

Some important Taylor polynomials centered at \(t = 0\) (also called Maclaurin polynomials):

Function	Taylor Polynomial (at \(t = 0\))
\(e^t\)	\(1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots + \frac{t^n}{n!}\)
\(\frac{1}{1+t}\)	\(1 - t + t^2 - t^3 + \cdots + (-1)^n t^n\)
\(\cos(t)\)	\(1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \frac{t^6}{6!} + \cdots\)
\(\sin(t)\)	\(t - \frac{t^3}{3!} + \frac{t^5}{5!} - \frac{t^7}{7!} + \cdots\)

9. Engineering Applications

9.1 Transfer Functions

In control systems and signal processing, transfer functions are often expressed as ratios of polynomials in the complex variable \(s\):

\[ H(s) = \frac{b_m s^m + b_{m-1} s^{m-1} + \ldots + b_1 s + b_0}{a_n s^n + a_{n-1} s^{n-1} + \ldots + a_1 s + a_0} \]

The roots of the denominator polynomial (called poles) determine the stability and transient response of the system, while the roots of the numerator polynomial (called zeros) affect the steady-state behavior.

Example: Low-Pass Filter

A first-order low-pass filter has the transfer function:

\[ H(s) = \frac{K}{s + \omega_c} \]

where \(K\) is the DC gain and \(\omega_c\) is the cutoff frequency.

The pole at \(s = -\omega_c\) determines the filter's time constant \(\tau = 1/\omega_c\).

9.2 Electrical Circuit Analysis

The impedance of electrical circuits often involves polynomials:

Example: RLC Circuit

The impedance of a series RLC circuit is:

\[ Z(s) = Ls + R + \frac{1}{Cs} = \frac{Ls^2 + Rs + \frac{1}{C}}{s} \]

The natural frequencies of the circuit are determined by the roots of the numerator polynomial \(Ls^2 + Rs + \frac{1}{C}\).

9.3 Differential Equations

Linear differential equations with constant coefficients lead to characteristic polynomials:

\[ a_n \frac{d^n y}{dt^n} + a_{n-1} \frac{d^{n-1} y}{dt^{n-1}} + \ldots + a_1 \frac{dy}{dt} + a_0 y = f(t) \]

The characteristic polynomial is:

\[ p(\lambda) = a_n \lambda^n + a_{n-1} \lambda^{n-1} + \ldots + a_1 \lambda + a_0 \]

The roots of this polynomial determine the form of the solution.

Example: Second-Order System

For the differential equation \(\frac{d^2 y}{dt^2} + 2\zeta \omega_n \frac{dy}{dt} + \omega_n^2 y = 0\)

The characteristic polynomial is \(\lambda^2 + 2\zeta \omega_n \lambda + \omega_n^2\)

Its roots are \(\lambda = -\zeta \omega_n \pm \omega_n\sqrt{\zeta^2 - 1}\)

For an underdamped system (\(0 < \zeta < 1\)), the roots are complex conjugates, leading to oscillatory behavior.

10. Problem-Solving Strategies

10.1 Finding Roots

Strategies for finding polynomial roots:

Rational Root Theorem: If \(p(z) = a_n z^n + \ldots + a_0\) with integer coefficients, then any rational root \(\frac{p}{q}\) must have \(p\) dividing \(a_0\) and \(q\) dividing \(a_n\)
Factoring by Grouping: Rearrange terms to identify common factors
Numerical Methods: Use algorithms like Newton-Raphson for approximating roots
Graphical Analysis: Sketch the polynomial to identify potential root locations

10.2 Working with Complex Roots

When dealing with complex roots:

Remember that complex roots of polynomials with real coefficients come in conjugate pairs
A pair of complex conjugate roots \(a \pm bi\) corresponds to a quadratic factor \(z^2 - 2az + (a^2 + b^2)\)
Use the quadratic formula to identify complex roots from quadratic factors

Example Problem: Complete Factorization

Find all roots and completely factorize \(p(z) = z^4 - z^3 - 7z^2 + z + 6\)

Step 1: Try some potential rational roots using the Rational Root Theorem

Potential rational roots are divisors of 6: \(±1, ±2, ±3, ±6\)

Testing \(z = 1\): \(p(1) = 1 - 1 - 7 + 1 + 6 = 0\) ✓

Testing \(z = -1\): \(p(-1) = 1 + 1 - 7 - 1 + 6 = 0\) ✓

Step 2: Use polynomial division to factorize

\(p(z) = (z - 1)(z + 1)(z^2 - z - 6)\)

Step 3: Factorize the quadratic term

\(z^2 - z - 6 = (z - 3)(z + 2)\)

Final factorization: \(p(z) = (z - 1)(z + 1)(z - 3)(z + 2)\)

Roots: \(z = 1, -1, 3, -2\)

Glossary of Terms

Polynomial: An expression consisting of variables and coefficients using only addition, subtraction, multiplication, and non-negative integer exponents
Degree: The highest power of the variable in a polynomial
Coefficient: The numerical factor in a term of a polynomial
Root (or Zero): A value of the variable that makes the polynomial equal to zero
Multiplicity: The number of times a particular root appears in the factorization of a polynomial
Discriminant: For a quadratic polynomial \(az^2 + bz + c\), the value \(b^2 - 4ac\) that determines the nature of the roots
Factor: A polynomial that divides another polynomial with zero remainder
Division Algorithm: A method for dividing one polynomial by another to obtain a quotient and remainder
Taylor Polynomial: A polynomial approximation of a function based on its derivatives at a specific point
Transfer Function: A mathematical representation of the relationship between input and output of a system, often expressed as a ratio of polynomials

Conclusion

Polynomials are powerful mathematical tools with applications across various engineering disciplines. From representing transfer functions in control systems to approximating complex behaviors through Taylor polynomials, they provide a foundation for much of mathematical modeling in engineering.

Understanding the properties of polynomials, especially their roots and factorization methods, enables engineers to analyze system behavior, stability, and response characteristics. This knowledge is essential for designing and optimizing systems in electrical, mechanical, and other engineering fields.