Engineering mathematics is fundamental in electronics engineering, enabling precise analysis and design of circuits and systems. Power series provide a powerful tool to represent complex functions as infinite sums of polynomial terms, facilitating approximation and solution of differential equations in signal processing, control systems, and semiconductor device modeling. Mastery of power series equips electronics engineers with analytical capabilities essential for advanced problem solving and innovation in Kenya’s dynamic technology sector.

1.1 Meaning of the Term Power Series

Power series are infinite sums of terms involving powers of a variable, typically used to express functions as polynomials of increasing degree. They are central in electronics engineering for approximating nonlinear functions and solving differential equations encountered in circuit analysis and signal processing. The general form of a power series centered at \(a\) is:

$$ \sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1 (x - a) + c_2 (x - a)^2 + c_3 (x - a)^3 + \cdots $$

where \(c_n\) are coefficients and \(x\) is the variable.

1.1.1 Definition and General Form

Power series represent functions as infinite sums of powers of a variable, allowing approximation near a specific point \(a\). The coefficients \(c_n\) determine the contribution of each term.

$$ f(x) = \sum_{n=0}^{\infty} c_n (x - a)^n $$

This series converges for values of \(x\) within a radius of convergence, beyond which it diverges.

Worked Examples

Example 1: Approximate a function near \(a=0\) using a power series with given coefficients.

Given: \(c_0 = 1\), \(c_1 = 2\), \(c_2 = 3\), approximate \(f(x)\) at \(x=0.1\).

$$ f(x) = c_0 + c_1 x + c_2 x^2 $$

$$ f(0.1) = 1 + 2(0.1) + 3(0.1)^2 $$

$$ = 1 + 0.2 + 0.03 $$

Answer: 1.23

Example 2: Given power series coefficients \(c_0=0\), \(c_1=1\), \(c_2=-1\), \(c_3=0.5\), find \(f(0.2)\).

$$ f(0.2) = 0 + 1(0.2) - 1(0.2)^2 + 0.5(0.2)^3 $$

$$ = 0.2 - 0.04 + 0.004 $$

$$ = 0.164 $$

Answer: 0.164

Example 3: Evaluate the power series sum up to \(n=4\) for \(c_n = \frac{1}{n!}\), \(a=0\), at \(x=1\).

$$ f(1) = \sum_{n=0}^4 \frac{1}{n!} (1)^n = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} $$

$$ = 1 + 1 + 0.5 + 0.1667 + 0.0417 $$

$$ = 2.7084 $$

Answer: 2.7084

1.1.2 Radius and Interval of Convergence

The power series converges only within a certain range of \(x\) values around \(a\), defined by the radius of convergence \(R\). The series converges absolutely for \(|x - a| < R\) and diverges for \(|x - a| > R\).

The radius of convergence can be found using the ratio test formula:

$$ R = \lim_{n \to \infty} | \frac{c_n}{c_{n+1}} | $$

Worked Examples

Example 1: Find the radius of convergence for the power series with coefficients \(c_n = \frac{1}{3^n}\).

$$ R = \lim_{n \to \infty} | \frac{c_n}{c_{n+1}} | = \lim_{n \to \infty} \frac{\frac{1}{3^n}}{\frac{1}{3^{n+1}}} = \lim_{n \to \infty} 3 = 3 $$

Answer: Radius of convergence \(R = 3\)

Example 2: Determine \(R\) for \(c_n = \frac{2^n}{n!}\).

$$ R = \lim_{n \to \infty} | \frac{c_n}{c_{n+1}} | = \lim_{n \to \infty} \frac{\frac{2^n}{n!}}{\frac{2^{n+1}}{(n+1)!}} = \lim_{n \to \infty} \frac{(n+1)!}{n!} \times \frac{1}{2} = \lim_{n \to \infty} \frac{n+1}{2} = \infty $$

Answer: Radius of convergence \(R = \infty\) (series converges for all \(x\))

Example 3: For \(c_n = n!\), find the radius of convergence.

$$ R = \lim_{n \to \infty} | \frac{n!}{(n+1)!} | = \lim_{n \to \infty} \frac{1}{n+1} = 0 $$

Answer: Radius of convergence \(R = 0\) (only converges at \(x=a\))

1.1.3 Power Series Representation of Elementary Functions

Many elementary functions used in electronics engineering can be expressed as power series, enabling approximation and analytical manipulation. Examples include the exponential, sine, and cosine functions.

• Exponential function:

$$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!} $$

• Sine function:

$$ \sin x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} $$

• Cosine function:

$$ \cos x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} $$

Worked Examples

Example 1: Approximate \(e^{0.5}\) using the first four terms of its power series.

$$ e^{0.5} \approx 1 + \frac{0.5}{1!} + \frac{0.5^2}{2!} + \frac{0.5^3}{3!} $$

$$ = 1 + 0.5 + \frac{0.25}{2} + \frac{0.125}{6} $$

$$ = 1 + 0.5 + 0.125 + 0.0208 $$

$$ = 1.6458 $$

Answer: 1.6458

Example 2: Approximate \(\sin 0.3\) using the first three terms.

$$ \sin 0.3 \approx 0.3 - \frac{0.3^3}{3!} + \frac{0.3^5}{5!} $$

$$ = 0.3 - \frac{0.027}{6} + \frac{0.00243}{120} $$

$$ = 0.3 - 0.0045 + 0.00002025 $$

$$ = 0.29552 $$

Answer: 0.29552

Example 3: Approximate \(\cos 0.4\) using the first three terms.

$$ \cos 0.4 \approx 1 - \frac{0.4^2}{2!} + \frac{0.4^4}{4!} $$

$$ = 1 - \frac{0.16}{2} + \frac{0.0256}{24} $$

$$ = 1 - 0.08 + 0.00107 $$

$$ = 0.92107 $$

Answer: 0.92107

1.1.4 Application of Power Series in Electronics Engineering

Power series are applied in electronics for analyzing nonlinear device characteristics, filter design, and signal approximations. For example, nonlinear diode current-voltage characteristics can be approximated by power series to simplify circuit analysis.

Worked Examples

Example 1: Approximate the diode current \(I\) given by \(I = I_s (e^{\frac{V}{nV_T}} - 1)\) near \(V=0\) using the power series for \(e^x\).

Given: \(I_s = 1 \times 10^{-12} A\), \(n=1\), \(V_T = 25 mV\), \(V = 0.05 V\).

Approximate \(e^{\frac{V}{V_T}}\) using first three terms:

$$ e^{\frac{V}{V_T}} \approx 1 + \frac{V}{V_T} + \frac{V^2}{2 V_T^2} $$

Substitute:

$$ = 1 + \frac{0.05}{0.025} + \frac{0.05^2}{2 \times 0.025^2} $$

$$ = 1 + 2 + \frac{0.0025}{0.00125} $$

$$ = 1 + 2 + 2 = 5 $$

Calculate current:

$$ I = I_s (5 - 1) = 1 \times 10^{-12} \times 4 = 4 \times 10^{-12} A $$

Answer: \(4 \times 10^{-12}\) A

Example 2: Use power series to approximate the transfer function \(H(s) = \frac{1}{1+sRC}\) for small \(s\).

Given: \(R = 1 k\Omega\), \(C = 1 \mu F\), \(s = j 1000\) rad/s.

Write as:

$$ H(s) = \frac{1}{1 + sRC} = 1 - sRC + (sRC)^2 - (sRC)^3 + \cdots $$

Calculate terms up to second order:

$$ sRC = j 1000 \times 1000 \times 1 \times 10^{-6} = j1 $$

$$ H(s) \approx 1 - j1 + (j1)^2 = 1 - j1 - 1 = 0 - j1 $$

Magnitude:

$$ |H(s)| = \sqrt{0^2 + (-1)^2} = 1 $$

Answer: Magnitude approximately 1, phase \(-90^\circ\)

Example 3: Approximate \(\ln(1+x)\) using power series for \(x=0.1\).

Power series:

$$ \ln(1+x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots $$

Using first three terms:

$$ = 0.1 - \frac{0.1^2}{2} + \frac{0.1^3}{3} = 0.1 - 0.005 + 0.00033 = 0.09533 $$

Answer: 0.09533

Chapter Summary

This chapter introduced the concept of a power series as an infinite sum of terms involving powers of a variable, providing a fundamental tool for representing functions. Taylor's theorem was presented as a method to approximate functions by polynomials expanded about a point, with clear formulations for the coefficients based on derivatives. From Taylor's theorem, Maclaurin's theorem was deduced as a special case where the expansion is centered at zero, simplifying the power series for many common functions. The chapter emphasized the practical application of these theorems in numerical computations, demonstrating how they can be used to approximate function values accurately. Worked examples illustrated the step-by-step development of power series expansions and their use in solving engineering problems. Understanding these series expansions equips students with essential techniques for analysis and approximation in various civil engineering contexts. The chapter reinforced the importance of power series in bridging theoretical mathematics and practical numerical methods.