Power series are essential tools in automotive mechatronics, offering methods to approximate complex functions that arise in control systems, sensor signal analysis, and engine performance modelling. Understanding power series allows automotive engineers to simplify nonlinear behaviours into manageable polynomial forms, facilitating easier computations and system design. This chapter covers the fundamental concept of power series and their use in representing functions relevant to automotive mechatronics.

1.1 Definition of the Term Power Series

Power series are infinite sums of terms involving powers of a variable, typically used to express functions as polynomials around a specific point. In automotive mechatronics, power series expansions help approximate sensor output signals or system responses near operating points, enabling precise control and diagnostics.

A power series centered at \(x = 0\) (also called a Maclaurin series) is expressed as:

$$ f(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots $$

where \(a_n\) are coefficients determined by the function and its derivatives at \(x=0\).

1.1.1 Concept and Formula of Power Series

The power series representation of a function \(f(x)\) about \(x=0\) is given by the Maclaurin series:

$$ f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $$

where \(f^{(n)}(0)\) is the \(n\)th derivative of \(f\) evaluated at zero.

Worked Examples

Example 1: Approximate \( \cos x \) near \(x=0\) to the fourth term.

Given: \(f(x) = \cos x\)

Calculate derivatives at \(x=0\):

\(f(0) = \cos 0 = 1\)

\(f'(x) = -\sin x \Rightarrow f'(0) = 0\)

\(f''(x) = -\cos x \Rightarrow f''(0) = -1\)

\(f'''(x) = \sin x \Rightarrow f'''(0) = 0\)

\(f^{(4)}(x) = \cos x \Rightarrow f^{(4)}(0) = 1\)

Formula:

$$ f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 $$

Substitute:

$$ = 1 + 0 \times x + \frac{-1}{2} x^2 + 0 \times x^3 + \frac{1}{24} x^4 $$

Calculate:

$$ = 1 - \frac{x^2}{2} + \frac{x^4}{24} $$

Answer:

\[ \boxed{ \cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} } \]

Example 2: Find the first four terms of the power series for \( e^x \) at \(x=0\).

Given: \(f(x) = e^x\)

Derivatives:

All derivatives of \(e^x\) are \(e^x\), so at \(x=0\), \(f^{(n)}(0) = 1\) for all \(n\).

Formula:

$$ f(x) = \sum_{n=0}^\infty \frac{1}{n!} x^n = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots $$

First four terms:

$$ 1 + x + \frac{x^2}{2} + \frac{x^3}{6} $$

Answer:

\[ \boxed{ e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} } \]

Example 3: Determine the first four terms of the power series for \( \ln(1+x) \) near \(x=0\).

Given: \(f(x) = \ln(1+x)\)

Derivatives:

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

\(f'(x) = \frac{1}{1+x} \Rightarrow f'(0) = 1\)

\(f''(x) = -\frac{1}{(1+x)^2} \Rightarrow f''(0) = -1\)

\(f'''(x) = \frac{2}{(1+x)^3} \Rightarrow f'''(0) = 2\)

\(f^{(4)}(x) = -\frac{6}{(1+x)^4} \Rightarrow f^{(4)}(0) = -6\)

Formula:

$$ f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 $$

Substitute:

$$ = 0 + \frac{1}{1} x + \frac{-1}{2} x^2 + \frac{2}{6} x^3 + \frac{-6}{24} x^4 $$

Calculate:

$$ = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$

Answer:

\[ \boxed{ \ln(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} } \]

1.1.2 Convergence Conditions of Power Series

For a power series to be useful in automotive mechatronics applications, it must converge to the function value within a certain interval around the expansion point. The convergence depends on the behaviour of the function and its derivatives at that point.

The main conditions for convergence at \(x=0\) are:

1. \(f(0)\) is finite (not infinite).

2. All derivatives \(f^{(n)}(0)\) exist and are finite.

3. The series terms approach zero fast enough so that the infinite sum converges.

Worked Examples

Example 1: Test convergence of the power series for \( \cos x \).

Given: \(f(x) = \cos x\)

Values at zero:

\(f(0) = 1\) finite

Derivatives at zero exist and are finite

The terms decrease as powers of \(x\) divided by factorials, which grow rapidly.

Answer: The power series for \( \cos x \) converges for all \(x\).

Example 2: Test convergence of the power series for \( \ln(1+x) \) at \(x=0\).

Given: \(f(x) = \ln(1+x)\)

\(f(0) = 0\) finite

Derivatives at zero exist and are finite

The radius of convergence is \(-1 < x \leq 1\) because \( \ln(1+x) \) is undefined for \(x \leq -1\).

Answer: The power series converges for \(-1 < x \leq 1\).

Example 3: Check if the power series for \( \ln x \) converges at \(x=0\).

Given: \(f(x) = \ln x\)

\(f(0) = \ln 0 = -\infty\) undefined

Derivatives at zero are infinite

Answer: The power series for \( \ln x \) does not converge at \(x=0\).

1.1.3 Applications of Power Series in Automotive Mechatronics

Power series expansions allow approximation of nonlinear sensor signals and actuator responses near operating points, enabling linear control system design and fault diagnosis.

Worked Examples

Example 1: Approximate the sensor output \(f(x) = \sin x\) near \(x=0\) up to the third term.

Given: \(f(x) = \sin x\)

Derivatives at zero:

\(f(0) = 0\)

\(f'(x) = \cos x \Rightarrow f'(0) = 1\)

\(f''(x) = -\sin x \Rightarrow f''(0) = 0\)

\(f'''(x) = -\cos x \Rightarrow f'''(0) = -1\)

Formula:

$$ f(x) = 0 + \frac{1}{1!} x + \frac{0}{2!} x^2 + \frac{-1}{3!} x^3 = x - \frac{x^3}{6} $$

Answer:

\[ \boxed{ \sin x \approx x - \frac{x^3}{6} } \]

Example 2: Linearise \(f(x) = e^{-x}\) near \(x=0\) to the second term for actuator modelling.

Given: \(f(x) = e^{-x}\)

Derivatives at zero:

\(f(0) = 1\)

\(f'(x) = -e^{-x} \Rightarrow f'(0) = -1\)

Formula:

$$ f(x) = 1 - x $$

Answer:

\[ \boxed{ e^{-x} \approx 1 - x } \]

Example 3: Approximate the function \(f(\theta) = \cos 2\theta\) near \(\theta=0\) up to the fourth term.

Use substitution in power series for \(\cos x\):

$$ \cos 2\theta = 1 - \frac{(2\theta)^2}{2!} + \frac{(2\theta)^4}{4!} = 1 - 2\theta^2 + \frac{2}{3} \theta^4 $$

Answer:

\[ \boxed{ \cos 2\theta \approx 1 - 2\theta^2 + \frac{2}{3} \theta^4 } \]

Chapter Summary

This chapter introduced the concept of a power series as an infinite sum of terms involving powers of a variable, providing a foundation for representing functions analytically. It explored Taylor's theorem, which allows a function to be expressed as a polynomial plus a remainder term based on its derivatives at a specific point. The chapter then deduced McLaurin's theorem as a special case of Taylor's theorem centered at zero, simplifying the power series expansion for many functions. Emphasis was placed on the practical application of both Taylor's and McLaurin's theorems in numerical methods, such as approximating function values and solving engineering problems. Through these developments, students gained tools for expressing complex functions in simpler polynomial forms, facilitating easier calculations and analysis. The chapter combined theoretical derivations with numerical applications, ensuring a comprehensive understanding of power series in engineering mathematics.