By the end of this chapter, you will be able to:
Mastering these skills will help you tackle complex engineering problems efficiently and with confidence in your trade.
In industrial control and installation, precise mathematical tools are essential for analyzing complex systems, modeling electrical circuits, and optimizing automation processes. Power series provide a powerful technique for representing and manipulating functions commonly encountered in engineering, such as trigonometric, exponential, and logarithmic functions. Understanding power series enables professionals to approximate values, solve differential equations, and analyze system behaviors with high accuracy, especially when exact formulas are impractical.
A power series is an infinite sum of terms, each involving powers of a variable \(x\), usually centered at a specific point. It is written as
$$ f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots $$
where \(a_0, a_1, a_2, \ldots\) are constants called coefficients. Power series allow engineers to express complicated functions as sums of simpler terms, which is useful for calculations in control systems.
Example 1: Write the first four terms of the power series for \(f(x) = 2 + 5x - 3x^2 + 4x^3\).
Given: Coefficients \(a_0 = 2\), \(a_1 = 5\), \(a_2 = -3\), \(a_3 = 4\).
$$ f(x) = 2 + 5x - 3x^2 + 4x^3 $$
Answer: \(f(x) = 2 + 5x - 3x^2 + 4x^3\)
Example 2: Express the function \(f(x) = 1 + 3x + 2x^2\) as a power series up to \(x^2\).
Given: \(a_0 = 1\), \(a_1 = 3\), \(a_2 = 2\).
$$ f(x) = 1 + 3x + 2x^2 $$
Answer: \(f(x) = 1 + 3x + 2x^2\)
Example 3: Expand \(f(x) = 4x^2 - 7x + 10\) as a power series up to \(x^2\).
Given: \(a_0 = 10\), \(a_1 = -7\), \(a_2 = 4\).
$$ f(x) = 10 - 7x + 4x^2 $$
Answer: \(f(x) = 10 - 7x + 4x^2\)
Example 4: Given \(f(x) = x^3 + 2x^2 - x + 5\), identify the coefficients in its power series.
Given: \(a_0 = 5\), \(a_1 = -1\), \(a_2 = 2\), \(a_3 = 1\).
$$ f(x) = 5 - x + 2x^2 + x^3 $$
Answer: \(f(x) = 5 - x + 2x^2 + x^3\)
Example 5: Write the power series for \(f(x) = 6x^3\) up to \(x^3\).
Given: \(a_0 = 0\), \(a_1 = 0\), \(a_2 = 0\), \(a_3 = 6\).
$$ f(x) = 0 + 0x + 0x^2 + 6x^3 $$
Answer: \(f(x) = 6x^3\)
The Maclaurin series is a power series expansion of a function about \(x = 0\). It allows functions to be represented as an infinite sum of their derivatives evaluated at zero:
$$ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots $$
This is especially useful for approximating functions in automation systems near \(x = 0\).
Example 1: Find the Maclaurin series for \(f(x) = e^x\) up to the \(x^3\) term.
Given: \(f(0) = 1\), \(f'(0) = 1\), \(f''(0) = 1\), \(f'''(0) = 1\).
$$ f(x) = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 $$$$ = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 $$
Answer: \(f(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3\)
Example 2: Determine the Maclaurin series for \(f(x) = \sin x\) up to the \(x^3\) term.
Given:
\(f(0) = 0\),
\(f'(0) = 1\),
\(f''(0) = 0\),
\(f'''(0) = -1\).
$$ f(x) = 0 + x + \frac{0}{2!}x^2 + \frac{-1}{3!}x^3 $$$$ = x - \frac{1}{6}x^3 $$
Answer: \(f(x) = x - \frac{1}{6}x^3\)
Example 3: Find the first four terms of the Maclaurin series for \(f(x) = \cos x\).
Given:
\(f(0) = 1\),
\(f'(0) = 0\),
\(f''(0) = -1\),
\(f'''(0) = 0\),
\(f^{iv}(0) = 1\).
$$ f(x) = 1 + 0x + \frac{-1}{2!}x^2 + \frac{0}{3!}x^3 + \frac{1}{4!}x^4 $$$$ = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 $$
Answer: \(f(x) = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4\)
Example 4: Expand \(f(x) = \ln(1 + x)\) up to the \(x^3\) term using Maclaurin series.
Given:
\(f(0) = 0\),
\(f'(0) = 1\),
\(f''(0) = -1\),
\(f'''(0) = 2\).
$$ f(x) = 0 + x + \frac{-1}{2!}x^2 + \frac{2}{3!}x^3 $$$$ = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 $$
Answer: \(f(x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3\)
Example 5: Find the Maclaurin series for \(f(x) = \tan^{-1}x\) up to the \(x^3\) term.
Given:
\(f(0) = 0\),
\(f'(0) = 1\),
\(f''(0) = 0\),
\(f'''(0) = -2\).
$$ f(x) = 0 + x + \frac{0}{2!}x^2 + \frac{-2}{3!}x^3 $$$$ = x - \frac{1}{3}x^3 $$
Answer: \(f(x) = x - \frac{1}{3}x^3\)
Power series are valid only under certain conditions. For Maclaurin's series, the following must hold at \(x = 0\):
Example 1: Check if the Maclaurin series can be used for \(f(x) = \ln x\).
Given: \(f(0) = \ln 0 = -\infty\)
Since \(f(0)\) is not finite, Maclaurin series cannot be used.
Answer: Not valid for Maclaurin series.
Example 2: Check if the Maclaurin series can be used for \(f(x) = \cos x\).
Given: \(f(0) = 1\), \(f'(0) = 0\), \(f''(0) = -1\), all finite.
The terms of the series become smaller for larger \(x\), and the sum approaches a limit.
Answer: Valid for Maclaurin series.
Example 3: For the series \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots\), determine if it converges.
Each term is half the previous term.
$$ \text{Sum} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots $$
The sum approaches 2 as more terms are added.
Answer: The series converges to 2.
Example 4: For \(f(x) = \tan x\), check if the Maclaurin series is valid at \(x = 0\).
Given: \(f(0) = 0\), \(f'(0) = 1\), \(f''(0) = 0\), \(f'''(0) = 2\), all finite.
The series converges for \(|x| < \frac{\pi}{2}\).
Answer: Valid for Maclaurin series within this interval.
Example 5: For \(f(x) = \frac{1}{1-x}\), determine the radius of convergence.
A geometric series converges when \(|x| < 1\).
Answer: The power series converges for \(|x| < 1\).
Power series are extensively used to approximate functions in automatic control, signal processing, and instrumentation. They simplify complex calculations in programmable logic controllers (PLCs), sensor calibration, and feedback loop analysis.
Example 1: Approximate \(e^{0.1}\) using the Maclaurin series up to the \(x^3\) term.
Given: \(x = 0.1\), \(f(x) = e^x\)
$$ f(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 $$$$ = 1 + 0.1 + \frac{1}{2}(0.1)^2 + \frac{1}{6}(0.1)^3 $$$$ = 1 + 0.1 + 0.005 + 0.0001667 $$$$ = 1.1051667 $$
Answer: \(e^{0.1} \approx 1.1052\)
Example 2: Estimate \(\sin(0.05)\) using the Maclaurin series up to the \(x^3\) term for a sensor signal.
Given: \(x = 0.05\)
$$ f(x) = x - \frac{1}{6}x^3 $$$$ = 0.05 - \frac{1}{6}(0.05)^3 $$$$ = 0.05 - \frac{1}{6}(0.000125) $$$$ = 0.05 - 0.00002083 $$$$ = 0.04997917 $$
Answer: \(\sin(0.05) \approx 0.04998\)
Example 3: Calculate the value of \(\cos(0.2)\) using the first three terms of the Maclaurin series for a PLC timer.
Given: \(x = 0.2\)
$$ f(x) = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 $$ First, calculate \(x^2\) and \(x^4\):
$$ x^2 = (0.2)^2 = 0.04 $$$$ x^4 = (0.2)^4 = 0.0016 $$
Plug into the formula:
$$ f(x) = 1 - \frac{1}{2}(0.04) + \frac{1}{24}(0.0016) $$$$ = 1 - 0.02 + 0.0000667 $$$$ = 0.9800667 $$
Answer: \(\cos(0.2) \approx 0.9801\)
Example 4: Approximate \(\ln(1.1)\) using the Maclaurin series up to the \(x^3\) term for a calibration routine.
Given: \(x = 0.1\)
$$ f(x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 $$$$ = 0.1 - \frac{1}{2}(0.1)^2 + \frac{1}{3}(0.1)^3 $$$$ = 0.1 - 0.005 + 0.0003333 $$$$ = 0.0953333 $$
Answer: \(\ln(1.1) \approx 0.0953\)
Example 5: Estimate \(\tan^{-1}(0.1)\) for a control angle calculation using Maclaurin up to \(x^3\).
Given: \(x = 0.1\)
$$ f(x) = x - \frac{1}{3}x^3 $$$$ = 0.1 - \frac{1}{3}(0.1)^3 $$$$ = 0.1 - \frac{1}{3}(0.001) $$$$ = 0.1 - 0.0003333 $$$$ = 0.0996667 $$
Answer: \(\tan^{-1}(0.1) \approx 0.0997\) radians
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free accountThis chapter introduced the concept of a power series, describing it as a mathematical tool for expressing functions as infinite sums of powers of a variable. Students explored Taylor's theorem, which provides a general formula for approximating functions near a particular point using derivatives. The chapter then demonstrated how Maclaurin's theorem is a special case of Taylor's theorem when the expansion point is zero, leading directly to the Maclaurin series for common functions. Through step-by-step worked examples, the deduction of Maclaurin's theorem was shown, making clear how the power series for several fundamental functions can be systematically derived. The practical use of Taylor's and Maclaurin's theorems in engineering calculations was emphasized, particularly for approximating values and simplifying complex expressions. The chapter included a progression of calculation-based examples to build student confidence in applying these theorems to real engineering problems. By the end, students gained both a theoretical understanding and practical skills for using power series expansions in numerical computations.
Given: \( f(x) = e^{x} \), expand up to \( x^3 \), use \( x = 0.2 \).
The Maclaurin series for \( e^{x} \) is:$$ e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} $$
For \( x = 0.2 \):$$ e^{0.2} \approx 1 + 0.2 + \frac{(0.2)^2}{2} + \frac{(0.2)^3}{6} $$$$ = 1 + 0.2 + \frac{0.04}{2} + \frac{0.008}{6} $$$$ = 1 + 0.2 + 0.02 + 0.001333 $$$$ = 1.221333 $$
Final answer: \( e^{0.2} \approx \mathbf{1.221} \)
The Maclaurin series for \( \sin x \) is:$$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} $$
For \( x = 0.1 \):$$ \sin(0.1) \approx 0.1 - \frac{(0.1)^3}{6} + \frac{(0.1)^5}{120} $$$$ = 0.1 - \frac{0.001}{6} + \frac{0.00001}{120} $$$$ = 0.1 - 0.0001667 + 0.0000000833 $$$$ = 0.0998334 $$
Final answer: \( \sin(0.1) \approx \mathbf{0.09983} \)
A technician at a Nairobi-based automation firm is calibrating a sensor whose output voltage \( V(x) \) can be expressed as a power series about \( x = 0 \). Define what is meant by a power series and give a practical example involving a sensor signal. (4 marks)
Write the general form of a power series centered at \( x = a \). Then, for the function \( f(x) = e^{x} \), write the first four terms of its Maclaurin series. (4 marks)
At the start of this chapter we promised you would be able to:
Tick each one you can genuinely do.
So, are you there yet?
You're competent when you can confidently do 50% or more of what this chapter promised.
Sign in to record how you're doing.