By the end of this chapter, you will be able to:
Mastering these skills will help you solve real engineering problems confidently and ensure your calculations support safe and effective designs.
Binomial expansion is a fundamental mathematical tool used extensively in Industrial Mechatronics Technology, especially when dealing with polynomial expressions in system modelling, control algorithms, and signal processing. Expanding binomials simplifies complex expressions, aiding in the analysis of mechanical linkages and electronic circuit behaviours under varying inputs. Mastering binomial expansion enables engineers to predict system responses efficiently without resorting to lengthy multiplications.
Binomial series provide a systematic method to expand powers of binomials, which is crucial in engineering calculations involving approximations and polynomial expressions. In Industrial Mechatronics, these expansions support the development of control system equations and the analysis of sensor data that follow polynomial trends. Understanding the binomial series aids in simplifying these expressions for practical computation and design optimization.
The binomial theorem describes the expansion of a binomial expression raised to a positive integer power. For any binomial \((a + b)^n\), the expansion is a sum of terms involving powers of \(a\) and \(b\) combined with binomial coefficients. The general formula is:
$$ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k $$
where \(\binom{n}{k}\) is the binomial coefficient.
Example 1: Calculate \((x + 2)^3\) for a mechatronics sensor output approximation.
Given: \(a = x\), \(b = 2\), \(n = 3\)
$$ (x + 2)^3 = \sum_{k=0}^3 \binom{3}{k} x^{3-k} 2^k $$
Calculate each term:
For \(k = 0\): \(\binom{3}{0} x^{3} 2^{0} = 1 \times x^{3} \times 1 = x^{3}\)
For \(k = 1\): \(\binom{3}{1} x^{2} 2^{1} = 3 \times x^{2} \times 2 = 6x^{2}\)
For \(k = 2\): \(\binom{3}{2} x^{1} 2^{2} = 3 \times x \times 4 = 12x\)
For \(k = 3\): \(\binom{3}{3} x^{0} 2^{3} = 1 \times 1 \times 8 = 8\)
Sum all terms:
$$ x^{3} + 6x^{2} + 12x + 8 $$
Answer: \(x^{3} + 6x^{2} + 12x + 8\)
Example 2: Expand \((3 + y)^4\) for an actuator displacement polynomial.
Given: \(a = 3\), \(b = y\), \(n = 4\)
$$ (3 + y)^4 = \sum_{k=0}^4 \binom{4}{k} 3^{4-k} y^{k} $$
Calculate terms:
\(k=0\): \(\binom{4}{0} 3^{4} y^{0} = 1 \times 81 \times 1 = 81\)
\(k=1\): \(\binom{4}{1} 3^{3} y^{1} = 4 \times 27 \times y = 108y\)
\(k=2\): \(\binom{4}{2} 3^{2} y^{2} = 6 \times 9 \times y^{2} = 54 y^{2}\)
\(k=3\): \(\binom{4}{3} 3^{1} y^{3} = 4 \times 3 \times y^{3} = 12 y^{3}\)
\(k=4\): \(\binom{4}{4} 3^{0} y^{4} = 1 \times 1 \times y^{4} = y^{4}\)
Sum all:
$$ 81 + 108y + 54y^{2} + 12y^{3} + y^{4} $$
Answer: \(81 + 108y + 54y^{2} + 12y^{3} + y^{4}\)
Example 3: Expand \((2x - 1)^5\) for a control system polynomial model.
Given: \(a = 2x\), \(b = -1\), \(n = 5\)
$$ (2x - 1)^5 = \sum_{k=0}^5 \binom{5}{k} (2x)^{5-k} (-1)^k $$
Calculate terms:
\(k=0\): \(\binom{5}{0} (2x)^5 (-1)^0 = 1 \times 32x^{5} \times 1 = 32x^{5}\)
\(k=1\): \(\binom{5}{1} (2x)^4 (-1)^1 = 5 \times 16x^{4} \times (-1) = -80 x^{4}\)
\(k=2\): \(\binom{5}{2} (2x)^3 (-1)^2 = 10 \times 8x^{3} \times 1 = 80 x^{3}\)
\(k=3\): \(\binom{5}{3} (2x)^2 (-1)^3 = 10 \times 4x^{2} \times (-1) = -40 x^{2}\)
\(k=4\): \(\binom{5}{4} (2x)^1 (-1)^4 = 5 \times 2x \times 1 = 10 x\)
\(k=5\): \(\binom{5}{5} (2x)^0 (-1)^5 = 1 \times 1 \times (-1) = -1\)
Sum all terms:
$$ 32x^{5} - 80x^{4} + 80x^{3} - 40x^{2} + 10x - 1 $$
Answer: \(32x^{5} - 80x^{4} + 80x^{3} - 40x^{2} + 10x - 1\)
Example 4: Expand \((a + b)^4\) for a polynomial used in Nairobi City Water and Sewerage Company flow modelling.
Given: \(a = a\), \(b = b\), \(n = 4\)
$$ (a + b)^4 = \sum_{k=0}^4 \binom{4}{k} a^{4-k} b^k $$
Calculate terms:
\(k=0\): \(\binom{4}{0} a^4 b^0 = 1 \times a^4 \times 1 = a^4\)
\(k=1\): \(\binom{4}{1} a^3 b^1 = 4 \times a^3 \times b = 4a^3b\)
\(k=2\): \(\binom{4}{2} a^2 b^2 = 6 \times a^2 \times b^2 = 6a^2b^2\)
\(k=3\): \(\binom{4}{3} a^1 b^3 = 4 \times a \times b^3 = 4ab^3\)
\(k=4\): \(\binom{4}{4} a^0 b^4 = 1 \times 1 \times b^4 = b^4\)
Sum all terms:
$$ a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 $$
Answer: \(a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\)
Context: Nairobi City Water and Sewerage Company uses such expansions in hydraulic modelling equations.
Binomial coefficients \(\binom{n}{k}\) determine the multipliers of each term in the binomial expansion. These coefficients can be calculated using factorial formulas or derived from Pascal’s triangle. In mechatronics computations, coefficients help quantify the influence of each term in polynomial approximations.
The binomial coefficient is given by:
$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$
where \(n!\) denotes factorial of \(n\).
Example 1: Calculate \(\binom{5}{2}\) for a polynomial term coefficient in a robotic arm kinematic model.
Given: \(n = 5\), \(k = 2\)
$$ \binom{5}{2} = \frac{5!}{2! \times (5-2)!} = \frac{120}{2 \times 6} $$
Calculate denominator:
$$ 2 \times 6 = 12 $$
Calculate fraction:
$$ \frac{120}{12} = 10 $$
Answer: 10
Example 2: Find \(\binom{7}{3}\) for use in sensor calibration polynomial.
Given: \(n = 7\), \(k = 3\)
$$ \binom{7}{3} = \frac{7!}{3! \times 4!} = \frac{5040}{6 \times 24} $$
Calculate denominator:
$$ 6 \times 24 = 144 $$
Calculate fraction:
$$ \frac{5040}{144} = 35 $$
Answer: 35
Example 3: Determine \(\binom{10}{4}\) for a polynomial coefficient in a mechatronic system stability analysis.
Given: \(n = 10\), \(k = 4\)
$$ \binom{10}{4} = \frac{10!}{4! \times 6!} = \frac{3628800}{24 \times 720} $$
Calculate denominator:
$$ 24 \times 720 = 17280 $$
Calculate fraction:
$$ \frac{3628800}{17280} = 210 $$
Answer: 210
Example 4: Calculate \(\binom{6}{2}\) for a polynomial term coefficient in a Kenya Power transformer design.
Given: \(n = 6\), \(k = 2\)
$$ \binom{6}{2} = \frac{6!}{2! \times 4!} = \frac{720}{2 \times 24} $$
Calculate denominator:
$$ 2 \times 24 = 48 $$
Calculate fraction:
$$ \frac{720}{48} = 15 $$
Answer: 15
Context: Kenya Power engineers use binomial coefficients in transformer winding calculations.
Pascal’s triangle is a triangular array of binomial coefficients that provides a quick reference for coefficients in binomial expansions without calculating factorials. Each number in the triangle is the sum of the two numbers directly above it. Pascal’s triangle is especially useful for engineers when expanding binomials up to moderate powers efficiently.
The triangle starts as:
\[ \begin{array}{cccccc} & & 1 & & \\ & 1 & & 1 & \\ 1 & & 2 & & 1 \\ & \vdots & & \vdots & \end{array} \]
Example 1: Use Pascal’s triangle to find coefficients for \((a + b)^4\).
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Coefficients: 1, 4, 6, 4, 1
Answer: Coefficients for \((a + b)^4\) are 1, 4, 6, 4, 1
Example 2: Find coefficients for \((x + y)^5\) using Pascal’s triangle.
Row 5: 1 5 10 10 5 1
Answer: Coefficients for \((x + y)^5\) are 1, 5, 10, 10, 5, 1
Example 3: Determine the coefficients for \((p + q)^6\).
Row 6: 1 6 15 20 15 6 1
Answer: Coefficients for \((p + q)^6\) are 1, 6, 15, 20, 15, 6, 1
Example 4: Use Pascal’s triangle to find coefficients for \((a + b)^5\) in a polynomial used by Safaricom PLC for network signal modelling.
Row 5: 1 5 10 10 5 1
Answer: Coefficients for \((a + b)^5\) are 1, 5, 10, 10, 5, 1
Context: Safaricom PLC applies binomial expansions in signal strength algorithms.
Expansion of binomials involves applying the binomial theorem or Pascal’s triangle coefficients to write the full polynomial expression. This process is essential in mechatronics when simplifying expressions for control algorithms or system response analysis.
The expansion formula is:
$$ (a + b)^n = \binom{n}{0} a^{n} b^{0} + \binom{n}{1} a^{n-1} b^{1} + \cdots + \binom{n}{n} a^{0} b^{n} $$
Example 1: Expand \((1 + x)^3\) for a feedback control measurement.
Given: \(a = 1\), \(b = x\), \(n = 3\)
Coefficients from Pascal’s triangle: 1, 3, 3, 1
Expansion:
$$ 1 \times 1^{3} \times x^{0} + 3 \times 1^{2} \times x^{1} + 3 \times 1^{1} \times x^{2} + 1 \times 1^{0} \times x^{3} $$
Simplify:
$$ 1 + 3x + 3x^{2} + x^{3} $$
Answer: \(1 + 3x + 3x^{2} + x^{3}\)
Example 2: Expand \((2 + 3x)^2\) for a motor torque polynomial.
Given: \(a = 2\), \(b = 3x\), \(n = 2\)
Coefficients: 1, 2, 1
Expansion:
$$ 1 \times 2^{2} \times (3x)^{0} + 2 \times 2^{1} \times (3x)^{1} + 1 \times 2^{0} \times (3x)^{2} $$
Calculate terms:
$$ 1 \times 4 \times 1 = 4 $$
$$ 2 \times 2 \times 3x = 12x $$
$$ 1 \times 1 \times 9x^{2} = 9x^{2} $$
Sum:
$$ 4 + 12x + 9x^{2} $$
Answer: \(4 + 12x + 9x^{2}\)
Example 3: Expand \((x - \frac{1}{2})^4\) for a sensor calibration curve.
Given: \(a = x\), \(b = -\frac{1}{2}\), \(n = 4\)
Coefficients: 1, 4, 6, 4, 1
Expansion:
$$ 1 \times x^{4} \times (-\frac{1}{2})^{0} + 4 \times x^{3} \times (-\frac{1}{2})^{1} + 6 \times x^{2} \times (-\frac{1}{2})^{2} + 4 \times x \times (-\frac{1}{2})^{3} + 1 \times (-\frac{1}{2})^{4} $$
Calculate powers of \(-\frac{1}{2}\):
\[ (-\frac{1}{2})^0 = 1 \]
\[ (-\frac{1}{2})^1 = -\frac{1}{2} \]
\[ (-\frac{1}{2})^2 = \frac{1}{4} \]
\[ (-\frac{1}{2})^3 = -\frac{1}{8} \]
\[ (-\frac{1}{2})^4 = \frac{1}{16} \]
Calculate terms:
$$ 1 \times x^{4} \times 1 = x^{4} $$
$$ 4 \times x^{3} \times (-\frac{1}{2}) = -2 x^{3} $$
$$ 6 \times x^{2} \times \frac{1}{4} = \frac{3}{2} x^{2} $$
$$ 4 \times x \times (-\frac{1}{8}) = -\frac{1}{2} x $$
$$ 1 \times \frac{1}{16} = \frac{1}{16} $$
Sum all terms:
$$ x^{4} - 2 x^{3} + \frac{3}{2} x^{2} - \frac{1}{2} x + \frac{1}{16} $$
Answer: \(x^{4} - 2 x^{3} + \frac{3}{2} x^{2} - \frac{1}{2} x + \frac{1}{16}\)
Example 4: Expand \((x + 3)^3\) for a polynomial used in Kenya Railways locomotive control.
Given: \(a = x\), \(b = 3\), \(n = 3\)
Coefficients: 1, 3, 3, 1
Expansion:
$$ 1 \times x^3 \times 3^0 + 3 \times x^2 \times 3^1 + 3 \times x^1 \times 3^2 + 1 \times x^0 \times 3^3 $$
Calculate terms:
$$ 1 \times x^3 \times 1 = x^3 $$$$ 3 \times x^2 \times 3 = 9x^2 $$$$ 3 \times x \times 9 = 27x $$$$ 1 \times 1 \times 27 = 27 $$
Sum:
$$ x^3 + 9x^2 + 27x + 27 $$
Answer: \(x^3 + 9x^2 + 27x + 27\)
Context: Kenya Railways uses binomial expansions in locomotive control algorithms.
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Create a free accountThis chapter introduced the concept of the binomial series and its use in expanding expressions raised to powers. It began by explaining binomial powers and how coefficients determine the terms in the expansion, highlighting the role of Pascal's triangle in finding these coefficients efficiently. The process of expanding binomial expressions was demonstrated, showing how each term is constructed. The binomial theorem was then presented for positive integer powers, providing a general formula for expansion. This was extended to include negative and fractional powers, allowing for more complex expansions such as roots and reciprocals. Finally, the chapter addressed the estimation of errors when using binomial expansions to approximate values, especially for small changes in variables. Through these topics, students gained a comprehensive understanding of how to carry out binomial expansions in various engineering mathematics contexts.
Expand the expression \((1 + x)^3\) using binomial expansion and find the coefficient of \(x^2\). (2 marks)
Using Pascal’s triangle, write the coefficients for the expansion of \((a + b)^5\). (2 marks)
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