By the end of this chapter, you will be able to: - correctly find the roots of numbers using the binomial theorem, - accurately calculate errors from small changes with the binomial theorem, - confidently derive power series by applying the binomial expansion method.
Mastering these skills will help you solve real engineering problems efficiently and improve your mathematical confidence in the workplace.
Industrial control and installation professionals in Kenya routinely encounter binomial expressions when dealing with electrical circuits, control systems, and measurement calculations. The binomial theorem is a powerful tool that simplifies the expansion and computation of expressions involving powers, which are common in engineering design and analysis. Mastering binomial expansion enables technicians and engineers to estimate roots of numbers, analyze tolerances, and solve practical problems in system calibration and component specification.
The binomial theorem allows the expansion of expressions of the form \((a + b)^n\), where \(n\) is a positive integer or, in some cases, a fraction or negative number. This is essential in industrial control when approximating values such as \(\sqrt{1+x}\), \((1+x)^n\), or cube roots, which appear in sensor calibration and system response calculations.
The binomial theorem states that:$$ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k $$ where \(\binom{n}{k}\) is the binomial coefficient.
Example 1:
Expand \((2 + x)^3\) as used in a control relay timing circuit.
Given: \(a = 2\), \(b = x\), \(n = 3\)$$ (2 + x)^3 = \binom{3}{0}2^3x^0 + \binom{3}{1}2^2x^1 + \binom{3}{2}2^1x^2 + \binom{3}{3}2^0x^3 $$$$ = 1 \times 8 \times 1 + 3 \times 4 \times x + 3 \times 2 \times x^2 + 1 \times 1 \times x^3 $$$$ = 8 + 12x + 6x^2 + x^3 $$ Answer: \(8 + 12x + 6x^2 + x^3\)
Example 2:
Expand \((1 - 3y)^4\) for a current divider network.
Given: \(a = 1\), \(b = -3y\), \(n = 4\)$$ (1 - 3y)^4 = \binom{4}{0}1^4(-3y)^0 + \binom{4}{1}1^3(-3y)^1 + \binom{4}{2}1^2(-3y)^2 + \binom{4}{3}1^1(-3y)^3 + \binom{4}{4}1^0(-3y)^4 $$$$ = 1 \times 1 \times 1 + 4 \times 1 \times (-3y) + 6 \times 1 \times (9y^2) + 4 \times 1 \times (-27y^3) + 1 \times 1 \times 81y^4 $$$$ = 1 - 12y + 54y^2 - 108y^3 + 81y^4 $$ Answer: \(1 - 12y + 54y^2 - 108y^3 + 81y^4\)
Example 3:
Expand \((3a - 2b)^2\), as used in a voltage drop calculation.
Given: \(a = 3a\), \(b = -2b\), \(n = 2\)$$ (3a - 2b)^2 = \binom{2}{0}(3a)^2(-2b)^0 + \binom{2}{1}(3a)^1(-2b)^1 + \binom{2}{2}(3a)^0(-2b)^2 $$$$ = 1 \times 9a^2 \times 1 + 2 \times 3a \times (-2b) + 1 \times 1 \times 4b^2 $$$$ = 9a^2 - 12ab + 4b^2 $$ Answer: \(9a^2 - 12ab + 4b^2\)
Example 4:
Expand \((x + 0.1)^4\) for a temperature sensor calibration.
Given: \(a = x\), \(b = 0.1\), \(n = 4\)$$ (x + 0.1)^4 = \binom{4}{0}x^4 + \binom{4}{1}x^3(0.1) + \binom{4}{2}x^2(0.1)^2 + \binom{4}{3}x(0.1)^3 + \binom{4}{4}(0.1)^4 $$$$ = 1 \times x^4 + 4 \times x^3 \times 0.1 + 6 \times x^2 \times 0.01 + 4 \times x \times 0.001 + 1 \times 0.0001 $$$$ = x^4 + 0.4x^3 + 0.06x^2 + 0.004x + 0.0001 $$ Answer: \(x^4 + 0.4x^3 + 0.06x^2 + 0.004x + 0.0001\)
Example 5:
Expand \((2x - 5)^3\) for a load cell output calculation.
Given: \(a = 2x\), \(b = -5\), \(n = 3\)$$ (2x - 5)^3 = \binom{3}{0}(2x)^3 + \binom{3}{1}(2x)^2(-5) + \binom{3}{2}(2x)(-5)^2 + \binom{3}{3}(-5)^3 $$$$ = 1 \times 8x^3 + 3 \times 4x^2 \times (-5) + 3 \times 2x \times 25 + 1 \times (-125) $$$$ = 8x^3 - 60x^2 + 150x - 125 $$ Answer: \(8x^3 - 60x^2 + 150x - 125\)
The binomial expansion can estimate roots such as \(\sqrt{1+x}\), \((1+x)^{1/3}\), or \((1+x)^{-1}\) by using fractional or negative indices. For small \(x\), the expansion is truncated after a few terms.
The general expansion for \((1+x)^n\) is:$$ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots $$
Example 1:
Estimate \(\sqrt{1.04}\) using binomial expansion up to the second term.
Given: \(n = \frac{1}{2}\), \(x = 0.04\)$$
(1 + 0.04)^{1/2} = 1 + \frac{1}{2} \times 0.04 + \frac{1}{2} \times \frac{-1}{2} \times \frac{(0.04)^2}{2}
$$
First term: \(1\)
Second term: \(\frac{1}{2} \times 0.04 = 0.02\)
Third term: \(\frac{1}{2} \times \frac{-1}{2} = -\frac{1}{4}\), so \(-\frac{1}{4} \times \frac{(0.04)^2}{2} = -\frac{1}{4} \times \frac{0.0016}{2} = -\frac{1}{4} \times 0.0008 = -0.0002\)
Add:$$
1 + 0.02 - 0.0002 = 1.0198
$$
Answer: \(1.0198\)
Example 2:
Estimate \((1.05)^{1/3}\) up to the third term for a cube-root calculation in a PLC timing function.
Given: \(n = \frac{1}{3}\), \(x = 0.05\)$$
(1 + 0.05)^{1/3} = 1 + \frac{1}{3} \times 0.05 + \frac{1}{3} \times \frac{-2}{3} \times \frac{(0.05)^2}{2} + \frac{1}{3} \times \frac{-2}{3} \times \frac{-5}{3} \times \frac{(0.05)^3}{6}
$$
First term: \(1\)
Second term: \(\frac{1}{3} \times 0.05 = 0.01667\)
Third term: \(\frac{1}{3} \times \frac{-2}{3} = -\frac{2}{9}\), so \(-\frac{2}{9} \times \frac{0.0025}{2} = -\frac{2}{9} \times 0.00125 = -0.000278\)
Fourth term: \(\frac{1}{3} \times \frac{-2}{3} \times \frac{-5}{3} = \frac{1}{3} \times -\frac{2}{3} = -\frac{2}{9}; -\frac{2}{9} \times \frac{-5}{3} = \frac{10}{27}\), so
\(\frac{10}{27} \times \frac{(0.05)^3}{6} = \frac{10}{27} \times \frac{0.000125}{6} = \frac{10}{27} \times 0.00002083 = 0.00000772\)
Add:$$
1 + 0.01667 - 0.000278 + 0.00000772 = 1.01640
$$
Answer: \(1.0164\)
Example 3:
Estimate \((0.98)^{-1}\) up to the second term, as used in feedback gain calculation.
Given: \(n = -1\), \(x = -0.02\)$$
(1 - 0.02)^{-1} = 1 + (-1) \times (-0.02) + (-1) \times (-2) \times \frac{(-0.02)^2}{2}
$$
First term: \(1\)
Second term: \(1 \times 0.02 = 0.02\)
Third term: \(1 \times 2 = 2\), so \(2 \times \frac{0.0004}{2} = 2 \times 0.0002 = 0.0004\)
Add:$$
1 + 0.02 + 0.0004 = 1.0204
$$
Answer: \(1.0204\)
Example 4:
Estimate \(\sqrt[4]{1.08}\) up to the third term for a fourth-root sensor application.
Given: \(n = \frac{1}{4}\), \(x = 0.08\)$$
(1 + 0.08)^{1/4} = 1 + \frac{1}{4} \times 0.08 + \frac{1}{4} \times \frac{-3}{4} \times \frac{(0.08)^2}{2}
$$
First term: \(1\)
Second term: \(0.25 \times 0.08 = 0.02\)
Third term: \(0.25 \times -0.75 = -0.1875\), so \(-0.1875 \times \frac{0.0064}{2} = -0.1875 \times 0.0032 = -0.0006\)
Add:$$
1 + 0.02 - 0.0006 = 1.0194
$$
Answer: \(1.0194\)
Example 5:
Estimate \((1.02)^{5}\) up to the third term, as used in calculating the effect of tolerance stacking in a control panel.
Given: \(n = 5\), \(x = 0.02\)$$
(1 + 0.02)^5 = 1 + 5 \times 0.02 + \frac{5 \times 4}{2} \times (0.02)^2
$$
First term: \(1\)
Second term: \(5 \times 0.02 = 0.1\)
Third term: \(20 \times 0.0004 = 0.008\)
Add:$$
1 + 0.1 + 0.008 = 1.108
$$
Answer: \(1.108\)
Binomial expansions are used to approximate component values, calibrate measurement devices, and estimate errors in control systems. For small deviations, the first few terms provide accurate results, saving time and resources in field calculations.
Example 1:
A resistance thermometer's output is given by \(R = R_0(1 + \alpha T)\), where for small \(\alpha T\), \(R_0 = 100 \Omega\), \(\alpha = 0.004\), and \(T = 5^\circ C\). Approximate \(R\) up to the second term.
Given: \(R_0 = 100\), \(\alpha T = 0.02\), expand \((1 + 0.02)^1\)$$ (1 + 0.02)^1 = 1 + 0.02 $$$$ R = 100 \times (1 + 0.02) = 100 \times 1.02 = 102 \Omega $$ Answer: \(102 \Omega\)
Example 2:
A sensor output follows \(V = V_0(1 + 0.03)^{1/2}\), where \(V_0 = 10\) V. Approximate \(V\) up to the second term.
Given: \(n = 1/2\), \(x = 0.03\)$$ (1 + 0.03)^{1/2} = 1 + \frac{1}{2} \times 0.03 = 1 + 0.015 = 1.015 $$$$ V = 10 \times 1.015 = 10.15 \text{ V} $$ Answer: \(10.15 \text{ V}\)
Example 3:
A load cell's characteristic is \(F = F_0(1 + 0.01)^{-1}\), where \(F_0 = 200\) N. Approximate \(F\) up to the second term.
Given: \(n = -1\), \(x = 0.01\)$$ (1 + 0.01)^{-1} = 1 - 0.01 = 0.99 $$$$ F = 200 \times 0.99 = 198 \text{ N} $$ Answer: \(198 \text{ N}\)
Example 4:
In a PLC, the cycle time is affected by \((1 - 0.04)^{-2}\). Calculate the expansion up to the second term.
Given: \(n = -2\), \(x = -0.04\)$$
(1 - 0.04)^{-2} = 1 + (-2) \times (-0.04) + \frac{(-2) \times (-3)}{2} \times (-0.04)^2
$$
First term: \(1\)
Second term: \(2 \times 0.04 = 0.08\)
Third term: \(6/2 = 3\), \(3 \times 0.0016 = 0.0048\)
Add:$$
1 + 0.08 + 0.0048 = 1.0848
$$
Answer: \(1.0848\)
Example 5:
An amplifier gain is given by \(G = (1 + 0.06)^2\). Approximate \(G\) up to the third term.
Given: \(n = 2\), \(x = 0.06\)$$
(1 + 0.06)^2 = 1 + 2 \times 0.06 + 2 \times (0.06)^2
$$
Second term: \(2 \times 0.06 = 0.12\)
Third term: \(2 \times 0.0036 = 0.0072\)
Add:$$
1 + 0.12 + 0.0072 = 1.1272
$$
Answer: \(1.1272\)
Pascal’s Triangle provides a quick way to find binomial coefficients for integer powers. Each row corresponds to the coefficients for \((a + b)^n\).
Example 1:
Find the coefficients for \((x + y)^5\) using Pascal’s Triangle.
Row for \(n = 5\) is:
\(1, 5, 10, 10, 5, 1\)
Answer: Coefficients are \(1, 5, 10, 10, 5, 1\)
Example 2:
Expand \((a + 2b)^3\) using Pascal’s Triangle.
Coefficients for \(n = 3\) are \(1, 3, 3, 1\):
$$ (a + 2b)^3 = 1 \times a^3 + 3 \times a^2 \times 2b + 3 \times a \times (2b)^2 + 1 \times (2b)^3 $$$$ = a^3 + 6a^2b + 12ab^2 + 8b^3 $$ Answer: \(a^3 + 6a^2b + 12ab^2 + 8b^3\)
Example 3:
Expand \((x - 1)^4\) using Pascal’s Triangle.
Coefficients for \(n = 4\) are \(1, 4, 6, 4, 1\):
$$ (x - 1)^4 = 1 \times x^4 + 4 \times x^3 \times (-1) + 6 \times x^2 \times 1 + 4 \times x \times (-1) + 1 \times 1 $$$$ = x^4 - 4x^3 + 6x^2 - 4x + 1 $$ Answer: \(x^4 - 4x^3 + 6x^2 - 4x + 1\)
Example 4:
Find the coefficient of \(x^3\) in \((2 + x)^5\).
Coefficients for \(n = 5\) are \(1, 5, 10, 10, 5, 1\):
The term with \(x^3\) is when \(k = 3\):
$$ \binom{5}{3} \times 2^{5-3} \times x^3 = 10 \times 4 \times x^3 = 40x^3 $$ Answer: Coefficient is 40
Example 5:
Expand \((a, b)^2\) using Pascal’s Triangle.
Coefficients for \(n = 2\) are \(1, 2, 1\):
$$ (a, b)^2 = 1 \times a^2 + 2 \times a \times (-b) + 1 \times b^2 $$$$ = a^2 - 2ab + b^2 $$ Answer: \(a^2 - 2ab + b^2\)
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Create a free accountThis chapter explored how the binomial theorem is applied in engineering mathematics, beginning with its use in determining roots of numbers through binomial expansion techniques. Students learned to approximate roots such as square roots and cube roots by expressing numbers in a suitable binomial form and expanding to the required degree of accuracy. The chapter then covered how small changes in variables can be analyzed using the binomial theorem to estimate errors, a valuable tool for assessing the impact of slight variations in engineering measurements. The process of constructing full binomial expansions was demonstrated, including the step-by-step calculation of each term for both positive and negative integer exponents. Attention was given to using the binomial theorem for deriving power series, allowing complex functions to be expressed as infinite sums for easier manipulation and approximation. Throughout, the chapter emphasized the practical application of these methods in engineering contexts, providing numerous worked examples and calculation exercises. By mastering these concepts, students are equipped to apply binomial expansion techniques confidently in a wide range of civil engineering problems.
Given: \( n = 3 \), \( x = 0.04 \)
Formula:$$ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 $$
Substitute values:$$ (1 + 0.04)^3 = 1 + 3 \times 0.04 + \frac{3 \times 2}{2} \times (0.04)^2 $$
Calculate step-by-step:$$ = 1 + 0.12 + 3 \times 0.0016 $$$$ = 1 + 0.12 + 0.0048 $$$$ = 1.1248 $$
Final Answer: \( \boxed{1.1248} \)
Let \( a = 64 \), \( x = 1 \), \( n = 0.5 \):
Rewrite:$$ \sqrt{65} = (64 + 1)^{0.5} = 64^{0.5} (1 + \frac{1}{64})^{0.5} $$
\( 64^{0.5} = 8 \)
Expand \( (1 + \frac{1}{64})^{0.5} \) up to the second term:$$ 1 + \frac{n x}{1} = 1 + 0.5 \cdot \frac{1}{64} = 1 + 0.0078125 = 1.0078125 $$
Multiply:$$ 8 \times 1.0078125 = 8.0625 $$
Final Answer: \( \boxed{8.0625} \)
A PLC technician at a cement manufacturing plant in Athi River needs to estimate \( (1.04)^{5} \) using the binomial theorem for quick calculations during calibration. Expand and calculate the value up to the third term. (4 marks)
Use the binomial theorem to approximate the value of \( \sqrt{1.08} \) up to three terms. (4 marks)
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