Applying binomial expansion is a fundamental skill for land surveyors when dealing with approximations and calculations involving powers of binomials in coordinate geometry, error analysis, and curve fitting. In surveying computations, it simplifies expressions that arise in distance calculations, leveling corrections, and area computations where exact expansion is impractical. Mastery of binomial expansion helps avoid tedious multiplication and supports efficient problem-solving in field measurements and data processing.

1.1 Expression expansion as per Binomial Theorem Expansion

Binomial theorem provides a systematic method to expand expressions raised to a power, such as \((a + b)^n\), where \(n\) is a non-negative integer. This theorem is especially useful in surveying calculations to approximate values and to expand polynomial expressions encountered in coordinate transformations and error propagation.

1.1.1 The Binomial Theorem Formula and General Expansion

The binomial theorem states that for any integer \(n \geq 0\),

$$ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} $$

where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) is the binomial coefficient representing the number of ways to choose \(k\) elements from \(n\).

Worked Examples

Example 1: Expand \((x + 2)^3\) using the binomial theorem.

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 terms:

$$ x^3 + 6x^2 + 12x + 8 $$

Answer: \(x^3 + 6x^2 + 12x + 8\)

Example 2: Expand \((3 + y)^4\).

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:

$$ 81 + 108y + 54 y^2 + 12 y^3 + y^4 $$

Answer: \(81 + 108y + 54 y^2 + 12 y^3 + y^4\)

Example 3: Expand \((2x - 1)^5\).

Given: \(a = 2x\), \(b = -1\), \(n = 5\)

$$ (2x - 1)^5 = \sum_{k=0}^5 \binom{5}{k} (2x)^{5-k} (-1)^k $$

Calculate terms:

\(k=0\): \(\binom{5}{0} (2x)^5 (-1)^0 = 1 \times (2x)^5 \times 1 = 32 x^5\)

\(k=1\): \(\binom{5}{1} (2x)^4 (-1)^1 = 5 \times 16 x^4 \times (-1) = -80 x^4\)

\(k=2\): \(\binom{5}{2} (2x)^3 (-1)^2 = 10 \times 8 x^3 \times 1 = 80 x^3\)

\(k=3\): \(\binom{5}{3} (2x)^2 (-1)^3 = 10 \times 4 x^2 \times (-1) = -40 x^2\)

\(k=4\): \(\binom{5}{4} (2x)^1 (-1)^4 = 5 \times 2 x \times 1 = 10 x\)

\(k=5\): \(\binom{5}{5} (2x)^0 (-1)^5 = 1 \times 1 \times (-1) = -1\)

Sum:

$$ 32 x^5 - 80 x^4 + 80 x^3 - 40 x^2 + 10 x - 1 $$

Answer: \(32 x^5 - 80 x^4 + 80 x^3 - 40 x^2 + 10 x - 1\)

1.1.2 Expansion of Binomials with Fractional and Negative Powers (Approximation)

In land surveying, it is often necessary to approximate expressions such as \((1 + x)^n\) where \(n\) is not a positive integer, for example in distance corrections or slope calculations. The binomial theorem for fractional or negative powers is expressed as an infinite series:

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

valid for \(|x| < 1\).

Worked Examples

Example 1: Approximate \(\sqrt{1 + 0.02}\) using the first 3 terms of the binomial expansion.

Given: \(n = \frac{1}{2}\), \(x = 0.02\)

$$ (1 + 0.02)^{\frac{1}{2}} = 1 + \frac{1}{2} \times 0.02 + \frac{\frac{1}{2}(\frac{1}{2} - 1)}{2} \times (0.02)^2 $$

Calculate each term:

First term: \(1\)

Second term: \(\frac{1}{2} \times 0.02 = 0.01\)

Third term numerator: \(\frac{1}{2} \times (\frac{1}{2} - 1) = \frac{1}{2} \times -\frac{1}{2} = -\frac{1}{4}\)

Third term denominator: \(2\)

Third term: \(-\frac{1}{4} \div 2 = -\frac{1}{8}\)

Multiply by \(x^2 = (0.02)^2 = 0.0004\):

\(-\frac{1}{8} \times 0.0004 = -0.00005\)

Sum:

$$ 1 + 0.01 - 0.00005 = 1.00995 $$

Answer: \(\approx 1.00995\)

Example 2: Approximate \((1 - 0.03)^{-1}\) to 3 terms.

Given: \(n = -1\), \(x = -0.03\)

$$ (1 - 0.03)^{-1} = 1 + (-1)(-0.03) + \frac{-1(-2)}{2} (-0.03)^2 $$

Calculate terms:

First term: \(1\)

Second term: \((-1)(-0.03) = 0.03\)

Third term numerator: \(-1 \times -2 = 2\)

Third term denominator: \(2\)

Coefficient: \(\frac{2}{2} = 1\)

\(x^2 = (-0.03)^2 = 0.0009\)

Third term: \(1 \times 0.0009 = 0.0009\)

Sum:

$$ 1 + 0.03 + 0.0009 = 1.0309 $$

Answer: \(\approx 1.0309\)

Example 3: Approximate \((1 + 0.05)^{-\frac{3}{2}}\) using first 4 terms.

Given: \(n = -\frac{3}{2}\), \(x = 0.05\)

Calculate coefficients:

Term 1: \(1\)

Term 2: \(n x = -\frac{3}{2} \times 0.05 = -0.075\)

Term 3:

$$ \frac{n(n-1)}{2} x^2 = \frac{-\frac{3}{2} \times (-\frac{3}{2} -1)}{2} \times (0.05)^2 $$

Calculate numerator:

\(-\frac{3}{2} - 1 = -\frac{3}{2} - \frac{2}{2} = -\frac{5}{2}\)

\(n(n-1) = -\frac{3}{2} \times -\frac{5}{2} = \frac{15}{4}\)

Divide by 2:

\(\frac{15}{4} \div 2 = \frac{15}{8}\)

Calculate \(x^2 = 0.05^2 = 0.0025\)

Term 3:

\(\frac{15}{8} \times 0.0025 = 0.0046875\)

Term 4:

$$ \frac{n(n-1)(n-2)}{6} x^3 = \frac{-\frac{3}{2} \times -\frac{5}{2} \times -\frac{7}{2}}{6} \times (0.05)^3 $$

Calculate numerator:

\(-\frac{3}{2} \times -\frac{5}{2} = \frac{15}{4}\)

\(\frac{15}{4} \times -\frac{7}{2} = -\frac{105}{8}\)

Divide by 6:

\(-\frac{105}{8} \div 6 = -\frac{105}{8} \times \frac{1}{6} = -\frac{105}{48} = -2.1875\)

Calculate \(x^3 = 0.05^3 = 0.000125\)

Term 4:

\(-2.1875 \times 0.000125 = -0.0002734375\)

Sum all terms:

$$ 1 - 0.075 + 0.0046875 - 0.0002734375 = 0.9294140625 $$

Answer: \(\approx 0.9294\)

1.1.3 Application of Binomial Expansion in Surveying Distance and Area Calculations

Surveyors often encounter expressions involving powers of sums or differences when calculating distances or areas with small corrections or in coordinate transformations. Using binomial expansion simplifies these computations, especially when one term is small compared to the other.

Worked Examples

Example 1: Calculate the approximate length of a line given coordinates \( (x_1, y_1) = (1000, 2000)\) and \( (x_2, y_2) = (1003, 2004)\) using the binomial expansion to approximate the square root.

Given:

\[ \Delta x = 1003 - 1000 = 3, \quad \Delta y = 2004 - 2000 = 4 \]

Distance:

$$ d = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$

Use binomial expansion to approximate \(\sqrt{25}\) by rewriting:

$$ d = 5 \sqrt{1 + \frac{9 + 16 - 25}{25}} = 5 \sqrt{1 + 0} = 5 $$

Since the inside term is zero, the exact distance is 5. To illustrate binomial expansion, consider a small change:

If \(\Delta x = 3\), \(\Delta y = 4 + 0.1 = 4.1\):

Calculate approximate distance:

$$ d = \sqrt{3^2 + 4.1^2} = \sqrt{9 + 16.81} = \sqrt{25.81} = ? $$

Rewrite:

$$ d = 5 \sqrt{1 + \frac{0.81}{25}} = 5 \sqrt{1 + 0.0324} $$

Use binomial expansion for \(\sqrt{1 + x} \approx 1 + \frac{x}{2} - \frac{x^2}{8}\) with \(x=0.0324\):

Calculate terms:

\[ 1 + \frac{0.0324}{2} - \frac{(0.0324)^2}{8} = 1 + 0.0162 - \frac{0.00104976}{8} = 1 + 0.0162 - 0.00013122 = 1.01606878 \]

Distance:

$$ d \approx 5 \times 1.01606878 = 5.0803439 $$

Exact value:

$$ \sqrt{25.81} = 5.0803 $$

Answer: \(d \approx 5.08\) units

Example 2: Approximate the area of a trapezoid with bases \(a = 50.2\) m and \(b = 49.8\) m, and height \(h = 30\) m, using binomial expansion for \((a+b)/2\).

Calculate average base:

$$ \frac{a + b}{2} = \frac{50.2 + 49.8}{2} = \frac{100}{2} = 50 $$

Express \(a+b\) as \(100 + x\) where \(x = 0\), so no correction needed.

If \(a = 50.2\), \(b = 49.7\), then:

$$ a + b = 99.9 = 100 - 0.1 $$

Approximate area:

$$ A = \frac{a + b}{2} \times h = \frac{100 - 0.1}{2} \times 30 = 50 \times 30 - 0.05 \times 30 = 1500 - 1.5 = 1498.5 \text{ m}^2 $$

No binomial expansion needed here, but if height had a small correction, binomial expansion would apply.

Example 3: Approximate \((1 + 0.01)^4\) to find the increase factor in area due to a 1% increase in length and width (area scales with square of length, but here assume fourth power for volume or other power).

Given: \(n=4\), \(x=0.01\)

Using binomial theorem:

$$ (1 + 0.01)^4 = 1 + 4 \times 0.01 + \frac{4 \times 3}{2} \times 0.01^2 + \frac{4 \times 3 \times 2}{6} \times 0.01^3 + 0.01^4 $$

Calculate terms:

Second term: \(4 \times 0.01 = 0.04\)

Third term: \(\frac{12}{2} \times 0.0001 = 6 \times 0.0001 = 0.0006\)

Fourth term: \(\frac{24}{6} \times 0.000001 = 4 \times 0.000001 = 0.000004\)

Fifth term: \(0.01^4 = 0.00000001\)

Sum:

$$ 1 + 0.04 + 0.0006 + 0.000004 + 0.00000001 = 1.04060401 $$

Answer: \(\approx 1.0406\)

Kenyan Context Example

Example 4: Kenya Railways surveyors are checking the length of a new track segment. The coordinates of the endpoints are measured as \((x_1, y_1) = (2500, 3200)\) and \((x_2, y_2) = (2503, 3204)\). Use binomial expansion to approximate the distance.

\[ \Delta x = 2503 - 2500 = 3, \quad \Delta y = 3204 - 3200 = 4 \]

Distance:

$$ d = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$

If \(\Delta y = 4.2\):

$$ d = \sqrt{3^2 + 4.2^2} = \sqrt{9 + 17.64} = \sqrt{26.64} $$

Rewrite:

$$ d = 5 \sqrt{1 + \frac{1.64}{25}} = 5 \sqrt{1 + 0.0656} $$

Using binomial expansion for \(\sqrt{1 + x} \approx 1 + \frac{x}{2} - \frac{x^2}{8}\), \(x = 0.0656\):

\[ 1 + \frac{0.0656}{2} - \frac{(0.0656)^2}{8} = 1 + 0.0328 - \frac{0.004302}{8} = 1 + 0.0328 - 0.0005378 = 1.0322622 \]

Distance:

$$ d \approx 5 \times 1.0322622 = 5.161311 $$

Answer: Distance \(\approx 5.16\) m (Kenya Railways context)

1.1.4 Using Binomial Expansion for Error Estimation and Correction in Surveying

In surveying, small measurement errors are often propagated through calculations that involve powers and roots. Binomial expansion helps approximate how errors affect final results, enabling surveyors to estimate corrections without complex calculations.

Worked Examples

Example 1: A staff reading is 2.00 m with an error of 0.01 m. Calculate the error in the square of the reading using binomial expansion.

Given: \(R = 2.00\) m, error \(x = \frac{0.01}{2.00} = 0.005\)

We want to approximate:

$$ (R + \Delta R)^2 = R^2 (1 + x)^2 $$

Use binomial expansion:

$$ (1 + x)^2 = 1 + 2x + x^2 $$

Calculate:

\(2x = 2 \times 0.005 = 0.01\)

\(x^2 = (0.005)^2 = 0.000025\)

Sum:

\(1 + 0.01 + 0.000025 = 1.010025\)

Calculate squared reading:

$$ R^2 = (2.00)^2 = 4.00 \text{ m}^2 $$

Approximate squared reading with error:

$$ 4.00 \times 1.010025 = 4.0401 \text{ m}^2 $$

Error in squared reading:

$$ 4.0401 - 4.00 = 0.0401 \text{ m}^2 $$

Answer: Error is approximately 0.04 m\(^2\)

Example 2: The distance measured is 100 m with an error of 0.5%. Estimate the error in the cube of the distance.

Given: \(D = 100\) m, \(x = 0.005\), \(n=3\)

Using binomial expansion:

$$ (1 + x)^3 = 1 + 3x + 3x^2 + x^3 $$

Calculate terms:

\(3x = 3 \times 0.005 = 0.015\)

\(3x^2 = 3 \times (0.005)^2 = 3 \times 0.000025 = 0.000075\)

\(x^3 = (0.005)^3 = 0.000000125\)

Sum:

$$ 1 + 0.015 + 0.000075 + 0.000000125 = 1.015075125 $$

Calculate cube of distance:

$$ D^3 = 100^3 = 1,000,000 \text{ m}^3 $$

Approximate cube with error:

$$ 1,000,000 \times 1.015075125 = 1,015,075.125 \text{ m}^3 $$

Error in volume:

$$ 1,015,075.125 - 1,000,000 = 15,075.125 \text{ m}^3 $$

Answer: Error is approximately 15,075 m\(^3\)

Example 3: A slope distance is measured as 150 m with a 1% error. Estimate the error in the square root of the distance using binomial expansion.

Given: \(D = 150\) m, error \(x = 0.01\), \(n = \frac{1}{2}\)

Approximate:

$$ D^{1/2} = (150)^{1/2} (1 + x)^{1/2} \approx \sqrt{150} (1 + \frac{1}{2} x - \frac{1}{8} x^2) $$

Calculate terms:

\(\frac{1}{2} x = 0.5 \times 0.01 = 0.005\)

\(-\frac{1}{8} x^2 = -0.125 \times 0.0001 = -0.0000125\)

Sum:

$$ 1 + 0.005 - 0.0000125 = 1.0049875 $$

Calculate \(\sqrt{150}\):

$$ \sqrt{150} \approx 12.247 $$

Approximate square root with error:

$$ 12.247 \times 1.0049875 = 12.308 $$

Calculate exact square root with error:

$$ \sqrt{150 \times 1.01} = \sqrt{151.5} \approx 12.311 $$

Error estimate is close.

Answer: Approximate square root with error is 12.31 m

Chapter Summary

This chapter focused on applying the Binomial Expansion theorem to expand algebraic expressions raised to any power, providing a systematic approach to handle complex polynomial expansions. It began with the method of expanding expressions using binomial coefficients, emphasizing the importance of identifying the correct terms and powers in each step. The chapter then explored finding roots of numbers, demonstrating how to approximate square roots and other roots using binomial expansions for practical calculations. Finally, it addressed errors caused by small changes in variables, explaining how to estimate these errors and their impact on results through differential approximations. The integration of these concepts equips students with powerful tools to simplify and accurately solve engineering problems involving approximations and expansions. Throughout, the chapter maintained a focus on mathematical rigor applied to civil engineering contexts, ensuring relevance and practical value for diploma students.