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Power series are essential mathematical tools in quantity surveying, enabling approximation of complex functions that arise in cost estimations, resource planning, and financial modeling. Taylor's theorem provides the foundation for expressing these functions as infinite sums of polynomial terms, simplifying calculations in construction project analyses. Mastery of power series through Taylor expansions allows quantity surveyors to predict and adjust project parameters accurately when exact computations are impractical.
Taylor's theorem expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point. This expansion is invaluable in quantity surveying when evaluating functions that model costs, quantities, or rates which are otherwise difficult to compute directly. The power series form derived from Taylor's theorem approximates these functions near a point, facilitating easier numerical evaluation and sensitivity analysis in project estimates.
Taylor series represents a function \( f(x) \) as an infinite sum of terms based on its derivatives at a point \( a \). The formula is:
$$ f(x) = f(a) + f'(a)(x, a) + \frac{f''(a)}{2!}(x, a)^2 + \frac{f^{(3)}(a)}{3!}(x, a)^3 + \cdots + \frac{f^{(n)}(a)}{n!}(x, a)^n + \cdots $$
where \( f^{(n)}(a) \) is the \( n^{th} \) derivative of \( f \) evaluated at \( a \), and \( n! \) is factorial of \( n \).
Example 1: Approximate the function \( f(x) = \ln(x) \) near \( a = 1 \) using the first three terms of its Taylor series to estimate \( \ln(1.1) \).
Given: \( f(x) = \ln(x) \), \( a = 1 \), \( x = 1.1 \)
Calculate derivatives:
Taylor series approximation:
$$ f(1.1) \approx f(1) + f'(1)(1.1 - 1) + \frac{f''(1)}{2}(1.1 - 1)^2 $$
$$ = 0 + 1 \times 0.1 + \frac{-1}{2} \times (0.1)^2 $$
$$ = 0.1 - 0.005 = 0.095 $$
Answer: \(\ln(1.1) \approx 0.095\)
Example 2: Use Taylor series to approximate \( e^x \) at \( a = 0 \) up to the fourth term for \( x = 0.2 \).
Given: \( f(x) = e^x \), \( a = 0 \), \( x = 0.2 \)
Derivatives of \( e^x \) are all \( e^x \), so:
Formula:
$$ f(0.2) \approx 1 + 1 \times 0.2 + \frac{1}{2} \times (0.2)^2 + \frac{1}{6} \times (0.2)^3 + \frac{1}{24} \times (0.2)^4 $$
Calculate each term:
$$ 1 + 0.2 + \frac{1}{2} \times 0.04 + \frac{1}{6} \times 0.008 + \frac{1}{24} \times 0.0016 $$
$$ = 1 + 0.2 + 0.02 + 0.001333 + 0.000067 $$
Sum:
$$ 1.2214 $$
Answer: \( e^{0.2} \approx 1.2214 \)
Example 3: Estimate \( \sin(x) \) near \( a = 0 \) using the first three non-zero terms of the Taylor series for \( x = 0.3 \) radians.
Given: \( f(x) = \sin x \), \( a = 0 \), \( x = 0.3 \)
Derivatives:
Taylor series terms (odd derivatives only):
$$ \sin x \approx x - \frac{x^3}{3!} + \frac{x^5}{5!} $$
Calculate:
$$ 0.3 - \frac{0.3^3}{6} + \frac{0.3^5}{120} $$
$$ 0.3 - \frac{0.027}{6} + \frac{0.00243}{120} $$
$$ 0.3 - 0.0045 + 0.00002 = 0.29552 $$
Answer: \( \sin(0.3) \approx 0.2955 \)
Maclaurin series is a specific Taylor series expanded around \( a = 0 \). It simplifies calculations when the function behavior near zero is required, common in surveying when small deviations or increments are analyzed.
Formula:
$$ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \cdots $$
Example 1: Find the Maclaurin series approximation for \( \cos x \) up to the fourth term and estimate \( \cos 0.2 \).
Given: \( f(x) = \cos x \), \( a = 0 \), \( x = 0.2 \)
Derivatives at 0:
Series terms:
$$ \cos x \approx 1 - \frac{x^2}{2!} + \frac{x^4}{4!} $$
Calculate:
$$ 1 - \frac{(0.2)^2}{2} + \frac{(0.2)^4}{24} $$
$$ 1 - \frac{0.04}{2} + \frac{0.0016}{24} $$
$$ 1 - 0.02 + 0.000067 = 0.980067 $$
Answer: \( \cos 0.2 \approx 0.9801 \)
Example 2: Approximate \( e^x \) at \( x = -0.1 \) using the first four Maclaurin terms.
Given: \( f(x) = e^x \), \( a = 0 \), \( x = -0.1 \)
Derivatives at 0 are all 1.
Series:
$$ e^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} $$
Calculate:
$$ 1 - 0.1 + \frac{0.01}{2} - \frac{0.001}{6} $$
$$ 1 - 0.1 + 0.005 - 0.000167 = 0.904833 $$
Answer: \( e^{-0.1} \approx 0.9048 \)
Example 3: Use Maclaurin series to approximate \( \ln(1 + x) \) at \( x = 0.05 \) up to the third term.
Given: \( f(x) = \ln(1 + x) \), \( a = 0 \), \( x = 0.05 \)
Series expansion:
$$ \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots $$
Calculate:
$$ 0.05 - \frac{(0.05)^2}{2} + \frac{(0.05)^3}{3} $$
$$ 0.05 - \frac{0.0025}{2} + \frac{0.000125}{3} $$
$$ 0.05 - 0.00125 + 0.000042 = 0.048792 $$
Answer: \( \ln(1.05) \approx 0.0488 \)
For a power series to be useful in quantity surveying calculations, it must converge within a certain range of \( x \) values. The radius of convergence \( R \) defines the interval \( (a - R, a + R) \) where the series converges to the function. Understanding this helps avoid errors in approximations beyond valid limits.
The radius of convergence is found by:
$$ \lim_{n \to \infty} | \frac{c_n}{c_{n+1}} | = R $$
where \( c_n \) are the coefficients of the series terms.
Example 1: Determine the radius of convergence for the power series of \( \frac{1}{1 - x} = \sum_{n=0}^\infty x^n \).
Given: Coefficients \( c_n = 1 \) for all \( n \)
Calculate:
$$ R = \lim_{n \to \infty} | \frac{1}{1} | = 1 $$
Answer: Radius of convergence \( R = 1 \)
Example 2: Find the radius of convergence for the series \( \sum_{n=1}^\infty \frac{x^n}{n} \).
Given: \( c_n = \frac{1}{n} \)
Calculate:
$$ R = \lim_{n \to \infty} | \frac{\frac{1}{n}}{\frac{1}{n+1}} | = \lim_{n \to \infty} \frac{n+1}{n} = 1 $$
Answer: Radius of convergence \( R = 1 \)
Example 3: For the series \( \sum_{n=0}^\infty \frac{n! x^n}{n^n} \), estimate the radius of convergence.
Given: \( c_n = \frac{n!}{n^n} \)
Use ratio test:
$$ L = \lim_{n \to \infty} | \frac{c_{n+1}}{c_n} | = \lim_{n \to \infty} \frac{(n+1)!}{(n+1)^{n+1}} \times \frac{n^n}{n!} = \lim_{n \to \infty} \frac{(n+1) n^n}{(n+1)^{n+1}} = ? $$
Rewrite denominator:
$$ (n+1)^{n+1} = (n+1)^n \times (n+1) $$
Simplify:
$$ L = \lim_{n \to \infty} \frac{n^n}{(n+1)^n} = \lim_{n \to \infty} ( \frac{n}{n+1} )^n $$
Since \( \lim_{n \to \infty} ( \frac{n}{n+1} )^n = e^{-1} \)
Radius of convergence:
$$ R = \frac{1}{L} = e $$
Answer: Radius of convergence \( R = e \approx 2.718 \)
Taylor series approximations assist quantity surveyors in estimating non-linear cost functions, interest calculations, and material quantity variations when exact formulas are complex. These approximations enable efficient budgeting and forecasting in construction projects by simplifying analysis of small changes in project variables.
Example 1: A quantity surveyor models the cost increase \( C(x) = 5000e^{0.02x} \), where \( x \) is the number of additional units built. Approximate the cost for \( x = 3 \) using Taylor series about \( x = 0 \).
Given: \( C(x) = 5000 e^{0.02x} \), \( a = 0 \), \( x = 3 \)
Derivatives:
Taylor series up to second derivative:
$$ C(3) \approx C(0) + C'(0)(3 - 0) + \frac{C''(0)}{2} (3 - 0)^2 $$
Calculate:
$$ 5000 + 100 \times 3 + \frac{2}{2} \times 9 = 5000 + 300 + 9 = 5309 $$
Answer: Estimated cost \( C(3) \approx \text{Ksh } 5309 \)
Example 2: Approximate \( \sqrt{100 + x} \) at \( x = 4 \) using Taylor series about \( a = 0 \).
Given: \( f(x) = \sqrt{100 + x} \), \( a = 0 \), \( x = 4 \)
Calculate derivatives:
Taylor series up to second order:
$$ f(4) \approx 10 + 0.05 \times 4 + \frac{-0.00025}{2} \times 4^2 $$
Calculate:
$$ 10 + 0.2 - 0.00025 \times 8 = 10.2 - 0.002 = 10.198 $$
Answer: \( \sqrt{104} \approx 10.198 \)
Example 3: A quantity surveyor uses the function \( f(x) = \ln(50 + x) \) to model price variation. Estimate \( f(2) \) using Taylor series about \( a = 0 \).
Given: \( f(x) = \ln(50 + x) \), \( a = 0 \), \( x = 2 \)
Derivatives:
Taylor series up to second derivative:
$$ f(2) \approx 3.912 + 0.02 \times 2 + \frac{-0.0004}{2} \times 2^2 $$
Calculate:
$$ 3.912 + 0.04 - 0.0004 \times 2 = 3.952 - 0.0008 = 3.9512 $$
Answer: \( \ln(52) \approx 3.951 \)
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Create a free accountThis chapter explored the concept of power series through the application of Taylor's Theorem, demonstrating how functions can be expressed as infinite sums of terms calculated from the derivatives at a single point. It then focused on Maclaurin's Theorem, which is a special case of Taylor's series centered at zero, showing how power series simplify when expanded around this particular point. The chapter further explained the process of obtaining power series based on these theorems, detailing the step-by-step method of deriving series expansions for various functions. Emphasis was placed on understanding the conditions under which these series converge and their practical use in approximating functions. Additionally, the chapter incorporated statistical calculations related to power series, highlighting their relevance in engineering problem-solving. Overall, the chapter provided a comprehensive foundation for applying power series in mathematical analysis and engineering contexts.
Find the 4th degree Taylor polynomial of the function \(f(x) = \ln(1+x)\) centered at \(x=0\), and use it to approximate \(f(0.1)\). (2 marks)
Using the Maclaurin series, find the first 5 non-zero terms of the power series expansion for \(f(x) = e^{-2x}\). (2 marks)
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