Power series expansions are essential tools for architects involved in structural analysis, formulating design curves, and approximating complex functions in computational models. In Kenya, where innovative architectural solutions must often adapt to irregular sites and materials, mastery of power series allows precise local approximations that improve design accuracy and efficiency. This chapter focuses on applying Taylor and Maclaurin series to expand functions commonly encountered in architectural engineering calculations.

4.1 Taylor Series

Taylor series provide a method to approximate differentiable functions by polynomials centered around a point \(a\). This is particularly useful in architectural computations where exact function values are complex or unavailable, such as calculating load distributions or material behaviors near a known state.

4.1.1 Expansion of functions around a point using Taylor’s theorem

The Taylor series expansion of a function \(f(x)\) about a point \(x = a\) is given by:

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

where \(f^{(n)}(a)\) is the \(n^{th}\) derivative of \(f(x)\) evaluated at \(x = a\).

Worked Examples

Example 1: An architect needs to approximate the function \(f(x) = \sqrt{x}\) near \(x = 4\) to estimate the square root of 4.1 for a curved façade design.

Given: \(f(x) = \sqrt{x}\), \(a = 4\), \(x = 4.1\)

First, calculate derivatives at \(a=4\):

\(f(4) = \sqrt{4} = 2\)

\(f'(x) = \frac{1}{2\sqrt{x}}\), so \(f'(4) = \frac{1}{2 \times 2} = \frac{1}{4}\)

\(f''(x) = -\frac{1}{4x^{3/2}}\), so \(f''(4) = -\frac{1}{4 \times 8} = -\frac{1}{32}\)

Use up to the second order:

$$ f(4.1) \approx f(4) + f'(4)(4.1 - 4) + \frac{f''(4)}{2}(4.1 - 4)^2 $$

$$ = 2 + \frac{1}{4} \times 0.1 + \frac{-\frac{1}{32}}{2} \times (0.1)^2 $$

$$ = 2 + 0.025 - \frac{1}{64} \times 0.01 $$

$$ = 2 + 0.025 - 0.00015625 $$

$$ = 2.02484375 $$

Answer: \(f(4.1) \approx 2.0248\)

Example 2: Approximate \(f(x) = \ln(x)\) near \(x = 1\) to find \(\ln(1.05)\) for natural lighting calculations.

Given: \(f(x) = \ln(x)\), \(a = 1\), \(x = 1.05\)

Derivatives:

\(f(1) = \ln(1) = 0\)

\(f'(x) = \frac{1}{x}\), so \(f'(1) = 1\)

\(f''(x) = -\frac{1}{x^2}\), so \(f''(1) = -1\)

\(f'''(x) = \frac{2}{x^3}\), so \(f'''(1) = 2\)

Use up to third order:

$$ f(1.05) \approx 0 + 1 \times 0.05 + \frac{-1}{2} \times (0.05)^2 + \frac{2}{6} \times (0.05)^3 $$

$$ = 0.05 - 0.00125 + \frac{2}{6} \times 0.000125 $$

$$ = 0.05 - 0.00125 + 0.00004167 $$

$$ = 0.04879167 $$

Answer: \(\ln(1.05) \approx 0.0488\)

Example 3: For a curved roof design, approximate \(f(x) = e^x\) near \(x = 0.5\) to estimate \(e^{0.55}\).

Given: \(f(x) = e^x\), \(a = 0.5\), \(x = 0.55\)

Since all derivatives of \(e^x\) are \(e^x\):

\(f(0.5) = e^{0.5} = 1.6487\) (approximate)

Use up to second order:

$$ f(0.55) \approx f(0.5) + f'(0.5)(0.55 - 0.5) + \frac{f''(0.5)}{2}(0.55 - 0.5)^2 $$

$$ = 1.6487 + 1.6487 \times 0.05 + \frac{1.6487}{2} \times (0.05)^2 $$

$$ = 1.6487 + 0.082435 + 0.00206 $$

$$ = 1.733195 $$

Answer: \(e^{0.55} \approx 1.7332\)

Example 4: An architect approximates the function \(f(x) = \sin x\) near \(x = \frac{\pi}{4}\) to estimate \(\sin(0.85)\) radians for window shading angles.

Given: \(f(x) = \sin x\), \(a = \frac{\pi}{4} \approx 0.7854\), \(x = 0.85\)

Derivatives:

\(f(\pi/4) = \sin(0.7854) = 0.7071\)

\(f'(x) = \cos x\), so \(f'(\pi/4) = 0.7071\)

\(f''(x) = -\sin x\), so \(f''(\pi/4) = -0.7071\)

Use up to second order:

$$ f(0.85) \approx 0.7071 + 0.7071 \times (0.85 - 0.7854) + \frac{-0.7071}{2} \times (0.85 - 0.7854)^2 $$

Calculate differences:

\(0.85 - 0.7854 = 0.0646\)

Square:

\(0.0646^2 = 0.00417\)

Substitute:

$$ 0.7071 + 0.7071 \times 0.0646 - 0.35355 \times 0.00417 $$

$$ = 0.7071 + 0.0457 - 0.00147 $$

$$ = 0.75133 $$

Answer: \(\sin(0.85) \approx 0.7513\)

Example 5: Approximate \(f(x) = \frac{1}{1-x}\) near \(x = 0.2\) to find \(f(0.25)\) for stress analysis on a cantilever beam.

Given: \(f(x) = \frac{1}{1-x}\), \(a=0.2\), \(x=0.25\)

Derivatives:

\(f(x) = (1-x)^{-1}\)

\(f'(x) = (1-x)^{-2}\)

\(f''(x) = 2(1-x)^{-3}\)

Calculate at \(x=0.2\):

\(f(0.2) = \frac{1}{0.8} = 1.25\)

\(f'(0.2) = \frac{1}{0.8^2} = \frac{1}{0.64} = 1.5625\)

\(f''(0.2) = 2 \times \frac{1}{0.8^3} = 2 \times \frac{1}{0.512} = 3.90625\)

Use up to second order:

$$ f(0.25) \approx 1.25 + 1.5625 \times (0.25 - 0.2) + \frac{3.90625}{2} \times (0.25 - 0.2)^2 $$

Calculate differences:

\(0.25 - 0.2 = 0.05\)

Square:

\(0.05^2 = 0.0025\)

Substitute:

$$ 1.25 + 1.5625 \times 0.05 + 1.953125 \times 0.0025 $$

$$ = 1.25 + 0.078125 + 0.004883 $$

$$ = 1.333008 $$

Answer: \(f(0.25) \approx 1.3330\)

Chapter Summary

This chapter focused on the application of power series in mathematical analysis, beginning with the Taylor series assessment where functions are expanded into infinite sums of terms calculated from the function's derivatives at a single point. It introduced the concept of expanding various functions into their Taylor series to facilitate easier computation and approximation. The Maclaurin series was presented as a special case of the Taylor series expanded specifically about the point \(x = 0\), simplifying the process for functions centered at the origin. The chapter then explored the important topic of power series convergence, explaining how to determine whether a given power series converges and the interval within which this convergence holds true. Finally, practical applications of power series were discussed, particularly how they can be used to approximate complex functions in physics and engineering contexts, making calculations more manageable and accurate. Through these topics, the chapter provided a comprehensive understanding of power series and their significance in mathematical modeling and problem-solving.