Problem 7 - Calculus: Early Transcendentals

Question

Use the definition of a Taylor series to find the first four nonzero terms of the series for \( f(x) \) centered at the given value of \( a. \) $$ f(x) = \sqrt [3]{x}, \quad a = 8 $$

Step-by-Step Solution

Answer

First four nonzero terms: \(2 + \frac{1}{12}(x-8) - \frac{1}{288}(x-8)^2 + \frac{5}{5184}(x-8)^3\).

Step 1: Recall the Taylor Series Definition

The Taylor series of a function \( f(x) \) centered at \( a \) is given by \( f(x) = \sum_{n=0}^{fty} \frac{f^{(n)}(a)}{n!} (x-a)^n \). We need to find the derivatives of \( f(x) = \sqrt[3]{x} \) and evaluate them at \( x = 8 \).

Step 2: Calculate the First Derivative

The first derivative of \( f(x) = x^{1/3} \) is \( f'(x) = \frac{1}{3} x^{-2/3} \). Evaluating this at \( x = 8 \) gives \( f'(8) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \).

Step 3: Calculate the Second Derivative

The second derivative \( f''(x) = -\frac{2}{9} x^{-5/3} \). Evaluating this at \( x = 8 \) gives \( f''(8) = -\frac{2}{9} \times \frac{1}{32} = -\frac{1}{144} \).

Step 4: Calculate the Third Derivative

The third derivative \( f'''(x) = \frac{10}{27} x^{-8/3} \). Evaluating at \( x = 8 \) gives \( f'''(8) = \frac{10}{27} \times \frac{1}{64} = \frac{5}{864} \).

Step 5: Calculate the Function Value

At \( x = 8 \), the function value is \( f(8) = \sqrt[3]{8} = 2 \).

Step 6: Compute the Taylor Series Terms

The Taylor series up to the third degree is given by: \[\begin{align*}f(x) & = f(8) + f'(8)(x-8) + \frac{f''(8)}{2!}(x-8)^2 + \frac{f'''(8)}{3!}(x-8)^3 \& = 2 + \frac{1}{12}(x-8) - \frac{1}{288}(x-8)^2 + \frac{5}{5184}(x-8)^3.\end{align*}\]These are the first four nonzero terms of the Taylor series.

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with change and motion. It is divided into two main branches: differential calculus and integral calculus. Differential calculus focuses on the concept of a derivative, which represents the rate of change of a quantity. Integral calculus, on the other hand, deals with accumulation and the area under curves. These concepts help us understand how changing quantities are related, through rates of change and accumulations.

To solve problems using calculus, understanding the relationship between functions, limits, and derivatives is essential. In the Taylor series context, it allows us to approximate complex functions using simpler polynomial expressions.

Derivatives

Derivatives are a fundamental concept in calculus, representing an instantaneous rate of change. For a function like \(f(x) = x^{1/3}\), derivatives tell us how the function's output changes as the input changes. The first derivative, denoted as \(f'(x)\), gives the slope of the tangent line at any point. Calculating derivatives involves rules such as the power rule, product rule, and chain rule.

In the problem, the derivative \(f'(x) = \frac{1}{3}x^{-2/3}\) shows how the cube root function changes, allowing approximation through a Taylor Series. Higher-order derivatives, like the second and third, provide further terms to refine the Taylor approximation and capture more details about the function's behavior.

Functions

Functions are mappings from a set of inputs to a set of possible outputs, typically described by equations like \(f(x) = \sqrt[3]{x}\). Functions can model real-world phenomena, from physical laws to economic principles. Within calculus, functions are analyzed using limits and derivatives to understand their behavior across different input values.

A Taylor series uses derivatives at a specific point to create a polynomial that approximates the function. This is particularly useful when the exact form of a function is complex but can be closely approximated near a specific value. This approximation helps mathematicians and scientists work with difficult functions using simpler tools.

Mathematical Series

A mathematical series is the sum of the terms of a sequence. In calculus, one of the most important series is the Taylor series, which approximates functions using the sum of their derivatives at a specific point. Writing a function as a series helps simplify calculations, especially when working with functions that are otherwise challenging to manage.

For example, the Taylor series for \(f(x) = \sqrt[3]{x}\) about \(a = 8\) is a sum of terms derived from the function’s derivatives at 8. These terms, \[2 + \frac{1}{12}(x-8) - \frac{1}{288}(x-8)^2 + \frac{5}{5184}(x-8)^3\], help us approximate the cube root function around that point. By adding more terms, the approximation becomes more accurate, capturing the function's behavior nearer to the point of interest.