Question

Find a formula for the \(n^{\text {th }}\) term of the Taylor series of \(f(x),\) centered at \(c,\) by finding the coefficients of the first few powers of \(x\) and looking for a pattern. (The formulas for several of these are found in Key Idea 8.8.1; show work verifying these formula.) $$f(x)=1 / x ; \quad c=1$$

Short answer

The formula for the nth term is \(T_n = \frac{(-1)^n}{n} (x-1)^n\).

Step-by-step solution

Understanding the Taylor Series Expansion

The Taylor series of a function \(f(x)\) centered at \(c\) is given by: \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n.\] Here, \(f^{(n)}(c)\) is the \(n^{\text{th}}\) derivative of \(f(x)\) evaluated at \(x = c\). Our goal is to find the general formula for the \(n^{\text{th}}\) term in the case of \(f(x) = \frac{1}{x}\) centered at \(c = 1\).

Compute First Few Derivatives

First, find the first few derivatives of \(f(x) = \frac{1}{x}\): - \(f(x) = x^{-1}\) - \(f'(x) = -x^{-2}\) - \(f''(x) = 2x^{-3}\) - \(f'''(x) = -6x^{-4}\) - \(f^{(4)}(x) = 24x^{-5}\) Continue identifying the pattern here to find the general expression for \(f^{(n)}(x)\).

Evaluate Derivatives at c=1

Evaluate the derivatives at \(x = 1\): - \(f(1) = 1\) - \(f'(1) = -1\) - \(f''(1) = 2\) - \(f'''(1) = -6\) - \(f^{(4)}(1) = 24\) The evaluated derivatives form the sequence: \(1, -1, 2, -6, 24, \ldots\). Notice the pattern resembling the factorial sequence. Specifically, \(f^{(n)}(1) = (-1)^n (n-1)!\).

Substitute into Taylor Series Formula

For the general term of the Taylor series: \[a_n = \frac{f^{(n)}(1)}{n!} (x-1)^n,\] Substitute \(f^{(n)}(1) = (-1)^n (n-1)!\) into the formula: \[a_n = \frac{(-1)^n (n-1)!}{n!} (x-1)^n.\] This simplifies to: \[a_n = \frac{(-1)^n}{n} (x-1)^n.\]

Write the Formula for the nth Term

The nth term of the Taylor series for \(f(x) = \frac{1}{x}\) at \(x = 1\) is: \[T_n = \frac{(-1)^n}{n} (x-1)^n.\] This is the formula that expresses the general \(n^{\text{th}}\) term of the Taylor series.

Key concepts

Power series

A power series is a way to represent a function as an infinite sum of terms that consist of powers of a variable. This is like having a very long polynomial, but with infinitely many terms. The general form of a power series centered at a point \(c\) is:

• \(\sum_{n=0}^{\infty} a_n (x-c)^n\)

Here, \(a_n\) represents the coefficients of the series, and each term consists of the coefficient multiplied by a power of \((x-c)\).
The power series allows us to approximate functions with an infinite number of terms, where the accuracy depends on how many terms we consider. Power series are especially useful in calculus because they simplify working with complicated functions.

Series expansion

Series expansion refers to expressing a mathematical function as a series. In calculus, this often involves expanding a function into a form where it is represented as a sum of infinitely many terms. Taylor series are a popular form of series expansion.
For example, the Taylor series expansion of a function \(f(x)\) centered at \(c\) is given by:

• \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n\)

This series allows us to approximate the function \(f(x)\) around the point \(c\) using polynomials. Each term of the series relies on the derivatives of the function at \(c\), which describes how the function changes at that point. Expansions like this one are powerful tools in mathematics because they make complex functions easier to handle.

Derivatives

Derivatives are fundamental concepts in calculus that measure how a function changes as its input changes. When you take the derivative of a function, you're finding the rate at which one quantity changes with respect to another. For example, if we have a function \(f(x) = \frac{1}{x}\), its derivatives at various orders can look like:

• First derivative: \(f'(x) = -x^{-2}\)
• Second derivative: \(f''(x) = 2x^{-3}\)
• Third derivative: \(f'''(x) = -6x^{-4}\)

In general, the \(n^{\text{th}}\) derivative of a function captures higher-order changes — how it curves or twists as opposed to just going straight. Derivatives are crucial in finding the Taylor series because each term of the series involves the \(n^{\text{th}}\) derivative evaluated at a specific point, describing how the function behaves around that point.

Factorial

The factorial of a non-negative integer \(n\) is the product of all positive integers less than or equal to \(n\). It is denoted as \(n!\). For example,

• \(0! = 1\) by definition
• \(1! = 1\)
• \(2! = 2 \times 1 = 2\)
• \(3! = 3 \times 2 \times 1 = 6\)
• \(4! = 4 \times 3 \times 2 \times 1 = 24\)

Factorials grow very quickly with larger numbers, and they play a crucial role in the Taylor series. For each term \(\frac{f^{(n)}(c)}{n!}\) in the series, the factorial (\(n!\)) in the denominator helps account for the number of ways the derivatives can be arranged, balancing out the growth of derivatives to produce a series that converges to the function value and describes its behavior around the point \(c\).