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.) $$f(x)=x /(x+1) ; \quad c=1$$

Short answer

The formula for the nth term is \( \frac{(-1)^{n+1}(n-1)!}{2^n n!}(x-1)^n \).

Step-by-step solution

Apply Taylor Series Formula

The general formula for the Taylor series of a function \( f(x) \) centered at \( c \) is given by \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n \). Our goal is to determine the coefficients \( \frac{f^{(n)}(c)}{n!} \).

Calculate derivatives at x=c

First, find the function's derivatives at \( x = c = 1 \). Begin by calculating: - \( f(x) = \frac{x}{x+1} \), so \( f(1) = \frac{1}{2} \). - First derivative: \( f'(x) = \frac{1}{(x+1)^2} \), then \( f'(1) = \frac{1}{4} \). - Second derivative: \( f''(x) = -\frac{2}{(x+1)^3} \), and \( f''(1) = -\frac{1}{4} \).

Generalize the Pattern

Continue calculating more derivatives to find a pattern. Compute the third derivative: - \( f'''(x) = \frac{6}{(x+1)^4} \) and \( f'''(1) = \frac{3}{8} \).After calculating a few derivatives, identify any trends or patterns in the coefficients: - Derivatives alternate signs and decrease in magnitude. The pattern seems to suggest coefficients in the form: \( \frac{(-1)^{n+1}(n-1)!}{2^n} \) for \( n \geq 1 \).

Formulate nth Term Expression

Using the pattern observed, each term in the expansion can be expressed as \[ a_n = \frac{(-1)^{n+1}(n-1)!}{2^n} \]. Therefore, the nth term of the Taylor series is \( \frac{(-1)^{n+1}(n-1)!}{2^n n!}(x-1)^n \).

Key concepts

Derivative Calculation

To find a Taylor series, we first need to calculate the derivatives of the function at a specific point—in this case, at \( x = 1 \). Calculating derivatives helps us understand how the function behaves near that point. The function given is \( f(x) = \frac{x}{x+1} \). Begin by calculating the function value at \( x = 1 \), which is \( f(1) = \frac{1}{2} \).

Next, find the first derivative, \( f'(x) = \frac{1}{(x+1)^2} \), and evaluate it at \( x = 1 \), giving \( f'(1) = \frac{1}{4} \). Continue by calculating higher-order derivatives:

• Second derivative: \( f''(x) = -\frac{2}{(x+1)^3} \), at \( x=1 \), \( f''(1) = -\frac{1}{4} \)
• Third derivative: \( f'''(x) = \frac{6}{(x+1)^4} \), at \( x=1 \), \( f'''(1) = \frac{3}{8} \)

Computing these derivatives helps establish a basis for finding patterns and forming each term of the series.

Series Expansion

The concept of a series expansion is fundamental for breaking down complex functions into simpler polynomials. The Taylor series expresses a function as an infinite sum of terms calculated from its derivatives at a single point. The general formula is:

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

This series provides a way to approximate the function near the central point \( c \). The series expansion relies on calculating the derivatives and using these to form the coefficients of the polynomial terms, which represent the behavior of the function around \( c \). Understanding the series expansion is crucial as it allows us to represent the function in terms of powers of \( x \), simplifying calculations for analysis or estimation.

Coefficient Pattern

Identifying a pattern in the coefficients is key to simplifying the Taylor series, especially for higher derivative terms. In the given exercise, the coefficients are derived from evaluating the derivatives at \( x = 1 \) and show specific behavior:

• Alternating in sign
• Decreasing in magnitude

The pattern observed was that the coefficients could be represented as \( \frac{(-1)^{n+1}(n-1)!}{2^n} \) for \( n \geq 1 \). Understanding such a pattern can drastically reduce computation time as it allows direct formulation of the coefficients without needing to calculate every higher-order derivative explicitly.

Mathematical Formula

The final mathematical formula represents the nth term of the Taylor series based on the observed pattern. After calculating the derivatives and recognizing the coefficient pattern, the nth term is expressed as:

\[ a_n = \frac{(-1)^{n+1}(n-1)!}{2^n} \]

Substituting this into the formula for the Taylor series term, we get:

\[ \frac{(-1)^{n+1}(n-1)!}{2^n n!}(x-1)^n \]

This formula captures the trend of the coefficients and combines it with the power series in \( (x-1)^n \). It's important to note that understanding this formula allows us to effectively compute any term in the Taylor series without recalculating all preceding derivatives.