Problem 22 - Calculus: Early Transcendentals

Question

Find the Taylor series for \( f(x) \) centered at the given value of \( a. \) [Assume that \( f \) has a power series expansion. Do not show that \( R_n (x) \to 0.\)] Also find the associated radius of convergence. \( f(x) = 1/x,\) \( a = -3 \)

Step-by-Step Solution

Answer

The Taylor series is \[ -\frac{1}{3} - \frac{1}{9}(x + 3) + \frac{1}{27}(x + 3)^2 - \frac{1}{81}(x + 3)^3 + \cdots \] with radius of convergence \( R = 3 \).

Step 1: Understand the Taylor Series Formula

A Taylor series of a function \( f(x) \) about a point \( a \) is expressed as: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(3)}(a)}{3!}(x-a)^3 + \cdots \]This means we need the derivatives of \( f(x) \) evaluated at \( x = a \).

Step 2: Calculate Derivatives of the Function

Given \( f(x) = \frac{1}{x} \), the derivatives are:- \( f(x) = x^{-1} \) - \( f'(x) = -x^{-2} \) (since \( \frac{d}{dx}(x^{-n}) = -nx^{-(n+1)} \))- \( f''(x) = 2x^{-3} \)- \( f^{(3)}(x) = -6x^{-4} \), and so on. These are calculated by following the pattern \( f^{(n)}(x) = (-1)^n n! x^{-(n+1)}. \)

Step 3: Evaluate Derivatives at \( a = -3 \)

Evaluate the derivatives at \( a = -3 \):- \( f(-3) = -\frac{1}{3} \)- \( f'(-3) = -\frac{1}{9} \)- \( f''(-3) = \frac{2}{27} \)- \( f^{(3)}(-3) = -\frac{2}{27} \cdot 3 = -\frac{6}{81} = -\frac{2}{27} \), and so on. Expand this pattern for further derivatives.

Step 4: Construct the Taylor Series

Substitute the derivatives at \( a = -3 \) into the Taylor series formula: \[ f(x) = -\frac{1}{3} -\frac{1}{9}(x + 3) + \frac{2}{27}\frac{(x + 3)^2}{2!} - \frac{2}{27}\frac{(x + 3)^3}{3!} + \cdots \]Simplify to get: \[ -\frac{1}{3} - \frac{1}{9}(x + 3) + \frac{1}{27}(x + 3)^2 - \frac{1}{81}(x + 3)^3 + \cdots \]

Step 5: Determine the Radius of Convergence

The radius of convergence \( R \) of a Taylor series can be found using the formula for the general term of the derivative function:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{R} \]Here, the terms come from \( a_n = \frac{(-1)^n}{3^{n+1}} \). So we find:\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{1}{3} \right| = \frac{1}{3} \]Thus, the radius of convergence is \( R = 3 \).

Key Concepts

Radius of Convergence

The radius of convergence is a key concept when working with Taylor series. It's the distance from the center of the series, denoted by \( a \), within which the series converges to the given function \( f(x) \).

In our exercise, we calculated the radius of convergence for the Taylor series expansion of \( f(x) = \frac{1}{x} \) centered at \( a = -3 \). To find this, we used the ratio test, which involves examining the limit of the absolute value of the ratio of successive terms:

\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{R} \]

We found that this ratio equals \( \frac{1}{3} \), giving us a radius of convergence \( R = 3 \).

Understanding the radius of convergence helps in determining the interval around \( x = -3 \) on which the series accurately approximates the function without diverging.

Power Series Expansion

Power series expansion is like building a bridge from a complex function to a simpler polynomial. It allows us to express \( f(x) \) as a sum of terms consisting of powers of \((x-a)\), the distance from the center.

The Taylor series is a type of power series centered at a particular point \( a \). For our function \( f(x) = \frac{1}{x} \), the power series expansion around \( a = -3 \) begins with \( f(a) \) then includes terms with increasing derivatives of \( f \) at \( a \), each multiplied by \((x+3)^n\).

The expansion looks like this:

• Constant term: \( f(a) \)
• Linear term: \( f'(a)(x-a) \)
• Quadratic term: \( \frac{f''(a)}{2!}(x-a)^2 \)
• Cubic term: \( \frac{f^{(3)}(a)}{3!}(x-a)^3 \)

This series continues infinitely. Each additional term uses a higher derivative and a higher power of \( (x-a) \), giving the series greater precision in approximating the function within its radius of convergence.

Derivative of a Function

To construct a Taylor series, it's essential to calculate derivatives of the function up to the desired degree. Derivatives capture how fast \( f(x) \) is changing at a certain point and how this change continues for higher orders.

For \( f(x) = \frac{1}{x} \), we applied the power rule, resulting in:\[ f'(x) = -x^{-2}, \quad f''(x) = 2x^{-3}, \quad f^{(3)}(x) = -6x^{-4}, \ldots \]

This pattern can continue indefinitely, providing insights into the curvature and behavior of the function near \( a \). Each derivative, when evaluated at a point \( a \), contributes to a term in the Taylor series.

Understanding derivatives is crucial; they form the building blocks of the power series and represent the function's behavior over its interval of convergence.