Problem 65 - Calculus Early Transcendentals

Question

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. In most cases you do not need to use the definition of the Taylor series coefficients. b. If possible, determine the radius of convergence of the series. $$f(x)=\left(1+x^{2}\right)^{-2 / 3}$$

Step-by-Step Solution

Answer

Short Answer: The first four nonzero terms of the Taylor series of the function $$f(x) = (1+x^2)^{-\frac{2}{3}}$$ centered at 0 are $$f(x)=1+5x^2$$. The radius of convergence is infinite.

Step 1: Evaluate the function, and its first three derivatives, at x=0

To find the nonzero terms of the Taylor series, we will start by evaluating the function at x=0, and then we will find its first three derivatives and evaluate each of them at x=0. Since the Taylor series centered at 0 starts with the function itself, we will first evaluate $$f(0)=(1+0)^{-\frac{2}{3}}=1$$ Now, let's find the first derivative of the function $$f'(x)$$: $$f'(x)= -\frac{2}{3}(1+x^2)^{-\frac{5}{3}}(2x)$$ Evaluate the first derivative at x=0: $$f'(0)=-\frac{2}{3}(1+0)^{-\frac{5}{3}}(2\cdot 0)=0$$ We will continue this process for the second and third derivatives: The second derivative is: $$ f''(x) = \frac{10}{3}(1+x^{2})^{-\frac{8}{3}}(2x) + \frac{2}{3}(5)(1+x^{2})^{-\frac{5}{3}}$$ Evaluate the second derivative at x=0: $$ f''(0) = \frac{10}{3}(1+0)^{-\frac{8}{3}}(2\cdot 0) + \frac{2}{3}(5)(1+0)^{-\frac{5}{3}}=10$$ The third derivative is: $$ f'''(x) = ... = -\frac{56}{3}(1+x^{2})^{-\frac{11}{3}}(2x) - \frac{4}{\sqrt[3]{1+x^2}}$$ Evaluate the third derivative at x=0: $$f'''(0) = ... = -\frac{56}{3}(1+0)^{-\frac{11}{3}}(2\cdot 0) - \frac{4}{\sqrt[3]{1+0}}=0$$

Step 2: Write out the Taylor series terms

Now that we have the evaluated function and its derivatives at x=0, we can use those values to construct the first four nonzero terms of the Taylor series. The Taylor series for the function $$f(x)$$ is given by: $$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$ So, the first four nonzero terms for the given function are: $$f(x)=1+0x+\frac{10}{2!}x^2+0x^3$$ Simplifying this expression, we get $$f(x)=1+5x^2$$ Now, let's determine the radius of convergence.

Step 3: Determine the radius of convergence

To find the radius of convergence of the Taylor series, we use the Ratio Test: $$\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$$ For this function, since the series converges only for the first four terms, the ratio is always less than 1 and the rest of the terms are 0. Thus, the Radius of Convergence is infinite. In conclusion, the first four nonzero terms of the Taylor series of the function $$f(x)=(1+x^2)^{-2/3}$$ centered at 0 are: $$f(x)=1+5x^2$$ And the radius of convergence of the series is infinite.

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when working with Taylor series. It tells us the interval over which the Taylor series converges to the function. To determine the radius of convergence, we can often use the Ratio Test, which is a method to examine the limit of the ratio of successive terms in a series.

For a series \( \sum a_n x^n \), the Ratio Test involves checking where

• \( \lim_{n \to \infty} \left| \frac{a_{n+1}x^{n+1}}{a_n x^n} \right| = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \cdot |x| < 1 \).

To find the radius of convergence, we look at the distance from the center of the series (often 0) within which this ratio stays less than 1. If the limit of the ratio is zero for terms beyond a finite number or if it inherently stays less than 1 for all terms, we say that the radius of convergence is infinite.

In the given problem, the series only contains a finite number of terms such as \(1 + 5x^2\) and the rest become zero, so the Radius of Convergence is infinite. This means the series accurately represents the function across all real numbers.

Derivatives

Derivatives are the backbone of forming Taylor series. By taking derivatives, we capture how a function changes and express it as a polynomial. The Taylor series uses these derivatives evaluated at a point to approximate the function around that point.

Let's break this down:

• The 0th derivative is just the function value itself, like \(f(0)\).
• The 1st derivative \(f'(x)\) shows how the function changes instantly. Evaluating \(f'(0)\) gives us an idea of the slope at the center point.
• Higher-order derivatives such as the 2nd and 3rd give more refined details about the function's curvature and shape near the center point.

In the exercise, by taking the derivatives up to the third and evaluating them as \(x = 0\), we got specific values such as \(f''(0) = 10\), while \(f'(0)\) and \(f'''(0)\) were zero. This determined which terms contribute to the Taylor series.

Analytical Methods

Using analytical methods in calculus helps us systematically break down complex problems into simpler pieces. For Taylor series, analytical methods involve computing derivatives efficiently and employing convergence tests correctly.

The process consists of:

• Identifying the function to be approximated, for example, \((1+x^2)^{-2/3}\).
• Calculating derivatives step by step and evaluating them at the chosen center (often zero), where the series is centered.
• Compiling these derivatives into a polynomial series which represents the function.

By using analytical techniques, we skip directly working with infinite series or overly complex expressions and instead focus only on meaningful terms in the function's expansion. This speeds up the solution process and uncovers the function's behavior accurately. In the context of the exercise, these methods enabled quick identification of nonzero terms in the Taylor series and the proper assessment of convergence, as deduced from the derivatives and evaluating them at specific points.