Problem 16 - Single Variable Calculus

Question

Use the power series representations of functions established in this section to find the Taylor series of \(f\) at the given value of \(c .\) Then find the radius of convergence of the series. \(f(x)=\frac{1}{4+x^{2}}, \quad c=0\)

Step-by-Step Solution

Answer

The Taylor series of \(f(x) = \frac{1}{4+x^2}\) at \(c=0\) is approximated by \(f(x) \approx \frac{1}{4} + \frac{1}{8}x^2\), and the radius of convergence is infinite.

Step 1: Basic understanding of the Taylor series formula

The Taylor series formula for a function \(f(x)\) at \(x=c\) is given by: \[f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + ...\] To find the Taylor series of \(f(x) = \frac{1}{4+x^2}\) at \(c=0\), we'll first have to find its derivatives and then compute their values at \(x=0\).

Step 2: Find the derivatives of \(f(x)\)

Let's find the first few derivatives of the function \(f(x) = \frac{1}{4+x^2}\) at \(x=0\). \(f(x) = (4+x^2)^{-1}\) \(f'(x) = \frac{d}{dx} (4+x^2)^{-1} = -1 (4+x^2)^{-2} (2x)\) \(f''(x) = -2(4+x^2)^{-3}(2x)^2 + 2(4+x^2)^{-2}\) \(f'''(x) = -6(4+x^2)^{-4} (2x)^3 + 12(4+x^2)^{-3}(2x)\) Now, let's compute their values at \(x=0\).

Step 3: Evaluate derivatives at \(c=0\)

Evaluate each derivative at \(c=0\). \(f(0) = \frac{1}{4}\) \(f'(0) = -1(4)^{-2}(2 \cdot 0) = 0\) \(f''(0) = -2(4)^{-3}(2 \cdot 0)^2 + 2(4)^{-2} = \frac{1}{8}\) \(f'''(0) = -6(4)^{-4}(2 \cdot 0)^3 + 12(4)^{-3}(2 \cdot 0) = 0\)

Step 4: Use the Taylor series formula

Using the Taylor series formula and the values of the derivatives computed at \(c=0\), we obtain the Taylor series approximation for \(f(x) = \frac{1}{4+x^2}\): \(f(x) \approx \frac{1}{4} + 0(x) + \frac{1}{8}(x)^2 + 0(x)^3 + ... = \frac{1}{4} + \frac{1}{8}x^2\)

Step 5: Find the radius of convergence

We'll use the ratio test to find the radius of convergence. For the series \(\sum_{n=0}^{\infty} a_n x^n\), the ratio test states: \(\lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} = L\) If \(L < 1\), the series converges; if \(L > 1\), the series diverges. For our Taylor series, we can see that only the \(n=0\) and \(n=2\) terms exist: \(a_0 = \frac{1}{4}\) and \(a_2 = \frac{1}{8}\). So, the ratio will be: \(L = \frac{1/8}{1/4} = \frac{1}{2}\) Since \(L < 1\), the series converges, and thus, the radius of convergence for the Taylor series of \(f(x) = \frac{1}{4+x^2}\) at \(c=0\) is infinite. To summarize, the Taylor series of \(f(x) = \frac{1}{4+x^2}\) at \(c=0\) is: \(f(x) \approx \frac{1}{4} + \frac{1}{8}x^2\) and the radius of convergence is infinite.

Key Concepts

Power Series Representations

Understanding power series representations is a foundation in calculus for exploring the behavior of functions near a given point. A power series is an infinite sum of the form:

\[\begin{equation}\text{Power Series: } \[\sum_{n=0}^{\infty} a_n (x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + ...\]\end{equation}\]
where each coefficient 'a_n' is a constant and 'c' is the center of the series. Taylor series are a specific type of power series where the coefficients 'a_n' are determined by the derivatives of the function at the center 'c'.

When we seek the Taylor series of a function at a certain point, we calculate these coefficients by evaluating the function's derivatives at the center as shown in the previous exercise.

Finding the Series

In the provided example, the Taylor series is formed by first finding the derivatives of the function \(f(x) = \frac{1}{4+x^2}\) and then evaluating at \(c=0\). The coefficients for the series are derived from the function values and its derivatives at zero: