Problem 3 - The Calculus with Analytic Geometry

Question

Use a binomial series to find the Maclaurin series for the given function. Determine the radius of convergence of the resulting series.\(f(x)=\left(4+x^{2}\right)^{-1}\)

Step-by-Step Solution

Answer

The Maclaurin series is \(1 - \frac{x^2}{4} + \frac{x^4}{32} - \frac{x^6}{256} + \ldots \), and the radius of convergence is \ |x| < 2 \.

Step 1 - Identify the Function Form

Notice that the given function is in the form of \( (a + b)^{-n} \). Here, \( a = 4 \), \( b = x^2 \), and \( n = 1 \). Therefore, \( f(x) = (4 + x^2)^{-1} \).

Step 2 - Write the Binomial Series Expansion

The binomial series expansion for \((1 - u)^{-n}\) is \[ 1 + n u + \frac{n(n+1)}{2!} u^2 + \frac{n(n+1)(n+2)}{3!} u^3 + \ldots \]. Here, let \( u = -\frac{x^2}{4} \) and \( n = 1 \).

Step 3 - Substitute and Simplify

Substitute \( n \) and \( u \) into the binomial series expansion. This gives: \[ 1 + \frac{-x^2}{4} + \frac{1(2)}{2!} \left( -\frac{x^2}{4} \right)^2 + \frac{1(2)(3)}{3!} \left( -\frac{x^2}{4} \right)^3 + \ldots \].

Step 4 - Expand and Simplify Further

Simplify each term to obtain: \[ 1 - \frac{x^2}{4} + \frac{x^4}{32} - \frac{x^6}{256} + \ldots \]. Thus, the Maclaurin series for \( (4 + x^2)^{-1} \) is \[ 1 - \frac{x^2}{4} + \frac{x^4}{32} - \frac{x^6}{256} + \ldots \].

Step 5 - Determine the Radius of Convergence

To find the radius of convergence, use the general term and apply the ratio test. The general term is \[ (-1)^k \frac{(x^2)^{k}}{4^k} \]. Apply the ratio test: \[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \left| \frac{(x^2)^{k+1}}{4^{k+1}} \cdot \frac{4^k}{(x^2)^k} \right| = \left| \frac{x^2}{4} \right| \]. Set this less than 1 for convergence: \[ \left| \frac{x^2}{4} \right| < 1 \Rightarrow |x^2| < 4 \Rightarrow |x| < 2 \]..

Key Concepts

Binomial Series Expansion

The binomial series helps us expand expressions involving roots or powers of a sum, \( (a + b)^n \). For any real number \( n \) and \( |u| < 1 \), the series expansion for \( (1 + u)^n \) is given by:
\[ 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots \]
In our example, the function \( (4 + x^2)^{-1} \) fits this form where \( a = 4, b = x^2, \) and \( n = -1 \).
To convert it into the binomial series form, we write:
\[ (4 + x^2)^{-1} = 4^{-1}(1 + \frac{x^2}{4})^{-1} = \frac{1}{4}(1 - \frac{x^2}{4})^{-1} \]
With \( u = -\frac{x^2}{4} \) and \( n = -1 \), our binomial series expansion becomes:
\[ \frac{1}{4}(1 - \frac{x^2}{4})^{-1} = \frac{1}{4}[ 1 + (-1) \frac{x^2}{4} + \frac{(-1)(-2)}{2!} \left(\frac{x^2}{4} \right)^2 + \ldots ] \]
Next we'll expand and simplify these terms.

Radius of Convergence

The radius of convergence determines how far we can extend our series for it to remain valid.
We use the ratio test to find this radius. If the general term of our series is \( a_k = (-1)^k \frac{(x^2)^k}{4^k} \), we apply the ratio test:
\[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \left| \frac{(x^2)^{k+1}}{4^{k+1}} \cdot \frac{4^k}{(x^2)^k} \right| = \left| \frac{x^2}{4} \right| \] We set this less than 1 for convergence:
\[ \left| \frac{x^2}{4} \right| < 1 \Leading to \left| x^2 \right| < 4 \Rightarrow \left| x \right| < 2 \]
Therefore, the radius of convergence of our series is 2.

Ratio Test

The ratio test helps to find the series' convergence by comparing the ratios of successive terms.
It states:

• If the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \), the series converges.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

For our series with general term \( a_k \frac{(x^2)^k}{4^k} \), substituting and simplifying gives:
\[ L = \left| \frac{x^2}{4} \right| \]
To ensure convergence, we set this less than 1:
\[ \left| \frac{x^2}{4} \right| < 1 \Leading to \left| x^2 \right| < 4 \Rightarrow \left| x \right| < 2 \]
This confirms our radius of convergence is 2.