Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x)=(1+x)^{1 / 2}+(1-x)^{1 / 2}\). Find the Maclaurin series for \(f\) and use it to find \(f^{(4)}(0)\) and \(f^{(51)}(0)\).

Step-by-Step Solution

Answer

\(f^{(4)}(0) = -\frac{15}{4}\), \(f^{(51)}(0) = 0\).

Step 1: Understand the Problem

We're given the function \(f(x) = (1+x)^{1/2} + (1-x)^{1/2}\). We need to find the Maclaurin series for \(f\), which is the Taylor series around \(x = 0\). Then, from this series, we need to find \(f^{(4)}(0)\) and \(f^{(51)}(0)\).

Step 2: Find the Maclaurin Series for Each Term

To find the Maclaurin series, we'll expand each term separately. For \((1+x)^{1/2}\) and \((1-x)^{1/2}\), use the binomial series expansion:\[(1+x)^n = \sum_{k=0}^{fty} \binom{n}{k} x^k\], where \( \binom{n}{k} = \frac{n(n-1)...(n-k+1)}{k!}\).Apply this with \(n = 1/2\).

Step 3: Expand \((1+x)^{1/2}\) and \((1-x)^{1/2}\)

First, expand \((1+x)^{1/2}\):\[1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots\]Now, expand \((1-x)^{1/2}\):\[1 - \frac{1}{2}x - \frac{1}{8}x^2 - \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots\].

Step 4: Add the Series Together

Add the two series expansions term by term:\[ (1+x)^{1/2} + (1-x)^{1/2} = (1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots) + (1 - \frac{1}{2}x - \frac{1}{8}x^2 - \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots)\].Simplifying, the Maclaurin series for \(f(x)\) is:\[ 2 - \frac{1}{4}x^2 - \frac{5}{64}x^4 + \cdots\].

Step 5: Determine \(f^{(4)}(0)\)

In the Maclaurin series, the coefficient of \(x^4\) is \( -\frac{5}{64}\). The coefficient in a Maclaurin series is \(\frac{f^{(n)}(0)}{n!}\) for \(x^n\). Therefore,\[f^{(4)}(0) = -\frac{5}{64} \times 4! = -\frac{5}{64} \times 24 = -\frac{15}{4}\].

Step 6: Determine \(f^{(51)}(0)\)

Since the Maclaurin series is constructed only up to \(x^4\), and all other coefficients are zero, there is no \(x^{51}\) term in the series of \((1+x)^{1/2}\) and \((1-x)^{1/2}\) combined. Therefore, \(f^{(51)}(0)\) is zero.

Key Concepts

Binomial Series Expansion

The binomial series expansion is a powerful tool in calculus that allows us to expand expressions of the form \((1+x)^n\) into an infinite series. This expansion is given by the formula:

• \((1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k\)
• Where \(\binom{n}{k} = \frac{n(n-1)...(n-k+1)}{k!}\) is known as the binomial coefficient.

When \(n\) is not a positive integer, the series becomes infinite. This is especially useful for approximating functions near \(x=0\) when \(\left|x\right| < 1\). In our specific problem, we used the binomial series to expand \((1+x)^{1/2}\) and \((1-x)^{1/2}\), starting with the first few terms. We plug \(n = 1/2\) into the formula to find each term of the series. For example:

• \((1+x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots\)
• \((1-x)^{1/2} = 1 - \frac{1}{2}x - \frac{1}{8}x^2 - \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots\)

These expansions help us find the Maclaurin series for more complex expressions, like the addition of two such series.

Taylor Series

The Taylor series is a representation of a function as an infinite sum of terms. Each term is derived from the various derivatives of the function, all evaluated at a single point. The Maclaurin series is a type of Taylor series centered at \(x = 0\). The general Taylor series formula is:

• \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\)

For a Maclaurin series, this becomes:

• \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\)

The coefficients \(\frac{f^{(n)}(0)}{n!}\) are key to understanding the Maclaurin series, as they connect derivatives evaluated at zero to the structure of the series. In the exercise, we constructed the Maclaurin series for the function \(f(x) = (1+x)^{1/2} + (1-x)^{1/2}\), resulting in:

• \(f(x) = 2 - \frac{1}{4}x^2 - \frac{5}{64}x^4 + \cdots\)

This simple form makes it easy to find derivatives, like \(f^{(4)}(0)\), by examining the coefficients of the polynomial terms.

Derivative Calculation

Calculating derivatives is fundamental in mathematics, particularly when working with Taylor and Maclaurin series. In these series, derivatives are crucial for finding the coefficients, as each coefficient is the derivative of the original function evaluated at the center point divided by the factorial of the term's degree. For example, for the term \(x^n\) in the series:

• The coefficient is \(\frac{f^{(n)}(0)}{n!}\).
• This means if you have a coefficient \(a_n\) at \(x^n\), the derivative can be found by rearranging: \(f^{(n)}(0) = a_n \cdot n!\).

In the given problem, after finding the Maclaurin series expansion, we needed to determine the derivatives \(f^{(4)}(0)\) and \(f^{(51)}(0)\). By noting that the coefficient at \(x^4\) was \(-\frac{5}{64}\), we computed:

• \(f^{(4)}(0) = -\frac{5}{64} \times 4! = -\frac{15}{4}\)

And since there was no term \(x^{51}\):

• \(f^{(51)}(0) = 0\)

These calculations show how derivatives play a critical role in understanding the series' behavior.