Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-18, find the terms through \(x^{5}\) in the Maclaurin series for \(f(x)\). Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=\frac{\cos x}{\sqrt{1+x}} $$

Step-by-Step Solution

Answer

The Maclaurin series up to \( x^5 \) is \( 1 - \frac{1}{2}x^2 - \frac{5}{16}x^3 + \frac{1}{8}x^4 - \frac{103}{256}x^5 \).

Step 1: Recall Maclaurin Series for Common Functions

We start by recalling the Maclaurin series of the known functions. The Maclaurin series for \( \cos x \) is:\[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + R_5(x) \]The Maclaurin series for \( (1+x)^{-1/2} \) via the binomial series is:\[ (1+x)^{-1/2} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \frac{35}{128}x^4 - \frac{63}{256}x^5 + R_6(x) \]

Step 2: Multiply the Series

To find the Maclaurin series for \( f(x) = \frac{\cos x}{\sqrt{1+x}} \), multiply the series for \( \cos x \) and \( (1+x)^{-1/2} \). First few terms needed:- \( \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} \)- \( (1+x)^{-1/2} = 1 - \frac{1}{2} x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \frac{35}{128}x^4 \)Multiply term by term and collect terms up to \( x^5 \):

Step 3: Calculate Specific Terms

Calculate the coefficients for specific powers of \( x \).- Constant term: \( 1 \times 1 = 1 \)- Coefficient of \( x \): \( 0 \)- Coefficient of \( x^2 \): \( - \frac{1}{2} \)- Coefficient of \( x^3 \): \( - \frac{5}{16} \)- Coefficient of \( x^4 \): \( \frac{3}{8} - \frac{1}{2}\cdot\frac{1}{2} = \frac{3}{8} - \frac{1}{4} = \frac{1}{8} \)- Coefficient of \( x^5 \): \( -\frac{63}{256} - \frac{1}{2}\cdot\frac{5}{16} = -\frac{63}{256} - \frac{5}{32} \) Calculate precise values for \( x^5 \):- \( \frac{5}{32} = \frac{40}{256} \)- \( -\frac{63}{256} - \frac{40}{256} = -\frac{103}{256} \)

Step 4: Aggregate the Series

Aggregate the terms to express the full series up to \( x^5 \):\[ f(x) \approx 1 - \frac{1}{2}x^2 - \frac{5}{16}x^3 + \frac{1}{8}x^4 - \frac{103}{256}x^5 \]

Key Concepts

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent play a crucial role in mathematics, especially within calculus. These functions help us describe relationships within triangles and model periodic phenomena. In physics and engineering, for example, they are essential to understand waves and oscillations.

The cosine function, specifically, is pivotal in this context. Its Maclaurin series expansion helps approximate its behavior for small values of \(x\). The basic formula for the cosine function is \[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \]

This series is infinite, but for practical purposes, we often limit it to a few terms. This simplifies calculations while maintaining a usable approximation, thus aiding in the solving of complex calculus problems.

Polynomial Approximation

Polynomial approximation is a technique used to simplify complex functions by approximating them with a polynomial. The idea is to use a simpler function that is still closely related in behavior to the more complex original.

The Maclaurin series is a common type of polynomial approximation where a function is expanded around zero. By using the points and derivatives of the original function at zero, we derive a polynomial that closely mirrors the function near this point.

In our problem, to approximate \(f(x) = \frac{\cos x}{\sqrt{1+x}}\), we utilize the Maclaurin series for both \(\cos x\) and \((1+x)^{-1/2}\). This approach transforms the task of approximating a challenging function into more manageable steps by calculating with polynomials.

Binomial Series

The binomial series is a powerful tool for expanding functions of the form \((1+x)^n\) where \(n\) is any number, including non-integers. This series generalizes the binomial theorem and can be used to express functions like \((1+x)^{-1/2}\).

The binomial series for \((1+x)^{-1/2}\) is given as:\[(1+x)^{-1/2} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \frac{35}{128}x^4 - \frac{63}{256}x^5 + \cdots\]

By applying this series, we can approximate the behavior of the square root function over small values of \(x\), which is particularly useful when combined with the series for \(\cos x\) to approximate \(f(x) = \frac{\cos x}{\sqrt{1+x}}\).

Calculus Techniques

Calculus techniques play an essential role in deriving and working with Maclaurin series. In this exercise, we use differentiation, series expansion, and multiplication of series, all of which are fundamental calculus operations.

1. **Differentiation:** Calculating derivatives allows us to find the coefficients for each term in the series, giving the polynomial its shape.2. **Series Expansion:** Expanding functions into their Maclaurin series forms simplifies them, providing a polynomial form that is easier to handle.3. **Multiplication of Series:** By multiplying the series of \(\cos x\) and \((1+x)^{-1/2}\), we combine the two approximations to arrive at the composite function's series.

These techniques let us convert a complex problem into manageable polynomial calculations, enabling the analysis of function behavior and facilitating practical problem-solving.