Problem 7 - 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, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=e^{x}+x+\sin x $$

Step-by-Step Solution

Answer

The Maclaurin series for \(f(x) = e^x + x + \sin x\) through \(x^5\) is \(1 + 2x + \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^5}{60}\).

Step 1: Recall the Maclaurin Series for Basic Functions

The Maclaurin series expansion for each function related to our problem is as follows:1. For \(e^x\):\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots\]2. For \(x\):\[x = x\]3. For \(\sin x\):\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\]

Step 2: Write the Series up to the Fifth Degree

Since we need terms up to \(x^5\), we'll write the expansions for each function up to that degree:- **\(e^x\):** \[e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \cdots \]- **\(x\):** \[x = x \]- **\(\sin x\):** \[\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} \]

Step 3: Add the Series Together

To find the Maclaurin series for \(f(x) = e^x + x + \sin x\), we sum the series expressions from Step 2:1. Combine the constant and linear terms: \[ (1) + (x + x) = 1 + 2x \]2. The quadratic term comes only from \(e^x\): \[ \frac{x^2}{2} \]3. Combine the cubic terms: \[ \frac{x^3}{6} - \frac{x^3}{6} = 0 \text{ (they cancel out)} \]4. The quartic term again comes only from \(e^x\): \[ \frac{x^4}{24} \]5. Combine the quintic terms: \[ \frac{x^5}{120} + \frac{x^5}{120} = \frac{x^5}{60} \]

Step 4: Write Final Series Expression

Combine all the terms to write the full Maclaurin series expansion of \(f(x)\):\[f(x) = 1 + 2x + \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^5}{60}\]This is the expression for \(f(x)\) up to the term \(x^5\).

Key Concepts

Series Expansion

The concept of series expansion is fundamental when it comes to understanding complex functions. It's a way of representing functions as an infinite sum of terms, allowing us to approximate them with a finite number of terms for practical calculations. In particular, the Maclaurin series is a type of series expansion for functions centered at zero. This is highly useful in calculus because it provides a polynomial approximation of a function, which is much simpler to work with.

With the Maclaurin series expansion, we can express complex functions as a sum of basic function series. This involves identifying the standard series for basic functions like exponentials and trigonometric functions. - For instance, the Maclaurin series for the exponential function is simply the sum of all powers of the variable, each divided by its factorial. - Trigonometric functions are expressed with alternating series due to their inherent oscillatory nature.

Through the addition and manipulation of these basic series, we can accurately model more complicated functions with ease.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, and integrals. It's crucial in understanding changes and motion. The role of calculus in series expansion, particularly the Maclaurin series, is to simplify and approximate calculations that would otherwise be rather complex.

When dealing with Maclaurin series, calculus helps us:

• Derive the coefficients for each term of the series, ensuring the approximation closely matches the actual function's behavior around zero.
• Understand the behavior of a function at and near a particular point, which is critical for analysis and problem-solving in physics, engineering, and other sciences.

In our case of expanding the function \(f(x) = e^{x} + x + \sin x\), calculus allows us to combine the known expansions of each component and evaluate the sum. This leads us to a polynomial that estimates the function closely for small values of \(x\).

Trigonometric Functions

Understanding trigonometric functions and their behaviors is key in many mathematical applications. In calculus, these functions often involve series due to their periodic nature.

Trigonometric functions, like \(\sin x\) and \(\cos x\), have specific Maclaurin series that account for their oscillatory patterns. For example:

• The Maclaurin series for \(\sin x\) is \(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\).
• This series alternates due to the sine wave’s rise and fall, reflecting the function's inherent nature.

In the problem at hand, employing the series of \(\sin x\) enables us to represent this function as a polynomial, easily combined with the other components of \(f(x)\).

Ultimately, through series expansion, calculus, and an understanding of trigonometric functions, complex computations become manageable, helping us to solve otherwise difficult problems efficiently.