Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

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, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=x \sec \left(x^{2}\right)+\sin x\)

Step-by-Step Solution

Answer

The terms through \( x^5 \) are \( 2x - \frac{x^3}{6} + \frac{61}{120} x^5 \).

Step 1: Identify Known Maclaurin Series

First, we'll identify the known Maclaurin series for functions involved in our problem: - The Maclaurin series for \( \sec(x) \) is \( 1 + \frac{x^2}{2!} + \frac{5x^4}{4!} + \cdots \).- The Maclaurin series for \( \sin(x) \) is \( x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \).- Since \( f(x) = x \sec(x^2) + \sin(x) \), we need \( \sec(x^2) \), so substitute \( x^2 \) into the series for \( \sec(x) \).

Step 2: Expand \( \sec(x^2) \) Series

Substitute \( x^2 \) into the known Maclaurin series for \( \sec(x) \):\[\sec(x^2) = 1 + \frac{(x^2)^2}{2!} + \frac{5(x^2)^4}{4!} + \cdots = 1 + \frac{x^4}{2} + \cdots \]We only keep terms through \( x^4 \) as we multiply by \( x \) in the next step.

Step 3: Multiply \( x \sec(x^2) \)

Multiply the series we found for \( \sec(x^2) \) by \( x \):\[x \sec(x^2) = x \left(1 + \frac{x^4}{2}\right) = x + \frac{x^5}{2}\]This keeps terms only through \( x^5 \).

Step 4: Add \( \sin(x) \) Series Terms

Now, add the terms from the \( \sin(x) \) series:\[\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120}\]Combine these with the terms from \( x \sec(x^2) \):\[f(x) = \left(x + \frac{x^5}{2}\right) + \left(x - \frac{x^3}{6} + \frac{x^5}{120}\right)\]

Step 5: Combine Like Terms

Combine all like terms from the series expansions:\[f(x) = (2x) - \frac{x^3}{6} + \left(\frac{x^5}{2} + \frac{x^5}{120}\right)\]Simplify the \( x^5 \) terms:\[f(x) = 2x - \frac{x^3}{6} + \frac{61x^5}{120}\]

Step 6: Simplify and Finalize

Now finalize the series for easy reading:\[f(x) = x \sec(x^2) + \sin(x) \approx 2x - \frac{x^3}{6} + \frac{61}{120} x^5\]This expression contains terms through \( x^5 \).

Key Concepts

Taylor Series

The Taylor series is a way to represent a function as an infinite sum of terms that are calculated from the values of its derivatives at a single point. It is a powerful tool used to approximate complex functions with polynomial expressions. The Maclaurin series is a special case of the Taylor series centered at zero. It provides a useful approximation for functions that can be expanded in this way.
The general form for a Taylor series of a function \( f(x) \) centered at \( a \) is:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \)

When using a Maclaurin series, \( a \) is \( 0 \), making it:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \)

The coefficients in these series are derived from the function's derivatives, and by using enough terms, you can closely approximate the function. This becomes particularly useful in calculus and physics when examining functions that are difficult to work with in their original forms.

Trigonometric Functions

Trigonometric functions are the core functions in trigonometry, dealing with the relationships between side lengths and angles of triangles. The main trigonometric functions include sine, cosine, and tangent, often abbreviated as \( \sin \), \( \cos \), and \( \tan \).
These functions are not only foundational in geometry and calculus but are also essential for solving problems involving waves, oscillations, and circular motion.In series expansions like the Maclaurin series, these trigonometric functions can be expressed as infinite series:

• \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \)
• \( \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \)

One key benefit of expressing trigonometric functions this way is that it simplifies complex trigonometric expressions into polynomial approximations, which are easier to handle analytically. For example, \( \tan(x) = \frac{\sin(x)}{\cos(x)} \) can be expanded by using the series of \( \sin(x) \) and \( \cos(x) \) to provide an approximation.

Function Expansions

Function expansions involve expressing complex functions as power series, which are essentially sums of infinite terms. This transforms functions into a more simplified form that’s easier to work with, especially for analytical and numerical computations.
By expanding a function, we can analyze its behavior by understanding each term's contribution to the function's shape and properties. Common expansions include those for exponential, logarithmic, and trigonometric functions.In the context of the Maclaurin series, we transform functions into a format where they can be easily evaluated for small values of \( x \). This is especially helpful when dealing with composite functions or functions involving multiplication and division of basic functions.

• For instance, if \( f(x) = x \sec(x^2) + \sin(x) \), expanding \( \sec(x^2) \) and \( \sin(x) \) using their Maclaurin series terms provides a manageable way to approximate \( f(x) \).

The process involves identifying known series and substituting and expanding them to approximate our function to a desired degree, often up to a certain power of \( x \). This technique is crucial in fields like engineering and physics, where function approximations can simplify problem-solving processes.