Problem 15 - 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, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=x \sec \left(x^{2}\right)+\sin x $$

Step-by-Step Solution

Answer

The Maclaurin series up to \(x^5\) for \(f(x) = x \sec(x^2) + \sin x\) is \(2x - \frac{x^3}{6} + \frac{61}{120}x^5\).

Step 1: Identify Known Maclaurin Series

The known Maclaurin series expansions that are useful include: 1. \( \sec(x) = 1 + \frac{1}{2}x^2 + \frac{5}{24}x^4 + \cdots \)2. \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \) We can use these expansions for finding the series up to the order of \(x^5\) in \(f(x)\).

Step 2: Substitute and Expand \(x \sec(x^2)\)

Substitute \(y = x^2\) into the \( \sec(x)\) formula:\[ \sec(y) = 1 + \frac{y^2}{2} + \frac{5}{24}y^4 + \cdots \]This gives \( \sec(x^2) = 1 + \frac{x^4}{2} + \cdots \).Multiply by \(x\):\[ x \sec(x^2) = x \left(1 + \frac{x^4}{2}\right) = x + \frac{x^5}{2} \]

Step 3: Expand \( \sin(x)\) to \(x^5\)

Using the Maclaurin series expansion of \(\sin x\), expand to the term where \(x^5\):\[ \sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} \]This expansion gives us the necessary terms to the order of \(x^5\).

Step 4: Add the Series

Add the results from the expansions above. Combine terms:\[ f(x) = (x + \frac{x^5}{2}) + (x - \frac{x^3}{6} + \frac{x^5}{120}) \]Combine like terms:\[ f(x) = 2x - \frac{x^3}{6} + \frac{61}{120}x^5 \]This gives the Maclaurin series expansion up to and including \(x^5\).

Step 5: Verify Combining of Like Terms

To ensure accuracy, carefully add the coefficients:- For \(x\): \(x + x = 2x\)- For \(x^3\): \(-\frac{x^3}{6}\)- For \(x^5\): \(\frac{x^5}{2} + \frac{x^5}{120} = \frac{60x^5}{120} + \frac{x^5}{120} = \frac{61x^5}{120}\).

Key Concepts

Secant Function Expansion

When tackling series expansion problems like the one with the function \( f(x) = x \sec(x^2) + \sin x \), it's crucial to start with well-known series expansions. For the secant function, the expansion starts with \( \sec(x) = 1 + \frac{1}{2}x^2 + \frac{5}{24}x^4 + \ldots \).

In the problem, we substitute \( x^2 \) into the secant expansion, forming \( \sec(x^2) = 1 + \frac{x^4}{2} + \ldots \). This transformation helps align the series to terms more applicable for multiplication or further operations. Remember, these expansions are only approximations, focusing on small values of \( x \).

Finding the expansion for \( x \sec(x^2) \) involves just multiplying \( x \) again with the expanded series. This results in \( x \sec(x^2) = x + \frac{x^5}{2} + \ldots \). Keeping each expansion at terms up to \( x^5 \) ensures consistency and accuracy.

Sine Function Expansion

The sine function is a staple in calculus for series expansions due to its straightforward Taylor series form. The Maclaurin series for \( \sin(x) \) is given by \( x - \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots \). These terms reveal the essential polynomial approximations of sine.

Knowing these basic relationships lets us apply those to the exercise at hand. Here, we isolate terms up to the order of \( x^5 \), which in this case are \( x - \frac{x^3}{6} + \frac{x^5}{120} \). Recognizing the factorial in each component helps acknowledge why these approximations work especially for small values of \( x \).

Building upon this, combining the sine expansion with other series terms allows us to match the entirety of the function given in \( f(x) \). Understanding sine's role in these expansions gives a clearer picture of series techniques used in calculus.

Series Expansion Techniques

Series expansions offer a powerful technique in calculus to approximate functions. By using series like Maclaurin's, we simplify complex trigonometric or exponential functions into polynomial expressions. This is particularly useful in solving calculus problems involving derivatives, integrals, and more.

The technique involves starting with known expansions. For this exercise, the individual Maclaurin series of \( \sec(x^2) \) and \( \sin(x) \) serve as stepping stones. Incorporating basic algebraic operations — such as substitutions, multiplications, and additions — aids in achieving the resulting polynomial.

Key steps include:

• Substitute appropriate values to align with the target function.
• Multiply and add series terms carefully to avoid losing terms or miscalculating coefficients.
• Ensure all polynomial terms are matched up to the desired order, in this case, \( x^5 \).

Mastering these techniques sharpens one's calculus problem-solving skills.

Calculus Problems

Calculus problems often require understanding series expansions to simplify and solve effectively. This particular exercise illustrates the utility of such expansions in breaking down a function like \( f(x) = x \sec(x^2) + \sin x \).

Problems in calculus will often ask for the expansion of more complex expressions into series forms, requiring expertise in

• Identifying related and known series expansions.
• Combining and simplifying series expansions accurately.
• Applying algebraic manipulation to reach the simplest form.

Solving these problems often enhances both computational skills and theoretical understanding in calculus.