Problem 26 - Essential Calculus

Question

Find the Maclaurin series for the function. (Use the table of power series for elementary functions.) $$ g(x)=2 \sin x^{3} $$

Step-by-Step Solution

Answer

\(g(x) = 2 [x^3 - 1/3! x^9 + 1/5! x^{15} - 1/7! x^{21} +\ldots]\)

Step 1: Substitute function into the Maclaurin series for sin(x)

The key to solve this problem is recognizing that \(g(x)\) is simply a transformation of the function \(\sin x\). To find the Maclaurin series for \(g(x)\), we have to start from the Maclaurin series for \(\sin x\) and replace \(x\) with \(x^3\) which gives us: \(2 \sin x^3 = 2 [x^3 - (x^3)^{3}/3! + (x^3)^{5}/5! - (x^3)^{7}/7! +\ldots]\)

Step 2: Simplify the series

Take out the powers, apply them to the x-values and simplify the series: \(2 \sin x^3 = 2x^3 - 2/3! x^9 + 2/5! x^{15} - 2/7! x^{21} +\ldots\)

Step 3: Factor out the coefficient 2

Finally, factor out the coefficient 2: \(g(x) = 2x^3 - 2/3! x^9 + 2/5! x^{15} - 2/7! x^{21} +\ldots = 2 [x^3 - 1/3! x^9 + 1/5! x^{15} - 1/7! x^{21} +\ldots]\) which is the Maclaurin series for \(g(x)=2 \sin x^{3}\)

Key Concepts

Power Series

A power series is an infinite series of the form \( a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots \). This means each term of a power series has a coefficient and is raised to a power based on its position in the series. Power series are incredibly powerful tools because they allow us to express functions as an infinite sum of terms, which can sometimes make complex functions easier to work with.

Power series expansions are useful for approximating functions, especially around a specific point, often 0. This is especially handy in calculus and physics, where understanding how functions behave in local regions is crucial. By using a power series, you can represent even undefined or strange functions around certain points by using only polynomials, which are simple and easy to handle.

• One common example is the Maclaurin series, which is a specific type of power series centered at 0.
• In our exercise, substituting into a power series helps transform the given function into an accessible form.

Elementary Functions

Elementary functions are the basic building blocks in mathematics, including functions like polynomials, exponents, logarithms, and trigonometric functions such as \( \sin x \), \( \cos x \), and \( \tan x \). These functions are called 'elementary' because they represent fundamental operations and transformations that most other functions derive from or can be reduced to.

In calculus, especially when dealing with power series like Maclaurin or Taylor series, elementary functions become integral to simplifying and solving equations.

• The Maclaurin series itself relies heavily on elementary functions since it provides a polynomial approximation of these functions around zero.
• In the exercise, \( \sin x \) as an elementary function is transformed by substituting \( x^3 \), showcasing the versatility and power of algebraic manipulation with elementary functions.

Sin Function

The \( \sin x \) function is one of the fundamental trigonometric functions in mathematics. It describes the y-coordinate of a point on the unit circle as that point moves around the circle. The sine function oscillates between -1 and 1, making it periodic. Its natural period is \( 2\pi \), which means that every \( 2\pi \) units along the x-axis, the sine function repeats its values.

When expanding \( \sin x \) into a series around 0, we get its Maclaurin series, which is expressed as:\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots\]This series expansion is infinitely long, making it a fantastic approximation tool for the sine function, especially around x = 0.

• In applied mathematics and physics, this series is incredibly useful for calculating values of \( \sin x \) for small angles.
• In our problem, substituting \( x^3 \) changes \( \sin x \) into \( \sin x^3 \), allowing the exploration of the behavior in a new dimension while still maintaining the periodic properties natural to the sine function.

The sine function, when employed in a series expansion like this, underscores both the beauty and the complexity of trigonometric functions in calculus.