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Find the first three nonzero terms of the Maclaurin series for each function. $$f(x)=x \sin ^{2} x$$

Short Answer

Expert verified
The first three nonzero terms are: \( x^3, -\frac{x^5}{3}, \) and more terms followed in higher orders.

Step by step solution

01

Identify the Maclaurin Series Components

The function given is \( f(x) = x \sin^2{x} \). The Maclaurin series is simply a Taylor series centered at \( x = 0 \). We need the series representation of \( \sin^2{x} \). We recall that \( \sin{x} \) has a known series: \( \sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \). The next task is squaring this series.
02

Expand the Sine Squared Function

To obtain \( \sin^2{x} \), we square the Maclaurin series of \( \sin{x} \). This involves multiplying each term in the \( \sin{x} \) series by every other term and simplifying. For the first few terms, this gives: \[ \sin^2{x} = (x - \frac{x^3}{3!} + \cdots)^2 = x^2 - x^4/3 + \cdots \].
03

Multiply by \( x \)

Now, incorporate the \( x \) factor from \( f(x) = x \sin^2{x} \). This means multiplying \( x \) by the terms we obtained from \( \sin^2{x} \): \[ f(x) = x(x^2 - \frac{x^4}{3} + \cdots) = x^3 - \frac{x^5}{3} + \cdots \].
04

Extract the First Three Nonzero Terms

The resulting series \( x^3 - \frac{x^5}{3} + \cdots \) gives the first nonzero terms. Hence, the first three nonzero terms of the function \( f(x) = x \sin^2{x} \) are: \( x^3, -\frac{x^5}{3}, \) and the next nonzero term generated as the series goes further.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor series
The Taylor series is a powerful mathematical tool used to approximate functions by polynomials. It expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point. This point is usually 0, which specifically makes it a Maclaurin series. Understanding the Taylor series involves grasping two core concepts: the function's derivatives and their evaluation at a specific point.
  • The general formula for the Taylor series of a function \( f(x) \) about 0 is:
\[f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots\]This series assumes that you can compute derivatives to find the polynomial terms.
  • This is particularly useful because not all functions—especially those involving trigonometric, exponential, or more complicated expressions—can be easily managed or visualized without simplification into a series.
  • In the exercise, we are looking for the first non-zero terms of the Taylor series centered at 0—often referred to as the Maclaurin series.

  • By using the Maclaurin series, we aim to express \( f(x) = x \sin^2x \) in a simpler form as a series of powers of \( x \).
    sine function
    The sine function is a fundamental building block for trigonometry, oscillations, and waves. Known for its periodic nature, the sine function can be expanded into a power series, which forms the basis of solving problems like the one in the exercise.
    • The sine function is ordinarily defined by its behavior, representing the y-coordinate on the unit circle for an angle \( x \).
    • Importantly, the sine function has a straightforward Taylor series expansion. This is given by:
    \[\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\]
  • The series above hints at an alternating pattern of terms including only odd powers of \( x \), showcasing the cyclical nature of the sine.
  • In our specific problem, the task involves calculating \( \sin^2{x} \). This means recognizing you must square the sine series to progress toward expanding \( f(x) \), resulting in another series confined to even powers after simplification.

  • The sine function's series is crucial, especially in forming solutions in physics and engineering, due to its natural occurrence in wave-like phenomena.
    series expansion
    Series expansion is a mathematical method used to express complex functions as the sum of simpler terms. This helps in understanding, analyzing, and solving extensive problems more efficiently.
    • In essence, series expansion aims to replace a difficult-to-handle function with a sum of terms—each understood through basic operations.
    • This process involves taking known expansions (like the sine function) and manipulating them to suit our needs.
    When squared as in \( \sin^2{x} \), the principles of algebra are applied to each series term, creating a fresh series:
    • We find products like \( (x - \frac{x^3}{3!} + \cdots)^2 \) and carefully calculate each result following patterns of distribution in polynomial multiplication.
    • In the provided exercise, expanding \( f(x) = x \sin^2{x} \) necessitates multiplying \( x \) with every term derived from squaring the sine series, resulting in heightened powers of \( x \).

    • Finally, by isolating low-degree terms, we effectively identify the significant nonzero terms: \( x^3 \) and \( -\frac{x^5}{3} \). This gives students insights into both the mechanism and simplicity afforded by series expansions.

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