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The Maclaurin series for \( \sin x \) is an expansion of the sine function at zero. It is derived from the Taylor series, specifically centered at \( x = 0 \). This series allows us to express \( \sin x \) as an infinite sum of terms involving powers of \( x \). The general form of the Maclaurin series for \( \sin x \) is:\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\]
The terms in this series alternate in sign. Notice that only odd powers of \( x \) appear here. The factorial in the denominator (\(3! = 6\), \(5! = 120\), for example) ensures that higher powers decrease rapidly, making the series converge quickly for values of \( x \) close to zero.
For the exercise, determining the first three nonzero terms of \( \sin x \) implies we initially focus on the first and second terms: \( x \) and \(-\frac{x^3}{6}\). These terms are critical in calculating the series for\( f(x) = (\sin x) \ln(1+x) \) when combined with the series for \( \ln(1+x) \).

The natural logarithm function, \( \ln (1+x) \), also has its Maclaurin series, offering a way to approximate \( \ln (1+x) \) near \( x = 0 \). The series expansion is expressed as:\[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\]
This series is alternating, just like the \( \sin x \) series, but it includes all integer powers of \( x \), not just the odd ones. Each term is increasingly smaller the further out we go due to both the increasing power and the factorial-like denominators. This property helps in ensuring the series converges for \( |x| < 1 \).
In the given problem, it's important to accurately extract the terms \( x \), \(-\frac{x^2}{2}\), and \( \frac{x^3}{3} \) from this series because these are pivotal in forming the terms of the function \( f(x) = (\sin x) \ln (1+x) \) up to the required order.

The Taylor series is a powerful tool in calculus used to approximate functions. It represents a function as an infinite sum of terms calculated from the values of the function’s derivatives at a single point. The Maclaurin series is a special case of the Taylor series where this point is 0.In general, the Taylor series for a function \( f(x) \) centered at \( a \) is given by:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]
The series outlines how well a polynomial can approximate a certain function around a particular point.For the exercise at hand, we are interested in the Maclaurin series, the specific Taylor expansion where \( a = 0 \). By multiplying the series for \( \sin x \) and \( \ln(1+x) \), we get:* Terms like \( x \cdot x \) and \( x \cdot -\frac{x^2}{2} \) combine to form the solution.* Identify the nonzero results: \( x^2 - \frac{x^3}{2} \) and simplify further.
Ultimately, the first three nonzero terms for \( f(x) = (\sin x) \ln(1+x) \) are \( x^2 \) and \( -\frac{x^3}{6} \), providing a simplified approximation of the function up to the third degree.