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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
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Tutti i libri di testoHass 15th EditionCh. 10 - Infinite Sequences and SeriesProblema 10.8.37
Capitolo 10, Problema 10.8.37

In Exercises 35–40, find the first three nonzero terms of the Maclaurin series for each function.
f(x) = (sin x) ln(1 + x)

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Recall that the Maclaurin series is the Taylor series expansion of a function about \(x = 0\). We want to find the first three nonzero terms of the Maclaurin series for \(f(x) = (\sin x) \ln(1 + x)\).
Write down the Maclaurin series expansions for each component function separately: \(\sin x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\) and \(\ln(1+x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\).
Multiply the two series together term-by-term, keeping track of powers of \(x\). Specifically, multiply each term of \(\sin x\) by each term of \(\ln(1+x)\) and combine like powers of \(x\).
Identify and collect the first three nonzero terms from the resulting series after multiplication. This involves adding coefficients of like powers of \(x\) and ignoring terms that are zero.
Write the resulting expression as the sum of these first three nonzero terms, which will be the beginning of the Maclaurin series for \(f(x) = (\sin x) \ln(1 + x)\).

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A Maclaurin series is a Taylor series expansion of a function about zero. It expresses the function as an infinite sum of terms involving derivatives evaluated at zero, allowing approximation of the function near x = 0.
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