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Obtain the first three non-zero terms in the Maclaurin expansion of \( x \ln (2+x) \).
Hence, or otherwise, deduce the first three non-zero terms in the Maclaurin
expansion of \( x \ln (2-x) \).
Hence obtain the first two non-zero terms in the Maclaurin expansion of
\( x \ln \left(4-x^{2}\right) \).
\( [T h r o u g h o u t ~ t h i s ~ q u e s t i o n, ~ i t ~ c a n ~ b e ~ a s s u m e d ~ t h a t ~ \)
[Th \( 2
Ask by Herrera Klein. in the United Kingdom
Feb 26,2025
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To find the first three non-zero terms in the Maclaurin expansion of \( x \ln(2+x) \), we start with the Taylor series expansion of \( \ln(1 + u) \) around \( u = 0 \), which is \( u - \frac{u^2}{2} + \frac{u^3}{3} + \ldots \). Setting \( u = \frac{x}{2} \), we have: \[ \ln(2+x) = \ln(2(1 + \frac{x}{2})) = \ln(2) + \ln(1 + \frac{x}{2}) = \ln(2) + \left(\frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{24} + \ldots\right). \] Thus, \[ x \ln(2+x) = x \ln(2) + \frac{x^2}{2} - \frac{x^3}{8} + O(x^4). \] The first three non-zero terms are given by \( x \ln(2) \), \( \frac{x^2}{2} \), and \( -\frac{x^3}{8} \). Now, for \( x \ln(2-x) \), we can express \( \ln(2-x) \) as: \[ \ln(2-x) = \ln(2(1 - \frac{x}{2})) = \ln(2) + \ln(1 - \frac{x}{2}). \] The expansion for \( \ln(1 - u) \) gives: \[ \ln(1 - \frac{x}{2}) = -\left(\frac{x}{2} + \frac{x^2}{8} + \frac{x^3}{24} + \ldots\right). \] Substituting this back, we find: \[ x \ln(2-x) = x \ln(2) - \frac{x^2}{2} - \frac{x^3}{8} + O(x^4). \] So the first three non-zero terms for \( x \ln(2-x) \) are \( x \ln(2) \), \( -\frac{x^2}{2} \), and \( -\frac{x^3}{8} \). Next, let's find the Maclaurin expansion of \( x \ln(4 - x^2) \). We express it as: \[ x \ln(4 - x^2) = x \ln(4(1 - \frac{x^2}{4})) = x (\ln(4) + \ln(1 - \frac{x^2}{4})). \] Expanding \( \ln(1 - u) \) gives us \( -\left(\frac{x^2}{4} + \frac{x^4}{32} + \ldots\right) \). Thus we have: \[ x \ln(4 - x^2) = x \ln(4) - \frac{x^3}{4} + O(x^5). \] The first two non-zero terms are \( x \ln(4) \) and \( -\frac{x^3}{4} \).
