T)(NG) 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

To solve the problem, we will follow these steps: 1. **Find the Maclaurin expansion of \( x \ln(2+x) \)**. 2. **Deduce the Maclaurin expansion of \( x \ln(2-x) \)** from the previous result. 3. **Obtain the Maclaurin expansion of \( x \ln(4-x^2) \)** using the results from steps 1 and 2. ### Step 1: Maclaurin Expansion of \( x \ln(2+x) \) The Maclaurin series expansion of a function \( f(x) \) is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \] Let's define \( f(x) = x \ln(2+x) \). #### Calculate \( f(0) \): \[ f(0) = 0 \cdot \ln(2+0) = 0 \] #### Calculate \( f'(x) \): Using the product rule: \[ f'(x) = \ln(2+x) + x \cdot \frac{1}{2+x} \] Now, evaluate \( f'(0) \): \[ f'(0) = \ln(2+0) + 0 \cdot \frac{1}{2+0} = \ln(2) \] #### Calculate \( f''(x) \): Differentiate \( f'(x) \): \[ f''(x) = \frac{1}{2+x} + \left( \frac{1}{2+x} - \frac{x}{(2+x)^2} \right) \] Now, evaluate \( f''(0) \): \[ f''(0) = \frac{1}{2+0} + \left( \frac{1}{2+0} - \frac{0}{(2+0)^2} \right) = \frac{1}{2} + \frac{1}{2} = 1 \] #### Calculate \( f'''(x) \): Differentiate \( f''(x) \): \[ f'''(x) = -\frac{1}{(2+x)^2} + \left( -\frac{1}{(2+x)^2} + \frac{2x}{(2+x)^3} \right) \] Now, evaluate \( f'''(0) \): \[ f'''(0) = -\frac{1}{(2+0)^2} + \left( -\frac{1}{(2+0)^2} + 0 \right) = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2} \] #### Construct the Maclaurin series: Now we can construct the series: \[ f(x) = 0 + \ln(2)x + \frac{1}{2} \cdot x^2 - \frac{1}{12} x^3 + \ldots \] Thus, the first three non-zero terms in the Maclaurin expansion of \( x \ln(2+x) \) are: \[ \ln(2)x + \frac{1}{2}x^2 - \frac{1}{12}x^3 \] ### Step 2: Deduce the Maclaurin Expansion of \( x \ln(2-x) \) Using the fact that \( \ln(2-x) = \ln(2) + \ln(1 - \frac{x}{2}) \), we can expand \( \ln(1 - u) \) where \( u = \frac{x}{2} \): \[ \ln(1 - u) = -\left( \frac{x}{2} + \frac{1}{2} \left( \frac{x}{2} \right)^2 + \frac{1}{3} \left( \frac{x}{2} \right)^3 + \ldots \right) \] Thus, \[ \ln(2-x) = \ln(2) - \left( \frac{x}{2} + \frac{x^2}{8} + \frac{x^3}{24} + \ldots \right) \] Now, multiplying by \( x \): \[ x \ln(2-x) = x \ln(2) - \left( \frac{x^2}{2} + \frac{x^3}{8} + \frac{x^4}{24} + \ldots \right) \] The first three non-zero terms in the Maclaurin expansion of \( x \ln(2-x) \) are: \[ \ln(2)x - \frac{1}{2}x^2 - \frac{1}{8}x^3 \] ### Step 3: Maclaurin Expansion of \( x \ln(4-x^2) \) Using the identity \( \ln(4-x^2) = \ln(4) + \ln(1 - \frac{x^2}{4}) \): \[ \ln(1 - u) = -\left( u + \frac{u^2}{2} + \frac{u^3}{3} + \ldots \right) \] where \( u = \frac{x^2}{4} \): \[ \ln(1 - \frac{x^2}{4}) = -\left( \frac{x^2}{4} + \frac{1}{2} \left( \frac{x^2}{4} \right)^2 + \ldots \right) \] Thus, \[ \ln(4-x^2) = \ln(4) - \left( \frac{x^2}{4} + \frac{x^4}{32} + \ldots \right) \] Now, multiplying by \( x \): \[ x \ln(4-x^2) = x \ln(4) - \frac{x^3}{4} - \frac{x^5}{32} - \ldots \] The first two non-zero terms in the Maclaurin expansion of \( x \ln(4-x^2) \) are: \[ \ln(4)x - \frac{1}{4}x^3 \] ### Summary of Results 1. First three non-zero terms in \( x \ln(2+x) \): \( \ln(2)x + \frac{1}{2}x^2 - \frac{1}{12}x^3 \) 2. First three non-zero terms in \( x \ln(2-x) \): \( \ln(2)x - \frac{1}{2}x^2 - \frac{1}{8}x^3 \) 3. First two non-zero terms in \( x \