Question 1 a. Give the difference between the Taylor series and a Maclaurin series b. Find the Maclaurin series for \( \ln (1+x) \) [. Proof that if a function \( f(x) \) has a power series representation centered at \( x=a \) that converges to \( f(x) \) on some open interval containing \( a \), then this power series a Taylor series for \( f(x) \) at \( x=a \).

### Question 1 #### a. Difference between Taylor series and Maclaurin series The **Taylor series** of a function \( f(x) \) centered at a point \( a \) is given by the formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n \] where \( f^{(n)}(a) \) is the \( n \)-th derivative of \( f \) evaluated at \( x = a \). The **Maclaurin series** is a special case of the Taylor series where the center \( a \) is specifically chosen to be 0. Thus, the Maclaurin series for \( f(x) \) is given by: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \] In summary, the main difference is the point about which the series is expanded: the Taylor series is centered at \( a \), while the Maclaurin series is centered at 0. #### b. Finding the Maclaurin series for \( \ln(1+x) \) To find the Maclaurin series for \( \ln(1+x) \), we will compute the derivatives of \( \ln(1+x) \) at \( x = 0 \). 1. **Function and its derivatives**: - \( f(x) = \ln(1+x) \) - \( f'(x) = \frac{1}{1+x} \) - \( f''(x) = -\frac{1}{(1+x)^2} \) - \( f'''(x) = \frac{2}{(1+x)^3} \) - \( f^{(4)}(x) = -\frac{6}{(1+x)^4} \) - Continuing this pattern, we can find the \( n \)-th derivative. 2. **Evaluating at \( x = 0 \)**: - \( f(0) = \ln(1+0) = 0 \) - \( f'(0) = \frac{1}{1+0} = 1 \) - \( f''(0) = -1 \) - \( f'''(0) = 2 \) - \( f^{(4)}(0) = -6 \) - The pattern for the \( n \)-th derivative evaluated at 0 can be observed as: - \( f^{(n)}(0) = (-1)^{n-1} (n-1)! \) for \( n \geq 1 \). 3. **Constructing the Maclaurin series**: - The Maclaurin series is given by: \[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \] - Substituting the derivatives: \[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} (n-1)!}{n!} x^n \] - Simplifying the factorial: \[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n \] Thus, the Maclaurin series for \( \ln(1+x) \) is: \[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n \quad \text{for } |x| < 1. \] #### c. Proof that if a function \( f(x) \) has a power series representation centered at \( x=a \) that converges to \( f(x) \) on some open interval containing \( a \), then this power series is a Taylor series for \( f(x) \) at \( x=a \). Let \( f(x) \) be a function that has a power series representation centered at \( x = a \): \[ f(x) = \sum_{n=0}^{\infty} c_n (x - a)^n \] where \( c_n \) are coefficients. We want to show that this power series is a Taylor series for \( f(x) \) at \( x = a \). 1. **Definition of Taylor series**: The Taylor