Sure, let's tackle each part of the question step-by-step.
### Part (a): Expand \(\( f(z) = \ln(z) \)\) Centered at \(\( c = 1 \)\) Using Taylor Series
To expand \(\( f(z) = \ln(z) \)\) around \(\( z = 1 \)\) using the Taylor series, we consider the series:
\(\[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (z - 1)^n \]\)
where \(\( f^{(n)} \)\) denotes the n-th derivative of \(\( f \)\) evaluated at \(\( z=1 \)\).
- For \(\( n = 0 \)\), \(\( f(1) = \ln(1) = 0 \)\)
- For \(\( n = 1 \)\), \(\( f'(z) = \frac{1}{z} \)\) and \(\( f'(1) = 1 \)\)
- For \(\( n = 2 \)\), \(\( f''(z) = -\frac{1}{z^2} \)\) and \(\( f''(1) = -1 \)\)
- For \(\( n = 3 \)\), \(\( f'''(z) = \frac{2}{z^3} \)\) and \(\( f'''(1) = 2 \)\)
- And so on...
Following this pattern, we can write the first few terms of the Taylor series expansion:
\(\[ \ln(z) \approx (z - 1) - \frac{(z - 1)^2}{2} + \frac{(z - 1)^3}{3} - \frac{(z - 1)^4}{4} + \cdots \]\)
The expansion up to the 9th term is:
\(\[ \ln(z) \approx -1 - \frac{(z - 1)^2}{2} + \frac{(z - 1)^3}{3} - \frac{(z - 1)^4}{4} + \frac{(z - 1)^5}{5} - \frac{(z - 1)^6}{6} + \frac{(z - 1)^7}{7} - \frac{(z - 1)^8}{8} + \frac{(z - 1)^9}{9} + z + O((z-1)^{10}) \]\)
The above expression provides us the Taylor series expansion of \(\( \ln(z) \)\) around \(\( z = 1 \)\).
### Part (b): Using Maclaurin Series for \(\( f(x) = \sin(x) \)\)
The Maclaurin series for a function \(\( f(x) \)\) is a special case of the Taylor series centered at \(\( x = 0 \)\). For \(\( f(x) = \sin(x) \)\), the series is:
\(\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \]\)
The sine function is an odd function, so only odd powers of \(\( x \)\) appear in its series expansion. Its derivatives alternate in a specific pattern due to the trigonometric properties:
- \(\( f(0) = \sin(0) = 0 \)\)
- \(\( f'(0) = \cos(0) = 1 \)\)
- \(\( f''(0) = -\sin(0) = 0 \)\)
- \(\( f'''(0) = -\cos(0) = -1 \)\)
- \(\( f^{(4)}(0) = \sin(0) = 0 \)\)
- And so on...
Thus, the Maclaurin series for \(\( \sin(x) \)\) is:
\(\[ \sin(x) \approx x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \frac{x^9}{362880} + O(x^{10}) \]\)
This expansion provides us with the Maclaurin series of \(\( \sin(x) \)\) up to the 9th power of \(\( x \)\).
Together, these expansions answer both parts of the question comprehensively.
