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