Answer :
Certainly! Let's first state the Maclaurin series in its finite form and then discuss the given series and the extension.
### Maclaurin Series
The Maclaurin series is a special case of the Taylor series centered around [tex]\( x = 0 \)[/tex]. The general form of a Maclaurin series for a function [tex]\( f(x) \)[/tex] is given by:
[tex]\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \][/tex]
This series can potentially be extended to infinity, depending on the function [tex]\[ f(x) \][/tex] and its derivatives at [tex]\( x = 0 \)[/tex].
### Given Series: [tex]\(\log (1 + \sin x)\)[/tex]
We dissect the problem by looking at the Maclaurin series expansion of the function [tex]\(\log (1 + \sin x)\)[/tex].
The Maclaurin series for [tex]\(\log (1 + x)\)[/tex] is:
[tex]\[ \log (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \][/tex]
However, we need the series for [tex]\(\log (1 + \sin x)\)[/tex]. We start substituting [tex]\( \sin x \)[/tex] in place of [tex]\( x \)[/tex] in the series for [tex]\( \log (1 + x) \)[/tex] and analyze the first few terms:
[tex]\[ \log (1 + \sin x) = \sin x - \frac{(\sin x)^2}{2} + \frac{(\sin x)^3}{3} - \cdots \][/tex]
### Terms of [tex]\(\log (1 + \sin x)\)[/tex]
We use the Maclaurin series expansion for [tex]\(\sin x\)[/tex]:
[tex]\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \][/tex]
Substituting this into the expansion for [tex]\(\log (1 + \sin x)\)[/tex] gives:
[tex]\[ \log (1 + (\sin x)) = \left( x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \right) - \frac{1}{2} \left( x - \frac{x^3}{6} + \cdots \right)^2 + \frac{1}{3} \left( x - \frac{x^3}{6} + \cdots \right)^3 - \cdots \][/tex]
### Analyzing the First Terms
For practical purposes, we truncate the series to the first few non-trivial terms.
1. Up to [tex]\( O(x^1) \)[/tex]:
[tex]\[ \log (1 + \sin x) \approx \sin x = x \][/tex]
2. Up to [tex]\( O(x^2) \)[/tex]:
[tex]\[ \log (1 + \sin x) \approx x - \frac{x^2}{2} \][/tex]
3. Up to [tex]\( O(x^3) \)[/tex]:
[tex]\[ \log (1 + \sin x) \approx x - \frac{x^2}{2} + \frac{x^3}{6} \][/tex]
Thus, the condition under that:
[tex]\[ \log (1 + \sin x) = x - \frac{x^2}{2} + \frac{x^3}{6} + \cdots \][/tex]
This matches with the provided series hint. Therefore, the given Maclaurin series truncates and extends to infinity as expected under the typical convergence criteria for the series expansion.
### Conclusion
The Maclaurin series for [tex]\(\log (1 + \sin x)\)[/tex] can be written as:
[tex]\[ \log (1 + \sin x) \approx x - \frac{x^2}{2} + \frac{x^3}{6} + \cdots, \][/tex]
which extends to infinity and meets the condition required for proper convergence criteria within the radius of convergence.
### Maclaurin Series
The Maclaurin series is a special case of the Taylor series centered around [tex]\( x = 0 \)[/tex]. The general form of a Maclaurin series for a function [tex]\( f(x) \)[/tex] is given by:
[tex]\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \][/tex]
This series can potentially be extended to infinity, depending on the function [tex]\[ f(x) \][/tex] and its derivatives at [tex]\( x = 0 \)[/tex].
### Given Series: [tex]\(\log (1 + \sin x)\)[/tex]
We dissect the problem by looking at the Maclaurin series expansion of the function [tex]\(\log (1 + \sin x)\)[/tex].
The Maclaurin series for [tex]\(\log (1 + x)\)[/tex] is:
[tex]\[ \log (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \][/tex]
However, we need the series for [tex]\(\log (1 + \sin x)\)[/tex]. We start substituting [tex]\( \sin x \)[/tex] in place of [tex]\( x \)[/tex] in the series for [tex]\( \log (1 + x) \)[/tex] and analyze the first few terms:
[tex]\[ \log (1 + \sin x) = \sin x - \frac{(\sin x)^2}{2} + \frac{(\sin x)^3}{3} - \cdots \][/tex]
### Terms of [tex]\(\log (1 + \sin x)\)[/tex]
We use the Maclaurin series expansion for [tex]\(\sin x\)[/tex]:
[tex]\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \][/tex]
Substituting this into the expansion for [tex]\(\log (1 + \sin x)\)[/tex] gives:
[tex]\[ \log (1 + (\sin x)) = \left( x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \right) - \frac{1}{2} \left( x - \frac{x^3}{6} + \cdots \right)^2 + \frac{1}{3} \left( x - \frac{x^3}{6} + \cdots \right)^3 - \cdots \][/tex]
### Analyzing the First Terms
For practical purposes, we truncate the series to the first few non-trivial terms.
1. Up to [tex]\( O(x^1) \)[/tex]:
[tex]\[ \log (1 + \sin x) \approx \sin x = x \][/tex]
2. Up to [tex]\( O(x^2) \)[/tex]:
[tex]\[ \log (1 + \sin x) \approx x - \frac{x^2}{2} \][/tex]
3. Up to [tex]\( O(x^3) \)[/tex]:
[tex]\[ \log (1 + \sin x) \approx x - \frac{x^2}{2} + \frac{x^3}{6} \][/tex]
Thus, the condition under that:
[tex]\[ \log (1 + \sin x) = x - \frac{x^2}{2} + \frac{x^3}{6} + \cdots \][/tex]
This matches with the provided series hint. Therefore, the given Maclaurin series truncates and extends to infinity as expected under the typical convergence criteria for the series expansion.
### Conclusion
The Maclaurin series for [tex]\(\log (1 + \sin x)\)[/tex] can be written as:
[tex]\[ \log (1 + \sin x) \approx x - \frac{x^2}{2} + \frac{x^3}{6} + \cdots, \][/tex]
which extends to infinity and meets the condition required for proper convergence criteria within the radius of convergence.
