Sure, let's expand the function
[tex]\[
\frac{1}{(1 + x)^{1/2}}
\][/tex]
using the Taylor series expansion around [tex]\( x = 0 \)[/tex]. The expansion of
[tex]\[
\frac{1}{(1 + x)^{1/2}}
\][/tex]
will look like a sum of terms of the form [tex]\( a_n x^n \)[/tex].
When we expand a function [tex]\( f(x) \)[/tex] as a Taylor series around [tex]\( x = 0 \)[/tex], we can express it as:
[tex]\[
f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots
\][/tex]
For the function
[tex]\[
f(x) = \frac{1}{(1 + x)^{1/2}}
\][/tex]
the Taylor series expansion is:
[tex]\[
\frac{1}{(1 + x)^{1/2}} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \frac{35}{128}x^4 - \frac{63}{256}x^5 + \frac{231}{1024}x^6 - \frac{429}{2048}x^7 + \frac{6435}{32768}x^8 - \frac{12155}{65536}x^9
\][/tex]
So, the detailed step-by-step expansion of the function
[tex]\[
\frac{1}{(1 + x)^{1/2}}
\][/tex]
up to the 9th power of [tex]\( x \)[/tex] is:
[tex]\[
\frac{1}{(1 + x)^{1/2}} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \frac{35}{128}x^4 - \frac{63}{256}x^5 + \frac{231}{1024}x^6 - \frac{429}{2048}x^7 + \frac{6435}{32768}x^8 - \frac{12155}{65536}x^9
\][/tex]
This series provides an approximation of the function [tex]\( \frac{1}{(1 + x)^{1/2}} \)[/tex] when [tex]\( x \)[/tex] is close to 0, and it allows us to understand the behavior and properties of the function around that point.
