31.tex

Video notes: vid19.mp4

\subsection*{Series expansion of 1/(1 - x)}
\begin{itemize}
\item{to show and prove:
\begin{align*}
\ln (\frac{1}{1 - x}) = x + 
\frac{x^2}{2} + \frac{x^{3}}{3} + \cdots \frac{x^{k}}{k} + \cdots
\end{align*}
}
\item{
\begin{align*}
f(x) &= f(0) + f'(0)x + \frac{1}{2!}x^2 + \cdots \\
f(0) &= \ln{ \frac{1}{1 - 0} } = ln{1} = 0 \\
f'(x) &=  \frac{[(1 - x)^{-1}]'}{(1 - x)^{-1}} = 
\frac{
    (-1)(1 - x)^{-2}(-1)
} {
    (1 - x)^{-1}
}
= (1 - x)^{-1} \stackrel{\longrightarrow}{x \rightarrow 0} 1 \\
f''(x) &= (-1)(1 - x)^{-2}(-1) = \frac{1}{(1 - x)^2} = 1 \\
f'''(x) &= [(1 - x)^2]' = (-2)(1 - x)^{-3}(-1) = 2(1 - x)^3 = 2 \\
f''''(x) &= 2 \cdot (1 - x)^{-3} = 2 \cdot(-3)(1 - x)^{-4}(-1)  \\
f^(k)(x) &= (k - 1)!, k = 2, \cdots \\
\end{align*} 
}
\item{
Now, lets plug in for the taylor expansion:
\begin{align*}
\ln{\frac{1}{1 - x}} &= f(0) + \frac{f'(0)x}{1!} + f''(0)\frac{x^2}{2!} + \cdots
&= 0 + 1 \cdot x + \frac{1 \cdot x^2}{2!} + \frac{(2!)x^3}{3!} + \cdots
&= x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots
\end{align*}
}
\end{itemize}