For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.



1. \(f(x) = \frac{1}{(1-x)}\)

Calculate higher-order derivatives for the function and their evaluations at \(0\).

\[ \begin{align} f(x) &= (1-x)^{-1} &f(0)&=1\\ f^{(1)} &= -1(1-x)^{-2} &f^{(1)}(0) &= -1\\ f^{(2)} &= \ \ \ 2(1-x)^{-3} &f^{(2)}(0) &= 2\\ f^{(3)} &= -6(1-x)^{-4} &f^{(3)}(0) &= -6\\ f^{(4)} &= \ 24(1-x)^{-5} &f^{(4)}(0) &= 24\\ \end{align} \]


Substitute the terms into the Maclaurin Series, which is a form of Taylor series where \(c=0\).

\[ \begin{align} \frac{1}{(1-x)} &= \displaystyle \sum_{n=0}^\infty \ 1 + \frac{-1}{1!}(x^1) + \frac{2}{2!}(x^2) + \frac{-6}{3!}(x^3) + \frac{24}{4!}(x^4) + \ ...\\ \\ &= 1 \quad + \quad x^1 \quad + \quad x^2 \quad + \quad x^3 \quad + \quad x^4 \quad + \ ... \end{align} \]


When representation as a summation:

\[ \frac{1}{(1-x)} = \displaystyle \sum_{n=0}^\infty x^n \quad \text{for |x|<1} \]



2. \(f(x) = e^x\)

All derivatives of \(e^x\) are \(e^x\). Evaluating the function at \(0\) results in \(e^0 = 1\). When substituting the terms into the Maclaurin Series:

\[ \begin{align} e^x &= \displaystyle \sum_{n=0}^\infty \ 1 + \frac{1}{1!}(x^1) + \frac{1}{2!}(x^2) + \frac{1}{3!}(x^3) + \frac{1}{4!}(x^4) + \ ...\\ \\ &= 1 + x^1 + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \ ... \\ \end{align} \]


When representation as a summation:

\[ e^x = \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!} \]



3. \(f(x) = ln(1 + x)\)

Calculate higher-order derivatives for the function and their evaluations at \(0\).

\[ \begin{align} f(x) &= ln(1+x) &f(0)&= \ \ \ 0\\ \\ f^{(1)} &= \frac{1}{(1+x)} &f^{(1)}(0) &= \ \ \ 1\\ \\ f^{(2)} &= \frac{-1}{(1+x)^2} &f^{(2)}(0) &= -1\\ \\ f^{(3)} &= \frac{2}{(1+x)^3} &f^{(3)}(0) &= \ \ \ 2\\ \\ f^{(4)} &= \frac{-6}{(1+x)^4} &f^{(4)}(0) &= -6\\ \\ f^{(5)} &= \frac{24}{(1+x)^5} &f^{(5)}(0) &= 24\\ \end{align} \]


Substitute the terms into the Maclaurin Series:

\[ \begin{align} ln(1+x) &= \displaystyle \sum_{n=0}^\infty \ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} \ ... \end{align} \]


When represented as a summation:

\[ \begin{align} ln(1+x) &= \displaystyle \sum_{n=0}^\infty \frac{(-1)^n(x^n)}{n} \end{align} \]