Taylor Series Expansions
1.
\[f(x)=\frac{1}{1-x}\]
calculate first 4 derivatives:
\[f^{(1)}(x) = \frac{1}{(1-x)^2}\] \[f^{(2)}(x) = -\frac{2}{(x-1)^3}\]
\[f^{(3)}(x) = \frac{6}{(x-1)^4}\]
\[f^{(4)}(x) = -\frac{24}{(x-1)^5}\]
plug these derivatives into Talor polynomial function, and set the \(c = 0\), it becomes
\[\begin{align*} \frac{1}{1-x} = & f(0) + f^{(1)}(0)x + \frac{f^{(2)}(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + ... \\ = & 1 + \frac{1}{(1-0)^2}x + \frac{-\frac{2}{(0-1)^3}}{2!}x^2 + \frac{\frac{6}{(0-1)^4}}{3!}x^3 + \frac{-\frac{24}{(0-1)^5}}{4!}x^4 + ... \\ = & 1+x+\frac{2}{2!}x^2+\frac{6}{3!}x^3+\frac{24}{4!}x^4+...\\ = & \sum_{n=0}^{\infty} x^n \end{align*}\]
2.
\[f(x)=e^x\]
the first 4 derivatives are:
\[f^{(1)}(x) = e^x\] \[f^{(2)}(x) = e^x\] \[f^{(3)}(x) = e^x\] \[f^{(4)}(x) = e^x\]
plug them into Taylor polynomial function, and centered at 0, it becomes:
\[\begin{align*} e^x= & f(0) + f^{(1)}(0)x + \frac{f^{(2)}(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + ... \\ = & 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\\ = & \sum_{n=0}^{\infty}\frac{x^n}{n!} \end{align*}\]
3.
\[f(x) = ln(1+x)\]
find the first 4 derivatives:
\[f^{(1)}(x) = \frac{1}{x+1}\] \[f^{(2)}(x) = -\frac{1}{(x+1)^2}\] \[f^{(3)}(x) = \frac{2}{(x+1)^3}\] \[f^{(4)}(x) = -\frac{6}{(x+1)^4}\]
plug them into Taylor polynomial function and centered to 0, the function becomes:
\[\begin{align*} ln(1+x)= & f(0) + f^{(1)}(0)x + \frac{f^{(2)}(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + ... \\ = & 0+x+\frac{-1}{2!}x^2+\frac{2}{3!}x^3+\frac{-6}{4!}x^4+...\\ = & x -\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+...\\ = & \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^n \end{align*}\]