This week, we’ll work out some Taylor Series expansions of popular functions.

• \(f(x) = \frac{1}{(1 − x)}\)
• \(f(x) = e^x\)
• \(f(x) = ln(1 + x)\)

For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion. Please submit your assignment as a R-Markdown document.

• \(f(x) = \frac{1}{(1 − x)}\)
\(f^{'}(x) = \frac{1}{(1 − x)^2}\), \(f^{'}(0) = 1\)
\(f^{''}(x) = \frac{2}{(1 − x)^3}\), \(f^{''}(0) = 2\)
\(f^{'''}(x) = \frac{6}{(1 − x)^4}\), \(f^{'''}(0) = 6\)
\(f^4(x) = \frac{24}{(1 − x)^5}\), \(f^4(0) = 24\)
. . .
\(f^n(x) = \frac{n!}{(1 − x)^{n+1}}\), \(f^n(0) = n!\)

Therefore based on Maclaurin series, \(\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^{\infty}\frac{n!}{n!}x^n = \sum_{n=0}^{\infty}x^n\)



• \(f(x) = e^x\)
\(f^{'}(x) = e^x\), \(f^{'}(0) = 1\)
\(f^{''}(x) = e^x\), \(f^{''}(0) = 1\)
\(f^{'''}(x) = e^x\), \(f^{'''}(0) = 1\)
\(f^4(x) = e^x\), \(f^4(0) = 1\)
. . .
\(f^n(x) = e^x\), \(f^n(0) = 1\)

Therefore based on Maclaurin series, \(\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^{\infty}\frac{1}{n!}x^n = \sum_{n=0}^{\infty}\frac{x^n}{n!}\)



• \(f(x) = ln(1 + x)\)
\(f^{'}(x) = \frac{1}{(1 + x)}\), \(f^{'}(0) = 1\)
\(f^{''}(x) = -\frac{1}{(1 + x)^2}\), \(f^{''}(0) = -1\)
\(f^{'''}(x) = \frac{2}{(1 + x)^3}\), \(f^{'''}(0) = 2\)
\(f^4(x) = -\frac{6}{(1 + x)^4}\), \(f^4(0) = -6\)
. . .
\(f^n(x) = (-1)^{(n+1)}\frac{(n-1)!}{(1 + x)^n}\), \(f^n(0) = (-1)^{(n+1)}(n-1)!\)

Therefore based on Maclaurin series, \(\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^{\infty}\frac{(-1)^{(n+1)}(n-1)!}{n!}x^n = \sum_{n=0}^{\infty}\frac{(-1)^{(n+1)}x^n}{n}\)