Taylor Series is defined as \(f(x) = \sum\limits_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n\).
Find first several derivatives.
\(f^0(c) = \frac{1}{(1-c)}\)
\(f'(c) = \frac{1}{(1-c)^2}\)
\(f''(c) = \frac{2}{(1-c)^3}\)
\(f'''(c) = \frac{6}{(1-c)^4}\)
\(f''''(c) = \frac{24}{(1-c)^5}\)
Per definition,
\[ \begin{split} f(x) &= \frac{1}{(1-c)0!}(x-c)^0 + \frac{1}{(1-c)^2 1!}(x-c)^1 + \frac{2}{(1-c)^3 2!}(x-c)^2 + \frac{6}{(1-c)^4 3!}(x-c)^3 + \frac{24}{(1-c)^5 4!}(x-c)^4 + ... \\ &= \frac{1}{(1-c)} + \frac{1}{(1-c)^2}(x-c) + \frac{2!}{(1-c)^3 2!}(x-c)^2 + \frac{3!}{(1-c)^4 3!}(x-c)^3 + \frac{4!}{(1-c)^5 4!}(x-c)^4 + ... \\ &= \frac{1}{(1-c)} + \frac{1}{(1-c)^2}(x-c) + \frac{1}{(1-c)^3}(x-c)^2 + \frac{1}{(1-c)^4}(x-c)^3 + \frac{1}{(1-c)^5}(x-c)^4 + ... \\ &= \sum\limits_{n=0}^{\infty} \frac{1}{(1-c)^{n+1}}(x-c)^n \end{split} \]
The Maclaurin Series of \(f(x)\), \(c=0\), \(f(x) = \sum\limits_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + x^4 + ...\).
Find first several derivatives.
\(f^0(c) = e^c\)
\(f'(c) = e^c\)
\(f''(c) = e^c\)
\(f'''(c) = e^c\)
\(f''''(c) = e^c\)
Per definition,
\[ \begin{split} f(x) &= \frac{e^c}{0!}(x-c)^0 + \frac{e^c}{1!}(x-c)^1 + \frac{e^c}{2!}(x-c)^2 + \frac{e^c}{3!}(x-c)^3 + ...\\ &= e^c + e^c(x-c) + e^c\frac{(x-c)^2}{2!} + e^c\frac{(x-c)^3}{3!} + ...\\ &= e^c \sum\limits_{n=0}^{\infty} \frac{(x-c)^n}{n!} \end{split} \]
The Maclaurin Series of \(f(x)\), \(c=0\), \(f(x) = \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + ...\).
Find first several derivatives.
\(f^0(c) = ln(1+c)\)
\(f'(c) = \frac{1}{c+1}\)
\(f''(c) = -\frac{1}{(c+1)^2}\)
\(f'''(c) = \frac{2}{(c+1)^3}\)
\(f''''(c) = -\frac{6}{(c+1)^4}\)
Per definition,
\[ \begin{split} f(x) &= \frac{ln(1+c)}{0!}(x-c)^0 + \frac{1}{(c+1)1!}(x-c)^1 - \frac{1}{(c+1)^2 2!}(x-c)^2 + \frac{2}{(c+1)^3 3!}(x-c)^3 - \frac{6}{(c+1)^4 4!}(x-c)^4 + ...\\ &= ln(1+c) + \frac{1}{(c+1)}(x-c) - \frac{1!}{(c+1)^2 2\times1!}(x-c)^2 + \frac{2!}{(c+1)^3 3\times2!}(x-c)^3 - \frac{3!}{(c+1)^4 4\times3!}(x-c)^4 + ...\\ &= ln(1+c) + \frac{1}{(c+1)}(x-c) - \frac{1}{2(c+1)^2}(x-c)^2 + \frac{1}{3(c+1)^3}(x-c)^3 - \frac{1}{4(c+1)^4}(x-c)^4 + ...\\ &= ln(1+c) + \sum\limits_{n=1}^{\infty} (-1)^{n+1}\frac{(x-c)^n}{n(c+1)^n} \end{split} \]
The Maclaurin Series of \(f(x)\), \(c=0\), \(f(x) = \sum\limits_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...\).