Taylor Series is defined as: \(f(x) = \sum\limits_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n\).
Find first several derivatives.
\(f(c) = e^{-c}\), At c = 0, \(f(0) = 1\)
\(f'(c) = -e^{-c}\), At c = 0, \(f'(c) = -1\)
\(f''(c) = e^{-c}\), At c = 0, \(f''(c) = 1\)
\(f'''(c) = -e^{-c}\), At c = 0, \(f''(c) = -1\)
\(f''''(c) = e^{-c}\), At c = 0, \(f''''(c) = 1\)
\(f'''''(c) = -e^{-c}\), At c = 0, \(f''''(c) = -1\)
puting the derivative into the taylor series,
\(f(x) = \frac{e^c}{0!}(x-c)^0 + \frac{-e^c}{1!}(x-c)^1 + \frac{e^c}{2!}(x-c)^2 + \frac{-e^c}{3!}x-c)^3 + \frac{e^c}{4!}(x-c)^4 + \frac{-e^c}{5!}(x-c)^5 + ...\)
At \(c = 0\), its refers to as Maclaurin series and given as:
\(f(x) = \frac{e^0}{0!}x^0 - \frac{e^0}{1!}x^1 + \frac{e^0}{2!}x^2 - \frac{e^0}{3!}x^3 + \frac{e^0}{4!}x^4 - \frac{e^0}{5!}x^5 + ...\)
\(f(x) = 1 - \frac{1}{1!}x^1 + \frac{1}{2!}x^2 - \frac{e^c}{3!}x^3 + \frac{1}{4!}x^4 - \frac{1}{5!}x^5 + ...\)
\(f(x) = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120} + ...\)
The Taylor Series of \(f(x)\), at \(c=0\), \(f(x) = \sum\limits_{n=0}^{\infty} (-1)^n\frac{x^n}{n!}\)