\[ \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n \]
\[ \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} (x)^n \]
Really interesting to see this play out given the HW exercise for \(e^x\). The spin for this problem is that the derivatives alternate positive and negative, changing the pattern and formula somewhat from the HW.
Original function: \(f(x) = e^{-x} ; c=0\)
First four derivatives:
\(f'(x) = -e^{-x},f''(x) = e^{-x},f'''(x) = -e^{-x},f''''(x) = e^{-x}\)
When x = 0 for the four derivatives:
\(=f^n(0) = (-1)+1+(-1)+1+...\)
Substitute from the Taylor equation:
\(f(x)= \sum_{n=0}^{\infty} \frac{f^{(n)}0}{n!} x^n\)
Simplify:
\(= (-x)+ \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - ...\)
\(= (-x)+ \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - ...\)
The series appears to summarize to:
\(= \sum_{n=0}^{\infty} \frac{(-1)x^{n}}{n!}\)