| Function and Derivatives | Function at \(x = 0\) |
|---|---|
| \(f(x) = e^{-x}\) | \(f(0) = e^{-0} = \frac{1}{e^0} = 1\) |
| \(f'(x) = -e^{-x}\) | \(f'(0) = -e^{-0} = -\frac{1}{e^0} = -1\) |
| \(f''(x) = e^{-x}\) | \(f''(0) = e^{-0} = \frac{1}{e^0} = 1\) | \(f'''(x) = -e^{-x}\) | \(f'''(0) = -e^{-0} = -\frac{1}{e^0} = -1\) |
| \(f^4(x) = e^{-x}\) | \(f^4(0) = e^{-0} = \frac{1}{e^0} = 1\) |
| \(f^5(x) = -e^{-x}\) | \(f^5(0) = -e^{-0} = -\frac{1}{e^0} = -1\) |
| \(f^6(x) = e^{-x}\) | \(f^6(0) = e^{-0} = \frac{1}{e^0} = 1\) | \(f^7(x) = -e^{-x}\) | \(f^7(0) = -e^{-0} = -\frac{1}{e^0} = -1\) |
| \(f^8(x) = e^{-x}\) | \(f^8(0) = e^{-0} = \frac{1}{e^0} = 1\) | \(f^9(x) = -e^{-x}\) | \(f^9(0) = -e^{-0} = -\frac{1}{e^0} = -1\) |
The series representation of this function is :
\(e^{-x} = \displaystyle\sum_{n=0}^{\infty} \frac{(-x)^n}{n!}\)
Use the Maclaurin series (Taylor series centered at point of zero):
\(e^{-x} = \displaystyle\sum_{n=0}^{\infty} \frac{(-x)^n}{n!} = f(0) + \frac{f^{'}(0)}{1!}x + \frac{f^{''}(0)}{2!}x^2 + \frac{f^{'''}(0)}{3!}x^3 + ... + \frac{f^{n}(0)}{n!}x^n\)
By substituting terms, we get the following series:
\(e^{-x} = \displaystyle\sum_{n=0}^{\infty} \frac{(-x)^n}{n!} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120} + \frac{x^6}{720} - \frac{x^7}{5040} + \frac{x^8}{40320} - \frac{x^9}{362880} + ...\)