Taylor Series expansions


1

\[ f(x) = \frac{1}{(1-x)} \] \[ f'(x) = \frac{1}{(1-x)^2} \] \[ f''(x) = \frac{2}{(1-x)^3} \] \[ f'''(x) = \frac{6}{(1-x)^4} \] \[ f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}} \] \[ f^{(n)}(0) = n! \] So, using Taylor Series formula…

\[ \frac{1}{(1-x)} = \sum \frac{f^{(n)}(0)}{n!}x^n \] \[ = \sum \frac{n!}{n!}x^n \] \[ = \sum_{n=0}^{\infty} x^n \]

2

\[ f(x) = e^x \] \[ f'(x) = e^x; f''(x) = e^x; f'''(x) = e^x \] \[ f^{(n)}(x) = e^x \] \[ f^{(n)}(0) = 1 \] So, using Taylor Series formula…

\[ e^x = \sum \frac{f^{(n)}(0)}{n!}x^n \] \[ = \sum \frac{1}{n!}x^n \]

3

\[ f(x) = ln(1 + x) \] \[ f'(x) = \frac{1}{(x + 1)} \] \[ f''(x) = \frac{-1}{(x + 1)^2} \] \[ f'''(x) = \frac{2}{(x + 1)^3} \] \[ f''''(x) = \frac{-6}{(x + 1)^4} \] \[ f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{(x + 1)^n} \] \[ f^{(n)}(0) = (-1)^{n-1}(n-1)! \] So, using Taylor Series formula…

\[ ln(x + 1) = \sum \frac{f^{(n)}(0)}{n!}x^n \] \[ = \sum \frac{(-1)^{n-1}(n-1)!}{n!}x^n \] And after a few steps…

\[ = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} ... \]