Taylor Series
For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.
The following is the formula for a Taylor Series expansion.
\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 +...+ \frac{f^n(a)}{n!}(x-a)^n\]
We first evaluate the first few derivatives of f(x).
\[ f'(x) = \frac{1}{(1-x)^2}\] \[ f''(x) = \frac{2}{(1-x)^3}\] \[ f'''(x) = \frac{6}{(1-x)^4}\] \[ f^n(x) = \frac{n!}{(1-x)^{n+1}}\]
Plugging in a = 0 into f(x), we get the following equation for the nth derivative:
\[ f^n(0) = n!\]
Using the standard formula for a taylor series expansion and the nth derivative formula for this function, we get the following series.
\[f(x) = 1 + x + x^2 +...+ x^{n}\]
We first evaluate the first few derivatives of f(x).
\[ f'(x) = e^x\] \[ f''(x) = e^x\]
\[ f'''(x) = e^x\] \[ f^n(x) = e^x\]
Plugging in a = 0 into f(x), we get the following equation for the nth derivative:
\[ f^n(0) = 1\]
Using the standard formula for a taylor series expansion and the nth derivative formula for this function, we get the following series.
\[ f^n(x) = 1 + x + \frac{x^2}{2!}+...+ \frac{x^n}{n!}\]
We first evaluate the first few derivatives of f(x).
\[ f'(x) = \frac{1}{x+1}\] \[ f''(x) = \frac{-1}{(x+1)^2}\] \[ f'''(x) = \frac{2}{(x+1)^3}\]
\[ f^n(x) = \frac{(-1)^{n+1}(n-1)!}{(x+1)^n}\]
Plugging in a = 0 into f(x), we get the following equation for the nth derivative:
\[ f^n(0) = (-1)^{n+1}(n-1)!\]
Using the standard formula for a taylor series expansion and the nth derivative formula for this function, we get the following series.
\[f(x) = x+ \frac{x^2}{2} +...+ \frac{(-1)^{n+1}}{n}x^n\]