This week, we’ll work out some Taylor Series expansions of popular functions.
For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.
To calculate the expansions, we will use the following formula:
\[ \Sigma_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^{n} \]
\[ f(x) \text{ is not defined at 1, so the valid range for the Taylor Series expansion is } -1 < x < 1.\\ \text{Find the first 3 derivatives of } f(x):\\ f(c) = \frac{1}{(1-c)}\\ f'(c) = \frac{1}{(1-c)^2}\\ f''(c) = \frac{2}{(1-c)^3}\\ f'''(c) = \frac{6}{(1-c)^4}\\ \text{Plug into the Taylor Series formula:}\\ \Sigma_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^{n} = \frac{1}{(1-c)0!}(x-c)^0 + \frac{1}{(1-c)^2 1!} (x-c)^1 +\frac{2}{(1-c)^3 2!}(x-c)^2 + \frac{6}{(1-c)^4 3!}(x-c)^3 + ...\\ \text{Sum notation for the expansion: }\\ f(x) \approx P(x) = \Sigma_{n=0}^{\infty}\frac{1}{(1-c)^{n+1}}(x-c)^n\\ \text{Plug in } 0 \text{ for c to get the Maclaurin Series of } f(x):\\ f(x) \approx P(x) = 1 + x + x^2 + x^3 + ... \text{ valid for } -1 \leq x \leq 1. \]
\[ e^x \text{ is valid for all x, so the valid range is } -\infty < x < \infty\\ \text{Find the first three derivatives of } f(x):\\ f(c) = e^c\\ f'(c) = e^c\\ f''(c) = e^c\\ f'''(c) = e^c\\ \text{Plug into the Taylor Series formula: }\\ \Sigma_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^{n} = \frac{e^c}{0!}(x-c)^0 + \frac{e^c}{1!}(x-c)^1 +\frac{e^c}{2!}(x-c)^2 + \frac{e^c}{3!}(x-c)^3 + ...\\ \text{Sum notation for the expansion: }\\ f(x) \approx P(x) = \Sigma_{n=0}^{\infty} \frac{e^c}{n!}(x-c)^n\\ \text{Plug in } 0 \text{ for c to get the Maclaurin Series of } f(x):\\ f(x) \approx P(x) = 1+ x +\frac{x^2}{2!}+\frac{x^3}{3!}+... \text{ valid for } -\infty < x < \infty. \] ### \(f(x) = ln(1+x)\)
\[ \text{f(x) is valid for the range } (-1, \infty)\\ \text{Find the first three derivatives of } f(x):\\ f(c) = ln(1+c)\\ f'(c) = \frac{1}{1+c}\\ f''(c) = \frac{-1}{(1+c)^2}\\ f'''(c) = \frac{2}{(1+c)^3}\\ \text{Plug into the Taylor Series formula: }\\ \Sigma_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^{n} = \frac{ln(1+c)}{0!}(x-c)^0 +\frac{1}{(1+c)1!}(x-c)^1 - \frac{1}{(1+c)^2 2!}(x-c)^2 + \frac{2}{(1+c)^3 3!} (x-c)^3\\ \text{Sum notation for the expansion: }\\ f(x) \approx P(x) = ln(1+c) + \Sigma_{n=1}^{\infty} (-1)^{n+1} \frac{(x-c)^n}{n(c+1)^n}\\ \text{Plug in } 0 \text{ for c to get the Maclaurin Series of } f(x):\\ f(x) \approx P(x) = x - \frac{x^2}{2} + \frac{x^3}{3} +... \text{ valid for } -1 < x < \infty. \]
\[ \text{f(x) is valid for all x } -\infty < x < \infty\\ f(c) = ln(1+c)\\ f'(c) = \frac{1}{1+c}\\ f''(c) = \frac{-1}{(1+c)^2}\\ f'''(c) = \frac{2}{(1+c)^3}\\ \text{Plug into the Taylor Series formula: }\\ \Sigma_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^{n} = \frac{ln(1+c)}{0}(x-c)^0 +\frac{1}{(1+c)1!}(x-c)^1 + \frac{-1}{(1+c)^2 2!}(x-c)^2 +\frac{2}{(1+c)^3 3!} (x-c)^3\\ \text{Sum notation for the expansion: }\\ f(x) \approx P(x) = ln(1+c) + \Sigma_{n=1}^{\infty} (-1)^{n+1} \frac{(x-c)^n}{n(c+1)^n}\\ \text{Plug in 0 for c to get the Maclaurin Series of f(x): }\\ f(x) \approx P(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ... \text{ valid for } -\infty < x < \infty. \]