This week, we’ll work out some Taylor Series expansions of popular functions. \[f(x) = \frac{1}{1-x}\] \[f(x) = e^x\] \[f(x) = ln(1 + x)\] For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion. Please submit your assignment as a R-Markdown document.
The general form of taylor series is given by:
\[f(x) = a + bx + cx^2 + dx^3 + ex^4 + ...\] or \[f(x) = f(0) + \frac{f'(0)x}{1!} + \frac{f''(0)x^2}{2!} + ... + \frac{f^n(0)n}{n!}\] ###1
\[f(x) = \frac{1}{1-x}, f(0) = 1\] \[f'(x) = \frac{-1}{(1-x)^2} * -1 = \frac{1}{(1-x)^2 , f'(0) = 1\] \[f''(x) = \frac{-2}{(1-x)^3} * -1 = \frac{1*2}{(1-x)^3 , f''(0) = 1*2\] \[f'''(x) = \frac{2*-3}{(1-x)^4} * -1 = \frac{2*3}{(1-x)^4 , f''(0) = 1*2*3\] \[f(x) = \frac{1}{1-x} = 1 + \frac{1*2}{1!}*x + \frac{1*2*3}{2!}*x^2+...\] \[f(x) = 1 + 2x + 3x^2 + 4x^3\] ###2 \[f(x) = e^x\] \[f(x) = f'(x) = f''(x) = f'''(x) = e^x\] \[f(0) = f'(0) = f''(0) = f'''(0) = e^0 = 1\] \[f(x) = e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} +...+ \frac{x^n}{n!}\]
\[f(x) = ln(1 + x), f(0) = ln(1) = 0\] \[f'(x) = \frac{1}{1+x}, f'(0) = 1\] \[f''(x) = \frac{-1}{(1+x)^2}, f''(0) = -1\] \[f'''(x) = \frac{2}{(1+x)^3}, f'''(0) = 2\] \[f(x) = ln(1 + x) = 0 + \frac{x}{1!} - \frac{x^2}{2!} + \frac{2x^3}{3!} +...\] \[f(x) = \frac{x}{1!} - \frac{x^2}{2!} + \frac{x^3}{3!} - \frac{x^4}{4!} + \frac{x^5}{5!} ...\]