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.
\[f(x) = \frac{1}{(1âx)} = \frac{1}{(1-a)}+\frac{(x-a)}{(1-a)^2}+\frac{(1-a)^2}{(1-a)^3}+\frac{(1-a)^3}{(1-a)^4}+...\] \[if a = 0\]
\[f(x) = 1+x+x^2+x^3+x^4+ ...\]
\[f(x)\quad =\sum _{ n=0 }^{ \infty }{ x^{ n },\quad x\quad \in \quad (-1,1) }\]
\(f(x) = e^x\)
\[f(x) = e^x = e^a + e^a (x-a) + e^a (x-a)^2 + e^a (x-a)^3+ ...\]
a=0
\[f(x) = e^x = e^0 + e^0 (x-0) + e^0 (x-0)^2 + e^0 (x-0)^3+ ...\] \[f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...\]
\[f(x)\quad =\sum _{ n=0 }^{ \infty }{ \frac { x^{ n } }{ n! } ,\quad x\quad \in \quad R }\]
\[f(x) = ln(1 + x) = ln(1+a) + \frac{(x-a)}{(1+a)}-\frac{(x-a)^2}{2!(1+a)^2}+\frac{(x-a)^3}{3!2(1+a)^3} - \frac{(x-a)^4}{4!(3)(2)(1+a)^3} + ...\]
a = 0
\[f(x) = x - \frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+ ...\] \[f(x)\quad =\sum _{ n=0 }^{ \infty }{ (-1)^{ n-1 }\frac { x^{ n } }{ n } ,\quad x\quad \in \quad (-1,1) }\]