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.

Solution:

\(f(x) = \frac{1}{(1-x)}\)

\[f(x)\quad =\sum _{ n=0 }^{ \infty } {\frac{f^{(n)} (a)}{n!}}(x-a)^{n}\] \[f^{(n)} (x) = {\frac{n!}{(1-x)^{(n+1)}}}\] \[f(a)\quad =\sum _{ n=0 }^{ \infty } {\frac{{\frac{n!}{(1-a)^{(n+1)}}}}{n!}}(x-a)^{n}\] \[f(a)\quad =\sum _{ n=0 }^{ \infty } {\frac{(x-a)^{n}}{(1-a)^{(n+1)}}}\] a = 0 \[f(x)\quad =\sum _{ n=0 }^{ \infty } {\frac{(x-0)^{n}}{(1-0)^{(n+1)}}}\] \[f(x)\quad =\sum _{ n=0 }^{ \infty }{ x^{ n },\quad x\quad \in \quad (-1,1) }\]

Solution:

\(f(x) = e^x\)

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

Solution:

\(f(x) = ln(1 + x)\)

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