This week, we’ll work out some Taylor Series expansions of popular functions.
\[f(x) = \frac {1}{(1−x)}\] \[f(x) = e^x\] \[f(x) = ln(1 + x)\] \[f(x) = x^{1/2}\]
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 an R-Markdown document.
Taylor’s Theorem:
\[f(x) = \sum_{n=0}^{\infty} \frac {f^{(n)}(c)}{n!} (x-c)^{n}\]
\[f(x) = \frac {1}{(1−x)}\]
Find the Taylor Series centered at c = 0:
\[f^{(0)}(c) = \frac {1}{(1-c)}\] \[f^{(0)}(c) = \frac {1}{(1-c)}\] \[f^{(2)}(c) = \frac {2}{(1-c)^{3}}\] \[f^{(3)}(c) = \frac {6}{(1-c)^{4}}\]
The expansion can be written as:
\[f(x) = \frac {1}{(1-c) 0!}(x-c)^{0} + \frac {1}{(1-c)^{2} 1!} (x-c)^{1} + \frac {2}{(1-c)^{3} 2!} (x-c)^{2} + \frac {6}{(1-c)^{3} 3!} (x-c)^{3} + ...\] \[f(x)=\sum_{n=0}^{\infty}\frac{1}{(1-c)^{(n+1)}}(x-c)^n\]
Setting c = 0:
\[1+x+x^2+x^3+...\]
\[f(x) = e^x\]
Find the Taylor Series centered at c = 0:
\[f^{(0)}(c)=e^c\] \[f(1)(c)=ec\] \[f^{(2)}(c)=e^c\] \[f^{(3)}(c) = e^{c}\]
The expansion can be written as: \[f(x) = \frac {e^{c}}{0!}(x-c)^{0} + \frac {e^{c}}{1!}(x-c)^{1} + \frac {e^{c}}{2!}(x-c)^{2} + \frac {e^{c}}{3!}(x-c)^{3} + ...\] \[f(x) = e^{c} \sum_{n=0}^{\infty} \frac {(x-c)^{n}}{n!}\]
Setting c = 0: \[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\]
\[f(x)=ln(1+x)\]
Find the Taylor Series centered at c = 0:
\[f^{(0)}(c)=ln(1+c)\] \[f^{(1)}(c)=\frac{1}{(c+1)}\] \[f^{(2)}(c) = -\frac {1}{(c+1)^{2}}\] \[f^{(3)}(c) = \frac {2}{(c+1)^{3}}\]
The expansion can be written as:
\[f(x) = \frac {ln(1+c)}{0!}(x-c)^{0} + \frac {1}{(c+1)1!}(x-c)^{1} - \frac {1}{(c+1)^{2}2!}(x-c)^{2} + \frac {2}{(c+1)^{3}3!}(x-c)^{3} - ...\]
\[f(x) = ln(1+c) + \sum_{n=1}^{\infty} (-1)^{n+1} \frac {(x-c)^{n}}{n(c+1)^{n}} \]
Setting c = 0:
\[x - \frac {x^{2}}{2} + \frac {x^{3}}{3} - \frac {x^{4}}{4} + ...\]
\[f(x) = x^{1/2}\]
Find the Taylor Series centered at c = 0:
\[f^{(0)}(c) = c^{1/2}\] \[f^{(1)}(c) = \frac {1}{2 (c^{1/2})}\] \[f^{(2)}(c) = -\frac {1}{4 (c^{3/2})}\] \[f^{(3)}(c) = \frac {3}{8 (c^{5/2})}\]
The expansion can be written as:
\[f(x) = c^{1/2} + \frac {1}{2 (c^{1/2}) 1!} (x-1)^{1} - \frac {1}{4 (c^{3/2}) 2!} (x-2)^{2} + \frac {3}{8 (c^{5/2}) 3!} (x-3)^{3} - ...\]
\[f(x) = c^{1/2} + \sum_{n=1}^{\infty} (-1)^{n+1} \binom{1/2}{n}(x-c)^{n} (c^{1/2-n})\]
Setting c = 0:
\[\sum_{n=1}^{\infty} (-1)^{n+1} \binom{1/2}{n}x^{n}\]