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. Please submit your assignment as a R-Markdown document.
Taylor series is defined as
\[f(x)\quad =\sum _{n=0 }^{\infty}{\frac {f^{n}(a)}{n!}}(x-a)^{n}\quad\]
Lets find derivatives,
\[f^0(c) = \frac{1}{(1-c)}\] \[f^{'}(c) = \frac{1}{(1-c)^2}\] \[f^{''}(c) = \frac{2}{(1-c)^3}\]
\[f^{'''}(c) = \frac{6}{(1-c)^4}\]
\[f^{''''}(c) = \frac{24}{(1-c)^5}\]
As per definition,
\[ f(x) = \frac{1}{(1-c)0!}(x-c)^0+\frac{1}{(1-c)^21!}(x-c)^1+\frac{2}{(1-c)^32!}(x-c)^2+\frac{6}{(1-c)^43!}(x-c)^3+ ...\] \[ = \frac{1}{(1-c)}+\frac{1}{(1-c)^2}(x-c)+\frac{2!}{(1-c)^32!}(x-c)^2+\frac{3!}{(1-c)^43!}(x-c)^3+ ...\] \[ = \frac{1}{(1-c)}+\frac{1}{(1-c)^2}(x-c)+\frac{1}{(1-c)^3}(x-c)^2+\frac{1}{(1-c)^4}(x-c)^3+ ...\] \[ =\sum _{n=0 }^{\infty}\frac {1}{(1-c)^{n+1}}(x-c)^n\]
Using Maclaurin series c = 0,
\[f(x) =\sum _{n=0 }^{\infty}x^n = 1+x+x^2+x^3+x^4+ ..... \]
The series converges in the interval (-1,1)
Lets find derivatives,
\[f^0(c) = e^c\] \[f^{'}(c) = e^c\] \[f^{''}(c) = e^c\] \[f^{'''}(c) = e^c\] \[f^{''''}(c) = e^c\]
As per definition,
\[ f(x) = \frac{e^c}{0!}(x-c)^0+\frac{e^c}{1!}(x-c)^1+\frac{e^c}{2!}(x-c)^2+\frac{e^c}{3!}(x-c)^4+ ...\]
\[ =e^c \sum _{n=0 }^{\infty}\frac {(x-c)^n}{n!}\]
Using Maclaurin series c = 0,
\[f(x) =\sum _{n=0 }^{\infty}\frac {x^n}{n!} = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+ ..... \]
Test ratio
\[ \frac {a_{n+1}}{a_n} = \frac {x^{n+1}}{(n+1)!} * \frac {n!}{x^n} = \frac {x}{n+1}\]
\[L = \lim_{n\to\infty} \frac {x}{n+1} = 0\]
Hence the series converges for \(\forall\) x in the range (-\(\infty\), \(\infty\))
Lets find derivatives,
\[f^0(c) = ln(1 + c)\]
\[f^{'}(c) = \frac{1}{(c+1)}\]
\[f^{''}(c) = -\frac{1}{(c+1)^2}\]
\[f^{'''}(c) = \frac{2}{(c+1)^3}\]
\[f^{''}(c) = -\frac{6}{(c+1)^4}\]
As per definition,
\[ f(x) = \frac{ln(1+c)}{0!}(x-c)^0+\frac{1}{(c+1)1!}(x-c)^1-\frac{1}{(c+1)^22!}(x-c)^2+\frac{2}{(c+1)^33!}(x-c)^3-\frac{6}{(c+1)^44!}(x-c)^4+ ...\]
\[= ln(1+c)+\frac{1}{(c+1)}(x-c)-\frac{1!}{(c+1)^22*1!}(x-c)^2+\frac{2!}{(c+1)^33*2!}(x-c)^3-\frac{3!}{(c+1)^44*3!}(x-c)^4+ ...\]
\[= ln(1+c)+\frac{1}{(c+1)}(x-c)-\frac{1}{2(c+1)^2}(x-c)^2+\frac{1}{3(c+1)^3}(x-c)^3-\frac{1}{4(c+1)^4}(x-c)^4+ ...\]
\[= ln(1+c)+\sum _{n=0 }^{\infty}(-1)^{n+1}\frac {(x-c)^n}{n(c+1)^n!}\]
Using Maclaurin series c = 0,
\[f(x) =\sum _{n=0 }^{\infty}(-1)^{n+1}\frac {x^n}{n} = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+ ..... \]
Test ratio
\[ \frac {a_{n+1}}{a_n} = \frac {(-1)^{n+1+1}x^{n+1}}{(n+1)} * \frac {n}{(-1)^{n+1}x^{n}} = \frac {-xn}{n+1}\]
\[L = \lim_{n\to\infty} | \frac {-xn}{n+1}| = |x|\]
Hence the series converges for x in the range (-1, 1).