Work out the Taylor Series expansions for the following popular functions.
For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.
Start with the formula:
\(\displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(c)}{n!}(x-c)^n\)
Obtain the first few derivatives of the function and evaluate at 0:
\(f(x)=\frac{1}{(1-x)};\ \ \
f(0)=1\)
\(f'(x)=\frac{1}{(1-x)^2};\ \ \
f'(0)=1\)
\(f''(x)=\frac{2}{(1-x)^3};\ \ \
f''(0)=2\)
\(f^{(3)}(x)=\frac{6}{(1-x)^4};\ \ \
f^{(3)}(0)=6\)
\(f^{(4)}(x)=\frac{24}{(1-x)^5};\ \ \
f^{(4)}(0)=24\)
Assume \(c=0\) and populate the first few terms of the Taylor polynomial:
\(\frac{1}{0!}x^0+\frac{1}{1!}x^1+\frac{2}{2!}x^2+\frac{6}{3!}x^3+\frac{24}{4!}x^4\)
The pattern for this is (note \(2!=2,\ \ \ 3!=6,\ \ \ 4!=24\)):
\(\frac{1}{(1-x)}=\displaystyle\sum_{n=0}^\infty x^n\)
with \(c=0\)
Start with the formula:
\(\displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(c)}{n!}(x-c)^n\)
Obtain the first few derivatives of the function:
\(f(x)=e^x\)
\(f'(x)=e^x\)
\(f''(x)=e^x\)
\(f^{(3)}(x)=e^x\)
\(f^{(4)}(x)=e^x\)
Populate the first few terms of the Taylor polynomial:
\(\frac{e^c}{0!}(x-c)^0+\frac{e^c}{1!}(x-c)^1+\frac{e^c}{2!}(x-c)^2+\frac{e^c}{3!}(x-c)^3+\frac{e^c}{4!}(x-c)^4\)
Assume c = 0:
\(\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\)
The pattern for this is:
\(e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}\)
with \(c=0\)
Start with the formula:
\(\displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(c)}{n!}(x-c)^n\)
Obtain the first few derivatives of the function and evaluate at 1:
\(f(x)=ln(x);\ \ \ f(1)=0\)
\(f'(x)=\frac{1}{x};\ \ \
f'(1)=1\)
\(f''(x)=-\frac{1}{x^2};\ \ \
f''(1)=-1\)
\(f^{(3)}(x)=\frac{2}{x^3};\ \ \
f^{(3)}(1)=2\)
\(f^{(4)}(x)=-\frac{6}{x^4};\ \ \
f^{(4)}(1)=-6\)
Assume \(c=1\) and populate the first few terms of the Taylor polynomial:
\(\frac{0}{0!}(x-1)^0+\frac{1}{1!}(x-1)^1-\frac{1}{2!}(x-1)^2+\frac{2}{3!}(x-1)^3-\frac{6}{4!}(x-1)^4\)
Which can be further reduced to: \(\frac{0!}{1!}(x-1)^1-\frac{1!}{2!}(x-1)^2+\frac{2!}{3!}(x-1)^3-\frac{3!}{4!}(x-1)^4\)
And then to: \(\frac{(x-1)^1}{1}-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\frac{(x-1)^4}{4}\)
The pattern for this is:
\(ln(x)=\displaystyle\sum_{n=1}^\infty(-1)^{n-1}\frac{(x-1)^n}{n}\)
with \(c=1\)
Start with the formula:
\(\displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(c)}{n!}(x-c)^n\)
Obtain the first few derivatives of the function and evaluate at 1:
\(f(x)=x^{(1/2)};\ \ \
f(1)=1\)
\(f'(x)=\frac{1}{2x^{(1/2)}};\ \ \
f'(1)=\frac{1}{2}\)
\(f''(x)=-\frac{1}{4x^{3/2}};\ \ \
f''(1)=-\frac{1}{4}\)
\(f^{(3)}(x)=\frac{3}{8x^{5/2}};\ \ \
f^{(3)}(1)=\frac{3}{8}\)
\(f^{(4)}(x)=-\frac{15}{16x^{7/2}};\ \ \
f^{(4)}(1)=-\frac{15}{16}\)
Assume \(c=1\) and populate the first few terms of the Taylor polynomial:
\(\frac{1}{0!}(x-1)^0+\frac{1}{2}\frac{1}{1!}(x-1)^1-\frac{1}{4}\frac{1}{2!}(x-1)^2+\frac{3}{8}\frac{1}{3!}(x-1)^3-\frac{15}{16}\frac{1}{4!}(x-1)^4\)
Which can be further reduced to: \(\frac{1}{0!}(x-1)^0+\frac{1}{2}\frac{1}{1!}(x-1)^1-\frac{1}{4}\frac{1}{2!}(x-1)^2+\frac{3}{8}\frac{1}{3!}(x-1)^3-\frac{15}{16}\frac{1}{4!}(x-1)^4\)
All of these are easy to breakdown except the pattern \(3\), \(3*5\), \(3*5*7\). If I remove the first two terms then I can capture that pattern with \(\frac{(2n+1)!}{2^nn!}\). And so this can be transformed a little closer to the final pattern thusly:
\(1+\frac{x-1}{2}-\frac{1!}{2^00!}\frac{1}{2^{(0+2)}}\frac{1}{(0+2)!}(x-1)^{(0+2)}+\frac{3!}{2^11!}\frac{1}{2^{(1+2)}}\frac{1}{(1+2)!}(x-1)^{(1+2)}-\frac{5!}{2^22!}\frac{1}{2^{(2+2)}}\frac{1}{(2+2)!}(x-1)^{2+2}\)
The pattern for this is:
\(x^{(1/2)}=\frac{x+1}{2}+\displaystyle\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)! }{2^nn!}\frac{(x-1)^{n+2}}{2^{n+2}(n+2)!}\)
with \(c=1\)