ASSIGNMENT 14 - TAYLOR SERIES
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 an R- Markdown document.
Utilize the Taylor Series Function: \[f(x) = \sum^{\infty}_{n=0}\frac{f^{(n)}(c)}{n!}(x-c)^n\]
Find the derivatives and evaluate at x = 0: \(f(x) = \frac{1}{(1-x)}\)
\[f(x) = \frac{1}{(1-x)} = 1\\ f'(x) = \frac{1}{(1-x)^2} = 1\\ f''(x) = \frac{2}{(1-x)^3} = 2\\ f'''(x) = \frac{6}{(1-x)^4} = 6 \]
McClaurin Series formula as:
\[1+\frac{1}{1!}x^1+\frac{2}{2!}x^2+\frac{6}{3!}x^3...\frac{n!}{n!}x^n\]
Which is simplified to:
\[1+x+x^2+x^3+...x^n\]
Making the Taylor Expansion of \(f(x) = \frac{1}{(1-x)} = \sum^\infty_{n=o}x^n\)
Get the derivatives and evaluate when x = 0 \(f(x) = e^x\)
This one is less involved since the derivative of \(e^x = e^x\) \[f(x) = e^x = 1\\ f'(x) = e^x = 1\\ f''(x) = e^x= 1\\ f'''(x) = e^x = 1 \]
Thus the McClaurin Formula and summation will look like:
\[Mclaurin=1+\frac{1}{1!}x^1+\frac{1}{2!}x^2+\frac{1}{3!}x^3...\frac{1}{n!}x^n\\ Summation= \sum^\infty_{n=o}\frac{x^n}{n!} \]
Find the derivatives and evaluate at x = 0 \(f(x) = ln(1 + x)\) \[ f(x) = ln(1 + x) = 0\\ f'(x) = \frac{1}{1+x}=1\\ f''(x) = -\frac{1}{(1+x)^2}=-1\\ f'''(x) = \frac{2}{(1+x)^3}=2\\ f''''(x) = -\frac{6}{(1+x)^4}=-6\\ \]
Thus the McClaurin Formula and summation will look like:
\[ Mclaurin=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}...\frac{1}{n!}x^n\\ Summation= \sum^\infty_{n=1}\frac{(-1)^{n-1}x^n}{n} \]
Get the derivatives and evaluate when x = 1, because x=0 will not work in this case \(f(x) = x^{\frac{1}{2}}\)
\[ f(x) = x^{\frac{1}{2}}=1\\ f'(x)=\frac{1}{2x^{\frac{1}{2}}}=\frac{1}{2}\\ f''(x)=-\frac{1}{4x^{\frac{3}{2}}}=-\frac{1}{4}\\ f'''(x)=\frac{3}{8x^{\frac{5}{2}}}=-\frac{3}{8} \]
Thus the McClaurin Formula and summation will look like:
Having some trouble discerning the McClaurin and summation formulas for this problem.