ASSIGNMENT 14 - TAYLOR SERIESIS 605 FUNDAMENTALS OF COMPUTATIONAL MATHEMATICS - 2014 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.
Setting c = 0 in the Taylor series makes it equivalent to the Maclaurin Series.
\[\sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n\]
is equivalent to
\[\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}(x)^n\]
\[\ f(x) = \frac{1}{1-x}\]
\[\ f(x) = \frac{1}{1-x} => f(0) = 1\]
\[\ f'(x) = \frac{1}{(1-x)^2} => f'(0) = 1\]
\[\ f''(x) = -\frac{2}{(1-x)^3} => f''(0) = -2\] \[\ f'''(x) = \frac{6}{(1-x)^4} => f'''(0) = 6\]
\[\ f^{4}(x) = -\frac{24}{(1-x)^5} => f^{4}(0) = -24\]
Substitute the derivatives into the formula
\[\ f(x) = 1 + x - x^2 + x^3 - x^4 ... = \sum_{n=0}^{\infty}x^n\]
\[\ f(x) = e^x\]
If \(\ f(x) = e^x\), then
\(\ f(x) = f'(x) = f''(x) = f^{n}(x) = e^x\)
substitute x = 0
\(\ f(0) = f'(0) = f''(0) = f^{n}(0) = e^0 = 1\)
Therefore we get ,
\[\sum_{n=0}^{\infty}\frac{1}{n!}x^n = \sum_{n=0}^{\infty}\frac{x^n}{n!}\]
\[\ f(x) = ln(1 + x)\]
\[\ f(x) = ln(1+x) => f(0) = 0\]
\[\ f'(x) = 1 + x => f'(0) = 1\]
\[\ f''(x) = -(1 + x )^2=> f''(0) = -1\]
\[\ f'''(x) = -2(1 + x )^3 => f'''(0) = -2\] \[\ f^{4}(x) = -6(1 + x )^4 => f^{4}(0) = -6\]
Therefore:
\[\ f(x) = x+ \frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+... \sum_{n=0}^{\infty}\frac{x^n}{n!}\]