1 This week, we’ll work out some Taylor Series expansions of popular functions.
\(f(x) = \frac{1}{(1−x)}\)
Given this is the approximation of the geometric series, we know that the first few terms of it are
\(1+x^2+x^3+x^4..x^n\) \(\\\) \(\Sigma_{n=0}^{\infty} x^n\)
Reviewing the derivatives to demonstrate the pattern
\(f(x)=\frac{1}{1-x}\) \(f'(x) = \frac{2*1}{(1-x)^2}\) \(f''(x) = \frac{3*2*1}{(1-x)^3}\) \(f'''(x) = \frac{4*3*2*1}{(1-x)^4}\)
Which can be represented as:
\(\sum_{n=0}^{\infty} \frac{1}{(x-c)^{n+1}}x^n; c=0 \sum_{n=0}^{\infty} x^n\)
\(f(x) = e^x\) \(f(x)=e^x\) \(f'(x) = e^x\) \(f''(x) = e^x\) \(f'''(x) = e^x\)
Based on the formula/theorom for the Taylor series expansion:
\(f(x) \sim \frac{e^c}{0!}(x-c)^0 + \frac{e^c}{1!}(x-c)^1 + \frac{e^c}{2!}(x-c)^2\)
\(\sum_{n=0}^{\infty} \frac{x^n}{n!}\)
\(f(x) = ln(1 + x)\)
\(f'(x)=\frac{1}{1+x}\) \(f''(x)=-\frac{1}{{1+x}^2}\) \(f'''(x)=\frac{2*1}{{1+x}^3}\) \(f''''(x)=-\frac{3*2*1}{{1+x}^4}\)
\(f(x) \sim 1 - \frac{1}{1!(1+x)}(x-c)^1 +\frac{2}{2!(1+x)}(x-c)^2\) \(\Sigma_{n=0}^{\infty} \frac{{-1}^{n}x^{n+1}}{n+1}\)
\(f(x)=x^{(1/2)}\)
\(f'(x)=\frac{x^{-\frac{1}{2}}}{2}\) \(f''(x)=-\frac{x^{-\frac{3}{2}}}{4}\) \(f'''(x)=\frac{3x^{-\frac{5}{2}}}{8}\)
The link indicates that the function provided in not differentiable at 0 and given the f(x) is not infinitely differentiable it cannot be approximated with Taylor Series approximation.