This week, we’ll work out some Taylor Series expansions of popular functions. \(f (x) = \frac1{(1−x)}\) \(f (x) = e^x\) \(f (x) = ln(1 + x)\) \(f(x)=x^{1/2}\)
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.
\(f (x) = \frac1{(1−x)}\)
\(f(x) = \frac1{(1−x)}\) \(f(0) = 1\)
$f’(x) = $ \(f'(0) = 1\)
\(f''(x) = -\frac2{(x-1)^3}\) \(f''(0) = 2\)
\(f'''(x) = \frac6{(x-1)^4}\) \(f'''(0) = 6\)
\(f''''(x) = -\frac{24}{(x-1)^5}\) \(f''''(0) = 24\)
\(f(x) = \frac1{1-x} = \sum_{n=0}^{\infty} x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5}\)
\(f(x) = \frac1{1-x} = \sum_{n=0}^{\infty} x^n\)
\(f(x) = e^x\) \(f(0) = 1\)
\(f'(x) = e^x\) \(f'(0) = 1\)
\(f''(x) = e^x\) \(f''(0) = 1\)
\(f'''(x) = e^x\) \(f'''(0) = 1\)
\(f(x) = e^x = \sum_{n=0}^{\infty} 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ....\)
\(f(x) = e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\)
\(f (x) = ln(1 + x)\)
\(f(x) = ln(1+x)\) \(f(0) = 0\)
\(f'(x) = \frac1{1+x}\) \(f'(0) = 1\)
\(f''(x) = -\frac1{(1+x)^2}\) \(f''(0) = -1\)
\(f'''(x) = \frac2{(1+x)^3}\) \(f'''(0) = 2\)
\(f''''(x) = \frac2{(1+x)^3}\) \(f''''(0) = -6\)
\(f(x) = ln(1+x) = \sum_{n=0}^{\infty} 0 + x - \frac{x^2}{2!} + \frac{2x^3}{3!} - \frac{6x^4}{4!} + ....\)
\(f(x) = ln(1+x) = \sum_{n=0}^{\infty} x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\)
\(f(x) = ln(1+x) = \sum_{n=0}^{\infty} \frac{-1^{n+1}x^n}{n}\)
\(f(x)=x^{1/2}\)
\(f(x) = x^{1/2}\) \(f(0) = 0\)
\(f'(x) = \frac{x^{1/2}}{2}\) \(f'(0) = Undefined\)
The Taylor series expansion for \(x^{1/2}\) is not possible as the first derivative of the function is undefined at f(0).