This week, we’ll work out some Taylor Series expansions of popular functions.
\(f(x) = \frac{1}{(1−x)}\)
\(f(x) = e^x\)
\(f(x) = ln(1 + x)\)
\(f(x)=x^{(1/2)}\)
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.
\(f(x) = \frac{1}{(1−x)}\)
Derivative-itive:
\(f(0) = 0\)
\(f'(x) = (1−x)^{-2} \rightarrow f'(0) = 1\)
\(f''(x) = 2(1−x)^{-3} \rightarrow f''(0) = 2\)
\(f''''(x) = 6(1−x)^{-4} \rightarrow f'''(0) = 6\)
\(f'''''(x) = 24(1−x)^{-5} \rightarrow f''''(0) = 24\)
The pattern is this:
\(1 + x^2 + x^3 + x^4\)
\(\sum_{n=0}^{\infty} x^n\)
Valid ranges from -1 to 1.
\(f(x) = e^x\)
\(f(0) = 1\)
\(f'(x) = e^x \rightarrow f'(0) = 1\)
\(f''(x) = e^x \rightarrow f''(0) = 1\)
\(f'''(x) = e^x \rightarrow f'''(0) = 1\)
\(f''''(x) = e^x \rightarrow f''''(0) = 1\)
Pattern is:
\(e^n (0) = e^x = 1\) for \(n = 0,1,2...\)
Taylor series for around \(x = 0\):
\(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\)
\(f(x) = ln(1 + x)\)
\(f(0) = 0\)
\(f'(x) = \frac{1}{1+x} \rightarrow f'(0) = 1\)
\(f''(x) = \frac{-1}{(1+x)^2} \rightarrow f''(0) = -1\)
\(f'''(x) = \frac{2}{(1+x)^3} \rightarrow f'''(0) = 2\)
\(f''''(x) = \frac{-6}{(1+x)^4} \rightarrow f''''(0) = -6\)
Pattern is:
\(\frac{0}{0!}*1 + \frac{1}{1!}*x^1 - \frac{1}{2!}*x^2 + \frac{2}{3!}*x^3 - \frac{6}{4!}*x^4\) …
Taylor series for range -1 to 1:
\(\sum_{n=0}^{\infty}(-1)^{n+1}\frac{x^n}{n}\)
\(f(x)=x^{(1/2)}\)
Since \(x^{(1/2)}\) doesn’t have a first derivative at x = 0, it’s not defined…I hope.