This week, we’ll work out some Taylor Series expansions of popular functions.
\(f(x) =\frac{1}{(1-x)}\)
\(f(x)=e^x\)
\(f(x)=ln(1+x)\)
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.
This function is not defined when x = 1.
Derivaties and evaluation at \(x=0\).
\(f(x) = \frac{1}{(1-x)} \rightarrow f(0) = 1\)
\(f'(x) = \frac{1}{(1-x)^2} \rightarrow f'(0) = 1\)
\(f''(x) = \frac{2}{(1-x)^3} \rightarrow f''(0) = 2\)
\(f'''(x) = \frac{6}{(1-x)^4} \rightarrow f'''(0) = 6\)
\(f^4(x) = \frac{24}{(1-x)^5} \rightarrow f^4(0) = 24\)
\(f^5(x) = \frac{120}{(1-x)^6} \rightarrow f^5(0) = 120\)
Use McClaurin Series formula:
\(1 + \frac{1}{1!} x^1 + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 + \frac{120}{5!}x^5 + ...\)
Simplifies to:
\(1 + x + x^2 + x^3 + x^4 + x^5 + ...+ x^n\)
In summation form:
Derivatives and evaluation at \(x=0\):
\(f(x)=e^x\) \(\rightarrow\) 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
\(f^4(x)=e^x\) \(\rightarrow\) \(f^4(0) = 1\)
\(f^5(x)=e^x\) \(\rightarrow\) \(f^5(0) = 1\)
Use McClaurin Series formula:
\(1\) + \(\frac{1}{1!}\)\(x^1\) + \(\frac{1}{2!}\)\(x^2\) + \(\frac{1}{3!}\)\(x^3\) + \(\frac{1}{4!}\)\(x^4\) + …
This simplifies to:
\(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + ... + \frac{x^n}{n!}\)
In summation form:
Derivatives and evaluation at \(x=0\).
\(f(x) = ln(1+x) \rightarrow f(0) = 0\)
\(f'(x) = \frac{1}{x+1} \rightarrow f'(0) = 1\)
\(f''(x) = \frac{-1}{(x+1)^2} \rightarrow f''(0) = -1\)
\(f'''(x) = \frac{2}{(x+1)^3} \rightarrow f'''(0) = 2\)
\(f^4(x) = \frac{-6}{(x+1)^4} \rightarrow f^4(0) = -6\)
\(f^5(x) = \frac{24}{(x+1)^5} \rightarrow f^5(0) = -24\)
Use McClaurin Series formula:
\(0 + \frac{1}{1!}x^1 - \frac{1}{2!}x^2 + \frac{2}{3!}x^3 - \frac{6}{4!}x^4+ \frac{24}{5!}x^5 + ...\)
Simplifies to:
\(x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5+ ... (-1)^{n+1}\frac{1}nx^n\)
In summation form: