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.
When \(1 - x = 0 \quad so \quad when \quad x = 1\) the function is not defined Getting the first 5 terms in the tylor serier
\[f(x) = \frac{1}{(1-x)} \quad then \quad f(0) = 1\]
\[f'(x) = \frac{1}{(1-x)^2} \quad then \quad f'(0) = 1\]
\[f^2(x) = \frac{2}{(1-x)^3} \quad then \quad f^2(0) = 2\] \[f^3(x) = \frac{6}{(1-x)^4} \quad then \quad f^3(0) = 6\] \[f^4(x) = \frac{24}{(1-x)^5} \quad then \quad f^4(0) = 24\] \[f^5(x) = \frac{120}{(1-x)^6} \quad then \quad f^5(0) = 120\]
We can apply the McClaurin Series formula \[1 + \frac{1}{1!} x^1 + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 + \frac{120}{5!}x^5 \]
To simplify, \[1 + x + x^2 + x^3 + x^4 + x^5\]
\[\sum_{n=0}^{\infty} x^n\]
Derivatives and evaluation at \(x = 0\) \[f^1(x)=e^x \quad then \quad f^1(0) = 1\]
\[f^2(x)=e^x \quad then \quad f^2(0) = 1\]
\[f^3(x)=e^x \quad then \quad f^3(0) = 1\]
\[f^4(x)=e^x \quad then \quad f^4(0) = 1\]
\[f^5(x)=e^x \quad then \quad f^5(0) = 1\]
\[1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + ... + \frac{x^n}{n!}\] The summation formula
\[\sum_{n=0}^{\infty} \frac{x^n}{n!}\]
Derivatives and evaluation at \(x = 0\)
\[f^1(x) = \frac{1}{x+1} \quad then \quad f^1(0) = 1\] \[f^2(x) = \frac{-1}{(x+1)^2} \quad then \quad f^2(0) = -1\] \[f^3(x) = \frac{2}{(x+1)^3} \quad then \quad f^3(0) = 2\] \[f^4(x) = \frac{-6}{(x+1)^4} \quad then \quad f^4(0) = -6\] \[f^5(x) = \frac{24}{(x+1)^5} \quad then \quad f^5(0) = 24\] after simplifying \[x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5+ ... (-1)^{n+1}\frac{1}nx^n\]
\[\sum_{n=0}^{\infty} (-1)^{n+1}\frac{1}nx^n\]