This week, we’ll work out some Taylor Series expansions of popular functions. 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.
Using Chain Rule
\[ \begin{aligned} f(x)&=\frac{1}{1-x} \\f'(x)&=\frac{1}{(1-x)} = (1-x)^{-1} \\f'(x) &= - 1 (1-x)^{-1-1} * \frac{d(1-x)}{dx} = -(1-x)^{-2} (-1) = (1-x)^{-2} \end {aligned} \]
\[ \begin{aligned} f''(x) &= 2! (1-x)^ {-3} \Rightarrow f''(0) = 2! \\ f'''(x) &= 3! (1-x)^{-4} \Rightarrow f ''' (0) = 3! \\f^{n} &= n!(1-x) ^{-n-1} \end {aligned} \]
According to Maclaurin Series
\[ \begin{aligned} P_n(x) &=a_{0} + a_1(x-0) + a_2(x-0) 2 + ... + a_n(x-0) n \\&= f(0) + \frac {f(0)}{1!x} + \frac {f''(0)}{2!x^2} + \frac {f'''(0)}{3!x^3}+ ... + \frac {f(n)}{n!x^n} \\&= 1+ \frac {1}{1!x} + \frac {2!}{2!x^2} + \frac {3!}{3!x^3} +... + \frac {n!}{n!x^n} \end {aligned} \]
Note: The derivative of \(e^x\) is \(e^x\) because the slope is the same as the function value.
\[ \begin{aligned} f(x) &= e^{x} \\f'(x) &= e^x \Rightarrow f(0) = 1 \\f''(x) &= e^x \Rightarrow f(0) = 1 \\f'''(x) &= e^x \Rightarrow f(0) = 1 \\f''''(x) &= e^x \Rightarrow f(0) = 1 \\[3ex] \\f(x) &= f(0) + \frac {f'(0)} {1!x} + \frac {f''(0)} {2!x^2} \\f(x) &=\sum_{n=0}^{\infty} a_n(x-c)^n \end {aligned} \]
\[ \begin{aligned} f(x) &= ln(1 + x) \\ f'(x) &= \frac{1}{1-(-x)} \\ln(1+x) &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ \frac{x^5}{5} -\frac{x^6}{6} +... \end {aligned} \]