Taylor Series expansions of popular functions.
To recall:
\[p_n = \dfrac{f(n)}{n!}\]
\[f(x) = \sum^{\infty}_{n=0}p_n(x-c)^n\] Question 1:
\[f(c)= \dfrac{1}{(1−c)} \]
\[f(c) = \dfrac{1}{(1−c)} ;\ f^0(0)= 1, \ p_0 = 1\]
\[f(c) = \dfrac{1}{(1−c)^2} ;\ f^1(0)= 1, \ p_1 = 1\]
\[f(c) = \dfrac{2}{(1−c)^3} ;\ f^2(0)= 2, \ p_2 = 1\]
\[f(c) = \dfrac{6}{(1−c)^4} ;\ f^3(0)= 6, \ p_3 = 1\]
\[f(c) = \dfrac{24}{(1−c)^5} ;\ f^3(0)= 24, \ p_4 = 1\]
Taylor Series expansion:
\[1 + \dfrac{1}{(1−c)} + (\dfrac{1}{(1−c)})^2 + (\dfrac{1}{(1−c)})^3 + ...\]
\[1 + x + x^2 + x^3 + ...\]
Question 2:
\[f(x)=e^x\]
\[f(c) = e^c ;\ f^0(0)= 1, \ p_0 = 1\]
\[f(c) = e^c ;\ f^1(0)= 1, \ p_1 = 1\]
\[f(c) = e^c ;\ f^2(0)= 1, \ p_2 = \dfrac{1}{2}\]
\[f(c) = e^c ;\ f^3(0)= 1, \ p_3 = \dfrac{1}{6}\]
\[f(c) = e^c ;\ f^4(0)= 1, \ p_4 = \dfrac{1}{24}\]
Taylor Series Expansion:
\[1 + e^c + \dfrac{1}{2}(e^c)^2 + \dfrac{1}{6}(e^c)^3 + ...\]
\[1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} ...\]
Question 3: \[f(x)=ln(1+x)\]
\[f(c) = ln(1+c) ;\ f^0(0)= 0, \ p_0 = 0\]
\[f(c) = \dfrac{1}{1+c} ;\ f^1(0)= 1, \ p_0 = 1\]
\[f(c) = \dfrac{-1}{(1+c)^2} ;\ f^2(0)= -1, \ p_0 = \dfrac{-1}{2}\]
\[f(c) = \dfrac{2}{(1+c)^3} ;\ f^3(0)= 2, \ p_0 = \dfrac{1}{3}\]
\[f(c) = \dfrac{-6}{(1+c)^4} ;\ f^4(0)= -6, \ p_0 = \dfrac{-1}{4}\]
Taylor Series Expansion:
\[ln(1+c) - \dfrac{1}{2}(ln(1+c))^2 + \dfrac{1}{3}(ln(1+c))^3 + ...\]
\[ x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} ...\]