Taylor Series Socratic Question
https://socratic.org/questions/how-do-you-find-the-first-3-non-zero-terms-in-the-taylor-series-of-the-function-
The formula for a third degree Taylor Series polynomial is:
#T_3 f(x) = f(a) + f'(a)(x-a) +f''(a)frac{(x-a)^2}{2!}+f'''(a)frac{(x-a)^3}{3!} #
So to find this, we need to determine the first, second, and third derivatives of #f(x)#.
#f(x) = cosh x = 1/2 e^x + 1/2 e^(-x)#
#f(ln8) = 1/2 e^(ln8) + 1/2 e^(ln(1/8)) = 4 + 1/16 = 65/16#
#f'(x) = 1/2 e^x - 1/2 e^(-x)#
#f'(ln8) = 1/2 e^(ln8) - 1/2 e^(ln(1/8)) = 4 -1/16 = 63/16#
#f''(x) = 1/2 e^x + 1/2 e^(-x)#
#f''(ln8) = 65/16#
#f'''(x) = 1/2 e^x - 1/2 e^(-x)#
#f'''(ln8) = 63/16#
#T_3 f(x) = f(ln8) + f'(ln8)(x-ln8) +f''(ln8)frac{(x-ln8)^2}{2!}+f'''(ln8)frac{(x-ln8)^3}{3!} #
#T_3 f(x) = frac{65}{16} + frac{63}{16}(x-ln8) +frac{65}{16}frac{(x-ln8)^2}{2!}+ frac{63}{16} * frac{(x-ln8)^3}{3!} #
Finally, use this Taylor Series approximation to estimate #cosh(2)#:
#T_3 f(2) = frac{65}{16} + frac{63}{16}(2-ln8) + frac{65*(2-ln8)^2}{16*2}+ frac{63(2-ln8)^3}{16*6} #
This would be tedious to simplify, so I'll plug this in to my calculator to get:
#T_3 f(2) approx 3.7667#
Compare this to the actual value: #cosh (2) approx 3.7622#. This shows that our approximation is accurate to the nearest hundreths place.
