Calculus of a Single Variable, Chapter 9, 9.7, Section 9.7, Problem 14

Maclaurin series is a special case of Taylor series that is centered at a=0 . The expansion of the function about 0 follows the formula:
f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n
or
f(x)= f(0)+(f'(0)x)/(1!)+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...
To determine the Maclaurin polynomial of degree n=5 for the given function f(x)=e^(-x) , we may apply the formula for Maclaurin series..
To list f^n(x) , we may apply derivative formula for exponential function: d/(dx) e^u = e^u * (du)/(dx) .
Let u =-x then (du)/(dx)= -1
Applying the values on the derivative formula for exponential function, we get:
d/(dx) e^(-x) = e^(-x) *(-1)
= -e^(-x)
Applying d/(dx) e^(-x)= -e^(-x) for each f^n(x) , we get:
f'(x) = d/(dx) e^(-x)
=-e^(-x)
f^2(x) = d/(dx) (- e^(-x))
=-1 *d/(dx) e^(-x)
=-1 *(-e^(-x))
=e^(-x)
f^3(x) = d/(dx) e^(-x)
=-e^(-x)
f^4(x) = d/(dx) (- e^(-x))
=-1 *d/(dx) e^(-x)
=-1 *(-e^(-x))
=e^(-x)
f^5(x) = d/(dx) e^(-x)
=-e^(-x)
Plug-in x=0 , we get:
f(0) =e^(-0) =1
f'(0) =-e^(-0)=-1
f^2(0) =e^(-0)=1
f^3(0) =-e^(-0)=-1
f^4(0) =e^(-0)=1
f^5(0) =-e^(-0)=-1
Note: e^(-0)=e^0 =1 .
Plug-in the values on the formula for Maclaurin series, we get:
f(x)=sum_(n=0)^5 (f^n(0))/(n!) x^n
= 1+(-1)/(1!)x+1/(2!)x^2+(-1)/(3!)x^3+1/(4!)x^4+(-1)/(5!)x^5
= 1-1/1x+1/2x^2-1/6x^3+1/24x^4-1/120x^5
= 1-x+x^2/2-x^3/6+x^4/24 -x^5/120
The Maclaurin polynomial of degree n=5 for the given function f(x)=e^(-x) will be:
P_5(x)=1-x+x^2/2-x^3/6+x^4/24 -x^5/120