Calculus of a Single Variable, Chapter 9, 9.7, Section 9.7, Problem 15

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=4 for the given function f(x)=e^(-x/2) , 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/2 then (du)/(dx)= -1/2
Applying the values on the derivative formula for exponential function, we get:
d/(dx) e^(-x/2) = e^(-x/2) *(-1/2)
= -e^(-x/2)/2 or -1/2e^(-x/2)
Applying d/(dx) e^(-x/2)= -e^(-x/2)/2 for each f^n(x) , we get:
f'(x) = d/(dx) e^(-x/2)
=-1/2e^(-x/2)
f^2(x) = d/(dx) (-1/2e^(-x/2))
=-1/2 *d/(dx) e^(-x/2)
=-1/2 *(-1/2e^(-x/2))
=1/4e^(-x/2)
f^3(x) = d/(dx) (1/4e^(-x/2))
=1/4 *d/(dx) e^(-x/2)
=1/4 *(-1/2e^(-x/2))
=-1/8e^(-x/2)
f^4(x) = d/(dx) (-1/8e^(-x/2))
=-1/8 *d/(dx) e^(-x/2)
=-1/8 *(-1/2e^(-x/2))
=1/16e^(-x/2)
Plug-in x=0 on each f^n(x) , we get:
f(0)=e^(-0/2) = 1
f'(0)=-1/2e^(-0/2) = -1/2
f^2(0)=1/4e^(-0/2)=1/4
f^3(0)=-1/8e^(-0/2)=-1/8
f^4(0)=1/16e^(-0/2)=1/16
Note: e ^(-0/2) = e^0 =1 .
Plug-in the values on the formula for Maclaurin series, we get:
f(x)=sum_(n=0)^4 (f^n(0))/(n!) x^n
= 1+(-1/2)/(1!)x+(1/4)/(2!)x^2+(-1/8)/(3!)x^3+(1/16)/(4!)x^4
=1-1/2x+1/8x^2-1/48x^3+1/384x^4
The Maclaurin polynomial of degree n=4 for the given function f(x)=e^(-x/2) will be:
P_4(x)=1-1/2x+1/8x^2-1/48x^3+1/384x^4