Calculus of a Single Variable, Chapter 9, 9.7, Section 9.7, Problem 13

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 4th Maclaurin polynomial from the given function f(x)=e^(4x) ,
we may apply derivative formula for exponential function: d/(dx) e^u = e^u * (du)/(dx)
Let u =4x then (du)/(dx)= 4
Applying the values on the derivative formula for exponential function, we get:
d/(dx) e^(4x) = e^(4x) *4
Applying d/(dx) e^(4x)= 4e^(4x) for each f^n(x) , we get:
f'(x) = d/(dx) e^(4x)
=e^(4x) * 4
= 4e^(4x)
f^2(x) = 4 *d/(dx) e^(4x)
= 4*4e^(4x)
=16e^(4x)
f^3(x) = 16*d/(dx) e^(4x)
= 16*4e^(4x)
=64e^(4x)
f^4(x) = 64*d/(dx) e^(4x)
= 64*4e^(4x)
=256e^(4x)
Plug-in x=0 , we get:
f(0) =e^(4*0) =1
f'(0) =4e^(4*0)=4
f^2(0) =16e^(4*0)=16
f^3(0) =64e^(4*0)=64
f^4(0) =2564e^(4*0)=256
Note: e^(4*0)=e^0 =1 .
Plug-in the values on the formula for Maclaurin series.
f(x)=sum_(n=0)^4 (f^n(0))/(n!) x^n
= 1+4/(1!)x+16x^2+64x^3+256/(4!)x^4
=1+ 4/1x +16/(1*2)x^2 + 64/(1*2*3)x^3 +256/(1*2*3*4)x^4
=1+ 4/1x +16/2x^2 + 64/6x^3 +256/24x^4
= 1+4x+ 8x^2 + 32/3x^3 + 32/3x^4
The 4th Maclaurin polynomial for the given function f(x)= e^(4x) will be:
e^(4x) =1+4x+ 8x^2 + 32/3x^3 + 32/3x^4
or P_4(x) =1+4x+ 8x^2 + 32/3x^3 + 32/3x^4