Maclaurin’s series

When x =0, f ′′′(0)=(3)(2)a3=3!a3, i.e. a3 = f ′′′(0) 3!

Continuing the same procedure gives a4 = f iv(0) 4!

,

a5= f v(0) 5!

, and so on.

Substituting for a0,a1,a2, . . . in equation (1) gives:

f (x) = f (0)+ f ′(0)x + f ′′(0) 2!