ABSTRACT
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!
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!