State and prove Taylors theorem,
Answers
Taylor’s Theorem. Let f be an (n + 1) times differentiable function on
an open interval containing the points a and x. Then
f(x) = f(a) + f
0
(a)(x − a) + f
00(a)
2! (x − a)
2 + . . . +
f
(n)
(a)
n!
(x − a)
n + Rn(x)
where
Rn(x) = f
(n+1)(c)
(n + 1)! (x − a)
n+1
for some number c between a and x.
The function Tn defined by
Tn(x) = a0 + a1(x − a) + a2(x − a)
2 + . . . + an(x − a)
n where ar =
f
(r)
(a)
r!
,
is called the Taylor polynomial of degree n of f at a. This can be thought
of as a polynomial which approximates the function f in some interval
containing a. The error in the approximation is given by the remainder
term Rn(x). If we can show Rn(x) → 0 as n → ∞ then we get a sequence
of better and better approximations to f leading to a power series expansion
f(x) = X
∞
n=0
f
(n)
(a)
n!
(x − a)
n
which is known as the Taylor series for f. In general this series will
converge only for certain values of x determined by the radius of
convergence of the power series (see Note 17). When the Taylor polynomials
converge rapidly enough, they can be used to compute approximate
values of the function.