Taylor's Theorem

Taylor's Theorem
Suppose we're working with a function f(x) that is continuous and has n + 1 continuous
derivatives on an interval about x = 0. We can approximate f near 0 by a polynomial Pn(x)
of degree n:

For n = 0, the best constant approximation near 0 is
P0(x) = f(0)
which matches f at 0.

For n = 1, the best linear approximation near 0 is
P1(x) = f(0) + f0(0)x:
Note that P1 matches f at 0 and P01 matches f0 at 0.

For n = 2, the best quadratic approximation near 0 is
P2(x) = f(0) + f0(0)x +
f00(0)
2!
x2:

Note that P2, P02 , and P00 2 match f, f0, and f00, respectively, at 0.
Continuing this process,
Pn(x) = f(0) + f0(0)x +
f00(x)
2!
x2 + : : : +
f(n)(0)
n!
xn:
This is the Taylor polynomial of degree n about 0 (also called the Maclaurin series
of degree n). More generally, if f has n + 1 continuous derivatives at x = a, the Taylor
series of degree n about a is
Xn
k=0
f(k)(a)
k!
(x - a)k = f(a) + f0(a)(x - a) +
f00(a)
2!
(x  a)2 + : : : +
f(n)(a)
n!
(x - a)n:
This formula approximates f(x) near a. Taylor's Theorem gives bounds for the error in this
approximation:

Taylor's Theorem
Suppose f has n +1 continuous derivatives on an open interval containing a. Then for each
x in the interval,
f(x) =
"
Xn
k=0
f(k)(a)
k!
(x - a)k
#
+ Rn+1(x)
where the error term Rn+1(x) satis es Rn+1(x) =
f(n+1)(c)
(n + 1)!
(x - a)n+1 for some c between a
and x.
This form for the error Rn+1(x), derived in 1797 by Joseph Lagrange, is called the Lagrange
formula for the remainder. The in nite Taylor series converges to f,
f(x) =
1X
k=0
f(k)(a)
k!
(x - a)k;
if and only if lim
n!1
Rn(x) = 0.
Examples of Taylor Series about 0
1. For f(x) = ex,
f(k)(x) = ex =) f(k)(0) = 1:
So
ex = 1 + x +
x2
2!
+
x3
3!
+ : : :
=
1X
k=0
xk
k!
which converges for all x since lim
n!1
Rn(x) = lim
n!1
ecx(n+1)
(n + 1)!
= 0 for all c between 0 and
x.
2. For f(x) = ln(1 + x),
f(x) = ln(1 + x)
f0(x) = 1
1+x
f00(x) = -1
(1+x)2
f000(x) = 2
(1+x)3
f(4)(x) = -3-2
(1+x)4
... 9

===========
f(0) = 0
f0(0) = 1
f00(0) = -1
f000(0) = 2
f(4)(x) = -6

So
ln(1 + x) = x - x2
2
+
x3
3 -
x4
4
+ : : :
=
1X
k=0
(-1)k xk+1
k + 1
which converges only for -1 < x - 1.
The Taylor Series in (x - a) is the unique power series in (x - a) converging to f(x) on an
interval containing a. For this reason,- By Example 1,
e-2x = 1 - 2x + 2x2 -
4
3
x3 + : : :
where we have substituted -2x for x.
- By Example 2, since
d
dx
[ln(1 + x)] =
1
1 + x
, we can di erentiate the Taylor series for
ln(1 + x) to obtain
1
1 + x
= 1 - x + x2 - x3 + : : : :
Substituting -x for x,
1
1 - x
= 1 + x + x2 + x3 + : : : :
In the Exploration, compare the graphs of various functions with their rst through fourth
degree Taylor polynomials.
Exploration
Key Concepts
Taylor's Theorem
Suppose f has n +1 continuous derivatives on an open interval containing a. Then for each
x in the interval,