9.9
Convergence of Taylor Series
54S
where for some c between 0 and 1,
I
Rn( D = ec
{ n + 1)!
<
( n + 1)!
e< < e* < 3
It is often possible to estimate Rn ( x ) as we did in Example 1. This method of estimation is so
convenient that we state it as a theorem for future reference.
hi
If there is a positive
t.
constant M such that
1 ^ M for all £ between x and <2, inclusive, then
the remainder term Rn ( x ) in Taylor's Theorem satisfies the inequality
'F!
i!,. 1 , '-.JiI'II.ip I I ■ ^ L: ..............................................................
- / V '
|x - a \ n + l
\ R n(x)\ < M
( n + 1)!
If this inequality holds for every n and the other conditions of Taylor's Theorem are
satisfied by /, then the series converges to /(x).
The next two examples use Theorem 24 to show that the Taylor series generated by the sine and
cosine functions do in fact converge to the functions themselves.
EXAMPLE 2
Solutk
Show that the Taylor series for sin x at x = 0 converges for all x.
The function and its derivatives are
sin x,
f(x) =
- sin x,
f"(x) =
cos x,
f'(x) =
/"'(*) =
f 2 k \ x ) = (-l)/csinx,
- cos x,
f 2 k + l \ x ) = (-l)/ccosx,
so
/(2/c)(0) - 0
f{2k+l)(0) =
and
The series has only odd-powered terms and, for n =
x3
x5
k
2k+1
(-l) x
2k
+ 1, Taylor's Theorem gives
sinx = * - — + — - . . . + ^j^TYjT +
^/(+iW.
All the derivatives of sin x have absolute values less than or equal to 1, so we can apply the
Remainder Estimation Theorem with M = 1 to obtain
\x\2k+2
+l(*)| ^ 1
|*2Jfc
( 2 k + 2)!
2k+2
From Theorem 5, Rule 6, we have (|x\
/(2k + 2 ) ! ) ^ 0 a s / c ^ oo? whatever the value of x, so
R2k+ \ ( x ) ~* 0 and the Maclaurin series for sin x converges to sin x for every x. Thus,
&mX=
+l2kTW
/c=0
=
X
~ V
+
5! " 7! + • • • •
(4)