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In this section we develop general methods for finding power series representations. Suppose that f (x) is represented by a power series centered at x = c on an interval (c − R, c + R) with R > 0:
According to
THEOREM 2 Term-by-Term Differentiation and Integration Assume that
has radius of convergence R > 0. Then F (x) is differentiable on (c − R, c + R) [or for all x if R =].
Furthermore, we can integrate and differentiate term by term. For x (c − R, c + R),
0
n
nn
F x a x c
1
0
( is any constant)1
nn
n
aF x dx A x c A
n
20 1 20
n
nn
F x a x c a a x c a x c
2 3
1 2 3 4
2 3
0 1
2 3
2
2 3 4 5
3
3
' 2 3
4
'' 2 2 3 3 4 4 5
''' 2 3
f x a a x c a x c a x c
f x a a x c a x c a x c
f x a a x c a x c a x c
f x a
24 5 2 3 4 3 4 5 a x c a x c
1! 2 3 1kk kf x k a k a x c
we can compute the derivatives of f (x) by differentiating the series expansion term by term:
In general,
Setting x = c in each of these series, we find that
We see that ak is the kth coefficient of the Taylor polynomial studied in Section 9.4
0 1 2 3, ' , '' 2 , ''' 2 3 , , !kkf c a f c a f c a f c a f c k a
!
k
k
f ca
k
Power Series
This proves the next theorem.
THEOREM 1 Taylor Series Expansion If f (x) is represented by a power series centered at c in an interval |x − c| < R with R > 0, then that power series is the Taylor series
In the special case c = 0, T (x) is also called the Maclaurin series:
0 !
nn
n
f cT x x c
n
4
2 3 4
0
0 " 0 ''' 0 00 ' 0
! 2! 3! 4!
nn
n
f f f ff x x f f x x x x
n
We define the nth Taylor polynomial centered at x = a as follows:
2' "
1! 2! !
nn
n
f a f a f aT x f a x a x a x a
n
Therefore f (x) = T (x), where T (x) is the Taylor series of f (x) centered at x = c:
2 3'' ''''
2! 3!
f c f cT x f c f c x c x c x c
4 5
3
' 3 , " 3 4 ,
and in general, 1 3 4 2nn n
f x x f x x
f x n x
Find the Taylor series for f (x) = x−3 centered at c = 1.
1Note that 3 4 2 2 !.
2n n
1 3 1 4
2 3 2 5
3 3 3 6
i.e. ' 1 3 3
" 1 3 2 2 12
''' 1 3 4 3 2 60
f x x x
f x x x
f x x x
1i.e. 3 4 3 2 3 2 !
2
11 1 2 !
2nnf n
0 !
nn
n
f cT x x c
n
2 ! 2 1 !n n n n
we can write the coefficients of the Taylor series as:
1 11
2
2 ! 1 2 1 !1 2 2! !
11
!
2
n n
n
n
n
n n n nfa
n n nn n
The Taylor series for f (x) = x −3 centered at c = 1 is
0 !
nn
n
f cT x x c
n
2 3
0
1 3 1 6 1 10 1
2 11 1
2n n
n
T x x x x
n nx
11 1 2 !
2nnf n
However, there is no guarantee that T(x) converges to f(x), even if T(x) converges. To study convergence, we consider the kth partial sum, which is the Taylor polynomial of degree k:
Theorem 1 tells us that if we want to represent a function f (x) by a power series centered at c, then the only candidate for the job is the Taylor series:
0 !
nn
n
f cT x x c
n
2"
'2! !
kk
k
f c f cT x f c f c x c x c x c
k
In Section 8.4, we defined the remainder as .k kR x f x T x
Since T(x) is the limit of the partial sums Tk(x), we see that
the Taylor series converges to f(x) .lim 0kk
R x
THEOREM 2 Let I = (c − R, c + R), where R > 0. Suppose there exists K > 0 such that all derivatives of f are bounded by K on I:
, 0 with kf x K k x I
Then f (x) is represented by its Taylor series in I:
0
, !
nn
n
f cf x x c x I
n
Taylor expansions were studied throughout the seventeenth and eighteenth centuries by Gregory, Leibniz, Newton, Maclaurin, Taylor, Euler, and others. These developments were anticipated by the great Hindu mathematician Madhava (c. 1340–1425), who discovered the expansions of sine and cosine and many other results two centuries earlier.
There is no general method for determining whether Rk(x) tends to zero, but the following theorem can be applied in some important cases.
Taylor Expansion
Expansions of Sine and Cosine Show that the following Maclaurin expansions are valid for all x.
2 1 3 5 7
0
sin 12 1 ! 3! 5! 7!
nn
n
x x x xx x
n
2 2 4 6
0
cos 1 12 ! 2! 4! 6!
nn
n
x x x xx
n
Recall that the derivatives of f (x) = sinx and their values at x = 0 form a repeating pattern of period 4:
For f (x) = cosx, the situation is reversed. The odd derivatives are zero and the even derivatives alternate in sign: f (2n)(0) = (−1)ncos0 = (−1)n. Therefore the nonzero Taylor coefficients for cosx are a2n = (−1)n/(2n)!.
In other words, the even derivatives are zero and the odd derivatives alternate in sign: f (2n+1)(0) = (−1)n. Therefore, the nonzero Taylor coefficients for sinx are
2 1
1
2 1 !
n
nan
We can apply Theorem 2 with K = 1 and any value of R because both sine and cosine satisfy |f (n)(x)| ≤ 1 for all x and n. The conclusion is that the Taylor series converges to f (x) for |x| < R. Since R is arbitrary, the Taylor expansions hold for all x.
0
Taylor Coefficients!
n
n
f c
n
THM 2
We have f (n)(c) = ec for all x, and thus
Taylor Expansion of f(x) = ex at x = c Find the Taylor series T(x) of f (x) = ex at x = c.
Because ex is increasing for all R > 0 we have |f (k)(x)| ≤ ec+R for x (c − R, c + R). Applying Theorem 2 with K = ec+R, we conclude that T(x) converges to f(x) for all x (c − R, c + R). Since R is arbitrary, the Taylor expansion holds for all x. For c = 0, we obtain the standard Maclaurin series
0 !
cn
n
eT x x c
n
2 3
12! 3!
x x xe x
0 !
nn
n
f cT x x c
n
Shortcuts to Finding Taylor SeriesThere are several methods for generating new Taylor series from known ones. First of all, we can differentiate and integrate Taylor series term by term within its interval of convergence, by Theorem 2 of Section 11.6. We can also multiply two Taylor series or substitute one Taylor series into another (we omit the proofs of these facts).
Find the Maclaurin series for f (x) = x2ex.
2 3 4 5
0
1
Standard Maclaurin Series for :
2! 3! 4! 5! !
nx
n
x
x x x x xe
n
e
x
4 5 6 7
2 3 4 52
2 3
2
2 12! 3! 4! 5!
2! 3! 4! 5! 2 !
x
n
n
x x x
x x x xx e x x
x xx x
n
2
2 3 42 2 2 222
0 0
4 6 82
11
! ! 2! 3! 4!
12! 3! 4!
n n nx
n n
x x x xxe x
n n
x x xx
Substitution Find the Maclaurin series for 2
.xe
Substitute −x2 for x in the Maclaurin series for ex.
2 3 4 5
0
1
Standard Maclaurin Series for :
2! 3! 4! 5! !
nx
n
x
x x x x xe
n
e
x
The Taylor expansion of ex is valid for all x, so this expansion is also valid for all x.
2 21n n nx x
2 3
0 1n
n
ccr c cr cr cr
r
Integration Integrate the geometric series with common ratio −x (valid for |x| < 1) and c = 1, to find the Maclaurin series for f (x) = ln(1 + x).
2 3
0
1 11
1 1n
n
x x x xx x
2 3 4
1
1
ln 12 3
141
nn
n
xdxx x
x
x x x
n
The constant of integration is zero because ln(1 + x) = 0 for x = 0. This expansion is valid for |x| < 1. It also holds for x = 1 (see Exercise 84).
2 3 4 5 2 3 2 4
1 1 12 6 24 120 2 6 2 24
x x x x x x x xx x x
In many cases, there is no convenient general formula for the Taylor coefficients, but we can still compute as many coefficients as desired.Multiplying Taylor Series Write out the terms up to degree five in the Maclaurin series for f (x) = ex cosx. 2 3 4 5
12! 3! 4! 5!
x x x x xe x
2 4 6
cos 12! 4! 6!
x x xx
We multiply the fifth-order Taylor polynomials of ex and cos x together, dropping the terms of degree greater than 5:
2 3 4 5 2 4
1 12 6 24 120 2! 4!
x x x x x xx
Distributing the term on the right (and ignoring terms of degree greater than 5), we obtain:
2 3 4 5 2 3 4 5 4 5
12 6 24 120 2 2 4 12 24 24
x x x x x x x x x xx
3 4 5
13 6 30
x x xx
In the next example, we express the definite integral of sin(x2) as an infinite series. This is useful because the integral cannot be evaluated explicitly. Figure 1 shows the graph of the Taylor polynomial T12(x) of the Taylor series expansion of the antiderivative.
Graph of T12(x) for the power series expansion of the antiderivative
2
0
sin .x
F x t dt
1
2
0
Let sin .J x dx (a) Express J as an infinite series.(b) Determine J to within an error less than 10−4.
2
4 2
0
1
2
0
1s
1s
i
in2 1
n2
!
1 !
n
n
n
n
n
n
x
n
x
x
n
x
We obtain an infinite series for J by integration:
1 12 4 2
00 0
14 3
0 00
1sin
2 1 !
1 1 1
2 1 ! 4 3 2 1 ! 4 3
n
n
n
n nn
n n
J x dx x dxn
x
n n n n
0
1 1
2 1 ! 4 3
1 1 1 1
3 42 1320 75,600
n
n n n
I can only get the first
3 fractions on my calc.
The infinite series for J is an alternating series with decreasing terms, so the sum of the first N terms is accurate to within an error that is less than the (N + 1)st term. The absolute value of the fourth term (1/75,600) is smaller than 10−4 so we obtain the desired accuracy using the first three terms of the series for J: 1 1 1
0.310283 42 1320
J The error satisfies
51 1 1 11.3 10
3 42 1320 75,600J
The percentage error is less than 0.005% with just three terms.
(b) Determine J to within an error less than 10−4.
Binomial SeriesIsaac Newton discovered an important generalization of the Binomial Theorem around 1665. For any number a (integer or not) and integer n ≥ 0, we define the binomial coefficient:
( ) the number of distinct ways items can be chosen from items.an n a
0
1 2 1( ) , ( ) 1
!a an
a a a a n
n
63
6 5 4( ) 20
3 2 1
4/33
4 1 243 3 3( )
3 2 1 81
Let
The Binomial Theorem of algebra states that for any whole number a,
Setting r = 1 and s = x, we obtain the expansion of f (x):
1a
f x x
1 2 2 11 2 1( ) ( ) ( )
a a a a a a a a aar s r r s r s rs s
2 11 2 11 1 ( ) ( ) ( )
a a a a a aax x x x x
We derive Newton’s generalization by computing the Maclaurin series of f (x) without assuming that a is a whole number. Observe that the derivatives follow a pattern:
In general, f (n)(0) = a (a − 1)(a − 2)…(a − (n − 1)) and
Hence the Maclaurin series for f (x) = (1 + x)a is the binomial series
1
2
3
1
' 1
'' 1 1
''' 1 2 1
a
a
a
a
f x x
f x a x
f x a a x
f x a a a x
0 1
' 0
" 0 1
''' 0 1 2
f
f a
f a a
f a a a
1 2 10( )
! !
nan
a a a a nf
n n
2 3
0
1 1 2( ) 1 ( )
2! 3!a n a nn n
n
a a a a ax ax x x x
0 !
nn
n
f cT x x c
n
THEOREM 3 The Binomial Series For any exponent a and for |x| < 1,
Find the terms through degree four in the Maclaurin expansion of
f (x) = (1 + x)4/3
2 31 1 21 1 ( )
1! 2! 3!a a n
n
a a a a aax x x x x
4/3 2 3 4
4 2 4 5 3 The binomial coefficients are: 1, , ,
4 2 4 51 1
3
, 3 9 81 24
9
3
1 2
8 43x x x
T M
x x
H
0 !
nn
n
f cT x x c
n
4 / 3a
!
Note: ( )! !
nk
n
n k k
To use " " on the calculator,
must be a whole number.
nCr
n
0 1 1 2 2 3 3 1 1 0
0 1 2 3 1( ) ( ) ( ) ( ) ( ) ( )
n
n n n n n n n n n n n nn n
a b
a b a b a b a b a b a b
Find the Maclaurin series for
First, let’s find the coefficients in the binomial series for (1 + x)−1/2:
2 3
1 1 3 1 3 5
1 1 21 1 ( )
1! 2! 3!The binomial coefficients are:
1 3 1 3 512 2 2 2 22, 1
2 ,
2 4,
1! 1 2 1 2 2 4 63
a a nn
a a a a aax x x x x
The general pattern is
1
2
1 3 5 2 12 2 2 2 1 3
1 2 3
5 2 11
2 4 6 2n
n
n
n
n
n
2
1
1f x
x
0 1 1 2 2 3 3 1 1 0
0 1 2 3 1
( ) ( ) ( ) ( ) ( ) (
)
n
n n n n n n n n n n n nn n
a b
a b a b
but n is not a whole
a b a b a b a
number
b
1/ 2a ?Patterns
Thus, the following binomial expansion is valid for |x| < 1:
2
1
1 3 5 2 11 1 1 31 1 1
2 4 6 2 2 2 41
n n
n
nx x x
nx
If |x| < 1, then |x|2 < 1, and we can substitute -x2 for x to obtain
22 4
21
11 3 5 2 1 31
2 2 4
111
2 4 6 21
n
n
nx
nx
xx
Find the Maclaurin series for
2
1
1f x
x
