maclaurin and taylor polynomials objective: improve on the local linear approximation for higher...
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Maclaurin and Taylor Polynomials
Objective: Improve on the local linear approximation for higher
order polynomials.
Local Quadratic Approximations• Remember we defined the local linear approximation
of a function f at x0 as . In this formula, the approximating function
is a first-degree polynomial. The local linear approximation of f at x0 has the property that its value
and the value of its first derivative match those of f at x0.
))(()()( 00/
0 xxxfxfxf
))(()()( 00/
0 xxxfxfxp
Local Quadratic Approximations• If the graph of a function f has a pronounced “bend”
at x0 , then we can expect that the accuracy of the local linear approximation of f at x0 will decrease rapidly as we progress away from x.
Local Quadratic Approximations• One way to deal with this problem is to approximate
the function f at x0 by a polynomial p of degree 2 with the property that the value of p and the values of its first two derivatives match those of f at x0. This ensures that the graphs of f and p not only have the same tangent line at x0 , but they also bend in the same direction at x0. As a result, we can expect that the graph of p will remain close to the graph of f over a larger interval around x0. The polynomial p is called the local quadratic approximation of f at x = x0.
Local Quadratic Approximations• We will try to find a formula for the local quadratic
approximation for a function f at x = 0. This approximation has the form where c0 , c1 , and c2 must be chosen so that the values of
and its first two derivatives match those of f at 0. Thus, we want
2210)( xcxccxp
2210)( xcxccxf
)0()0(),0()0(),0()0( ////// fpfpfp
Local Quadratic Approximations• The values of p(0), p /(0), and p //(0) are as follows:
2210)( xcxccxp 0)0( cp
xccxp 21/ 2)(
2// 2)( cxp
1/ )0( cp
2// 2)0( cp
Local Quadratic Approximations• Thus it follows that
• The local quadratic approximation becomes
)0(0 fc )0(/1 fc 2)0(//
2fc
2210)( xcxccxp 2
210)( xcxccxf
2//
/
2)0()0()0()( xfxffxf
Example 1• Find the local linear and local quadratic
approximations of ex at x = 0, and graph ex and the two approximations together.
Example 1• Find the local linear and local quadratic
approximations of ex at x = 0, and graph ex and the two approximations together.
• f(x)= ex, so
• The local linear approximation is
• The local quadratic approximation is
1)0( 0/ ef1)0( 0 ef 1)0( 0// ef
xex 1
21
2xxe x
Example 1• Find the local linear and local quadratic
approximations of ex at x = 0, and graph ex and the two approximations together.
• Here is the graph of the three functions.
Maclaurin Polynomials• It is natural to ask whether one can improve the
accuracy of a local quadratic approximation by using a polynomial of degree 3. Specifically, one might look for a polynomial of degree 3 with the property that its value and the values of its first three derivatives match those of f at a point; and if this provides an improvement in accuracy, why not go on to polynomials of higher degree?
Maclaurin Polynomials• We are led to consider the following problem:
Maclaurin Polynomials• We will begin by solving this problem in the case
where x0 = 0. Thus, we want a polynomial
• such that)0()0(),...,0()0(),0()0(),0()0( ////// nn pfpfpfpf
nnxcxcxcxccxp ...)( 3
32
210
Maclaurin Polynomials• But we know that
nn
nn
nn
nn
nn
cnnnxp
xcnnncxp
xcnnxccxp
xncxcxccxp
xcxcxcxccxp
)...2)(1()(
.
.)2)(1(...23)(
)1(...232)(
...32)(
...)(
33
///
232
//
12321
/
33
2210
Maclaurin Polynomials• But we know that
• Thus we need
33//////
2////
1//
0
!323)0()0(
2)0()0(
)0()0(
)0()0(
ccpf
cpf
cpf
cpf
nn
nn
nn
nn
nn
cnnnxp
xcnnncxp
xcnnxccxp
xncxcxccxp
xcxcxcxccxp
)...2)(1()(
.
.)2)(1(...23)(
)1(...232)(
...32)(
...)(
33
///
232
//
12321
/
33
2210
Maclaurin Polynomials• This yields the following values for the coefficients of
p(x)
!)0(
.
.!3)0(
!2)0(
)0(
)0(
///
3
/
2
/1
0
nfc
fc
fc
fc
fc
n
n
Maclaurin Polynomials• This leads us to the following definition.
Example 2
• Find the Maclaurin Polynomials p0 , p1, p2 , p3, and pn for ex.
Example 2
• Find the Maclaurin Polynomials p0 , p1, p2 , p3, and pn for ex.
• We know that
• so
1)0()0()0()0()0( ////// nfffff
!;!
...2
1)(
621
!3)0(
2)0()0()0()(
21
!2)0()0()0()(
1)0()0()(
1)0()(
2
32///2///
3
22///
2
/1
0
memorizenxxxxp
xxxfxfxffxp
xxxfxffxp
xxffxp
fxp
n
n
Example 2
• Find the Maclaurin Polynomials p0 , p1, p2 , p3, and pn for ex.
• The graphs of ex and all four approximations are shown.
Taylor Polynomials• Up to now we have focused on approximating a
function f in the vicinity of x = 0. Now we will consider the more general case of approximating f in the vicinity of an arbitrary domain value x0.
Taylor Polynomials• Up to now we have focused on approximating a
function f in the vicinity of x = 0. Now we will consider the more general case of approximating f in the vicinity of an arbitrary domain value x0.
• The basic idea is the same as before; we want to find an nth-degree polynomial p with the property that its value and the values of its first n derivatives match those of f at x0. However, rather than expressing p(x) in powers of x, it will simplify the computation if we express it in powers of x – x0; that is
nn xxcxxcxxccxp )(...)()()( 0
202010
Taylor Polynomial• This leads to the following definition:
Example 3• Find the first four Taylor Polynomials for lnx about
x = 2.
Example 3• Find the first four Taylor Polynomials for lnx about
x = 2.• Let f(x) = lnx. Thus
3///
2//
/
/2)(
/1)(
/1)(
ln)(
xxf
xxf
xxf
xxf
4/1)2(
4/1)2(
2/1)2(
2ln)2(
///
//
/
f
f
f
f
Example 3• Find the first four Taylor Polynomials for lnx about
x = 2.• This leads us to:
32412
81
21
3
281
21
2///
2
21/
1
0
)2()2()2(2ln)(
)2()2(2ln2)2()2()2)(2()2()(
)2(2ln)2)(2()2()(
2ln)2()(
xxxxp
xxxfxffxp
xxffxp
fxp 4/1)2(
4/1)2(
2/1)2(
2ln)2(
///
//
/
f
f
f
f
Example 3• Find the first four Taylor Polynomials for lnx about
x = 2.• The graphs of all five polynomials are shown.
Sigma Notation• Frequently we will want to express a Taylor
Polynomial in sigma notation. To do this, we use the notation f k(x0) to denote the kth derivative of f at x = xo , and we make the connection that f 0(x0) denotes f(x0). This enables us to write
nn
kn
k
k
xxnxfxxxfxfxx
kxf )(
!)(...))(()()(
!)(
00
00/
000
0
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)xcosxsinx11
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(a)
xcosxsinx11
xxf
xxf
xxf
xxf
cos)(
sin)(
cos)(
sin)(
///
//
/
1)0(
0)0(
1)0(
0)0(
///
//
/
f
f
f
f
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(a) This leads us to:
xcosxsinx11
!50
!300)(
0!3
00)(
!300)(
00)(0)(0)(
53
5
3
4
3
3
2
1
0
xxxxp
xxxp
xxxp
xxpxxp
xp
1)0(
0)0(
1)0(
0)0(
///
//
/
f
f
f
f
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(a) This leads us to:
xcosxsinx11
)!12()1(...
!7!5!3)(
12753
kxxxxxxpk
k
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(a) The graphs are shown.
xcosxsinx11
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(b) We start with:
xcosxsinx11
xxf
xxf
xxf
xxf
sin)(
cos)(
sin)(
cos)(
///
//
/
0)0(
1)0(
0)0(
1)0(
///
//
/
f
f
f
f
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(b) This leads us to:
xcosxsinx11
!6!4!21)(
!4!21)(
!21)(
1)(
642
6
42
4
2
2
0
xxxxp
xxxp
xxp
xp
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(b) This leads us to:
xcosxsinx11
)!2()1(...
!6!4!21)(
2642
kxxxxxpk
k
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(b) The graphs are shown.
xcosxsinx11
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
You need to memorize these!
xcosxsinx11
)!12()1(...
!7!5!3sin
12753
kxxxxxxk
k
)!2()1(...
!6!4!21cos
2642
kxxxxxk
k
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
You need to memorize these!
xcosxsinx11
)!12()1(...
!7!5!3sin
12753
kxxxxxxk
k
)!12()2()1(...
!7)2(
!5)2(
!3)2(22sin
12753
kxxxxxx
kk
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
You need to memorize these!
xcosxsinx11
)!12()1(...
!7!5!3sin
12753
kxxxxxxk
k
)!12()()1(...
!7)(
!5)(
!3)(sin
12272523222
kxxxxxx
kk
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(c) We start with:
xcosxsinx11
24)(;6)(
;2)0(;1)0(;1)0(///////
///
xfxf
fff
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(c) We start with:
xcosxsinx11
15////
4///
3//
2/
)1(!)(;
)1(234)(;
)1(23)(
;)1(
2)(;)1(
1)(;11)(
kk
xkxf
xxf
xxf
xxf
xxf
xxf
24)0(;6)0(
;2)0(;1)0(;1)0(///////
///
ff
fff
Example 4• Find the nth Maclaurin polynomials for
(a) (b) (c)
(c) This leads us to
• You should memorize this as well.
xcosxsinx11
nxxxx
...111 2
The nth Remainder• It will be convenient to have a notation for the error
in the approximation . Accordingly, we will let denote the difference between f(x) and its nth Taylor polynomial: that is
• This can also be written as:
)(xRn
)()( xpxf n
The nth Remainder• The function is called the nth remainder for the
Taylor series of f, and the formula below is called Taylor’s formula with remainder.
)(xRn
The nth Remainder• Finding a bound for gives an indication of the
accuracy of the approximation . The following theorem provides such a bound.
• The bound, M, is called the Lagrange error bound.
)(xRn)()( xfxpn
Example 6• Use an nth Maclaurin polynomial for to
approximate e to five decimal places.
xe
Example 6• Use an nth Maclaurin polynomial for to
approximate e to five decimal places. • The nth Maclaurin polynomial for is
• from which we have
xe
!...
!21
!
2
0 kxxx
kx kn
k
k
xe
!1...
!2111
!10
1
nkee
n
k
k
Example 6• Use an nth Maclaurin polynomial for to
approximate e to five decimal places.• Our problem is to determine how many terms to
include in a Maclaurin polynomial for to achieve five decimal-place accuracy; that is, we want to choose n so that the absolute value of the nth remainder at x = 1 satisfies
xe
000005.|)1(| nR
xe
Example 6• Use an nth Maclaurin polynomial for to
approximate e to five decimal places.• To determine n we use the Remainder Estimation
Theorem with , and I being the interval [0, 1]. In this case it follows that
• where M is an upper bound on the value of for x in the interval [0, 1].
0,1,)( 0 xxexf x
)!1(|)1(|
nMRn
xe
xn exf )(1
Example 6• Use an nth Maclaurin polynomial for to
approximate e to five decimal places.• However, is an increasing function, so its maximum
value on the interval [0, 1] occurs at x = 1; that is, on this interval. Thus, we can take M = e to
obtaineex
)!1(|)1(|
neRn
xe
xe
Example 6• Use an nth Maclaurin polynomial for to
approximate e to five decimal places.• Unfortunately, this inequality is not very useful
because it involves e, which is the very quantity we are trying to approximate. However, if we accept that e < 3, then we can use this value. Although less precise, it is more easily applied.
)!1(3|)1(|
n
Rn
xe
Example 6• Use an nth Maclaurin polynomial for to
approximate e to five decimal places.• Thus, we can achieve five decimal-place accuracy by
choosing n so that or
• This happens when n = 9.
000005.)!1(
3
n
xe
000,600)!1( n
Homework
• Page 684• 1, 3, 7, 9, 11, 15-21 odd• 31, 32
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