maclaurin and taylor polynomials
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Maclaurin and Taylor Polynomials. Objective: Improve on the local linear approximation for higher order polynomials. Local Quadratic Approximations. - PowerPoint PPT PresentationTRANSCRIPT
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(//2
fc
2210)( xcxccxp 2
210)( xcxccxf
2//
/
2
)0()0()0()( x
fxffxf
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
2xxex
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
n
fc
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
memorizen
xxxxp
xxx
fxfxffxp
xx
xfxffxp
xxffxp
fxp
n
n
Example 2
• Why do I need to memorize this?
• They may ask you to find a Maclaurin polynomial to approximate xex. This is what they want you to do:
!;!
...2
1)(2
memorizen
xxxxp
n
n
!...
2
!...
21
132
2
n
xxxxxe
n
xxxe
nx
nx
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 4
• Find the first four Taylor Polynomials for lnx about x = 2.
Example 4
• 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 4
• 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
xxx
fxffxp
xxffxp
fxp 4/1)2(
4/1)2(
2/1)2(
2ln)2(
///
//
/
f
f
f
f
Example 4
• 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
xxn
xfxxxfxfxx
k
xf)(
!
)(...))(()()(
!
)(0
000
/00
0
0
Example 5
• Find the nth Maclaurin polynomials for
(a) (b) (c)xcosxsinx11
Example 5
• 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 5
• 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
xxp
xxp
xp
1)0(
0)0(
1)0(
0)0(
///
//
/
f
f
f
f
Example 5
• Find the nth Maclaurin polynomials for
(a) (b) (c)
(a) This leads us to:
xcosxsinx11
)!12()1(...
!7!5!3)(
12753
k
xxxxxxp
kk
Example 5
• Find the nth Maclaurin polynomials for
(a) (b) (c)
(a) The graphs are shown.
xcosxsinx11
Example 5
• 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 5
• 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 5
• Find the nth Maclaurin polynomials for
(a) (b) (c)
(b) This leads us to:
xcosxsinx11
)!2()1(...
!6!4!21)(
2642
k
xxxxxp
kk
Example 5
• Find the nth Maclaurin polynomials for
(a) (b) (c)
(b) The graphs are shown.
xcosxsinx11
Example 5
• Find the nth Maclaurin polynomials for
(a) (b) (c)
You need to memorize these!
xcosxsinx11
)!12()1(...
!7!5!3sin
12753
k
xxxxxx
kk
)!2()1(...
!6!4!21cos
2642
k
xxxxx
kk
Example 5
• Find the nth Maclaurin polynomials for
(a) (b) (c)
You need to memorize these!
xcosxsinx11
)!12()1(...
!7!5!3sin
12753
k
xxxxxx
kk
)!12(
)2()1(...
!7
)2(
!5
)2(
!3
)2(22sin
12753
k
xxxxxx
kk
Example 5
• Find the nth Maclaurin polynomials for
(a) (b) (c)
You need to memorize these!
xcosxsinx11
)!12()1(...
!7!5!3sin
12753
k
xxxxxx
kk
)!12(
)()1(...
!7
)(
!5
)(
!3
)(sin
12272523222
k
xxxxxx
kk
Example 5
• 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 5
• 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)(;
1
1)(
kk
x
kxf
xxf
xxf
xxf
xxf
xxf
24)0(;6)0(
;2)0(;1)0(;1)0(///////
///
ff
fff
Example 5
• Find the nth Maclaurin polynomials for
(a) (b) (c)
(c) This leads us to
• You should memorize this as well.
xcosxsinx11
nxxxx
...11
1 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 7
• Use an nth Maclaurin polynomial for to approximate e to five decimal places.
xe
Example 7
• 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 k
xxx
k
x kn
k
k
xe
!
1...
!2
111
!
1
0
1
nkee
n
k
k
Example 7
• 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 7
• 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(|
n
MRn
xe
xn exf )(1
Example 7
• 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 obtain
eex
)!1(|)1(|
n
eRn
xe
xe
Example 7
• 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 7
• 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
Example
• Use an nth Maclaurin polynomial for to approximate to three decimal places.
xsin6sin
Example
• Use an nth Maclaurin polynomial for to approximate to three decimal places.
• To determine n we use the Remainder Estimation Theorem with , and I being the interval . In this case it follows that
• where M is an upper bound on the value of for x in the interval [0, 1]. What is the upper bound on the derivative?
0,,sin)( 06 xxxxf
)!1(
|0||)(|
16
6
n
MR
n
n
xsin
xxxf n cos/sin)(1
6sin
],0[ 6
Example
• Use an nth Maclaurin polynomial for to approximate to three decimal places.
0005.
)!1(
1
6
x
x
xsin6sin
Example
• Use an nth Maclaurin polynomial for to approximate to three decimal places.
0005.
)!1(
1
6
x
x
xsin6sin
41
6 105)!1(
x
x
Example
• Use an nth Maclaurin polynomial for to approximate to three decimal places.
!36
sin3
66
0005.
)!1(
1
6
x
x
xsin6sin
Example
• Use an nth Maclaurin polynomial for to approximate to four decimal places.
5
1
12
105
00005.)!1(
x
x
xcos
12cos
Example
• Use an nth Maclaurin polynomial for to approximate to four decimal places.
!4!2
1cos4
122
1212
xcos
12cos
5
1
12
105
00005.)!1(
x
x
Homework
• Section 9.7• Page 658• 1, 3, 7, 9, 11, 15-21 odd• 31, 32