in section 4.1, we used the linearization l(x) to approximate a function f (x) near a point x =...
TRANSCRIPT
In Section 4.1, we used the linearization L(x) to approximate a function f (x) near a point x = a:
We refer to L(x) as a “first-order” approximation to f (x) at x = a because f (x) and L(x) have the same value and the same first derivative at x = a:
, ' 'L a f a L a f a
A first-order approximation is useful only in a small interval around x = a. In this section we learn how to achieve greater accuracy over larger intervals using the higher-order approximations.
1 1y y m x x
In what follows, assume that f (x) is defined on an open interval I and that all higher derivatives f (k)(x) exist on I. Let a .IWe say that two functions f (x) and g (x) agree to order n at x = a if their derivatives up to order n at x = a are equal:
, ' ' , " " , '" '" ,
, n n
f a g a f a g a f a g a f a g a
f a g a
g(x) “approximates f (x) to order n” at x = a.
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
Before proceeding to the examples, we write Tn(x) in summation notation:
0 !
jnj
nj
f aT x x a
j
0
1
2
2
2 3
3
'
1' "
21 1
' " "'2 6
T x f a
T x f a f a x a
T x f a f a x a f a x a
T x f a f a x a f a x a f a x a
The first few Taylor polynomials are:
2' "
1! 2! !
nn
n
f a f a f aT x f a x a x a x a
n
Note that T1(x) is the linearization of f (x) at a. Note also that Tn(x) is obtained from Tn−1(x) by adding on a term of degree n:
1 !
nn
n n
f aT x T x x a
n
'L x f x x a f a
0 !
jnj
nj
f aT x x a
j
THEOREM 1 The polynomial Tn(x) centered at a agrees with f (x) to order n at x = a, and it is the only polynomial of degree at most n with this property.
2
2 2
2 2
2 2
1' " ,
2' ' " , ' '
" " , " "
T x f a f a x a f a x a T a f a
T x f a f a x a T a f a
T x f a T a f a
This shows that the value and the derivatives of order up to n = 2 at x = a are equal.
By convention, we regard f (x) as the zeroeth derivative, and thus f (0)(x) is f (x) itself. When a = 0, Tn(x) is also called the nth Maclaurin polynomial.
The 4th Taylor Polynomial
of 1f x x
Maclaurin Polynomials for ex Plot the third and fourth Maclaurin polynomials for f (x) = ex. Compare with the linear approximation.
2 3
3
2 3 444
1 1' " '''
2 61 1 1
' " '''2 6 24
T x f a f a x a f a x a f a x a
T x f a f a x a f a x a f a x a f a x a
2 33
2 3 44
1 11
2 61 1 1
12 6 24
T x x x x
T x x x x x
0 !
jnj
nj
f aT x x a
j
1 !
nn
n n
f aT x T x x a
n
0a
Taylor
Polynomials
'L x f a f a x a
1L x x
2 3 444
1 1 1' " '''
2 6 24T x f a f a x a f a x a f a x a f a x a
Computing Taylor Polynomials Compute the Taylor polynomial T4(x) centered at a = 3 for 1.f x x
2 3 4
4
1 1 1 152 3 3 3 3
4 64 512 49,152T x x x x x
0 !
jnj
nj
f aT x x a
j
Keep practicing this pattern!
1/ 2 3/ 2
5/ 2 7 / 24 4
1 1' 1 , ; " 1 , ;
2 43 15
''' 1 , ; 1 , 8 1
1 1' 3 " 3
4 323 15
''' 3 3256 20486
f x x f x x
f x
f f
fx f x x f
A factor of each coefficient is the output of ...jf a
Linear Approximation The 4th Taylor Polynomial
k
1 1
1 the coefficient of 1 in are
1 ,
:!
1 11
!
!
k
n
k k
fx T x
k
kk
k
k
11 1 !kk kf x k x
4
2 3 4
Evaluate the derivatives up to degree at 1 and look for a pattern:
1 1 2 6ln , ' , " , '''
? ?
,
? a
f x x f x f x f x f xx x x x
Find the Taylor polynomials Tn(x) of f (x) = ln x centered at a = 1.
1
2 3 11 11 1 1 1
2 3
nn
nT x x x x xn
0! 1! 1
0 !
jnj
nj
f aT x x a
j
1 ln 1 0f a f
The derivatives form a repeating pattern of period 4:
0 1
22 3
2 44 5
1
11
2!1 1
1 Do you see a pattern?2! 4!
T x T x
T x T x x
T x T x x x
Cosine Find the Maclaurin polynomials of f (x) = cos x.
4
Evaluate the derivatives up to degree :
cos , ' sin , " cos , ''' sin ,
???
cosf x x f x x f x x f x x f x x
f (j)(x)= f (j+4)(x) . The derivatives at x = 0 also form a pattern:
4 5 6 7 80 ' 0 '' 0 ''' 0 0 0 0 0 0 f f f f f f f f f
1 0 1 0 1 0 1 0 1
2 4 6 22 2 1
1 1 1 11 1 , in 0,1, 2,...
2! 4! 6! 2 !n n
n nT x T x x x x x nn
Coefficients!
2' "
1! 2! !
nn
n
f a f a f aT x f a x a x a x a
n
1 !
nn
n n
f aT x T x x a
n
Scottish mathematician Colin Maclaurin (1698–1746) was a professor in Edinburgh. Newton was so impressed by his work that he once offered to pay part of Maclaurin’s salary.
Would you like to see the accuracy of the
Maclaurin polynomial approximations of cos ?y x
Horizon
2 2
2
1 21 2 1 2
2!
2 2
R R R
R h R h R h
h h
R h R h
How far is the horizon? Valerie is at the beach, looking out over the ocean. How far can she see? Use Maclaurin polynomials to estimate the distance d, assuming that Valerie’s eye level is h = 1.7 m above ground. What if she looks out from a window where her eye level is 20 m? Valerie can see a distance d = Rθ, the length
of the circular arc AH.
22
1cos 1
2!
RT
R h
Our key observation is that θ is close to zero (both θ and h are much smaller than shown in the figure), so we lose very little accuracy if we replace cos θ by its second Maclaurin polynonomial 2
2
1cos 1
2!x T x x
Not drawn
to scale...
, in radians!d r
MaclaurinPolynomials
How far is the horizon? Valerie is at the beach, looking out over the ocean. How far can she see? Use Maclaurin polynomials to estimate the distance d, assuming that Valerie’s eye level is h = 1.7 m above ground. What if she looks out from a window where her eye level is 20 m? Valerie can see a distance d = Rθ, the length
of the circular arc AH.
22
1cos 1
2!
RT
R h
2 2
2
1 21 2 1 2
2!
2 2
R R R
R h R h R h
h h
R h R h
Furthermore, h is very small relative to R, so we may replace R + h by R to obtain
22
hd R R Rh
R
Valerie can see a distance d = Rθ, the length of the circular arc AH.
22
1cos 1
2!
RT
R h
2 2
2
1 21 2 1 2
2!
2 2
R R R
R h R h R h
h h
R h R h
Furthermore, h is very small relative to R, so we may replace R + h by R to obtain
The earth’s radius is approximately R ≈ 6.37 × 106 m, so
2 3569 md Rh h
In particular, we see that d is proportional to .h
Valerie’s eye level is h = 1.7 m 3569 1. 4567 3 md
3569 20 15,961 md
22
hd R R Rh
R
THEOREM 2 Error Bound Assume that f (n+1)(x) exists and is continuous. Let K be a number such that |f (n+1)(u)| ≤ K for all u between a and x. Then
where Tn(x) is the nth Taylor polynomial centered at x = a.
1
1 !
n
n
x af x T x K
n
Using the Error Bound Apply the error bound to |ln 1.2 − T3(1.2)|where T3(x) is the third Taylor polynomial for f (x) = lnx at a = 1. Check your result with a calculator.
2 3
3
1 11 1 1
2 3T x x x x
2
3
1'
1"
2'''
f xx
f xx
f xx
0 1 2 3, , , T x T x T x T x
Using the Error Bound Apply the error bound to
|ln 1.2 − T3(1.2)|
where T3(x) is the third Taylor polynomial for f (x) = ln x at a = 1.Step 1. Find a value of K.
Therefore, we find a value of K such that |f (4)(u)| ≤ K for all u between a = 1 and x = 1.2. As we computed earlier, f (4)(x) = −6x−4
1
1 !
n
n
x af x T x K
n
The size of the error depends on the size of the 1 derivative.st
n
4
2 3 4
1 1 2 6ln , ' , " , ''' , f x x f x f x f x f x
x x x x
6K Step 2. Apply the error bound.
1 4
3 0.01.2 1
ln 1.2 1.2 61
0! !
44
0n
x aT K
n
Step 3. Check the result.
0.0004 3 1.2T
Observe that ln x and T3(x) are indistinguishable near x = 1.2.
Approximating with a Given Accuracy Let Tn(x) be the nth Maclaurin polynomial for f (x) = cos x. Find a value of n such that
| cos 0.2 − Tn(0.2)| < 10−5
Step 1. Find a value of K.
0 1 1nf x K Step 2. Find a value of n.
1
1 !
n
n
x af x T x K
n
1 10.2 0 0.2
cos0.2 0.2 11 ! 1 !
n n
nT Kn n
1
50.210
1 !
n
n
It’s not possible to solve this inequality for n, but we can find a suitable n by checking several values:
2 3 4n
10.2
1 !
n
n
0.0013 56.67 10 6 52.67 10 10 4n
n nR x f x T x The error in Tn(x) is the absolute value |Rn(x)|.
Rn(x) can be represented as an integral.
Taylor’s Theorem: Version I Assume that f (n+1)(x) exists and is continuous. Then
11
!
xn n
n n
a
R x f x T x x u f u dun
Taylor’s Theorem: Version II (Lagrange) Assume f (n+1)(x) exists and is continuous. Then
11,
1 !
for some between and .
nn
n n
f cR x f x T x x a
n
c a x
The th remainder of the Taylor Polynomial...n
For reference, we will include a table of standard Maclaurin and Taylor polynomials.
