Remainder Theorem
The n-th Talor polynomial
The polynomial
n
pn ( x)  
k 0
f ( k ) (c)
( x  c) k
k!
is called the n-th Taylor polynomial for f about c
The n-th Maclaurin polynomial
The polynomial
n
pn ( x)  
k 0
f ( k ) (0) k
x
k!
is called the n-th Maclaurin polynomial for f
Taylor formula for f with the
Remainder
The difference
Rn ( x)  f ( x)  Pn ( x)
n
 f ( x)  
k 0
f
(k )
(c )
k
( x  c)
k!
is called the n-th remainder for the Taylor series of f
about c.
Thr Remainder Theorem
Assume that all the first n+1 derivatives exist on
an interval l containing c and
Sup{ f ( n 1) ( x) : x  I }   ,
Then
Rn ( x) 

(n  1)!
xc
n 1
; xI
for some positive number 
Example (1)
Approximate e to five decimal places
We have,
The Maclaurin polynomial for e x is
1 2 1 3 1 4
1 n
pn ( x)  1  x  x  x  x   x
2!
3!
4!
n!
n
1 k

x
k  0 k!
The Maclaurin polynomial for e x is
n
1 k
x
k
!
k 0
1
1
1
1
 1  x  x 2  x 3  x 4   x n
2!
3!
4!
n!
The theorem says
pn ( x )  
Rn (1) 

(n  1)!
1 0
n 1
where  is a small enough upper bound for the set
{ f ( n 1) ( x)  e x : x  (1 , 1 
1
)}
6
10
for example   3
Notice that sup{ f ( n 1) ( x)  e x : x  (1 , 1 
and that I  (1, 1 
1
) is an int erval containing 0 and 1
6
10
Thus,
Rn (1) 
1
)}  3
10 6
3
n 1
1 0
(n  1)!
The accuracy we need  0.000005
Let
3
 0.000005
(n  1)!
 (n  1)! 600000
Choose n  9Why ?
 an approximation of e accurate to 5 decimal places is :
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9
e  1  1  (1)  (1)  (1)  (1)  (1)  (1)  (1)  (1)
2!
3!
4!
5!
6!
7!
8!
9!
 2.71828
Why we chose n=9 ?
Let
3
 0.000005
(n  1)!
 (n  1)! 600000
We have,
9! 9(8)(7)(6)(5)( 4)(3)( 2)(1)  362880
10! 3628800
9! 362880  600000  3628800  10!
Thus, the smallest n satisfying
3
 0.000005 is
(n  1)!
n9
Example (2)
Approximate sin3○ to five decimal places
We have,
The Maclaurin polynomial for sin x is
2 n 1
x
sin x   (1) n
(2n  1)!
n 1
1 3
1 5
1 7
 0  x  0  x  0  x  0  x  
3!
5!
7!

The Maclaurin polynomial for sin x is
x 2 k 1
pn ( x)   (1)
xk
(2k  1)!
k 0
1
1
1
 0  x  0  x 3  0  x 5  0  x 7  
3!
5!
7!
The theorem says
n
k
Rn (

60
)


(n  1)! 60
n 1
0


(n  1)!
(

60
) n 1
where  is any upper bound for the set { f ( n 1) ( x) x  (1 ,
Because f ( n 1) ( x) is either equal to sin x or cos x
and sin x , cos x  1, we choose   1
Rn (

60
)
1

( ) n 1
(n  1)! 60

50
)}
The error tolerated  0.000005
Let
1
 n 1
( )  0.000005
(n  1)! 60
 The smallest n satisfying the above inequality is n  3
 an approximation of sin 3 accurate to 5 decimal places is :

1  3

sin 3  0 
0 ( )
60
31 60
 0.5234
Approximating the sum of a
convergent alternating series
Theorem
Let T be the sum of an alternating series satisfying
the main convergence test for an alternating
series. Then:
1. T lies between any two successive partial sums
of the series
2. If T is approximated by a partial sum Tn, then:
a. |T - Tn | ≤ an+1
b. The sign of the error is the same as that of the
coefficient of an+1
Using power series to
approximate definite integrals
Example
Give an estimation of
the integral on the
right accurate to 3
decimal places

1
0
e
 x2
dx
A power series for the given
integral
1
e
 x2
0
dx
4
6
8
n 2n
x
x
x
(

1
)
x
2
  [1  x      
 ] dx
0
2! 3! 4!
n!
1
3
5
7
9
x
x
x
x
 [x



]
1
3 (5) 2! (7) 3! (9) 4!
1
1
1
1
 1 


 
3 (5) 2! (7) 3! (9) 4!
(1) n

n  0 ( 2 n  1) n!

0
Approximating the integral
For any partial sum T n of the series
1
0 e dx T n  a n1 (2n  3)(n  1)!
To get the required accuracy 0.0005, we let
1
 x2
1
 0.0005
(2n  3)( n  1)!
 the smallest n satisfying that, is n  5
1
 e
 x2
0
 0.747
1
1
1
1
1
dx  1  



3 (5) 2! (7) 3! (9) 4! (9) 5!
Homework
(1) Approximate sin 2  to four decimal places
1
(2). Approximate sin( ) within 0.001
2
(3) Approximate 10 e by p 4 (0.1), and then calculate the max imum
error of such approximation.
(4) Approximate ln( 1.4) within .001
(1)
sin 2
2
2 correspond s to
 2  0.0349

360
1 3
1 5
sin x  0  x  0  x  0  x 
3!
5!
The theorem says

Rn (0.0349) 

0.0349  0
n 1
(n  1)!
where  is any small enough upper bound for the set
{ f ( n 1) ( x) : x  (1,1) }
for example   1 (Why ?)
Thus,
1
n 1
Rn (0.0349) 
0.0349
(n  1)!
We need that the remainder Rn (0.0349) be less than 0.00005
1
n 1
0.0349  0.00005
(n  1)!
1
3
because R2 (0.0349)  0.0349  0.00004
3!
Then we need n  2
We find n, such that
Thus
p2 (0.0349)  0  (0.0349)  0
 0.0349
Notice that
p2 ( x )  0  x  0  x
( 2)
sin
1
2
1 3 1 5
sin x  x  x  x  
3!
5!
The theorem says
n 1
1
 1
Rn ( ) 
0
2
(n  1)! 2
where  is a small enough upper bound for the set
{ f ( n 1) ( x) : x  (0,1) }
for example   1
Why ?
Thus,
1
1 1
Rn ( ) 
2
(n  1)! 2
n 1

1
(n  1)! 2 n 1
We need n such that,
1
 0.001
n 1
(n  1)! 2
n4
1
1
Thus p 4 ( ) is the approximation of sin within the given accuracy
2
2
Notice that
1
p 4 ( x)  0  x  0  ( x) 3  0
3!
1
1
1 1
1 1 23
sin  0  ( )  0  ( ) 3  0  

2
2
3! 2
2 48 48
Question :
Why we picked n  4 ?
1
1
1


 0.001
4 1
5
3840
(4  1)! 2
5! 2
(4)
ln
14
10
1 2 1 3 1 4 1 5
ln( 1  x)  x  x  x  x  x  
2
3
4
5
ln( 1  0.4)
1
1
1
1
 0.4  (0.4) 2  (0.4) 3  (0.4) 4  (0.4) 5  
2
3
4
5

n 1
  (1) (0.4) n
n
n 1
The series satisfies the main test for the convergence of alternating series
The theorem of the sum of alternating series says that :
The sum lies between any two cos ucative partial sums
1
The first term less than 0.01 is :  (0.4) 4  0.0064
4
1
1
1
14
1
1
 0.4  (0.4) 2  (0.4)3  (0.4) 4  ln
 0.4  (0.4) 2  (0.4)3
2
3
4
10
2
3
14
 0.335  ln
 0.341
10
Thus an approximation satisfying the required accuracy is , for example :
14
ln
 0.34
10