9.7 and 9.10
Taylor Polynomials and
Taylor Series
Suppose we wanted to find a fourth degree polynomial of
the form:
P  x   a0  a1 x  a2 x2  a3 x3  a4 x 4
that approximates the behavior of
f  x   ln  x  1 at x  0
If we make f (0) = P(0), f’(0) = P’(0), f’’(0) = P’’(0), and so on,
then we would have a pretty good approximation.
P  x   a0  a1 x  a2 x2  a3 x3  a4 x 4
f  x   ln  x  1
f  x   ln  x  1
P  x   a0  a1 x  a2 x2  a3 x3  a4 x 4
f  0  ln 1  0
P  0   a0
P  x   a1  2a2 x  3a3 x2  4a4 x3
1

f  x 
1 x
1
f  0   1
1
f   x   
a0  0
P  0  a1
1
1  x 
1

f  0     1
1
2
a1  1
P  x   2a2  6a3 x  12a4 x2
P  0  2a2
1
a2  
2
f   x   2 
P  x   6a3  24a4 x
1
1  x 
3
P  0  6a3
f   0  2
f
 4
f
 x   6
4
1
1  x 
 0  6
4
P
P
4
 4
2
a3 
6
 x   24a4
 0  24a4
6
a4  
24
1 2 2 3 6 4
P  x   0  1x  x  x 
x
2
6
24
This is called the Taylor Polynomial of degree 4.
x 2 x3 x 4
P  x  0  x   
 ln  x  1
2 3 4
If we plot both functions, we see
that near zero the functions match
very well!
5
4
3
2
0.5
f  x
1
-5 -4 -3 -2 -1 0
-1
1
1
2
3
4
5
-1
-2
-0.5
0
0.5
-0.5
-3
-4
P  x
-5
Our polynomial
has the form:
or:
0  1x 
-1
1 2 2 3 6 4
x  x 
x
2
6
24
f   0  2 f   0  3 f  4  0  4
f  0  f   0 x 
x 
x 
x
2
6
24
f  0 f   0
f   0  2 f   0  3 f  4  0  4

x
x 
x 
x
0!
1!
2!
3!
4!
1
Theorem

n
a
(
x

c
)
If f (x) is represented by a power series  n
n 0
f ( n ) (c )
for all x in an open interval I containing c, then an 
n!
Definition

f ( n ) (c )
n
(
x

c
)
The series of the form  n !
n 0
f c
f   c 
f   c 
2
3
 f c 
 x  c 
 x  c 
 x  c    
1!
2!
3!
is called the Taylor Series for f (x) at c. If c = 0, then the
series is called the Maclaurin series for f .
Example
Find the Maclaurin series for
f  x   cos x
f   x    sin x
f ( x)  cos x
f  0  1
f   x   sin x f   0  0
f   0  0
f
f   x    cos x f   0   1
4
 4
x

cos
x
f  0  1
 
1
-5
-4
1 2 0 3 1 4
cos  x   1  0 x  x  x  x  
2!
3!
4!
x 2 x 4 x 6 x8 x10
cos x  1 

  

2! 4! 6! 8! 10!
-3
-2
-1
0
-1
1
2
3
4
5
Example
Find the Maclaurin series for f ( x)  cos  2 x 
Rather than start from scratch, we can use the function
that we already know:
x 2 x 4 x 6 x8 x10
cos  x   1 

  

2! 4! 6! 8! 10!
cos  2 x   1 
 2x
2!
2

 2x
4!
4

 2x
6!
6
 2x
8

8!
 2x 
10

10!

Example

Find the Taylor series for y  cos  x  at x 
2
 
f  x   cos x
 


f  x   sin x f    1
2
f  0
2


f   x    sin x f     1
2
f

 4
f   x    cos x f     0
2
 x   cos x f
 4  
 0  1 

P  x   0  1 x     x     x    
2  2! 
2  3! 
2

2

3
3

5


x

x 




2 
2

P  x    x    

 
2
3!
5!


 0
2
Example
Find the Maclaurin series for
f  n  x 
f  n  0
sin  x 
0
cos  x 
1
 sin  x 
0
 cos  x 
1
sin  x 
0
f ( x)  sin x
0 2 1 3 0 4
sin  x   0  1x  x  x  x  
2!
3!
4!
x3
x5 x 7
sin  x   x  
 
3!
5! 7!
Example
1
f
(
x
)

Find the Maclaurin series for
1 x
f  n  0
f  n  x 
1  x 
1
1  x 
2
2 1  x 
3
6 1  x 
4
24 1  x 
5
1
1
2
6  3!
24  4!
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
1
2
3!
4!
 1  1x  x 2  x 3  x 4  
1 x
2!
3!
4!
1
 1  x  x 2  x3  x 4  
1 x
This is a geometric series with
a = 1 and r = x.
Example
Find the Maclaurin series for f ( x)  ln 1  x 
f  n  x 
f  n  0
ln 1  x 
0
1  x 
1
1
 1  x 
2 1  x 
2
3
6 1  x 
4
1
2
6  3!
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
1 2 2 3 3! 4
ln 1  x   0  1x  x  x 
x  
2!
3!
4!
x 2 x3 x 4
ln 1  x   x     
2 3 4
Example
Find the Maclaurin series for
f  n  x 
f  n  0
ex
1
f ( x)  e x
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
1 2 1 3 1 4
e  1  1x  x  x  x  
2!
3!
4!
x
ex
1
ex
1
ex
1
ex
1
x 2 x3 x 4
e  1  x     
2! 3! 4!
x
Euler’s Formula
An amazing use for infinite series:
2
3
4
x
x
x
e x  1  x     
2! 3! 4!
Substitute xi for x.
 xi 2  xi 3  xi 4  xi 5  xi 6
e  1  xi 




 
2!
3!
4!
5!
6!
xi
x 2i 2 x3i 3 x 4i 4 x5i 5 x 6i 6
e  1  xi 




 
2!
3!
4!
5!
6!
xi
x 2 x3i x 4 x5i x 6
e  1  xi  
 
   Factor out the i terms.
2! 3! 4! 5! 6!
xi
2
4
6
x
x
x


x3 x5
xi
e  1       i  x     
2! 4! 6!
3! 5!


2
4
6
x
x
x


x3 x5
xi
e  1       i  x     
2! 4! 6!
3! 5!


This is the series
for cosine.
e xi  cos  x   i sin  x 
This is the series
for sine.
Let
x 
ei  cos    i sin  
ei  1  i  0
i
e 1  0
This amazing identity contains
the five most famous numbers
in mathematics, and shows
that they are interrelated.
Taylor series are used to estimate the value of functions (at
least theoretically - nowadays we can usually use the
calculator or computer to calculate directly.)
An estimate is only useful if we have an idea of how
accurate the estimate is.
When we use part of a Taylor series to estimate the value
of a function, the end of the series that we do not use is
called the remainder. If we know the size of the remainder,
then we know how close our estimate is.
1
2
4
6
Ex:
Use 1  x  x  x to approximate
over  1,1 .
1  x2
8
x
the truncation error is x8  x10  x12   , which is
.
2
1 x
When you “truncate” a number, you drop
off the end.
Taylor’s Theorem with Remainder
If f has derivatives of all orders in an open interval I
containing a, then for each positive integer n and for each x
in I:
f   a 
f n  a 
2
f  x   f  a   f   a  x  a  
 x  a   
 x  a n  Rn  x 
2!
n!
Lagrange Form of the Remainder
Rn  x  
 c  x  a n1


 n  1!
f
 n 1
Remainder after
partial sum Sn
where c is between
a and x.
This is also called the remainder of order n or the error term.
Lagrange Form of the Remainder Note that this looks just
Rn  x  
like the next term in the
 c  x  a n1


 n  1!
f
 n 1
series, but “a” has been
replaced by the number
“c” in f
If M is the maximum value of
between a and x, then:
f
 n 1
 x
 n 1
c
.
on the interval
Taylor’s Inequality
M
n 1
Rn  x  
xa
 n  1!
This is called Taylor’s
Inequality.
x2
Find the Lagrange Error Bound when x 
is used
2
to approximate ln 1  x  and x  0.1 .
x2
f  x   0  x   R2  x 
2
f  x   ln 1  x 
1

f  x   1  x 
f   x    1  x 2
f   x   2 1  x 3
2
On the interval  .1,.1 , 1  x 3 decreases, so
its maximum value occurs at the left end-point.
M
Remainder after 2nd order term
2
1  .13
2

 2.74348422497
.9 3
f   0
f   0  2
f  x   f  0 
x
x  R2  x 
1
2!
x ln 1  x 
2.7435 .1 3
Rn  x  
3!
 0.000457
Lagrange Error Bound
x2
x
2
error
.095
.000310
.1
.0953102
.1
.1053605 .105
.000361