9.2: Taylor Series
Brook Taylor was an
accomplished musician and
painter. He did research in a
variety of areas, but is most
famous for his development of
ideas regarding infinite series.
Brook Taylor
1685 - 1731
Greg Kelly, Hanford High School, Richland, Washington
Suppose we wanted to find a fourth degree polynomial of
the form:
P  x   a0  a1 x  a2 x2  a3 x3  a4 x4
that approximates the behavior of
f  x   ln  x  1 at x  0
If we make P  0  f  0, and the first, second, third and fourth
derivatives the same, then we would have a pretty good
approximation.

P  x   a0  a1 x  a2 x2  a3 x3  a4 x4
f  x   ln  x  1
f  x   ln  x  1
P  x   a0  a1 x  a2 x2  a3 x3  a4 x4
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

P  x   a0  a1 x  a2 x2  a3 x3  a4 x4
f   x   
P  x   2a2  6a3 x  12a4 x2
1
1  x 
2
P  0  2a2
1
f   0     1
1
f   x   2 
3
P  0  6a3
f   0  2
f
 4
 x   6
1
1  x 
f  4  0   6
1
a2  
2
P  x   6a3  24a4 x
1
1  x 
f  x   ln  x  1
4
2
a3 
6
P 4  x   24a4
P
 4
 0   24a4
6
a4  
24

P  x   a0  a1 x  a2 x2  a3 x3  a4 x4
f  x   ln  x  1
1 2 2 3 6 4
P  x   0  1x  x  x 
x
2
6
24
x 2 x3 x 4
P  x  0  x   
2 3 4
If we plot both functions, we see
that near zero the functions match
very well!
5
4
3
f  x
1
1
2
3
4
5
-2
-5
-1
-0.5
0
0.5
1
-0.5
-3
-4
1
0.5
2
-5 -4 -3 -2 -1 0
-1
f  x   ln  x  1
P  x
-1

Our polynomial:
has the form:
or:
0  1x 
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!
This pattern occurs no matter what the original function was!

Maclaurin Series:
(generated by f at
x0 )
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
If we want to center the series (and it’s graph) at some
point other than zero, we get the Taylor Series:
Taylor Series:
(generated by f at
xa )
f   a 
f   a 
2
3
P  x   f  a   f   a  x  a  
x

a

x

a



  
2!
3!

example:
y  cos x
f  x   cos x
f   x    sin x
f  0  1
f   x   sin x f   0  0
f   0  0
f  4  x   cos x f  4  0   1
f   x    cos x f   0   1
1x 2 0 x3 1x 4 0 x5 1x 6
P  x  1 0x 




 
2!
3!
4!
5!
6!
x 2 x 4 x 6 x8 x10
P  x  1    

2! 4! 6! 8! 10!

x 2 x 4 x 6 x8 x10
P  x  1    

2! 4! 6! 8! 10!
y  cos x
1
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
The more terms we add, the better our approximation.
Hint: On the TI-89, the factorial symbol is:


example:
y  cos  2 x 
Rather than start from scratch, we can use the function
that we already know:
2x   2x   2x   2x   2x 

P  x  1




2!
4!
6!
8!
10!
2
4
6
8
10


example:
y  cos  x  at x 
f  x   cos x

2
 
f  0
2
 


f  x   sin x f    1
2


f   x    sin x f     1
2
f

 4
f   x    cos x f     0
2
 x   cos x f
 4  

 0
2
 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!


There are some Maclaurin series that occur often enough
that they should be memorized. They are on your
formula sheet.

When referring to Taylor polynomials, we can talk about
number of terms, order or degree.
x2 x4
cos x  1  
2! 4!
This is a polynomial in 3 terms.
It is a 4th order Taylor polynomial, because it was
found using the 4th derivative.
It is also a 4th degree polynomial, because x is raised
to the 4th power.
x2
The 3rd order polynomial for cos x is 1 
, but it is
2!
degree 2.
x3 term
dropsrequired
out when
the to
third
derivative.
AThe
recent
AP exam
theusing
student
know
the
difference
order
degree.
Thisbetween
is also the
2ndand
order
polynomial.

The TI-89 finds Taylor Polynomials:
taylor (expression, variable, order, [point])
F3
taylor
taylor
taylor
9
 cos  x  , x, 6 
 x6 x 4 x 2
  1
720 24 2
 cos  2 x  , x, 6 
4 x 6 2 x 4

 2 x2  1
45
3
 cos  x  , x,5,  / 2 
  2x   
5
3840
2x   


3
48
2x  

2
