BC 2-3 Taylor Series Name: ____________________
In class and on the homework we have determined the following Maclaurin series and intervals of convergence for the given functions: e x
1 x x
2
 x
3
2 !
3 !

L
 n



0 x n n !
for x

(–

,

) sin
  x
3 x
5 x
7 x x
 
3 !
5 !
7 !
L cos
  x
   x
2
 x
3
 x
2 3 4
4
L n



0

1 n 
2 n

1
2 n

1 !
x
2 x
4 x
6
1
   
2 !
4 !
6 !
L n



0

1 n  x
2 n
2 n !

 n

1
  n

1  x n n for x

(–

,

) for x

(–

,

) for x

(–1, 1] tan

1
  x
3 x
5 x
7 x x
 
3 5 7

L


 n

0
1
1
 x
1 x x
2  x
3
L n



0 x n

1 n 
2 n

1
2 n

1 for x

[–1, 1] for x

(–1, 1)
Suppose you want to determine the Maclaurin series for the function find the first few derivatives of
   e
2 x
. Clearly you could
 
, evaluate each at x = 0, generate the first few terms of the
Maclaurin series, and then look for a pattern in these terms so that you extend these first few terms to an infinite series. However, one would think there must be an easier way given that we already have a
Maclaurin series for
 
 e x
. And of course there is.
Since
   g
 
, the Maclaurin series for
Maclaurin series for
 
. Thus,
 
can be generated by substituting 2 x for x in the e
2 x
1 2 x

2 x
2

2 x
3
2 !
3 !

L 1 2 x

4 x
2

8 x
2 !
3 !
2

L
 n



0
  n n !
Also, since the series for the series for
 
 e x
converges for –

< x <

,
   e
2 x
will converge for –

< 2 x <

or –

< x <

.
Taylor Series 1 F12

7.
Use substitution and an appropriate Maclaurin series to determine power series that represents each of the following functions. (DO NOT use derivatives!) Then determine the interval of convergence for each of these series.
1.
   interval of convergence:
2.
1 e x
2
 interval of convergence:
3. x
 cos

 x
4
3



4.
1
2
 x
3
 interval of convergence:
 x
  interval of convergence: interval of convergence:
8.
1
1
 x
2
 interval of convergence:
9.
4
2
 x
 interval of convergence:
Taylor Series 2 Fall 12

Taylor Series Error :
Remember that if f is differentiable on an interval ( , ) containing x
 x
0
, then the tangent line is the first order Taylor Polynomial,
If x is near x
0
we expect that
1
( )

( )
0
 
( )(
0 x
 x
0
P x
1
 f x . But how close??
) .
Let’s do some “error analysis” using an old favorite – The Mean Value Theorem.
If x

( , ) , we use the mean value theorem as follows.
Suppose that ,

and f

are continuous on [ , ] and let ( )

( )

( )

( )(
 x ) .
Differentiating, we get ( )
Thus, ( )
  f

( )(
 a )
 
.

( )
 
( )
 f

( )(
 x )
  f

( )(
 x ) .
Now, let ( )

( )
 

 b
 x
 2
. It’s easy to check that ( )

( )

0 .
Therefore, g satisfies the conditions of Rolle’s Theorem (look it up – it’s just a special case of the
MVT).
So, there exists a c

( , ) such that ( )

0 . Looking at how we defined g we see that:
 
2(

( )
2
)
( )

F a
 
(

)
2
2(

)
  
(

)
2(

)
[
 f

( )(

)]

(

)
2
2 f

However, ( )

( )

( )
 
( )(
 x )

( )

( )

( )
 
( )(

) .
Combining the last two results give: ( )

( )
 
( )(

)

(

)
2 f

( )
. Here’s the rub:
2
There exists a c

( , ) , such that the error when using the tangent line at a to approximate f at b is
(

2
)
2 f

( ) !
The way we will use this is as follows:
Error Theorem: Suppose that ,

and f

are continuous on ( , ) with a
 x
0
 b . If
( )

( )
 
( )( x
 x
0
) is the first order Taylor Polynomial, then for all x

( , ) , let the
( ) error in approximating f with P
1
be given by E x
 f x

P x
1
( )

( )
. Then

M t

M
2
( x
 x
0
)
2 x
0 x . where M is any bound on f

( ) between x
0
and x . That is
Taylor Series 3 Fall 12

Example: Find a bound on the error when using P x
1
( )
 x to approximate ( )
 sin on [ .2,.2] .
Solution: We note that f

( )
  sin
 
So ( )
 sin x x
M
2
(.2 0)
2 
1
2
(.04)

.02
for x [ .2,.2] .
Your turn.
(1) Find a bound on the error when using the tangent line to ( )
 e x
on [ 0.1,0.1] to approximate f .
Now, suppose that

( x ) has derivatives of orders up to at least n + 1 in an open interval ( a , b ) that contains x
0
. Let P x n
( ) be the n th degree Taylor polynomial expanded about x = x
0
, that is n
( )

( )
0
 
( )(
0 x
 x
0
)

Then for every value of x in the interval ( a , f

( )
0 b
2!
( x
 x
0
) we have
)
2   f x
0 ( x
 x
0 n

!
( )
 n
( ) n
( ,
0
)
) n
.
. In other words, the exact value of f ( x ) is the approximation p ( x ) plus an error term R x x n
( ,
0
) , called the n th remainder at x (for the polynomial expanded about x
0
).
Taylor’s Theorem tells us how to compute this error term: R n
( x , x
0
)  f ( n  1) ( c )
( n  1)!
( x  x
0
) n  1
, where c is some (unknown) value between x
0
and x .
Taylor’s Theorem tells the exact error made using the n th-order Taylor polynomial at x
0
to approximate the value of f ( x ). The problem is that we usually have no way of knowing c .
We often approximate the size of the error by choosing a c to make | R n
( x , x
0
)| as large as possible. If we overestimate the size of the error, we know we are safe (we can say that the error is no more than our estimate, and will usually be even less).
Note: the above version of Taylor’s Theorem is slightly different than the one in the text
(page 526). In the text, they don’t give the exact version, they only give the overestimate of the size of the error. Their M is the same as our choosing c to make the error as large as possible.
Taylor Series 4 Fall 12

1. If | x | < 2, find an upper bound for the error made in approximating cos( x ) using the fourth degree Maclaurin polynomial: a. by using R
4
( x , 0) b. by using R
5
( x , 0) c. Compare these results. Which is better? Why? Is it mathematically correct to use either of these remainders to estimate the error? Why or why not?
2. Find an upper bound for the error made in approximating tan

1
 
0.8

using the first seven nonzero terms of its Maclaurin series.
3. When approximating tan

1
 
0.8

, how many nonzero terms of its Maclaurin series do you have to use in order to have an error that is less than 0.00001?
Taylor Series 5 Fall 12

Practice :
1. What are the possible values for x if we want the error in approximating cos( x ) using the first six nonzero terms of its Maclaurin series to be less than 0.0005?
2. When approximating cos(–2.75), how many nonzero terms of its Maclaurin series do you have to use in order to have an error that is less than 0.00001?
3. If x > 0, what are the possible values for x if we want the error in approximating ln(1 + x ) using the first seven nonzero terms of its Maclaurin series to be less than 0.0005?
Taylor Series 6 Fall 12

4. When approximating ln(1.75), how many nonzero terms of its Maclaurin series do you have to use in order to have an error that is less than 0.00001?
5. Find an upper bound for the error made in approximating e
 1
2 terms of its Maclaurin series?
using the first six nonzero
6. If x < 0, what are the possible values for x if we want the error in approximating ex using the first seven nonzero terms of its Maclaurin series to be less than 0.0005?
Taylor Series 7 Fall 12

7. Use Maclaurin series to write power series expanded about x
0

0 for each of the following functions. a.

3 x x b.
 x
3 e
2 x

1
(Be careful with this one!)
8. Use series to approximate

0
0.9
cos( x 2 ) dx with an error less than .005.
9. Suppose f
  
1, f
   
2, f
   
4, f
   
18, and f
  
6 for all x and for n > 3. a. Write the first four nonzero terms of the Taylor series for

expanded about x
0

2 . b. Find an upper bound for the error made in approximating the value of f using the first four nonzero terms of Taylor series for f expanded about x
0

2 from part (a).
Taylor Series 8 Fall 12

10. Use Maclaurin series to evaluate lim x

0
  x x
 

.
11. Suppose you are given the power series n



0
1
 x n n
. a. If x < 0, what are the possible values for x if we want the error in approximating the value of the series using the first 15 nonzero terms of the series to be less than
0.0005? b. When approximating the value of the series for x = –0.9, how many nonzero terms of the series do you have to use in order to have an error that is less than 0.0001?
Taylor Series 9 Fall 12

12. We can show that the series n


0 n
2 
converges by the Ratio Test, but to what value does it
2 n converge? To find out, complete the following steps.
1
Express
1
 x
as a series:
1
1
 x

Differentiate both sides.
Multiply both sides by x .
Differentiate both sides (again).
Multiply both sides by x (again) and express the right-hand side in simplified

-notation
Substitute
1
2
for
How cool is that? x and simplify both sides of your equation. What result do you get?
Taylor Series 10 Fall 12

13. Use ideas similar to those in problem 12 to find the following sums: a.
1

   
4 4
2

  3

 
4
4
= __________________ b.
n

0
 n
3 n

1 
= __________________
 c.
n

 1
= __________________

1
2 n  n !
14) The Maclaurin series for some function is given by
( )

1

2 x

3 x
2

4 x
3

5 x
4

6 x
5

7 x
6

8 x
2!
3!
4!
5!
6!
7!
8!
9!
7
..........
a.
Find ƒ'(0). b.
Find ƒ (60) (0).
Taylor Series 11 Fall 12

15) Use series to approximate


0
1 sin x dx with an error of less than 0.001. (This integral is not x improper at the origin, so don’t worry about that.) Explain how you know your answer has the desired accuracy. Is your approximation an overestimate or an underestimate? How do you know? [Do not add up the terms of the series].
16) (AP 2011 #6) Let f x
 sin( x
2
)
 x The graph of y
 f
5
( ) is shown above.
(a) Write the first four nonzero terms of the Taylor series for sin x about x

0 , and write the first four nonzero terms of the Taylor series for sin( x
2
) about x

0 .
(b) Write the first four nonzero terms of the Taylor series for cos x about x

0 . Use this series and the series for sin( x
2
) , found in part (a), to write the first four nonzero terms of the Taylor series for f about x = 0.
(c) Find the value of
(d) Let f
(6)
(0) .
( ) be the fourth-degree Taylor polynomial for f about x

0 . Using information from the graph of y
 f
5
( ) shown above, show that P
4
1
 
 f
 
 

1
4 3000
.
Taylor Series 12 Fall 12