Taylor’s Theorem: Error Analysis for Series
Tacoma Narrows Bridge: November 7, 1940
Taylor series are used to estimate the value of functions (at
least theoretically - now days 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.

For a geometric series, this is easy:
ex. 2:
1
 1,1.
Use 1  x  x  x to approximate
2 over
1 x
2
4
6
Since the truncated part of the series is: x8  x10  x12   ,
x8
the truncation error is x  x  x   , which is
.
2
1 x
8
10
12
When you “truncate” a number, you drop
off the end.
Of course this is also trivial, because we have a formula
that allows us to calculate the sum of a geometric series
directly.

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.

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.
Note that this looks just like the next term in the series, but
“a” has been replaced by the number “c” in f  n 1  c  .
This seems kind of vague, since we don’t know the value of c,
but we can sometimes find a maximum value for f  n 1  c  .

Lagrange Form of the Remainder
Rn  x  
 c  x  a n1


 n  1!
f
 n 1
If M is the maximum value of
between a and x, then:
f
 n 1
 x
on the interval
Remainder Estimation Theorem
M
n 1
Rn  x  
xa
 n  1!
We call this the
Remainder Estimation
Theorem.

Prove that


k 0
 1
k
x 2 k 1
 2k  1!
, which is the Taylor
series for sin x, converges for all real x.
Since the maximum value of sin x or any of it’s
derivatives is 1, for all real x, M = 1.
 Rn  x  
1
 n  1!
x0
n 1

Remainder Estimation Theorem
M
n 1
Rn  x  
xa
 n  1!
n 1
x
 n  1!
n 1
x
lim
0
n  
n  1!
so the series converges.

Since each term of a convergent alternating series
moves the partial sum a little closer to the limit:
Alternating
Theorem
A quick
review of Series
an errorEstimation
bound formula
that we
already
so we alternating
can see how
it fitsthe
in with
For a know
convergent
series,
truncation
LaGrange…
error is less than the first missing term[, and is the
same sign as that term].

  1
n 1
n 1
1
1 1 1 1
 1   
n
2 3 4 5
How far off is this from the actual sum?
1 1 1
...    ...
6 7 8
1 1 111 111 1 11 1
1 11
... 
error
error
 
    ...
...  
error

6 66
6 6 767 788 8 66 6
This is a good tool to remember, because it is easier
than the LaGrange Error Bound.

Since each term of a convergent alternating series
moves the partial sum a little closer to the limit:
Alternating Series Estimation Theorem
For a convergent alternating series, the truncation
error is less than the first missing term[, and is the
same sign as that term].
One more thing about this theorem:
Are there any Taylor or MacClaurin Series for which
we can use this instead of the LaGrange Error Bound?
Think about it… sin x and cos x…but why?
Think about it…

Recall that the Taylor Series for ln
x centered at x = 1 is given by…
n

( x  1) 2 ( x  1)3 ( x  1) 4
(
x

1
)
f ( x)  ln x  ( x  1) 


...   (1) n1
2
3
4
n
n 1
Find the maximum error bound for each approximation.
Because the series is alternating, we can start with…
3
(
1
.
5

1
)
3  error 
 0.0417
ln
3
2
actual error | ln( 1.5)  0.625 | 0.03047
3
(
0
.
5

1
)
1  error 
 0.0417
ln
3
2
actual error | ln( 0.5)  (0.625) | 0.068147
Wait! How is the actual error bigger than the error bound for ln 0.5?
3
(
0
.
5

1
)
1  error 
 0.0417
ln
3
2
actual error | ln( 0.5)  (0.625) | 0.068147
Wait! How is the actual error bigger than the error bound for ln 0.5?
Plugging in ½ for x makes each term of the series positive and therefore it is
no longer an alternating series. So we need to use the La Grange Error
Bound.
3
2
c
f (c)
(0.125)
error 
(0.5  1) 3 
3!
3!
The third derivative
of ln x at x = c
What value of c will give us the maximum error?
Normally, we wouldn’t care about the actual value of c but in this case, we
can graph 2c–3 and use the graph to find where the maximum value is.
3
2
c
f (c)
(0.125)
error 
(0.5  1) 3 
3!
3!
The third derivative
of ln x at x = c
The question is what value of c between x and a will give us the maximum
error?
So we are looking for a number for c between 0.5 and 1. Let’s look at a
graph of 2x–3.
Since the function is decreasing
between 0.5 and 1, it’s maximum is at
0.5 and therefore…
c = 0.5














3
(
0
.
5

1
)
1  error 
 0.0417
ln
3
2
actual error | ln( 0.5)  (0.625) | 0.068147
Which is larger than
the actual error!
Wait! How is the actual error bigger than the error bound for ln 0.5?
Plugging in ½ for x makes each term of the series positive and therefore it is
no longer an alternating series. So we need to use the La Grange Error
Bound.
3
2
c
f (c)
(0.125)
error 
(0.5  1) 3 
3!
3!
What value of c will give us the maximum error?
Since we’ve established that c = 0.5, we now know that…
2(0.5) 3
16 1
error 
(0.125) 

3!
6 8
1
3
The third derivative
of ln x at x = c
ex. 5:
x2
Find the Lagrange Error Bound when x 
is used
2
to approximate ln 1  x  and x  0.1 .
f  x   ln 1  x 
f   x   1  x 1
f   x    1  x 2
f   x   2 1  x 3
x2
f  x   0  x   R2  x 
2
2
On the interval  .1,.1 , 1  x 3 decreases, so
its maximum value occurs at the left end-point.
M
2
1  .13
2

3  2.74348422497
.9 
f  0
f   0  2
f  x   f  0 
x
x  R2  x 
1
2!
Remainder after 2nd order term

ex. 5:
x2
Find the Lagrange Error Bound when x 
is used
2
to approximate ln 1  x  and x  0.1 .
Remainder Estimation Theorem
M
n 1
Rn  x  
xa
 n  1!
2
On the interval  .1,.1 , 1  x 3 decreases, so
its maximum value occurs at the left end-point.
2.7435 .1
Rn  x  
3!
3
Rn  x   0.000457
Lagrange Error Bound
M
2
1  .13
2

3  2.74348422497
.9 
x2
x
error
2
Error is
x
ln 1  x 
.1
.0953102
.1
.1053605 .105
.095
.000310
.000361
less than
error
bound. p