Taylor’s Polynomials & LaGrange Error Review
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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!

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


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.

There are some Maclaurin series that occur often enough
that they should be memorized. They are on your
formula sheet, but today we are going to look at where
they come from.
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!

1
1
 1  x 
1 x
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
f  n  x 
1  x 
1
1  x 
2
2 1  x 
3
6 1  x 
4
24 1  x 
5
List the function and its
derivatives.
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
1
1
 1  x 
1 x
f  n  x 
1  x 
1
1  x 
2
2 1  x 
3
6 1  x 
4
24 1  x 
5
f  n  0 
List the function
Evaluate
and
column
its one
1 derivatives.
1 for x = 0. 2 2 3! 3 4! 4
 1  1x  x  x  x  
1 x
2!
3!
4!
1
2
6  3!
24  4!
1
 1  x  x 2  x3  x 4  
1 x
This is a geometric series with
a = 1 and r = x.
1
We could generate this same series for
with
1 x
polynomial long division:
1  x  x  x 
2
1  x 
3
1
1 x
x 2
xx
2
x
x 2  x3
3
x

f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
1
1
 1  x 
1 x
f  n  0 
f  n  x 
1  x 
1
 1  x 
2 1  x 
2
3
1
1
2
6 1  x  6  3!
4
24 1  x 
5
24  4!
1
2 2 3! 3 4! 4
 1  1x  x 
x  x  
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.

We wouldn’t expect to use the previous two series to
evaluate the functions, since we can evaluate the
functions directly.
They do help to explain where the formula for the sum of
an infinite geometric comes from.
We will find other uses for these series, as well.
A more impressive use of Taylor series is to
evaluate transcendental functions.
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
cos  x 
f  n  x 
f  n  0 
cos  x 
1
 sin  x 
0
 cos  x 
1
sin  x 
0
cos  x 
1
1 2 0 3 1 4
cos  x   1  0 x  x  x  x  
2!
3!
4!
x2
x4 x6
cos  x   1  
 
2!
4! 6!
Both sides are even functions.
Cos (0) = 1 for both sides.

f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
sin  x 
f  n  x 
f  n  0 
sin  x 
0
cos  x 
1
 sin  x 
0
 cos  x 
1
sin  x 
0
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!
Both sides are odd functions.
Sin (0) = 0 for both sides.

1
1  x2
If we start with this function:
and substitute
x
2
for
x
1
 1  x  x 2  x3  x 4  
1 x
1
2
4
6
8

1

x

x

x

x
 
, we get: 1  x 2
This is a geometric series with a = 1 and r = -x2.
If we integrate both sides:
1
2
4
6
8
dx

1

x

x

x

x
  dx
 1  x2

x3 x5 x 7
tan  x   x     
3 5 7
1
This looks the same as the
series for sin (x), but without
the factorials.

f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
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
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
2
6  3!

e
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
x
f  n  x 
f  n  0 
ex
1
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
p
Finding Truncation Error in a
Taylor Polynomial
Graph the function y1 = ln (1 + x) and it’s corresponding
Taylor Polynomial
x2
x3 x4
x5
y2 = x +
+
2
3
4
5
Finding Truncation Error in a
Taylor Polynomial
Graph the function y1 = ln (1 + x) and it’s corresponding
2
3
4
5
Taylor Polynomial y 2 = x - x + x - x - x
2
3
4
5
Finding Truncation Error in a
Taylor Polynomial
To find the error between the functions, graph y3 = abs (y2 – y1).
Finding Truncation Error in a
Taylor Polynomial
To find the error between the functions, graph y3 = abs (y2 – y1).
Where is the error the smallest?
Use Table to see where truncation
error is least on (-1,1)
Finding Error in a Taylor
Polynomial
Graph the function y1 = sin x and it’s corresponding Taylor
Polynomial y = x - x6 and find the interval in which the
3
2
Taylor polynomial is accurate to the thousandths place.
Finding Error in a Taylor
Polynomial
The third order Taylor Polynomial for y = sin x is accurate to
the thousandths place on the interval (-.65, .65). Graph
y3 = abs (y1(x) – y2(x)), and y4 = .001.
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:
Taylor’s Inequality
M
n 1
Rn  x  
xa
 n  1!
f
 n 1
 x
on the interval
Note
this
is not
This
is that
called
Taylor’s
the formula that is in
Inequality.
our book. It is from
another textbook.

ex. 2:
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
Taylor’s Inequality
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.

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:
2
x
Find the Lagrange Error Bound when x 
is used to
2
approximate ln 1  x  and x  0.1 .
Taylor’s Inequality
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.

Example using Taylor’s Theorem
with Remainder
For approximately
what values of x can you replace sin x by
3
x with an error magnitude no greater than 1 x 10-3 ?
x3!
Since
f  5  c  = cos c
 1, then
5
R 4 (x)

x
5!
x
x
5
 1 x 10 - 3
 .12
 .65
Taylor’s Theorem with
Remainder
Taylor’s Theorem with Remainder works well with the
functions y = sin x and y = cos x because |f (n+1)(c)| ≤ 1.
lim R n (x) =
f
c  x 
 n + 1!
 n+1
n 
n 1
n 1
x
 lim
 0
n  (n  1)!
Note: Factorial growth in the denominator is larger than
the exponential growth in the numerator.
100n
100
Ex : lim
=
n   n+1 !
1
100
2
100
...
3
100
100
100
101
...
100
...
7
10
Since | Rn(x)|  0 then the Taylor Series converges as n →
