Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
8.8
Approximations by Taylor’s Polynomials
In this and the next section we are interested in approximating function
values by using polynomials which are easy to compute. The polynomials
used in the process are referred to as Taylor’s polynomials.
Let f (x) be a function which has derivatives f 0 (a), f 00 (a), f 000 (a), · · · , f (n) (a).
We want to approximate f (x) by a polynomial of degree n as follows.
f (x) ≈ c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + · · · + cn (x − a)n .
Note that f (a) ≈ c0 . Now if we take the first derivative of f (x) we find
f 0 (x) ≈ c1 + 2c2 (x − a) + 3c3 (x − a)2 + 4c4 (x − a)3 + · · · + ncn (x − a)n .
From this we see that f 0 (a) ≈ c1 . Next, take the second derivative of f (x)
to obtain
f 00 (x) ≈ 2 · 1c2 + 3 · 2c3 (x − a) + 4 · 3c4 (x − a)2 + · · · + n · (n − 1)cn (x − a)n−2
This implies that f 00 (a) ≈ 2 · 1c2 = 2!c2 or c2 ≈
third order derivative of f (x) we find
f 00 (a)
2! .
Again, taking the
f 000 (x) ≈ 3 · 2 · 1c3 + 4 · 3 · 2c4 (x − a) + · · · + n · (n − 1) · (n − 2)cn (x − a)n−3 .
It follows that f 000 (a) ≈ 3 · 2 · 1c3 = 3!c3 or c3 ≈
f 000 (a)
3! . Continuing this
f (4) (a)
f (n) (a)
4! , · · · , cn ≈
n! .
process of taking successive derivatives we find c4 ≈
Hence, f (x) can be approximated by the polynomial
f (x) ≈ f (a)+
f 0 (a)
f 00 (a)
f 000 (a)
f (n) (a)
(x−a)+
(x−a)2 +
(x−a)3 +· · ·+
(x−a)n .
1!
2!
3!
n!
We define the nth Taylor approximation of f (x) at x = a to be the
polynomial
Tn (x) = f (a)+
f 00 (a)
f 000 (a)
f (n) (a)
f 0 (a)
(x−a)+
(x−a)2 +
(x−a)3 +· · ·+
(x−a)n .
1!
2!
3!
n!
Example 8.8.1
Use the process of successive differentiation to write the polynomial
p(x) = x2 + x + 2
in the form
p(x) = c0 + c1 (x − 2) + c2 (x − 2)2 .
1
Solution.
We have
p(x) =x2 + x + 2,
c0 =p(2) = 8
0
p (x) =2x + 1,
c1 =p0 (2) = 5
p00 (x) =2,
c2 =
p00 (2)
=1
2!
Thus,
p(x) = 8 + 5(x − 2) + (x − 2)2
Example 8.8.2
Find the fourth degree Taylor polynomial approximating
f (x) =
1
1+x
near a = 0.
Solution.
We have
1
,
x+1
f 0 (x) = − (x + 1)−2 ,
f (x) =
c0 = f (0) = 1
c1 = f 0 (0) = −1
f 00 (0)
2
= =1
2!
2
f 000 (0)
c3 =
= −1
3!
f (4) (0)
=1
c4 =
4!
f 00 (x) =2(x + 1)−3 ,
c2 =
f 000 (x) = − 6(x + 1)−4 ,
f (4) (x) =24(x + 1)−5 ,
Thus,
T4 (x) = 1 − x + x2 − x3 + x4
Example 8.8.3
Find the third degree Taylor polynomial approximating
f (x) = arctan x,
near a = 0.
2
Solution.
We have
f (x) = arctan x,
1
,
f 0 (x) =
1 + x2
c0 = f (0) = 0
c1 = f 0 (0) = 1
f 00 (x) = − (1 + x2 )−2 (2x),
f 00 (0)
=0
2!
f 000 (0)
1
c3 =
=−
3!
3
c2 =
f 000 (x) =2(1 + x2 )−3 (4x2 ) − 2(1 + x2 )−2 ,
Thus,
1
T3 (x) = x − x3
3
Example 8.8.4
Find the fifth degree Taylor polynomial approximating
f (x) = ln (1 + x),
near a = 0.
Solution.
We have
f (x)
f 0 (x)
f 00 (x)
= ln (1 + x), c0 =
1
=
c1 =
1+x ,
1
= − (1+x)2 , c2 =
f 000 (x)
=
f (4) (x) =
f (5) (x) =
2
,
(1+x)3
6
− (1+x)
4,
24
,
(1+x)5
c3 =
c4 =
c5 =
f (0) =
0
0
f (0) =
1
f 00 (0)
= − 12
2!
f 000 (0)
3!
f (4) (0)
4!
f (5) (0)
5!
=
=
=
Thus,
1
1
1
1
T5 (x) = x − x2 + x3 − x4 + x5
2
3
4
5
Example 8.8.5
Find the fourth degree Taylor polynomial approximating
f (x) = sin x,
near a = π2 .
3
1
3
1
−4
1
5
Solution.
We have
f (x)
f 0 (x)
=
=
f 00 (x)
= − sin x, c2 =
f 000 (x)
= − cos x, c3 =
f (4) (x) =
sin x,
cos x,
sin x,
c0 =
c1 =
c4 =
f ( π2 ) =
f 0 ( π2 ) =
f 00 ( π2 )
2!
f 000 ( π2 )
3!
f (4) ( π2 )
4!
1
0
= − 12
=
0
=
1
24
Thus,
1
π
1
π
T4 (x) = 1 − (x − )2 + (x − )4
2
2
24
2
Example 8.8.6
Suppose that the function f (x) is approximated near x = 0 by a sixth degree
Taylor polynomial
T6 (x) = 3x − 4x3 + 5x6 .
Find the value of the following:
(a) f (0) (b) f 0 (0) (c) f 000 (0) (d) f (5) (0)
(e) f (6) (0)
Solution.
If
T6 (x) = c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + c5 x5 + c6 x6
then c0 = 0, c1 = 3, c2 = 0, c3 = −4, c4 = c5 = 0, and c6 = 5.
(a) f (0) = c0 = 0, (b) f 0 (0) = c1 = 3, (c) f 000 (0) = 3!c3 = −24, (d)
f (5) (0) = 5!c5 = 0, (e) f (6) (0) = 6!c6 = 3600
Example 8.8.7
Let g(x) be a function such that g(5) = 3, g 0 (5) = −1, g 00 (5) = 1 and g 000 (5) =
−3.
(a) What is the Taylor polynomial of degree 3 for g(x) near 5?
(b) Use (a) to approximate g(4.9).
Solution
00
(a) We have: c0 = g(5) = 3, c1 = g 0 (5) = −1, c2 = g 2!(5) = 12 , and c3 =
g 000 (5)
= − 12 . Thus, T3 (x) = 3 − (x − 5) + 21 (x − 5)2 − 12 (x − 5)3 .
3!
(b) g(4.9) = 3 − (4.9 − 5) + 21 (4.9 − 5)2 − 21 (4.9 − 5)3 = 3.1675
Example 8.8.8
Use Taylor’s theorem to approximate sin 3◦ to four decimal places accuracy;
that is, the magnitude of the error is less than 0.5 × 10−4 .
4
Solution.
Since sin 3◦ is close to 0, we approximate sin x by a Taylor polynomial near
x = 0. Thus, by Taylor Theorem we have sin x = Tn (x) + Rn (x). Replace x
π
by 60
= 3◦ to obtain
π π n+1
1
·
< 0.5 × 10−4 .
≤
Rn
60
60
(n + 1)!
Solving for n we find n = 2. Hence, T2 (x) = x and
π
π
sin 3◦ = sin
≈
≈ 0.0523
60
60
5