4.7. TAYLOR AND MACLAURIN SERIES
105
The Taylor’s inequality states the following: If |f (n+1) (x)| ≤ M for
|x − a| ≤ d then the reminder satisfies the inequality:
|Rn (x)| ≤
M
|x − a|n+1
(n + 1)!
for |x − a| ≤ d .
Example: Find the third degree Taylor approximation for sin x at
x = 0, use it to find an approximate value for sin 0.1 and estimate its
difference from the actual value of the function.
Answer : We already found
T3 (x) = x −
x3
,
6
and
T3 (0.1) = 0.099833333 . . .
Now we have f (4) (x) = sin x and | sin x| ≤ 1, hence
1 4
0.1 = 0.0000041666 · · · < 0.0000042 = 4.2 · 10−6 .
4!
In fact the estimation is correct, the approximate value differs from the
actual value in
|R3 (0.1)| =≤
|T3 (0.1) − sin 0.1| = 0.000000083313 · · · < 8.34 · 10−8 .
4.7.3. Taylor Series. If the given function has derivatives of all
orders and Rn (x) → 0 as n → ∞, then we can write
f (x) =
∞
!
f (n) (a)
n=0
n!
(x − a)n
= f (a) + f # (a)(x − a) +
f ## (a)
(x − a)2 +
2!
f (n) (a)
··· +
(x − a)n + · · ·
n!
The infinite series to the right is called Taylor series of f (x) at x = a.
If a = 0 then the Taylor series is called Maclaurin series.
Example: The Taylor series of f (x) = ex at x = 0 is:
1+x+
∞
! xn
xn
x2
+ ··· +
+ ··· =
2
n!
n!
n=0
4.7. TAYLOR AND MACLAURIN SERIES
106
For |x| < d the remainder can be estimated taking into account that
f (n) (x) = ex and |ex | < ed , hence
|Rn (x)| <
ed
|x|n+1 .
(n + 1)!
We know that limn→∞ xn /n! = 0, so
ed
|x|n+1 = 0
n→∞ (n + 1)!
lim
hence Rn (x) → 0 as n → ∞. Consequently we can write:
x
e =
∞
!
xn
n=0
n!
=1+x+
x2 x3
xn
+
+ ··· +
+ ···
2!
3!
n!
For x = 1 this formula provides a way of computing number e:
∞
!
1
1
1
1
= 1 + 1 + + + · · · + + · · · = 2.718281828459 . . .
e=
n!
2! 3!
n!
n=0
The following are Maclaurin series of some common functions:
x
e =
∞
!
xn
n=0
n!
=1+x+
∞
!
x2 x3 x4
+
+
+ ···
2!
3!
4!
x3 x5 x7
x2n+1
=x−
+
−
+ ···
sin x =
(−1)
(2n + 1)!
3!
5!
7!
n=0
n
∞
!
x2 x4 x6
x2n
cos x =
(−1)n
=1−
+
−
+ ···
(2n)!
2!
4!
6!
n=0
ln (1 + x) = −
α
(1 + x) =
∞
!
xn
x2 x3 x4
=x−
+
−
+ ···
(−1)n
n
2
3
4
n=1
! "α #
n=0
" #
" #
α 2
α 3
x = 1 + αx +
x +
x + ···
n
2
3
n
" #
α(α − 1)(α − 2) . . . (α − n + 1)
α
.
where
=
n!
n
!
1
= (1 + x)−1 =
(−1)n xn = 1 − x + x2 − x3 + · · ·
1+x
n=0
4.7. TAYLOR AND MACLAURIN SERIES
tan−1 x =
107
!
x2n+1
x3 x5 x7
(−1)n
=x−
+
−
+ ···
(2n
+
1)!
3
5
7
n=0
Remark : By letting x = 1 in the Taylor series for tan−1 x we get
the beautiful expression:
∞
!
1 1 1
π
1
= 1 − + − + ··· =
(−1)n
.
4
3 5 7
2n
+
1
n=0
Unfortunately that series converges too slowly for being of practical use
in computing π. Since the series for tan−1 x converges more quickly for
small values of x, it is more convenient to express π as a combination
of inverse tangents with small argument like the following one:
π
1
1
= 4 tan−1 − tan−1
.
4
5
239
That identity can be checked with plain trigonometry. Then the inverse
tangents can be computed using the Maclaurin series for tan−1 x, and
from them an approximate value for π can be found.
4.7.4. Finding Limits with Taylor Series. The following example shows an application of Taylor series to the computation of limits:
ex − 1 − x
Example: Find lim
.
x→0
x2
Answer : Replacing ex with its Taylor series:
2
3
4
(1 + x + x2 + x6 + x24 + . . . ) − 1 − x
ex − 1 − x
lim
= lim
x→0
x→0
x2
x2
x2
x3
x4
+ 6 + 24 + . . .
= lim 2
x→0
x2
$
%
1 x x2
1
= lim
+ +
+ ... =
.
x→0
2 6 24
2
4.8. APPLICATIONS OF TAYLOR POLYNOMIALS
108
4.8. Applications of Taylor Polynomials
4.8.1. Applications to Physics. Here we illustrate an application of Taylor polynomials to physics.
Consider the following formula from the Theory of Relativity for
the total energy of an object moving at speed v:
m0 c2
E=&
,
2
1 − vc2
where c is the speed of light and m0 is the mass of the object at rest.
Let’s rewrite the formula in the following way:
"
#−1/2
v2
2
E = m0 c 1 − 2
.
c
Now we expand the expression using the power series of the binomial
function:
" #
" #
! "α#
α 2
α 3
α
n
(1 + x) =
x = 1 + αx +
x +
x + ··· ,
n
2
3
n=0
which for α = −1/2 becomes:
" 1#
" 1#
1
−2 2
−2 3
x +
x + ···
(1 + x)
=1− x+
2
2
3
3
1
5
= 1 − x + x2 − x3 + · · · ,
2
8
16
2 2
hence replacing x = −v /c we get the desired power series:
"
#
1 v2 3 v4
5 v6
2
E = m0 c 1 + 2 + 4 +
+ ··· .
2c
8c
16 c6
−1/2
If we subtract the energy at rest m0 c2 we get the kinetic energy:
" 2
#
1v
3 v4
5 v6
2
2
K = E − m0 c = m0 c
+
+
+ ··· .
2 c2 8 c4 16 c6
For low speed all the terms except the first one are very small and can
be ignored:
" 2#
1
1v
2
K ≈ m0 c
= m0 v 2 .
2
2c
2
That is the expression for the usual (non relativistic or Newtonian)
kinetic energy, so this tells us at low speed the relativistic kinetic energy
is approximately equal to the non relativistic one.