# Formulas for the Remainder Term in Taylor Series

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Formulas for the Remainder Term in Taylor Series
In Section 10.7 we considered functions f with derivatives of all orders and their Taylor
series
f na
n! x an
n0
The nth partial sum of this Taylor series is the nth-degree Taylor polynomial of f at a:
Tnx f a f a
f a
f na
x a x a2 x an
1!
2!
n!
We can write
f x Tnx Rnx
where Rnx is the remainder of the Taylor series. We know that f is equal to the sum of
its Taylor series on the interval x a R if we can show that lim n l Rnx 0 for
x a R.
Here we derive formulas for the remainder term Rnx. The first such formula involves
an integral.
1
Theorem If f n1 is continuous on an open interval I that contains a, and x is
in I , then
Rnx 1
n!
y
x
a
x t n f n1t dt
Proof We use mathematical induction. For n 1,
R1x f x T1x f x f a f ax a
and the integral in the theorem is xax x t f t dt. To evaluate this integral we integrate
by parts with u x t and dv f t dt, so du dt and v f t. Thus
y
x
a
x
x t f t dt x t f t tx
ta y f t dt
a
0 x a f a f x f a
(by FTC 2)
f x f a f ax a R1x
The theorem is therefore proved for n 1.
Now we suppose that Theorem 1 is true for n k, that is,
Rk x 1
k!
y
x
a
x tk f k1t dt
We want to show that it’s true for n k 1, that is
Rk1 x 1
k 1!
y
x
a
x tk1 f k2t dt
1
2
❙❙❙❙
FORMULAS FOR THE REMAINDER TERM IN TAYLOR SERIES
Again we use integration by parts, this time with u x t k1 and dv f k2t. Then
du k 1x t k dt and v f k1t, so
1
k 1!
y
x
a
x tk1 f k2t dt
1
x t k1 f k1t
k 1!
0
tx
ta
k1
k 1!
1
1
x a k1 f k1a k 1!
k!
x
y x t
k
a
x
y x t
a
k
f k1t dt
f k1t dt
f k1a
x a k1 Rk x
k 1!
f x Tkx f k1a
x a k1
k 1!
f x Tk1x Rk1 x
Therefore, (1) is true for n k 1 when it is true for n k. Thus, by mathematical
induction, it is true for all n.
To illustrate Theorem 1 we use it to solve Example 4 in Section 10.7.
EXAMPLE 1 Find the Maclaurin series for sin x and prove that it represents sin x for all x.
SOLUTION We arrange our computation in two columns as follows:
f x sin x
f 0 0
f x cos x
f 0 1
f x sin x
f 0 0
f x cos x
f 0 1
f 4x sin x
f 40 0
Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as
follows:
f 0 f 0
f 0 2
f 0 3
x
x x 1!
2!
3!
x
x3
x5
x7
x 2n1
1n
3!
5!
7!
2n 1!
n0
With a 0 in Theorem 1, we have
Rnx 1
n!
y
x
0
x tn f n1t dt
FORMULAS FOR THE REMAINDER TERM IN TAYLOR SERIES
❙❙❙❙
3
Since f n1t is sin t or cos t, we know that f n1t 1 for all t. We use the fact
(see Property 9 in Section 6.3) that, for a b,
y
b
a
f t dt y
b
a
f t dt
Thus, for x 0,
1
R x n! y
n
1
n!
x
0
y
x
0
x tn f n1 t dt x tn dt 1
n!
y
x
0
x tn f n1 t dt
1 x n1
x n1
n! n 1
n 1!
For x 0 we can write
Rnx so
Rnx 1
n!
y 0
x
xt
n
1
n!
y
0
x
x tn f n1t dt
f n1 t dt 1
n!
y
0
x
t xn dt x n1
n 1!
Thus, in any case, we have
x
R x n 1!
n1
n
The right side of this inequality approaches 0 as n l (see Equation 10.7.10), so
Rnx l 0 by the Squeeze Theorem. It follows that Rnx l 0 as n l , so sin x is
equal to the sum of its Maclaurin series.
For some purposes the integral formula in Theorem 1 is awkward to work with, so we
are going to establish another formula for the remainder term. To that end we need to prove
the following generalization of the Mean Value Theorem for Integrals (see Section 7.5).
2 Weighted Mean Value Theorem for Integrals If f and t are continuous on a, b and
t does not change sign in a, b, then there exists a number c in a, b such that
y
b
a
b
f xtx dx f c y tx dx
a
Proof Because t doesn’t change sign, either tx 0 or tx 0 for a x b. For the
sake of definiteness, let’s assume that tx 0.
By the Extreme Value Theorem (5.2.3), f has an absolute minimum value m and an
absolute maximum value M , so m f x M for a x b. Since tx 0, we have
mtx f xtx Mtx
a
x
b
4
❙❙❙❙
FORMULAS FOR THE REMAINDER TERM IN TAYLOR SERIES
and so
b
b
b
m y tx dx y f xtx dx M y tx dx
3
a
a
a
If xab tx dx 0, these inequalities show that xab f xtx dx 0 and so Theorem 2 is
true because both sides of the equation are 0. If xab tx dx 0, it must be positive and
we can divide by xab tx dx in (3):
m
xab f xtx dx
xab tx dx
M
Then, by the Intermediate Value Theorem (2.5.9), there exists a number c in a, b such
that
f c 4
xab f xtx dx
xab tx dx
y
and so
b
a
b
f xtx dx f c y tx dx
a
Theorem If f n1 is continuous on an open interval I that contains a, and x is
in I , then there exists a number c between a and x such that
Rnx f n1c
x a n1
n 1!
Proof The function tt x t n doesn’t change sign in the interval from a to x, so the
Weighted Mean Value Theorem for Integrals gives a number c between a and x such that
y
x
a
x
x t n f n1t dt f n1c y x t n dt
a
f
x t n1
c
n1
n1
tx
f n1c
ta
x a n1
n1
Then, by Theorem 1,
Rnx 1
n!
y
x
a
x t n f n1t dt
1 n1
x a n1
f n1c
f
c
x a n1
n!
n1
n 1!
The formula for the remainder term in Theorem 4 is called Lagrange’s form of the
remainder term. Notice that this expression
Rnx f n1c
x a n1
n 1!
is very similar to the terms in the Taylor series except that f n1 is evaluated at c instead
of at a. All we can say about the number c is that it lies somewhere between x and a.
In the following example we show how to use Lagrange’s form of the remainder term
as an alternative to the integral form in Example 1.
FORMULAS FOR THE REMAINDER TERM IN TAYLOR SERIES
❙❙❙❙
EXAMPLE 2 Prove that Maclaurin series for sin x represents sin x for all x.
SOLUTION Using the Lagrange form of the remainder term with a 0, we have
Rnx f n1c n1
x
n 1!
where f x sin x and c lies between 0 and x. But f n1c is sin c or cos c. In any
case, f n1c 1 and so
f
c
x
R x n 1! x n 1!
n1
n1
n1
n
By Equation 10.7.10 the right side of this inequality approaches 0 as n l , so
R x l 0 by the Squeeze Theorem. It follows that R x l 0 as n l , so sin x is
n
n
equal to the sum of its Maclaurin series.
EXAMPLE 3
3
(a) Approximate the function f x s
x by a Taylor polynomial of degree 2 at a 8.
(b) How accurate is this approximation when 7 x 9?
SOLUTION
3
f x s
x x 13
(a)
f 8 2
f x 13 x23
f 8 121
f x 29 x53
1
f 8 144
83
f x 10
27 x
Thus the second-degree Taylor polynomial is
T2 x f 8 f 8
f 8
x 8 x 82
1!
2!
1
2 121 x 8 288
x 82
The desired approximation is
1
3
x T2 x 2 121 x 8 288
x 82
s
(b) Using the Lagrange form of the remainder term we can write
R2 x 3
f c
5x 83
83 x 8
x 83 10
27 c
3!
3!
81c 83
where c lies between 8 and x. In order to estimate the error we note that if 7 x 9,
then 1 x 8 1, so x 8 1 and therefore x 8 3 1. Also, since x 7,
we have
c 83 7 83 179
and so
5 x8
R x 81c 3
2
83
51
0.0004
81 179
Thus if 7 x 9, the approximation in part (a) is accurate to within 0.0004.
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