Math 3210-3
HW 21
Solutions
Taylor’s Theorem
1. √
Use the example from class to approximate the following: (Note: The example dealt with f (x) =
1 + x.)
√
(a) 3
√
√
2
3
We will use p3 (x) = 1 + x2 − x8 + x16 to approximate 3. So 3 ∼ p3 (2) = 2.
√
(b) 1.2
√
1.2 ∼ p2 (0.2) = 1.0955.
2. Assuming we know the familiar derivative formulas for the trigonometric functions, let f (x) = sin x.
(a) Find p6 for f at x = 0.
x3
x5
p6 (x) = x −
+ .
3!
5!
(b) How accurate is this on the interval [−1, 1]?
7
For x ∈ [−1, 1], there is some c ∈ [−1, 1] such that |R6 (x)| = − cos(c)
x
≤
7!
1
7!
= 1.984 × 10−4 .
3. ♣ Show that if x ∈ [0, 1], then
x−
x2
x3
x4
x2
x3
+
−
≤ log(1 + x) ≤ x −
+ .
2
3
4
2
3
x3
x2
+
and
2
3
x4
− 4 . Also we have for x ∈ [0, 1] there is some c ∈ [0, 1] such that R3 (x) =
When we compute the derivatives of f (x) = log(1 + x) at x = 0, we get p3 (x) = x −
p4 (x) = x −
x2
2
+
x3
3
and we see R3 (x) ≤ 0, thus f (x) < p3 (x). Similarly we have R4 (x) = 24(1+c)
5!
x3
x4
x2
x3
x2
+
−
≤ log(1 + x) ≤ x −
+ .
which implies that f (x) > p4 (x). Therefore x −
2
3
4
2
3
−6(1+c)−4 4
x ,
4!
−5
x5 ≥ 0
The Riemann Integral
4. ♣ Theorem 33 revised says: Let S and T be nonempty subsets of R with s ≤ t for all s ∈ S and t ∈ T .
Then sup S ≤ inf T . Use this theorem to give another proof of Theorem 96. Specify the sets S and T
that you use.
Proof: Let f : [a, b] → R, and let S = {L(f, P ) : P is a partition of [a, b]} and T = {U (f, P ) :
P is a partition of [a, b]}. In a proof we did in class, we mentioned that for any partition Q of [a, b],
L(f, P ) ≤ U (f, Q) for all partitions P of [a, b]. Since this holds for every partition Q of [a, b], we have
s ≤ t for all s ∈ S and t ∈ T . Thus by Theorem 33, sup S = L(f ) ≤ inf T = U (f ).
˜
n−1
1 2
,1 .
5. ♣ Let f (x) = x for x ∈ [0, 1]. Find L(f ) and U (f ) using the partitions Pn = 0, , , . . . ,
n n
n
n2 (n + 1)2
.
Hint: You may use the formula 13 + 23 + · · · + n3 =
4
3
3
i
and mi (f, Pn ) =
Proof: Given the above partitions Pn for n ∈ N, notice that Mi (f, Pn ) =
n
3
1
i−1
and ∆xi = . Thus we have
n
n
U (f, Pn ) =
n
X
Mi (f, Pn )∆xi
i=1
=
=
=
So U (f ) ≤ lim U (f, Pn ) =
n→∞
X i 3 1 n
n
1 X 3
i
n4
n2 (n + 1)2
4n4
1
.
4
On the other hand we have
L(f, Pn ) =
n
X
mi (f, Pn )∆xi
i=1
=
=
=
So L(f ) ≥ lim L(f, Pn ) =
n→∞
1
4.
X i − 1 3 1 n
n
1 X
(i − 1)3
n4
(n − 1)2 n2
4n4
1
. By Theorem 96, we know
4
1
4
≤ L(f ) ≤ U (f ) ≤ 41 . Thus U (f ) = L(f ) =
˜
6. Let f be integrable on [a, b] and suppose that g is a function on [a, b] such that g(x) = f (x) except for
Rb
Rb
finitely many x ∈ [a, b]. Show that g is integrable and that a f = a g.
Proof: If we can prove this is true for g(x) = f (x) for all x ∈ [a, b] except at one u ∈ [a, b], then a
simple induction argument will give us the actual result we want. Since f and g are integrable on [a, b]
there is some B ∈ R with B > 0 such that |f (x)| ≤ B and |g(x)| ≤ B for all x ∈ [a, b]. Let ǫ > 0. Then
ǫ
there exists a partition P = {x0 , x1 , . . . , xn } of [a, b] such that U (f, P ) − L(f, P ) < 3ǫ and ∆xi < 12B
for all i = 1, . . . n. Since u ∈ [a, b], u belongs to at most two intervals of the form [xi−1 , xi ], so
n
X
4Bǫ
ǫ
[Mi (g, P ) − Mi (f, P )]∆xi ≤ 2[B + B] · max{∆xi : i = 1, 2 . . . , n} <
= .
U (g, P ) − U (f, P ) =
12B
3
i=1
Likewise
So
ǫ
L(g, P ) − L(f, P ) < .
3
U (g, P ) − L(g, P ) = U (g, P ) − U (f, P ) − L(g, P ) + L(f, P ) + U (f, P ) − L(f, P )
≤ |U (g, P ) − U (f, P ) − L(g, P ) + L(f, P ) + U (f, P ) − L(f, P )|
≤ |U (g, P ) − U (f, P )| + |L(g, P ) − L(f, P )| + |U (f, P ) − L(f, P )|
ǫ
ǫ
ǫ
+ +
<
3 3 3
= ǫ.
b
ǫ
Hence by Theorem 97 g is integrable. Furthermore we have
g ≤ U (g, P ) < U (f, P ) + < L(f, P ) +
3
aZ
Z b
Z b
Z b
Z b
b
2ǫ
2ǫ
2ǫ
≤
f + . Similarly we have
g>
f − . Thus
g=
f.
3
3
3
a
a
a
a
a
Z
˜
7. ♣ Show that if f is integrable on [a, b], then f is integrable on every interval [c, d] ⊆ [a, b].
Proof: We will show if a < c < b and f is integrable on [a, b] then f is integrable on [a, c] and [c, b].
Then applying this result twice, we would get f is integrable on any [c, d] ⊆ [a, b].
If ǫ > 0 there exists a partition P = {t1 , t2 , . . . , tn } of [a, b] such that U (f, P ) − L(f, P ) < ǫ. We might
as well assume c = tj ∈ P . (Otherwise let Q be the partition of [a, b] given by Q = P ∪ {c}. Then
U (f, Q) − L(f, Q) ≤ U (f, P ) − L(f, P ) < ǫ.)
Now P ′ = {t1 , . . . tj } is a partition of [a, c] and P ′′ = {tj , . . . tn } is a partition of [c, b]. Since
L(f, P ) =
L(f, P ′ ) + L(f, P ′′ ) and
U (f, P ) =
U (f, P ′ ) + U (f, P ′′ )
we have [U (f, P ′ ) − L(f, P ′ )] + [U (f, P ′′ ) − L(f, P ′′ )] = U (f, P ) − L(f, P ) < ǫ. Since each term inside
the brackets is nonnegative, each term in the brackets must be less then ǫ. Thus f is integrable on
[a, c] and [c, b].
˜