MA 1: PROBLEM SET NO. 7 SOLUTIONS
1.
√
(1) Obtain the number r = 15 − 3 as an approximation to the
nonzero root of the equation x2 = sin(x) by using the cubic
Taylor polynomial approximation to sin(x).
Proof. The cubic Taylor polynomial approximation of sin(x)
(around 0) is
T3,0 sin(x) = x −
x3
.
3!
Thus, to find a nonzero root of the equation x2 = sin(x) using
this approximation, we need to find solutions to
x3
1
+ x2 − x = x(x2 + 6x − 6).
3!
6
Using the quadratic formula, the quadratic polynomial has roots
√
√
−6 ± 36 + 24
x=
= −3 ± 15.
2
√
In particular, r = 15 − 3 is such a root.
(2) Show that the approximation in the previous part satisfies the
inequality
| sin(r) − r2 | <
1
,
200
√
given that 15 − 3 < 0.9. Is the difference (sin(r) − r2 ) positive
or negative? Give full details of your reasoning.
Proof. Since the fourth derivative of sin(x) at zero is zero, the
third and fourth degree Taylor polynomials of sin(x) around
zero coincide. Therefore, using the error for the Taylor expansion, we have sin(r)−T3,0 sin(r) = sin(r)−T4,0 sin(r) = E4 (r) =
1
sin(5) (c) 5
r
5!
for some number c between 0 and r. Hence:
| sin(r) − r2 | = |T3,0 sin(r) + E4 (r) − r2 |
= |E4 (r)|(since r is chosen to be a root of T3,0 sin(r) − r2 )
sin(5) (c) 5
r |
=|
5!
(0.9)5
1
<
<
.
5!
200
We have sin(r) − r2 > 0 sine the error E4 (r) is a positive
number for r < 0.9.
2.
Use proper Taylor polynomial approximations to evaluate the
following limits:
(1) limx→0
sin(ax)
sin(bx)
= ab .
Solution. Using the Taylor polynomial approximation at 0, for
any constant c, we have
sin(cx) = cx −
(cx)3 (cx)5
+
− ···
3!
5!
and so
ax(1 − (ax)2 /3! + (ax)4 /5! − · · · )
sin(ax)
= lim
x→0
x→0 sin(bx)
bx(1 − (bx)2 /3! + (bx)4 − · · · )
a
= .
b
lim
(2) limx→0
sin(2x)−2x
x3
= − 43 .
Solution. We have
sin(2x) − 2x
−(2x)3 /3! + (2x)5 /5! − (2x)7 /7! + · · ·
=
lim
x→0
x→0
x3
x3
−8
=
6
4
=− .
3
lim
(3) limx→0
ax −1
2x −1
=
log a
.
log 2
2
Solution. We have
ax − 1 = ex log a − 1
= −1 + 1 + x log a + (x log a)2 /2! + (x log a)3 /3! + · · ·
= (x log a) + (x log a)2 /2! + (x log a)3 /3! + · · ·
Thus,
ax − 1
(x log a) + (x log a)2 /2! + (x log a)3 /3! + · · ·
lim
= lim
x→0 2x − 1
x→0 (x log 2) + (x log 2)2 /2! + (2 log a)3 /3! + · · ·
log a
.
=
log 2
(4) limx→0
cosh(x)−cos(x)
x2
= 1.
Solution. Recall that
cosh(x) = 1 +
x 2 x4
+
+ ···
2!
4!
cos(x) = 1 −
x2 x4
+
− ···
2!
4!
and
Therefore,
cosh(x) − cos(x)
(2x2 )/2! + 0 + 2x6 /6! + · · ·
=
lim
x→0
x→0
x2
x2
= lim 1 + 0 + 2x4 /6 + · · ·
lim
x→0
= 1.
1
(5) limx→1 x 1−x = e−1 .
1
Solution. We have x1/(1−x) = exp( 1−x
log x). Note that
lim
x→1
log x
= −1
1−x
3
(say, by L’Hopital’s rule). Thus,
log x
1/(1−x)
lim x
= lim exp
x→1
x→1
1−x
2
1
log x
log x
1−x
= lim 1 +
+
+ ···
x→1
1−x
2!
= 1 + (−1) + (−1)2 /2! + (−1)3 /3! + · · ·
= exp(−1).
3.
(1) Suppose f and g are two functions. Give a definition of the
function h(x) := f (x)g(x) whenever this makes sense. If f, g are
differentiable, compute the derivative of h wherever it exists.
(2) Use part (a) to differentiate the following functions:
• 3x
• (log x)log x
Proof.
(1) For h(x) being well defined, f (x) must be positive. Then
we can consider a function H(x) = ln(h(x)) which is equal to
0 (x)
ln(f (x)g(x) ) = g(x)ln(f (x)). We can easily see H 0 (x) = hh(x)
=
0
(x)
g 0 (x)ln(f (x)) + g(x) ff (x)
.
0
(x)
by mulTherefore h0 (x) = g 0 (x)h(x)ln(f (x)) + h(x)g(x) ff (x)
tiplying h(x) on both sides.
(2) • 3x By using the formula in (1), the derivative of this function is 3x ln(3)
• (logx)logx The derivative of this function is
ln(logx)(logx)logx (logx)logx logx
+
x
xlogx
4. Let f : (a, b) → R be a continuous and strictly monotone function.
Show that
f ((a, b)) = {f (x) : x ∈ (a, b)}
is an open interval (possibly unbounded). (Hint: distinguish between
the cases f is bounded and f is not bounded, and use the intermediate
value theorem.
4
Proof. Denote A = f ((a, b)). Now let’s first prove the lemme.
Lemma. If m, n ∈ A and m < n, then (m, n) ∈ A.
Proof. Since m, n ∈ A, there are x, y ∈ (a, b) satisfying f (x) = m, f (y) =
n. If f is strictly increasing, x < y. For arbitrary c between m and
n, by intermediate value theorem, there exists z ∈ (x, y) satisfying
f (z) = c. Therefore c ∈ A, hence (m, n) ∈ A. (When f is strictly decreasing, there is z ∈ (y, x) satisfying f (z) = c from the intermediate
value theorem.)
Now let’s consider supremum of A and infimum of A. There are four
possible cases:
(1) supA = ∞, inf A = t for some t ∈ R
(2) supA = ∞, inf A = −∞
(3) supA = s for some s ∈ R, inf A = t for some t ∈ R
(4) supA = s for some s ∈ R, inf A = −∞
Now let’s focus on the first case. Let’s show that A = (t, ∞) which
is an open interval. For arbitrary c ∈ (t, ∞), since t is infimum, there
is some d ∈ A satisfying t < d < c. Also there is some M ∈ A such
that c < M . From the lemma, c ∈ (d, M ) ∈ A. Therefore c ∈ A for
arbitrary c. This means (t, ∞) ∈ A.
Conversely, from the definition of supremum and infimum, A ∈
[t, ∞). Moreover if t ∈ A, there exits x ∈ (a, b) satisfying f (x) = t.
If f is increasing f ( a+x
) ∈ A but it is smaller than f (x) = t. If f
2
b+x
is decreasing f ( 2 ) ∈ A but it is smaller than f (x) = t. Therefore
A ∈ (t, ∞).
Conclusively, A = (t, ∞). Other three cases can be proved similarly.
5. In the following, let f be a real-valued function defined in some
open neighborhood of 0. Let α > 0 be a rational number.
(1) Prove that if f satisfies |f (x)| ≤ |x|α locally about 0 for some
α > 1, then f is differentiable at 0.
(2) Prove that if f satisfies |f (x)| ≥ |x|α locally about 0 for some
0 < α < 1 and f (0) = 0, then f is not differentiable at 0.
Proof.
(1) If we put x = 0 on the given inequality, |f (0)| ≤ 0, hence
(0)
f (0) = 0. Now let’s show that limx→0 f (x)−f
= limx→0 f (x)
=
x−0
x
0.
f (x)
|x|α
|
|≤
= |x|α−1
x
|x|
5
Now taking limx→0 of both sides, the right hand side goes to
zero, hence limx→0 f (x)
= 0.
x
(2)
f (x)
|x|α
1
|
|≥
= |x|α−1 = 1−α
x
|x|
|x|
Since 1 − α > 0, as |x| goes to zero, |x|1−α goes to zero,
therefore the right hand side diverges to infinity. Therefore
(0)
= limx→0 f (x)
does not exist.
limx→0 f (x)−f
x−0
x
6