MATH 5105 Differential and Integral Analysis 2017 :
Solution Sheet 3
The deadline is 5pm on Thursday 2nd February. Please hand in your coursework
(solution to the feedback question) to the green coursework box on the basement. Coursework will be returned during the exercise class following the deadline.
Exercise for Feedback
1. The functions sinh(x) and cosh(x) are given by
1
sinh(x) = (ex − e−x )
2
&
1
cosh(x) = (ex + e−x ).
2
(a) Prove that sinh and cosh are differentiable and that
d
sinh(x) = cosh(x),
dx
&
d
cosh(x) = sinh(x).
dx
Proof. The exponential function is differentiable and therefore sinh and cosh
d
are differentiable. Using dx
exp = exp together with the chain rule, the
derivatives follow immediately.
(b) Show that the function
f (x) = cosh2 (x) − sinh2 (x)
is a constant.
Proof. Since f is differentiable and f 0 (x) = 2 cosh(x) sinh(x)−2 sinh(x) cosh(x) =
0 by the chain rule/product rule. This shows that f is constant. As f (0) =
cosh2 (0) − sinh2 (0) = 1 this shows that cosh2 (x) − sinh2 (x) = 1.
(c) Show that sinh is invertible.
Proof. Recall that exp(y) > 0 for all y ∈ R. This follows that cosh(x) > 0
for all x ∈ R. Hence sinh0 (x) = cosh(x) > 0 for all x ∈ R, therefore sinh is
strictly increasing and hence invertible.
1
(d) Show that sinh(R) = R.
Proof. If x > 0 then we have that exp(x) > 1 + x and exp(−x) < 1 so that
sinh(x) >
x
.
2
Let c > 0 then
sinh(0) = 0 < c < sinh(2c)
it follows by the intermediate value theorem applied to the interval [0, 2c] that
there exists an x ∈ (0, 2c) such that sinh(x) = c. Therefore R+ ⊂ sinh(R).
Now let c < 0. Since R+ ⊂ sinh(R) we can find y ∈ R+ such that sinh(y) = −c
since sinh is an odd function (that is sinh(−x) = sinh(−x)∀x ∈ R). We see
that sinh(−y) = − sinh(y) = −(−c) = c. Therefore R− ⊂ sinh(R).
Altogether we have shown that R = R− ∪ R+
0 ⊂ sinh(R) and hence that
sinh(R) = R.
(e) Prove that arcsinh(x) = sinh−1 (x) is differentiable and that
d
1
arcsinh(x) = √
dx
1 + x2
Proof. Now sinh0 (x) = cosh(x) for all x ∈ R therefore by the inverse function
theorem arcsinh is differentiable and we have
d
1
arcsinh(x) =
.
dx
cosh(arcsinh(x))
Now cosh(x) =
q
1 + sinh2 (x) (as cosh(x) ≥ 0) so that
d
1
arcsinh(x) = √
.
dx
1 + x2
Extra Exercises
2. Let g = arctan be the inverse of the function f (x) = tan(x), x ∈ − π2 , π2 . Applying
the inverse function theorem for one variable, find a formula for the derivative g 0 (y)
in terms of y.
Proof. As f (x) = tan(x), x ∈ − π2 , π2 = I and f (I) = (−∞, +∞) = R . We define
the inverse as g(y) = arctan(y), g : R → − π2 , π2 . Note that f 0 (x) = sec2 (x) > 0 so
2
f is differentiable and strictly increasing. Applying the inverse function theorem
we get for y = f (x)
g 0 (y) =
1
f 0 (x)
=
1
.
sec2 (x)
Now we have the identity cos2 (x) + sin2 (x) = 1, dividing through by cos2 (x) 6= 0
as x ∈ − π2 , π2 , we get
tan2 (x) + 1 = sec2 (x).
Therefore
g 0 (y) =
1
1
.
=
2
1 + sec (x)
1 + y2
3. (a) Find a bijective, continuously differentiable function f : R → R with f 0 (0) = 0
and a continuous inverse.
Proof. We take f : R → R given by f (x) = x3 . Now f is differentiable with
a continuous derivative f 0 (x) = 3x2 with f 0 (0) = 0. The inverse is given by
f −1 (x) = x1/3 .
As f is strictly increasing on R f is bijective and f (R) = R implies f is
surjective and hence f is bijective.
As f is differentiable, it is continuous therefore f −1 is also continuous.
(b) Let f : R+
0 → R be differentiable and decreasing. Prove or disprove : if
limx→0 f (x) = 0 then limx→0 f 0 (x) = 0.
Proof. Let f : R → R be given by f (x) = −x. Then f is differentiable and
f 0 (x) = −1 for all x and limx→0 f (x) = 0 but limx→0 f 0 (x) = −1.
4. Using the Intermediate Value Theorem, prove that a continuous function maps
intervals to intervals.
Proof. We use the following characterisation of an interval: I ⊆ R is an interval if
and only if for all x1 , x2 ∈ I with x1 < x2 we get
x1 < c < x2 =⇒ c ∈ I.
Let J = f (I) with y1 < y2 . Then we need to show that J is an interval i.e. that
for all y1 , y2 ∈ J with y1 < y2 then y1 < c < y2 =⇒ c ∈ J.
3
To this end, let us consider y1 < y2 . Then there exists x1 < x2 ∈ I such that
y1 = f (x1 ), y2 = f (y2 ). As y1 6= y2 it follows that x1 6= x2 so that either x1 < x2
or x2 > x1 . WLOG consider the case x1 < x2 . By assumption f is a continuous
function on I, so it is a continuous function on [x1 , x2 ].
Hence by the intermediate value theorem, for all c with y1 < c < y2 there exists
an a ∈ [x1 , x2 ] such that f (a) = c. This implies that c ∈ J.
5. Let f be a differentiable function on R and let
a = sup{|f 0 (x)| | x ∈ R} < 1.
Let x0 ∈ R and define recursively sn = f (sn−1 ), n ≥ 1. Prove that {sn } is convergent sequence and determine its limit.
Proof. Let a = sup{|f 0 (x)| | x ∈ R} < 1. By MVT, ∀x ∈ (a, b) =⇒
|f (b) − f (a)| ≤ |f 0 (ξ)||b − a|
or
f (x) − f (b)| ≤ a|x − y|.
Therefore f is a contractive mapping, that is it shrinks the distance between points.
Let sn = f (sn−1 ), s0 ∈ R. Then we will show that {sn } is a Cauchy sequence.
Consider
|sn+1 − sn | = |f (sn ) − f (sn−1 )|
≤ a|sn − sn−1 |
≤ an |s1 − s0 |.
Let n > m then
|sm − sn | ≤ |sm − sm−1 + sm−1 − sm−2 + · · · + sn+1 − sn |
≤ |sm − sm−1 | + |sm−1 − sm−2 | + · · · + |sn+1 − sn |
≤ (am−(n+1) + · · · + 1)|sn+1 − sn |
an (am−(n+1) + · · · + 1)|s1 − s0 |
an
.
≤
1−a
Since a < 1, limn→∞
an
1−a
= 0, then there exists N |
|sm − sn | < ε,
∀m, n > N.
Therefore {sn } is a Cauchy sequence. As the limit exists, we apply the limit laws
to conclude
s = lim sn = lim f (sn−1 ) = f lim sn−1 = f (s).
n→∞
n→∞
n→∞
4
6. Show that sin(x) ≤ x for all x ≥ 0.
Proof. Let g(x) = sin x − x. Then g(0) = 0 and g 0 (x) = cos x − 1. As cos(x) ≤ 1
we see g 0 (x) ≤ 0 that is g(x) is monotone decreasing. Hence
sin x − x = g(x) ≤ g(0) ≤ 0,
5
∀x > 0 =⇒ sin(x) ≤ x.