Chapter 2
Differential Calculus of Functions of One
Variable
2.1 Differentiable Functions of One Variable
Definition 2.1 Assume that f : T → R is a function and let t ∈ Tκ . We define f ∆ (t)
to be the number, provided it exists, with the property that for any ε > 0, there exists
a neighbourhood U of t, U = (t − δ, t + δ) ∩ T for some δ > 0, such that
| f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)| ≤ ε|σ (t) − s| for all s ∈ U.
We call f ∆ (t) the delta or Hilger derivative of f at t. We say that f is delta or
Hilger differentiable, shortly differentiable, in Tκ if f ∆ (t) exists for all t ∈ Tκ . The
function f ∆ : T → R is said to be the delta derivative or Hilger derivative, shortly
derivative, of f in Tκ .
Remark 2.2 If T = R, then the delta derivative coincides with the classical
derivative.
Theorem 2.3 The delta derivative is well defined.
Proof Let t ∈ Tκ . Suppose f 1∆ (t) and f 2∆ (t) are such that
| f (σ (t)) − f (s) − f 1∆ (t)(σ (t) − s)| ≤
ε
|σ (t) − s|
2
| f (σ (t)) − f (s) − f 2∆ (t)(σ (t) − s)| ≤
ε
|σ (t) − s|
2
and
for any ε > 0 and any s belonging to a neighbourhood U of t, U = (t − δ, t + δ) ∩ T
for some δ > 0. Hence, if s = σ (t), then
© Springer International Publishing Switzerland 2016
M. Bohner and S.G. Georgiev, Multivariable Dynamic Calculus on Time Scales,
DOI 10.1007/978-3-319-47620-9_2
23
24
2 Differential Calculus of Functions of One Variable
f (σ (t)) − f (s)
f (σ (t)) − f (s)
+
− f 2∆ (t)
| f 1∆ (t) − f 2∆ (t)| = f 1∆ (t) −
σ (t) − s
σ (t) − s
f (σ (t)) − f (s) f (σ (t)) − f (s)
∆
+
≤ f 1∆ (t) −
−
f
(t)
2
σ (t) − s
σ (t) − s
=
| f (σ (t)) − f (s) − f 1∆ (t)(σ (t) − s)| | f (σ (t)) − f (s) − f 2∆ (t)(σ (t) − s)|
+
|σ (t) − s|
|σ (t) − s|
≤
ε
ε
+
2 2
= ε.
Since ε > 0 was arbitrarily chosen, we conclude that
f 1∆ (t) = f 2∆ (t).
⊔
⊓
This completes the proof.
Remark 2.4 Let us assume that sup T < ∞ and f ∆ (t) is defined at a point t ∈ T \ Tκ
with the same definition as given in Definition 2.1. Then the unique point t ∈ T \ Tκ is
sup T. Hence, for any ε > 0, there is a neighbourhood U = (t − δ, t + δ) ∩ (T \ Tκ ),
for some δ > 0, such that
f (σ (t)) = f (s) = f (σ (sup T)) = f (sup T), s ∈ U.
Therefore, for any α ∈ R and s ∈ U , we have
| f (σ (t)) − f (s) − α(σ (t) − s)| = | f (sup T) − f (sup T) − α(sup T − sup T)|
≤ ε|σ (t) − s|,
i.e., any α ∈ R is the delta derivative of f at the point t ∈ T \ Tκ .
Example 2.5 Let f (t) = α ∈ R. We will prove that f ∆ (t) = 0 for any t ∈ Tκ .
Indeed, for t ∈ Tκ and for any ε > 0, s ∈ (t − 1, t + 1) ∩ T implies
| f (σ (t)) − f (s) − 0(σ (t) − s)| = |α − α| = 0 ≤ ε|σ (t) − s|.
Example 2.6 Let f (t) = t, t ∈ T. We will prove that f ∆ (t) = 1 for any t ∈ Tκ .
Indeed, for t ∈ Tκ and for any ε > 0, s ∈ (t − 1, t + 1) ∩ T implies
2.1 Differentiable Functions of One Variable
25
| f (σ (t)) − f (s) − 1(σ (t) − s)| = |σ (t) − s − (σ (t) − s)|
= 0 ≤ ε|σ (t) − s|.
Example 2.7 Let f (t) = t 2 , t ∈ T. We will prove that f ∆ (t) = σ (t) + t, t ∈ Tκ .
Indeed, for t ∈ Tκ and for any ε > 0, s ∈ (t − ε, t + ε) ∩ T implies |t − s| < ε and
| f (σ (t)) − f (s) − (σ (t) + t)(σ (t) − s)|
= |(σ (t))2 − s 2 − (σ (t) + t)(σ (t) − s)|
= |(σ (t) − s)(σ (t) + s) − (σ (t) + t)(σ (t) − s)|
= |σ (t) − s||t − s|
≤ ε|σ (t) − s|.
Exercise 2.8 Let f (t) =
√
t, t ∈ T, t > 0. Prove that
f ∆ (t) = √
t+
1
√
for all t ∈ Tκ ∩ (0, ∞).
σ (t)
Exercise 2.9 Let f (t) = t 3 , t ∈ T. Prove that
f ∆ (t) = (σ (t))2 + tσ (t) + t 2 for all t ∈ Tκ .
Theorem 2.10 Assume f : T → R is a function and let t ∈ Tκ . Then we have the
following.
1. If f is differentiable at t, then f is continuous at t.
2. If f is continuous at t and t is right-scattered, then f is differentiable at t with
f ∆ (t) =
f (σ (t)) − f (t)
.
µ(t)
3. If t is right-dense, then f is differentiable at t iff the limit
lim
s→t
f (t) − f (s)
t −s
exists as a finite number. In this case,
26
2 Differential Calculus of Functions of One Variable
f ∆ (t) = lim
s→t
f (t) − f (s)
.
t −s
4. If f is differentiable at t, then the “simple useful formula”
f (σ (t)) = f (t) + µ(t) f ∆ (t)
holds.
Proof 1. Assume that f is differentiable at t ∈ Tκ . Let ε ∈ (0, 1) be arbitrarily
chosen. Set
ε
.
ε∗ =
∆
1 + | f (t)| + 2µ(t)
Since f is differentiable at t, there exists a neighbourhood U of t such that
| f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)| ≤ ε∗ |σ (t) − s|.
Hence, for all s ∈ U ∩ (t − ε∗ , t + ε∗ ), we have
| f (t) − f (s)| = | f (t) + f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)
− f (σ (t)) + f ∆ (t)(σ (t) − s)|
≤ | f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)|
+| f (σ (t)) − f (t) − f ∆ (t)(σ (t) − s)|
≤ ε∗ |σ (t) − s| + | f (σ (t)) − f (t) − f ∆ (t)(σ (t) − t)
+ f ∆ (t)(s − t)|
≤ ε∗ |σ (t) − s| + | f (σ (t)) − f (t) − f ∆ (t)(σ (t) − t)|
+| f ∆ (t)||s − t|
≤ ε∗ |σ (t) − s| + ε∗ µ(t) + ε∗ | f ∆ (t)|
= ε∗ |σ (t) − s| + µ(t) + | f ∆ (t)|
= ε∗ |σ (t) − t + t − s| + µ(t) + | f ∆ (t)|
2.1 Differentiable Functions of One Variable
27
≤ ε∗ σ (t) − t + |t − s| + µ(t) + | f ∆ (t)|
= ε∗ 2µ(t) + |t − s| + | f ∆ (t)|
≤ ε∗ (2µ(t) + ε∗ + | f ∆ (t)|)
≤ ε∗ (1 + 2µ(t) + | f ∆ (t)|)
= ε,
which completes the proof.
2. Assume that f is continuous at t and t is right-scattered. By continuity, we have
lim
s→t
f (σ (t)) − f (t)
f (σ (t)) − f (s)
=
σ (t) − s
σ (t) − t
=
f (σ (t)) − f (t)
.
µ(t)
Therefore, for any ε > 0, there exists a neighbourhood U of t such that
f (σ (t)) − f (s)
f (σ (t)) − f (t)
−
≤ε
σ (t) − s
µ(t)
for all s ∈ U , i.e.,
f (σ (t)) − f (s) − f (σ (t)) − f (t) (σ (t) − s) ≤ ε|σ (t) − s|
µ(t)
for all s ∈ U . Hence,
f ∆ (t) =
f (σ (t)) − f (t)
.
µ(t)
3. Assume that t is right-dense. Let ε > 0 be arbitrarily chosen. Then f is differentiable at t iff there is a neighbourhood U of t such that
| f (t) − f (s) − f ∆ (t)(t − s)| ≤ ε|t − s| for all s ∈ U,
i.e., iff
i.e., iff
f (t) − f (s)
∆
− f (t) ≤ ε for all s ∈ U,
t −s
28
2 Differential Calculus of Functions of One Variable
lim
s→t
f (t) − f (s)
= f ∆ (t).
t −s
4. Assume that f is differentiable at t.
a. If t is right-dense, then σ (t) = t, µ(t) = 0 and
f (σ (t)) = f (t) = f (t) + µ(t) f ∆ (t).
b. If t is right-scattered, then
f ∆ (t) =
f (σ (t)) − f (t)
,
µ(t)
whereupon
f (σ (t)) = f (t) + µ(t) f ∆ (t).
⊔
⊓
This completes the proof.
Example 2.11 Let
T=
1
: n ∈ N0 ∪ {0}
2n + 1
and f (t) = σ (t) for t ∈ T. We will find f ∆ (t) for t ∈ Tκ = T \ {1}. For
t ∈ Tκ \ {0}, t =
1−t
1
, n=
, n ∈ N,
2n + 1
2t
we have
σ (t) = inf s ∈ T : s >
i.e., any point t =
1
,
2n+1
=
1
2n − 1
=
2 1−t
2t
=
t
> t,
1 − 2t
1
2n + 1
1
−1
n ∈ N, is right-scattered. At these points,
2.1 Differentiable Functions of One Variable
29
f ∆ (t) =
f (σ (t)) − f (t)
σ (t) − t
=
σ (σ (t)) − σ (t)
σ (t) − t
=2
(σ (t))2
(1 − 2σ (t))(σ (t) − t)
t 2
= 2
1−
=2
=2
=
Let now t = 0. Then
1−2t
t
2t
1−2t
1−2t
−t
t2
(1−2t)2
1−4t 2t 2
1−2t 1−2t
t2
2t 2 (1 − 4t)
1
.
1 − 4t
σ (0) = inf {s ∈ T : s > 0} = 0.
Consequently, t = 0 is right-dense. Also,
σ (h) − σ (0)
= lim
h→0
h→0
h
lim
h
1−2h
h
Therefore, f ∆ (0) = 1. Altogether, f ∆ (t) =
−0
= lim
h→0
1
1−4t
1
= 1.
1 − 2h
for all t ∈ Tκ .
Example 2.12 Let T = {n 2 : n ∈ N0 } and f (t) = t 2 , g(t) = σ√
(t) for t ∈ T. We will
find f ∆ (t) and g ∆ (t) for t ∈ Tκ = T. For t ∈ T, t = n 2 , n = t, n ∈ N0 , we have
σ (t) = inf{l 2 : l 2 > n 2 , l ∈ N0 } = (n + 1)2 = (1 +
√
t)2 > t.
Therefore, all points of T are right-scattered. We note that f and g are continuous
functions in T. Hence,
f ∆ (t) =
=
f (σ (t)) − f (t)
σ (t) − t
(σ (t))2 − t 2
σ (t) − t
= σ (t) + t
30
2 Differential Calculus of Functions of One Variable
√ 2
= 1+ t +t
√
= t +2 t +1+t
√
= 1 + 2 t + 2t
and
g ∆ (t) =
=
=
=
=
=
=
g(σ (t)) − g(t)
σ (t) − t
σ (σ (t)) − σ (t)
σ (t) − t
√
(1 + σ (t))2 − σ (t)
σ (t) − t
√
σ (t) + 2 σ (t) + 1 − σ (t)
σ (t) − t
√
1 + 2 σ (t)
σ (t) − t
√
1 + 2(1 + t)
√ 2
(1 + t) − t
√
3+2 t
√ .
1+2 t
√
Example 2.13 Let T = 4 2n + 1 : n ∈ N0 and f (t) = t 4 for t ∈ T. We will find
√
4
f ∆ (t) for t ∈ T. For t ∈ T, t = 4 2n + 1, n = t 2−1 , n ∈ N0 , we have
√
√
√
4
4
4
σ (t) = inf{ 2l + 1 : 2l + 1 > 2n + 1, l ∈ N0 }
=
=
√
4
4
2n + 3
t 4 + 2 > t.
Therefore, every point of T is right-scattered. We note that the function f is continuous in T. Hence,
2.1 Differentiable Functions of One Variable
f ∆ (t) =
=
31
f (σ (t)) − f (t)
σ (t) − t
(σ (t))4 − t 4
σ (t) − t
= (σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3
=
Exercise 2.14 Let T =
for t ∈ Tκ .
Solution 1 +
5
4
(t 4 + 2)3 + t 2 t 4 + 2 + t t 4 + 2 + t 3 .
4
√
5
n + 1 : n ∈ N0 and f (t) = t + t 3 for t ∈ T. Find f ∆ (t)
√
5
(t 5 + 1)2 + t t 5 + 1 + t 2 .
Example 2.15 Let T = Z and f be differentiable at t. Note that all points of t are
right-scattered and σ (t) = t + 1. Therefore,
f ∆ (t) =
=
f (σ (t)) − f (t)
σ (t) − t
f (t + 1) − f (t)
t +1−t
= f (t + 1) − f (t)
= ∆f (t),
where ∆ is the usual forward difference operator.
Exercise 2.16 Let T = hZ for some h > 0. Prove that
t=
1 ∆
h
f 2 (t) − ,
2
2
t2 =
h
1
1 ∆
f 3 (t) − f 2∆ (t) + h 2 ,
3
2
6
t3 =
1 ∆
1
1
f (t) − h f 3∆ (t) + h 2 f 2∆ (t),
4 4
2
4
t4 =
1
1
1
1 ∆
f (t) − h f 4∆ (t) + h 2 f 3∆ (t) − h 4 ,
5 5
2
3
30
t5 =
1 ∆
1
5
1
f (t) − h f 5∆ (t) + h 2 f 4∆ (t) − h 4 f 2∆ (t),
6 6
2
12
12
where f i (t) = t i , i = 1, . . . , 6, t ∈ T.
32
2 Differential Calculus of Functions of One Variable
Solution We note that all points of T are right-scattered and σ (t) = t + h, µ(t) = h,
t ∈ T. Then
1 ∆
h
1 (σ (t))2 − t 2
h
f 2 (t) − =
−
2
2
2 σ (t) − t
2
=
1 (σ (t) − t)(σ (t) + t)
h
−
2
σ (t) − t
2
=
1
h
(σ (t) + t) −
2
2
=
1
h
(2t + h) −
2
2
=t+
h
h
−
2
2
= t,
1 ∆
h
1
1 (σ (t))3 − t 3
h (σ (t))2 − t 2
1
f 3 (t) − f 2∆ (t) + h 2 =
−
+ h2
3
2
6
3 σ (t) − t
2 σ (t) − t
6
=
1 (σ (t) − t)((σ (t))2 + tσ (t) + t 2 )
3
σ (t) − t
−
h (σ (t) − t)(σ (t) + t)
1
+ h2
2
σ (t) − t
6
=
h
1
1
((σ (t))2 + tσ (t) + t 2 ) − (σ (t) + t) + h 2
3
2
6
=
h
1
1
(t + h)2 + t (t + h) + t 2 − (t + h + t) + h 2
3
2
6
=
1 2
h
1
(t + 2ht + h 2 + t 2 + ht + t 2 ) − (2t + h) + h 2
3
2
6
=
h2
1
1 2
(3t + 3ht + h 2 ) − ht −
+ h2
3
2
6
= t 2 + ht +
h2
h2
− ht −
3
3
2.1 Differentiable Functions of One Variable
= t 2,
1 ∆
1
1
1 (σ (t))4 − t 4
1 (σ (t))3 − t 3
f 4 (t) − h f 3∆ (t) + h 2 f 2∆ (t) =
− h
4
2
4
4 σ (t) − t
2
σ (t) − t
+
=
=
1 (σ (t) − t)((σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3 )
4
σ (t) − t
−
1 (σ (t) − t)((σ (t))2 + tσ (t) + t 2 )
h
2
σ (t) − t
+
1 2 (σ (t) − t)(σ (t) + t)
h
4
σ (t) − t
1
1
((σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3 ) − h((σ (t))2 + tσ (t) + t 2 )
4
2
+
=
1 2
1
h(t + 2ht + h 2 + t 2 + ht + t 2 ) + h 2 (2t + h)
2
4
1 3
(3t + 4t 2 h + 3th 2 + h 3 + t 3 + 2t 2 h + th 2 )
4
−
=
1
1
h (t + h)2 + t (t + h) + t 2 + h 2 (t + h + t)
2
4
13
t + 3t 2 h + 3th 2 + h 3 + t (t 2 + 2th + h 2 ) + t 3 + ht 2 + t 3
4
−
=
1 2
h (σ (t) + t)
4
1
(t + h)3 + t (t + h)2 + t 2 (t + h) + t 3
4
−
=
1 2 (σ (t))2 − t 2
h
4
σ (t) − t
1
1
1
h(3t 2 + 3ht + h 2 ) + th 2 + h 3
2
2
4
1 3
3
3
1
1
1
(4t + 6t 2 h + 4th 2 + h 3 ) − ht 2 − h 2 t − h 3 + h 2 t + h 3
4
2
2
2
2
4
33
34
2 Differential Calculus of Functions of One Variable
= t3 +
3 2
1
3
3
1
1
1
ht + h 2 t + h 3 − ht 2 − h 2 t − h 3 + h 2 t + h 3
2
4
2
2
2
2
4
= t 3,
1 ∆ h ∆ 1 2 ∆
1 (σ (t))5 − t 5
1 4
f5 − f4 + h f3 −
h =
5
2
3
30
5 σ (t) − t
−
=
=
=
h (σ (t))4 − t 4
1 (σ (t))3 − t 3
1 4
+ h2
−
h
2 σ (t) − t
3
σ (t) − t
30
1 (σ (t) − t)((σ (t))4 + t (σ (t))3 + t 2 (σ (t))2 + t 3 σ (t) + t 4 )
5
σ (t) − t
−
h (σ (t) − t)((σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3 )
2
σ (t) − t
+
1 2 (σ (t) − t)((σ (t))2 + tσ (t) + t 2 )
1 4
h
−
h
3
σ (t) − t
30
1
(t + h)4 + t (t + h)3 + t 2 (t + h)2 + t 3 (t + h)+ t 4
5
−
h
(t + h)3 + t (t + h)2 + t 2 (t + h) + t 3
2
+
1 4
1 2
h (t + h)2 + t (t + h) + t 2 −
h
3
30
1 4
t + 4t 2 h 2 + h 4 + 4t 3 h + 2t 2 h 2 + 4th 3 + t (t 3 + 3t 2 h + 3th 2 + h 3 )
5
+ t 2 (t 2 + 2th + h 2 ) + t 4 + ht 3 + t 4
=
−
h3
t + 3t 2 h + 3th 2 + h 3 + t (t 2 + 2th + h 2 ) + t 3 + ht 2 + t 3
2
+
1 2 2
1 4
h (t + 2th + h 2 + t 2 + ht + t 2 ) −
h
3
30
1 4
(t + 6t 2 h 2 + h 4 + 4t 3 h + 4th 3 + t 4 + 3t 3 h + 3t 2 h 2 + th 3
5
2.1 Differentiable Functions of One Variable
35
+ t 4 + 2t 3 h + t 2 h 2 + 2t 4 + ht 3 )
=
−
h 3
(3t + 4t 2 h + 3th 2 + h 3 + t 3 + 2t 2 h + th 2 )
2
+
1 4
1 2 2
h (3t + 3ht + h 2 ) −
h
3
30
h
1 4
(5t + 10t 3 h + 10t 2 h 2 + 5th 3 + h 4 ) − (4t 3 + 6t 2 h + 4th 2 + h 3 )
5
2
+ h2t 2 + h3t +
1 4
1 4
h −
h
3
30
= t 4 + 2t 3 h + 2t 2 h 2 + th 3 +
+ h2t 2 + h3t +
1 4
h4
h − 2ht 3 − 3t 2 h 2 − 2th 3 −
5
2
1 4
1 4
h −
h
3
30
= t 4.
The last equality is left to the reader for exercise.
Exercise 2.17 Let T = q N for some q > 1. Prove
t=
1
f ∆ (t),
1+q 2
t2 =
1
f ∆ (t),
1 + q + q2 3
t3 =
1
f ∆ (t),
1 + q + q2 + q3 4
t4 =
1
f ∆ (t),
1 + q + q2 + q3 + q4 5
t5 =
1
f ∆ (t),
1 + q + q2 + q3 + q4 + q5 6
where f i (t) = t i , i = 1, . . . , 6, t ∈ T.
Theorem 2.18 Assume f, g : T → R are differentiable at t ∈ Tκ . Then we have the
following.
36
2 Differential Calculus of Functions of One Variable
1. The sum f + g : T → R is differentiable at t with
( f + g)∆ (t) = f ∆ (t) + g ∆ (t).
2. For any constant α, α f : T → R is differentiable at t with
(α f )∆ (t) = α f ∆ (t).
3. The product f g : T → R is differentiable at t, and the “product rule”
( f g)∆ (t) = f ∆ (t)g(t) + f (σ (t))g ∆ (t) = f (t)g ∆ (t) + f ∆ (t)g(σ (t))
holds.
4. If g(t)g(σ (t)) = 0, then the quotient gf : T → R is differentiable at t, and the
“quotient rule”
f ∆
f ∆ (t)g(t) − f (t)g ∆ (t)
(t) =
g
g(t)g(σ (t))
holds.
Proof 1. Let ε > 0 be arbitrarily chosen. Since f and g are differentiable at t, there
exist neighbourhoods U1 and U2 of t so that
| f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)| ≤
ε
|σ (t) − s| for all s ∈ U1
2
|g(σ (t)) − g(s) − g ∆ (t)(σ (t) − s)| ≤
ε
|σ (t) − s| for all s ∈ U2 .
2
and
Hence, for s ∈ U1 ∩ U2 , we have
| f (σ (t)) + g(σ (t)) − f (s) − g(s) − ( f ∆ (t) + g ∆ (t))(σ (t) − s)|
≤ | f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)| + |g(σ (t)) − g(s) − g ∆ (t)(σ (t) − s)|
≤
ε
ε
|σ (t) − s| + |σ (t) − s|
2
2
= ε|σ (t) − s|,
which completes the proof.
2. Let α = 0. Assume that ε > 0 is arbitrarily chosen. Since f is differentiable at
t, there exists a neighbourhood U of t such that
2.1 Differentiable Functions of One Variable
37
| f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)| ≤
ε
|σ (t) − s| for all s ∈ U.
|α|
Hence, for s ∈ U , we have
|α f (σ (t)) − α f (s) − α f ∆ (t)(σ (t) − s)|
= |α|| f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)|
≤ |α|
ε
|α|
= ε,
which completes the proof.
3. Let ε > 0 be arbitrarily chosen. Let also [a, b] ⊂ T be such that t, σ (t) ∈ [a, b].
Set
M = max | f (t)|
t∈[a,b]
and
ε∗ =
ε
.
1 + M + |g(σ (t))| + |g ∆ (t)|
Since f is differentiable at t, there exists a neighbourhood U1 of t such that
| f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)| ≤ ε∗ |σ (t) − s| for any s ∈ U1 .
Since g is differentiable at t, there exists a neighbourhood U2 of t such that
|g(σ (t)) − g(s) − g ∆ (t)(σ (t) − s)| ≤ ε∗ |σ (t) − s| for any s ∈ U2 .
Since f is differentiable at t, f is continuous at t. Therefore, there exists a
neighbourhood U3 of t such that
| f (t) − f (s)| ≤ ε∗ for any s ∈ U3 .
Then, for any s ∈ U1 ∩ U2 ∩ U3 ∩ [a, b], we get
| f (σ (t))g(σ (t)) − f (s)g(s) − ( f ∆ (t)g(σ (t)) + f (t)g ∆ (t))(σ (t) − s)|
= | f (σ (t))g(σ (t)) − f (s)g(σ (t)) − f ∆ (t)g(σ (t))(σ (t) − s)
+ f (s)g(σ (t)) + f ∆ (t)g(σ (t))(σ (t) − s) + g ∆ (t) f (s)(σ (t) − s)
38
2 Differential Calculus of Functions of One Variable
− ( f ∆ (t)g(σ (t)) + f (t)g ∆ (t))(σ (t) − s) − f (s)g(s) − g ∆ (t) f (s)(σ (t) − s)|
= |( f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s))g(σ (t))
+ f (s)(g(σ (t)) − g(s) − g ∆ (t)(σ (t) − s)) − ( f (t) − f (s))g ∆ (t)(σ (t) − s)|
≤ |g(σ (t))|| f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)|
+ | f (s)||g(σ (t)) − g(s) − g ∆ (t)(σ (t) − s)| + | f (t) − f (s)||g ∆ (t)||σ (t) − s|
≤ ε∗ |g(σ (t))||σ (t) − s|
+ ε∗ | f (s)||σ (t) − s| + ε∗ |g ∆ (t)||σ (t) − s|
≤ ε∗ (|g(σ (t))| + M + |g ∆ (t)|)|σ (t) − s|
≤ ε|σ (t) − s|,
which completes the proof.
4. Let ε > 0 be arbitrarily chosen. Since f and g are differentiable at t, they are
continuous at t. Because g(t) = 0, there exists a neighbourhood U of t such that
|g(s)| ≥ m 1 > 0 for all s ∈ U,
for some constant m 1 . Let [a, b] ⊂ T be such that σ (t), t ∈ [a, b]. Set
M1 = sup | f (t)|,
t∈[a,b]
and
ε∗ = ε
M2 = sup |g(t)|
t∈[a,b]
m 1 |g(σ (t))g(t)|
.
1 + 2M1 M2 + |g ∆ (t)|
Since f and g are continuous at t, there exists a neighbourhood U1 of t such that
| f (t)g(s) − f (s)g(t)| ≤ ε∗ and |g(s) − g(t)| ≤ ε∗
for any s ∈ U1 . Since f is differentiable at t, there exists a neighbourhood U2
of t such that
| f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)| ≤ ε∗ |σ (t) − s| for any s ∈ U2 .
Since g is differentiable at t, there exists a neighbourhood U3 of t such that
2.1 Differentiable Functions of One Variable
|g(σ (t)) − g(s) − g ∆ (t)(σ (t) − s)| ≤ ε∗ |σ (t) − s| for any s ∈ U3 .
Hence, for s ∈ U1 ∩ U2 ∩ U3 ∩ U ∩ [a, b], we get
f (σ (t))
f (s)
f ∆ (t)g(t) − f (t)g ∆ (t)
(σ (t) − s)
g(σ (t)) − g(s) −
g(t)g(σ (t))
=
1
f (σ (t))g(t)g(s) − f (s)g(t)g(σ (t))
|g(σ (t))g(s)g(t)|
− ( f ∆ (t)g(t) − f (t)g ∆ (t))g(s)(σ (t) − s)
=
1
( f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s))g(t)g(s)
|g(σ (t))g(s)g(t)|
+ f (s)g(t)g(s) + f ∆ (t)g(t)g(s)(σ (t) − s)
− (g(σ (t)) − g(s) − g ∆ (t)(σ (t) − s)) f (s)g(t)
− g(s) f (s)g(t) − f (s)g ∆ (t)g(t)(σ (t) − s)
− f ∆ (t)g(t)g(s)(σ (t) − s) + f (t)g ∆ (t)g(s)(σ (t) − s)
=
1
( f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s))g(t)g(s)
|g(σ (t))g(s)g(t)|
− (g(σ (t)) − g(s) − g ∆ (t)(σ (t) − s)) f (s)g(t)
+ g ∆ (t)( f (t)g(s) − f (s)g(t))(σ (t) − s)
≤
1
| f (σ (t)) − f (s) − f ∆ (t)(σ (t) − s)||g(t)||g(s)|
|g(σ (t))g(s)g(t)|
+ |g(σ (t)) − g(s) − g ∆ (t)(σ (t) − s)|| f (s)||g(t)|
39
40
2 Differential Calculus of Functions of One Variable
+ |g ∆ (t)|| f (t)g(s) − f (s)g(t)||σ (t) − s|
≤
1
(ε∗ M1 M2 |σ (t) − s| + ε∗ |σ (t) − s|M1 M2
|g(σ (t))g(s)g(t)|
+ |g ∆ (t)|ε∗ |σ (t) − s|)
≤ ε∗
1
(1 + 2M1 M2 + |g ∆ (t)|)|σ (t) − s|
|g(σ (t))g(s)g(t)|
≤ ε∗
1
(1 + 2M1 M2 + |g ∆ (t)|)|σ (t) − s|
m 1 |g(σ (t))g(t)|
= ε|σ (t) − s|,
⊔
⊓
which completes the proof.
Example 2.19 Let f, g, h : T → R be differentiable at t ∈ Tκ . Then
( f gh)∆ (t) = (( f g)h)∆ (t)
= ( f g)∆ (t)h(t) + ( f g)(σ (t))h ∆ (t)
= ( f ∆ (t)g(t) + f (σ (t))g ∆ (t))h(t) + f σ (t)g σ (t)h ∆ (t)
= f ∆ (t)g(t)h(t) + f σ (t)g ∆ (t)h(t) + f σ (t)g σ (t)h ∆ (t).
Example 2.20 Let f : T → R be differentiable at t ∈ Tκ . Then
f2
∆
(t) = ( f f )∆ (t)
= f ∆ (t) f (t) + f (σ (t)) f ∆ (t)
= f ∆ (t)( f σ (t) + f (t)).
Also,
2.1 Differentiable Functions of One Variable
41
( f 3 )∆ (t) = ( f f 2 )∆ (t)
= f ∆ (t) f 2 (t) + f (σ (t))( f 2 )∆ (t)
= f ∆ (t) f 2 (t) + f σ (t) f ∆ (t)( f σ (t) + f (t))
= f ∆ (t)( f 2 (t) + f (t) f σ (t) + ( f σ )2 (t)).
We assume that
( f n )∆ (t) = f ∆ (t)
n−1
f k (t)( f σ )n−1−k (t)
k=0
for some n ∈ N. We will prove that
(f
n+1 ∆
∆
) (t) = f (t)
n
f k (t)( f σ )n−k (t).
k=0
Indeed,
( f n+1 )∆ (t) = ( f f n )∆ (t)
= f ∆ (t) f n (t) + f σ (t)( f n )∆ (t)
= f ∆ (t) f n (t) + f ∆ (t) f σ (t)( f n−1 (t) + f n−2 (t) f σ (t)
+ · · · + f (t)( f σ )n−2 (t) + ( f σ )n−1 (t)) f σ (t)
= f ∆ (t) f n (t) + f n−1 (t) f σ (t) + f n−2 (t)( f σ )2 (t) + · · · + ( f σ )n (t)
∆
= f (t)
n
f k (t)( f σ )n−k (t).
k=0
Example 2.21 Now, we consider f (t) = (t − a)m for a ∈ R and m ∈ N. We set
h(t) = t − a.
Then
h ∆ (t) = 1.
42
2 Differential Calculus of Functions of One Variable
By Example 2.20, we get
h
m ∆
∆
(t) = h (t)
=
Let now g(t) =
1
.
f (t)
m−1
k=0
m−1
h k (t)(h σ )m−1−k (t)
k=0
(t − a)k (σ (t) − a)m−1−k .
Then
g ∆ (t) = −
f ∆ (t)
,
f (σ (t)) f (t)
whereupon
1
hm
∆
(t) = −
=−
m−1
1
(t − a)k (σ (t) − a)m−1−k
(σ (t) − a)m (t − a)m k=0
m−1
k=0
1
1
.
(t − a)m−k (σ (t) − a)k+1
Exercise 2.22 Let t ∈ T be right-scattered. Find f ∆ (t) for
f (t) = 2 sin t + t 2 − 3t 3 .
Solution We have
f ∆ (t) = 2
(σ (t) − t)(σ (t) + t)
sin σ (t) − sin t
+
µ(t)
µ(t)
−3
=4
=4
=4
(σ (t) − t)((σ (t))2 + tσ (t) + t 2 )
µ(t)
sin
σ (t)−t
2
sin
µ(t)
2
sin
µ(t)
2
cos σ (t)+t
2
+ σ (t) + t − 3(σ (t))2 − 3tσ (t) − 3t 2
µ(t)
cos σ (t)+t
2
+ σ (t) + t − 3(σ (t))2 − 3tσ (t) − 3t 2
µ(t)
cos σ (t)+t
2
− 3(σ (t))2 + σ (t)(1 − 3t) + t − 3t 2 .
µ(t)
Exercise 2.23 Let t ∈ 2N0 . Find f ∆ (t) for
2.1 Differentiable Functions of One Variable
f (t) =
t 3 + t 2 − 2t
.
2t 2 + 3t + 1
Solution Here, σ (t) = 2t, µ(t) = t, t ∈ T. Then
f ∆ (t) =
(σ (t))2 + tσ (t) + t 2 + t + σ (t) − 2 (2t 2 + 3t + 1)
−
(2t 2 + 3t + 1)(2(σ (t))2 + 3σ (t) + 1)
(t 3 + t 2 − 2t)(2σ (t) + 2t + 3)
(2t 2 + 3t + 1)(2(σ (t))2 + 3σ (t) + 1)
=
(4t 2 + 2t 2 + t 2 + t + 2t − 2)(2t 2 + 3t + 1) − (t 3 + t 2 − 2t)(4t + 2t + 3)
(2t 2 + 3t + 1)(8t 2 + 6t + 1)
=
(7t 2 + 3t − 2)(2t 2 + 3t + 1) − (t 3 + t 2 − 2t)(6t + 3)
(2t 2 + 3t + 1)(8t 2 + 6t + 1)
=
14t 4 + 21t 3 + 7t 2 + 6t 3 + 9t 2 + 3t − 4t 2 − 6t − 2
(2t 2 + 3t + 1)(8t 2 + 6t + 1)
−
=
6t 4 + 3t 3 + 6t 3 + 3t 2 − 12t 2 − 6t
(2t 2 + 3t + 1)(8t 2 + 6t + 1)
8t 4 + 18t 3 + 21t 2 + 3t − 2
.
(2t 2 + 3t + 1)(8t 2 + 6t + 1)
Exercise 2.24 Let T = 3N0 . Find f ∆ (t) for
f (t) =
1 4 1 3
t − t − t 2 + 1.
4
3
Solution Here, σ (t) = 3t, µ(t) = 2t, t ∈ T. Then
f ∆ (t) =
=
1 (σ (t))3 − t 3
(σ (t))2 − t 2
1 (σ (t))4 − t 4
−
−
4 σ (t) − t
3 σ (t) − t
σ (t) − t
1
((σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3 )
4
1
− ((σ (t))2 + tσ (t) + t 2 ) − (σ (t) + t)
3
=
1
1
(27t 3 + 9t 3 + 3t 3 + t 3 ) − (9t 2 + 3t 2 + t 2 ) − 4t
4
3
= 10t 3 −
13 2
t − 4t.
3
43
44
2 Differential Calculus of Functions of One Variable
Exercise 2.25 Let T = Z. Find f ∆ (t) for
f (t) =
Solution t 2 − 4t −
1 3 5 2
t − t − 4t.
3
2
37
.
6
√
Exercise 2.26 Let T = { 4 n : n ∈ N}. Find f ∆ (t) for
f (t) =
Solution Let t ∈ T, t =
√
4
1 3
t − 2t 2 − 3t + 2.
3
n for some n ∈ N. Then n = t 4 and
√
√
√
4
4
σ (t) = inf{ l : l > 4 n, l ∈ N}
=
=
√
4
n+1
4
t 4 + 1.
Hence,
f ∆ (t) =
1 (σ (t))3 − t 3
(σ (t))2 − t 2
−2
−3
3 σ (t) − t
σ (t) − t
1
((σ (t))2 + tσ (t) + t 2 ) − 2(σ (t) + t) − 3
3
1 4
4
4
=
t + 1 + t t4 + 1 + t2 − 2
t4 + 1 + t − 3
3
=
=
1
t −2
3
4
t4 + 1 +
1 4
1
t + 1 + t 2 − 2t − 3.
3
3
2
Definition 2.27 Let f : T → R and t ∈ (Tκ )κ = Tκ . We define the second derivative of f at t, provided it exists, by
∆
2
2
f ∆ = f ∆∆ = f ∆ : Tκ → R.
n
n
Similarly, we define higher-order derivatives f ∆ : Tκ → R.
Example 2.28 Let T = Z. We will find f ∆∆ (t) for
f (t) = t 2 − 3t + 1.
We have σ (t) = t + 1 and
2.1 Differentiable Functions of One Variable
f ∆ (t) =
45
σ (t) − t
(σ (t))2 − t 2
−3
σ (t) − t
σ (t) − t
= σ (t) + t − 3
= t +1+t −3
= 2t − 2.
Hence,
f ∆∆ (t) = 2.
3
Example 2.29 Let T = 3N0 . We will find f ∆ (t) for
f (t) = t 3 + 3t 2 + t + 1.
Here σ (t) = 3t, t ∈ T. Then
f ∆ (t) =
(σ (t))2 − t 2
σ (t) − t
(σ (t))3 − t 3
+3
+
σ (t) − t
σ (t) − t
σ (t) − t
= (σ (t))2 + tσ (t) + t 2 + 3(σ (t) + t) + 1
= 9t 2 + 3t 2 + t 2 + 12t + 1
= 13t 2 + 12t + 1,
f ∆∆ (t) = 13
σ (t) − t
(σ (t))2 − t 2
+ 12
σ (t) − t
σ (t) − t
= 13(σ (t) + t) + 12
= 52t + 12.
Hence,
3
f ∆ (t) = 52.
Example 2.30 Let T = 2N0 . We will find f ∆∆ (t) for
f (t) =
t2 + 1
.
t +1
46
2 Differential Calculus of Functions of One Variable
Here σ (t) = 2t, t ∈ T. Then
∆
f (t) =
(σ (t))2 −t 2
(t
σ (t)−t
+ 1) − (t 2 + 1) σσ (t)−t
(t)−t
(t + 1)(2t + 1)
=
(σ (t) + t)(t + 1) − t 2 − 1
2t 2 + t + 2t + 1
=
3t (t + 1) − t 2 − 1
2t 2 + 3t + 1
=
3t 2 + 3t − t 2 − 1
2t 2 + 3t + 1
=
2t 2 + 3t − 1
.
2t 2 + 3t + 1
Hence,
f ∆∆ (t) =
1
2
(2t + 3t + 1)(8t 2 + 6t + 1)
−(2t 2 + 3t − 1) 2
2
σ (t) − t
(σ (t))2 − t 2
(2t 2 + 3t + 1)
+3
σ (t) − t
σ (t) − t
σ (t) − t
(σ (t))2 − t 2
+3
σ (t) − t
σ (t) − t
=
(2(σ (t) + t) + 3)(2t 2 + 3t + 1) − (2t 2 + 3t − 1)(2(σ (t) + t) + 3)
(2t 2 + 3t + 1)(8t 2 + 6t + 1)
=
(6t + 3)(2t 2 + 3t + 1) − (2t 2 + 3t − 1)(6t + 3)
(2t 2 + 3t + 1)(8t 2 + 6t + 1)
=
(6t + 3)(2t 2 + 3t + 1 − 2t 2 − 3t + 1)
(2t 2 + 3t + 1)(8t 2 + 6t + 1)
=
6(2t + 1)
(2t 2 + 3t + 1)(8t 2 + 6t + 1)
.
Exercise 2.31 Let T = 3N0 . Find f ∆∆ (t) for
f (t) =
t +2
.
t +3
Solution − 49 (t+3)(3t12 +4t+1) .
Theorem 2.32 (Leibniz Formula) Let Sk(n) be the set consisting of all possible strings
of length n, containing exactly k times σ and n − k times ∆. If
2.1 Differentiable Functions of One Variable
47
f Λ exists for all Λ ∈ Sk(n) ,
then
( f g)
∆n
n
=
k=0
⎛
⎝
f
Λ∈Sk(n)
⎞
Λ⎠
k
g∆ .
(2.1)
Proof We will use induction.
1. n = 1. Then
S0(1) = ∆, S1(1) = σ.
Hence,
1
k=0
⎛
⎝
Λ∈Sk(1)
⎞
k
f Λ⎠ g∆ =
f Λg +
Λ∈S0(1)
f Λ g∆
Λ∈S1(1)
= f ∆ g + f σ g∆
= ( f g)∆ ,
i.e., the assertion holds for n = 1.
2. Assume that the assertion is valid for some n ∈ N. We will prove that
( f g)∆
n+1
=
n+1
k=0
⎛
⎝
Λ∈Sk(n+1)
⎞
k
f Λ⎠ g∆ .
Using (2.1), we have
( f g)∆
n+1
n ∆
= ( f g)∆
⎛
=⎝
=
=
n
k=0
⎛
⎝
Λ∈Sk(n)
n
⎛⎛
n
⎛⎛
k=0
k=0
⎝⎝
Λ∈Sk(n)
⎝⎝
Λ∈Sk(n)
⎞
k
⎞∆
⎞
k
⎞∆
f Λ⎠ g∆ ⎠
f Λ⎠ g∆ ⎠
⎞
f Λσ ⎠ g ∆
k+1
⎛
+⎝
Λ∈Sk(n)
⎞
⎞
k
f Λ∆ ⎠ g ∆ ⎠
48
2 Differential Calculus of Functions of One Variable
=
n
=
n+1
k=0
k=1
⎛
⎝
Λ∈Sk(n)
⎞
f Λσ ⎠ g ∆
k+1
+
n
k=0
⎞
⎛
⎛
⎝
Λ∈Sk(n)
⎞
f Λ∆ ⎠ g ∆
k
⎞
⎛
n
k
k
⎟
⎜
⎝
f Λ∆ ⎠ g ∆
f Λσ ⎠ g ∆ +
⎝
k=0
(n)
Λ∈Sk−1
Λ∈Sk(n)
⎛
⎞
⎞
⎛
n
n+1
⎜
⎟ k
=
f Λσ ⎠ g ∆ + ⎝
f Λσ ⎠ g ∆
⎝
k=1
+
(n)
Λ∈Sk−1
n
k=1
⎛
⎝
Λ∈Sn(n)
Λ∈Sk(n)
⎞
⎛
k
f Λ∆ ⎠ g ∆ + ⎝
⎞
Λ∈S0(n)
⎞
⎛
n
⎟ k
⎜
f Λσ +
f Λ∆ ⎠ g ∆
=
⎝
k=1
(n)
Λ∈Sk−1
⎛
+⎝
=
⎛
=
n+1
⎛
k=1
k=0
Λ∈Sk(n)
f Λσ g ∆
n+1
Λ∈Sn(n)
n
⎝
⎝
f Λ∆ ⎠ g
+
⎞
k
⎞
k
f Λ⎠ g∆
Λ∈Sk(n+1)
Λ∈S0(n)
⎛
⎞
f Λ∆ g ⎠
⎜
+⎝
(n+1)
Λ∈Sn+1
⎞
⎛
⎟ n+1
f Λ⎠ g∆ + ⎝
Λ∈S0(n+1)
Λ∈Sk(n+1)
f Λ⎠ g
⊔
⊓
Example 2.33 Let f and µ be differentiable in Tκ . Then
2
f ∆σ =
fσ − fσ
µσ
2
and f σ ∆ =
fσ − fσ
.
µ
Therefore,
Also,
⎞
f Λ⎠ g∆ .
The proof is complete.
f σ∆ =
f ∆σ µ(1 + µ∆ )
f ∆σ µσ
=
= (1 + µ∆ ) f ∆σ .
µ
µ
2.1 Differentiable Functions of One Variable
3
fσ
2
∆
f
σ ∆σ
=
fσ − fσ
µ
49
2
3
2
and f σ ∆σ =
fσ − fσ
.
µσ
Therefore,
2
2
2
fσ ∆
f σ ∆µ
f σ ∆µ
=
=
=
,
µσ
µ(1 + µ∆ )
1 + µ∆
whereupon
fσ
2
∆
= (1 + µ∆ ) f σ ∆σ .
We have
3
2
f ∆σ =
fσ − fσ
µσ 2
2
2
=
f σ ∆µ
µσ 2
=
µf σ ∆
µ(1 + µ∆ )(1 − µ∆σ )
=
fσ ∆
,
(1 + µ∆ )(1 + µ∆σ )
2
2
from where
fσ
2
∆
2
= (1 + µ∆ )(1 + µ∆σ ) f ∆σ .
Example 2.34 Let
T = tn =
1
2
2n
: n ∈ N0 ∪ {0, −1}.
Then
σ (tn ) → 0 = σ (0) as n → ∞.
Therefore, σ is continuous at 0. Also,
√
σ (σ (s)) − σ (s)
σ (s) − σ (s)
= lim
lim
√
s→0
s→0
σ (s) − s
s− s
√
√
s− 4s
= lim
√
s→0 s −
s
50
2 Differential Calculus of Functions of One Variable
1
= lim √
√
s→0
s+ 4s
= ∞.
Consequently, σ is not differentiable at 0.
2.2 Mean Value Theorems
Let T be a time scale and a, b ∈ T, a < b. Let f : T → R be a function.
Theorem 2.35 If f is delta differentiable at t, then there exists a function g, defined
in a neighbourhood U of t with
lim g(s) = g(t) = 0,
s→t
such that
for all s ∈ U .
f (σ (t)) = f (s) + f ∆ (t) + g(s) (σ (t) − s)
(2.2)
Proof Define
g(s) =
⎧ f (σ (t))− f (s)
∆
⎪
⎨ σ (t)−s − f (t)
⎪
⎩
0,
if s = σ (t), t ∈ T
(2.3)
if s = σ (t), t ∈ T.
Solving (2.3) for f (σ (t)) yields (2.2). Since f is differentiable at t, we have that f
is continuous at t. Hence, g is continuous at t and
lim g(s) = g(t) = 0.
s→t
⊔
⊓
The proof is complete.
Theorem 2.36 Suppose that f has a delta derivative at each point of [a, b]. If
f (a) = f (b),
then there exist points ξ1 , ξ2 ∈ [a, b] such that
f ∆ (ξ2 ) ≤ 0 ≤ f ∆ (ξ1 ).
2.2 Mean Value Theorems
51
Proof Since f is delta differentiable at each point of [a, b], f is continuous on [a, b].
Therefore, there exist ξ1 , ξ2 ∈ [a, b] such that
m = min f (t) = f (ξ1 ) and M = max f (t) = f (ξ2 ).
t∈[a,b]
t∈[a,b]
Because f (a) = f (b), we assume that ξ1 , ξ2 ∈ [a, b).
1. If σ (ξ1 ) > ξ1 , then
f ∆ (ξ1 ) =
f (σ (ξ1 )) − f (ξ1 )
≥ 0.
σ (ξ1 ) − ξ1
2. If σ (ξ1 ) = ξ1 , then
f ∆ (ξ1 ) = lim
t→ξ1
f (t) − f (ξ1 )
≥ 0.
t − ξ1
3. If σ (ξ2 ) > ξ2 , then
f ∆ (ξ2 ) =
f (σ (ξ2 )) − f (ξ2 )
≤ 0.
σ (ξ2 ) − ξ2
4. If σ (ξ2 ) = ξ2 , then
f ∆ (ξ2 ) = lim
t→ξ2
f (t) − f (ξ2 )
≤ 0.
t − ξ2
⊔
⊓
This completes the proof.
2
Example 2.37 Let T = Z, f (t) = t . We will find ξ1 , ξ2 ∈ (−3, 3) such that
f ∆ (ξ2 ) ≤ 0 ≤ f ∆ (ξ1 ).
We have σ (t) = t + 1, µ(t) = 1, f (−3) = f (3) = 9 and
f ∆ (t) = σ (t) + t = 2t + 1.
Hence,
f ∆ (ξ2 ) = 2ξ2 + 1 ≤ 0 and f ∆ (ξ1 ) = 2ξ1 + 1 ≥ 0
hold for ξ1 ∈ {0, 1, 2} and ξ2 ∈ {−2, −1}.
t 2 + 2. We will find ξ1 , ξ2 ∈ (1, 8) such
Example 2.38 Let T = 2N0 , f (t) = t 3 − 73
9
that
f ∆ (ξ2 ) ≤ 0 ≤ f ∆ (ξ1 ).
We have
52
2 Differential Calculus of Functions of One Variable
f (1) = 1 −
and
f (8) = 512 −
73
46
73
+2=3−
=−
9
9
9
72 − 73
64
46
73
64 + 2 =
64 + 2 = − + 2. = − .
9
9
9
9
Therefore, f (1) = f (8). Also, σ (t) = 2t, µ(t) = t and
f ∆ (t) = (σ (t))2 + tσ (t) + t 2 −
= 4t 2 + 2t 2 + t 2 −
= 7t 2 −
73
(σ (t) + t)
9
73
(3t)
9
73
t.
3
Hence,
f ∆ (ξ2 ) = 7ξ22 −
73
ξ2 ≤ 0 for ξ2 = 2
3
f ∆ (ξ1 ) = 7ξ12 −
73
ξ1 ≥ 0 for ξ1 = 4.
3
and
Example 2.39 Let T = 3N0 ∪ {0}, f (t) = t 2 − 9t + 3. We will find ξ1 , ξ2 ∈ (0, 9)
so that
f ∆ (ξ2 ) ≤ 0 ≤ f ∆ (ξ1 ).
Here, σ (t) = 3t, t = 0, σ (0) = 1,
f (0) = 3 and f (9) = 3.
Then
f ∆ (t) = σ (t) + t − 9 = 3t + t − 9 = 4t − 9.
Hence,
f ∆ (ξ2 ) = 4ξ2 − 9 ≤ 0 for ξ2 = 1
and
f ∆ (ξ1 ) = 4ξ1 − 9 ≥ 0 for ξ1 = 3.
Exercise 2.40 Let T = Z, f (t) = t 3 − 16t + 1. Find ξ1 , ξ2 ∈ (−4, 4) so that
f ∆ (ξ2 ) ≤ 0 ≤ f ∆ (ξ1 ).
2.2 Mean Value Theorems
53
Solution ξ2 ∈ {−2, −1, 0, 1} and ξ1 ∈ {−3, 2, 3}.
Theorem 2.41 (Mean Value Theorem) Suppose that f is continuous on [a, b] and
has a delta derivative at each point of [a, b). Then there exist ξ1 , ξ2 ∈ [a, b) such
that
(2.4)
f ∆ (ξ1 )(b − a) ≤ f (b) − f (a) ≤ f ∆ (ξ2 )(b − a).
Proof Consider the function φ defined on [a, b] by
φ(t) = f (t) − f (a) −
f (b) − f (a)
(t − a).
b−a
Then φ is continuous on [a, b] and has a delta derivative at each point of [a, b). Also,
φ(a) = φ(b) = 0. Hence, there exist ξ1 , ξ2 ∈ [a, b) such that
φ ∆ (ξ1 ) ≤ 0 ≤ φ ∆ (ξ2 ),
i.e.,
f ∆ (ξ1 ) −
f (b) − f (a)
f (b) − f (a)
≤ 0 ≤ f ∆ (ξ2 ) −
,
b−a
b−a
and thus, we get (2.4).
Example 2.42 Let T = Z, f (t) = t 2 + 2t, [a, b] = [−2, 4]. Then σ (t) = t + 1,
f ∆ (t) = σ (t) + t + 2 = t + 1 + t + 2 = 2t + 3.
Since
f (4) = 24 and f (−2) = 0,
we have
2ξ1 + 3 ≤
24
≤ 2ξ2 + 3,
6
i.e.,
2ξ1 + 3 ≤ 4 and 2ξ2 + 3 ≥ 4,
so that
ξ1 ≤
1
1
and ξ2 ≥ ,
2
2
and therefore
ξ1 ∈ {−2, −1, 0} and ξ2 ∈ {1, 2, 3}.
Example 2.43 Let T = Z, f (t) = et . Then σ (t) = t + 1 and
f ∆ (t) =
eσ (t) − et
= et+1 − et .
σ (t) − t
⊔
⊓
54
2 Differential Calculus of Functions of One Variable
Hence, for every t > 1, there exist ξ1 , ξ2 ∈ (0, t) such that
eξ1 +1 − eξ1 ≤
et − 1
≤ eξ2 +1 − eξ2 .
t
Example 2.44 Let
T = 2N0 ∪ {0} and f (t) = sin t.
Then, for every t > 0, we have σ (t) = 2t and
f ∆ (t) =
f (σ (t)) − f (t)
σ (t) − t
=
sin(2t) − sin t
2t − t
=
t
3t
2
sin cos .
t
2
2
Hence, for every t > 1, there exist ξ1 , ξ2 ∈ (0, t) such that
3ξ1
sin t
2
3ξ2
2
ξ1
ξ2
≤
≤
.
sin cos
sin cos
ξ1
2
2
t
ξ2
2
2
Exercise 2.45 Let T = 2N0 ∪ {0}. Prove that, for every t > 1, there exist ξ1 , ξ2 ∈
(0, t) so that
7ξ12 + 3ξ1 ≤ t 2 + t ≤ 7ξ22 + 3ξ2 .
Solution Use the function f (t) = t 3 + t 2 .
Corollary 2.46 Let f be a continuous function on [a, b] that has a delta derivative
at each point of [a, b). If f ∆ (t) = 0 for all t ∈ [a, b), then f is a constant function
on [a, b].
Proof For every t ∈ [a, b], using (2.4), we have that there exist ξ1 , ξ2 ∈ [a, b) such
that
0 = f ∆ (ξ1 )(t − a) ≤ f (t) − f (a) ≤ f ∆ (ξ2 )(t − a) = 0,
i.e., f (t) = f (a).
⊔
⊓
Corollary 2.47 Let f be a continuous function on [a, b] that has a delta derivative
at each point of [a, b). Then f is increasing, decreasing, nondecreasing and nonincreasing on [a, b] if f ∆ (t) > 0, f ∆ (t) < 0, f ∆ (t) ≥ 0, f ∆ (t) ≤ 0 for all t ∈ [a, b),
respectively.
Proof Let f ∆ (t) > 0 for any t ∈ [a, b]. Then, for any t1 , t2 ∈ [a, b], t1 < t2 , there
exists ξ ∈ (t1 , t2 ) such that
2.2 Mean Value Theorems
55
f (t1 ) − f (t2 ) ≤ f ∆ (ξ )(t1 − t2 ) < 0,
i.e., f (t1 ) < f (t2 ). The cases f ∆ (t) < 0, f ∆ (t) ≥ 0 and f ∆ (t) ≤ 0 are left to the
reader for exercise.
⊓
⊔
Example 2.48 Let T = Z and f (t) = t 3 − 2t 2 − t. Then σ (t) = t + 1 and
f ∆ (t) = (σ (t))2 + tσ (t) + t 2 − 2(σ (t) + t) − 1
= (t + 1)2 + t (t + 1) + t 2 − 2(t + 1 + t) − 1
= t 2 + 2t + 1 + t 2 + t + t 2 − 4t − 2 − 1
= 3t 2 − t − 2.
Hence,
2
f (t) = 0 iff t ∈ − , 1 .
3
∆
Therefore,
f ∆ (t) ≥ 0 for t ∈ −∞, −
and
2
∪ [1, ∞)
3
2
f (t) ≤ 0 for t ∈ − , 1 .
3
∆
Consequently, f is increasing in (−∞, −1] ∪ [1, +∞) and f is decreasing in
[−1, 1].
Example 2.49 Let T = 2N0 . We will investigate the monotonicity of the function
f (t) =
1 − 2t
.
1+t
Here, σ (t) = 2t. Then
f ∆ (t) =
=
−2(1 + t) − (1 − 2t)
(1 + t)(1 + 2t)
−2 − 2t − 1 + 2t
(1 + t)(1 + 2t)
=−
3
.
(1 + t)(1 + 2t)
56
2 Differential Calculus of Functions of One Variable
Therefore, the function f is decreasing for all t ∈ T.
Example 2.50 Let T = Z. We will investigate where the function
f (t) = 4t 3 − 21t 2 + 18t + 20
is increasing and decreasing. Here, σ (t) = t + 1 and
f ∆ (t) = 4((σ (t))2 + tσ (t) + t 2 ) − 21(σ (t) + t) + 18
= 4((t + 1)2 + t (t + 1) + t 2 ) − 21(t + 1 + t) + 18
= 4(t 2 + 2t + 1 + t 2 + t + t 2 ) − 21(2t + 1) + 18
= 12t 2 + 12t + 4 − 42t − 21 + 18
= 12t 2 − 30t + 1.
Hence,
∆
f (x) = 0 for x ∈
√
√
15 + 213 15 − 213
.
,
12
12
Therefore, f ∆ (t) ≥ 0 for t ∈ (−∞, 0] ∪ [3, ∞) and f ∆ (t) ≤ 0 for t ∈ [1, 2].
Exercise 2.51 Let T = Z. Investigate where the function
f (t) = t 3 − 4t 2 − 4t + 5
is increasing and decreasing.
Solution f is increasing for t ∈ (−∞, −1] ∪ [3, ∞) and f is decreasing for t ∈
[0, 2].
2.3 Chain Rules
Theorem 2.52 (Chain Rule) Assume g : R → R is continuous, g : T → R is delta
differentiable on Tκ , and f : R → R is continuously differentiable. Then there exists
c ∈ [t, σ (t)] with
(2.5)
( f ◦ g)∆ (t) = f ′ (g(c))g ∆ (t).
Proof Fix t ∈ Tκ .
2.3 Chain Rules
57
1. If t is right-scattered, then
( f ◦ g)∆ (t) =
f (g(σ (t))) − f (g(t))
.
µ(t)
If g(t) = g(σ (t)), then
( f ◦ g)∆ (t) = 0 and g ∆ (t) = 0,
and so (2.5) holds for any c ∈ [t, σ (t)]. Assume that g(σ (t)) = g(t). Then, by
the mean value theorem,
( f ◦ g)∆ (t) =
f (g(σ (t))) − f (g(t)) g(σ (t)) − g(t)
g(σ (t)) − g(t)
µ(t)
= f ′ (ξ )g ∆ (t),
where ξ is between g(t) and g(σ (t)). Since g : R → R is continuous, there
exists c ∈ [t, σ (t)] such that g(c) = ξ .
2. If t is right-dense, then
( f ◦ g)∆ (t) = lim
s→t
= lim
s→t
= lim
s→t
f (g(t)) − f (g(s))
t −s
f (g(t)) − f (g(s)) g(t) − g(s)
g(t) − g(s)
t −s
f ′ (ξs )
g(t) − g(s)
,
t −s
where ξs is between g(s) and g(t). By the continuity of g, we get that lims→t ξs =
g(t). Therefore,
( f ◦ g)∆ (t) = f ′ (g(t))g ∆ (t).
⊔
⊓
This completes the proof.
Example 2.53 Let T = Z, f (t) = t 3 + 1, g(t) = t 2 . We have that g : R → R is
continuous, g : T → R is delta differentiable on Tκ , f : R → R is continuously
differentiable, σ (t) = t + 1. Then
g ∆ (t) = σ (t) + t
and
( f ◦ g)∆ (1) = f ′ (g(c))g ∆ (1) = 3g 2 (c)(σ (1) + 1) = 9c4 .
Here, c ∈ [1, σ (1)] = [1, 2]. Also,
(2.6)
58
2 Differential Calculus of Functions of One Variable
( f ◦ g)(t) = f (g(t)) = g 3 (t) + 1 = t 6 + 1
so that
( f ◦ g)∆ (t) = (σ (t))5 + t (σ (t))4 + t 2 (σ (t))3 + t 3 (σ (t))2 + t 4 σ (t) + t 5
and
( f ◦ g)∆ (1) = (σ (1))5 + (σ (1))4 + (σ (1))3 + (σ (1))2 + σ (1) + 1 = 63.
By (2.6), we get
63 = 9c4 , so c4 = 7, so c =
√
4
7 ∈ [1, 2].
Example 2.54 Let T = {2n : n ∈ N0 }, f (t) = t + 2, g(t) = t 2 − 1. We note that
g : T → R is delta differentiable, g : R → R is continuous and f : R → R is continuously differentiable. For t ∈ T, t = 2n , n ∈ N0 , n = log2 t, we have
σ (t) = inf 2l : 2l > 2n , l ∈ N0 = 2n+1 = 2t > t.
Therefore, all points of T are right-scattered. Since sup T = ∞, we have that Tκ = T.
Also, for t ∈ T, we have
( f ◦ g)(t) = f (g(t)) = g(t) + 2 = t 2 − 1 + 2 = t 2 + 1
and
( f ◦ g)∆ (t) = σ (t) + t = 2t + t = 3t.
Hence,
( f ◦ g)∆ (2) = 6.
(2.7)
Now, using Theorem 2.52, we get that there exists c ∈ [2, σ (2)] = [2, 4] such that
( f ◦ g)∆ (2) = f ′ (g(c))g ∆ (2) = g ∆ (2) = σ (2) + 2 = 4 + 2 = 6.
(2.8)
From (2.7) and (2.8), we find that for every c ∈ [2, 4],
( f ◦ g)∆ (2) = f ′ (g(c))g ∆ (2).
2
Example 2.55 Let T = 3n : n ∈ N0 , f (t) = t 2 + 1, g(t) = t 3 . We note that g :
R → R is continuous, g : T → R is delta differentiable and f : R → R is continu
1
2
ously differentiable. For t ∈ T, t = 3n , n ∈ N0 , n = log3 t 2 , we have
2.3 Chain Rules
59
2
2
2
σ (t) = inf 3l : 3l > 3n , l ∈ N0
2
= 3(n+1)
2
= 3 · 3n · 32n
1
2
= 3t32(log3 t ) > t.
Consequently, all points of T are right-scattered. Also, sup T = ∞. Then Tκ = T.
Hence, for t ∈ T, we have
( f ◦ g)(t) = f (g(t)) = g 2 (t) + 1 = t 6 + 1
and
( f ◦ g)∆ (t) = (σ (t))5 + t (σ (t))4 + t 2 (σ (t))3 + t 3 (σ (t))2 + t 4 σ (t) + t 5 .
Thus,
( f ◦ g)∆ (1) =(σ (1))5 + (σ (1))4 + (σ (1))3 + (σ (1))2 + σ (1) + 1
=35 + 34 + 33 + 32 + 3 + 1
(2.9)
=364.
From Theorem 2.52, it follows that there exists c ∈ [1, σ (1)] = [1, 3] such that
( f ◦ g)∆ (1) = f ′ (g(c))g ∆ (1) = 2g(c)g ∆ (1) = 2c3 g ∆ (1).
Because all points of T are right-scattered, we have
g ∆ (1) = (σ (1))2 + σ (1) + 1 = 9 + 3 + 1 = 13.
By (2.10), we find
( f ◦ g)∆ (1) = 26c3 .
From the last equation and from (2.9), we obtain
364 = 26c3 , so c3 =
√
364
3
= 14, so c = 14.
26
(2.10)
60
2 Differential Calculus of Functions of One Variable
Exercise 2.56 Let T = Z, f (t) = t 2 + 2t + 1, g(t) = t 2 − 3t. Find a constant c ∈
[1, σ (1)] such that
( f ◦ g)∆ (1) = f ′ (g(c))g ∆ (1).
Solution Any c ∈ [1, 2].
Theorem 2.57 (Chain Rule) Let f : R → R be continuously differentiable and suppose g : T → R is delta differentiable. Then f ◦ g : T → R is delta differentiable,
and the formula
∆
( f ◦ g) (t) =
1
′
f (g(t) + hµ(t)g (t))dh g ∆ (t)
0
∆
(2.11)
holds.
Proof Note that
f (g(σ (t))) − f (g(s)) =
g(σ (t))
f ′ (y)dy
g(s)
= (g(σ (t)) − g(s))
1
0
f ′ (h(g(σ (t))) + (1 − h)g(s))dh.
Let ε > 0 and t ∈ Tκ . Set
ε∗ =
ε
1+2
1
0
|
f ′ (hg(σ (t))
.
+ (1 − h)g(t))|dh
Since g is differentiable at t, there exists a neighbourhood U1 of t such that
|g(σ (t)) − g(s) − g ∆ (t)(σ (t) − s)| ≤ ε∗ |σ (t) − s| for all s ∈ U1 .
Since f ′ is continuous on R, using that
|hg(σ (t)) + (1 − h)g(s) − (hg(σ (t)) + (1 − h)g(t))| = (1 − h)|g(t) − g(s)|
≤ |g(t) − g(s)|
for h ∈ [0, 1], there exists a neighbourhood U2 of t such that
| f ′ (hg(σ (t)) + (1 − h)g(s)) − f ′ (hg(σ (t)) + (1 − h)g(t))| ≤
for all s ∈ U2 . Let
U = U1 ∩ U2 , s ∈ U.
2(ε∗
ε
+ |g ∆ (t)|)
2.3 Chain Rules
61
We put
α = hg(σ (t)) + (1 − h)g(s) and β = hg(σ (t)) + (1 − h)g(t).
Then, using the choice of ε, we have
| f ′ (α) − f ′ (β)| < 1, i.e., | f ′ (α)| < 1 + | f ′ (β)|.
Also,
( f ◦ g)(σ (t)) − ( f ◦ g)(s) − (σ (t) − s)g ∆ (t)
1
0
1
= (g(σ (t)) − g(s))
′
f (β)dh
′
∆
f (α)dh − (σ (t) − s)g (t)
0
= (g(σ (t)) − g(s) − (σ (t) − s)g ∆ (t))
0
1
1
f (β)dh
′
f ′ (α)dh
0
+ (σ (t) − s)g ∆ (t)
1
0
( f ′ (α) − f ′ (β))dh
≤ |g(σ (t)) − g(s) − (σ (t) − s)g ∆ (t)|
+ |σ (t) − s||g ∆ (t)|
1
≤ ε∗ |σ (t) − s|
1
≤ ε∗ |σ (t) − s|
0
0
1
| f ′ (α)|dh
1
| f ′ (α) − f ′ (β)|dh
| f ′ (α)|dh + |σ (t) − s||g ∆ (t)|
0
0
0
1
| f ′ (α) − f ′ (β)|dh
| f ′ (α)|dh
!
"
+ ε∗ + |g ∆ (t)| |σ (t) − s|
1
0
| f ′ (α) − f ′ (β)|dh
62
2 Differential Calculus of Functions of One Variable
≤
ε
ε
|σ (t) − s| + |σ (t) − s|
2
2
= ε|σ (t) − s|.
Therefore, f ◦ g is differentiable at t, and its derivative satisfies (2.11).
⊔
⊓
1
Example 2.58 Let T = Z, f (x) = 1+x
2 , g(t) = t + 1. Note that f : R → R is continuously differentiable and g : T → R is delta differentiable. We have
f ′ (x) = −
2x
, µ(t) = 1, g ∆ (t) = 1
(1 + x 2 )2
and
g(t) + hµ(t)g ∆ (t) = t + 1 + h.
Hence,
f ′ (g(t) + hµ(t))g ∆ (t) = f ′ (t + 1 + h)
=−
2(t + 1 + h)
.
(1 + (t + 1 + h)2 )2
From Theorem 2.57, we conclude that f ◦ g : T → R is delta differentiable and
( f ◦ g)∆ (t) = −
1
0
2(t + 1 + h)
dh
(1 + (t + 1 + h)2 )2
1
d(t + 1 + h)2
2 2
0 (1 + (t + 1 + h) )
h=1
1
=
1 + (t + 1 + h)2 h=0
=−
=
1
1
−
1 + (t + 2)2
1 + (t + 1)2
=
(t + 1)2 − (t + 2)2
(t 2 + 4t + 5)(t 2 + 2t + 3)
=
t 2 + 2t + 1 − t 2 − 4t − 4
(t 2 + 4t + 5)(t 2 + 2t + 3)
=
−2t − 3
.
(t 2 + 4t + 5)(t 2 + 2t + 3)
2.3 Chain Rules
63
Example 2.59 Let T = 2N0 , f (x) = sin x, g(t) = t 2 + 1. We have that f : R → R
is continuously differentiable, g : T → R is delta differentiable, σ (t) = 2t, µ(t) = t,
and
g ∆ (t) = σ (t) + t = 2t + t = 3t
so that
g(t) + hµ(t)g ∆ (t) = t 2 + 1 + ht (3t) = t 2 + 1 + 3t 2 h.
Moreover,
f ′ (x) = cos x,
and thus
f ′ (g(t) + hµ(t)g ∆ (t)) = cos(t 2 + 1 + 3t 2 h).
By Theorem 2.57, we conclude that f ◦ g : T → R is delta differentiable and
∆
( f ◦ g) (t) =
=
=
1
0
3t
3t 2
cos(t 2 + 1 + 3t 2 h)dh(3t)
1
0
cos(t 2 + 1 + 3t 2 h)d(t 2 + 1 + 3t 2 h)
h=1
1
sin(t 2 + 1 + 3t 2 h)
h=0
t
=
1
(sin(4t 2 + 1) − sin(t 2 + 1))
t
=
3t 2
5t 2
2
sin
cos
+1 .
t
2
2
Example 2.60 Let
T = 3N0 ,
f (x) = log(1 + x 2 ), g(t) = t 3 − 2t.
We have that f : R → R is continuously differentiable, g : T → R is delta differentiable, σ (t) = 3t, µ(t) = 2t, and
g ∆ (t) = (σ (t))2 + tσ (t) + t 2 − 2
= 9t 2 + 3t 2 + t 2 − 2
= 13t 2 − 2,
64
2 Differential Calculus of Functions of One Variable
so that
g(t) + hµ(t)g ∆ (t) = t 3 − 2t + h(2t)(13t 2 − 2)
= t 3 − 2t + (26t 3 − 4t)h.
Moreover,
f ′ (x) =
2x
,
1 + x2
and therefore
f ′ (g(t) + hµ(t)g ∆ (t)) =
2(t 3 − 2t) + 2(26t 3 − 4t)h
1 + (t 3 − 2t + (26t 3 − 4t)h)2
=2
t 3 − 2t + (26t 3 − 4t)h
.
1 + (t 3 − 2t + (26t 3 − 4t)h)2
By Theorem 2.57, we conclude that f ◦ g is delta differentiable and
t 3 − 2t + (26t 3 − 4t)h
dh(13t 2 − 2)
3
3
2
0 1 + (t − 2t + (26t − 4t)h)
13t 2 − 2 1
t 3 − 2t + (26t 3 − 4t)h
=
d(t 3 − 2t + (26t 3 − 4t))h
13t 3 − 2t 0 1 + (t 3 − 2t + (26t 3 − 4t)h)2
( f ◦ g)∆ (t) = 2
1
1 1 d(t 3 − 2t + (26t 3 − 4t)h)2
2t 0 1 + (t 3 − 2t + (26t 3 − 4t)h)2
h=1
1
log(t 3 − 2t + (26t 3 − 4t)h)
=
h=0
2t
=
=
1
log(27t 3 − 6t) − log(t 3 − 2t)
2t
=
27t 2 − 6
1
log 2
.
2t
t −2
Exercise 2.61 Let
T = Z,
f (x) = cos x, g(t) = t 2 + t.
Using Theorem 2.57, find ( f ◦ g)∆ (t).
Solution −2 sin(t + 1) sin((t + 1)2 ).
Theorem 2.62 (Chain Rule) Assume v : T → R is strictly increasing and T̃ = v(T)
˜
is a time scale. Let w : T̃ → R. If v∆ (t) and w∆ (v(t)) exist for t ∈ Tκ , then
2.3 Chain Rules
65
˜
(w ◦ v)∆ = (w∆ ◦ v)v∆ .
Proof Let ε ∈ (0, 1) be arbitrarily chosen. We put
ε∗ =
ε
1+
|v∆ (t)|
+ |w∆˜ (v(t))|
.
Note that 0 < ε∗ < 1. Since w is differentiable at t, there exists a neighbourhood U1
of t such that
|v(σ (t)) − v(t) − (σ (t) − s)v∆ (t)| ≤ ε∗ |σ (t) − s| for all s ∈ U1 .
Since w is differentiable at v(t), there exists a neighbourhood U2 of v(t) such that
˜
|w(σ̃ (v(t))) − w(r ) − (σ̃ (v(t)) − r )w∆ (v(t))| ≤ ε∗ |σ̃ (v(t)) − r |
for all r ∈ U2 . Let U = U1 ∩ v−1 (U2 ) and let s ∈ U . Then s ∈ U1 , v(s) ∈ U2 , and
˜
|w(v(σ (t))) − w(v(s)) − (σ (t) − s)w∆ (v(t))v∆ (t)|
˜
= |w(v(σ (t))) − w(v(s)) − (σ̃ (v(t)) − v(s))w∆ (v(t))
˜
+ [(σ̃ (v(t)) − v(s)) − (σ (t) − s)v∆ (t)]w∆ (v(t))|
˜
≤ |w(v(σ (t))) − w(v(s)) − (σ̃ (v(t)) − v(s))w∆ (v(t))|
˜
+ |(σ̃ (v(t)) − v(s)) − (σ (t) − s)v∆ (t)||w∆ (v(t))|
˜
≤ ε∗ |σ̃ (v(t)) − v(s)| + ε∗ |σ (t) − s||w∆ (v(t))|
= ε∗ |σ̃ (v(t)) − v(s) − (σ (t) − s)v∆ (t) + (σ (t) − s)v∆ (t)|
˜
+ ε∗ |σ (t) − s||w∆ (v(t))|
≤ ε∗ |σ̃ (v(t)) − v(s) − (σ (t) − s)v∆ (t)| + ε∗ |σ (t) − s||v∆ (t)|
˜
+ ε∗ |σ (t) − s||w∆ (v(t))|
66
2 Differential Calculus of Functions of One Variable
˜
≤ ε∗ ε∗ |σ (t) − s| + |σ (t) − s||v∆ (t)| + |σ (t) − s||w∆ (v(t))|
˜
= ε∗ (ε∗ + |v∆ (t)| + |w∆ (v(t))|)|σ (t) − s|
˜
≤ ε∗ (1 + |v∆ (t)| + |w∆ (v(t))|)|σ (t) − s|
= ε|σ (t) − s|,
⊔
⊓
which completes the proof.
Example 2.63 Let T = 22n : n ∈ N0 , v(t) = t 2 , w(t) = t 2 + 1. Then v : T → R
is strictly increasing and T̃ = v(T) = 24n : n ∈ N0 is a time scale. For t ∈ T, t =
22n , n ∈ N0 , we have
σ (t) = inf 22l : 22l > 22n , l ∈ N0 = 22n+2 = 4t
and
v∆ (t) = σ (t) + t = 5t.
For t ∈ T̃, t = 24n , n ∈ N0 , we have
σ̃ (t) = inf 24l : 24l > 24n , l ∈ N0 = 24n+4 = 16t.
Also, for t ∈ T, we have
(w ◦ v)(t) = w(v(t)) = (v(t))2 + 1 = t 4 + 1
and
(w ◦ v)∆ (t) = (σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3
= 64t 3 + 16t 3 + 4t 3 + t 3
= 85t 3 .
Thus,
2.3 Chain Rules
67
˜
w∆ ◦ v (t) = σ̃ (v(t)) + v(t)
= 16v(t) + v(t)
= 17v(t)
= 17t 2
and
˜
w∆ ◦ v (t)v∆ (t) = 17t 2 (5t) = 85t 3 .
Consequently,
˜
(w ◦ v)∆ (t) = w∆ ◦ v (t)v∆ (t), t ∈ Tκ .
Example 2.64 Let T = {n + 1 : n ∈ N0 }, v(t) = t 2 , w(t) = t. Then v : T → R is
strictly increasing and T̃ = {(n + 1)2 : n ∈ N0 } is a time scale. For t ∈ T, t = n + 1,
n ∈ N0 , we have
σ (t) = inf{l + 1 : l + 1 > n + 1, l ∈ N0 } = n + 2 = t + 1
and
v∆ (t) = σ (t) + t = t + 1 + t = 2t + 1.
For t ∈ T̃, t = (n + 1)2 , n ∈ N0 , we have
σ̃ (t) = {(l + 1)2 : (l + 1)2 > (n + 1)2 , l ∈ N0 } = (n + 2)2
√
= (n + 1)2 + 2(n + 1) + 1 = t + 2 t + 1.
Hence, for t ∈ T, we get
˜
˜
(w∆ ◦ v)(t) = 1, (w∆ ◦ v)(t)v∆ (t) = 1(2t + 1) = 2t + 1,
and thus
(w ◦ v)(t) = v(t) = t 2 , (w ◦ v)∆ (t) = σ (t) + t = 2t + 1.
Consequently,
˜
(w ◦ v)∆ (t) = w∆ ◦ v (t)v∆ (t), t ∈ Tκ .
68
2 Differential Calculus of Functions of One Variable
Example 2.65 Let T = {2n : n ∈ N0 }, v(t) = t, w(t) = t 2 . Then v : T → R is
strictly increasing and v(T) = T. For t ∈ T, t = 2n , n ∈ N0 , we have
σ (t) = inf{2l : 2l > 2n , l ∈ N0 } = 2n+1 = 2t
and
v∆ (t) = 1.
Moreover,
(w ◦ v)(t) = w(v(t)) = (v(t))2 = t 2 ,
and thus
(w ◦ v)∆ (t) = σ (t) + t = 2t + t = 3t.
Therefore,
(w∆ ◦ v)(t) = σ (v(t)) + v(t) = 2v(t) + v(t) = 3v(t) = 3t,
so that
(w∆ ◦ v)(t)v∆ (t) = 3t.
Consequently,
˜
(w ◦ v)∆ (t) = (w∆ ◦ v)(t)v∆ (t), t ∈ Tκ .
Exercise 2.66 Let T = 23n : n ∈ N0 , v(t) = t 2 , w(t) = t. Prove
˜
(w ◦ v)∆ (t) = (w∆ ◦ v)(t)v∆ (t), t ∈ Tκ .
Theorem 2.67 (Derivative of the Inverse) Assume v : T → R is strictly increasing
and T̃ = v(T) is a time scale. Then
˜
(v−1 )∆ ◦ v (t) =
1
v∆ (t)
for any t ∈ Tκ such that v∆ (t) = 0.
Proof Let w = v−1 : T̃ → T in Theorem 2.62.
⊔
⊓
Example 2.68 Let T = N and v(t) = t 2 + 1. Then σ (t) = t + 1, v : T → R is
strictly increasing, and
v∆ (t) = σ (t) + t = 2t + 1.
Hence,
−1 ∆˜
◦ v (t) =
v
1
1
=
.
v∆ (t)
2t + 1
2.3 Chain Rules
69
Example 2.69 Let T = {n + 3 : n ∈ N0 }, v(t) = t 2 . Then v : T → R is strictly increasing, σ (t) = t + 1, and
v∆ (t) = σ (t) + t = 2t + 1.
Hence,
−1 ∆˜
◦ v (t) =
v
1
1
=
.
v∆ (t)
2t + 1
2
Example 2.70 Let T = 2n : n ∈ N0 , v(t) = t 3 . Then v : T → R is strictly
1
2
increasing, and for t ∈ T, t = 2n , n ∈ N0 , n = log2 t 2 , we have
2
1
2
2
2
2
2
σ (t) = inf 2l : 2l > 2n , l ∈ N0 = 2(n+1) = 2n 22n+1 = t22(log2 t ) +1 .
Then
v∆ (t) = (σ (t))2 + tσ (t) + t 2 = t 2 24(log2 t )
1
2
+2
+ t 2 22(log2 t )
1
2
+1
+ t 2.
Hence,
−1 ∆˜
◦ v (t) =
v
1
1
2
t 2 24(log2 t ) +2
.
1
+
2
t 2 22(log2 t ) +1
+
t2
˜
∆
Exercise 2.71 Let T = {n + 5 : n ∈ N0 }, v(t) = t 2 + t. Find v−1 ◦ v (t).
Solution
1
.
2t+2
2.4 One-Sided Derivatives
Definition 2.72 If f is defined on [t0 , b) ⊂ T, then the right-hand derivative of f
at t0 is defined to be
f (σ (t0 )) − f (t)
,
f +∆ (t0 ) = lim
t→t0 +
σ (t0 ) − t
while if f is defined on (a, t0 ] ⊂ T, then the left-hand derivative of f at t0 is defined
by
f (σ (t0 )) − f (t)
.
f −∆ (t0 ) = lim
t→t0 −
σ (t0 ) − t
Remark 2.73 f is differentiable at t0 if and only if f +∆ (t0 ) and f −∆ (t0 ) exist and
f ∆ (t0 ) = f −∆ (t0 ) = f +∆ (t0 ).
70
2 Differential Calculus of Functions of One Variable
Example 2.74 Consider
f (t) =
⎧
⎪
⎨t + 1
⎪
⎩
t2 + t
for t ∈ [−1, 1),
for t ∈ {1, 2, 3},
where [−1, 1) is the real-valued interval. Note that f is continuous on
[−1, 1) ∪ {1, 2, 3}.
Also,
f −∆ (1) = lim
t→1−
= lim
t→1−
f (σ (1)) − f (t)
σ (1) − t
f (2) − f (t)
2−t
6 − (t + 1)
t→1−
2−t
= lim
5−t
t→1− 2 − t
= lim
=4
and
f +∆ (1) = lim
t→1+
= lim
t→1+
f (σ (1)) − f (t)
σ (1) − t
f (2) − f (t)
2−t
6 − t2 − t
t→1+
2−t
= lim
= lim
t→1+
(2 − t)(t + 3)
2−t
= lim (t + 3)
t→1+
= 4.
2.4 One-Sided Derivatives
71
Therefore,
f −∆ (1) = f +∆ (1).
Hence, f is differentiable at t = 1.
Example 2.75 Consider
f (t) =
We have
⎧
⎪
⎨t + 3
⎪
⎩
t2 + t
f −∆ (2) = lim
t→2−
for t ∈ [−2, 2),
for t ∈ {2, 4, 8}.
f (σ (2)) − f (t)
σ (2) − t
= lim
f (4) − f (t)
4−t
= lim
20 − (t + 3)
4−t
t→2−
t→2−
17 − t
t→2− 4 − t
= lim
=
15
2
and
f +∆ (t) = lim
t→2+
= lim
t→2+
f (σ (2)) − f (t)
σ (2) − t
f (4) − f (t)
4−t
20 − t 2 − t
t→2+
4−t
= lim
= 7.
Therefore,
f −∆ (2) = f +∆ (2).
Hence, f is not differentiable at t = 2.
Example 2.76 Consider
72
2 Differential Calculus of Functions of One Variable
f (t) =
⎧
⎪
⎨3
for t ∈ [0, 3),
⎪
⎩
1
for t ∈ {3, 5, 7, 9},
where [0, 3) is the real-valued interval. We have
f −∆ (3) = lim
t→3−
f (σ (3)) − f (t)
σ (3) − t
= lim
f (5) − f (t)
5−t
= lim
1−3
5−t
t→3−
t→3−
−2
t→3− 5 − t
= lim
= −1
and
f +∆ (3) = lim
t→3+
f (σ (3)) − f (t)
σ (3) − t
= lim
f (5) − f (t)
5−t
= lim
1−1
5−t
t→3+
t→3+
= 0.
Consequently,
f −∆ (3) = f +∆ (3).
Hence, f is not differentiable at 3.
Exercise 2.77 Consider
f (t) =
⎧
7
⎪
⎨t + 27
⎪
⎩
t 2 + 4t − 5
for t ∈ [−3, 0),
for t ∈ {0, 1, 2, 3, 4},
where [−3, 0) is the real-valued interval. Investigate if f is differentiable at t = 0.
Solution No, since it is not continuous at t = 0.
2.5 Nabla Derivatives
73
2.5 Nabla Derivatives
We define the backward graininess
ν(t) = t − ρ(t).
If T has a right-scattered minimum m, then we put Tκ = T \ {m}. Otherwise, Tκ = T.
Definition 2.78 (The Nabla Derivative) A function f : T → R is said to be nabla
differentiable at t ∈ Tκ if
1. f is defined in a neighbourhood U of t,
2. f is defined at ρ(t),
3. there exists a unique real number f ∇ (t), called the nabla derivative of f at t,
such that for each ε > 0, there exists a neighbourhood N of t with N ⊆ U and
| f (ρ(t)) − f (s) − (ρ(t) − s) f ∇ (t)| ≤ ε|ρ(t) − s| for all s ∈ N .
Similar to the proofs of Theorems 2.3, 2.10, and 2.18, one can prove the following
theorem.
Theorem 2.79 Let f, g : T → R be functions and let t ∈ Tκ . Then we have the
following.
1. The nabla derivative is well defined.
2. If f is nabla differentiable at t, then f is continuous at t.
3. If f is continuous at t and t is left-scattered, then f is nabla differentiable at t
with
f (t) − f (ρ(t))
.
f ∇ (t) =
ν(t)
4. If t is left-dense, then f is nabla differentiable at t iff the limit
lim
s→t
f (t) − f (s)
t −s
exists as a finite number. In this case,
f ∇ (t) = lim
s→t
f (t) − f (s)
.
t −s
5. If f is differentiable at t, then
f (ρ(t)) = f (t) + ν(t) f ∇ (t).
6. If f and g are nabla differentiable at t, then
74
2 Differential Calculus of Functions of One Variable
a. the sum f + g : T → R is nabla differentiable at t with
( f + g)∇ (t) = f ∇ (t) + g ∇ (t).
b. For any constant α, α f : T → R is nabla differentiable at t with
(α f )∇ (t) = α f ∇ (t).
c. The product f g : T → R is nabla differentiable at t with
( f g)∇ (t) = f ∇ (t)g(t) + f (ρ(t))g ∇ (t) = f (t)g ∇ (t) + f ∇ (t)g(ρ(t)).
d. If g(t)g(ρ(t)) = 0, then
f
g
∇
f
g
: T → R is nabla differentiable at t with
(t) =
f ∇ (t)g(t) − f (t)g ∇ (t)
.
g(t)g(ρ(t))
Definition 2.80 Let f : T → R and let t ∈ (Tκ )κ = Tκ 2 . We define the second nabla
derivative of f at t, provided it exists, by
∇
f ∇∇ = f ∇ : Tκ 2 → R.
n
Similarly, we define higher-order nabla derivatives f ∇ : Tκ n → R.
Example 2.81 Let T = Z and f : T → R be a function. Then
f ∇ (t) = f (t) − f (t − 1) for any t ∈ T.
Example 2.82 Let T = 2N0 and f (t) = t 3 + t 2 + t + 1, t ∈ Tκ . We will find f ∇ (t)
for t ∈ Tκ . We have that Tκ = T \ {1} and ρ(t) = 2t , t ∈ Tκ . Hence,
f ∇ (t) = (ρ(t))2 + tρ(t) + t 2 + ρ(t) + t + 1
=
t2
t2
t
+ + t2 + + t + 1
4
2
2
=
7 2 3
t + t + 1, t ∈ Tκ .
4
2
Exercise 2.83 Let T = 2N0 . Find f ∇ (t) for t ∈ Tκ , where
f (t) =
t 2 + 2t − 3
.
t −7
2.5 Nabla Derivatives
Solution
75
t 2 −21t−22
.
(t−7)(t−14)
2.6 Extreme Values
Assume that f is defined on D f ⊂ T. Let t0 ∈ T.
Definition 2.84 We say that f (t0 ) is a local maximum value of f if there exists
δ > 0 so that
f (t) ≤ f (t0 ) for all t ∈ (t0 − δ, t0 + δ) ∩ D f
and f (ρ(t0 )), f (σ (t0 )) ≤ f (t0 ), or a local minimum value of f if there exists δ > 0
such that
f (t) ≥ f (t0 ) for all t ∈ (t0 − δ, t0 + δ) ∩ D f
and f (ρ(t0 )), f (σ (t0 )) ≥ f (t0 ). The point t0 is called a local extreme point of f ,
more specifically, a local maximum or local minimum point of f .
Theorem 2.85 Let f be delta and nabla differentiable in a neighbourhood (t0 −
δ, t0 + δ) of t0 . If f ∆ (t) ≤ 0 in [t0 , t0 + δ) and f ∇ (t) ≥ 0 in (t0 − δ, t0 ], then t0 is a
local maximum point of f .
Proof 1. Let t0 be an isolated point. Then ρ(t0 ) < t0 < σ (t0 ) and
f (σ (t0 )) − f (t0 )
= f ∆ (t0 ) ≤ 0,
σ (t0 ) − t0
f (t0 ) − f (ρ(t0 ))
= f ∇ (t0 ) ≥ 0.
t0 − ρ(t0 )
Therefore, f (t0 ) ≥ f (σ (t0 )) and f (t0 ) ≥ f (ρ(t0 )). Also, there exists δ1 > 0
such that f (t) ≤ f (t0 ) for all t ∈ (t0 − δ1 , t0 + δ1 ). Consequently, t0 is a local
maximum point.
2. Let t0 be left-dense and right-scattered. Since
f (σ (t0 )) − f (t0 )
= f ∆ (t0 ) ≤ 0,
σ (t0 ) − t0
we have that f (σ (t0 )) ≤ f (t0 ).
a. Let f ∇ (t0 ) = 0. Then there exists δ1 > 0 such that f (t) = f (t0 ) for any
t ∈ (t0 − δ1 , t0 + δ1 ).
b. Let f ∇ (t0 ) > 0. Then, for every ε ∈ 0, f ∇ (t0 ) , there exists δ1 > 0 such
that
| f (t0 ) − f (t) − (t0 − t) f ∇ (t0 )| ≤ ε|t0 − t|
for any t ∈ (t0 − δ1 , t0 + δ1 ) and [t0 , t0 + δ1 ) = {t0 }. If t ∈ (t0 − δ1 , t0 ), then
we get
76
2 Differential Calculus of Functions of One Variable
−ε(t0 − t) ≤ f (t0 ) − f (t) − (t0 − t) f ∇ (t0 ) ≤ ε(t0 − t),
i.e.,
f ∇ (t0 ) − ε (t0 − t) ≤ f (t0 ) − f (t) ≤ ε + f ∇ (t0 (t0 − t),
whereupon
f (t) ≤ f (t0 )
for any t ∈ (t0 − δ1 , t0 + δ1 ), i.e., t0 is a local maximum point.
3. Let t0 be left-scattered and right-dense. Since
f (t0 ) − f (ρ(t0 ))
= f ∇ (t0 ) ≥ 0,
t0 − ρ(t0 )
we have that f (ρ(t0 )) ≤ f (t0 ).
a. Let f ∆ (t0 ) = 0. Then there exists δ1 > 0 such that f (t) = f (t0 ) for any
t ∈ (t0 − δ1 , t0 + δ1 ).
b. Let f ∆ (t0 ) < 0. Then, for every ε ∈ 0, − f ∆ (t0 ) , there exists δ1 > 0 such
that
| f (t) − f (t0 ) − (t − t0 ) f ∆ (t0 )| ≤ ε|t0 − t|
for any t ∈ (t0 − δ1 , t0 + δ1 ) and (t0 − δ1 , t0 ] = {t0 }. If t ∈ (t0 , t0 + δ1 ), then
we get
−ε(t − t0 ) ≤ f (t) − f (t0 ) − (t − t0 ) f ∆ (t0 ) ≤ ε(t − t0 ),
i.e.,
f ∆ (t0 ) − ε (t − t0 ) ≤ f (t) − f (t0 ) ≤ ε + f ∆ (t0 ) (t − t0 ),
whereupon
f (t) ≤ f (t0 )
for any t ∈ (t0 − δ1 , t0 + δ1 ), i.e., t0 is a local maximum point.
4. Let t0 be dense.
a. Assume f ∆ (t0 ) = f ∇ (t0 ) = 0. Then there exists δ1 > 0 such that f (t) =
f (t0 ) for any t ∈ (t0 − δ1 , t0 + δ1 ).
b. Assume f ∆ (t0 ) = 0 and f ∇ (t0 ) > 0. Then, for any ε ∈ 0, f ∇ (t0 ) , there
exists δ1 > 0 such that f (t) = f (t0 ) for t ∈ [t0 , t0 + δ1 ) and
−ε(t0 − t) ≤ f (t0 ) − f (t) − (t0 − t) f ∇ (t0 ) ≤ ε(t0 − t)
2.6 Extreme Values
77
for t ∈ (t0 − δ1 , t0 ]. Hence, f (t) ≤ f (t0 ) for any t ∈ (t0 − δ1, t0 + δ1 ).
c. Assume f ∆ (t0 ) < 0 and f ∇ (t0 ) = 0. Then, for every ε ∈ 0, − f ∆ (t0 ) ,
there exists δ1 > 0 such that f (t) = f (t0 ) for any t ∈ (t0 − δ1 , t0 ], and for
any t ∈ [t0 , t0 + δ1 ),
f ∆ (t0 ) − ε (t − t0 ) ≤ f (t) − f (t0 ) ≤ ε + f ∆ (t0 ) (t − t0 ),
whereupon
f (t) ≤ f (t0 )
for any t ∈ (t0 − δ1 , t0 + δ1 ).
d. Assume f ∆ (t0 ) < 0 and f ∇ (t0 ) > 0. Then, for every
ε ∈ 0, min{− f ∆ (t0 ), f ∇ (t0 )} ,
there exists δ1 > 0 such that, for any t ∈ [t0 , t0 + δ1 ),
and
f ∆ (t0 ) − ε (t − t0 ) ≤ f (t) − f (t0 ) ≤ ε + f ∆ (t0 ) (t − t0 )
−ε(t0 − t) ≤ f (t0 ) − f (t) − (t0 − t) f ∇ (t0 ) ≤ ε(t0 − t)
for t ∈ (t0 − δ1 , t0 ]. Hence, f (t) ≤ f (t0 ) for any t ∈ (t0 − δ1 , t0 + δ1 ). ⊓
⊔
As in the proof of Theorem 2.85, one can prove the following theorem.
Theorem 2.86 Let f be delta and nabla differentiable in a neighbourhood (t0 −
δ, t0 + δ) of t0 . If f ∆ (t) ≥ 0 in [t0 , t0 + δ) and f ∇ (t) ≤ 0 in (t0 − δ, t0 ], then t0 is a
local minimum point of f .
Example 2.87 Let T = Z. Consider the function
f (t) = t 2 − 5t + 4.
Then
f ∆ (t) = σ (t) + t − 5 = t + 1 + t − 5 = 2t − 4
and
f ∇ (t) = ρ(t) + t − 5 = t − 1 + t − 5 = 2t − 6.
Hence,
f ∆ (t) ≤ 0 and f ∇ (t) ≥ 0
iff
2t − 4 ≤ 0 and 2t − 6 ≥ 0
78
2 Differential Calculus of Functions of One Variable
iff
t ≤ 2 and t ≥ 3.
Therefore, f has no local maximum points. Also,
f ∆ (t) ≥ 0 and f ∇ (t) ≤ 0
iff
2t − 4 ≥ 0 and 2t − 6 ≤ 0
iff
t ≥ 2 and t ≤ 3.
Consequently, t = 2 and t = 3 are local minimum points. We have
f (2) = f (3) = −2.
Example 2.88 Let T = Z. We will find the local extreme values of the function
f (t) =
t +1
.
t2 + 1
Here, σ (t) = t + 1, ρ(t) = t − 1. Also,
f ∆ (t) =
t 2 + 1 − (t + 1)(σ (t) + t)
(t 2 + 1)((t + 1)2 + 1)
=
t 2 + 1 − (t + 1)(t + 1 + t)
(t 2 + 1)(t 2 + 2t + 2)
=
t 2 + 1 − (t + 1)(2t + 1)
(t 2 + 1)(t 2 + 2t + 2)
=
t 2 + 1 − (2t 2 + t + 2t + 1)
(t 2 + 1)(t 2 + 2t + 2)
=
t 2 + 1 − 2t 2 − 3t − 1
(t 2 + 1)(t 2 + 2t + 2)
=
(t 2
=−
and
−t 2 − 3t
+ 1)(t 2 + 2t + 2)
t (t + 3)
(t 2 + 1)(t 2 + 2t + 2)
2.6 Extreme Values
79
f ∇ (t) =
t 2 + 1 − (t + 1)(ρ(t) + t)
(t 2 + 1)((t − 1)2 + 1)
=
t 2 + 1 − (t + 1)(2t − 1)
(t 2 + 1)(t 2 − 2t + 2)
=
t 2 + 1 − (2t 2 − t + 2t − 1)
(t 2 + 1)(t 2 − 2t + 2)
=
t 2 + 1 − 2t 2 − t + 1
(t 2 + 1)(t 2 − 2t + 2)
=
−t 2 − t + 2
(t 2 + 1)(t 2 − 2t + 2)
t2 + t − 2
+ 1)(t 2 − 2t + 2)
=−
(t 2
=−
(t + 2)(t − 1)
.
(t 2 + 1)(t 2 − 2t + 2)
Hence,
f ∆ (t) ≤ 0 and f ∇ (t) ≥ 0
iff
−
(t 2
t (t + 3)
≤ 0 and
+ 1)(t 2 + 2t + 2)
−
(t 2
(t + 2)(t − 1)
≥0
+ 1)(t 2 − 2t + 2)
iff
t (t + 3) ≥ 0 and (t − 1)(t + 2) ≤ 0
so that
t = 0 and t = 1.
Therefore,
f max = f (0) = f (1) = 1.
Also,
f ∆ (t) ≥ 0 and f ∇ (t) ≤ 0
iff
−
(t 2
t (t + 3)
≥ 0 and
+ 1)(t 2 + 2t + 2)
−
(t 2
(t + 2)(t − 1)
≤0
+ 1)(t 2 − 2t + 2)
iff
t (t + 3) ≤ 0 and (t − 1)(t + 2) ≥ 0
80
2 Differential Calculus of Functions of One Variable
so that
t = −2 and t = −1.
Consequently,
f min = f (−2) =
1
−2 + 1
=−
4+1
5
and
f min = f (−1) = 0.
Example 2.89 Let T = 2N0 . We will find the extreme values of the function
f (t) =
t2 + 2
for t ≥ 4.
t +2
Here, σ (t) = 2t, ρ(t) = 21 t for all t ∈ T and t ≥ 4. Then, for t ≥ 4, we have
f ∆ (t) =
(σ (t) + t)(t + 2) − (t 2 + 2)
(t + 2)(2t + 2)
=
3t (t + 2) − (t 2 + 2)
2(t + 1)(t + 2)
=
3t 2 + 6t − t 2 − 2
2(t + 1)(t + 2)
=
t 2 + 3t − 1
(t + 1)(t + 2)
and
f ∇ (t) =
=
=
=
=
(ρ(t) + t)(t + 2) − (t 2 + 2)
(t + 2) 21 t + 2
3
t (t
2
+ 2) − t 2 − 2
(t + 2) 21 t + 2
3 2
t
2
+ 3t − t 2 − 2
(t + 2) 21 t + 2
1 2
t
2
+ 3t − 2
(t + 2) 21 t + 2
t 2 + 6t − 4
.
(t + 2)(t + 4)
2.6 Extreme Values
81
Note that f ∆ (t) ≥ 0 and f ∇ (t) ≥ 0 for all t ≥ 4. Therefore, the function f has no
local extreme values.
Exercise 2.90 Let T = Z. Find the local extreme values of the function
f (t) = t 3 − 3t 2 + 4.
Solution f max = f (0) = 4 and f min = f (2) = 0.
Definition 2.91 Suppose that f is ∆-differentiable and ∇-differentiable at t0 . We
say that t0 is a critical point of f if
f ∆ (t0 ) ≤ 0 and f ∇ (t0 ) ≥ 0
or
f ∆ (t0 ) ≥ 0 and f ∇ (t0 ) ≤ 0.
Exercise 2.92 Let T = 2N0 . Prove that
t2 + 2
3
≥
for all t ∈ T.
t +3
4
Solution We have σ (t) = 2t for all t ∈ T and
f ∆ (t) =
(t + σ (t))(t + 3) − (t 2 + 2)
(t + 3)(2t + 3)
=
3t (t + 3) − t 2 − 2
(t + 3)(2t + 3)
=
3t 2 + 9t − t 2 − 2
(t + 3)(2t + 3)
=
2t 2 + 9t − 2
≥0
(t + 3)(2t + 3)
for all t ∈ T. Consequently, f is increasing in T. Hence
f (t) ≥ f (1) =
3
for all t ∈ T.
4
Exercise 2.93 Let T = 3N0 . Find a positive constant a such that
1 + a log t ≤ t 2 for all t ∈ T.
Solution Let
f (t) = t 2 − a log t − 1, t ∈ T.
82
2 Differential Calculus of Functions of One Variable
Here, σ (t) = 3t for all t ∈ T and
f ∆ (t) = σ (t) + t − a
= 3t + t − a
= 4t − a
Since
log σ (t) − log t
σ (t) − t
log(3t) − log t
3t − t
log 3
for all t ∈ T.
2t
log 3
log 3
≤
for all t ∈ T,
2t
2
we conclude that
4t − a
Hence, if 0 < a <
8
,
log 3
log 3
log 3
≥4−a
for all t ∈ T.
2t
2
then f is increasing in T. From here,
f (t) ≥ f (1) = 0 for all t ∈ T and for 0 < a <
8
.
log 3
2.7 Convex and Concave Functions
Suppose that f : T → R.
Definition 2.94 The function f is called convex if for all t1 , t2 ∈ T and for all λ ∈
[0, 1], the inequality
f (λt1 + (1 − λ)t2 ) ≤ λ f (t1 ) + (1 − λ) f (t2 )
holds.
Definition 2.95 The function f is called strictly convex if for all t1 , t2 ∈ T with
t1 = t2 and for all λ ∈ (0, 1), the inequality
f (λt1 + (1 − λ)t2 ) < λ f (t1 ) + (1 − λ) f (t2 )
holds.
Definition 2.96 The function f is said to be (strictly) concave if − f is (strictly)
convex.
2.7 Convex and Concave Functions
83
Theorem 2.97 Let f be twice delta differentiable on (a, b) and f ∆∆ (t) ≥ 0 for all
t ∈ (a, b). Then f is convex.
Proof Let t1 , t2 ∈ T, t1 < t2 , and λ ∈ (0, 1). Then
λ f (t1 ) + (1 − λ) f (t2 ) − f (λt1 + (1 − λ)t2 )
= λ f (t1 ) + (1 − λ) f (t2 ) − (1 − λ + λ) f (λt1 + (1 − λ)t2 )
= (1 − λ)( f (t2 ) − f (λt1 + (1 − λ)t2 )) − λ( f (λt1 + (1 − λ)t2 ) − f (t1 )).
(2.12)
By the mean value theorem (Theorem 2.41), it follows that there exist
ξ1 ∈ (t1 , λt1 + (1 − λ)t2 ) and ξ2 ∈ (λt1 + (1 − λ)t2 , t2 )
so that
f (λt1 + (1 − λ)t2 ) − f (t1 ) ≤ f ∆ (ξ1 )(λt1 + (1 − λ)t2 − t1 )
= (1 − λ) f ∆ (ξ1 )(t2 − t1 )
and
f (t2 ) − f (λt1 + (1 − λ)t2 ) ≥ f ∆ (ξ2 )(t2 − λt1 − (1 − λ)t2 )
= λ f ∆ (ξ2 )(t2 − t1 ).
By (2.12), we obtain
λ f (t1 ) + (1 − λ) f (t2 ) − f (λt1 + (1 − λ)t2 )
≥ (1 − λ)λ f ∆ (ξ2 )(t2 − t1 ) − λ(1 − λ) f ∆ (ξ1 )(t2 − t1 )
(2.13)
= λ(1 − λ)( f ∆ (ξ2 ) − f ∆ (ξ1 ))(t2 − t1 ).
By the mean value theorem (Theorem 2.41), it follows that there exists ξ3 ∈ (ξ1 , ξ2 )
so that
f ∆ (ξ2 ) − f ∆ (ξ1 ) ≥ f ∆∆ (ξ3 )(ξ2 − ξ1 ).
From the last inequality and from (2.13), we obtain
84
2 Differential Calculus of Functions of One Variable
λ f (t1 ) + (1 − λ) f (t2 ) − f (λt1 + (1 − λ)t2 ) ≥ λ(1 − λ) f ∆∆ (ξ3 )(ξ2 − ξ1 )(t2 − t1 )
≥ 0,
which completes the proof.
As in Theorem 2.97, one can prove the following theorem.
Theorem 2.98 Let f be twice delta differentiable on (a, b) and f ∆∆ (t) ≤ 0 for all
t ∈ (a, b). Then f is concave.
Example 2.99 Let T = Z. Consider
f (t) = t 3 − 7t 2 + t − 10.
Here, σ (t) = t + 1 and
f ∆ (t) = (σ (t))2 + tσ (t) + t 2 − 7(σ (t) + t) + 1
= (t + 1)2 + t (t + 1) + t 2 − 7(t + 1 + t) + 1
= t 2 + 2t + 1 + t 2 + t + t 2 − 14t − 7 + 1
= 3t 2 − 11t − 5,
f ∆∆ (t) = 3(σ (t) + t) − 11
= 3(t + 1 + t) − 11
= 6t − 8.
Hence,
f ∆∆ (t) ≥ 0 for t ≥ 2 and f ∆∆ (t) ≤ 0 for t ≤ 1.
Therefore, f is convex for t ≥ 2 and concave for t ≤ 1.
Example 2.100 Let T = 2N0 . Consider
f (t) = t 4 − t 3 − t 2 − t.
Here, σ (t) = 2t and
2.7 Convex and Concave Functions
85
f ∆ (t) = (σ (t))3 + t (σ (t))2 + t 2 σ (t) + t 3
−((σ (t))2 + tσ (t) + t 2 ) − (σ (t) + t) − 1
= 8t 3 + 4t 3 + 2t 3 + t 3 − (4t 2 + 2t 2 + t 2 ) − (2t + t) − 1
= 15t 3 − 7t 2 − 3t − 1,
f ∆∆ (t) = 15((σ (t))2 + tσ (t) + t 2 ) − 7(σ (t) + t) − 3
= 15(4t 2 + 2t 2 + t 2 ) − 7(2t + t) − 3
= 105t 2 − 21t − 3.
Hence, f ∆∆ (t) > 0 for all t ∈ T. Therefore, the function f is strictly convex in T.
Example 2.101 Let T = 3N0 . Consider the function
f (t) =
t −3
.
t +2
We have σ (t) = 3t and
f ∆ (t) =
t + 2 − (t − 3)
(t + 2)(3t + 2)
=
5
3t 2 + 2t + 6t + 4
=
3t 2
5
,
+ 8t + 4
f ∆∆ (t) = −5
(3t 2
3(σ (t) + t) + 8
+ 8t + 4)(3(σ (t))2 + 8σ (t) + 4)
= −5
(3t 2
12t + 8
+ 8t + 4)(27t 2 + 24t + 4)
= −20
(3t 2
3t + 2
+ 8t + 4)(27t 2 + 24t + 4)
< 0 for all t ∈ T.
Therefore, f is a strictly concave function in T.
86
2 Differential Calculus of Functions of One Variable
Exercise 2.102 Let T = Z. Find the intervals of convexity and concavity of the
following functions.
1. f (t) = t 3 − 6t 2 + 12t + 4,
1
.
2. f (t) = t+3
Solution 1. f is convex for t ≥ 1 and concave for t ≤ 1.
2. f is convex for t ≥ −4 and concave for t ≤ −4.
Exercise 2.103 Let f and g be convex functions. Prove that so are
m(t) = max{ f (t), g(t)} and h(t) = f (t) + g(t).
2.8 Completely Delta Differentiable Functions
Definition 2.104 A function f : T → R is called completely delta differentiable at
a point t 0 ∈ Tκ if there exist constants A1 and A2 such that
f (t 0 ) − f (t) = A1 (t 0 − t) + α(t 0 − t) for all t ∈ Uδ (t 0 )
(2.14)
f (σ (t 0 )) − f (t) = A2 (σ (t 0 ) − t) + β(σ (t 0 ) − t) for all t ∈ Uδ (t 0 ),
(2.15)
and
where Uδ (t 0 ) is a δ-neighbourhood of t 0 and
α = α(t 0 , t) and β = β(t 0 , t)
are equal to zero for t = t 0 such that
lim α(t 0 , t) = lim β(t 0 , t) = 0.
t→t 0
t→t 0
Theorem 2.105 Let a function f : T → R be continuous and have the first-order
delta derivative f ∆ in some δ-neighbourhood Uδ (t 0 ) of the point t 0 ∈ Tκ . If f ∆ is
continuous at the point t 0 , then f is completely delta differentiable at t 0 .
Proof Using the definition of delta derivative, we have that
f (σ (t 0 )) − f (t) = f ∆ (t 0 )(σ (t 0 ) − t) + β(σ (t 0 ) − t),
(2.16)
where β = β(t 0 , t) and β → 0 as t → t 0 , i.e., (2.15) holds. Now, we will prove
(2.14).
1. Let t 0 ∈ T be isolated. Then (2.14) is satisfied independently of A1 and α because
in this case, Uδ (t 0 ) consists of the single point t 0 for sufficiently small δ > 0.
2.8 Completely Delta Differentiable Functions
87
2. Let t 0 be right-dense. In this case, σ (t 0 ) = t 0 , and (2.16) coincides with (2.14).
3. Let t 0 be left-dense and right-scattered. Then for sufficiently small δ > 0, any
point t ∈ Uδ (t 0 ) \ {t 0 } must satisfy t < t 0 . By Theorem 2.41, we obtain
f ∆ (ξ1 )(t 0 − t) ≤ f (t 0 ) − f (t) ≤ f ∆ (ξ2 )(t 0 − t),
where ξ1 , ξ2 ∈ [t, t 0 ). Since ξ1 , ξ2 → t 0 as t → t 0 and f ∆ is continuous at t 0 ,
we have
f (t 0 ) − f (t)
= f ∆ (t 0 ).
lim
0
t0 − t
t→t
Therefore,
f (t 0 ) − f (t)
= f ∆ (t 0 ) + α,
t0 − t
where α = α(t 0 , t) and α → 0 as t → t 0 . Thus, (2.14) holds for
A1 = f ∆ (t 0 ).
⊔
⊓
2.9 Geometric Sense of Differentiability
In this section, we consider the geometric sense of complete delta differentiability
in the case of single variable functions on time scales (see also [30]). Let T be a
time scale with the forward jump operator σ and the delta differentiation operator
∆. Consider a real-valued continuous function
u = f (t) for t ∈ T.
(2.17)
Let Γ be the “curve” represented by the function (2.17), that is, the set of points
{(t, f (t)) : t ∈ T} in the x y-plane. Let t 0 be a fixed point in Tκ . Then P0 = (t 0 , f (t 0 ))
is a point on Γ .
Definition 2.106 A line L0 passing through the point P0 is called the delta tangent
line to the curve Γ at the point P0 if
1. L0 passes also through the point P0σ = (σ (t 0 ), f (σ (t 0 )));
2. if P0 is not an isolated point of the curve Γ , then
lim
P→P0
P= P0
d(P, L0 )
= 0,
d(P, P0 )
(2.18)
where P is the moving point of the curve Γ , d(P, L0 ) is the distance from the
point P to the line L0 , and d(P, P0σ ) is the distance from the point P to the
point P0σ .
88
2 Differential Calculus of Functions of One Variable
Theorem 2.107 If the function f is completely delta differentiable at the point t 0 ,
then the curve represented by this function has the uniquely determined delta tangent
line at the point P0 = (t 0 , f (t 0 )) specified by the equation
y − f (t 0 ) = f ∆ (t 0 )(x − t 0 ),
(2.19)
where (x, y) is the current point of the line.
Proof Let f be a completely delta differentiable function at a point t 0 ∈ Tκ , Γ
be the curve represented by this function, and L0 be the line described by (2.19).
Let us show that L0 passes also through the point P0σ . Indeed, if σ (t 0 ) = t 0 , then
P0σ = P0 and the statement is true. Let now σ (t 0 ) > t 0 . Substituting the coordinates
(σ (t 0 ), f (σ (t 0 ))) of the point P0σ into (2.19), we get
"
!
f (σ (t 0 )) − f (t 0 ) = f ∆ (t 0 ) σ (t 0 ) − t 0 ,
which is obviously true by virtue of the continuity of f at t 0 . Now, we check condition
(2.18). Assume that P0 is not an isolated point of the curve Γ (note that if P0 is an
isolated point of Γ , then from P → P0 we get P = P0 and the left-hand side of (2.18)
becomes meaningless). The variable point P ∈ Γ has the coordinates (t, f (t)). As
is known from analytic geometry, the distance of the point P from the line L0 with
(2.19) is expressed by the formula
d(P, L0 ) =
where
1
f (t) − f (t 0 ) − f ∆ (t 0 ) t − t 0 ,
M
#
!
"2
M = 1 + f ∆ (t 0 ) .
Hence, by the differentiability condition (2.14) in which we have A = f ∆ (t 0 ) due
to the other differentiability condition (2.15),
d(P, L0 ) =
Next,
d(P, P0 ) =
Therefore
#
1
1
|α| t − t 0 .
α(t − t 0 ) =
M
M
2 !
"2
t − t 0 + f (t) − f (t 0 ) ≥ t − t 0 .
1
d(P, L0 )
|α| → 0 as P → P0 .
≤
d(P, P0 )
M
Thus we have proved that the line specified by (2.19) is the delta tangent line to Γ
at the point P0 .
Now, we must show that there are no other delta tangent lines to Γ at the point
P0 distinct from L0 . If P0 = P0σ , then the delta tangent line (provided it exists) is
2.9 Geometric Sense of Differentiability
89
unique as it passes through the distinct points P0 and P0σ . Let now P0 = P0σ so that
P0 is nonisolated. Suppose that there is a delta tangent line L to Γ at the point P0
described by an equation
!
"
a(x − t 0 ) − b y − f (t 0 ) = 0 with a 2 + b2 = 1.
(2.20)
Let P = (t, f (t)) be a variable point on Γ . Using (2.20), we have
!
"
d(P, L ) = a t − t 0 − b f (t) − f (t 0 ) .
Hence, by the differentiability condition (2.14) with A = f ∆ (t 0 ), the latter being a
result of the condition (2.15),
!
"
d(P, L ) = a − b f ∆ (t 0 ) + α t − t 0 .
Next, by the same differentiability condition,
So, we have
#
!
"2
+ f (t) − f (t 0 )
#
2 !
"2
t − t 0 + f ∆ (t 0 ) + α (t − t 0 )2
=
#
!
"2
= 1 + f ∆ (t 0 ) + α t − t 0 .
d(P, P0 ) =
t − t0
2
!
"
a − b f ∆ (t 0 ) + α
d(P, L )
= #
!
"2 .
d(P, P0 )
1 + f ∆ (t 0 ) + α
Passing here to the limit as t → t 0 and taking into account that the left-hand side (by
the definition of delta tangent line) and α tend to zero, we obtain
a − b f ∆ (t 0 ) = 0.
We now see that b = 0 for if otherwise, we would have a = b = 0. Hence, the line
⊔
⊓
L is described by (2.19).
Remark 2.108 If P0 is an isolated point of the curve Γ (hence P0 = P0σ ), then there
exists a delta tangent line at the point P0 to the curve Γ that coincides with the unique
line through the points P0 and P0σ .
Remark 2.109 If P0 is not an isolated point of the curve Γ and if Γ has a delta tangent
line at the point P0 , then the line P P0 , where P ∈ Γ (and P = P0 ), approaches this
tangent line as P → P0 . Conversely, if the line P P0 approaches as P → P0 some
line L0 passing through the point P0σ , then this limiting line is a delta tangent line
at P0 . For the proof it is sufficient to note that if ϕ is the angle between the lines L0
90
2 Differential Calculus of Functions of One Variable
u
P0
P
P0σ
0
t0
0
σ (t 0 )
t
Fig. 2.1 T consists of two separate real number intervals. Accordingly, the (time scale) curve Γ
consists of two arcs of usual curves. In order the curve Γ to have a delta tangent line L0 at the point
P0 , there must exist the usual left-sided tangent line to Γ at P0 and, moreover, that line must pass
through the point P0σ
and P P0 , then (see Figure 2.1)
d(P, L0 )
= sin ϕ.
d(P, P0 )
Example 2.110 Let T = Z, f (t) = t 3 − 3t 2 + 2t, t ∈ T. We will find the tangent
line of the curve {(t, f (t)) : t ∈ T} at the point t 0 = 1, where f (t) = t 3 − 3t 2 + 2t.
We have
σ (t) = t + 1, µ(t) = 1, t ∈ T.
Hence,
f (1) = 0,
f ∆ (t) = (σ (t))2 + tσ (t) + t 2 − 3(σ (t) + t) + 2
= (t + 1)2 + t (t + 1) + t 2 − 3(t + 1 + t) + 2
= t 2 + 2t + 1 + t 2 + t + t 2 − 6t − 3 + 2
2.9 Geometric Sense of Differentiability
91
= 3t 2 − 3t,
f ∆ (1) = 0.
Thus, y = 0 is the tangent line of the considered curve at t 0 = 1.
Example 2.111 Let T = 2N0 , f (t) = t 2 − 7t + 10, t ∈ T. We will find the tangent
line of the curve {(t, f (t)) : t ∈ T} at t 0 = 4, where f (t) = t 2 − 7t + 10. Here,
σ (t) = 2t, µ(t) = t, t ∈ T.
Hence,
f (4) = 16 − 28 + 10 = −2,
f ∆ (t) = σ (t) + t − 7 = 3t − 7,
f ∆ (4) = 12 − 7 = 5.
Therefore,
y + 2 = 5(x − 4), i.e., y = 5x − 22
is the tangent line of the considered curve at t = 4.
√
Example 2.112 Let T = 3N0 , f (t) = t − t 3 , t ∈ T. We will find
√the tangent line
of the curve {(t, f (t)) : t ∈ T} at the point t 0 = 1, where f (t) = t − t 3 . Here,
σ (t) = 3t, µ(t) = 2t, t ∈ T.
Hence,
f (1) = 0,
f ∆ (t) = √
= √
=
f ∆ (1) =
1
√ − (σ (t))2 + tσ (t) + t 2
σ (t) + t
1
3t +
√ − (9t 2 + 3t 2 + t 2 )
t
1
√ √ − 13t 2 ,
(1 + 3) t
1
√ − 13
3
1+
92
2 Differential Calculus of Functions of One Variable
=
=
Therefore,
√
3−1
− 13
2
√
3 27
− .
2
2
$√
%
3 27
y=
−
(x − 1)
2
2
is the tangent line of the considered curve at the point t 0 = 1.
Exercise 2.113 Let T = 2N0 , f (t) = 4t 2 − 7t + 1, t ∈ T. Find the tangent line of
the curve {(t, f (t)) : t ∈ T} at the point t 0 = 2.
Solution y = 17x − 31.
2.10 Advanced Practical Problems
Problem 2.114 Let T = {n 3 : n ∈ N0 }, f (t) = t 2 + 2t, t ∈ T. Find f ∆ (t), t ∈ Tκ .
√ 3
Solution 2 + t + 1 + 3 t .
Problem 2.115 Let T = 2N0 . Prove
t + t2 + t3 =
1
1 ∆
1 ∆
f (t) + f 3∆ (t) +
f (t).
3 2
7
15 4
Problem 2.116 Let T = 2N . Find f ∆ (t), where
1. f (t) = t 2 − 3t + 2,
3
−t 2
,
2. f (t) = t t+1
.
3. f (t) = t−1
t+1
Solution 1. 3t − 3,
6t 3 +5t 2 −3t
2. (t+1)(2t+1)
,
2
.
3. (t+1)(2t+1)
Problem 2.117 Let T = Z. Find f ∆∆ (t), where
f (t) =
Solution
2
.
(t+1)(t+2)(t+3)
1
.
t +1
2.10 Advanced Practical Problems
93
Problem 2.118 Let T = {n + 2 : n ∈ N0 }, f (t) = t 2 + 2, g(t) = t 2 . Find a constant
c ∈ [2, σ (2)]
such that
( f ◦ g)∆ (2) = f ′ (g(c))g ∆ (2).
Solution c =
#
13
.
2
Problem 2.119 Let T = N, f (t) = et , g(t) = t 3 . Using Theorem 2.57, find ( f ◦
g)∆ (t).
3
3
Solution e(t+1) − et .
Problem 2.120 Let T = 24n+2 : n ∈ N0 , v(t) = t 3 , w(t) = t 2 + t. Prove
˜
(w ◦ v)∆ (t) = w∆ ◦ v (t)v∆ (t), t ∈ Tκ .
Problem
˜ 2.121
Let
−1 ∆
v
◦ v (t).
Solution
T = {n + 9 : n ∈ N0 },
v(t) = t 2 + 7t + 8.
Find
1
.
2t+8
Problem 2.122 Consider
f (t) =
⎧
⎪
⎨2
⎪
⎩
4t + 215
for t ∈ [−3, 3),
for t ∈ {3, 9, 27},
where [−3, 3) is the real-valued interval. Investigate whether f is differentiable at
t = 3.
Solution No.
Problem 2.123 Let T = 2N0 ∪ {0}, f (t) = t 3 − 16t + 1. Find ξ1 , ξ2 ∈ (0, 4) so that
f ∆ (ξ2 ) ≤ 0 ≤ f ∆ (ξ1 ).
Solution ξ1 ∈ {2, 3}, ξ2 = 1.
Problem 2.124 Let T = 3N0 ∪ {0}. Prove that for every t > 1, there exist ξ1 , ξ2 ∈
(0, t) such that
40ξ13 + 4ξ1 ≤ t 3 + t ≤ 40ξ23 + 4ξ2 .
Solution Use the function f (t) = t 4 + t 2 .
94
2 Differential Calculus of Functions of One Variable
Problem 2.125 Let T = 2N0 . Investigate where the function
f (t) = t 4 − 3t 2 − 7
is increasing and decreasing.
Solution f is increasing for all t ∈ T.
Problem 2.126 Let T = 2N0 . Find the local extreme values of the function
f (t) = t 3 + t 2 + t + 1, t ≥ 4.
Problem 2.127 Let T = 2N0 . Find the intervals of convexity and concavity of the
following functions.
1. f (t) = t 3 − 2t 2 − 2t − 2,
1
2. f (t) = t+1
.
Solution 1. f is convex in T.
2. f is convex in T.
Problem 2.128 Let T = 3N0 . Find y ∆ (t) in terms of y(t) and y σ (t) if
y 3 + y 2 t + t 2 − t 4 y = 3.
Solution
40yt 3 −4t−y 2
.
y 2 +yy σ +(y σ )2 +3t (y+y σ )−81t 4
2.11 Notes and References
This chapter deals with differential calculus for single-variable functions on time
scales. The basic definition of delta differentiation is due to Hilger. Numerous examples on differentiation on various time scales are included. The Leibniz formula for
the nth derivative of a product of two functions is given in Theorem 2.32, and it can
be found together with its proof in [21, Theorem 1.32]. Mean value results are presented that will be used in the multivariable case. Several versions of the chain rule
are included, e.g., Theorem 2.57 is due to Christian Pötzsche [38] and it also appears
in Keller [35]. It can be found together with its proof in [21, Theorem 1.90]. The
concept of nabla derivative due to Ferhan Atıcı and Gusein Guseinov [3] is briefly
discussed. Throughout the book, results are given in terms of delta derivatives, but
all results may also be formulated with nabla instead. New sufficient conditions for
a local maximum and minimum are given in Theorems 2.85 and 2.86, respectively.
Moreover, new sufficient conditions for convexity and concavity of single-variable
2.11 Notes and References
95
functions are presented in Theorems 2.97 and 2.98, respectively. In Theorem 2.105,
a sufficient condition for complete delta differentiability of a single-variable function is given. The section on the geometric sense of differentiability is extracted from
Bohner and Guseinov [8]. Aside from this last section and the new parts that have not
been published before, all material in this chapter is taken from Bohner and Peterson
[21, 25].
Martin Bohner