Discrete Fractional Calculus (2015)
1. Basic Difference Calculus
1.2. Delta Exponential Function
In this section we want to study the delta exponential function that plays a similar role in the delta calculus on that the exponential function e pt , , does in the continuous calculus. Keep in mind that when p is a constant, x(t) = e pt is the unique solution of the initial value problem
For the delta exponential function we would like to consider functions in the set of regressive functions defined by
Some of the results that we give will be true if in the definition of regressive functions we consider complex-valued functions instead of real-valued functions. We leave it to the reader to note when this is true.
We then define the delta exponential function corresponding to a function , based at , to be the unique solution (why does guarantee uniqueness?), e p (t, s), of the initial value problem
(1.4)
(1.5)
Theorem 1.11.
Assume and Then
(1.6)
Here, by a standard convention on products, it is understood that for any function h that
Proof.
We solve the IVP (1.4), (1.5) to get a formula for e p (t, s). Solving (1.4) for x(t + 1) we get
(1.7)
Letting t = s in (1.7) and using the initial condition (1.5) we get
Next, letting t = s + 1 in (1.7) we get
Proceeding in this fashion we get
(1.8)
for . In the product in (1.8), it is understood that the index τ takes on the values By convention . Next assume . Solving (1.7) for x(t) we get
(1.9)
Letting t = s − 1 in (1.9), we get
Next, letting t = s − 2 in (1.9), we get
Continuing in this manner we get
□
Theorem 1.11 gives us the following example.
Example 1.12.
If p(t) = p is a constant with p ≠ − 1 (note this constant function is in ), then from (1.6)
Example 1.13.
Find e p (t, 1) if p(t) = t − 1, First note that 1 + p(t) = t ≠ 0 for , so . From (1.6) we get
for .
It is easy to prove the following theorem.
Theorem 1.14.
If , then a general solution of
is given by
where c is an arbitrary constant.
The following example is an interesting application using an exponential function.
Example 1.15.
According to folklore, Peter Minuit in 1626 purchased Manhattan Island for goods worth $24. If at the beginning of 1626 the $24 could have been invested at an annual interest rate of 7% compounded quarterly, what would it have been worth at the end of the year 2014. Let y(t) be the value of the investment after t quarters of a year. Then y(t) satisfies the equation
Thus y is a solution of the IVP
Using Theorem 1.14 and the initial condition we get that
It follows that
(about 11.8 trillion dollars!).
We now develop some properties of the (delta) exponential function e p (t, a). To motivate our later results, consider, for the product
Hence we get the law of exponents
holds for provided
Theorem 1.16.
If we define the circle plus addition , ⊕, on by
then , ⊕ is an Abelian group.
Proof.
First to see the closure property is satisfied, note that if , then 1 + p(t) ≠ 0 and 1 + q(t) ≠ 0 for It follows that
for and hence .
Next the zero function as 1 + 0 = 1 ≠ 0. Also
so the zero function 0 is the additive identity element in
To show that every element in has an additive inverse let . Then set and note that since
for , so and we also have that
so q is the additive inverse of p. For we use the following notation for the additive inverse of p:
(1.10)
The fact that the addition ⊕ is associative and commutative is Exercise 1.19. □
We can now define circle minus subtraction on in the standard way that subtraction is defined in terms of addition.
Definition 1.17.
We define circle minus subtraction on by
It can be shown (Exercise 1.18) that if then
The next theorem gives us several properties of the exponential function e p (t, s), based at .
Theorem 1.18.
Assume and . Then
(i)
e 0 (t,s) = 1 and e p (t,t) = 1;
(ii)
(iii)
if 1 + p > 0, then e p (t,s) > 0;
(iv)
(v)
(vi)
e p (t,s)e p (s,r) = e p (t,r);
(viii)
e p (t,s)e q (t,s) = e p⊕q (t,s);
(viii)
(ix)
(x)
Proof.
We prove many of these properties when s = a and leave it to the reader to show that the same results hold for any . By the definition of the exponential we have that (i) and (iv) hold. To see that (ii) holds when s = a note that since , 1 + p(t) ≠ 0 for and hence we have that
for . The proof of (iii) is similar to the proof of (ii).
Since
we have that (v) holds when s = a.
We only show (vi) holds when t ≥ s ≥ r and leave the other cases to the reader. In particular, we merely observe that
We proved (vii) holds when with s = a, earlier to motivate the definition of the circle plus addition. To see that (viii) holds with s = a note that
Since
we have (ix) holds when s = a. Since
we have that (x) holds. □
Before we derive some other properties of the exponential function we give another example where we use an exponential function.
Example 1.19.
Assume initially that the number of bacteria in a culture is P 0 and after one hour the number of bacteria present is . Find the number of bacteria, P(t), present after t hours. How long does it take for the number of bacteria to triple? Experiments show that P(t) satisfies the IVP (why is this plausible?)
Solving this IVP we get from Theorem 1.14 that
Using the fact that we get . It follows that
Let t 0 be the amount of time it takes for the population of the bacteria to triple. Then
which implies that
The set of positively regressive functions, , is defined by
Note that by Theorem 1.18, part (iii), we have that if , then e p (t, a) > 0 for . It is easy to see (Exercise 1.20) that is a subgroup of .
We next define the circle dot scalar multiplication ⊙ on
Definition 1.20.
The circle dot scalar multiplication, ⊙, is defined on by
Theorem 1.21.
If and , then
for
Proof.
Consider
This completes the proof. □
The following lemma will be used in the proof of the next theorem.
Lemma 1.22.
If and
then p = q.
Proof.
Assume and e p (t, a) = e q (t, a) for It follows that
Dividing by e p (t, a) = e q (t, a) we get that p = q. □
Theorem 1.23.
The set of positively regressive functions , with the addition ⊕, and the scalar multiplication ⊙ is a vector space.
Proof.
We just prove two of the properties of a vector space and leave the rest of the proof (see Exercise 1.25) to the reader. First we show that the distributive law
holds for , This follows from
and an application of Lemma 1.22. Next we show that 1 ⊙ p = p for all This follows from
and an application of Lemma 1.22. □