DHARM

30 MATHEMATICAL FOUNDATION OF COMPUTER SCIENCE

Similarly we define numeric function S–I an, i.e.,

S–I an = an + I for n ≥ 0

For example, let an = RS^1010211 forfor ≤≤≥

T

n

n

Then, S^5 an =

00 14

5 5

for 5

for

≤≤− =

≥

R

S

T −

n

ann

()

Now determine an–5, for n ≥ 5, so there exists two cases,

• Since an = 1 (independent of n) for 0 ≤ n ≤ 10, now replace n by n – 5, therefore

an (^) –5 = 1 for 0 ≤ n – 5 ≤ 10 or 5 ≤ n ≤ 15.

• Similarly, an–5 = 2 for n – 5 ≥ 11 or n ≥ 16.

Hence, S^5 an =

004

1515

216

for

for

for

≤≤

≤≤

≥

R

S

|

T|

n

n

n

and S–5 an = an+5 for n ≥ 0

To determine an+5, for n ≥ 0, we obtain

S–5 an = an+5 =

1051005 0

25116

for or since

for or

≤+≤ ≤≤ ≥

+≥ ≥

R

S

T

nnn

nn

()

Example 2.7. Let an be a numeric function such that

an = 2for0n3

25forn4n

≤≤

+≥

R
S
T
−
Find S^2 an and S–2 an.
Sol. Using definition S^2 an, we have

S^2 an =

00121

2 2

for

for

≤≤=−

≥

RS

T −

n

ann

()

Now to determine numeric function an–2, for n ≥ 2 we have,

• For 2 ≤ n ≤ 3, an = 2 (independent of n) therefore an–2 = 2 for 2 ≤ n – 2 ≤ 3 or 4 ≤ n ≤ 5.

• For n ≥ 4, an = 2–n + 5, so an (^) –2 = 2–(n – 2) + 5 for n – 2 ≥ 4 or n ≥ 6.
Therefore,
S^2 an =
001
22
23
245
252 6
for
for
for
for
for
≤≤

=
≤≤
+≥
R
S
|
|
T
|
|−−
n
n
n
n
()n n
From the definition of S–2 an, we have
S–2 an = an+2 for n ≥ 0
To determine the numeric function an+2 , for n ≥ 0 we have,

• Since an = 2, for 0 ≤ n ≤ 3, therefore an+2 = 2, for 0 ≤ n + 2 ≤ 3 or 0 ≤ n ≤ 1


• For n ≥ 4, an = 2–n + 5, therefore an+ 2 = 2–(n + 2) + 5, for n + 2 ≥ 4 or n ≥ 2.

Hence, S–2 an =

201

252 2

for

for

≤≤

+≥

R

S

T

−+

n

()n n