Solve the first-order linear recurrence relation: Sn+1 = Sn + 2, with S0=1.
Solve the first-order linear recurrence relation: Sn+1 = Sn + 2, with S0=1.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Solve the first-order linear recurrence relation: Sn+1 = Sn + 2, with S0=1. You may use the general solution given on P.342.
![342
8 Sequences and Series
is given in two parts:
if a = 1,
Sn =I+nc
for V n E N;
if a + 1,
Sn = a"A+
1
for Vn E N.
a
When a = 1, any particular solution is obtained by determining a specific,
numerical value for I. In fact, a particular solution is determined by a specific,
numerical value J for any (particular) entry, S;. Solving the equation
J =I+ jc for I,
I = J - jc.
// since S; = I+ jc
// where So =I
we get
// One particular “particular solution" has I = 0.
When a + 1, any particular solution is obtained by determining a specific,
numerical value for A; if the starting value I is given, then A = I –
In fact,
1 - a
a particular solution is determined by a specific, numerical value J for any
(particular) entry, S;. Solving the equation
J = Ad +
for A,
1 - a
we get
A =
// But what if a = 0?
1
// One particular “particular solution" has A = 0.
Example 8.2.1: The Towers of Hanoi
The recurrence equation for the number of moves in the Towers of Hanoi
problem is a first-order linear recurrence equation:
Tn = 2T,-1+ 1.
1
-1, and any sequence T that satisfies
2
Here a =
2 and c =
1, so
1
- a
1
this RE is given by the formula
Tn = 2" [1 – (–1)] +(-1)
= 2"[I + 1] – 1.
Assuming T has domain N and denoting T, by I, we saw at the beginning of this
chapter several particular solutions:
if I = 0, then
if I = 2, then
if I = 4, then
if I %3D-1, then Т%3D(-1,-1,-1, -1,-1,...). /Т, %3 2"|-1+1]— 1%3
= (0, 1,3,7, 15, 31,...);
T = (2,5, 11, 23, 47, 95, ...);
T = (4,9, 19, 39, 79, 159,...); // T, = 2" [4 + 1] – 1= 5 x 2" – 1.
// T, = 2" (0+1] –1= 2"
// T = 2" [2+ 1]
- 1.
- 1 = 3 × 2" – 1.
-1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4792360-22b4-40c0-a48a-02ce03c9f6f7%2F18e91af7-0c5b-44c7-8ac6-67ad5df84e15%2F8391nh8_processed.png&w=3840&q=75)
Transcribed Image Text:342
8 Sequences and Series
is given in two parts:
if a = 1,
Sn =I+nc
for V n E N;
if a + 1,
Sn = a"A+
1
for Vn E N.
a
When a = 1, any particular solution is obtained by determining a specific,
numerical value for I. In fact, a particular solution is determined by a specific,
numerical value J for any (particular) entry, S;. Solving the equation
J =I+ jc for I,
I = J - jc.
// since S; = I+ jc
// where So =I
we get
// One particular “particular solution" has I = 0.
When a + 1, any particular solution is obtained by determining a specific,
numerical value for A; if the starting value I is given, then A = I –
In fact,
1 - a
a particular solution is determined by a specific, numerical value J for any
(particular) entry, S;. Solving the equation
J = Ad +
for A,
1 - a
we get
A =
// But what if a = 0?
1
// One particular “particular solution" has A = 0.
Example 8.2.1: The Towers of Hanoi
The recurrence equation for the number of moves in the Towers of Hanoi
problem is a first-order linear recurrence equation:
Tn = 2T,-1+ 1.
1
-1, and any sequence T that satisfies
2
Here a =
2 and c =
1, so
1
- a
1
this RE is given by the formula
Tn = 2" [1 – (–1)] +(-1)
= 2"[I + 1] – 1.
Assuming T has domain N and denoting T, by I, we saw at the beginning of this
chapter several particular solutions:
if I = 0, then
if I = 2, then
if I = 4, then
if I %3D-1, then Т%3D(-1,-1,-1, -1,-1,...). /Т, %3 2"|-1+1]— 1%3
= (0, 1,3,7, 15, 31,...);
T = (2,5, 11, 23, 47, 95, ...);
T = (4,9, 19, 39, 79, 159,...); // T, = 2" [4 + 1] – 1= 5 x 2" – 1.
// T, = 2" (0+1] –1= 2"
// T = 2" [2+ 1]
- 1.
- 1 = 3 × 2" – 1.
-1.
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