Compute the values: D2-2D1, D3-3D2, D4-4D3, D5-5D4, …, D10-10D9, where the D’s are the derangements values described on P.335. What is the pattern?
Compute the values: D2-2D1, D3-3D2, D4-4D3, D5-5D4, …, D10-10D9, where the D’s are the derangements values described on P.335. What is the pattern?
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Compute the values: D2-2D1, D3-3D2, D4-4D3, D5-5D4, …, D10-10D9, where the D’s are the derangements values described on P.335. What is the pattern?
![8.1
Examples Defined by Recurrence Equations
335
Using Eq. 8.1.2 together with the values of D, and D2, we can evaluate D, for
// in principle
any value of n:
D3
D4 =
D5
D6
D7
D8
D9
D10
(3 – 1){D2 +D1} = 2{1+0}
(4 – 1){D3 +D2} = 3{2+1}
(5 – 1){D4+D3} = 4{9+2}
(6 – 1){D5+D4} = 5{44+9}
(7 – 1){D6+D5} = 6{265 +44}
(8 – 1){D7+D6} = 7{1854 +265}
(9 – 1){D3 +D7} = 8{14833 +1854}
(10 – 1){D9+Ds} = 9{133496 +14833}
2
%3D
44
265
1 854
14 833
133 496
1 334 961
// It's strange that 1 334 961 = 10 × (133 496) + 1. Or is it?
// Is there a (convenient and compact) formula for D, that we can use to calculate
// its values?
The
sequence on P defined by Sn=A × n! where A is any real number satisfies
the recurrence equation (8.1.2). If n >= 3 then
(n – 1){Sn-2+Sn–1}
- (п — 1){A(п — 2)! + A(n - 1)}
(п 1)4(п — 2){1+ [n - 1]}
3D А(n — 1) (п — 2){n}
= A x n!
= Sn.
// But will this "formula" apply when n =
// Does there exist a real number
// No, because
// and
1 or n =
2?
such that Dn
A(n!) when n = 1 or n = 2?
if 0 = D1 = A(1!), then A must equal 0,
if 1 = D2 = A(2!), then A must equal ½.
We can however use this formula to prove that D, is O(n!) by proving
|| || || || || || ||](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4792360-22b4-40c0-a48a-02ce03c9f6f7%2Fb386ee9b-9957-408e-a054-5b32ed3148d5%2Fx4z4u9_processed.png&w=3840&q=75)
Transcribed Image Text:8.1
Examples Defined by Recurrence Equations
335
Using Eq. 8.1.2 together with the values of D, and D2, we can evaluate D, for
// in principle
any value of n:
D3
D4 =
D5
D6
D7
D8
D9
D10
(3 – 1){D2 +D1} = 2{1+0}
(4 – 1){D3 +D2} = 3{2+1}
(5 – 1){D4+D3} = 4{9+2}
(6 – 1){D5+D4} = 5{44+9}
(7 – 1){D6+D5} = 6{265 +44}
(8 – 1){D7+D6} = 7{1854 +265}
(9 – 1){D3 +D7} = 8{14833 +1854}
(10 – 1){D9+Ds} = 9{133496 +14833}
2
%3D
44
265
1 854
14 833
133 496
1 334 961
// It's strange that 1 334 961 = 10 × (133 496) + 1. Or is it?
// Is there a (convenient and compact) formula for D, that we can use to calculate
// its values?
The
sequence on P defined by Sn=A × n! where A is any real number satisfies
the recurrence equation (8.1.2). If n >= 3 then
(n – 1){Sn-2+Sn–1}
- (п — 1){A(п — 2)! + A(n - 1)}
(п 1)4(п — 2){1+ [n - 1]}
3D А(n — 1) (п — 2){n}
= A x n!
= Sn.
// But will this "formula" apply when n =
// Does there exist a real number
// No, because
// and
1 or n =
2?
such that Dn
A(n!) when n = 1 or n = 2?
if 0 = D1 = A(1!), then A must equal 0,
if 1 = D2 = A(2!), then A must equal ½.
We can however use this formula to prove that D, is O(n!) by proving
|| || || || || || ||
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