Using Eq. 8.1.2 together with the values of D, and D2, we can evaluate D, for // in principle any value of n: (3 – 1){D2 +D1} = 2{1+0} (4 – 1){D3 +D2} = 3{2+1} (5 – 1){D4+D3} = 4{9+2} (6 – 1){Ds +D4} = 5{44+9} (7 – 1){D6+D5} = 6{265+44} (8 – 1){D7+D6} = 7{1854 +265} (9 – 1){D8 +D7} = 8{14833 +1854} (10 – 1){D9+Dg}= 9{133496 +14833} D3 2 D4 = 9. D5 = 44 D6 265 1 854 D7 D8 D9 D10 14 833 133 496 1 334 961 // It's strange that 1 334 961 // Is there a (convenient and compact) formula for D, that we can use to calculate // its values? = 10 × (133 496) + 1. Or is it? The sequence on P defined by S, = A × n! where A is any real number satisfies the recurrence equation (8.1.2). If n >= 3 then (n – 1){S„-2+Sn-1} = (n – 1){A(n – 2)! + A(n – 1)!} 3 (п — 1)4(п — 2){1+(n - 1)} %3D A(n - 1) (п — 2)n} = A x n! = Sn- // But will this "formula" apply when n = 1 or n = 2? // Does there exist a real number A such that D, // No, because if 0 = D1 = A(1!), then A must equal 0, // and A(n!) when n = 1 or n = 2? 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 I| || || |||l || |
Using Eq. 8.1.2 together with the values of D, and D2, we can evaluate D, for // in principle any value of n: (3 – 1){D2 +D1} = 2{1+0} (4 – 1){D3 +D2} = 3{2+1} (5 – 1){D4+D3} = 4{9+2} (6 – 1){Ds +D4} = 5{44+9} (7 – 1){D6+D5} = 6{265+44} (8 – 1){D7+D6} = 7{1854 +265} (9 – 1){D8 +D7} = 8{14833 +1854} (10 – 1){D9+Dg}= 9{133496 +14833} D3 2 D4 = 9. D5 = 44 D6 265 1 854 D7 D8 D9 D10 14 833 133 496 1 334 961 // It's strange that 1 334 961 // Is there a (convenient and compact) formula for D, that we can use to calculate // its values? = 10 × (133 496) + 1. Or is it? The sequence on P defined by S, = A × n! where A is any real number satisfies the recurrence equation (8.1.2). If n >= 3 then (n – 1){S„-2+Sn-1} = (n – 1){A(n – 2)! + A(n – 1)!} 3 (п — 1)4(п — 2){1+(n - 1)} %3D A(n - 1) (п — 2)n} = A x n! = Sn- // But will this "formula" apply when n = 1 or n = 2? // Does there exist a real number A such that D, // No, because if 0 = D1 = A(1!), then A must equal 0, // and A(n!) when n = 1 or n = 2? 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 I| || || |||l || |
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
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?
![Using Eq. 8.1.2 together with the values of D1 and D2, we can evaluate D, for
any value of n:
// in principle
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){D8 +D7} = 8{14833 +1854}
(10 – 1){D9+D3} = 9{133496 +14833}
2
9.
44
265
1 854
14 833
133 496
1 334 961
// It's strange that 1 334 961 =
// Is there a (convenient and compact) formula for D, that we can use to calculate
// its values?
10 x (133 496) + 1. Or is it?
The sequence on P defined by S,= 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} = (n – 1){A(n – 2)! +A(n – 1)!}
= (n – 1)A(n – 2)!{1+[n – 1]}
= A(n – 1)(n – 2)!{n}
= Ax n!
= Sn.
// But will this "formula" apply when n = 1 or n =
// Does there exist a real number A such that Dn = A(n!) when n =
// No, because if 0 = D1 = A(1!), then A must equal 0,
// and
2?
1 or n = 2?
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
Theorem 8.1.1: For all n >= 2, (1/3)n! <= D, <= (1/2)n!
Consider the table of values
n
(1/3)n!
Dn
(1/2)n!
1
1/3
1/2
2/3
1
1 = 2/2
3
6/3 = 2
3 = 6/2
24/3 = 8
9.
12 = 24/2
120/3 = 40
44
60
= 120/2
6.
720/3 = 240
265
360 = 720/2
|| || || || || || ||
IL || || | || || || ||
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Transcribed Image Text:Using Eq. 8.1.2 together with the values of D1 and D2, we can evaluate D, for
any value of n:
// in principle
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){D8 +D7} = 8{14833 +1854}
(10 – 1){D9+D3} = 9{133496 +14833}
2
9.
44
265
1 854
14 833
133 496
1 334 961
// It's strange that 1 334 961 =
// Is there a (convenient and compact) formula for D, that we can use to calculate
// its values?
10 x (133 496) + 1. Or is it?
The sequence on P defined by S,= 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} = (n – 1){A(n – 2)! +A(n – 1)!}
= (n – 1)A(n – 2)!{1+[n – 1]}
= A(n – 1)(n – 2)!{n}
= Ax n!
= Sn.
// But will this "formula" apply when n = 1 or n =
// Does there exist a real number A such that Dn = A(n!) when n =
// No, because if 0 = D1 = A(1!), then A must equal 0,
// and
2?
1 or n = 2?
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
Theorem 8.1.1: For all n >= 2, (1/3)n! <= D, <= (1/2)n!
Consider the table of values
n
(1/3)n!
Dn
(1/2)n!
1
1/3
1/2
2/3
1
1 = 2/2
3
6/3 = 2
3 = 6/2
24/3 = 8
9.
12 = 24/2
120/3 = 40
44
60
= 120/2
6.
720/3 = 240
265
360 = 720/2
|| || || || || || ||
IL || || | || || || ||
ส์ร์ร์ ร์
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