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 RecurrenceEquationsExample 8.1.1:Imagine a party where couples arrive together, but at the endof the evening, every person leaves with a new partner. For each ne P, let D, denote the number of different ways n couples can beDerangements"deranged" - that is, rearranged in couples so no one is pairedwith the person they arrived with.// one couple cannot be deranged./| 3 one and only one way to derange two couples.// If the couples arrive as Aa, Bb, and Cc,// then A would be paired with b or c.//Then D1 = 0D2 = 1D3 = 2If A is paired with b, C must be paired with a (not c),//If A is paired with c, B must be paired with a (not b),//and then B with c.//and then C with b.// How big are Dand DD5'? How can we calculate104'them? Is there a formula?Let's develop a strategy for counting derangements when n>= 4. Suppose the n women are A,, A2, A3, ... An,and each Ajarrives with man a ¡.Woman A, may be "re-paired" with any of the n - 1 men a, oraz or ... or a ni; let's say she's paired with a k where 2 <= k <= n.Now let's consider a k's original partner, woman Ak: she mighttake a, or she might refuse a, and take someone else.If A, is paired with a kand A kis paired with a,there are n – 2couples left to derange, and that may be done in exactly Dn-2different ways.If A, is paired with a k but A k refuses a,, we could pretend thatAkand a, arrived together, so now, there aren - 1 couples left toderange which may be done in exactly Dn-1 different ways.For each of the n – 1 men that A, might choose, there are{Dn-2 + Dn-1} different ways to complete the derangement.Therefore, when n >= 4,Dn = (n – 1) {Dn-2 + Dn-1} ·(8.1.2)// This also applies when n = 3.// Equation 8.1.2 is an example of a second-order recurrenceequation,// where the generic entry is expressed in terms the previous%3Dtwo entries.Using Eq. 8.1.2 together with the values of D, andwe canD2'evaluate Dn for any value of n:// inprincipleD3 = (3 – 1) {D2 + D¡} = 2{1+ 0}2|| // where the generic entry is expressed in terms the previoustwo entries.Using Eq. 8.1.2 together with the values of D, and D2,// inwe canPg 335evaluate D, for any value of n:principleD3 = (3 – 1) {D2 + D¡} = 2{1 + 0}D4 = (4 – 1) {D3 + D2} = 3 {2 + 1}D5 = (5 – 1) {D4 + D3} = 4 {9 + 2}(6 – 1) {D5 + D4} = 5 {44 + 9}D7 = (7 – 1) {D6 + D5} = 6 {265 + 44}29.44D62651 854%3DD3 = (8 – 1) {D7 + D6} = 7{1854 + 265}14 833D9 = (9 – 1) {D8 + D7} = 8 {14833 + 1854}133 496%3|D10 = (10 – 1){D9 + Dg} = 9{133496 + 14833}1 334 961%D// It's strange that 1 334 961 = 10 × (133 496) + 1. Or is it?// Is there a (convenient and compact) formula for Dn that wecan use to calculate// its values?The sequence on P defined by Sn = A x n! where A is any realnumber satisfies the recurrence equation (8.1.2). If n >= 3 then: (п — 1){А (п — 2)! + A (п — 1)!}= (n – 1) A (n – 2)! {1 + [n – 1]}= A (n – 1) (n – 2)! {n}= A x n!(n – 1) {S„-2 + Sn-1}= Sµ.// But will this "formula" apply when n = 1 or n = 2?// Does there exist a real number A such that D,= A(n!) whenn = 1 or n = 2?if o = D, = A(1!), then A must equal o,if 1= D2 = A(2!), then A must equal 2.// No, because// andWe can however use this formula to prove that D, is O(n!) byprovingTheorem 8.1.1:For all n >= 2, (1/3) n ! <= Dn<= (1/2) n!Consider the table of valuesn (1/3) n!Dn(1/2) n!1 1/31/22 2/31 = 2/23 6/3 = 23 = 6/24 24/3 = 8912 = 24/25 120/3 = 404460 = 120/26 720/3 = 240 265 360 = 720/2Proof.I| || || || ||||

Name: Compute the values: D2-2D1, D3-3D2, D4-4D3, D5-5D4, …, D10-10D9, where
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Description: 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 RecurrenceEquationsExample 8.1.1:Imagine a party where couples arrive together

The values of D1, D2 , D3, …. , Dn are given and pattern is also given for the Dn. Now, according…

Answered: Compute the values: D2-2D1, D3-3D2,…

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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?

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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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
Example 8.1.1:
Imagine a party where couples arrive together, but at the end
of the evening, every person leaves with a new partner. For each n
e P, let D, denote the number of different ways n couples can be
Derangements
"deranged" - that is, rearranged in couples so no one is paired
with the person they arrived with.
// one couple cannot be deranged.
/| 3 one and only one way to derange two couples.
// If the couples arrive as Aa, Bb, and Cc,
// then A would be paired with b or c.
//
Then D1 = 0
D2 = 1
D3 = 2
If A is paired with b, C must be paired with a (not c),
//
If A is paired with c, B must be paired with a (not b),
//
and then B with c.
//
and then C with b.
// How big are D
and D
D
5'
? How can we calculate
10
4'
them? Is there a formula?
Let's develop a strategy for counting derangements when n
>= 4. Suppose the n women are A,, A2, A3, ... An,
and each Aj
arrives with man a ¡.
Woman A, may be "re-paired" with any of the n - 1 men a, or
az or ... or a ni; let's say she's paired with a k where 2 <= k <= n.
Now let's consider a k's original partner, woman Ak: she might
take a, or she might refuse a, and take someone else.
If A, is paired with a kand A kis paired with a,
there are n – 2
couples left to derange, and that may be done in exactly Dn-2
different ways.
If A, is paired with a k but A k refuses a,, we could pretend that
Akand a, arrived together, so now, there aren - 1 couples left to
derange which may be done in exactly Dn-1 different ways.
For each of the n – 1 men that A, might choose, there are
{Dn-2 + Dn-1} different ways to complete the derangement.
Therefore, when n >= 4,
Dn = (n – 1) {Dn-2 + Dn-1} ·
(8.1.2)
// This also applies when n = 3.
// Equation 8.1.2 is an example of a second-order recurrence
equation,
// where the generic entry is expressed in terms the previous
%3D
two entries.
Using Eq. 8.1.2 together with the values of D, and
we can
D2'
evaluate Dn for any value of n:
// in
principle
D3 = (3 – 1) {D2 + D¡} = 2{1+ 0}
2
||

// where the generic entry is expressed in terms the previous
two entries.
Using Eq. 8.1.2 together with the values of D, and D2,
// in
we can
Pg 335
evaluate D, for any value of n:
principle
D3 = (3 – 1) {D2 + D¡} = 2{1 + 0}
D4 = (4 – 1) {D3 + D2} = 3 {2 + 1}
D5 = (5 – 1) {D4 + D3} = 4 {9 + 2}
(6 – 1) {D5 + D4} = 5 {44 + 9}
D7 = (7 – 1) {D6 + D5} = 6 {265 + 44}
2
9.
44
D6
265
1 854
%3D
D3 = (8 – 1) {D7 + D6} = 7{1854 + 265}
14 833
D9 = (9 – 1) {D8 + D7} = 8 {14833 + 1854}
133 496
%3|
D10 = (10 – 1){D9 + Dg} = 9{133496 + 14833}
1 334 961
%D
// It's strange that 1 334 961 = 10 × (133 496) + 1. Or is it?
// Is there a (convenient and compact) formula for Dn that we
can use to calculate
// its values?
The sequence on P defined by Sn = A x n! where A is any real
number satisfies the recurrence equation (8.1.2). If n >= 3 then
: (п — 1){А (п — 2)! + A (п — 1)!}
= (n – 1) A (n – 2)! {1 + [n – 1]}
= A (n – 1) (n – 2)! {n}
= A x n!
(n – 1) {S„-2 + Sn-1}
= Sµ.
// But will this "formula" apply when n = 1 or n = 2?
// Does there exist a real number A such that D,= A(n!) when
n = 1 or n = 2?
if o = D, = A(1!), then A must equal o,
if 1= D2 = A(2!), then A must equal 2.
// No, because
// and
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 ! <= Dn<= (1/2) n!
Consider the table of values
n (1/3) n!
Dn
(1/2) n!
1 1/3
1/2
2 2/3
1 = 2/2
3 6/3 = 2
3 = 6/2
4 24/3 = 8
9
12 = 24/2
5 120/3 = 40
44
60 = 120/2
6 720/3 = 240 265 360 = 720/2
Proof.
I| || || || ||||

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