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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 || |

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 67E
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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?

### Educational Resource: Exploring Recurrence Relations and Factorials **Overview:** This resource explores mathematical relationships and sequences through the lens of recurrence equations and factorials. It builds upon equation (8.1.2) to calculate specific values within a sequence, and examines the behavior of these sequences under certain conditions. **Content:** 1. **Equation Use and Evaluation of \( D_n \):** Using Eq. 8.1.2 alongside previously defined values of \( D_1 \) and \( D_2 \), the values of \( D_n \) are determined for various \( n \). \[ \begin{align*} D_3 &= (3 - 1)\{D_2 + D_1\} = 2\{1 + 0\} = 2 \\ D_4 &= (4 - 1)\{D_3 + D_2\} = 3\{2 + 1\} = 9 \\ D_5 &= (5 - 1)\{D_4 + D_3\} = 4\{9 + 2\} = 44 \\ D_6 &= (6 - 1)\{D_5 + D_4\} = 5\{44 + 9\} = 265 \\ D_7 &= (7 - 1)\{D_6 + D_5\} = 6\{265 + 44\} = 1\,854 \\ D_8 &= (8 - 1)\{D_7 + D_6\} = 7\{1854 + 265\} = 14\,833 \\ D_9 &= (9 - 1)\{D_8 + D_7\} = 8\{14833 + 1854\} = 133\,496 \\ D_{10} &= (10 - 1)\{D_9 + D_8\} = 9\{133496 + 14833\} = 1\,334\,961 \\ \end{align*} \] 2. **Question Exploration:** - Exploration of the unexpected result \( 1\,334\,961 = 10 \times (133\,496) + 1 \). - Inquiry into a unified
Transcribed Image Text:### Educational Resource: Exploring Recurrence Relations and Factorials **Overview:** This resource explores mathematical relationships and sequences through the lens of recurrence equations and factorials. It builds upon equation (8.1.2) to calculate specific values within a sequence, and examines the behavior of these sequences under certain conditions. **Content:** 1. **Equation Use and Evaluation of \( D_n \):** Using Eq. 8.1.2 alongside previously defined values of \( D_1 \) and \( D_2 \), the values of \( D_n \) are determined for various \( n \). \[ \begin{align*} D_3 &= (3 - 1)\{D_2 + D_1\} = 2\{1 + 0\} = 2 \\ D_4 &= (4 - 1)\{D_3 + D_2\} = 3\{2 + 1\} = 9 \\ D_5 &= (5 - 1)\{D_4 + D_3\} = 4\{9 + 2\} = 44 \\ D_6 &= (6 - 1)\{D_5 + D_4\} = 5\{44 + 9\} = 265 \\ D_7 &= (7 - 1)\{D_6 + D_5\} = 6\{265 + 44\} = 1\,854 \\ D_8 &= (8 - 1)\{D_7 + D_6\} = 7\{1854 + 265\} = 14\,833 \\ D_9 &= (9 - 1)\{D_8 + D_7\} = 8\{14833 + 1854\} = 133\,496 \\ D_{10} &= (10 - 1)\{D_9 + D_8\} = 9\{133496 + 14833\} = 1\,334\,961 \\ \end{align*} \] 2. **Question Exploration:** - Exploration of the unexpected result \( 1\,334\,961 = 10 \times (133\,496) + 1 \). - Inquiry into a unified
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