Содержание
- 2. Recurrences Let (x0, x1, x2, ...) be an infinite sequence of numbers in that several initial
- 3. Examples 1) Initial element: x0 = 1, recurrence: xn = xn -1 + 1 for n
- 4. 2) Initial element: x0 = 1, recurrence: xn = 2xn -1 for n > 0. We
- 5. First order linear recurrences The general linear recurrence of the first order has the form where
- 6. Any element xn, n > 0, is uniquely defined by a, b, and x0. Can we
- 7. 1. a = 1. The equation has the form We can find ... Obviously, This sequence
- 8. 2. b = 0. The equation has the form We can find ... Obviously, This sequence
- 9. Now we consider the general case First we shall reduce the equation to a simplified form
- 10. Let us select such s that i.e. Note that for a = 1 this expression is
- 11. This equation has a simplest form (case 2), its solution is Because of we obtain and
- 12. We don't recommend you to remember this formula. It is rather the solution method. It includes
- 13. Note that the solution has the form where c1 and c2 are some constants. We see
- 14. Example: Towers of Hanoi The French mathematician Edouard Lucas invents in 1883 the following problem. Eight
- 15. Let us consider this problem in a general form when there are n disks. Let Tn
- 16. We can transpose three disks in the following manner: 1) move two disks as above (3
- 17. In general case we must transpose n – 1 smaller disks before we move the largest
- 18. We solve this equation in three described stage. 1. Reduction. then Let Take s = −
- 19. Second order linear recurrences General linear recurrence equation of second order has the form where a,
- 20. If we know the two initial elements x0 and x1 then we are able to compute
- 21. We shall look for a solution in the exponential form: where α is unknown constant (the
- 22. Thus α must be a root of the equation called a characteristic equation. There are the
- 23. A. The characteristic equation has two distinct roots α1 and α2. Then both sequences and satisfy
- 24. The homogeneous linear equation has two following important properties: 1) if a sequence satisfies the equation
- 25. both satisfy the equation (3) then the sequence also satisfies this equation. 2) if sequences and
- 26. Due to properties 1), 2) we may affirm that the sequence satisfies the equation (3) for
- 27. Thus we get a system of two linear equations with two unknowns c1, c2: The determinant
- 28. B. The characteristic equation has a single root α. In this case the general solution has
- 29. So we have the following algorithm for solving of a linear homogeneous recurrence equation of the
- 30. Examples 1) The characteristic equation: Roots: The general solution: Equations for c1, c2: Solution of the
- 31. Examples 2) The characteristic equation: Unique root: The general solution: Equations for c1, c2: Solution of
- 32. Inhomogeneous equation An inhomogeneous linear recurrence of the second order may be reduced to homogeneous equation
- 33. If a + b ≠ 1 then s exist and we obtain a homogeneous equation, solve
- 34. Fibonacci numbers Leonardo Fibonacci (1170 – 1250) also known as Leonardo Pisano, was an Italian mathematician.
- 35. The Fibonacci numbers are elements of the sequence 0, 1, 1, 2, 3, 5, 8, 13,
- 36. If we denote the n-th element of the sequence by Fn (n = 0, 1, 2,
- 37. The characteristic equation for this recurrence is It has two roots The general solution of the
- 38. Now we use the initial values for writing the equations for the constants c1 and c2:
- 39. A binary word containing no two consecutive 1’s will be called a sparse word. For example,
- 40. For small n we have: 0 1 2 3 4 Sparse words n Un λ 0
- 41. What is U5 ? If α is a sparse word of length 5 and the first
- 42. If the first letter in α is 1 then the second letter must be 0 and
- 43. In general case let α be a sparse word of the length n. If the first
- 45. Скачать презентацию