Constant Coefficients - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

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CS 173:

Discrete Mathematical Structures

Recurrences

Linear NONhomogeneous recurrence relations with constant coefficients.

c 0 an + c 1 an-1 + c 2 an-2 + … + c (^) k an-k = f(n),

Where f(n) is constant polynomial in n cn^ for some constant c cn^ · polynomial(n)

This approach is different than the one in your text. Easier and more general.

Recurrences

Sequence operators:

Constant multiplication c·〈an〉 defined to be 〈c·an〉 Ex: 3·〈 2 n〉 = 〈3·2 n〉 = 3, 6, 12, 24, 48, …

Shift “E” E〈an〉 = 〈an+1〉 shifts sequence to left Ex: E〈 2 n〉 = 〈 2 n+1〉 = 2, 4, 8, 16, … Ex: E〈3n + 1〉 = 〈3(n+1) + 1〉 = 〈3n + 4〉

Recurrences

Combining operators:

If A,B are seq ops, then A+B is a seq op: (A+B)〈an〉 defined to be A〈an〉 + B〈an〉 Ex: (E+2)〈 2 n〉 = E〈 2 n〉 + 2〈 2 n〉 = 〈 2 n+1〉 + 〈2·2 n〉 = 〈 2 n+1〉 + 〈 2 n+1〉 = 〈 2 n+1^ + 2 n+1〉 = 〈2·2 n+1〉 = 〈 2 n+2〉 If A,B are seq ops, then AB is a seq op: (AB)〈an〉 defined to be A(B〈an〉) Ex: E^3 〈an〉 = E·E·E〈an〉 = E(E(E〈an〉)) = 〈an+3〉

Recurrences

For any constant sequence 〈c〉:

(E-1)〈c〉 = 〈 0 〉

TRICK: to solve a NONhomogeneous linear recurrence with constant coefficients, turn it into a homogeneous recurrence by applying operators to annihilate the right side.

(E-1) “annihilates” 〈c〉

Recurrences

Example: solve an = 5an-1 - 6an-2 + 4

Rewrite: an - 5an-1 + 6an-2 = 4 Rewrite again so n is smallest index: an+2 - 5an+1 + 6an = 4 Rewrite again as a sequence: 〈an+2 - 5an+1 + 6an〉 = 〈 4 〉 Rewrite again using operators: (E 2 - 5E + 6)〈an〉 = 〈 4 〉 (E-1) “annihilates” 〈c〉

Recurrences

Example: solve an - 6an-1 + 11an-2 - 6an-3 = 0

Characteristic equation: (r^3 - 6r^2 + 11r - 6) = 0 (r-1)(r-2)(r-3) = 0

General solution: an = A 1 + A 22 n^ + A 33 n

Homogeneous!!!

Recurrences

Linear NONhomogeneous recurrence relations with constant coefficients.

c 0 an + c 1 an-1 + c 2 an-2 + … + c (^) k an-k = f(n),

Where f(n) is constant polynomial in n cn^ for some constant c cn^ · polynomial(n)

This is approach is different than the one in your text. Easier and more general.

Recurrences

Example: solve an = 5an-1 - 6an-2 + 4

(E 2 - 5E + 6)〈an〉 = 〈 4 〉 Annihilate right side: (E-1)(E^2 - 5E + 6)〈an〉 = (E-1)〈 4 〉 (E^3 - 6E^2 + 11E - 6)〈an〉 = 〈 0 〉 But that’s just: an+3 - 6an+2 + 11an+1 - 6an = 0 an - 6an-1 + 11an-2 - 6an-3 = 0

(E-1) “annihilates” 〈c〉

Homogeneous!!!

Recurrences

Example: solve an - 6an-1 + 11an-2 - 6an-3 = 0

Characteristic equation: (r^3 - 6r^2 + 11r - 6) = 0 (r-1)(r-2)(r-3) = 0

General solution: an = A 1 + A 22 n^ + A 33 n

Homogeneous!!!

Not a coincidence that characteristic equation looks the same as the annihilating operator: f(r)=0 corresponds exactly to g(E)〈an〉 = 〈 0 〉

Recurrences

Linear NONhomogeneous recurrence relations with constant coefficients.

c 0 an + c 1 an-1 + c 2 an-2 + … + c (^) k an-k = f(n),

Where f(n) is constant polynomial in n cn^ for some constant c cn^ · polynomial(n)

So far we know how to annihilate 〈c〉 for constant c.

Use operator (E - 1) “Discrete derivative”

Recurrences

Linear NONhomogeneous recurrence relations with constant coefficients.

c 0 an + c 1 an-1 + c 2 an-2 + … + c (^) k an-k = f(n),

Where f(n) is constant polynomial in n cn^ for some constant c cn^ · polynomial(n)

Group challenge: find the annihilators for the remaining function types.

Use operator (E - 1) “Discrete derivative”

(E - 1)〈a (^) n〉 = E〈a (^) n〉 - 〈a (^) n〉 = 〈a (^) n+1〉 - 〈a (^) n〉 = 〈a (^) n+1 - a (^) n〉

Recurrences

Consider 〈 3 n〉: E〈 3 n〉 = 3 〈 3 n〉 =

So (E-3)〈 3 n〉 = 〈 0 〉

How do you annihilate 〈c n〉?

3, 9, 27, … 3, 9, 27, …

In general (E-c) annihilates 〈c n〉

Recurrences

Table of handy annihilators:

〈c〉 E-

〈polynomial(n) of degree ≤ k〉 (E-1)k+

〈c n〉 E-c 〈c n·polynomial(n) of degree ≤ k〉 (E-c) k+

Sequence Annihilator