Chapter 4 Discrete Mathematics and Combinatorics.pdf
- 1. Discrete Mathematics and Combinatorics Instructor: Solomon Ketema (PhD) Email :[email protected] Telegram :0911980245
- 2. A recurrence relation for the sequence 𝑎𝑛 is an equation that expresses 𝑎𝑛 in terms of one or more of the previous terms 𝑎0, 𝑎1 , … , 𝑎𝑛−1 for all integers n with n≥𝑛0. The general solution of the recurrence relation 𝑎𝑛+1= 𝑑𝑎𝑛 where n≥0, d is a constant, and 𝑎0=A, is unique and is given by 𝑎𝑛= 𝐴 𝑑𝑛 , n≥0. Thus the solution 𝑎𝑛= 𝐴 𝑑𝑛 , n≥0 defines a discrete function whose domain is the set of all nonnegative integers. Chapter 4: Recurrence Relations
- 3. Example Solve the recurrence relation 𝑎𝑛+1=5 𝑎𝑛 for n≥0 given that 𝑎0=2. Solution Since 𝑎𝑛+1=5 𝑎𝑛, we have 𝑎1= 5 𝑎0 𝑎2= 5(5 𝑎0)= 52 𝑎0 𝑎3= 5𝑎2= 5(52 𝑎0)= 53 𝑎0 Therefore the general solution is 𝑎𝑛= 5𝑛 𝑎0. 2(5𝑛) is the initial condition.
- 4. Example Solve the recurrence relation 4𝑎𝑟 −5 𝑎𝑟−1=0, for r≥1 given that 𝑎0=1. Solution We have 4𝑎𝑟 −5 𝑎𝑟−1=0, 4𝑎𝑟 =5 𝑎𝑟−1 𝑎𝑟= ( 5 4 )𝑎𝑟−1 The general solution is 𝑎𝑟= ( 5 4 )𝑎𝑟−1 𝑎0=1 is the initial condition.
- 5. Example Solve the recurrence relation 𝑎𝑛= 𝑛𝑎𝑛−1 for n≥1 given that 𝑎0=1. Solution (Ex!) Example If 𝑎𝑛 is a solution of 𝑎𝑛+1= 𝑘𝑎𝑛 for n≥0 and 𝑎3= 153 49 and 𝑎5= 1377 2401 , then find the value of 𝑘. Solution (Ex!)
- 6. Example Solve the recurrence relation 𝑎𝑛= 2𝑎𝑛−1+1 with 𝑎1=7 for n≥1, by substitution method. Solution Given 𝑎𝑛= 2𝑎𝑛−1+1; 𝑎1=7 𝑎2= 2 𝑎1 + 1= 2 7 +1 𝑎3= 2𝑎2+1= 2 (2 7 +1 +1= 22 (7)+2+1 and so on. 𝑎𝑛= (7)2𝑛−1 +2𝑛−2 +2𝑛−3 +∙ ∙ ∙ +2+1 𝑎𝑛= (7)2𝑛−1 +1+2∙ ∙ ∙ +2𝑛−2 𝑎𝑛= (7)2𝑛−1 + ( 2𝑛−1−1 2−1 ) 𝑎𝑛= (7)2𝑛−1 + 2𝑛−1 − 1 for r > 1
- 7. If 𝑎𝑛, n≥0, 𝐶𝑛≠0 is a discrete function then 𝐶𝑛𝑎𝑛+𝐶𝑛−1𝑎𝑛−1+∙ ∙ ∙ +𝐶𝑛−𝑘 𝑎𝑛−𝑘 = 𝑓(𝑛), n≥k (k positive integer) is a linear recurrence relation (with constant coefficient) of order k. when 𝑓(𝑛)=0, for all n ≥0, the relation is called homogeneous, otherwise it is nonhomogeneous. 𝐶𝑛𝑎𝑛+𝐶𝑛−1𝑎𝑛−1 +𝐶𝑛−2 𝑎𝑛−2=0, n ≥2 is homogeneous relation of order 2. Substituting 𝑎𝑛= 𝐴 𝑑𝑛 in to the above equation, we obtain, 𝐶𝑛𝐴 𝑑𝑛 +𝐶𝑛−1𝐴 𝑑𝑛−1 +𝐶𝑛−2 𝐴 𝑑𝑛−2 =0 with A,d ≠0 it becomes 𝐶𝑛𝑑𝑛 +𝐶𝑛−1 𝑑𝑛−1 +𝐶𝑛−2 𝑑𝑛−2 =0, a quadratic equation which is called the characteristic equation.
- 8. Method of characteristic roots Example Solve the recurrence relation 𝑎𝑟 − 7𝑎𝑟−1+12𝑎𝑟−2=0 Solution The characteristic equation of the given equation is 𝑘2 −7𝑘 + 12 = 0 (𝑘 − 3)(𝑘 − 7) = 0. The characteristic roots are 𝑘 = 3 𝑎𝑛𝑑 𝑘 = 7. The solution of the given recurrence relation is 𝑎𝑟 = 𝐴1 3𝑟 + 𝐴2 4𝑟
- 9. Method of characteristic roots Example Solve the recurrence relation 𝑎𝑟 − 7𝑎𝑟−1+10𝑎𝑟−2=0 Solution (Ex!) Example Solve the recurrence relation 𝑎𝑛 + 𝑎𝑟−1 − 6𝑎𝑛−2=0 Solution (Ex!)