Topic: Solving Recurrence Relations
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How to Solve Recurrence Relations
A recurrence relation is an equation that defines each term of a sequence as a function of the previous term. It may also be known as a recursive equation, and even a simple one can exhibit complex behavior. The study of recurrence relation... Read More »
Source: http://www.ehow.com/how_5257039_solve-recurrence-relations.html
How to solve following recurrence relations?
1) Assuming log base 2... Writing n = 2^k for some non-negative integer k yields T(2^k) = 4T(2^k/2) + 2^k * log(2^k) .........= 4T(2^(k-1)) + k * 2^k .........= 4[4T(2^(k-1)/2) + 2^(k-1) * log(2^(k-1))] + k * 2^k .........= 4^2 T(2^(k-2)) +... Read More »
Source: http://answers.yahoo.com/question/index?qid=20111129232251AAFTlnl
How to solve recurrence Relation for Big-Theta?
T(n) = n + 2T(n/3) T(1) = 1 + 2T(0) =1 So T(0)=0 T(2) = 2 + 2T(2/3)= 2 + 2T(0) =2+0=2 T(3) = 3 + 2T(3/3) = 3 + 2T(1) = 3 + 2(1) = 3 + 2=5 T(4) = 4 + 2T(4/3) =4 + 2T(1)=4 + 2(1)=6 T(5) = 5 + 2T(5/3) = 5 + 2T(1) = 5 + 2(1) =7 T(6) = 6 + 2T(6/... Read More »
Source: http://answers.yahoo.com/question/index?qid=20120425215057AAxWh0F
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Solving Recurrence Relations
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Let n = length of string, and f(n) = number of such strings. f(0) = 0 f(1) = 0 f(2) = 1 That's the easy part. Let a string with a 00 in it be a z string, or just a "z", for short. Now consider strings of length n. There are 3^n strings in t...
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Source: http://answers.yahoo.com/question/index?qid=20110215003544AAoQcO8
here is a hint that would allow you to solve t he question rather easily. assume that you change veriables from a and b to some c and d, where c and d are just linear combinations of a and b (for example, c_n = alpha *a_n + beta * b_n, with...
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Source: http://answers.yahoo.com/question/index?qid=20100823001734AAwx5x5
( ri′kər·əns ri′lā·shən ) (mathematics) An equation relating a term in a sequence to one or more of its predecessors in the sequence.
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Source: http://www.answers.com/topic/recurrence-relation
for r0=r1=1 Rn = (((-1)^n+2*2^n)/3)^2 is correct ((-1)^0 + 2*2^0)/3 = 1 ((-1)^1 + 2*2^1)/3 = 1
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Source: http://www.experts-exchange.com/Q_20928349.htm
A recurrence system can be solved by finding and solving its closed form. A closed form is easily found for simple arithmetic or geometric recurrence systems, but may be hard to find for recurrence systems of a more complex nature. In this ...
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Source: http://wiki.answers.com/Q/How_do_you_Solve_recurrence_equations
Well, it just so happens that I figured this out for whoever was wondering... When n is even, and n+1 is odd (obviously) Sn+1 = Sn + (n+2)(3n+2)/4 When n is odd, and n+1 is even (obviously) Sn+1 = Sn + (n+1)(3n+5)/4 These formulas are for t...
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Source: http://wiki.answers.com/Q/What_is_the_recurrence_relation_of_the_...
