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notes-091015

# notes-091015 - MATH 681 1 Notes Combinatorics and Graph...

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MATH 681 Notes Combinatorics and Graph Theory I 1 Recurrence relations, continued yet again One last lousy class of recurrences we should be able to solve: 1.1 Systems of recurrence relations Sometimes multiple recurrences working in tandem are more effective than a single recurrence. Let’s see how such a thing might be of assistance. Question 1: Let a n represent the number of strings from { 0 , 1 , 2 } of length n with an even number of 0s. Find a recurrence for a n . Answer 1: Note that recurrences are not actually the best way to solve this particular problem: it would be a lot easier to do it with the exponential generating function e x + e - x 2 e x e x = e 3 x + e x 2 = n =0 3 n +1 2 x n n ! to get a n = 3 n +1 2 . But, armed with that knowledge, let’s look at how we might tackle it as a recurrence We introduce an auxiliary sequence b n , which counts the bitstrings of length n with an odd number of zeroes. Now, we know: a n = 2 a n - 1 + b n - 1 b n = a n - 1 + 2 b n - 1 and a 0 = 1 and b 0 = 0 . This uniquely determines a n , but solving it might be tricky. We can have recourse to either OGFs or EGFs to solve it, depending whether you prefer algebra or differential equations. Let’s suppose we use an OGF; letting f ( x ) = n =0 a n x n and g ( x ) = n =0 b n x n . Then the first equation in the recurrence can be simplified to f ( x ) - a 0 = 2 xf ( x ) + xg ( x ) ; the second will become g ( x ) - b 0 = xf ( x ) + 2 xg ( x ) . Collecting terms, we’ll get the system: f ( x ) = xg ( x ) + 1 1 - 2 x g ( x ) = xf ( x ) 1 - 2 x so, solving for f ( x ) in terms of itself, f ( x ) = x xf ( x ) 1 - 2 x + 1 1 - 2 x = x 2 f ( x ) - 2 x + 1 (1 - 2 x ) 2 and thus 1 - x 2 (1 - 2 x ) 2 f ( x ) = 1 - 2 x (1 - 2 x ) 2 so f ( x ) =

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notes-091015 - MATH 681 1 Notes Combinatorics and Graph...

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