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The (inescapable) p -adics Alex J. Best 5/5/2018 BU Math Retreat 2018 Linear recurrence sequences Definition (Linear recurrence sequence) A linear recurrence sequence , is a sequences whose n th term is the linear combination of the previous k

  1. The (inescapable) p -adics Alex J. Best 5/5/2018 BU Math Retreat 2018

  2. Linear recurrence sequences Definition (Linear recurrence sequence) A linear recurrence sequence , is a sequences whose n th term is the linear combination of the previous k terms (for all n ≥ k ) 1

  3. Linear recurrence sequences Definition (Linear recurrence sequence) A linear recurrence sequence , is a sequences whose n th term is the linear combination of the previous k terms (for all n ≥ k ) Example (Fibonacci) a 0 = 0 , a 1 = 1 and a n = a n − 1 + a n − 2 for n ≥ k = 2: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377 , 610 , 987 , 1597 , 2584 , 4181 , 6765 1

  4. Linear recurrence sequences Definition (Linear recurrence sequence) A linear recurrence sequence , is a sequences whose n th term is the linear combination of the previous k terms (for all n ≥ k ) Example (Fibonacci) a 0 = 0 , a 1 = 1 and a n = a n − 1 + a n − 2 for n ≥ k = 2: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377 , 610 , 987 , 1597 , 2584 , 4181 , 6765 a n grows exponentially. 1

  5. Linear recurrence sequences Definition (Linear recurrence sequence) A linear recurrence sequence , is a sequences whose n th term is the linear combination of the previous k terms (for all n ≥ k ) Example (A periodic sequence) a 0 = 1 , a 1 = 0 with a n = − a n − 1 − a n − 2 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , 1

  6. Linear recurrence sequences Definition (Linear recurrence sequence) A linear recurrence sequence , is a sequences whose n th term is the linear combination of the previous k terms (for all n ≥ k ) Example (A periodic sequence) a 0 = 1 , a 1 = 0 with a n = − a n − 1 − a n − 2 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , a n is periodic now. 1

  7. Linear recurrence sequences Definition (Linear recurrence sequence) A linear recurrence sequence , is a sequences whose n th term is the linear combination of the previous k terms (for all n ≥ k ) Example (Natural numbers interlaced with zeroes) a 0 = 1 , a 1 = 0 , a 2 = 2 , a 3 = 0 with a n = 2 a n − 2 − a n − 4 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , 9 , 0 , 10 , 0 , 11 , 0 , 12 , 0 , 13 , 0 , 14 , 0 , 15 , 0 1

  8. Linear recurrence sequences Definition (Linear recurrence sequence) A linear recurrence sequence , is a sequences whose n th term is the linear combination of the previous k terms (for all n ≥ k ) Example (Natural numbers interlaced with zeroes) a 0 = 1 , a 1 = 0 , a 2 = 2 , a 3 = 0 with a n = 2 a n − 2 − a n − 4 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , 9 , 0 , 10 , 0 , 11 , 0 , 12 , 0 , 13 , 0 , 14 , 0 , 15 , 0 not periodic but the zeroes do have a regular repeating pattern. 1

  9. The ultimate question Question What possible patterns are there for the zeroes of a linear recurrence sequence? 2

  10. The ultimate question Question What possible patterns are there for the zeroes of a linear recurrence sequence? Observation A linear recurrence sequence is the Taylor expansion around 0 of a rational function a 1 + a 2 x + · · · + a ℓ x ℓ b 1 + b 2 x · · · + b k x k with b 1 � = 0 (so that the expansion makes sense). 2

  11. Linear recurrence sequences Example x 1 − x − x 2 . ↔ Fibonacci 3

  12. Linear recurrence sequences Example x 1 − x − x 2 . ↔ Fibonacci 1 1 + x + x 2 . ↔ 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , 3

  13. Linear recurrence sequences Example x 1 − x − x 2 . ↔ Fibonacci 1 1 + x + x 2 . ↔ 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , 1 ( 1 − x 2 ) 2 . ↔ 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , 9 , 0 , 10 , 0 , 11 , 0 , 12 , 0 , 13 , 0 , 14 3

  14. Linear recurrence sequences Example x 1 − x − x 2 . ↔ Fibonacci 1 1 + x + x 2 . ↔ 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , − 1 , 1 , 0 , 1 ( 1 − x 2 ) 2 . ↔ 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , 9 , 0 , 10 , 0 , 11 , 0 , 12 , 0 , 13 , 0 , 14 ( 1 + x ) 3 − x 3 ( 1 + x ) 5 − x 5 ↔ 1 , − 2 , 3 , − 5 , 10 , − 20 , 35 , − 50 , 50 , 0 , − 175 , 625 , − 1625 , 3625 , − 7250 , 13125 , − 21250 , 29375 , − 29375 , 0 , 106250 , − 384375 , 1006250 , − 2250000 , 4500000 , − 8140625 , 13171875 , − 18203125 , 18203125 , 0 , − 65859375 , 238281250 3

  15. Consequences Observation The set of all linear recurrence sequences is a vector space! Hard to tell how the rule changes. 4

  16. Consequences Observation The set of all linear recurrence sequences is a vector space! Hard to tell how the rule changes. We can always mess up a finite amount of behaviour. So assume a n has infinitely many zeroes, what is the structure of the zero set? 4

  17. Linear recurrence sequences Example 1 ( 1 − x 2 ) 2 − ( 1 − x + 2 x 2 + 3 x 4 + 4 x 6 ) ↔ 0 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , 5

  18. Linear recurrence sequences Example 1 ( 1 − x 2 ) 2 − ( 1 − x + 2 x 2 + 3 x 4 + 4 x 6 ) ↔ 0 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , Interlacing with 0 and shifting correspond to plugging in x 2 and multiplying by x respectively in the rational functions 5

  19. Linear recurrence sequences Example 1 ( 1 − x 2 ) 2 − ( 1 − x + 2 x 2 + 3 x 4 + 4 x 6 ) ↔ 0 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , Interlacing with 0 and shifting correspond to plugging in x 2 and multiplying by x respectively in the rational functions 1 ( 1 − x ) 2 ↔ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 5

  20. Linear recurrence sequences Example 1 ( 1 − x 2 ) 2 − ( 1 − x + 2 x 2 + 3 x 4 + 4 x 6 ) ↔ 0 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , Interlacing with 0 and shifting correspond to plugging in x 2 and multiplying by x respectively in the rational functions 1 ( 1 − x ) 2 ↔ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 1 ( 1 − x 2 ) 2 ↔ 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , 9 , 0 , 10 , 0 , 11 , 0 , 12 , 0 , 13 5

  21. Linear recurrence sequences 1 ( 1 − x ) 2 ↔ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 1 ( 1 − x 2 ) 2 ↔ 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , 9 , 0 , 10 , 0 , 11 , 0 , 12 , 0 , 13 1 ( 1 − x 4 ) 2 ↔ 1 , 0 , 0 , 0 , 2 , 0 , 0 , 0 , 3 , 0 , 0 , 0 , 4 , 0 , 0 , 0 , 5 , 0 , 0 , 0 , 6 , 0 , 0 , 0 , 7 , 0 , 0 , 6

  22. Linear recurrence sequences 1 ( 1 − x ) 2 ↔ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 1 ( 1 − x 2 ) 2 ↔ 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , 9 , 0 , 10 , 0 , 11 , 0 , 12 , 0 , 13 1 ( 1 − x 4 ) 2 ↔ 1 , 0 , 0 , 0 , 2 , 0 , 0 , 0 , 3 , 0 , 0 , 0 , 4 , 0 , 0 , 0 , 5 , 0 , 0 , 0 , 6 , 0 , 0 , 0 , 7 , 0 , 0 , x ( 1 − x 4 ) 2 ↔ 0 , 1 , 0 , 0 , 0 , 2 , 0 , 0 , 0 , 3 , 0 , 0 , 0 , 4 , 0 , 0 , 0 , 5 , 0 , 0 , 0 , 6 , 0 , 0 , 0 , 7 , 0 , 6

  23. Linear recurrence sequences 1 ( 1 − x ) 2 ↔ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 1 ( 1 − x 2 ) 2 ↔ 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , 9 , 0 , 10 , 0 , 11 , 0 , 12 , 0 , 13 1 ( 1 − x 4 ) 2 ↔ 1 , 0 , 0 , 0 , 2 , 0 , 0 , 0 , 3 , 0 , 0 , 0 , 4 , 0 , 0 , 0 , 5 , 0 , 0 , 0 , 6 , 0 , 0 , 0 , 7 , 0 , 0 , x ( 1 − x 4 ) 2 ↔ 0 , 1 , 0 , 0 , 0 , 2 , 0 , 0 , 0 , 3 , 0 , 0 , 0 , 4 , 0 , 0 , 0 , 5 , 0 , 0 , 0 , 6 , 0 , 0 , 0 , 7 , 0 , 1 + 2 x ( 1 − x 4 ) 2 ↔ 1 , 2 , 0 , 0 , 2 , 4 , 0 , 0 , 3 , 6 , 0 , 0 , 4 , 8 , 0 , 0 , 5 , 10 , 0 , 0 , 6 , 12 , 0 , 0 , 7 , 14 6

  24. Linear recurrence sequences 1 ( 1 − x ) 2 ↔ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , 1 ( 1 − x 2 ) 2 ↔ 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , 5 , 0 , 6 , 0 , 7 , 0 , 8 , 0 , 9 , 0 , 10 , 0 , 11 , 0 , 12 , 0 , 13 1 ( 1 − x 4 ) 2 ↔ 1 , 0 , 0 , 0 , 2 , 0 , 0 , 0 , 3 , 0 , 0 , 0 , 4 , 0 , 0 , 0 , 5 , 0 , 0 , 0 , 6 , 0 , 0 , 0 , 7 , 0 , 0 , x ( 1 − x 4 ) 2 ↔ 0 , 1 , 0 , 0 , 0 , 2 , 0 , 0 , 0 , 3 , 0 , 0 , 0 , 4 , 0 , 0 , 0 , 5 , 0 , 0 , 0 , 6 , 0 , 0 , 0 , 7 , 0 , 1 + 2 x ( 1 − x 4 ) 2 ↔ 1 , 2 , 0 , 0 , 2 , 4 , 0 , 0 , 3 , 6 , 0 , 0 , 4 , 8 , 0 , 0 , 5 , 10 , 0 , 0 , 6 , 12 , 0 , 0 , 7 , 14 Still has periodic zero set, all n congruent to 2 , 3 modulo 4. 6

  25. Approach Expand into partial fractions n j m p ( x ) r ij � � q ( x ) = ( 1 − α i x ) j i = 1 j = 1 7

  26. Approach Expand into partial fractions n j m p ( x ) r ij � � q ( x ) = ( 1 − α i x ) j i = 1 j = 1 do some math:   n j m ∞ � n + j − 1 � � � � α n  x n r ij  i j − 1 n = 0 i = 1 j = 1 7

  27. Approach Expand into partial fractions n j m p ( x ) r ij � � q ( x ) = ( 1 − α i x ) j i = 1 j = 1 do some math:   n j m ∞ � n + j − 1 � � � � α n  x n r ij  i j − 1 n = 0 i = 1 j = 1 Upshot: there are polynomials A i ( n ) such that m � A i ( n ) α n a n = i . i = 1 Like that formula for Fibonacci with the golden ratio in. 7

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