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For Problem 6 given in the OP, the claim is false: Notation: We borrow subscript and superscript notation from D. Knuth's TeX. So, a with a subscript j is wri

by graycat 11y ago

For Problem 6 given in the OP, the claim is false: Notation: We borrow subscript and superscript notation from D. Knuth's TeX. So, a with a subscript j is written a_j. Consider the simple graph, based just on typing, below. The graph shows the first four values of a specific sequence a_j. This sequence is a contradiction to claim of Problem 6. In this graph, the values of j = 1, 2, ... are plotted on the horizontal axis with 1, 2, ..., and the values of the a_j are plotted on the vertical axis from 1 to 2015. We plot a value of a_j as just 'a'. Due to constraint (ii), we indicate by 'x' points that, due to prior values of a_j, cannot have an 'a'. So, the graph is: 2015 . . . a_j . 5 4 3 a a 2 x x ... 1 axax 123456789 j Then the first two terms of the sequence a_j are: a_1 = 3 a_2 = 1 and generally for j = 1, 2, ..., a_j is: / | 3 if j is odd | a_j = < | | 1 if j is even \ Then for any positive integer b and j = 1, 2, ... (a_j - b) + (a_{j + 1} - b) >= 2 More generally for positive integers m and k, sum_{j = m + 1}^{m + 2k} (a_j - b) >= 2 k Then for any positive integer N, we can set m = N and n = m + 2(1007^2) and get | sum_{j=m+1}^n (a_j - b) | = sum_{j=m+1}^n (a_j - b) = 2 (1007^2) > 1007^2 showing that the claim is false.

paulfr 11y ago
There are no absolute values in the summands, so in your example pairs of consecutive terms sum to 0 when you choose b = 2. The theorem seems entirely correct to me. You can prove it with these sub-steps: (1) the set of all j + a_j is the set of nonnegative integers minus a finite number of gaps (2) thus for large enough n you can express \sum_{j=1}^n (j + a_j) as a quadratic function of n, plus a residual term e(n) of magnitude at most 1007^2/2 (3) more precisely, \sum_{j=m+1}^n a_j = g (n-m) + e(n) - e(m) where g is the number of gaps in (1) Then choosing b = g solves the problem. Hope that helps.

by graycat 11y ago

paulfr 11y ago

There are no absolute values in the summands, so in your example pairs of consecutive terms sum to 0 when you choose b = 2. The theorem seems entirely correct to me. You can prove it with these sub-steps: (1) the set of all j + a_j is the set of nonnegative integers minus a finite number of gaps (2) thus for large enough n you can express \sum_{j=1}^n (j + a_j) as a quadratic function of n, plus a residual term e(n) of magnitude at most 1007^2/2 (3) more precisely, \sum_{j=m+1}^n a_j = g (n-m) + e(n) - e(m) where g is the number of gaps in (1) Then choosing b = g solves the problem. Hope that helps.