By John Earman
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Extra resources for A primer on determinism
To illustrate further the method of induction, we now apply it to two worked examples; the ﬁrst concerns the sum of the squares of the ﬁrst n natural numbers. IProve that the sum of the squares of the ﬁrst n natural numbers is given by X n r2 = 16 n(n + 1)(2n + 1). 60) r=1 As previously we start by assuming the result is true for n = N. Then it follows that X N+1 r=1 X N r2 = r2 + (N + 1)2 r=1 = 16 N(N + 1)(2N + 1) + (N + 1)2 = 16 (N + 1)[N(2N + 1) + 6N + 6] = 16 (N + 1)[(2N + 3)(N + 2)] = 16 (N + 1)[(N + 1) + 1][2(N + 1) + 1].
Clearly, it could be argued that any mathematical result can be so expressed but, if the proposition is no more than a guess, the chances of success are negligible. 7 SOME PARTICULAR METHODS OF PROOF are either the result of true inspiration or, much more normally, yet another reworking of an old chestnut! The essence of the method is to exploit the fact that mathematics is required to be self-consistent, so that, for example, two calculations of the same quantity, starting from the same given data but proceeding by diﬀerent methods, must give the same answer.
We next examine Q(N + 1), writing each of its terms as a binomial expansion: Q(N + 1) = (N + 1)4 + 2(N + 1)3 + 2(N + 1)2 + (N + 1) = (N 4 + 4N 3 + 6N 2 + 4N + 1) + 2(N 3 + 3N 2 + 3N + 1) + 2(N 2 + 2N + 1) + (N + 1) = (N 4 + 2N 3 + 2N 2 + N) + (4N 3 + 12N 2 + 14N + 6). Now, by our assumption, the group of terms within the ﬁrst parentheses in the last line is divisible by 6 and clearly so are the terms 12N 2 and 6 within the second parentheses. Thus it comes down to deciding whether 4N 3 + 14N is divisible by 6 – or equivalently, whether R(N) = 2N 3 + 7N is divisible by 3.
A primer on determinism by John Earman