Mathematical Methods for Physics and Engineering - Matematica.NET

1.7 SOME PARTICULAR METHODS OF PROOF1.7.1 Proof by inductionThe proof of the binomial expansion given in subsection 1.5.2 and the identityestablished in subsection 1.6.1 have already shown the way in which an inductiveproof is carried through. They also indicated the main limitation of the method,namely that only an initially supposed result can be proved. Thus the methodof induction is of no use for deducing a previously unknown result; a putativeequation or result has to be arrived at by some other means, usually by noticingpatterns or by trial and error using simple values of the variables involved. Itwill also be clear that propositions that can be proved by induction are limitedto those containing a parameter that takes a range of integer values (usuallyinfinite).For a proposition involving a parameter n, the five steps in a proof usinginduction are as follows.(i) Formulate the supposed result for general n.(ii) Suppose (i) to be true for n = N (or more generally for all values ofn ≤ N; seebelow),whereN is restricted to lie in the stated range.(iii) Show, using only proven results and supposition (ii), that proposition (i)is true for n = N +1.(iv) Demonstrate directly, and without any assumptions, that proposition (i) istrue when n takes the lowest value in its range.(v) It then follows from (iii) and (iv) that the proposition is valid for all valuesof n in the stated range.(It should be noted that, although many proofs at stage (iii) require the validityof the proposition only for n = N, some require it for all n less than or equal to N– hence the form of inequality given in parentheses in the stage (ii) assumption.)To illustrate further the method of induction, we now apply it to two workedexamples; the first concerns the sum of the squares of the first n natural numbers.◮Prove that the sum of the squares of the first n natural numbers is given byn∑r 2 = 1 n(n + 1)(2n +1). (1.60)6r=1As previously we start by assuming the result is true for n = N. Then it follows that∑N+1N∑r 2 = r 2 +(N +1) 2r=1r=1= 1 N(N + 1)(2N +1)+(N +1)26= 1 (N +1)[N(2N +1)+6N +6]6= 1 (N + 1)[(2N +3)(N +2)]6= 1 (N +1)[(N +1)+1][2(N +1)+1].631