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We test the conjecture for the case ra = 3 and possibly others and then we prove the conjecture. Figure 12.6 shows a diagrammatic proof. Note that, at each stage, the square of size m is increased in size by adding 2m + 1 dots, as indicated by the boxes1. Typically, inductive reasoning is not so straightforward. More often than not, the conjectures we make are unfounded. They do not stand up to proof and have to be discarded or, at best, modified in some way. In order to improve the effectiveness of inductive reasoning, it is important to limit the amount of guesswork, reducing induction as far as possible to deduction. Of course, it is never possible to eliminate guesswork altogether otherwise the creative element of inductive reasoning would be eliminated, and that is too much to expect.

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The effect of the parameter C on the symmetry of E is illustrated in Fig. 5.16. Again, the investment made earlier, in structuring the problem in simple units, means that all that needs to be done in order to specify all of the quantities in the differential equation (5.23) exactly, is to evaluate the integrals Ij with the new form for r". Consider I1 in the usual plane polar coordinates, i.e.

Therefore:

The principle of mathematical induction provides a method of proving that a property P predicated on natural numbers is true for all natural numbers2. An example is the predicate S defined by S.n = < I k : l < k ^ n : k > = An(n+l)

5

As previously, this can be integrated by parts. Choosing:

(12.6)

then:

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(Using the dotdotdot notation: S.n = l + 2 + . . . + n = | n ( n + l ) . We prefer to use the quantifier notation in order to be precise and unambiguous. In particular, the quantifier notation makes it clear that S.O is well defined, whereas the dotdotdot notation appears to exclude the case that n equals 0.)

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Therefore:

The leftmost square in the diagram has zero dots, but you cannot see them! The inability to handle important special cases here, the case m = 0 is a major drawback of diagrams. 2 Recall that a natural number is a non-negative integer. So the natural numbers are the numbers 0, 1, 2, etc.

and so:

1 72

6

leading finally to:

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The essence of the principle of mathematical induction is that an arbitrary property P of the natural numbers is provably true for all natural numbers, n, if it is possible to prove (i) P.O is true, and (ii) for all n, P.(n+l) follows from the assumption that P.n is true. Example 12.7. To illustrate the principle let us apply it to the predicate S defined in (12.6). We begin by proving S.O. This first step is called the basis of the proof. We have

Secondly:

0 =0

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With this final substitution, i.e. the use of z", then this form resembles the 3D version from earlier, and use can be made of some of the results. For example, using equation (5.80), then:

definition } empty-range rule to simplify the summation, arithmetic for the right side of the equality }

Returning to z' and substituting r" for rL, then obtain:

reflexivity of equality }

B.1.5

As the values r" cancel, the integration then becomes trivial, and indeed:

The next step, called the induction step, is to show that S.(n+l) follows from the assumption S.n. We have { { { definition } range splitting applied to the summation } assume S.n . That is, assume that

In order to evaluate I3, it is again necessary to deduce V2e-r/A first. In a similar manner to equation (5.102), then:

( Z k : l^k^n+l : k ) = f ( n + l ) ( ( n + l ) + l)

7

Considering z" as the direct analogy of z' for the 3D case, then use can be made of the earlier result in equation (5.82), hence:

arithmetic }

Using the equivalent forms for and summing gives:

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The crucial step in this calculation is the bulleted step in which the assumption is made that S.n is true. This assumption is called the induction hypothesis. The final step is to cite the principle of mathematical induction to combine the basis and the induction step in the conclusion that property S.n is true for all natural numbers n. D

also given in equation (5.82)

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