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Fibonacci Numbers, Binomial Coefficients and Mathematical Induction

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Fibonacci Numbers, Binomial Coefficients and Mathematical Induction Math 163 - Introductory Seminar Lehigh University Spring 2008 Notes on Fibonacci numbers, binomial coe±cients and mathematical induction. These are mostly notes from a previous class and thus include some material not covered in Math 163. For completeness this extra material is left in the notes. Observe that these notes are somewhat informal. Not all terms are de¯ned and not all proofs are complete. The material in these notes can be found (with more detail in many cases) in many di®erent textbooks. Fibonacci Numbers What are the Fibonacci numbers? The Fibonacci numbers are de¯ned by the simple recurrence relation Fn = Fn¡1 + Fn¡2 for n ¸ 2 with F0 = 0;F1 = 1: This gives the sequence F0;F1;F2;::: = 0; 1; 1; 2; 3; 5; 8; 13; 21; 34; 55; 89; 144; 233;:::. Each number in the sequence is the sum of the previous two numbers. We read F0 as `F naught'. These numbers show up in many areas of mathematics and in nature. For example, the numbers of seeds in the outermost rows of sunflowers tend to be Fibonacci numbers. A large sunflower will have 55 and 89 seeds in the outer two rows. Can we easily calculate large Fibonacci numbers without ¯rst calculating all smaller values using the recursion? Surprisingly, there is a simple and non-obvious formula for the Fibonacci numbers. It is: à p !n à p !n 1 1 + 5 ¡1 1 ¡ 5 Fn = p + p : 5 2 5 2 p 1¡ 5 It is not immediately obvious that this should even give an integer. Since ¡1 < 2 < 0 (it is approximately ¡:618) the second term approaches 0 as n gets large. Thus the ¯rst term is a good approximation of the Fibonacci numbers. In fact à p !n 1 1 + 5 Fn is the integer closest to p : 5 2 The Golden Ratio p 1+ 5 The number 2 shows up in many places and is called the Golden ratio or the Golden mean. For one example, consider a rectangle with height 1 and width x. If a vertical line is drawn in the middle so that the left side is a square and the right side is a smaller rectangle proportional to the original then x is the golden ratio. To see this, note that for the rectangles to be proportional, the ratios of the longer sides to the smaller are equal. That is x=1 = 1=(x ¡ 1). So x(x ¡ 1) = 1 ) x2 ¡ x ¡ 1. Using 1 p p the quadratic formula x must be (1 + 5)=2 or (1 ¡ 5)=2. Since the second is negative, x is the ¯rst term, the golden ratio. This proportion for rectangles shows up for example in the dimensions of the Parthenon and in art. There is no evidence that the ancient Greek used the golden ratio in planning the parthenon. Rather the shape seems to be aesthetically pleasing. Da Vinci did sketches of the golden ration in relation to the human body. There is debate as to whether or not he incorporated these ideas into his paintings. Dali explicitly used the golden ratio in some of his paintings. Other examples of recurrences Consider two other examples of simple recurrences If An = 2An¡1 ¡ 2An¡2 for n ¸ 2 with A0 = 2 and A1 = 2 we get the sequence 2; 2; 0; ¡4; ¡8; ¡6; 0; 16; 32; 32; 0;:::. A number in this sequence is twice the previous number minus twice the number preceding the previous number. The general expression for An is n n evenp more surprising than that for Fibonacci numbers: An = (1 + i) + (1 ¡ i) where i = ¡1. A simpler example that will be useful for illustration is Bn = Bn¡1 + 6Bn¡2 for n ¸ 2 with B0 = 1 and B1 = 8. This recurrence gives the sequence 1; 8; 14; 62; 146; 518;:::. The general n n formula is Bn = 2 ¢ 3 + (¡1)(¡2) . Mathematical Induction Later we will see how to easily obtain the formulas that we have given for Fn;An;Bn. For now we will use them to illustrate the method of mathematical induction. We can prove these formulas correct once they are given to us even if we would not know how to discover the formulas. First we give a proof using the idea of contradiction and that of a minimal counterexample. This is essentially the same as what we will do with induction but using slightly di®erent language. Proposition: If Bn = Bn¡1 + 6Bn¡2 for n ¸ 2 with B0 = 1 and B1 = 8 then Bn = 2 ¢ 3n + (¡1)(¡2)n. Proof (using the method of minimal counterexamples): We prove that the formula is correct by contradiction. Assume that the formula is false. Then there is some smallest value of n for which it is false. Calling this value k we are assuming that the formula fails for k k k but holds for all smaller values. That is, we assume that Bk 6= 2 ¢ 3 + (¡1)(¡2) but that k¡1 k¡1 k¡2 k¡2 0 0 Bk¡1 = 2¢3 +(¡1)(¡2) , Bk¡2 = 2¢3 +(¡1)(¡2) ;:::;B0 = 2¢3 +(¡1)(¡2) . Note 0 0 ¯rst that substituting n = 0 and n = 1 into the formula we get B0 = 2 ¢ 3 + (¡1)(¡2) = 1 1 1 and B1 = 2 ¢ 3 + (¡1)(¡2) = 8. So the formula works for n = 0; 1. Thus k ¸ 2 and we can apply the recursion to Bk. Then using the recursion and the assumptions of the correctness 2 of the formulas for Bk¡1 and Bk¡2 we get B = B + 6B (1) k £ k¡1 k¡2 ¤ £ ¤ = 2 ¢ 3k¡1 + (¡1)(¡2)k¡1 + 6 2 ¢ 3k¡2 + (¡1)(¡2)k¡2 (2) = 2 (3 + 6) 3k¡2 + (¡1) (¡2 + 6) (¡2)k¡2 (3) = 2 ¢ 32 ¢ 3k¡2 + (¡1) ¢ (¡2)2 ¢ (¡2)k¡2 (4) = 2 ¢ 3k + (¡1)(¡2)k (5) Thus the formula holds for k, contradicting the assumption that k is the smallest number for which the formula fails. So there can be no such number: the formula holds for all n = 0; 1; 2;:::. 2 The key to the proof was showing that if the formula is correct for Bk¡2 and Bk¡1 then it is correct for Bk. These are the displayed computations above. Those computations work for any k. So we could use them along with the easily checked fact that the formula is correct for B0 and B1 to show that it is correct for B2. Then since it is correct for B1 and B2 we use the computations above to see that the formula is correct for B3. Continuing this way we see that we can build up to showing that the formula is correct for any Bn. We display this as follows: Formula holds for B0 and B1 ) (using equations (1)-(5) ) formula holds for B2) Formula holds for B1 and B2 ) (using equations (1)-(5) ) formula holds for B3) Formula holds for B2 and B3 ) (using equations (1)-(5) ) formula holds for B4) . Formula holds for Bk¡2 and Bk¡1 ) (using equations (1)-(5) ) formula holds for Bk) . It is apparent that this shows that this shows that the formula works for all n = 0; 1; 2;:::. The language used by mathematicians to describe and present this process is called math- ematical induction. This provides a succinct way of stating the ideas described above. Technically it is equivalent to the minimal counterexample proof we gave above but in many situations the presentation is cleaner and shorter. We present the same proof using the terminology of mathematical induction. Proposition: If Bn = Bn¡1 + 6Bn¡2 for n ¸ 2 with B0 = 1 and B1 = 8 then Bn = 2 ¢ 3n + (¡1)(¡2)n. Proof (using mathematical induction): We prove that the formula is correct using mathe- 0 0 1 1 matical induction. Since B0 = 2 ¢ 3 + (¡1)(¡2) = 1 and B1 = 2 ¢ 3 + (¡1)(¡2) = 8 the formula holds for n = 0 and n = 1. For n ¸ 2, by induction B = B + 6B n £ n¡1 n¡2 ¤ £ ¤ = 2 ¢ 3n¡1 + (¡1)(¡2)n¡1 + 6 2 ¢ 3n¡2 + (¡1)(¡2)n¡2 = 2 (3 + 6) 3n¡2 + (¡1) (¡2 + 6) (¡2)n¡2 = 2 ¢ 32 ¢ 3n¡2 + (¡1) ¢ (¡2)2 ¢ (¡2)n¡2 = 2 ¢ 3n + (¡1)(¡2)n 3 Hence the formula holds for all n = 0; 1;:::. 2 The words `by induction' (sometimes `by the induction hypothesis' is used) are shorthand for the idea described above that we have already proved the statement for smaller values of n and are presenting the method to get to the next value and that this can be repeated to show the statement for all n. Induction and the well ordering principle Formal descriptions of the induction process can appear at ¯rst very abstract and hide the simplicity of the idea. For completeness we give a version of a formal description of mathematical induction and also that of the well ordering principle on which the minimal counterexample proof was based. The two are equivalent in that if either is assumed as a basic axiom then the other can be shown to follow from the axiom. Well ordering principle - Every non-empty subset of integers contains a smallest element. Principle of mathematical induction - If Sn is a statement about the positive integer n such that S1 is true and Sk is true whenever Sk¡1 is true then Sn is true for all positive integers. Solving linear recurrences The example above with the Golden ratio and rectangles involved the quadratic x2¡x¡1 = 0.
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