Fibonacci-Like Sequences and Pell Equations

Transcription

1 n local maxima for f (x corresponds to an inflection point for F(x that does not admit a Descartes line. Hence the polynomial F(x of degree n + has a Descartes number of n. Note that given the restrictions on F(x of symmetry increasing values and Descartes lines with zero slopes the solution above is unique up to multiplication by a positive constant and the choice of values a. Every cubic has exactly one inflection point which admits a Descartes line and so every cubic has a Descartes number of. Moreover the polynomial x n+ has a Descartes number of for all n. Is there for each n a polynomial of degree greater (less than n + with a Descartes number of n? The reader may thin of other questions regarding Descartes lines but we will pose just two more in this paper. How does a vertical shift either up or down of f (x affect F(x? In particular how does such a shift affect the slope of the Descartes lines? Is there another family of continuous curves such that for each n there is a curve that admits a Descartes number of n? The students solution to the project relied heavily on Mathematica for generating examples and providing graphical insight. Judging from their written explanations of the role that concavity played in their solutions the students thereby gained a better understanding of concavity especially inflection points. They understood how their visual impressions of a curve bending up or down was related to the tangent line lying below or above the curve. Their appreciation for the relationship between the first and second derivatives of a given function and the shape of the corresponding curve was also enhanced. All who solved the extra credit problem conjectured that some quintic should have two Descartes lines and then used graphical reasoning; that is examining the concavity and slope and placing tangent lines in the plot of a candidate quintic until they found one that wored. One student first constructed a quartic and too the antiderivative. See [] for a website from which you can download two Mathematica noteboos one illustrating solutions to the Descartes tangent number problem and the other containing the calculus project. Also see [3] for more on Descartes method. References. William Barnier Allan Cruse and Millianne Granberg Lectures on Freshman Calculus Addison-Wesley Jeff Suzui The Lost Calculus ( : Tangency and Optimization without Limits Mathematics Magazine 78 ( Fibonacci-Lie Sequences and Pell Equations Ayoub B. Ayoub (aba@ .psu.edu Pennsylvania State University Abington College Abington PA 900 Pell equations although they are not as widely nown as the Pythagorean equation x + y = z belong to the fascinating area of Diophantine equations in elementary number theory. The Pell equations are of the form x dy =± ( VOL. 38 NO. JANUARY 007 THE COLLEGE MATHEMATICS JOURNAL 49

2 where d is a non-square positive integer and solutions are also to be positive integers. An example of such an equation is x 8y =. It is easy to see that the smallest positive solution is x = 3andy =. From this the equation x n + 8 y n = ( x + 8 y n gives the nth solution; thus (x y = (7 6 (x 3 y 3 = (99 35and so on. The solution (3 which generates all other integral solutions is called the fundamental solution. Another way to solve the equation x 8y = is to expand 8 in its continued fraction that is 8 = The solutions to our equation (3 (7 6 ( are then obtained from the even numbered terms of the sequence of truncated values (called convergents of the continued fraction. Both of these methods can be used to solve the general Pell equation (; see []and[3]. In this note we present a different method one involving Fibonacci-lie sequences to solve special Pell equations of the form x ( + 4y =±. ( In doing this we first obtain a larger set of solutions some of which are non-integral. Rational solutions. For a fixed positive integer we define the -Fibonacci sequence by a 0 = 0 a = and a n+ = a n + a n for n. Straightforward induction can be used to verify the following explicit formula for the terms of this sequence: a n = ( n ( + 4 n n. + 4 This formula can be derived by applying the techniques of homogeneous second order difference equations to a n+ a n a n = 0; see []. Now let x n = (a n+ + a n and y n = a n. (3 We will show that this pair is a solution to (. A little algebra yields + n + 4 y n = + 4 and x n ( n + 4 y n = + 4 (4 50 c THE MATHEMATICAL ASSOCIATION OF AMERICA

3 which when multiplied gives the result x n ( + 4y n = ( n =±. For example the -Fibonacci sequence starting with a 0 is so from (3 we get these solutions to our equations x 8y =±: ( (3 (7 5 9 (7 6 (4 ( Similarly the 3-Fibonacci sequence begins with and the corresponding solutions to the equations x 3y =±are ( 3 ( 3 ( 9 ( ( ( Note that in the first example (where is even alternate solutions are integral while in the second (where is odd every third solution is integral. It turns out that this is no coincidence. Integral solutions. We consider the cases of odd and even separately. Case. even: It follows from the definition of the -Fibonacci sequence that the even-numbered terms are even and the odd-numbered ones are odd. Therefore x n and y n are integers if and only if n is even. Consequently (x r y r is a solution to x ( + 4y =+. Note that in this case the other equation x ( + 4y = has no integer solutions. For if there were one then x (mod4 and this cannot be. For example consider the case = 4. The 4-Fibonacci sequence is Then (x y = (9 and (x 4 y 4 = (6 36 are integer solutions to x 0y =. Case. odd: Since a n+ a n = a n + a n (a n a n = ( a n + ( + a n the right hand side is even. This implies that a n+ and a n have the same parity. Since the first three terms of the -Fibonacci sequence are and + the first two terms are odd and the third is even. Hence x n and y n are integers if and only if n is a multiple of 3. Consequently (x 3r y 3r is a solution to x ( + 4y =±. For example if = 5 the 5-Fibonacci sequence is Then (x 3 y 3 = (70 3 and (x 6 y 6 = ( are solutions to x 9y =±. The Fibonacci sequence itself that is when = is Applying the above result gives integral solutions to x 5y = ± and the first three are ( (9 4and(38 7. Now we show that the approach used here produces all the integral solutions including the fundamental solutions of the special family of Pell equations. As we saw in (4 the joint equation of (x n y n + n + 4 y n = + 4 gives rational solutions to xn ( + 4yn =±. However for the integral solutions the cases of even and odd needed to be considered separately. VOL. 38 NO. JANUARY 007 THE COLLEGE MATHEMATICS JOURNAL 5

4 If is even the integral solutions are given by + n + 4 y n = + 4 ; or equivalently ( y n = This equation gives all the integral solutions of xn ( + 4yn = if ( + (x y = ( n. (5 is the fundamental solution. To show that this is indeed the case let = a. Then the continued fraction of + 4is[; a a...] which is of period two. Hence its second convergent (a + /a corresponds to the first solution of the equation which is the smallest one (see [3]. If we replace a by wegettheconvergent Thus ( / +. ( + is the fundamental solution. If is odd the integral solutions are given by + 3n + 4 y n = + 4 which is the same as ( y n = + ( n (6 This equation gives all the integral solutions of xn ( + 4yn =±if ( is the fundamental solution. Let = a + ; then the continued fraction of + 4is [; a a a a...] which is of period five. Then the fifth convergent (a + a + /(a + a + corresponds to the first solution of xn ( + 4yn = (see [3] and hence the smallest one. If we replace a by we get the fundamental solution ( c THE MATHEMATICAL ASSOCIATION OF AMERICA

5 References. Underwood Dudley Elementary Number Theory nd ed. Freeman Ronald E. Micens Difference Equations nd ed. Van Nostrand Reinhold Ivan Niven Herbert Zucerman and Hugh Montgomery An Introduction to the Theory of Numbers 5th ed. Wiley 99. Tennis with Marov Roman Wong and Megan Zigarovich edu Washington and Jefferson College Washington PA 530 In his article A Geometric Series from Tennis in the May 005 issue of this Journal Sandefur [] showed that if the probability that player A wins a point against player B has a constant value p then the probability that A will win a game from deuce (by the required two points is P(Awins deuce = p (p( p n = n=0 p p + p. ( This result established by the same method appeared earlier in Ian Stewart s boo [3] Game Set and Math pp Furthermore Stewart gave a complete analysis not only for a single game but for an entire best-two-of-three-sets match with possible tie-brea. More recently the result ( was illustrated using a matrix approach by Hodgson and Bure []. This method was extended to a more formal Marov chain approach by the second author in a capstone course at Washington and Jefferson College. This note describes that wor. Marov chains and stochastic matrices A Marov chain is a sequence of random values whose probabilities at the next states depend only on the state at the time and no prior history. The controlling factor in a Marov chain is the transition probability. It is a conditional probability for the system to go to a particular new state given the current state of the system. Following Sandefur we assume that p is the probability that a player wins the next point in a tennis match. This assumption means that serving is not an advantage in winning a point as seems to be the case in women s tennis. In our tennis problem (or any win-by--points game the five states that a player can reach after the deuce position are ( loss ( advantage-out (3 deuce (4 advantage-in and (5 win. By our assumption the probability of maing the transition between states remains constant from point to point. For instance the probability of going from deuce to advantage-in in one service point is always p. That gives us the following transition matrix P: P = from Loss Ad Out Deuce Ad In Win to Loss Ad Out Deuce Ad In Win p 0 p p 0 p p 0 p VOL. 38 NO. JANUARY 007 THE COLLEGE MATHEMATICS JOURNAL 53