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Transcript
ON A SET OF NEW MATHEMATICAL CONSTANTS It is well known that the ratio R[n+1]=F[n+1]/F[n] of the Fibonacci numbers F[n] approaches the Golden Ratio- 1 5 1.61803398874989484820458683437 2 as n gets large. This number was already known to the ancient Greeks and has been given considerable attention in a variety of areas. The book “The Golden Ratio” by Mario Livio gives an entertaining discussion of this constant not only in connection with its role in mathematics but also in the arts, literature, and designs found in nature. What has apparently not been recognized is that is only one of an infinite number of related mathematical constants which are derivable from Fibonacci type number sequences. It is our purpose here to show how these new mathematical constants are arrived at. We begin with the standard Fibonacci sequence as first proposed by the early Italian mathematician Leonardo of Pisa (1170-1250). It readsF[n]={ 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 … } This sequence follows the generating formulaF[n+1]=F[n]+F[n-1], which may also be written as F[n+1]=2F[n]-F[n-2] If one looks at the ratio R[n+1]=F[n+1]/F[n], one getsR[n]= { 2,3/2, 5/3,8/5,13/8, 21/13, 34/21,55/34,89/55,144/89,233/144,..} This implies as n we get R[n]= as already known to Johannes Kepler (1571-1630). , who believed the golden ratio to be a divine number. In the 18 hundreds the French mathematician Edouward Lucas (1842-1891) showed that the Fibonacci sequence will look quite different if one replaces the first two integers by other numbers. Thus a Lucas sequence might readL[n]= { 2 1 3 4 7 11 18 29 47 76 123 199 323 521 843 1364 2207 .. } . What is here especially interesting is that the ratio L[n+1]/L[n] heads toward the same Golden Ratio as n gets large. This fact clearly shows that the staring values L[1] and L[2] in the Lucas-Fibonacci sequence do not influence the final ratio R[n] as n. We can write a generic form of the Fibonacci Sequence starting with F[1]=a and F[2]=b. this produces – G[n]= { a b (a+b) (a+2b) (2a+3b) (3a+5b) (5a+8b) .. } Also G[n+1]=2G[n]-G[n-2]. From this we can surmise that- lim aG[n ] bG[n 1] n aG[n 1] bG[n ] If we next let R[n+1]=G[n+1]/G[n], we get the interesting result that- R[n 2] 2 1 R[n ]R[n 1] Letting n approach infinity, the three values of R shown should all converge to the same value x. As a result we obtain the cubic algebraic equation- x3 2x2 1 which is no longer dependent on either a or b but rather offers the unique solutionsx= { 1 , 1.618033989, -0.6180339890 }={ 1, , -} The solution of interest to us is =1.6118033.. . The other solution -=-(-1) is the negative of the compliment to . A plot of y=x3-2x2+1 follows- The three real roots are indicated. One can also divide out the obvious root of x=1 to get the quadratic- x2 x 1 This equation has the closed form solutions- x 1 5 2 and x 1 5 2 In view of the discussions above concerning the standard Fibonacci sequence and its connection to the Golden Ratio, we are now in a position to define some new mathematical constants. Take first the sequence of numbersS={ 1 2 3 6 11 20 37 68 125 230 423 378 1431 263 ..} This sequence follows the ruleq[n+1]=q[n]+q[n-1]+q[n-2] or the equivalent q[n+1]=2q[n]-q[n-3] This time the ratio R[n+1]=q[n+1]/q[n] seems to be going toward a constant near x=1839… as n gets large and no longer toward . The solution for this limit is governed by the following 4th order algebraic equation- x4 2x3 1 Eliminating the trivial solution of x=1, we are left with solving- x3 x2 x 1 This equation has the solutionsx=1.8392867552141611325, and x=-.419643377I 0.606290729 Again the real solution1.839…is the one of interest to us. Continuing on, we can write down a new sequenceT= { 1 2 3 4 10 19 36 69 134 258 497 958 … } in which the first four terms shown in blue are given. We see the ratio T[n+1]/T[n] appears to approach another constant near x=1.927. Looking at the x equations in the two previous cases , we suspect that solving- x4 x3 x2 x 1 will yield a new mathematical constant.. Using our PC to do so, we get the one root of interest among a total of four roots to bex= 1.9275619754829253043… It is now clear if one generates a Fibonacci like sequence with the first N values specified , the ratio term will tend to a value obtained by solvingN 1 xN xn n 0 But we recognize the right hand side of this expression to be the finite geometric series. This fact allows us to sum things to obtain- 1 xN x 1 x N or its equivalent x N 1 (2 x ) This last equality can be thought of as the generating function for a new set of mathematical constants. It can be solved by a simple one-line iteration programx[1]:=2; for n from 1 to 30 do x[n+1]:=evalf (2-1/x[n]^N, 25) od; Here is a short table giving the value of x for integer N accurate to 25 places- N 1 2 3 4 5 6 x 1 1.6180339887498948482045868 1.8392867552141611325518525 1.9275619754829253042619058 1.9659482366454853371899373 1.9835828434243263303856292 These mathematical constants all stem from the same algebraic equation. It is clear that as N gets large the constant value approaches x=2. The value of x at N=2 is just the Golden Ratio . All the listed values beyond N=2 are new universal mathematical constants. A wealth of mathematical identities may be generated involving these constants. Consider first the constant x=1.839.. which we will designate below as . It occurs at N=3 and has the properties- 3 1 2 1 =6.222262515 2 Solving for 2, we find- 2 1 3.38297576... 1 Also we find- 4 , 2 5 2 (2 ) and 6 1 (2 ) 2 These results may be generalized to yield- n 3 n (2 ) Furthermore – 1 n 0 n ( 1) and ( 1) n n 0 n ( 1) as can be derived via the geometric series. Combining these sums, we have n 0 1 2n 1 1 2 1 4 1 6 ... ( 1) 1.419643378... 2 and ( 1) n n 0 2n 2 0.7718445064 (1 2 ) The tau constant also lends itself to a geometrical interpretation via the following right triangle- The angles are unique and the three sides satisfy the Pythagorean Theorem. A related triangle applicable to the Golden Ratio was first found by Johannes Kepler of astronomy fame. We can continue on with the N=4 case denoting the value x=1.927..by . We have the identity- 4 1 3 2 1 (2 ) From it one has- 4 n n (2 ) with 4 m 1 (2 )m Thus we find- 1 12 (2 ) (2 )4 16 One may also generate finite and infinite sums with . Consider the finite series- 1 1 N 1 1 1 1 1 1 2 ... N n 1 n 0 1 N The infinite series version has the even simpler form 1 n 0 n ( 1) 2.07809507766784... This time a right triangle can be formed using the form- 1 2 2 1 ( 2 )(1 ) 2 n also look at the constants corresponding to N=5, 6, 7,..etc.These also will satisfy their own equalities. We will not pursue these here except to point out that one uses similar arguments as those used for finding the and identities. In particular one has the starting definition involving powers of the constants. It readsN 1 x N m x nm n 0 April 25, 2015