## Tuesday, August 22, 2017

### Polynomial Equations and Associated Number Sequences (1)

As is well known the Fibonacci sequence is

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987,…

This in turn is related to the simple binomial equation, x2 – x – 1 = 0.

The positive real valued solution to this equation is given by the important constant
ϕ (phi) = (1 + √5)/2  = 1.618033….

The Fibonacci sequence itself can be obtained in the following manner.

The general binomial equation can be given as ax2 + bx + c = 0.

Where a = 1, this simplifies to x2+ bx + c = 0.

Then starting with the two numbers 0 and 1 we can obtain the Fibonacci sequence in the following manner.
To obtain the next number add 1 *(– b) + 0 (– c).

So in this specific case we get 1 * (1) + 0 * (1)  =  1.

Thus the next term in the Fibonacci sequence = 1.

So we now have the terms 0, 1, 1.

Then to obtain the next term we again multiply this latest generated term (i.e. 1) by – b, before adding it to the previous terms (i.e. 1) multiplied by – c.

So we thereby obtain in this case 1 * (1) + 1 * (1)  =  2.

We now have the terms 0, 1, 1, 2.

Thus once more to generate the next term we again multiply the latest term generated (i.e. 2) by – b, before again combining it with the previous term (i.e. 1) multiplied by – c.

This now gives 1 * (2) + 1 * (1)  =  3 which is the next term in the sequence.

Therefore, continuing on in this fashion, we can generate as many terms as we wish with respect to the infinite Fibonacci sequence.

Now the interesting thing is that we can then approximate the value of ϕ (phi) by dividing the latest term in the sequence by the previous term or in more general terms the
nth/(n – 1)th term.

Now if we attempt this from the limited number of terms already generated i.e.

0, 1, 1, 2, 3 we will get 3/2 = 1.5 which is not a very good approximation to the true value 1.618033…

However when we divide the 16th term listed above (i.e. 987) by the previous term (610) we obtain 1.618032… which is already correct to 5 decimal places!

So the approximation improves rapidly through using later terms in the series.

However, though the use of this approximation approach is indeed well known with respect to the Fibonacci equation (and corresponding sequence), what is not equally emphasised is that this same general procedure can in principle be used with respect to any polynomial equation (with one variable).

Thus associated with every polynomial equation (with one variable) is a unique infinite sequence of digits, which can then be used to approximate all the real valued solutions - if indeed such solutions exist - for the equation in question.

Now in applying this approach xn, the term of highest degree should be expressed with unitary coefficient.

Therefore, for example the equation 3x2 + 2x – 5 = 0, would be expressed as

x2 + 2/3x – 5/3 = 0, before continuing in the previous manner.

Indeed one interesting class of equation relates to the extension of the Fibonacci equation to dimensions (> 2).

For example the Tribonacci equation - where the highest term is of degree 3 - would be given as

x3 x2 – x – 1 = 0.

Now the general equation for equations of degree 3 would be expressed as

x3 + bx2 + cx + d = 0.

Then to generate the corresponding Tribonacci sequence we now start with the 3 numbers 0, 0, 1 (Though the two 0’s do not strictly comprise terms in the sequence, they are necessary so as to generate initial terms in the sequence in a correct manner.
And the correct number of 0’s involved will always be 1 less that the highest power of x in the polynomial equation!)

So in this case we add {– b * (1)} + {– c * (0)} + {– d * (0)}  which in the case of the Tribonacci equation = {1 * (1)} + {1 * (0)} + {1 * (0)} = 1.

Thus the sequence is now 0, 0, 1, 1,

Continuing on in this manner, the next term = {1 * (1)} + {1 * (1)} + {1 * (0)} = 2.

We now have 0, 0, 1, 1, 2,

Then the following term = {1 * (2)} + {1 * (1)} + {1 * (1)} = 4.

So we now have 0, 0, 1, 1, 2, 4,

And if continued to generate fresh terms, we would obtain the Tribonacci sequence

0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927,…

And once again the real valued solution to the Tribonacci equation i.e. 1.839286… can be approximated through calculation of the nth/(n – 1)th term.

So taking the last two terms of the sequence above, this would give 927/504 = 1.839285… which again gives the true result correct to 5 decimal places.

And of course this Fibonacci type procedure can be extended to any required value of n.

So on general we have xn xn – 1 – …– x – 1 = 0.

Where n = 4, we get the Tetranacci equation (with associated Tetranacci numbers in the associated sequence).

Where n = 5 we get the Pentanacci, with 6 the Hexanacci, with 7 the Heptanacci, with 8 the Octonacci and with 9, the Nonacci equations and associated numbers respectively.

And in generating all these number sequences we would keep adding 4, 5, 6, 7, 8 and 9 numbers respectively to generate the next number in the corresponding sequence.

One fascinating use of the positive real valued solutions to these equations (which can be conveniently approximated through the associated number sequences) is in providing approximations to the sum of the Riemann zeta function (for positive integer values).