Polynomial Equations and Associated Number Sequences (2)

In the last entry I mentioned that the unique number sequence associated with a given polynomial equation can be used to approximate the real roots of that equation.

When there is more than one real solution to the equation, the initial root approximated from the unique number sequence will relate to the largest valued root (in absolute terms).

For example, to illustrate, let us take the simple binomial equation x² + 2x – 3 = 0.

Then starting with 0, 1 we add (– 2 * 1) + (3 * 0) = – 2, which is then the next term in the unique number sequence of this equation.

So we now have 0, 1, – 2.

We now add  (– 2 * – 2)  + (3 * 1) = 7.

This gives us 0, 1, – 2, 7.

Continuing again in this manner, we obtain the next number in the sequence as

(– 2 * 7) + (3 * – 2) =  – 20.

So the unique number sequence is now 0, 1, – 2, 7, – 20.

Even with these few terms we can approximate the largest root (ignoring the sign) as

–        20/7 = – 2.857…

Thus it already appears that the value of this root is quickly converging to – 3.

Now of course, because this is a binomial equation, this implies that the other root is also real and in this case as the product of the two roots = – 3 (as the value of c in the general form of the equation) the other root therefore =  1.

However there is a procedure for approximating the other real roots, which could be useful for polynomial equations of degree > 2).

So we multiply the 1^st hidden term of the sequence i.e. 0 by – 3 = 0. We then subtract this from the next term 1 to obtain 1, which is then the 1st term of a new number sequence..

We then multiply the next term of the original sequence (i.e. 1) =  – 3 and then subtract this from the next term in the original sequence (i.e. – 2).

Thus we obtain – 2 – (– 3) = – 2 + 3 = 1.

So 1 is again the 2^nd term in the new number sequence.

Continuing on, we multiply the 3^rd term of the original series (i.e. 7) again by the estimated root (i.e. – 3) = – 21. So we then subtract this from the next number in the original series (i.e. – 20) to obtain – 20  – (– 21) = – 20 + 21 = 1.

So 1 is again the 3^rd term in the new sequence.

And if we continued on in this manner, 1 will always be the next number generated in the new number sequence.

So now with respect to this new number sequence 1, 1, 1,… the estimated value of the second root = 1/1/ = 1.

Thus the two roots are – 3 and 1.

And though we have illustrated this with respect to two real roots in principle the same approach can be extended for any number of real roots.

Of course the values of the roots will not always exist as simple rational numbers.

However, where irrational values are involved, their value can in principle be approximated to any required degree of accuracy by calculating a sufficient amount of terms in the unique number sequence (associated with the polynomial equation in question) and then applying the procedures illustrated.

We can also reverse the approach so that starting with a given sequence of numbers we can then attempt to construct the unique polynomial equation to which they relate.

However for a meaningful finite equation to result it is important that this sequence of digits to be ordered in a consistent manner. Where this is the case, a finite number of steps is required to generate the digits in question.

However when not properly ordered one can keep generating potentially an unlimited number of steps in the attempt to match the given sequence of digits.

So I will briefly illustrate here for both cases.

The first sequence relates to what are known as the Pell numbers, i.e.

1, 2, 5, 12, 29, 70, 169,…

Now initially - even if it is in fact an ordered sequence - we do not know how many terms are involved.

However seeing as the next term (after 1) is 2 then this means that the coefficient of the

x^{n – 1} term is – 2. So we now multiply 2 by 2 and see what adjustment has to made to derive 5 (which is the next term in the sequence). So clearly we need to add 1 in this case, which implies that the coefficient of the x^{n – 1} term is – 1. So we now combine (5 * 2) + (2 * 1) = 12 indicating that no adjustment using coefficients of further terms is necessary.

So n = 2 in this case, so that the equation which uniquely generates the Pell numbers is x² – 2x – 1 = 0.

It bears a relationship to the Fibonacci equation, except that in this case we keep combining twice the last term in the sequence derived with the previous term!

As the positive real-valued root of this equation = √2 + 1, we can use its associated unique number sequence to approximate √2.

So taking the last two terms the ratio is 169/70 = 2.414285…

Therefore the approximation for √2 = 1.414285… (which is already correct to the first 4 significant figures).  

However the well known sequence of odd integers, i.e. 1, 3, 5, 7, 9,…is not so well ordered in this case.

The coefficient of the x^{n – 1} term is – 3. Then we multiply 3 by 3 and add it to x * 1 to obtain the next term in the sequence i.e. 5. Therefore x = – 4, so that the coefficient of the  x^{n – 2} term is 4.

Then (3 * 5) + (– 4 * 3) + (x * 1) = 7 (as the next term).

This entails that x = 4, so that the coefficient of the x^{n – 3}  term is – 4.

If we stopped here, we would get the equation

x³ – 3x² + 4x – 4.

Now, we know that the ratio of the n_th to the previous term for the odd integers converges to 1. However strictly when we divide these numbers there will always be a remainder of 2.

It is fascinating therefore in this context that when we substitute 1 in the equation (of degree 3) above that we get an answer of – 2.

And in absolute terms we will always obtain the same number no matter how many terms we include in the equation.

In fact a definite pattern emerges whereby the coefficient of each successive term keeps alternating as between + 4 and – 4. Therefore when we substitute 1 in the polynomial equation involved the answer in absolute terms = 2 (continually alternating as between + 2 and – 2).

This is a pattern (i.e. alternating as between + and –) that - as we will see - is directly associated with imaginary numbers.

In fact if we take the earlier binomial expression of the odd number integer sequence, it will be given as x² – 3x + 4 = 0.

When we let x = 1 (as indicated by the ratio of successive higher numbered terms), we get the result of + 2.

And there are no real valued solutions to this equation, with the two solutions imaginary.  

However in the case of the odd integers there is clearly a discernible pattern to the numbers.

However there are a great many possible infinite number sequences which offer no discernible pattern and therefore cannot be successfully associated with polynomial equations of a finite size.