Short & Sweet Math Challenge #21: Powers that be

[Valentin Albillo]

 
Hi, Mike:

(11-10-2016 09:51 AM)Mike (Stgt) Wrote:  So the reason for the almost integer effect of increasing powers is just the "symetry" of roots r1 and r2 with abs(r1)<1 and abs(r2)>1 ?

The reason for the almost-integer increasing powers is, in short:

- all symmetric functions of the roots of any polynomial can be expressed in terms of the coefficients of the polynomial, where symmetric means that the function remains the same and has the same value for every permutation of the roots. For instance, the sum of the Nth-powers of the roots of a quintic polynomial is symmetric because:

         S = x1^5 + x2^5 + x3^5 + x4^5 + x5^5    (initial permutation)
           = x1^5 + x2^5 + x3^5 + x5^5 + x4^5    (second permutation)
                        ...
           = x5^5 + x4^5 + x3^5 + x2^5 + x1^5    (last permutation)

so you see, no matter how you permute the roots the sum stays the same.

- if the polynomial is monic (leading coefficient is 1) and has integer coefficients, then any symmetric function of the roots is integer as well because it can be computed from the integer coefficients using just addition, subtraction and multiplication of their integer values. Thus, in particular, the sum of the roots raised to any integer power is mandatorily integer as well, regardless of the absolute values of the roots or whether they're real and/or complex.

-now, if any number is >1 in absolute value, its increasing powers will grow ever bigger, while if the number is <1 in absolute value its increasing powers will grow ever smaller, tending to 0.

- so, for those monic polynomials which have just the one root >1 in absolute value while all other roots are <1 in absolute value, the sum of the powers of the roots will essentially equal the power of the large root alone plus change, and as the sum has to mandatorily be integer, then the power of the larger root tends to be integer, almost-integer so to say. The only difference with an integer is the sum of the powers of all other roots, whichs tends to 0 for each root and so for the sum of all of them.

I hope this explanation makes it clear for you, thanks a lot for your interest.

Reegards.
V.