Comments on: What's the next polynomial in this series?

Question: What's the next polynomial in this series?

Mapes — Tue, 27 Nov 2007 07:23:17 -0800

What's the next polynomial in this series: x⁴+2x²+16, x⁸+40x⁶+1128x⁴+2560x²+65536, ?

I'm doing some work in elasticity and have found an analytical solution to a tricky problem. The solution is a converging infinite series, and I've been able to work out the first two terms (with great difficulty). It may be easier to guess the additional terms than work them out. The next polynomial almost certainly has seven terms, including x¹² and 2,176,782,336 (6¹²). Any guesses as to what the other coefficients are, and what the general pattern is?

(The sum converges because the denominators are (4+x²)^5/2, (16+x²)^9/2, ... and every other term is subtracted.)

By: Johnny Assay

Johnny Assay — Tue, 27 Nov 2007 07:31:16 -0800

Have you tried the online encyclopedia of integer sequences?

By: Mapes

Mapes — Tue, 27 Nov 2007 08:25:56 -0800

Yep, nothing came up.

Not sure if it's significant that 40=2³+2⁵ and 1128 = 2³+2⁵+2⁶+2¹⁰.

By: Khalad

Khalad — Tue, 27 Nov 2007 08:29:13 -0800

I've got nothing... That 1128 is ugly. If it helps anyone, the prime factorization of the polynomials is:

x⁴ + 2x² + 2⁴
x⁸ + 5·2³x⁶ + 47·3·2³x⁴ + 5·2⁹x² + 2¹⁶

By: Wolfdog

Wolfdog — Tue, 27 Nov 2007 08:45:05 -0800

Not sure if it's significant that 40=23+25 and 1128 = 23+25+26+210.
It is not uncommon that numbers can be written as sums of powers of 2.

By: unSane

unSane — Tue, 27 Nov 2007 08:57:39 -0800

ALL numbers can be written as powers of 2.

It's called binary arithmetic.

in this case the coefficients are as follows (in binary)

first polynomial

second polynomial





           1

      101000

10001101000

101000000000

100000000000000000

By: mhum

mhum — Tue, 27 Nov 2007 10:33:58 -0800

Hmm. It'll be tough to extrapolate the sequence from just the two first entries. Any hints about how you're generating these polynomials?

By: Mapes

Mapes — Tue, 27 Nov 2007 11:17:40 -0800

It's from an approach called the method of images. Briefly, I'm trying to calculate the displacement caused by a certain load on an object, and to meet the boundary conditions I'm using reflected virtual loads that extend to infinity. Each new polynomial represents an additional virtual load. They grow successively harder to calculate, unfortunately.

The reason I thought there might be a simple pattern is that when x=0, the terms reduce into an alternating harmonic series, which is proportional to simply ln(2). For arbitrary x, the best answer I can get right now is that the sum is bounded by T₁ln(2) and T₁+T₂ln(2), where T₁ and T₂ are the two terms described above.

By: Mapes

Mapes — Tue, 27 Nov 2007 16:32:18 -0800

The third polynomial is x¹² + 162x¹⁰ + 13,824x⁸ + 514,656x⁶ + 17,708,544x⁴ + 33,219,072x² + 2,176,782,336.

By: christonabike

christonabike — Tue, 27 Nov 2007 18:11:50 -0800

factorization for the 3rd one:

x¹² + 3⁴·2x¹⁰ + 3³·2⁹x⁸ + 1787·3²·2⁵x⁶ +61·7·3⁴·2⁹x⁴ + 89·3⁶·2⁹x² + 3¹²·2¹²

certainly ain't obvious

By: number9dream

number9dream — Tue, 27 Nov 2007 20:11:41 -0800

Obviously there are no real roots. Have you tried to factor them over the complex numbers?

By: unSane

unSane — Tue, 27 Nov 2007 20:52:57 -0800

In the first equation, let u=x^2

then you can rewrite it as:

u^2 + 2u + 16

which has complex roots

((-1 + sqr(4 - 4*1*16))/2) = -0.5 + sqr(15)*i

and

((-1 - sqr(4 - 4*1*16))/2) = - 0.5 - sqr(15)*i

so the first equation factors to

(X^2 + 0.5 - sqr(15)*i) * (x^2 + 0.5 + sqr(15)*i)

at which point I am not at all confident that simple answers are going to drop out of this.

By: Mapes

Mapes — Wed, 28 Nov 2007 06:00:56 -0800

Factoring the first one isn't bad, but the complex factors of the second one take up two pages in Mathematica. So that looks like a dead end.

By: Mapes

Mapes — Wed, 28 Nov 2007 07:55:01 -0800

Hooray! I found an algorithm for calculating these polynomials in the literature (specifically, Fabrikant, "Tangential contact problem for a transversely isotropic elastic layer bonded to a rigid foundation"). Now I see why it was so challenging to guess the pattern. We need to find

-e^-2w[1+2e^-2w+(1-2w)²]Sum[-(2+4w²)e^-2w-e^-4w]ⁿ

with n ranging from 0 to infinity, and collect terms of equal wⁿ, which correspond to n^th partial derivatives of g(s) = 1/(s²+1). So, for example, the first polynomial was related to

-2g(s) - 4sg'(s) - 4s²g''(s),

the second to

2g(s) + 8sg'(s) + 16s²g''(s) + 16s³g'''(s) + 16s⁴g''''(s),

the third to

-2g(s) - 12sg'(s) - 36s²g''(s) - 64s³g'''(s) - 96s⁴g''''(s) - 64s⁵g'''''(s) - 64s⁶g''''''(s),

and so on. Wow. No wonder I was stumped. Truly a job for the computer.

Thanks everyone. I marked unSane's answer as best because I thought the idea of looking for patterns in another base was clever.

By: mhum

mhum — Mon, 03 Dec 2007 17:12:24 -0800

I found an algorithm for calculating these polynomials in the literature

Just as a weird aside, if you look up an abstract for that paper, you'll find that the author's contact info is given as "Prisoner #167932D, Archambault jail". That's because the author is Valery Fabrikant, a former professor at Concordia University in Montreal who killed four people in a shooting spree in 1992. Apparently, he continues to publish from behind bars.