# What's the next polynomial in this series?

November 27, 2007 7:23 AM Subscribe

What's the next polynomial in this series: x

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 x

(The sum converges because the denominators are (4+x

^{4}+2x^{2}+16, x^{8}+40x^{6}+1128x^{4}+2560x^{2}+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 x

^{12}and 2,176,782,336 (6

^{12}). Any guesses as to what the other coefficients are, and what the general pattern is?

(The sum converges because the denominators are (4+x

^{2})

^{5/2}, (16+x

^{2})

^{9/2}, ... and every other term is subtracted.)

Response by poster: Yep, nothing came up.

Not sure if it's significant that 40=2

posted by Mapes at 8:25 AM on November 27, 2007

Not sure if it's significant that 40=2

^{3}+2^{5}and 1128 = 2^{3}+2^{5}+2^{6}+2^{10}.posted by Mapes at 8:25 AM on November 27, 2007

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

x

x

posted by Khalad at 8:29 AM on November 27, 2007

x

^{4}+ 2x^{2}+ 2^{4}x

^{8}+ 5·2^{3}x^{6}+ 47·3·2^{3}x^{4}+ 5·2^{9}x^{2}+ 2^{16}posted by Khalad at 8:29 AM on November 27, 2007

*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.

posted by Wolfdog at 8:45 AM on November 27, 2007

Best answer: 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

posted by unSane at 8:57 AM on November 27, 2007

It's called binary arithmetic.

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

first polynomial

1

10

1000

second polynomial

1

101000

10001101000

101000000000

100000000000000000

posted by unSane at 8:57 AM on November 27, 2007

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

posted by mhum at 10:33 AM on November 27, 2007

posted by mhum at 10:33 AM on November 27, 2007

Response by poster: 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 T

posted by Mapes at 11:17 AM on November 27, 2007

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 T

_{1}ln(2) and T_{1}+T_{2}ln(2), where T_{1}and T_{2}are the two terms described above.posted by Mapes at 11:17 AM on November 27, 2007

Response by poster: The third polynomial is x

posted by Mapes at 4:32 PM on November 27, 2007

^{12}+ 162x^{10}+ 13,824x^{8}+ 514,656x^{6}+ 17,708,544x^{4}+ 33,219,072x^{2}+ 2,176,782,336.posted by Mapes at 4:32 PM on November 27, 2007

factorization for the 3rd one:

x

certainly ain't obvious

posted by christonabike at 6:11 PM on November 27, 2007

x

^{12}+ 3^{4}·2x^{10}+ 3^{3}·2^{9}x^{8}+ 1787·3^{2}·2^{5}x^{6}+61·7·3^{4}·2^{9}x^{4}+ 89·3^{6}·2^{9}x^{2}+ 3^{12}·2^{12}certainly ain't obvious

posted by christonabike at 6:11 PM on November 27, 2007

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

posted by number9dream at 8:11 PM on November 27, 2007

posted by number9dream at 8:11 PM on November 27, 2007

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.

posted by unSane at 8:52 PM on November 27, 2007

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.

posted by unSane at 8:52 PM on November 27, 2007

Response by poster: 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.

posted by Mapes at 6:00 AM on November 28, 2007

posted by Mapes at 6:00 AM on November 28, 2007

Response by poster: 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

with n ranging from 0 to infinity, and collect terms of equal w

-2g(s) - 4sg'(s) - 4s

the second to

2g(s) + 8sg'(s) + 16s

the third to

-2g(s) - 12sg'(s) - 36s

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.

posted by Mapes at 7:55 AM on November 28, 2007 [1 favorite]

-e

^{-2w}[1+2e^{-2w}+(1-2w)^{2}]Sum[-(2+4w^{2})e^{-2w}-e^{-4w}]^{n}with n ranging from 0 to infinity, and collect terms of equal w

^{n}, which correspond to n^{th}partial derivatives of g(s) = 1/(s^{2}+1). So, for example, the first polynomial was related to-2g(s) - 4sg'(s) - 4s

^{2}g''(s),the second to

2g(s) + 8sg'(s) + 16s

^{2}g''(s) + 16s^{3}g'''(s) + 16s^{4}g''''(s),the third to

-2g(s) - 12sg'(s) - 36s

^{2}g''(s) - 64s^{3}g'''(s) - 96s^{4}g''''(s) - 64s^{5}g'''''(s) - 64s^{6}g''''''(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.

posted by Mapes at 7:55 AM on November 28, 2007 [1 favorite]

*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.

posted by mhum at 5:12 PM on December 3, 2007

This thread is closed to new comments.

posted by Johnny Assay at 7:31 AM on November 27, 2007 [1 favorite]