Comments on: What's the next polynomial in this series?
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series/
Comments on Ask MetaFilter post What's the next polynomial in this series?Tue, 27 Nov 2007 07:31:16 -0800Tue, 27 Nov 2007 07:31:16 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: What's the next polynomial in this series?
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series
What's the next polynomial in this series: x<sup>4</sup>+2x<sup>2</sup>+16, x<sup>8</sup>+40x<sup>6</sup>+1128x<sup>4</sup>+2560x<sup>2</sup>+65536, ? <br /><br /> 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<sup>12</sup> and 2,176,782,336 (6<sup>12</sup>). Any guesses as to what the other coefficients are, and what the general pattern is?<br>
<br>
(The sum converges because the denominators are (4+x<sup>2</sup>)<sup>5/2</sup>, (16+x<sup>2</sup>)<sup>9/2</sup>, ... and every other term is subtracted.)post:ask.metafilter.com,2007:site.77206Tue, 27 Nov 2007 07:23:17 -0800MapespolynomialsseriessuminfiniteconvergecoefficientsBy: Johnny Assay
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1146821
Have you tried the <a href="http://www.research.att.com/~njas/sequences/">online encyclopedia of integer sequences</a>?comment:ask.metafilter.com,2007:site.77206-1146821Tue, 27 Nov 2007 07:31:16 -0800Johnny AssayBy: Mapes
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1146877
Yep, nothing came up. <br>
<br>
Not sure if it's significant that 40=2<sup>3</sup>+2<sup>5</sup> and 1128 = 2<sup>3</sup>+2<sup>5</sup>+2<sup>6</sup>+2<sup>10</sup>.comment:ask.metafilter.com,2007:site.77206-1146877Tue, 27 Nov 2007 08:25:56 -0800MapesBy: Khalad
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1146884
I've got nothing... That 1128 is ugly. If it helps anyone, the prime factorization of the polynomials is:<br>
<br>
x<sup>4</sup> + 2x<sup>2</sup> + 2<sup>4</sup><br>
x<sup>8</sup> + 5·2<sup>3</sup>x<sup>6</sup> + 47·3·2<sup>3</sup>x<sup>4</sup> + 5·2<sup>9</sup>x<sup>2</sup> + 2<sup>16</sup>comment:ask.metafilter.com,2007:site.77206-1146884Tue, 27 Nov 2007 08:29:13 -0800KhaladBy: Wolfdog
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1146916
<i>Not sure if it's significant that 40=23+25 and 1128 = 23+25+26+210.</i><br>
It is not uncommon that numbers can be written as sums of powers of 2.comment:ask.metafilter.com,2007:site.77206-1146916Tue, 27 Nov 2007 08:45:05 -0800WolfdogBy: unSane
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1146932
ALL numbers can be written as powers of 2.<br>
<br>
It's called binary arithmetic.<br>
<br>
in this case the coefficients are as follows (in binary)<br>
<br>
first polynomial<br>
<code><br>
1<br>
10<br>
1000</code><br>
<br>
second polynomial<br>
<code><br>
<br>
1<br>
101000<br>
10001101000<br>
101000000000<br>
100000000000000000<br>
</code>comment:ask.metafilter.com,2007:site.77206-1146932Tue, 27 Nov 2007 08:57:39 -0800unSaneBy: mhum
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1147118
Hmm. It'll be tough to extrapolate the sequence from just the two first entries. Any hints about how you're generating these polynomials?comment:ask.metafilter.com,2007:site.77206-1147118Tue, 27 Nov 2007 10:33:58 -0800mhumBy: Mapes
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1147183
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.<br>
<br>
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<sub>1</sub>ln(2) and T<sub>1</sub>+T<sub>2</sub>ln(2), where T<sub>1</sub> and T<sub>2</sub> are the two terms described above.comment:ask.metafilter.com,2007:site.77206-1147183Tue, 27 Nov 2007 11:17:40 -0800MapesBy: Mapes
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1147594
The third polynomial is x<sup>12</sup> + 162x<sup>10</sup> + 13,824x<sup>8</sup> + 514,656x<sup>6</sup> + 17,708,544x<sup>4</sup> + 33,219,072x<sup>2</sup> + 2,176,782,336.comment:ask.metafilter.com,2007:site.77206-1147594Tue, 27 Nov 2007 16:32:18 -0800MapesBy: christonabike
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1147715
factorization for the 3rd one:<br>
<br>
x<sup>12</sup> + 3<sup>4</sup>·2x<sup>10</sup> + 3<sup>3</sup>·2<sup>9</sup>x<sup>8</sup> + 1787·3<sup>2</sup>·2<sup>5</sup>x<sup>6</sup> +61·7·3<sup>4</sup>·2<sup>9</sup>x<sup>4</sup> + 89·3<sup>6</sup>·2<sup>9</sup>x<sup>2</sup> + 3<sup>12</sup>·2<sup>12</sup><br>
<br>
certainly ain't obviouscomment:ask.metafilter.com,2007:site.77206-1147715Tue, 27 Nov 2007 18:11:50 -0800christonabikeBy: number9dream
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1147845
Obviously there are no real roots. Have you tried to factor them over the complex numbers?comment:ask.metafilter.com,2007:site.77206-1147845Tue, 27 Nov 2007 20:11:41 -0800number9dreamBy: unSane
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1147892
In the first equation, let u=x^2<br>
<br>
then you can rewrite it as:<br>
<br>
u^2 + 2u + 16<br>
<br>
which has complex roots<br>
<br>
((-1 + sqr(4 - 4*1*16))/2) = -0.5 + sqr(15)*i<br>
<br>
and<br>
<br>
((-1 - sqr(4 - 4*1*16))/2) = - 0.5 - sqr(15)*i<br>
<br>
so the first equation factors to<br>
<br>
(X^2 + 0.5 - sqr(15)*i) * (x^2 + 0.5 + sqr(15)*i)<br>
<br>
at which point I am not at all confident that simple answers are going to drop out of this.comment:ask.metafilter.com,2007:site.77206-1147892Tue, 27 Nov 2007 20:52:57 -0800unSaneBy: Mapes
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1148140
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.comment:ask.metafilter.com,2007:site.77206-1148140Wed, 28 Nov 2007 06:00:56 -0800MapesBy: Mapes
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1148213
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<br>
<br>
-e<sup>-2w</sup>[1+2e<sup>-2w</sup>+(1-2w)<sup>2</sup>]Sum[-(2+4w<sup>2</sup>)e<sup>-2w</sup>-e<sup>-4w</sup>]<sup>n</sup><br>
<br>
with n ranging from 0 to infinity, and collect terms of equal w<sup>n</sup>, which correspond to n<sup>th</sup> partial derivatives of g(s) = 1/(s<sup>2</sup>+1). So, for example, the first polynomial was related to<br>
<br>
-2g(s) - 4sg'(s) - 4s<sup>2</sup>g''(s),<br>
<br>
the second to<br>
<br>
2g(s) + 8sg'(s) + 16s<sup>2</sup>g''(s) + 16s<sup>3</sup>g'''(s) + 16s<sup>4</sup>g''''(s),<br>
<br>
the third to<br>
<br>
-2g(s) - 12sg'(s) - 36s<sup>2</sup>g''(s) - 64s<sup>3</sup>g'''(s) - 96s<sup>4</sup>g''''(s) - 64s<sup>5</sup>g'''''(s) - 64s<sup>6</sup>g''''''(s),<br>
<br>
and so on. Wow. No wonder I was stumped. Truly a job for the computer.<br>
<br>
Thanks everyone. I marked unSane's answer as best because I thought the idea of looking for patterns in another base was clever.comment:ask.metafilter.com,2007:site.77206-1148213Wed, 28 Nov 2007 07:55:01 -0800MapesBy: mhum
http://ask.metafilter.com/77206/Whats-the-next-polynomial-in-this-series#1155087
<i>I found an algorithm for calculating these polynomials in the literature</i><br>
<br>
Just as a weird aside, if you look up an <a href="http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=279830">abstract</a> 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 <a href="http://en.wikipedia.org/wiki/Valery_Fabrikant"> Valery Fabrikant</a>, a former professor at Concordia University in Montreal who killed four people in a <a href="http://thelink.concordia.ca/view.php?aid=38639">shooting spree</a> in 1992. Apparently, he continues to publish from behind bars.comment:ask.metafilter.com,2007:site.77206-1155087Mon, 03 Dec 2007 17:12:24 -0800mhum