What bounds can we establish on coefficients of Swinnerton-Dyer polynomials? - MathOverflow most recent 30 from http://mathoverflow.net2013-06-19T08:41:38Zhttp://mathoverflow.net/feeds/question/106907http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/106907/what-bounds-can-we-establish-on-coefficients-of-swinnerton-dyer-polynomialsWhat bounds can we establish on coefficients of Swinnerton-Dyer polynomials?swinnly2012-09-11T12:46:51Z2013-06-17T09:54:57Z
<p>The maximum coefficient of the n<sub>th</sub> <a href="http://mathworld.wolfram.com/Swinnerton-DyerPolynomial.html" rel="nofollow">Swinnerton-Dyer polynomial</a> seems to grow very fast with n. These are the maximum absolute values of the first 6 polynomials,</p>
<ol>
<li>2</li>
<li>10</li>
<li>960</li>
<li>13950764</li>
<li>255690851718529024</li>
<li>1771080720430629161685158978892152599456</li>
</ol>
<p>What bounds can we establish on the absolute value of coefficients in the n<sub>th</sub> Swinnerton-Dyer polynomial? A very trivial bound appears to be $B(n) = 2^{2^n} n \sqrt{p_n}$ but this doesn't take into account cancellation of any of the terms. Is it possible to do better?</p>
http://mathoverflow.net/questions/106907/what-bounds-can-we-establish-on-coefficients-of-swinnerton-dyer-polynomials/133757#133757Answer by Fredrik Johansson for What bounds can we establish on coefficients of Swinnerton-Dyer polynomials?Fredrik Johansson2013-06-14T15:52:45Z2013-06-17T09:54:57Z<p>There is a missing exponent in your trivial bound. Letting $u_n = \sum_{k=1}^n \sqrt{p_k}$, we have</p>
<p>$$B_0(n) \equiv \max_i \; \left| [x^i] S_n \right|$$
$$\le B_1(n) \equiv \max_i \; [x^i] (x+u_n)^{2^n} = \max_i {2^n \choose i} u_n^i$$
$$\le B_2(n) \equiv { 2^n \choose 2^{n-1} } u_n^{2^n}$$
$$\le B_3(n) \equiv 2^{2^n} (n \sqrt{p_n})^{2^n}.$$</p>
<p>Of course, in the last step we can get a better bound by estimating $u_n$ less crudely. It should also be possible to find an analytic bound for $B_1(n)$ that is less crude than $B_2(n)$ (regardless, $B_1(n)$ is easy to evaluate numerically).</p>
<p>For reference, I have computed numerical approximations of the actual value $B_0(n)$ up to $n = 20$ (correct up to rounding in the last digit):</p>
<pre><code>0: 1
1: 2
2: 10
3: 960
4: 13950764
5: 2.55690851718529e+17
6: 1.77108072043063e+39
7: 8.57834471403602e+86
8: 4.69693103314689e+187
9: 3.24515842436673e+401
10: 8.31078370973184e+853
11: 4.18601441612784e+1793
12: 1.37441368638541e+3755
13: 7.32398012717744e+7815
14: 4.7364530607185e+16172
15: 8.41442697691365e+33355
16: 6.50154879984207e+68684
17: 2.98829955473397e+141188
18: 7.8161464597922e+289271
19: 2.1050522533847e+591950
20: 2.92232330678161e+1209132
</code></pre>
<p>Here is a comparison of the bounds:</p>
<p><img src="http://fredrikj.net/math/sdpoly.png"/></p>
<p>It appears that $B_3$, $B_2$ and $B_1$ asymptotically overestimate the number of bits in $B_0$ less than by a respective factor 2.20, 2.02 and 1.76.</p>
<p>As you say, it should possible to do better by taking into account the cancellation that occurs. Perhaps by expressing the coefficients in terms of elementary symmetric polynomials? A lower bound would also be interesting.</p>