[math-fun] Hypergeometric solution of the quintic
Stillwell's paper `Eisenstein's Footnote' (link below) gives a lucid explanation of how Eisenstein (at the tender age of 14) obtained a solution y for y^5+y=x as an infinite series in x of the form y=x-x^5+(10*x^9)/2!-(14*15*x^13)/3!+.... Furthermore, the Bring-Jerrard Theorem asserts that a general quintic can be reduced to one of the above form, with x a radical expression of the coefficient of said quintic. http://web.a.ebscohost.com/ehost/pdfviewer/pdfviewer?sid=9dd1fb0a-49ef-42b5-... Nonetheless, I don't think this gives the whole picture since the range of convergence for x isn't that big. Namely, the above series is exactly x*4F3(1/5, 2/5, 3/5, 4/5; 5/4, 2/4, 3/4|5^5*x^4/4^4), which converges (absolutely) for |x| less than or equal to (0.8)*(0.2)^(1/4), which is approximately 0.53499. So, if one is interested in solving y^5+y=1, say, then the series above won't converge. Is there an analytic continuation of the 4F3 that would be defined for large values of x and which would still give a solution of that quintic, or is there another way to get around that? Cheers, Ahmad
By coincidence, half an hour ago I was reading Beukers, Frits. Hypergeometric functions, how special are they?Notices Amer. Math. Soc. 61 (2014), no. 1, 48--56. MR3137256 which discusses solving this quintic. I don't think it answers your question, but it is a very nice paper! There are further references in http://oeis.org/A002294 Neil On Wed, Mar 12, 2014 at 4:42 AM, Ahmad El-Guindy <a.elguindy@gmail.com>wrote:
Stillwell's paper `Eisenstein's Footnote' (link below) gives a lucid explanation of how Eisenstein (at the tender age of 14) obtained a solution y for y^5+y=x as an infinite series in x of the form
y=x-x^5+(10*x^9)/2!-(14*15*x^13)/3!+....
Furthermore, the Bring-Jerrard Theorem asserts that a general quintic can be reduced to one of the above form, with x a radical expression of the coefficient of said quintic.
http://web.a.ebscohost.com/ehost/pdfviewer/pdfviewer?sid=9dd1fb0a-49ef-42b5-...
Nonetheless, I don't think this gives the whole picture since the range of convergence for x isn't that big. Namely, the above series is exactly x*4F3(1/5, 2/5, 3/5, 4/5; 5/4, 2/4, 3/4|5^5*x^4/4^4), which converges (absolutely) for |x| less than or equal to (0.8)*(0.2)^(1/4), which is approximately 0.53499.
So, if one is interested in solving y^5+y=1, say, then the series above won't converge. Is there an analytic continuation of the 4F3 that would be defined for large values of x and which would still give a solution of that quintic, or is there another way to get around that?
Cheers,
Ahmad _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Dear Friends, I have now retired from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
* Neil Sloane <njasloane@gmail.com> [Mar 12. 2014 17:19]:
By coincidence, half an hour ago I was reading
Beukers, Frits. Hypergeometric functions, how special are they?Notices Amer. Math. Soc. 61 (2014), no. 1, 48--56. MR3137256
http://www.ams.org/notices/201401/
which discusses solving this quintic. I don't think it answers your question, but it is a very nice paper!
There are further references in http://oeis.org/A002294
Apparently including one by me: pp.337-338 of the book shows hypergeometric forms for series reversions of x^k - x and a closed form (rel.15.5-6) that surprised me a bit back then.
Neil
[...]
Of course this message doesn't answer your question either 8-) Best regards, jj
* Ahmad El-Guindy <a.elguindy@gmail.com> [Mar 12. 2014 10:55]:
[...]
So, if one is interested in solving y^5+y=1, say, then the series above won't converge. Is there an analytic continuation of the 4F3 that would be defined for large values of x and which would still give a solution of that quintic, or is there another way to get around that?
It looks like http://arxiv.org/abs/1403.2858 is spot on. (Outside my turf, so I cannot rate quality.)
Cheers,
Ahmad _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Thanks to all for the many useful replies and references, that clarified a lot :). I was playing around with a slightly different aspect of this and a cute little result popped out: If q is a prime power, then (by the freshman's dream) y=\sum (-1)^k x^(q^k) gives a solution for y^q+y=x (the convergence, with respect to the absolute value coming from the `infinte prime' in the PID F_q[t], is for all deg(x)<0 in F_q(t), or even F_q((1/t)) and reasonable extensions thereof.) The method of Eisenstein, as Joerg remarked, would also give z=\sum_{i=0}^\infty \binomial{qi}{i} (-x^(q-1))^i/(1+(q-1)i) as a series solution with integer coefficients, which hence can also be reduced in F_q. It follows that, for instance \binomial{9i,i}*/(1+8i) is 0 mod 3, except when 8i+1=9^k, in which case it is 1 mod 3. Examples of the latter happens when i=1, 10, 91, 820, 7381, etc, which are (of course) precisely the numbers with all ones in their base 9 representation. I am sure there are other ways to prove results like that, but I thought the connection to series solutions was interesting. Best, Ahmad On Thu, Mar 13, 2014 at 11:11 AM, Joerg Arndt <arndt@jjj.de> wrote:
* Ahmad El-Guindy <a.elguindy@gmail.com> [Mar 12. 2014 10:55]:
[...]
So, if one is interested in solving y^5+y=1, say, then the series above won't converge. Is there an analytic continuation of the 4F3 that would
be
defined for large values of x and which would still give a solution of that quintic, or is there another way to get around that?
It looks like http://arxiv.org/abs/1403.2858 is spot on. (Outside my turf, so I cannot rate quality.)
Cheers,
Ahmad _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Neil Sloane