On Sat, Mar 15, 2014 at 2:41 AM, Bill Gosper <billgosper@gmail.com> wrote:
On Wed, Mar 12, 2014 at 5:24 PM, Bill Gosper <billgosper@gmail.com> wrote:
WDS>
On 3/12/14, Warren D Smith <warren.wds@gmail.com> wrote:
Of course 4F3(x) and indeed any pFq(x) with p<=q+1, is defined for all complex x except for a set of measure zero, via analytic continuation. And this particular 4F3 is of course an algebraic function which makes that obvious.
--and http://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/ gives explicit formulas such as integral representations. Pade and Borel summation should also work. <wds
For more ideas, this ancient Macsyma Newsletter: http://gosper.org/newsrope.pdf
(which, of course, isn't restricted to quintics.)
--rwg
Mma's InverseSeries should be able to do some of this. Then
FindSequenceFunction of a bleepload of terms might give you some
interesting identities.
Sadly, no. Mathematica's series stuff is still decades behind Macsyma's. Expansions in negative powers of (n+1/2) (like for the fixed points of Tan) are illegal. SeriesCoefficient[Log[n!], {n, \[Infinity], k}] can't recite Stirling's formula. FindSequenceFunction doesn't know Bernoullis. But it did find Log[Binomial[2*n, n]] == 2*n*Log[2] + (1/2)*Log[1/(n*Pi)] + Sum[((-2 + 2^(1 - k))*n^(1 - k)*BernoulliB[k])/ ((-1 + k)*k), {k, 2, Infinity}] after I divided out (and later restored) the Bernoullis. It won't even let you tell it the coefficients: In[340]:= Series[Sum[a[k]*x^k, {k, \[Infinity]}], {x, 0, 4}]
Out[340]= \!\( \*UnderoverscriptBox[\(\[Sum]\), \(k\), \(\[Infinity]\)]\( \*SuperscriptBox[\(x\), \(k\)]\ a[k]\)\)
OtOH, FindSequenceFunction miraculously fit the crazy coefficients of the root near 1 (for small y) of x^5-x = y: {1, 1/4, -(5/32), 5/32, -(385/2048), 1/4, -(23205/ 65536), 2145/4096, -(6705765/8388608), 5/4, -(534734915/ 268435456), 3380195/1048576, -(90759684405/17179869184), 35/4,...],
producing a huge pile of Gammas that Macsyma simplified to
Out[376]= 1 + y/4 + Sum[(5*(-1)^(1 + n)*y^n*Gamma[(5/4)*(-1 + n)])/ (4*Gamma[(1/4)*(-1 + n)]*Gamma[1 + n]), {n, 2, Infinity}]
on which Series malfunctions: In[377]:= Series[%, {y, 0, 9}]
Out[377]= SeriesData[y, 0, {1, 1/4}, 0, 10, 1] + Sum[(5*(-1)^(1 + n)*y^n*Gamma[(5/4)*(-1 + n)])/ (4*Gamma[(1/4)*(-1 + n)]*Gamma[1 + n]), {n, 2, Infinity}]
In[378]:= %/.\[Infinity]->9 Out[378]= 1+y/4-(5 Gamma[5/4] y^2)/(8 Gamma[1/4])+(5 y^3)/32-(5 Gamma[15/4] y^4)/(96 Gamma[3/4])+y^5/4-(Gamma[25/4] y^6)/(576 Gamma[5/4])+(2145 y^7)/4096-(Gamma[35/4] y^8)/(32256 Gamma[7/4])+(5 y^9)/4+O[y]^10 In[380]:= FullSimplify[%377^5-%377/.\[Infinity]->11] Out[380]= y+O[y]^10
I.e.x^5-x=y. --rwg I was thinking of sending the Mma developers the
http://gosper.org/newsrope.pdf URL.
But it seems I already did years ago.
This actually gives all five roots--the original one
-3 + 4 n Gamma[-4 + 5 n] y Sum[-(-------------------------), {n, ∞}] Gamma[n] Gamma[-2 + 4 n] and the other four, k n 1 5 n n I Gamma[-(-) + ---] y 1 4 4 - Sum[-------------------------, {n, 0, ∞}] 4 3 n Gamma[- + -] Gamma[1 + n] 4 4 -------------------------------------------------- k I k ∈ 0..3 all five based on the five roots of x^5-x=0 . (I:=√-1) Test that they satisfy the symmetric functions: In[506]:= FullSimplify[Series[Product[t-ReleaseHold[%/.∞->29],{k,4}]* ReleaseHold[t-%%/.∞->8],{y,0,29}]] Out[506]= (-t+t^5)-y+O[y]^30 --rwg