Re: Re: limit
- To: mathgroup at smc.vnet.net
- Subject: [mg78719] Re: [mg78652] Re: limit
- From: "Oleksandr Pavlyk" <pavlyk at gmail.com>
- Date: Sat, 7 Jul 2007 06:12:36 -0400 (EDT)
- References: <f6d9si$2l5$1@smc.vnet.net>
Yes, my post got scrambled somehow. However, it does represent a proof. The recurrence equation I was using is exactly derivable, I just did not present the proof, so I will try to rectify the situation here. Let us start with the recurrence equation for a function h[e, k, z] defined as follows: In[1]:= h[e_, k_, z_] := HypergeometricPFQRegularized[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/ 2, 1 + k, 1}, z] It satisfies the following recurrence equatio In[2]:= eq[e_, k_, z_] = -(-1 + z) h[e, k, z] + 1/2 (-13 + (15 + 8 e) z + k (-6 + 8 z)) h[e, k + 1, z] + 1/4 (86 + k^2 (12 - 24 z) - (139 + 114 e + 24 e^2) z - 2 k (-32 + 3 (19 + 8 e) z)) h[e, k + 2, z] + 1/8 (-252 + (779 + 800 e + 276 e^2 + 32 e^3) z + 8 k^3 (-1 + 4 z) + 4 k^2 (-19 + 3 (23 + 8 e) z) + 8 k (-30 + (100 + 69 e + 12 e^2) z)) h[e, k + 3, z] - 1/8 (3 + e + k) (7 + 2 e + 2 k)^3 z h[e, k + 4, z]; Which can be directly verified as follows: In[3]:= Series[eq[e, k, z], {z, 0, 10}] // Simplify // FullSimplify Out[3]= (SeriesData[z, 0, {}, 11, 11, 1]) or derived by representing the h[e,k,z] as inifnite series in z around the origin and doing manipulations term-wise. Because In[27]:= Limit[(z - 1) h[e, -1, z], z -> 1, Assumptions -> e > 0] Out[27]= 0 we consider In[4]:= RecEq = eq[e, -1, 1] Out[4]= ( 4 e HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 1}, 1])/Sqrt[\[Pi]] + ((-65 - 114 e - 24 e^2 + 2 (-32 + 3 (19 + 8 e))) HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 2}, 1])/( 4 Sqrt[\[Pi]]) + ((503 + 800 e + 276 e^2 + 32 e^3 - 8 (70 + 69 e + 12 e^2) + 4 (-19 + 3 (23 + 8 e))) HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 3}, 1])/( 16 Sqrt[\[Pi]]) - ((2 + e) (5 + 2 e)^3 HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 4}, 1])/(48 Sqrt[\[Pi]]) This is the recurrence equation I was using to get to the result: In[5]:= r[ e_] = -2^(-1 - 4 e) Gamma[ 1 - 2 e] Gamma[1/2 - e]^2 HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 1}, 1] Sin[e Pi]/Pi; In[6]:= q[e_] = r[e] /. First[ Solve[RecEq == 0, HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 1}, 1]]] Out[6]= -(1/(e Sqrt[\[Pi]])) 2^(-3 - 4 e) Gamma[1 - 2 e] Gamma[ 1/2 - e]^2 (-(((-65 - 114 e - 24 e^2 + 2 (-32 + 3 (19 + 8 e))) HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 2}, 1])/( 4 Sqrt[\[Pi]])) - ((503 + 800 e + 276 e^2 + 32 e^3 - 8 (70 + 69 e + 12 e^2) + 4 (-19 + 3 (23 + 8 e))) HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 3}, 1])/( 16 Sqrt[\[Pi]]) + ((2 + e) (5 + 2 e)^3 HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 4}, 1])/( 48 Sqrt[\[Pi]])) Sin[e \[Pi]] In[7]:= Collect[q[e], _HypergeometricPFQ, Limit[#, e -> 0, Direction -> -1] &] /. e -> 0 Out[7]= -(1/8) Hope this derivation is clear and rigorous enough. Oleksandr Pavlyk Special Functions Developer Wolfram Research On 7/6/07, DrMajorBob <drmajorbob at bigfoot.com> wrote: > I thought you could get from the first RecEq to the second one this way: > > 15/ > 2 e HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, > 1 - e}, {1/2, 1, > 1}, 1] - > 45/16 (3 + 14 e + 8 e^2) HypergeometricPFQ[{1/2 - e, 1/2 - e, > 1/2 - e, 1 - e}, {1, 1, 3/2}, 1] + > 5/16 (65 + 188 e + 132 e^2 + 32 e^3) HypergeometricPFQ[{1/2 - e, > 1/2 - e, 1/2 - e, 1 - e}, {1, 1, 5/2}, 1] - > 1/2 (2 + e)^3 (3 + 2 e) HypergeometricPFQ[{1/2 - e, 1/2 - e, > 1/2 - e, 1 - e}, {1, 1, 7/2}, 1]; > RecEq = Distribute[16/5 # &[%]] > > (snip) > > But no, that's not the same thing, even though the first term looks right. > (No clue where the first came from; I assume Oleksandr was writing stuff > at random and got lucky. The second doesn't follow from the first in any > obvious (to me) way, so he got lucky twice!) > > The first RecEq is near-zero at s=1, like the second, but it doesn't work > in the subsequent code for simplifying the OP's expression. Only the > second recurrence relation has that nice property! > > Anyway, that's no more a symbolic proof than any of the others. It depends > (AFAICT) on the numerical value of RecEq appearing to be zero, just like > the OP's approximations tending toward -1/8. > > RecEq // FullSimplify > > 2/5 (192 e HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, > 1, 1}, 1] - > 36 (5 + 22 e + 8 e^2) HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, > 1 - e}, {1/2, 1, 2}, 1] + > 3 (143 + 4 e (86 + e (45 + 8 e))) HypergeometricPFQ[{1/2 - e, > 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 3}, > 1] - (2 + e) (5 + 2 e)^3 HypergeometricPFQ[{1/2 - e, 1/2 - e, > 1/2 - e, 1 - e}, {1/2, 1, 4}, 1]) > > And don't forget... in addition to lacking a rule that makes this zero, > Mathematica incorrectly simplified Daniel's denominator derivative. > > Bobby > > On Fri, 06 Jul 2007 02:23:16 -0500, sashap <pavlyk at gmail.com> wrote: > > > Hi, > > > > I actually found a way to symbolically prove the result. > > One has to use different recurrence equation with respect > > to lower integer parameter: > > > > In[189]:= Clear[q, r, RecEq]; > > > > In[190]:= > > r[e_] = -2^(-1 - 4 e) Gamma[ > > 1 - 2 e] Gamma[1/2 - e]^2 HypergeometricPFQ[{1/2 - e, 1/2 - e, > > 1/2 - e, 1 - e}, {1/2, 1, 1}, 1] Sin[e Pi]/Pi; > > > > In[191]:= > > RecEq = 15/ > > 2 e HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, > > 1}, 1] - > > 45/16 (3 + 14 e + 8 e^2) HypergeometricPFQ[{1/2 - e, 1/2 - e, > > 1/2 - e, 1 - e}, {1, 1, 3/2}, 1] + > > 5/16 (65 + 188 e + 132 e^2 + 32 e^3) HypergeometricPFQ[{1/2 - e, > > 1/2 - e, 1/2 - e, 1 - e}, {1, 1, 5/2}, 1] - > > 1/2 (2 + e)^3 (3 + 2 e) HypergeometricPFQ[{1/2 - e, 1/2 - e, > > 1/2 - e, 1 - e}, {1, 1, 7/2}, 1]; > > > > In[192]:= > > RecEq = 24 e HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/ > > 2, 1, 1}, 1] + > > 1/4 (-90 - 396 e - 144 e^2) HypergeometricPFQ[{1/2 - e, 1/2 - e, > > 1/2 - e, 1 - e}, {1/2, 1, 2}, 1] + > > 1/8 (429 + 1032 e + 540 e^2 + 96 e^3) HypergeometricPFQ[{1/2 - e, > > 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 3}, 1] - > > 1/8 (2 + e) (5 + 2 e)^3 HypergeometricPFQ[{1/2 - e, 1/2 - e, > > 1/2 - e, 1 - e}, {1/2, 1, 4}, 1]; > > > > In[193]:= RecEq /. e -> 1`15*^-5 > > > > Out[193]= 0.*10^-15 > > > > In[194]:= > > q[e_] = r[e] /. > > First[Solve[RecEq == 0, > > HypergeometricPFQ[{1/2 - e, 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 1}, > > 1]]] > > > > Out[194]= -(1/(3 e \[Pi])) > > 2^(-4 - 4 e) > > Gamma[1 - 2 e] Gamma[ > > 1/2 - e]^2 (-(1/ > > 4) (-90 - 396 e - 144 e^2) HypergeometricPFQ[{1/2 - e, 1/2 - e, > > 1/2 - e, 1 - e}, {1/2, 1, 2}, 1] - > > 1/8 (429 + 1032 e + 540 e^2 + 96 e^3) HypergeometricPFQ[{1/2 - e, > > 1/2 - e, 1/2 - e, 1 - e}, {1/2, 1, 3}, 1] + > > 1/8 (2 + e) (5 + 2 e)^3 HypergeometricPFQ[{1/2 - e, 1/2 - e, > > 1/2 - e, 1 - e}, {1/2, 1, 4}, 1]) Sin[e \[Pi]] > > > > In[195]:= > > Collect[q[e], _HypergeometricPFQ, > > Limit[#, e -> 0, Direction -> -1] &] /. e -> 0 > > > > Out[195]= -(1/8) > > > > Oleksandr Pavlyk > > Special Functions Developer > > Wolfram Research Inc. > > > > On Jul 5, 3:34 am, DrMajorBob <drmajor... at bigfoot.com> wrote: > >> That seems plausible, of course, but numerical results contradict it, > >> both > >> for the original expression and the denominator derivative. > >> > >> Start by taking apart the OP's expression: > >> > >> Clear[expr, noProblem, zero, infinite] > >> expr[s_] = -((2^(-2 + s)*Cos[(1/4)*Pi*(1 + s)]*Gamma[(1 + s)/4]^2* > >> Gamma[(1 + > >> s)/2]* > >> HypergeometricPFQ[{1/4 + s/4, 1/4 + s/4, 1/4 + s/4, > >> 3/4 + s/4}, > >> {1/2, 1, 1}, 1])/Pi); > >> List @@ % /. s -> 1 > >> > >> {-1, 1/2, 1/\[Pi], 0, \[Pi], 1, \[Infinity]} > >> > >> noProblem[s_] = expr[s][[{1, 2, 3, 5, 6}]] > >> zero[s_] = expr[s][[4]] > >> infinite[s_] = expr[s][[-1]] > >> > >> -(2^(-2 + s) Gamma[(1 + s)/4]^2 Gamma[(1 + s)/2])/\[Pi] > >> > >> Cos[1/4 \[Pi] (1 + s)] > >> > >> HypergeometricPFQ[{1/4 + s/4, 1/4 + s/4, 1/4 + s/4, 3/4 + s/4}, {1/2, > >> 1, 1}, 1] > >> > >> Here's essentially the same result you have: > >> > >> D[zero[s], s] /. s -> 1 > >> D[1/infinite[s], s] /. s -> 1 > >> > >> -\[Pi]/4 > >> 0 > >> > >> But the latter derivative is clearly wrong (probably due to some > >> simplification rule, applied without our knowledge, that isn't correct > >> at > >> s=1): > >> > >> Clear[dy, infInverse] > >> dy[f_, dx_, digits_] := N[(f[1 - dx] - f[1 - 2 dx])/dx, digits] > >> infInverse[s_] = 1/infinite[s]; > >> > >> Table[dy[infInverse, 10^-k, 25], {k, 1, 10}] > >> > >> {-1.923427116516395532881478, -2.983463609047004067632190, \ > >> -3.125313973445201143419500, -3.139959997528752908922902, \ > >> -3.141429339969442289280062, -3.141576321747481534963308, \ > >> -3.141591020400759167851555, -3.141592490270841802264744, \ > >> -3.141592637257897614551351, -3.141592651956603671268599} > >> > >> Table[Pi + dy[infInverse, 10^-k, 30], {k, 10, 20}] > >> > >> {1.63318956719404435692*10^-9, 1.6331895676743358738*10^-10, > >> 1.633189567722365025*10^-11, 1.63318956772716794*10^-12, > >> 1.6331895677276482*10^-13, 1.633189567727696*10^-14, > >> 1.63318956772770*10^-15, 1.6331895677277*10^-16, > >> 1.633189567728*10^-17, 1.63318956773*10^-18, 1.6331895677*10^-19} > >> > >> So the denominator limit is (almost certainly) -Pi, and the original > >> limit > >> is > >> > >> noProblem[1] D[zero[s], s]/-Pi /. s -> 1 > >> % // N > >> > >> -1/8 > >> -0.125 > >> > >> which agrees perfectly with the OP's calculations and with these: > >> > >> Table[N[1/8 + expr[1 - 10^-k], 20], {k, 1, 10}] > >> > >> {-0.031802368727291105600, -0.0029147499047748859136, \ > >> -0.00028902994360074334672, -0.000028878738171604118632, \ > >> -2.8876314481087565099*10^-6, -2.8876072131304841863*10^-7, \ > >> -2.8876047896519244498*10^-8, -2.8876045473042611497*10^-9, \ > >> -2.8876045230694967464*10^-10, -2.8876045206460203254*10^-11} > >> > >> I don't know how to prove the result symbolically, however. > >> > >> Bobby > >> > >> On Wed, 04 Jul 2007 04:28:36 -0500, dh <d... at metrohm.ch> wrote > >> > >> > >> > >> > >> > >> > Hi Dimitris, > >> > >> > I think the limit is -Infinity. > >> > >> > consider the following trick : > >> > >> > f1=2^(-2+s)*Cos[(1/4)*Pi*(1+s)]*Gamma[(1+s)/4]^2*Gamma[(1+s)/2] > >> > >> > f2= 1/HypergeometricPFQ[{1/4 + s/4, 1/4 + s/4, 1/4 + s/4, 3/4 + s/4}, > >> > >> > {1/2, 1, 1}, 1] > >> > >> > then we are interessted in the limit of f1/f2. As both these > >> expressions > >> > >> > are 0 for s=1, we can take the quotient of the drivatives: > >> > >> > D[f1,s]=-\[Pi]^2/8 > >> > >> > D[f2,s]= 0 > >> > >> > therefore we get -Infinity > >> > >> > hope this helps, Daniel > >> > >> > dimitris wrote: > >> > >> >> Hello. > >> > >> >> Say > >> > >> >> In[88]:= > >> > >> >> o = -((2^(-2 + s)*Cos[(1/4)*Pi*(1 + s)]*Gamma[(1 + s)/4]^2*Gamma[(1 + > >> > >> >> s)/2]* > >> > >> >> HypergeometricPFQ[{1/4 + s/4, 1/4 + s/4, 1/4 + s/4, 3/4 + s/4}, > >> > >> >> {1/2, 1, 1}, 1])/Pi); > >> > > > > > >> >> I am interested in the value at (or as s->) 1. > >> > >> >> I think there must exist this value (or limit) at s=1. > >> > >> >> In[89]:= > >> > >> >> (N[#1, 20] & )[(o /. s -> 1 - #1 & ) /@ Table[10^(-n), {n, 3, 10}]] > >> > >> >> Out[89]= > >> > >> >> > >> {-0.12528902994360074335,-0.12502887873817160412,-0.12500288763144810876,-0.\ > >> > >> >> > >> 12500028876072131305,-0.12500002887604789652,-0.12500000288760454730, > >> -0.\ > >> > >> >> 12500000028876045231,-0.12500000002887604521} > >> > >> >> However both > >> > >> >> o/.s->1 > >> > >> >> and > >> > >> >> Limit[o,s->1,Direction->1 (*or -1*)] > >> > >> >> does not produce anything. > >> > >> >> Note also that > >> > >> >> In[93]:= > >> > >> >> 2^(-2 + s)*Cos[(1/4)*Pi*(1 + s)]*Gamma[(1 + s)/4]^2*Gamma[(1 + s)/ > >> > >> >> 2] /. s -> 1 > >> > >> >> HypergeometricPFQ[{1/4 + s/4, 1/4 + s/4, 1/4 + s/4, 3/4 + s/4}, {1/2, > >> > >> >> 1, 1}, 1] /. s -> 1 > >> > >> >> Out[93]= > >> > >> >> 0 > >> > >> >> Out[94]= > >> > >> >> Infinity > >> > >> >> So am I right and the limit exist (if yes please show me a way to > >> > >> >> evaluate it) or not > >> > >> >> (in this case explain me why; in either case be kind if I miss > >> > >> >> something fundamental!) > >> > >> >> Thanks > >> > >> >> Dimitris > >> > >> -- > >> > >> DrMajor... at bigfoot.com > > > > > > > > > > > > -- > DrMajorBob at bigfoot.com >