Re: limit
- To: mathgroup at smc.vnet.net
- Subject: [mg78702] Re: limit
- From: dimitris <dimmechan at yahoo.com>
- Date: Sat, 7 Jul 2007 06:03:43 -0400 (EDT)
- References: <f6d9si$2l5$1@smc.vnet.net><f6kr4o$kvn$1@smc.vnet.net>
sashap :
> 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
Thank you very much to all that participating in this thread!
Cheers
Dimitris