Re: Re: limit
- To: mathgroup at smc.vnet.net
- Subject: [mg78730] Re: [mg78652] Re: limit
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Sun, 8 Jul 2007 06:05:21 -0400 (EDT)
- References: <f6d9si$2l5$1@smc.vnet.net> <14396775.1183711623875.JavaMail.root@m35> <op.tu2q58jfqu6oor@monster.ma.dl.cox.net> <30802604.1183786642129.JavaMail.root@m35>
- Reply-to: drmajorbob at bigfoot.com
Thanks!!
Such things are not really my area of interest (to the extent I have one),
but I like to learn something new every day.
Bobby
On Sat, 07 Jul 2007 00:29:44 -0500, Oleksandr Pavlyk <pavlyk at gmail.com>
wrote:
> 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
>>
>
--
DrMajorBob at bigfoot.com