MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: limit

  • To: mathgroup at smc.vnet.net
  • Subject: [mg78718] Re: [mg78652] Re: limit
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Sat, 7 Jul 2007 06:12:05 -0400 (EDT)
  • References: <f6d9si$2l5$1@smc.vnet.net> <14396775.1183711623875.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

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 s1, 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


  • Prev by Date: Re: Re: Save As HTML does not produce any output
  • Next by Date: Re: Re: limit
  • Previous by thread: Re: limit
  • Next by thread: Re: Re: limit