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Re: ForAll testing equality, and Limit evaluating wrong

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104585] Re: [mg104512] ForAll testing equality, and Limit evaluating wrong
  • From: Rui Rojo <rui.rojo at gmail.com>
  • Date: Wed, 4 Nov 2009 01:37:33 -0500 (EST)
  • References: <200911030750.CAA00981@smc.vnet.net>

Yeah, the first xTransf[f] slipped when copying by hand, and the second
thing I don't see it.  The FullSimplify does what I wanted like you said.
Thanks :)

Any comments on the Limit issue?
Rui Rojo

On Tue, Nov 3, 2009 at 6:34 PM, DrMajorBob <btreat1 at austin.rr.com> wrote:

> As is, you posted a scramble for two reasons:
>
> 1) The second definition should start with xTransf[f_] (WITH the
> underscore).
>
> 2) The right hand side includes r[Pi] in the second Cos argument,
> indicating that r is a function. In that case, what does it mean for r to
> approach f in the Limit?
>
> If I rewrite to eliminate those problems, Simplify doesn't get the result
> you want, but FullSimplify does:
>
> Clear[xTransf, xTransf2]
> xTransf2[f_] := 36 Sinc[6 Pi f]^2;
> xTransf[f_] =
>  Limit[(E^(-24 I \[Pi] r) (-Cos[12 \[Pi] r] + Cos[24 r \[Pi]] +
>      12 \[Pi] r Sin[24 \[Pi] r]))/(2 \[Pi]^2 r^2), r -> f];
> diff[i_] = xTransf2[i/24] - xTransf[i/24];
> one = Simplify[diff@i, {i > 0, i \[Element] Integers}]
>
> 36 ((8 (-1 + (-1)^i Cos[(i \[Pi])/2]))/(i^2 \[Pi]^2) +
>   Sinc[(i \[Pi])/4]^2)
>
> FullSimplify[diff@i, {i > 0, i \[Element] Integers}]
>
> 0
>
> One way to work that out almost by hand is:
>
> two = Expand[(i^2 Pi^2)/(8*36) one /. Sinc[any_] :> Sin@any/any]
>
> -1 + (-1)^i Cos[(i \[Pi])/2] + 2 Sin[(i \[Pi])/4]^2
>
> (to eliminate Sinc)
>
> doubleAngle = ReplaceAll[#, Cos[x_] :> Cos[x/2]^2 - Sin[x/2]^2] &;
> three = MapAt[doubleAngle, two, 2] // Expand
>
> -1 + (-1)^i Cos[(i \[Pi])/4]^2 +
>  2 Sin[(i \[Pi])/4]^2 - (-1)^i Sin[(i \[Pi])/4]^2
>
> (to have all the Sin and Cos terms squared)
>
> four = three /. Cos[any_]^2 :> 1 - Sin[any]^2 // Expand // Factor
>
> -(-1 + (-1)^i) (-1 + 2 Sin[(i \[Pi])/4]^2)
>
> Separate into factors:
>
> {minus, five, six} = List @@ four
>
> {-1, -1 + (-1)^i, -1 + 2 Sin[(i \[Pi])/4]^2}
>
> "five" is zero when i is even:
>
> Reduce[five == 0]
>
> (-1)^i == 1
>
> "six" is zero when i is odd:
>
> Reduce[six == 0]
>
> Sin[(i \[Pi])/4] == -(1/Sqrt[2]) || Sin[(i \[Pi])/4] == 1/Sqrt[2]
>
> So the product is zero for all i. (NON-ZERO i, that is, since we assumed
> Sinc was defined.)
>
> Bobby
>
>
> On Tue, 03 Nov 2009 01:50:55 -0600, Rui <rui.rojo at gmail.com> wrote:
>
>  I want to prove that xTransf2[f]==xTransf[f] for all f multiple of
>> 1/24.
>> xTransf2[f_]:=36 Sinc[6 Pi f]^2 and
>> xTransf[f]:=Limit[(E^(-24 I r \[Pi]) (-Cos[12 r \[Pi]] + Cos[24 r \
>> [Pi]] +
>>   12 r \[Pi] Sin[24 r \[Pi]]))/(2 r^2 \[Pi]^2), r->f]
>>
>> If I do
>> ForAll[f \[Element] Integers, YTransf[f/24] == YTransf2[f/24]]
>> I don't get a result... I can't find a way.
>> In fact, I get
>> (144 E^(-I f \[Pi]) (-2 Cos[(f \[Pi])/2] + 2 Cos[f \[Pi]] +
>>    f \[Pi] Sin[f \[Pi]]))/(f^2 \[Pi]^2) == 36 Sinc[(f \[Pi])/4]^2
>>
>> They are clrealy equal, at least on the 48 points closest to 0,
>> because if I do
>> And @@ ((xTransf[1/24 #] == xTransf2[1/24 #]) & /@ Range[-24, 24])
>> I get "True"
>>
>> Any pretty way to be certain?
>>
>> I've also realised that Mathematica has evaluated Limits with
>> variables, making the "Limit" disappear when for some values of the
>> variables I could get an indetermined result with the evaluated
>> version. For example, the Limit in xTransf
>> xTransf[f]
>> I get
>> (E^(-24 I f \[Pi]) (-Cos[12 f \[Pi]] + Cos[24 f \[Pi]] +
>>   12 f \[Pi] Sin[24 f \[Pi]]))/(2 f^2 \[Pi]^2)
>> without the Limit.
>> So, if I do
>> xTransf[f]/.f->0
>> I get errors but if I do xTransf[0] I get 36
>> ...
>> Hope you can help :)
>>
>>
>
> --
> DrMajorBob at yahoo.com
>


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