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Re: ForAll testing equality, and Limit evaluating wrong

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104670] Re: [mg104512] ForAll testing equality, and Limit evaluating wrong
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 6 Nov 2009 05:17:51 -0500 (EST)
  • References: <200911030750.CAA00981@smc.vnet.net>

On 3 Nov 2009, at 16:50, Rui 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 :)
>


You can easily prove that they are equal with Mathematica. Define  
xTransf and xTransf2 in a slighlty different way:

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/24];

xTransf2[f_] := 36 Sinc[ Pi f/4]^2


And now:

FullSimplify[xTransf[n] - xTransf2[n], Element[n, Integers]]

0

That's all.


Andrzej Kozlowski


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