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 >

**References**:**ForAll testing equality, and Limit evaluating wrong***From:*Rui <rui.rojo@gmail.com>