Re: ForAll testing equality, and Limit evaluating wrong

*To*: mathgroup at smc.vnet.net*Subject*: [mg104595] Re: [mg104512] ForAll testing equality, and Limit evaluating wrong*From*: Rui Rojo <rui.rojo at gmail.com>*Date*: Wed, 4 Nov 2009 01:39:26 -0500 (EST)*References*: <200911030750.CAA00981@smc.vnet.net>

Thanks a lot ;) On Tue, Nov 3, 2009 at 9:24 PM, DrMajorBob <btreat1 at austin.rr.com> wrote: > If you don't want the Limit computed (and done away with) when f is > symbolic, define > > Clear[xTransf] > xTransf[f_?NumericQ] := > > 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] > > This means, however, that every single time you compute xTransf[aNumber], > the Limit is computed from scratch. > > Here's a timing for 201 calculations: > > {First@#, Length@Last@#} &@Timing@Table[xTransf@f, {f, 0, 10, .05}] > > {7.98758, 201} > > Here's a more efficient definition, with timing for the same evaluated > points: > > Clear[xTransf] > xTransf[0 | 0.] = > > 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 -> 0]; > > 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]; > {First@#, Length@Last@#} &@Timing@Table[xTransf@f, {f, 0, 10, .05}] > > {0.003726, 201} > > That's a big difference in speed: > > 7.98758/.003726 > > 2143.74 > > Bobby > > > On Tue, 03 Nov 2009 18:01:46 -0600, Rui Rojo <rui.rojo at gmail.com> wrote: > > I see why it happens, but I don't get why Mathematica removes the "Limit" >> from xTransf when it evaluates it at "f"... Or how I can avoid it, so I >> don't get those kinds of mistakes >> >> On Tue, Nov 3, 2009 at 8:33 PM, DrMajorBob <btreat1 at austin.rr.com> wrote: >> >> r[Pi] occurs because of copy/paste from e-mail to a notebook, when the >>> line >>> breaks badly like this: >>> >>> >>> 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] >>> >>> So that's an e-mail artifact. >>> >>> You defined xTransf this way: >>> >>> >>> >>> 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] >>> >>> and noticed that this evaluates fine: >>> >>> xTransf[0] >>> >>> 36 >>> >>> but this does not (error messages omitted): >>> >>> xTransf[f] /. f -> 0 >>> >>> Indeterminate >>> >>> It's easy to see why the error occurs, since this is undefined at f=0: >>> >>> xTransf[f] >>> >>> >>> (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) >>> >>> (We can't divide by zero.) >>> >>> But the limit as f->0 exists, so xTransf[0] can be calculated, and 36 is >>> the result. >>> >>> Bobby >>> >>> >>> On Tue, 03 Nov 2009 16:33:07 -0600, Rui Rojo <rui.rojo at gmail.com> wrote: >>> >>> 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 >>>>> >>>>> >>>>> >>> -- >>> DrMajorBob at yahoo.com >>> >>> > > -- > DrMajorBob at yahoo.com >

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

**Re: ForAll testing equality, and Limit evaluating wrong**

**easy way to represent polynomials (v 7.0)**

**Re: ForAll testing equality, and Limit evaluating wrong**

**Re: ForAll testing equality, and Limit evaluating wrong**