RE: RE: complex analysis problem in mathematica 3.0
- To: mathgroup at smc.vnet.net
- Subject: [mg48492] RE: [mg48476] RE: [mg48459] complex analysis problem in mathematica 3.0
- From: "David Park" <djmp at earthlink.net>
- Date: Wed, 2 Jun 2004 04:21:56 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
I suspected I would get some alternative methods to simplify, but I don't think that they are all that obvious, and therefore not quicker. I still think a MapLevelParts command would be useful. It complements MapAt and gives much better direct control at manipulating expressions. David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: Andrzej Kozlowski [mailto:akoz at mimuw.edu.pl] To: mathgroup at smc.vnet.net A quicker way to get the same answer is: step1 = 1/(s^3 + 2*s^2 + 2*s + 1) /. s -> I*w Collect[ComplexExpand[step1] /. I -> i, i, Simplify] /. i -> I (1 - 2*w^2)/(w^6 + 1) + (I*w*(w^2 - 2))/(w^6 + 1) P.S. Although it is true one can't use j, one can use the character entered as escape j j escape, which looks like a fancy j. Andrzej Kozlowski Chiba, Japan http://www.mimuw.edu.pl/~akoz/ On 1 Jun 2004, at 16:02, David Park wrote: > You can't use j. Use the Mathematica I. > > step1 = 1/(s^3 + 2*s^2 + 2*s + 1) /. s -> I*w > > Then you can separate into real and imaginary parts by using > ComplexExpand. > > step2 = ComplexExpand[step1] > 1/((1 - 2*w^2)^2 + (2*w - w^3)^2) - > (2*w^2)/((1 - 2*w^2)^2 + (2*w - w^3)^2) + > I*(-((2*w)/((1 - 2*w^2)^2 + (2*w - w^3)^2)) + > w^3/((1 - 2*w^2)^2 + (2*w - w^3)^2)) > > But how do we further simplify that? We can't use Simplify or Together > on > the entire expression because that mixes the real and imaginary parts > together again. We can get a partial simplification by mapping > Simplify onto > the parts. > > step3 = Simplify /@ step2 > 1/(1 + w^6) - (2*w^2)/(1 + w^6) + (I*w*(-2 + w^2))/ > (1 + w^6) > > But the expression could be further simplified by using Together on the > first two terms. Unfortunately Mathematica doesn't provide any direct > method > of doing that other than resorting to a rule that duplicates the parts > on > the lhs. I think that Mathematica needs an additional routine that > would > complement MapAt. I call it MapLevelParts and it maps a function to a > subset > of level parts in an expression. The common application would be to a > subset > of terms in a Plus expression or a subset of factors in a Times > expression. > > MapLevelParts::usage = > "MapLevelParts[function, {topposition, levelpositions}][expr] will > map \ > the function onto the selected level positions in an expression. \ > Levelpositions is a list of the selected parts. The function is > applied to \ > them as a group and they are replaced with a single new expression. > Other \ > parts not specified in levelpositions are left unchanged.\nExample:\na > + b + > \ > c + d + e // MapLevelParts[f, {{2,4,5}}] -> a + c + f[b + d + e]"; > > MapLevelParts[func_, > part : {toppart___Integer?Positive, > subp : {_Integer?Positive, > eprest__Integer?Positive}}][expr_] := > Module[{work, subparts, npos, null, i, nnull = Length[{eprest}]}, > work = func@Part[expr, Sequence @@ part]; > subparts = Thread[{toppart, subp}]; > newparts = {work, Table[null[i], {i, 1, nnull}]} // Flatten; > npos = Partition[Range[nnull + 1], 1]; > ReplacePart[expr, newparts, subparts, npos] /. null[_] -> > Sequence[] > ] > > Now we can use Together on just the first two terms of the sum. > > step3 // MapLevelParts[Together, {{1, 2}}] > (1 - 2*w^2)/(1 + w^6) + (I*w*(-2 + w^2))/(1 + w^6) > > David Park >