Re: Re: Simplify with NestedLessLess?

*To*: mathgroup at smc.vnet.net*Subject*: [mg106585] Re: [mg106531] Re: [mg106487] Simplify with NestedLessLess?*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Sun, 17 Jan 2010 07:11:40 -0500 (EST)*References*: <201001141049.FAA19892@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

Oops... Normal was extraneous there. I had another solution using Series, but I switched to Limit. Clear[h] h[expr_] /; FreeQ[expr, C] || FreeQ[expr, Cf] := expr h[expr_] := Limit[expr /. C -> delta Cf, delta -> 0] expr = -Cf^2 L2^2 Rg^2 Vg^4 + 3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg); vars = Complement[Variables@expr, {C, Cf}]; Collect[expr, vars, h] -Cf^2 L2^2 Rg^2 Vg^4 + Rg^2 (12 Cf Vd^2 + 6 C Vd Vg) Bobby On Sat, 16 Jan 2010 23:00:46 -0600, DrMajorBob <btreat1 at austin.rr.com> wrote: >> Vg is potentially much larger than Vd (last two terms). In other words, >> the magnitude of Vd, Vg and L2 could conspire together to make the last >> term significant even though C<<Cf. > > The same can be said for ANY term you want to ignore, can it not? > > But never mind; I think you want to apply the technique I just showed > you, but only to individual coefficients in > > Collect[-Cf^2 L2^2 Rg^2 Vg^4 + > 3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg), {L2, Rg, Vg}] > > -Cf^2 L2^2 Rg^2 Vg^4 + Rg^2 (12 C Vd^2 + 12 Cf Vd^2 + 6 C Vd Vg) > > ...but ONLY if C and Cf both appear in the same coefficient. > > That's a strange set of conditions, in my opinion, but here's a way to > do it: > > Clear[h] > h[expr_] /; FreeQ[expr, C] || FreeQ[expr, Cf] := expr > h[expr_] := Normal[Limit[expr /. C -> delta Cf, delta -> 0]] > > expr = -Cf^2 L2^2 Rg^2 Vg^4 + > 3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg); > vars = Complement[Variables@expr, {C, Cf}]; > Collect[expr, vars, h] > > -Cf^2 L2^2 Rg^2 Vg^4 + Rg^2 (12 Cf Vd^2 + 6 C Vd Vg) > > % == -Cf^2 L2^2 Rg^2 Vg^4 + > 3 (4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg) // Simplify > > True > > Bobby > > On Sat, 16 Jan 2010 21:09:38 -0600, Dave Bird <dbird at ieee.org> wrote: > >> Yes, but when we make delta->0 we eliminate _all_ C terms, thus missing >> the correct answer of >> >> -Cf^2 L2^2 Rg^2 Vg^4+3 (4 Cf Rg^2 Vd^2+2 C Rg^2 Vd Vg) >> >> Vg is potentially much larger than Vd (last two terms). In other words, >> the magnitude of Vd, Vg and L2 could conspire together to make the last >> term significant even though C<<Cf. >> >> Dave >> >> DrMajorBob wrote: >>> You said C<<Cf, so let's say C = delta CF, where delta is small. We >>> compute a series around delta=0 and, after giving it a look, we set >>> delta = 0. >>> >>> expr = >>> Normal@Series[-Cf^2 L2^2 Rg^2 Vg^4 + >>> 3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg) /. >>> C -> delta Cf, {delta, 0, 6}] >>> firstTry = expr /. delta -> 0 >>> >>> 12 Cf Rg^2 Vd^2 - Cf^2 L2^2 Rg^2 Vg^4 + >>> delta (12 Cf Rg^2 Vd^2 + 6 Cf Rg^2 Vd Vg) >>> >>> 12 Cf Rg^2 Vd^2 - Cf^2 L2^2 Rg^2 Vg^4 >>> >>> In a more complicated situation (if a few Series terms were >>> unconvincing), we might use Limit, instead: >>> >>> expr = -Cf^2 L2^2 Rg^2 Vg^4 + >>> 3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg) /. C -> delta Cf >>> secondTry = Limit[expr, delta -> 0] >>> >>> -Cf^2 L2^2 Rg^2 Vg^4 + >>> 3 (4 Cf Rg^2 Vd^2 + 4 Cf delta Rg^2 Vd^2 + 2 Cf delta Rg^2 Vd Vg) >>> >>> Cf Rg^2 (12 Vd^2 - Cf L2^2 Vg^4) >>> >>> In this case, both answers are the same: >>> >>> firstTry == secondTry // Simplify >>> >>> True >>> >>> Bobby >>> >>> On Sat, 16 Jan 2010 13:38:22 -0600, Dave Bird <dbird at ieee.org> wrote: >>> >>>> Interesting! But, I don't think I am correctly communicating what I'm >>>> after yet. (Although, I admit that I am struggling some to keep up >>>> with you guys in your Mathematica replies due to my inexperience.) >>>> >>>> The original expression that I put up for illustration is: >>>> >>>> -Cf^2 L2^2 Rg^2 Vg^4+3 (4 C Rg^2 Vd^2+4 Cf Rg^2 Vd^2+2 C Rg^2 Vd Vg) >>>> >>>> We compare 4 C Rg^2 Vd^2 to 4 Cf Rg^2 Vd^2 because the two terms >>>> share common coefficients so that they "reduce" to (4 Rg^2 Vd^2+4 >>>> Rg^2 Vd^2) (C+Cf) . Thus it becomes obvious that C may be discarded >>>> w.r.t. Cf. >>>> >>>> Please forgive if I have missed the correct application of your >>>> suggestion, and thanks for the interest. >>>> >>>> Dave >>>> >>>> DrMajorBob wrote: >>>>> Series[-Cf^2 L2^2 Rg^2 Vg^4 + >>>>> 3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg), {C, 0, >>>>> 5}] // Simplify >>>>> >>>>> SeriesData[C, 0, { >>>>> Cf Rg^2 (12 Vd^2 - Cf L2^2 Vg^4), 6 Rg^2 Vd (2 Vd + Vg)}, 0, 6, 1] >>>>> >>>>> Bobby >>>>> >>>>> On Fri, 15 Jan 2010 02:21:09 -0600, Dave Bird <dbird at ieee.org> wrote: >>>>> >>>>>> Not infinitesimals. I'm working in analog circuit design/analysis. I >>>>>> have a 3 pole symbolic circuit response (third order) which is not >>>>>> easily separable. I can use Mathematica to find the three roots of >>>>>> the >>>>>> response. But, the roots are, of course, very messy. I know that >>>>>> certain >>>>>> elements in the circuit are orders of magnitude larger than other >>>>>> like >>>>>> elements - capacitors in this case. For example, one small section >>>>>> of >>>>>> one root is >>>>>> >>>>>> -Cf^2 L2^2 Rg^2 Vg^4+3 (4 C Rg^2 Vd^2+4 Cf Rg^2 Vd^2+2 C Rg^2 Vd Vg) >>>>>> >>>>>> I know that C<<Cf. By careful inspection, I can see that the first >>>>>> term >>>>>> in the parens will drop out compared to the second term in the >>>>>> parens. I >>>>>> would like Mathematica to do this without my having to examine it so >>>>>> closely since there are many other like situations. >>>>>> >>>>>> This kind of situation occurs in many other engineering situations. >>>>>> >>>>>> Hope this helps clarify. >>>>>> >>>>>> Thanks for the interest. >>>>>> >>>>>> Dave >>>>>> >>>>>> >>>>>> >>>>>> >>>>>> Daniel Lichtblau wrote: >>>>>>> Dave Bird wrote: >>>>>>>> Thanks Daniel for the observation. I forgot to add that both a, >>>>>>>> and b >>>>>>>> are real positive. That, of course would have to be added to the >>>>>>>> assumptions. >>>>>>>> >>>>>>>> Dave >>>>>>> >>>>>>> It's still not obvious what you are wanting to do. I have the idea >>>>>>> you >>>>>>> are working in some sense with infinitesmals. If so, I doubt >>>>>>> Simplify >>>>>>> would be the best tool for removing them; it really can only do >>>>>>> that >>>>>>> if it is told, in some way, to replace them with zero. How might >>>>>>> one >>>>>>> instruct Simplify to figure that out? >>>>>>> >>>>>>> Daniel >>>>>>> >>>>>>> >>>>>>>> Daniel Lichtblau wrote: >>>>>>>>> dbird wrote: >>>>>>>>>> Please excuse if this has been answered before, but I can't >>>>>>>>>> find it. >>>>>>>>>> >>>>>>>>>> Is there some way to do a Simplify with assumptions using a >>>>>>>>>> NestedLessLess or something similar? For example: >>>>>>>>>> >>>>>>>>>> d=a+b >>>>>>>>>> Simplify[d,NestedLessLess[a,b]] >>>>>>>>>> >>>>>>>>>> Answer is: >>>>>>>>>> a+b >>>>>>>>>> >>>>>>>>>> Answer should be: >>>>>>>>>> b >>>>>>>>>> >>>>>>>>>> Thanks, >>>>>>>>>> >>>>>>>>>> Dave >>>>>>>>> >>>>>>>>> I fail to see why the result should be b. >>>>>>>>> >>>>>>>>> Daniel Lichtblau >>>>>>>>> Wolfram Research >>>>>>>>> >>>>>>>>> >>>>>>> >>>>>>> >>>>>> >>>>> >>>>> >>> >>> > > -- DrMajorBob at yahoo.com

**References**:**Simplify with NestedLessLess?***From:*dbird <dbird@ieee.org>

**Re: Re: Simplify with NestedLessLess?**

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**Re: Re: Simplify with NestedLessLess?**

**Re: Re: Simplify with NestedLessLess?**