Re: Re: Simplify with NestedLessLess?

*To*: mathgroup at smc.vnet.net*Subject*: [mg106584] Re: [mg106531] Re: [mg106487] Simplify with NestedLessLess?*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Sun, 17 Jan 2010 07:11:29 -0500 (EST)*References*: <201001141049.FAA19892@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

> 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?**

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

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

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