Re: Re: Re: Simplify with NestedLessLess?

*To*: mathgroup at smc.vnet.net*Subject*: [mg106609] Re: [mg106594] Re: [mg106531] Re: [mg106487] Simplify with NestedLessLess?*From*: "David Park" <djmpark at comcast.net>*Date*: Mon, 18 Jan 2010 02:35:39 -0500 (EST)*References*: <201001141049.FAA19892@smc.vnet.net> <4B4F39E7.1070002@wolfram.com> <4B4FAC81.7000108@ieee.org> <4B4FB26F.7050702@wolfram.com> <201001150821.DAA29881@smc.vnet.net> <op.u6k8eysrtgfoz2@bobbys-imac.local> <120224.1263732054539.JavaMail.root@n11>

I think I'm beginning to see what you want to do. You want to factor terms that can be factored into the form factor(C+Cf) and retain all other terms. Then you can spot these terms and make suitable simplifications. Here is a somewhat more complicated expression, just to give a more general test example. 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))/( 1 - Sqrt[4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg + extra]); ExpandAll[expr] //. a_ C + a_ Cf + terms_. -> a HoldForm[(C + Cf)] + terms // Simplify % /. C + Cf -> C // ReleaseHold (Rg^2 (-6 C Vd Vg+Cf^2 L2^2 Vg^4-12 Vd^2 (C+Cf)))/(-1+Sqrt[extra+2 C Rg^2 Vd Vg+4 Rg^2 Vd^2 (C+Cf)]) (Rg^2 (-12 C Vd^2-6 C Vd Vg+Cf^2 L2^2 Vg^4))/(-1+Sqrt[extra+4 C Rg^2 Vd^2+2 C Rg^2 Vd Vg]) We expanded everything, used a factoring rule, putting C+Cf in a HoldForm to protect it, and then used Simplify. You could then replace C+Cf with C. David Park djmpark at comcast.net http://home.comcast.net/~djmpark/ From: Dave Bird [mailto:dbird at ieee.org] 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 >>>>> >>>>> >>> >>> >> > >

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

**Re: Simplify with NestedLessLess?***From:*Dave Bird <dbird@ieee.org>

**Re: Simplify with NestedLessLess?**

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

**Re: Simplify with NestedLessLess?**

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