Re: Re: Simplify with NestedLessLess?
- To: mathgroup at smc.vnet.net
- Subject: [mg106557] Re: [mg106531] Re: [mg106487] Simplify with NestedLessLess?
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Sat, 16 Jan 2010 06:11:51 -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>
Dave Bird 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 > [...] Yes, that clarifies. This is along the lines of what I had meant by infinitesmals; you want to drop terms of certain order in certain settings. One respondant suggested using Series and then Normal to remove terms in that way. I think that is probably a good way to go about it, and I have seen that approach in the past. In the setting of your example above, what you might do is replace each small variable such as C by C*t (and if, say, L2 was much smaller than C, maybe use L2*t^2). For a series expansion in powers of t, dropping all terms beyond a certain point. If the "large" terms have varying orders of magnitude, that could be emulated by multiplying them by powers of 1/t. Another approach would use polynomial reduction. Short examples of each method can be found at the location below. http://forums.wolfram.com/mathgroup/archive/2007/Dec/msg00476.html Daniel Lichtblau Wolfram Research
- References:
- Simplify with NestedLessLess?
- From: dbird <dbird@ieee.org>
- Re: Simplify with NestedLessLess?
- From: Dave Bird <dbird@ieee.org>
- Simplify with NestedLessLess?