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Re: Re: Simplify with NestedLessLess?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg106602] Re: [mg106531] Re: [mg106487] Simplify with NestedLessLess?
  • From: Dave Bird <dbird at ieee.org>
  • Date: Mon, 18 Jan 2010 02:34:21 -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> <4B5215AE.6020402@ieee.org> <op.u6niq5nhtgfoz2@bobbys-imac.local> <4B527F72.9010000@ieee.org> <op.u6nuvknatgfoz2@bobbys-imac.local> <op.u6nvhla9tgfoz2@bobbys-imac.local>
  • Reply-to: dbird at ieee.org

This works for the test expression, obviously. But, can it be made to 
work for a more complex expression containing say a radical?

I don't understand it well enough to critique at this point; I'll have 
to study it more tomorrow.

Thanks,

Dave

DrMajorBob wrote:
> 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
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>
>>>>>>
>>>>
>>>>
>>
>>
>
>


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