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Re: Simplify with NestedLessLess?

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  • Subject: [mg106712] Re: Simplify with NestedLessLess?
  • From: dbird <dbird at>
  • Date: Thu, 21 Jan 2010 04:56:58 -0500 (EST)
  • References: <> <> <> <> <> <op.u6k8eysrtgfoz2@bobbys-imac.local> <> <op.u6niq5nhtgfoz2@bobbys-imac.local> <> <op.u6nuvknatgfoz2@bobbys-imac.local> <op.u6nvhla9tgfoz2@bobbys-imac.local> <> <op.u6pmhgrmtgfoz2@bobbys-imac.local> <hj40mc$sdo$>

I would like to thank all who have contributed to this thread, and put 
up with my lack of knowledge of Mathematica. During my studies of your 
proposals, I have found what to me is a useful, and sufficiently 
cautious procedure which accomplishes, pretty much, what I was looking 
for, even though I was unaware at the beginning how to precisely 
describe it.

For others who may be following this with interest, here it is:

First perform the following on the expression "expr", where eps<<1

Collect[ExpandAll[expr /. C -> Cf*eps], Cf, Simplify]

Second inspect the output from the above. Look for forms of (_a eps^_n 
\[PlusMinus] _b) or similar which can be safely changed to \[PlusMinus] 
b as follows:

% /. {a_ eps^2 - 1 -> -1, a_ eps^2 + 1 -> 1, (2 eps - 1) -> -1}

For me it is best at this point to add the /. rules one at a time and 
inspect after each application to be sure there are no surprises from 
whatever wildcards have been applied. As a novice wildcards are still an 

After satisfaction that no more eliminations are wanted, the final 
expression is restored similarly to the following:

FullSimplify[ExpandAll[% /. eps -> C/Cf], {assumptions}]

Thanks again for everyone's patience.


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