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simplifying expressions - What is the difference between a few simplification techniques?



I am trying to understand the difference between Refine, Simplify and FullSimplify, and when it's more appropriate to use a particular one. The help files on this are not entirely clear. For example, under the help for Refine it says



Use Simplify for more simplification rules:


Refine[Sqrt[x^2 + 2 x y + y^2], x + y >= 0]
(* Sqrt[x^2 + 2 x y + y^2] *)

Simplify[Sqrt[x^2 + 2 x y + y^2], x + y >= 0]
(* x + y *)

From this, I can't imagine why anyone would ever use Refine, if Simplify will get it simpler. Yet, it keeps turning up in answers on this site (e.g. here, here, or here), where Simplify would also work.



Then, regarding Simplify vs FullSimplify, the help file states:



Use FullSimplify to simplify expressions involving special functions:



Is there a guideline for which sorts of functions are considered "special functions"?


I realize that (based on the "More Information" section and personal experience) FullSimplify may take longer, but under what circumstances? Is a good general approach to try Simplify first, and then only if one is unhappy with the results, then try FullSimplify?



Answer



The primary difference between Refine and the two *Simplify functions is that Refine only evaluates the expression according to the assumptions given. It might so happen to be the simplest form when evaluated, but it does not check to see if it is indeed the simplest possible form. You should use Refine when your goal is not to simplify the expression but to just see how the assumptions transform it (e.g., square root of a positive quantity).


Simplify, on the other hand, performs basic algebraic simplifications and transformations to arrive at the "simplest" result. Refine is one among them, and is also mentioned in its doc page. Here, "simplest" might not necessarily fit your definition of simple. It is what appears simple to Mathematica, and that is defined by LeafCount. Here's an example showing the difference between the two:


Refine[(x - 1)/Sqrt[x^2] + 1/x, x > 0]

(* Out= 1/x + (-1 + x)/x *)

Simplify[(x - 1)/Sqrt[x^2] + 1/x, x > 0]
(* Out= 1 *)

FullSimplify behaves the same as Simplify, except that it also does manipulations and simplifications when it involves special functions. It is indeed slower, as a result, because it has to try all the available rules. The list of special functions is found in guide/SpecialFunctions and it's not a non-standard usage of the term and you can also read about it on Wikipedia. So in all cases not involving special functions, you should use Simplify. You can certainly give FullSimplify a try if you're not satisfied with Simplify's result, but it helps to not start with it if you don't need it.


Here's an example showing the difference between Simplify and FullSimplify:


Simplify[BesselJ[n, x] + I BesselY[n, x]]
(* BesselJ[n, x] + I BesselY[n, x] *)


FullSimplify[BesselJ[n, x] + I BesselY[n, x]]
(* HankelH1[n, x] *)


A few more notes on Simplify and FullSimplify:


As you have noted, FullSimplify is slow — sometimes it can take hours on end to arrive at the answer. The default for TimeConstrained, which is an option, is Infinity, which means that FullSimplify will take its sweet time to expand/transform until it is satisfied. However, it could very well be the case that a bulk of the time is spent trying out various transformations (which might eventually be futile) and the actual simplification step is quick. It helps to try out with a shorter time, and the documentation has a good example that shows this. This holds for Simplify too.


Note that setting the option TimeConstrained -> t does not mean that you'll get your answer in under t seconds. Rather, it means that Mathematica is allowed to spend at most t seconds for a single transformation/simplification step.


Similarly, you can exclude certain functions from being simplified using ExcludedForms or include other custom transformations using TransformationFunctions. You can even change the default measure of "simplicity" using ComplexityFunction, and here is an answer that uses this. However, these options are not available in Refine.


These are actually well documented in the documentation for both functions, but is not widely known and can often be the key to getting the result quickly or in the form you want.


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