# Replacement rules do not apply for an expression inside a root

Say, you have an expression which looks like this:

myexpr = (mess)+(some other mess)/( (mess)^3+(some other mess)^2 )^(1/3)


Here the 'mess' in the name of terms stands for what it is - a mess. To make it look nicer I use replacement rules and substitute (mess)->u, (some other mess)->v as follows

myexpr /. {(mess)->u,(some other mess)->v}


so that the output would look as nice as

u+v/(u^3+v^2)^(1/3)


Now, I actually don't get this result because when done as above the rules are not applied inside the cubic root, so the output I get instead is

u+v/( (mess)^3+(some other mess)^2 )^(1/3)


The question is, naturally, how can the full replacement be achieved with minimum pain? Thanks.

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• We will probably need to see the messes you are dealing with to help you find your problem. Sometimes problems like this are caused by a rule being applied to an outer expression, and not to an inner one. You might look at ReplaceRepeated Aug 8, 2015 at 12:52

Your have to use RuleDelayed (:>) in such a case, so that the evaluation is not done immediately:

myexpr = (mess) + (someOtherMess)/((mess)^3 + (someOtherMess)^2)^(1/3)


and then:

 myexpr /. {mess :> x, someOtherMess :> y}


Then you get what you want.

• @Patrick: Thanks for removing my typo - I didn´t see it.. Aug 8, 2015 at 15:14
• I'm not sure that I follow your reasoning here. RuleDelayed will delay the evaluation of $x$ and $y$ in your example, but won't affect the (mess) expressions. Since I expect that $x$ and $y$ might not even have been assigned a value in this case, I don't see the difference between this and a simple Rule. The opposite behavior might be required, i.e. we might have to prevent evaluation of mess with one of the Hold* functions so it can be matched literally to what is in the original expression. Of course, it's hard to be specific without seeing the OP's actual expressions. Aug 8, 2015 at 15:24
• I agree. In most cases ReplaceRepeated would be the right choice. But without having the original code, it´s hard to find an appropriate solution Aug 8, 2015 at 16:03

Thanks to mfvonh for the answer - using ReplaceRepeated (//.) as shown below replaces everything.

myexpr //. {(mess)->u,(some other mess)->v}


To those in power: I unfortunately cannot accept the answer properly because after setting up the account the question is not listed as mine (and can't leave a comment because low rep = having no rights).