In[1]:= Function[Solve[Sqrt[a + b] == #, a]][#] /. # -> a

Out[1]= {{a -> a^2 - b}}

The syntax coloring will look a bit odd when you try it out, but it works.

While that maybe true in general, let me worry about other underlining issues for the time being because answering this particular question also helps me understand what can and can not be done using pure functions in Mathematica. Right now, my principle goal is to get code that produces my desired results, followed by learning more programming strategies, followed by the efficiency of the code. If it helps, the topology of the space I'm working in is important, not the metric (I'm free to use any yardstick I wish).

If the only way to get the result I desire is to explicitly declare a variable "s", please say so.

And Rohit, I'm trying to be polite but it is getting difficult as I find it extremely offensive when people try to second guess me. I appreciate the help but please answer the question asked.

Hi Rohit,

Still not the correct answer as the output for

Solve[Sqrt[a+b]==??,a]???->a

must be

{{a->a^2-b}}.

What I'd like to do (avoid explicitly using s) may not be allowed, in which case I'll just have to use Solve[Sqrt[a+b]==s, a]/.s->a (= {{a->s^2-b}}/.s->a = {{a->a^2-b}}).

Also, I am using Remove frequently but that still does not fix the memory problem as Mathematica seems to insists on remembering every calculation that my program (which is 664 KB of text) does in a way I don't know how to prevent. Everything is purely symbolic and currently written procedurally, with a lot of modules (defined with := , as = appears not to work correctly) and very large paralleltables (I've tried breaking the tables up but that causes other problems). I'm a novice at writing code using the "pure function" formate, which various sources say uses less memory and is faster, but that is the direction I'd like to move to.

In the end, it is should not be necessary for others to understand why I wish to do what I wish to do in order to answer this question.

Thank you.

In[1]:= Function[Solve[Sqrt[a + b] == #, #]][a]

Out[1]= {{a -> 1/2 (1 - Sqrt[1 + 4 b])}, {a -> 1/2 (1 + Sqrt[1 + 4 b])}}

Hi Rohit,

Neither of your examples. Perhaps it would have been better if I had written

Solve[exp==s, variable]/.s->variable

such as

Solve[Sqrt[a+b]==s, a]/.s->a

Giving

{{a -> a^2 - b }}

My variables are arbitrary and I can replace them at will so long as I keep all the relationships the same; it allows me to greatly simplify my problem.