about plotting process — a further question about “A problem in Mathematica 8 with function declaration”

问题

Related A problem in Mathematica 8 with function declaration

Clear["Global`*"]

model = 4/Sqrt[3] - a1/(x + b1) - a2/(x + b2)^2 - a3/(x + b3)^4;
fit = {a1 -> 0.27, a2 -> 0.335, a3 -> -0.347, b1 -> 4.29, b2 -> 0.435,
b3 -> 0.712};

functionB1[x_] = model /. fit;
functionB2[x_] := model /. fit;

The evaluation difference between functionB1 and functionB2 can be revealed by Trace command in mma, as below:

functionB1[Sqrt[0.2]] // Trace
functionB2[Sqrt[0.2]] // Trace

I have no question about functionB1. what puzzles me is that because functionB2[Sqrt[0.2]] doesn't even gives a numeric result but gives a function of x 4/Sqrt[3] - 0.335/(0.435 + x)^2 + 0.347/(0.712 + x)^4 - 0.27/( 4.29 + x), and then how its plot Plot[functionB2[Sqrt[x]], {x, 0, 1}] is possible?

I mean when you run Plot[functionB2[Sqrt[x]], {x, 0, 1}], what happens inside mma is:

x takes a number, say, 0.2, then 0.2 is finally passed to functionB2, but functionB2 gives a function, not a number. Then how is the following figure generated?

And its trace result ( Plot[functionB2[Sqrt[x]], {x, 0, 1}] // Trace ) seems very unreadable. I wonder the clear plotting process of functionB2. Can anybody show it?

thanks~ :)

回答1:

SetDelayed acts as a scoping construction. Arguments are localized if necessary. Any variables that explicitly match the arguments are bound within this scope, others are not.

In[78]:= a[x_] := x^2 + b
         b = x^4;

(* the first x^2 is explicitly present and bound to the argument. 
   The x in x^4 present via b is not bound *)

In[80]:= a[x]

Out[80]= x^2 + x^4 (* this is what you would expect *)

In[81]:= a[y]

Out[81]= x^4 + y^2 (* surprise *)

In[82]:= a[1]

Out[82]= 1 + x^4 (* surprise *)

So, what you could do is one of two things:

Use Evaluate: functionB2[x_] := Evaluate[model /. fit];
Make dependence of model on x explicit:

In[68]:= model2[x_] = 4/Sqrt[3] - a1/(x + b1) - a2/(x + b2)^2 - a3/(x + b3)^4;

In[69]:= functionB3[x_] := model2[x] /. fit;

In[85]:= functionB3[Sqrt[0.2]]

Out[85]= 2.01415

Edit because of question update
Because of your definition of functionB2 any argument value yields the same result, as explained above:

In[93]:= functionB2[1]

Out[93]= 4/Sqrt[3] - 0.335/(0.435 + x)^2 + 0.347/(0.712 + 
   x)^4 - 0.27/(4.29 + x)

In[94]:= functionB2["Even a string yields the same ouput"]

Out[94]= 4/Sqrt[3] - 0.335/(0.435 + x)^2 + 0.347/(0.712 + 
   x)^4 - 0.27/(4.29 + x)

However, this expression contains x's and therefore it can get a numerical value if we provide a numerical value for x:

In[95]:= functionB2["Even a string yields the same ouput"] /. x -> 1

Out[95]= 2.13607

Well, this basically what Plot does too. This is why you still get a plot.

回答2:

The definition:

functionB2[x_] := model /. fit

is an instruction to Mathematica to replace all future occurrences of an expression that looks like functionB2[x_] with the result of substituting the value of the argument for every occurrence of x in the expression model /. fit. But there are no occurrences of x in model /. fit: the only symbols in that expression are model and fit (and, technically, ReplaceAll). Therefore, the definition returns a fixed result, model /. fit, irrespective of the argument. Indeed, the definition could just simply be:

functionB2a[] := model /. fit

If you plot functionB2a[], you will get the same result as if you plotted functionB2[anything]. Why? Because Plot will evaluate that expression while varying the symbol x over the plot range. It so happens that model /. fit evaluates to an expression involving that symbol, so you get the exhibited plot.

Now consider functionB1:

functionB1[x_] = model /. fit

It too says to replace all occurrences of x on the right-hand side -- but this time the right-hand side is evaluated before the definition is established. The result of evaluating model /. fit is an expression that does contain the symbol x, so now the definition is sensitive to the passed argument value. The net result is as if the function were defined thus:

functionB1a[x_] := 4/Sqrt[3]-0.335/(0.435+x)^2+0.347/(0.712+x)^4-0.27/(4.29+x)

So, if you plot functionB1[Sqrt[x]], the Plot command will see the expression:

4/Sqrt[3]-0.335/(0.435 +Sqrt[x])^2+0.347/(0.712 +Sqrt[x])^4-0.27/(4.29 +Sqrt[x])

Formal Symbols

When establishing definitions using SetDelayed, the name of the formal argument (x in this case) is independent of any occurrences of the same symbol outside of the definition. Such definitions can use any other symbol, and still generate the same result. On the other hand, definitions established using Set (such as functionB1) rely on the result of evaluating the right-hand side containing the same symbol as the formal argument. This can be a source of subtle bugs as one must take care not use symbols that accidentally have pre-existing down-values. The use of formal symbols (described in Letters and Letter-like Forms) for argument names can help manage this problem.

回答3:

You can understand what is going on by trying:

Table[functionB2[Sqrt[y]],{y,0.5,.5,.5}]
Table[functionB2[Sqrt[x]],{x,0.5,.5,.5}]
(*
{4/Sqrt[3] - 0.335/(0.435+ x)^2 + 0.347/(0.712+ x)^4 - 0.27/(4.29+ x)}
{2.03065}
*)

What is getting replaced is the x inside the definition of functionB2, not a formal argument.

Edit

The plot you are getting is not what you want. The Sqrt[x] is disregarded in functionB2[...] and the implicit x is replaced, as you can see here:

来源：https://stackoverflow.com/questions/6058281/about-plotting-process-a-further-question-about-a-problem-in-mathematica-8-w

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