Variables Inside Functions Not Evaluating

Question

x = 5*y
function1[y_] := x
function1[5]

Am I wrong in thinking that function1[5] should equal 25? Instead this comes out.

5y
5y

Answer 1

Without going into too much detail, this is a result of the way pattern matching is carried out. When you write the rule f[y_]:={x,y} for instance, it tells the compiler that it should match anything and call it y, then replace any instance of y on the right side with the match. But x doesn't evaluate until after this has already happend. A good way to see what goes on is to use Trace

x = 5 y;
f[y_] := {x, y}
f[3] // Trace

(*=> {f[3],{x,3},{x,5 y},{5 y,3}} *)

The way to get around this is to get x to evaluate before you define the rule, which you can do for instance by passing it through an anonomous function:

(f[y_] := {#, y})&@x
f[3] // Trace
(*=>  {f[3],{5 3,3},{5 3,15},{15,3}} *)

If you want to see why this works out you can see that at the time the rule definition is evaluated x has been fully replace by using Trace on the code defining the function

 (f[y_] := {#, y}) &@x // Trace

(*=> {{x,5 y},((f[y_]:={#1,y})&)[5 y],f[y_]:={5 y,y},Null} *)

Answer 2

As already explained the named pattern y_ is only substituted for explicit instances of y in the right-hand side.

As a point of reference this works:

x = 5*y;
function1[y_] = x;
function1[5]

25

The value of x is hard coded into function1:

Definition[function1]

function1[y_] = 5 y

An approach that will let you change the value of x is to use Block:

function2[yy_] := Block[{y = yy}, x];

function2[5]

25

x = 7 y;

function2[5]

35

Answer 3

What you want do is this:

function1[y_] := 5*y;

The way you define functions in Mathematica is by defining patterns, which is indicated here by y_, which matches any expression. So function1[y] is always replaced by 5 times whatever y is when evaluated. What you did is define a symbol called x, then make a delayed definition called function1[y] which equals x. But function1[5] has never been defined.