Functions with functions as arguments

Questions
1

Hi everyone. This doubt of mine does seem stupid, but I can’t seem to figure it out.

I’m 100% sure that I’ve wrote this simple (working) code for the usage of Simpson’s rule before, on a Numerical Analysis project, like the following

simpson[f_, nos_] := 
 Block[{h = nos[[2]] - nos[[1]], 
   n = Length[nos]}, (h/3) (f[nos[[1]]] + f[nos[[n]]] + 
     4 Sum[f[nos[[2 k]]], {k, (n - 1)/2}] + 
     2 Sum[f[nós[[2 k + 1]]], {k, (n - 3)/2}])]

Bare in mind that “nos” is the vector with nodal values for x, i.e. the x_i’s.
However, I’ve started using Mathematica again for another course, and this code right now isn’t working as before (I thinkkkk).

I thought of doing something like

function[ff_,a_]:=Block[{f},
    f[x_]:=ff;
    f[a]]

which also used to work for me before, but running " function[x,3] " also yields “x”.
Can anyone help me with this? There must be something stupid I’m doing. I’ll add that, interestingly enough, using f for integration does work on these kinds of functions; the problem seems to happen only when I need to evaluate the argument on a given point ( i.e. obtain f[x] ).

Finally, what would be your preferred ways of handling functions with other functions as arguments?

Thank you all. :slight_smile:

EDIT:

An even simpler example, running this code

function[f_]:=f[3]
function[x]

yields the result “x[3]”, however, my goal is to be able to obtain “3” with it. How could I do it?

2

You are using n = Length[nos], but in Simpson’s rule n should be the number of intervals, not the number of nodes.

3

Hi Gustavo

This works fine for me.

function[f_]:=f[3]
function[x]
(* x[3] *)

If the result for you is x, most likely you have x[3] = x defined. Try again after evaluating ClearAll[x, f, function]

4

Yes, thank you about that, but my actual problem is about being able to evaluate the argument (i.e. be able to use f[x] and such) inside the function.

5

Indeed, it yields “x[3]”, that’s my bad, however, this code

ClearAll[x, f, function]
function[ff_] := Block[{f},
  f[x_] := ff;
  f[3]]
function[x]

is yielding “x”. How could I write my function as to be able to obtain “3” (i.e. f evaluated at 3)?

6

Not sure what you are trying to accomplish. x is the result of f evaluated at 3. The function f[x_] := ff does not depend on x, it evaluates to ff no matter what argument it is passed. So function evaluates to whatever argument it is passed.

7

Basically, I’m trying to understand how to create a function which can receive the explicit form of another function, such that I can use it inside the original function.
i.e. create “result[f_]” where “f” is a function of x, for example, and be able to work with “f[x]” inside the code of “result”.

I mean, how else would you define (for example) a function “trapezoidal” which applies the trapezoidal method for integrating a given function “f” numerically in the interval [a,b] ?

8

Something like this?

integrate[func_, range_] := Integrate[func[x], {x, Splice@range}]

integrate[Sin, {0, Pi}]
(* 2 *)

integrate[Cos, {0, Pi/2}]
(* 1 *)
9

Something unexpected happens when you write f[x_]:=ff inside Block:

function[ff_] := Block[{f}, f[x_] := ff;
  f[3]]
function[x] // Trace

You will see that f[x_] becomes f[x$_]. This way the symbol x in function[x] does not match the pattern x$_ . To get the expected result you have to use x$ in your input:

In[5]:= function[x$]

Out[5]= 3

There is probably a good reason for this behaviour, but I am not an expert.

10

Nice. I guess I’ll be using “$” then.

In the meantime, I’ve discovered another way to do what I wanted

funçãocomoutrafunçãocomoargumentoMATHEMATICA.PNG
funçãocomoutrafunçãocomoargumentoMATHEMATICA.PNG1043×355 31.6 KB

in another discussion ( How do I fix this trapezoidal rule input? ).
Note that the important part of that code for this discussion is "func /. var → " ).