6
$\begingroup$

Simple question. How can we construct functions like Mathematica does?

For example, let's say that we want to create our own D function.

The D function works like:

In[1]:= D[x^2, x]
Out[1]= 2 x

where we can put a function in x and the function will treat this internally in a correct way.

How can we create our D1 function with the same behavior?

Attempts:

In[1]:= D1[f_, x_] := Limit[(-f[x] + f[x + a])/a, a -> 0]


In[2]:= D1[x^2, x]

Out[2]= Limit[(-(x^2)[x] + (x^2)[a + x])/a, a -> 0]


In[3]:= D1[Function[x, x^2], x]

Out[3]= 2 x


In[4]:= D1[#^2 &, x]

Out[4]= 2 x

As we can see, the second and third example work, but I'd like to create a function that would work like in the first example and like the D function.

How can we do this?

$\endgroup$
  • $\begingroup$ Such function would consist with a large set of rules: how to treat constants, sums, products, power functions, etc. $\endgroup$ – yarchik Jun 5 at 19:43
  • $\begingroup$ It is not necessary to have all behavior. It is more about how to use the arguments without using Function[x, ...] or #...&. $\endgroup$ – GarouDan Jun 5 at 19:55
6
$\begingroup$
ClearAll[D1]

D1[f : _Function | _Symbol, x_] := Block[{a}, Limit[(-f[x] + f[x + a])/a, a -> 0]]

D1[f_, x_] := D1[Function[x, f], x]

Now these all return 2 x:

D1[x^2, x]

D1[#^2 &, x]

fun[z_] := z^2; D1[fun, x]

D1[fun[x], x]
| improve this answer | |
$\endgroup$
  • $\begingroup$ This solution is very interesting, indeed. Is it possible to solve the problem pointed by @yarchik? D1[x^3,x^2] $\endgroup$ – GarouDan Jun 5 at 21:18
  • $\begingroup$ @GarouDan Pardon me, I'm not familiar with this syntax. D[x^3, x^2] yields General::ivar: x^2 is not a valid variable. >> $\endgroup$ – Mr.Wizard Jun 5 at 21:26
  • 3
    $\begingroup$ Yes, it should give an error in this case. That is what it does. $\endgroup$ – yarchik Jun 5 at 21:43
6
$\begingroup$

EDIT:

Based on the suggestion by Bob Hanlon, this code should be better as it isolates a from the context of the notebook.

D1[f_, x_] := Module[{a}, Limit[(-f + (f /. {x -> (x + a)}))/a, a -> 0]]

I've tried this on a few functions and it seems to return the same as D, but there are probably edge cases I haven't considered.

D1[f_, x_] := Limit[(-f + (f /. {x -> (x + a)}))/a, a -> 0]
D1[x^2, x]
D1[3 x^3 + y^2, x]
D1[3 Sin[2 x] y^2 + 14, x]

$2x$

$9x^2$

$6y^2\cos{2x}$

| improve this answer | |
$\endgroup$
  • $\begingroup$ Cool, this seems to be what I was looking for. I'll give it a try and return. $\endgroup$ – GarouDan Jun 5 at 19:55
  • 2
    $\begingroup$ I would use Module[{a}, Limit[ ... ]] to isolate a from the Global context. $\endgroup$ – Bob Hanlon Jun 5 at 20:13
  • $\begingroup$ @BobHanlon Great idea, thank you! $\endgroup$ – MassDefect Jun 5 at 20:16
  • 2
    $\begingroup$ The second argument needs to be a valid variable. Current version gives zero for D1[x^3,x^2]. $\endgroup$ – yarchik Jun 5 at 20:40
  • $\begingroup$ @yarchik I guess it should be set to fail and issue a warning message like D[x^3, x^2] does. Do you think D1[f_, x_Symbol]:=... would be sufficient? $\endgroup$ – MassDefect Jun 5 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.