3
$\begingroup$

I type the following command line

Table[D[#1,varx[[i]]] &,{i,3}]

where varx is a vector of dummy variables

varx = {x1, x2, x3}

According to this link NonCommutativeMultiply, I want to obtain a vector like this

$\left\{\frac{\partial }{\partial \text{x1}},\frac{\partial}{\partial \text{x2}},\frac{\partial}{\partial \text{x3}}\right\}$

But I got this

enter image description here

It seems like the index i in the command Table does not work. I don't know why this happens. I try not to use the placeholder '#1' like this

Table[D[2 x1^2 + 4 x2^3 + x3^4, varb[[i]]], {i, n}]

And this works all fine

{4 x1, 12 x2^2, 4 x3^3}

If I want to keep the placeholder, what should I do?

$\endgroup$
0

2 Answers 2

5
$\begingroup$

Use With:

Table[With[{v = varx[[i]]}, D[#1, v] &], {i, 3}]`
(* {D[#1, x1] & , D[#1, x2] & , D[#1, x3] & } *)

See the section "Scope" of the documentation page for With. Note that Function (&) has the attribute HoldAll, so that the value of varx[[i]] needs to be inserted into the function.

The above gives a list of operators. An alternate interpretation of what is sought is a single operator that evaluates to a list. Here's a way:

grad = Block[{D}, Evaluate[Table[D[#1, varx[[i]]], {i, 3}]] &]
(* {D[#1, x1], D[#1, x2], D[#1, x3]} & *)

Here's another method for this particular example, based on behavior of D[f, {{x1, x2, ...}}]:

grad = With[{v = varx}, D[#, {v}] &]

Both yield

grad[x1 + x2^2 x3]
(* {1, 2 x2 x3, x2^2} *)
$\endgroup$
1
$\begingroup$

You can use Slot (#) but the pure function (&) should be at a different position, i.e.

varx = {x1, x2, x3};

Table[List[#1, varx[[i]]], {i, 3}] & @@@ {{f, g}, {h, i}}

{{{f, x1}, {f, x2}, {f, x3}}, {{h, x1}, {h, x2}, {h, x3}}}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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