How to force Mathematica to output an expression such as
$a1u1+a2u2+a3u3$
as a Dot product like this one:
{a1,a2,a3}.{u1,u2,u3}
or A.U
$\begingroup$
$\endgroup$
7
-
$\begingroup$ Didn't you just accept an answer a couple of hours ago ? $\endgroup$– b.gates.you.know.whatCommented Sep 13, 2012 at 15:43
-
1$\begingroup$ Tell me, why shouldn't I close this question as a duplicate of this? $\endgroup$– J. M.'s missing motivation ♦Commented Sep 13, 2012 at 15:46
-
$\begingroup$ @J.M May be he wants to visualize them in traditional form. SergeyFomin are you asking about the display formatting? Then rephrase your question. $\endgroup$– PlatoManiacCommented Sep 13, 2012 at 15:48
-
$\begingroup$ Many thanks for coefficient extraction you have provided in my first question, but now I ask you how to force mma to present my expression in vector form. $\endgroup$– Igor FomenkoCommented Sep 13, 2012 at 16:00
-
1$\begingroup$ This question seems to need additional assumptions and constraints, because it has no unique answer: $a_1u_1 + a_2u_2 + a_3u_3$ = $(a_1,a_2,a_3)\cdot(u_1,u_2,u_3)$ = $(u_1,a_2,a_3)\cdot(a_1,u_2,u_3)$ etc. = $(a_2,a_3,a_1)\cdot(u_2,u_3,u_1)$ etc. = $(1,1,1)\cdot(a_1u_1 a_2u_2,a_3u_3)$ etc. $\endgroup$– whuberCommented Sep 13, 2012 at 17:01
|
Show 2 more comments
2 Answers
$\begingroup$
$\endgroup$
6
I'm going to answer this in the spirit of the question, making a few reasonable assumptions:
- that the two underlying vectors in $x = \sum_i a_i u_i$ are $\mathbf{a}=\{a_1,...,a_n\}$ and $\mathbf{u}=\{u_1,...,u_n\}$, thus giving $x = \mathbf{a}\cdot\mathbf{u}$. If not, there are several possibilities as in whuber's comment.
- The corresponding elements of the vectors $\mathbf{a}$ and $\mathbf{u}$ are ordered identically for all elements. In other words, for some ordering function $f$, $f(a_i,u_i)$ is the same for all $a_i$ and $u_i$. This is so that we aren't affected by the
Orderlessattribute ofTimes(in other words, don't try this for something like $\mathbf{a}=\{b, p, z\}$ and $\mathbf{u}=\{e,g,l\}$). - There are no numerals involved (i.e. this is purely symbolic) and the primary intent is to be able to display the vectors in the desired form.
With the above, the following is a very simple way to achieve the output with a few replacements:
expr = a1 u1 + a2 u2 + a3 u3;
expr /. Times -> List /. List -> CenterDot /. Plus -> List
(* {a1, a2, a3}·{u1, u2, u3} *)
-
$\begingroup$ As an added note, it is very instructive to break the replacements into a sequence of replacements via
% /. ...to see what it is they're doing. (+1, btw ... and enough with the edits!) $\endgroup$– rcollyerCommented Sep 13, 2012 at 18:23 -
$\begingroup$ @rcollyer I agree... I'll leave that to the reader — it's informative to do the replacements one at a time to see how it works, and it must be done in that order. $\endgroup$– rm -rf ♦Commented Sep 13, 2012 at 18:24
-
$\begingroup$ I forgot to add "left as an exercise for the reader," as that's why I left the note. Of course, the first one is the tricky one ... $\endgroup$– rcollyerCommented Sep 13, 2012 at 18:28
-
$\begingroup$ You took my expression literally,your way doesn`t work for more comlex case, for example expr = (a1 + a2) u1 + a2 u2 + a3 u3; $\endgroup$ Commented Sep 13, 2012 at 18:53
-
2$\begingroup$ Well, then you, as the question asker, should think of all complicated cases and present them in the question instead of making us guess what you might or might not have in mind... For this example, you can easily change
a1 + a2to, say,a4and then use the above. In the end, changea4back toa1 + a2$\endgroup$– rm -rf ♦Commented Sep 13, 2012 at 18:56
$\begingroup$
$\endgroup$
1
How about
expr = a1 u1 + a2 u2 + a3 u3;
HoldForm[#1.#2] & @@ Transpose[Apply[List, expr, {0, 1}]]
(* {a1,a2,a3}.{u1,u2,u3} *)
answered Sep 13, 2012 at 16:03
-
1$\begingroup$ easier,no?It`s so complex at first glance. And this way is very artificial,because you know result $\endgroup$ Commented Sep 13, 2012 at 16:12