Why doesn't symprod recognize indexed variables?

3 Ansichten (letzte 30 Tage)
Sree
Sree am 26 Dez. 2018
Kommentiert: Sree am 27 Dez. 2018
I was expecting to get: k1*k2*k3*k4. Instead . . . .

Akzeptierte Antwort

Stephan
Stephan am 26 Dez. 2018
Bearbeitet: Stephan am 26 Dez. 2018
Hi,
try:
syms k(n)
symprod(k(n),n,1,4)
this should give you the expected result.
Best regards
Stephan
  1 Kommentar
Walter Roberson
Walter Roberson am 27 Dez. 2018
Bearbeitet: Walter Roberson am 27 Dez. 2018
However, going from the symbolic function calls k(1)*k(2)*k(3)*k(4) to k1*k2*k3*k4 becomes tricky. And the k(1) and so on in the symprod result will be function calls, not indexing.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Walter Roberson
Walter Roberson am 27 Dez. 2018
MATLAB does not permit symbolic variables to be indices because it does not know how to index a definite matrix at an unresolved location.
When you define k as a matrix and try
symprod(k(n), n, 1, 4)
then that is equivalent to
TTTTTT = k(n);
symprod(TTTTTT, n, 1, 4)
because argument evaluation is always done first before the function itself is invoked. So the k(n) part has to be meaningful outside the context of the symprod() or symsum()... but it isn't.
When you want to do a symprod or symsum that would call for the dummy variable to be used as an index, then what you need to do is to replace the symprod or symsum with creation of a vector of definite values that are then prod() or sum(). For example,
TTTTTT = k(1:4);
prod(TTTTTT)
or more compactly,
prod(k(1:4))
or even
prod(k)
since you defined k as being only 4 long.
If you had something more complicated such as
symprod(sin(k(n)^n), n, 1, 4)
you would again proceed with definite values:
prod( sin(k(1:4)).^(1:4) )
  3 Kommentare
Stephan
Stephan am 27 Dez. 2018
Since this answer is meeting your requirements much better than mine, you should unaccept my answer and accept this one.
Sree
Sree am 27 Dez. 2018
No need for that change! Your answer was immediately useful and solved my problem (which was a bit more involved than the example I had included in my question). The other answer was philosophically instructive, for the long term.

Melden Sie sich an, um zu kommentieren.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by