Suppose I have an expression like (actually mine is much more complex, thousands of characters)

expr:a+b*c+b*c*d;

and I want to replace an internal sub-expression with a symbol (useful to avoid recomputation of common subexpressions), say k in place of b*c:

subst(b*c=k,expr);

returns

k+b*c*d+a

How I can make Maxima calculate the "right" substitution so to return (apart from obviuos simplification, here)

k+k*d+a

?

有帮助吗?

解决方案

Take a look at let and letsimp. E.g.:

(%i2) expr : a + b*c + b*c*d;
(%o2) b*c*d+b*c+a
(%i3) let (b*c, k);
(%o3) b*c --> k
(%i4) letsimp (expr);
(%o4) d*k+k+a

letsimp differs from subst and tellsimp or defrule in that those other functions make only formal substitutions, i.e., replacing subexpressions which are exactly the same as some pattern.

其他提示

You can try optimize

http://maxima.sourceforge.net/docs/manual/en/maxima_6.html#IDX219

(%i14) example(optimize);

(%i15) diff(exp(y+x^2)/(y+x),x,2)
                        2            2              2            2
               2   y + x        y + x          y + x        y + x
            4 x  %e         2 %e         4 x %e         2 %e
(%o15)      ------------- + ---------- - ------------ + ----------
                y + x         y + x               2             3
                                           (y + x)       (y + x)
(%i16) optimize(%)
                                                 2         y + %2       1
(%o16) block([%1, %2, %3, %4], %1 : y + x, %2 : x , %3 : %e      , %4 : --, 
                                                                        %1
                                                                 4 x %3   2 %3
                                          4 %2 %4 %3 + 2 %4 %3 - ------ + ----)
                                                                    2       3
                                                                  %1      %1
许可以下: CC-BY-SA归因
不隶属于 StackOverflow
scroll top