假设我有以下表述:与价值的替代变量,但不评价
(%i1) (8*x)*(log(x)/log(10));
(%i2) X^2;
现在,因为我想找出恒定值,我可以选择作出声明%i1 is O(%i2)
真的,我评估他们在像这样一个循环:
for a:1 thru 10 do print(%i1, "=", ev(%i1, x=a), %i2, "=", ev(%i2, numer, x=a));
输出是:
8 x log(x) 2
---------- = 0.0 , x = 1
log(10)
8 x log(x) 2
---------- = 4.816479930623698 , x = 4
log(10)
8 x log(x) 2
---------- = 11.45091011327189 , x = 9
log(10)
8 x log(x) 2
---------- = 19.26591972249479 , x = 16
log(10)
8 x log(x) 2
---------- = 27.95880017344075 , x = 25
log(10)
8 x log(x) 2
---------- = 37.35126001841489 , x = 36
log(10)
8 x log(x) 2
---------- = 47.32549024079837 , x = 49
log(10)
8 x log(x) 2
---------- = 57.79775916748438 , x = 64
log(10)
8 x log(x) 2
---------- = 68.70546067963139 , x = 81
log(10)
8 x log(x) 2
---------- = 80.0 , x = 100
log(10)
余万吨至使输出更容易眼球,是这样的:
8 1 log(1) 2
---------- = 0.0 , 1 = 1
log(10)
8 2 log(2) 2
---------- = 4.816479930623698 , 2 = 4
log(10)
8 3 log(3) 2
---------- = 11.45091011327189 , 3 = 9
log(10)
[snip]
8 10 log(10) 2
---------- = 80.0 , 10 = 100
log(10)
我怎么能告诉Maxima在循环的每次迭代替代的a
的价值x
不计算表达式?
我查过手册,但没有发现任何看似相关的东西。