pari-gp - 如何在 GP 中计算和评估复合函数？

Question

我找到了一种解决方法来制作复合功能，但我相信应该有更好的方法来做到这一点：

? f = x^2
%1 = x^2
? g = x^3      
%2 = x^3
? x = g
%3 = x^3
? fog = eval(f) 
%4 = x^6
? x = 2
%5 = 2
? result = eval(fog)
%6 = 64

在这种方法中，我需要x多次分配，我不想使用eval函数。代码不可读和不可维护。

Answer 1

1. You can simplify Piotr's nice answer to

   comp(f, g) = x->f(g(x));
   

Indeed, you do not need to assign to the (global) variable h in the comp function itself. Also, the braces are not necessary for a single-line statement, and neither are type annotations (which are meant to optimize the byte compiler output or help gp2c; in this specific case they do not help). Finally the parentheses around the argument list are optional in the closure definition when there is a single argument, as (x) here.

2. I would modify the examples as follows

   f(x) = x^2;
   g(x) = x^3;
   h = comp(f, g);
   
   ? h('x)     \\ note the backquote
   %1 = x^6
   ? h(2)
   %2 = 64
   

The backquote in 'x makes sure we use the formal variable x and not whatever value was assigned to the GP variable with that name. For the second example, there is no need to assign the value 2 to x, we can call h(2) directly

P.S. The confusion between formal variables and GP variables is unfortunate but very common for short names such as x or y. The quote operator was introduced to avoid having to kill variables. In more complicated functions, it can be cumbersome to systematically type 'x instead of x. The idiomatic construct to avoid this is my(x = 'x). This makes sure that the x GP variable really refers to the formal variable in the current scope.

Answer 2

PARI/GP 支持匿名闭包。所以你可以像这样自己定义函数组合：

comp(f: t_FUNC, g: t_FUNC) = {
    h = (x) -> f(g(x))
};

然后您的代码可以转换为更易读的形式：

f(x) = x^2;
g(x) = x^3;

h = comp(f, g);
h(x)
? x^6

x = 2; h(x)
? 64

希望这可以帮助。