J中总和（1 /（1 + x）^ y）的默认形式是什么

Question

作为一个初学练习，我试着用J计算下面的和， sum(1/(1+0.03)^n for n = 1 to 30使用+/%(1 + 0.03)^ >:i.30 。我怎么能把它写成一个简单的默认形式？我尝试过的所有内容都比上面的显式形式更加丑陋>:@[ (+/&:%)@:^ >:&i.@]

Answer 1

You could start with

+/@:%@((1 + 0.03) ^ >:@i.) 30

You can make the 0.03 a left argument using a fork, but using a hook can be cleaner

(1 + 0.03) +/@:%@([ ^ >:@i.@]) 30   NB. use fork
(1 + 0.03) +/@:%@(^ >:@i.) 30       NB. use hook

The same operation (increment) is being performed on both the left and right arguments to ^ . That is a hint that & ( Compose ) may be useful.

0.03 +/@:%@(^&>: i.) 30         NB. apply increment to both left & right arg

Answer 2

When I want a tacit function I often let 13 : bang it out for me. In this case, some variations:

   13 : '+/ %((1+0.03)^1+i.y)'
[: +/ [: % 1.03 ^ 1 + i.

   13 : '+/ %((1+0.03)^>:i.y)'
[: +/ [: % 1.03 ^ [: >: i.

And with 1+0.03 or whatever as a leftargument:

   13 : '+/ %(x^1+i.y)'
[: +/ [: % [ ^ 1 + [: i. ]

   13 : '+/ %(x^>:i.y)'
[: +/ [: % [ ^ [: >: [: i. ]

There are way too many caps ( [: ) to call it less ugly, though, but that's a start.