[ferret_users] How to do @rsum and @iin in reverse?

["William S. Kessler" <William.S.Kessler@xxxxxxxx>]

This may be a really stupid question. How does one perform @rsum or  
@iin in reverse (i.e. sum from the high-index end of the list)?

The answer I would have given before I actually looked at it in  
detail would have been: subtract the running sum or integral from the  
definite sum over the whole range. And indeed I have done this many  
times. But the result is not quite as expected, as the following  
example shows:

yes? let test = {1,2,3,4,5,6}
yes? list test, test[i=@rsum], test[i=@sum] - test[i=@rsum]
             X: 0.5 to 6.5
Column  1: TEST is {1,2,3,4,5,6}
Column  2: TEST[X=@RSU] is {1,2,3,4,5,6} (running sum on X)
Column  3: EX#3 is TEST[I=@SUM]-TEST[I=@RSUM]
          TEST   TEST   EX#3
1   / 1:  1.000   1.00  20.00
2   / 2:  2.000   3.00  18.00
3   / 3:  3.000   6.00  15.00
4   / 4:  4.000  10.00  11.00
5   / 5:  5.000  15.00   6.00
6   / 6:  6.000  21.00   0.00

The second column does what we want in a running sum: the first value  
is the value of the function over interval 1. The second value is the  
sum of the function at intervals 1+2. The last value is the sum over  
the whole range, as it should be. But I had hoped and expected that  
the third column would be the reverse: The last value (i=6) should be  
the last value of the function (6), the second-to-last should be the  
sum of the two last values (6+5=11), and the first value should be  
the same sum as if the operation were done the other way (21). But it  
isn't. The whole thing is shifted by 1 interval, the value in  
position 6 is 0, and the value in position 1 is not the whole sum.

Am I missing something obvious???? What is the proper way to take a  
running sum or integral in reverse? It should be possible to get the  
same qualities of the answer doing the operation in either direction.

Billy K