Hi Billy & Steve,
| I have just a few minutes, but maybe can shed a little light on
| this. I think this is the deal:
|
| The underlying math is simply that
| INTEGRAL[x to b] = INTEGRAL[a to b] - INTEGRAL[a to x]
|
| The thing is, that this statement is true where x is a continuous variable.
| Here we are on a discretized axis, where each value on our axis actually
| represents a cell. As you move from low to high along the axis the running
| sum is really the integral across that full cell. But the coordinate point
| lies in the center of the cell (ignoring the complexities of irregular cell
| sizes for this discussion). I've attempted to draw an axis to illustrate.
| The carets ("^") show where the values is located on the axis in each case.
|
| direction of integration ==>
| | 1 | 2 | 3 | 4 | 5 | 6 |
| @RSUM ^ ^ ^ ^ ^ ^
| "correct" ^ ^ ^ ^ ^ ^
[. . .]
I was about to send the following message, which (I think) says
exactly the same thing as yours. But, I got stuck at the end.
--------------------------------
I think the result Ferret gives is consistent "in a sense".
Look at the attached plot.
Let's assume that your function f(x) is defined as
f(0.5) = 1
f(1.5) = 2
f(2.5) = 3
f(3.5) = 4
f(4.5) = 5
f(5.5) = 6
Then its integral F(x) from x = 0 to x = x is
(or can be defined as)
F(0) = 0
F(1) = 1
F(2) = 3
F(3) = 6
F(4) = 10
F(5) = 15
F(6) = 21
(F(i) = sum_{1 <= j <= i} f(j-0.5) delta-x .)
Then, the most natural reverse integral G(x) = F(6) - F(x) is
G(0) = 21
G(1) = 20
G(2) = 18
G(3) = 15
G(4) = 11
G(5) = 6
G(6) = 0
This is what you got, except that you don't have a gridpoint at x = 0.
But . . . this makes me wonder: Shouldn't Ferret define @IIN at the
right edges of the original gridpoints? In fact, @IIN is defined
on the same gridpoints as the integrand.
---------------------
Here's where I got stuck. So, my tentative conclusion is
that in this sense, @RSUM and @IIN aren't quite equivalent.
Ryo
Attachment:
reverse-integral.eps
Description: PostScript document