[Hans Blom, 951003b]

(Bill Powers (951003.0500 MDT))

I appreciate your efforts to clarify my thinking about the

strategy of your model, but you're still giving me too much

credit for familiarity with your methods. What I'm trying to do

is eliminate all the statistical aspects of the strategy and

just see what is going on in the absence of noise.

You're loosing me here. Modelling is curve fitting. How can I

describe curve fitting without at least some statistics?

This is useful:

delta_K := delta_K + constant1 * estimate (dK/dt) * (x-y)

delta_D := delta_D + constant2 * estimate (dD/dt) * (x-y)

The constants, I presume, are to prevent computational

oscillations -- to keep the corrections from undershooting and

overshooting.

Yes, the equations implement a kind of hill-climbing toward the

"optimal" values of delta_K and delta_D. Not knowing where the

mountain top is and not knowing exactly how large your steps will

turn out to be (after all, you only have, maybe quite noisy,

_estimates_ of dK/dt and dD/dt), the most reliable strategy is to go

slow: pick small constants. But that may be too slow in practice,

especially if delta_K and delta_D keep varying -- the mountain top

keeps shifting around. So you'll have to find some practical compro-

mise.

In general, x will be some function f, with d being added to x

(because if d is an unknown waveform, it doesn't matter whether

it is passed through some function before having its effect on

x):

x = f(u) + d

In the second step of your method, it seems to me that it's

necessary to compute

-1 u = f (r - d)

What this step entails is that you want to pick that value of u that

brings the predicted x to r, i.e. solve for

r = x = f(u) + d

There are a variety of ways to do this. One way is by simulation:

vary u from a minimum to a maximum value and use the value of u that

brings f(u) + d as closely to r as you can discover. Binary search

over the range of u may be faster. Varying u, not from some min to

some max but in the vicinity of its previous value, is another way.

These methods work with any (known) function f.

If f is invertible, finding u is much simpler:

f (u) = r - d, i.e.

u = f^-1 (r - d)

which I presume your equation above ought to have been.

Hope this helps,

Hans