Donald Burrill wrote:

> If you're dealing with a growth curve, is it possible that an
> exponential curve would fit?  (I know that for some segments of an
> organism's history the expected growth rate follows an exponential
> function, which implies a constant doubling time when the predictor is
> the age of the organism, until constraints of available space (or of
> maturation, I suppose) prevent the exponential increase from
> continuing.)
>
> Your predictor, I gather, is fertilizer concentration, not time;
>  I can't tell whether an exponential (or logarithmic) function is at all
> reasonable, but such models might be worth trying.  I'd be inclined to
> distrust a polynomial of any order, particularly > 2, because I should
> think the underlying trend should be monotonically increasing, and a
> polynomial is not "naturally" monotonic, so to speak:  I suspect the
> 6th-order polynomial is useful mainly in sharpening the corner, so to
> speak, between slower growth rates at lower concentrations and faster
> growth rates at higher concentrations.  (Certainly a similar effect
> occurred with some data I once had occasion to work with, dry mass of
> organism as a function of time.  An exponential nicely fit the data;  it
> took a 4th-order polynomial to represent adequately the shape of the
> empirical curve, mainly due to having to sharpen that lower-right corner
> much more than one would get from a parabola.)
>
> For fitting Y as a function of X, if the response curve is exponential,
> one expected form of relationship would be  Y = a*e^(b*X).  Taking
> logarithms (natural logs, here),  log (Y) = log(a) + b*X
>  which is a nicely linear function.
>
> If plotting  log(Y)  as a function of  X  does not produce a (nearly)
> straight line in the scatterplot, you might also try  log(Y) vs. log(X)
> and/or Y vs. log(X).  (I have no theoretical reasons for these
> suggestions, only looking for a functionality that looks reasonably
> linear:  when one has found such a thing, one can then worry about
> whether it can be justified on grounds other than sheer empiricism.)
>
> And of course if  Y vs. X  is quadratic in X, you might try plotting
>  sqrt(Y) vs. X.

Don't forget reciprocal relationships: y = a/x (or y = a/(x-b) ). The latter

is tricky because if you have to estimate b from the data, you no longer
have a linear relationship in the coefficients, but the former can be used
in the straightforward manner.

.
.
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