Getting the coefficients of Y for many powers takes work, but for a rough approximation relatively little data is needed.
Y is odd - therefore in the neightborhood Y = 0, F(Y) = dF/dy(0) * Y. Furthermore, we know that F(X) = .5. So we can create a third order expression of F using these two known datapoints.
F(Y) = dF/dy(0) * Y + B * Y^3 F(X) = dF/dy(0) * X + B * X ^ 3 .5 = dF/dy(0) * X + B * X ^ 3 B = ( .5 - dF/dy(0) * X ) * X ^ -3 F(Y) = dF/dy(0) * Y + (.5-dF/df(0) * X ) * (Y/X)^3
So how do we procede from here? Well, (RS+e, RS-e) when rotated yields X = RS* 2^.5, Y = e * 2^.5. So, in this neighborhood,
dF/dy = [ W(RS+e,RS-e) - W(RS-e,RS+e) ] / 2*2^.5e = W(RS+e,RS-e) / 2^.5e
So to predict W(RS,RA), find Z = 2^-0.5(RS+RA), find a close match for W(Z+e,Z-e), calculate dF/dy, then find the third order coefficient B. Plug and chug.
June 25, 2004 11:18 PM
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