Curve fitting to non-linear distribution and transposition to a standard curve Dear stackoverflow community,
I acquired data from a chemical reaction and used PRISM 6 to fit a non-linear distribution (Total binding - One-Site)
Data:
x:
0.00000
0.03125
0.06250
0.12500
0.25000
0.50000
1.00000
2.00000

y: 
0.1557
0.2339
0.2827
0.3646
0.5196
0.6186
0.6710
0.7300

$y = \frac{(Bx)}{K+x} + Nx + G$
R-square is 0.9957
B = 0.7037
K = 0.2476
N = -0.02626
G = 0.1511
Then I tried to transpose the equiation to use it as a standard curve.
$y = \frac{(Bx)}{K+x} + Nx + G$
multiplied with (K+x)
$yK + yx = Bx + NKx+ Nx^2 + GK + Gx$
then substracted yK and yx
$0 = Bx + NKx+ Nx^2 + GK + Gx -yK - yx$
reorder and exclude x
$0 = Nx^2 + (B + NK+ G - y)x  -yK + GK$
Then used N as a, (B + NK+ + G - y) as b and (-yK + GK) as c in a quadratic equation
$x = \frac{-(B + NK+ G - y)+\sqrt{(B + NK+ G - y)^2 - 4N(GK-yK)}}{2N}$
If I use this formula as a standard curve, then my predictions are horrible.
Where is my mistake?
Thank you in advance,
Anton
 A: You have no mistake. I agree with your equation $x = \frac{-(B + NK+ G - y)+\sqrt{(B + NK+ G - y)^2 - 4N(GK-yK)}}{2N}$
Or equivalently $x = \frac{-(B + NK+ G - y)+\sqrt{(B - NK + G - y)^2 + 4NKB}}{2N}$
$$x=C_0\left(C_1+y+\sqrt{(C_2+y)^2+C_3}\right)$$
$$C_0=\frac{1}{2N}\quad;\quad
C_1=-NK-B-G\quad;\quad
C_2=NK-B-G\quad;\quad
C_3=4NKB$$
Non-linear regression leads to :
$C_0\simeq -23.1\quad;\quad C_1\simeq -0.850 \quad;\quad C_2\simeq -0.862 \quad;\quad C_3\simeq -0.0164$
Then
$N=\frac{1}{2C_0}\simeq -0.0216 \quad;\quad K=C_0(C_2-C1)\simeq 0.277 \quad;\quad B=\frac{C3}{2(C_2-C_1)}\simeq 0.683 \quad;\quad G=-\frac{C_1+C_2}{2}-\frac{C_3}{2(C_2-C_1)}\simeq 0.173$
The fitting is not bad :

The numerical values found for the parameters $N,K,B,G$ are not exactly whose previously found from non-linear regression wrt $y(x)$. This is not surprising since fitting for LMSE wrt $y(x)$ is not the same as fitting for LMSE wrt $x(y)$.
You wrote : " If I use this formula as a standard curve, then my predictions are horrible ". You formula is correct. Thus the trouble comes from the use of your non-linear regression software.
Without knowing what is the algorithm in your software (certainly iterative) and how it is initialised one cannot say why something went wrong.
Be carefull that all along the iterative process there is never negative argument in the square root of the function $x(y)$ leading to complex numbers. The solftware should have some implemented attribut to circumvent the aeras where the function is not real.
