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Transforming a Nonlinear Model to a Linear One

The following is a comprehensive list of transforming a nonlinear model to a linear one.

Nonlinear Linear Relationships
$y = \frac{\textstyle ax}{\textstyle 1+bx}$ $\underbrace{\frac{\textstyle 1}{\textstyle y}}_Y =
\underbrace{\frac{\textstyle...
 ...xtstyle 1}{\textstyle x} +
\underbrace{\frac{\textstyle b}{\textstyle a}}_\beta$ $a = \frac{\textstyle 1}{\textstyle \alpha}$,$b = \frac{\textstyle \beta}{\textstyle \alpha}$
$y = \frac{\textstyle a}{\textstyle x+b} $ $\underbrace{\frac{\textstyle 1}{\textstyle y}}_Y =
\underbrace{\frac{\textstyle...
 ...\textstyle a}}_\alpha x +
\underbrace{\frac{\textstyle b}{\textstyle a}}_\beta $ $a = \frac{\textstyle 1}{\textstyle \alpha}$,$b = \frac{\textstyle \beta}{\textstyle \alpha}$
$y = \frac{\textstyle ax}{\textstyle b^2+x^2} $ $\underbrace{\frac{\textstyle 1}{\textstyle y}}_Y =
\underbrace{\frac{\textstyle...
 ...ce{\frac{\textstyle b^2}{\textstyle a}}_\beta
\frac{\textstyle 1}{\textstyle x}$ $a = \frac{\textstyle 1}{\textstyle \alpha}$,$b^2=\frac{\textstyle \beta}{\textstyle \alpha}$
y = axb $\underbrace{\ln y}_Y = 
\underbrace{b}_\alpha \ln x + 
\underbrace{\ln a}_\beta$ $a = e^\beta$, $b=\alpha$
$y = \frac{\textstyle 1}{\textstyle 1+ax^b}$ $\underbrace{\ln \left(\frac{\textstyle 1-y}{\textstyle y}\right)}_Y =
\underbrace{b}_\alpha \ln x + \underbrace{\ln a}_\beta$ $a = e^\beta$,$b=\alpha$
$y = \frac{\textstyle 1}{\textstyle 1+\exp \left( \frac{\textstyle ax}{\textstyle b+x} \right)}$ $\underbrace{\left[\ln \left(\frac{\textstyle 1-y}{\textstyle y}\right)\right]^{...
 ...tstyle 1}{\textstyle x} + 
\underbrace{\frac{\textstyle 1}{\textstyle a}}_\beta$ $a = \frac{\textstyle 1}{\textstyle \beta}$,$b=\frac{\textstyle \alpha}{\textstyle \beta}$
$y = \ln a + x - \ln (e^x+b)$ $\underbrace{e^{-y}}_Y =
\underbrace{\frac{\textstyle b}{\textstyle a}}_\alpha e^{-x} + 
\underbrace{\frac{\textstyle 1}{\textstyle a}}_\beta$ $a = \frac{\textstyle 1}{\textstyle \beta}$,$b=\frac{\textstyle \alpha}{\textstyle \beta}$
$\frac{\textstyle x^2}{\textstyle a^2} + \frac{\textstyle y^2}{\textstyle b^2} = 1$ $\underbrace{y^2}_Y =
\underbrace{-\frac{\textstyle b^2}{\textstyle a^2}}_\alpha x^2 + 
\underbrace{b^2}_\beta $ $a^2 = -\frac{\textstyle \beta}{\textstyle \alpha}$,$b^2 = \beta$
$y = a \exp \left\{-\left(\frac{\textstyle x-c}{\textstyle b} \right)^2\right\}$ $\underbrace{\ln y}_Y = 
\underbrace{-\frac{\textstyle 1}{\textstyle b^2}}_\alph...
 ...}}_\beta x + 
\underbrace{\ln a - \frac{\textstyle c^2}{\textstyle b^2}}_\gamma$ $a=\exp \left(\gamma-\frac{\textstyle \beta^2}{\textstyle 4}\right)$,$b=\pm \sqrt{-\frac{\textstyle 1}{\textstyle \alpha}}$,$c=-\frac{\textstyle \beta}{\textstyle 2 \alpha}$
$y = \frac{\textstyle a}{\textstyle \sqrt{(1+bx^2)^2+c}}$ $\frac{\textstyle 1}{\textstyle y^2}=
\underbrace{\frac{\textstyle b^2}{\textsty...
 ...tyle a^2}}_\beta x^2+
\underbrace{\frac{\textstyle c+1}{\textstyle a^2}}_\gamma$ $a = \pm \frac{\textstyle \sqrt{4\alpha}}{\textstyle \beta}$,$b = \frac{\textstyle 2 \alpha}{\textstyle \beta}$,$c = \frac{\textstyle 4 \alpha \gamma}{\textstyle \beta^2}-1$


 

J.-S. Roger Jang
6/12/1998