Originally Posted by

**romsek** Ok I guess you've suffered enough. You really should be able to solve this though.

solving I

$\dfrac{dP_L}{dt}=k(M-P_L)$

$\dfrac{dP_L}{M-P_L}=k dt$

$-\ln\left|M-P_L\right| = k t + C_1$

$M-P_L=Ce^{-kt}$ Note that $M\geq P_L$ so I can omit the $||'s$. $C=e^{C_1}$

$P_L=M-Ce^{-kt}$

This is exactly the behavior we want from our learning function.

now III

$\dfrac{dP_L}{dt}=k\sqrt{M-P_L}$

$\dfrac{dP_L}{\sqrt{M-P_L}}=k dt$

$-2\sqrt{M-P_L}=kt + C$

$\sqrt{M-P_L}=-\dfrac{kt+C}{2}$

$M-P_L=\left(-\dfrac{kt+C}{2}\right)^2$

$P_L=M-\left(\dfrac{kt+C}{2}\right)^2$

This is an inverted parabola that reaches a maximum of M (or less) but then begins to decrease as time continues. This is not the behavior we want.

So only I models the situation correctly.