I am a molecular biologist and reading a book on mathematical modeling. In this book I encountered the following algebraic conversion and could not figure out how the conversion is performed. Please let me know the steps in such kind of conversions and also recommend me some short text/easy literature to know, how one can perform this.
Following expression $$ \frac{dP}{dt}=(m*(F-c*P)-n)*P $$ was changed into $$ \frac{dP}{dt}=(m*F-n)*[1-((m*c)/(m*F-n))*P]*P $$ Best Regards,
Mazhar Hussain
On
$$\frac{dP}{dt} = \left(m( F - cP) - n \right) P = \left(mF - mcP - n \right) P = \left( mF- n - mcP \right) P = \left((mF - n) \left( 1 - \frac{mcP}{mF - n} \right) \right) P = (mF - n)\left( 1 - \frac{mc}{mF - n} P \right) P$$
On
\begin{align*} \frac{\mathrm{d}P}{\mathrm{d}t} &= (m*(F-c*P)-n)*P \\ &= (m*F - m*c*P - n)*P \\ &= ((m*F - n) - m*c*P)*P \\ &= (m*F - n) * \left[\left(1 - \frac{m*c}{m*F - n}\right)*P\right]*P. \end{align*}
The expression was most likely rewritten to account for a carrying capacity. May I ask what book you're reading?
This is a simple factoring. Omitting the factor $P$ at the end (which stays the same)
$$ m*(F-c*P)-n = m*F -m*c*P -n = (m*F-n) - m*c*P = (m*F-n)*\left[1 - \dfrac{m*c*P}{m*F-n}\right] $$
What happened here is something like this:
$$ x - a = x \left(1 - \dfrac{a}{x}\right) $$