Can I approximate log(1-x) with -x, if my $\displaystyle x \leq 0.5$ ?
If i have an optimization problem: maximize A*x + B*log(1-x)
can I exchange with a problem: maximize A*x + B*(-x) ?
Can I approximate log(1-x) with -x, if my $\displaystyle x \leq 0.5$ ?
If i have an optimization problem: maximize A*x + B*log(1-x)
can I exchange with a problem: maximize A*x + B*(-x) ?
I suggest you use a truncated version of the MacLaurin Series $\displaystyle \displaystyle \begin{align*} \ln{\left(1 - x\right)} = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots \end{align*} $ which is convergent when $\displaystyle \displaystyle \begin{align*} -1 \leq x < 1 \end{align*} $.
So theoretically, yes you could truncate it after the first term, but it won't be very accurate. I would suggest truncating it after three or four terms.
Prove it has given you the series of expansion. For $\displaystyle |x| < 1$ increasing powers of x will get smaller so you can simply say high powers are roughly 0 compared to low powers and discard the former.
At which value of x this applies depends on how accurate your function needs to be.
Sure, i understand that.
My values of x are 0<x<0.25
I can then approximate: $\displaystyle \log(1-x)= - x - \frac{x^2}{2} $
My question was, can I approximate it with something like:
$\displaystyle \log(1-x)= - x - f(x) $, where f(x) is a linear function of x that is smaller than $\displaystyle \frac{x^2}{2}$?