Linear Approximation of x/ (1-x) I am trying to linearize the following function, but, having difficulties.
Let,
$x = \frac{l}{m},$
where $l,m \in R^+$ and $l<m$
Assume $l$ is a variable, while, $m$ is a constant (parameter), which makes $x$ a variable. I want to find a linear approximation for the following:
$f(x) = \frac{x}{2m(1-x)}$
In other words, as $2$ and $m$ are constant, I am interested in
$g(x)=\frac{x}{1-x}$
I plotted the graph, but, it did not really help to derive something useful. Any help is appreciated.
 A: If, over a range $a\leq x \leq b$, you want the best linear approximation $A+Bx$ of 
$$f(x) = \frac{x}{2m(1-x)}$$ the solution is to minimize the norm
$$F=\int_{a}^{b} \left(A+Bx-\frac{x}{2m(1-x)} \right)^2$$ with respect to parameters $A$ and $B$.
Integrating and then computing the partial derivatives $\frac{\partial F}{\partial  A}$ and $\frac{\partial F}{\partial  B}$ and setting them equal to $0$ leads to two linear equations in $(A,B)$ 
$$2m(b-a) A + m (b^2-a^2)B+(b-a)+\log \left(\frac{1-b}{1-a}\right)=0$$
$$6m(b^2-a^2)A+4m(b^3-a^3)B+3 (b-a) (a+b+2)+6\log \left(\frac{1-b}{1-a}\right)=0$$ which you can simplify factoring $(b-a)$ in several places.
I let to you the pleasure of finding $A,B$ (but this is simple).
Around $x=0$, for $a=-b$, using Taylor series, we have
$$A=\frac{b^2}{6 m}+\frac{b^4}{10 m}+O\left(b^6\right)\qquad  B=\frac{1}{2 m}+\frac{3 b^2}{10 m}+\frac{3 b^4}{14 m}+O\left(b^6\right)\qquad F=\frac{2 b^5}{45 m^2}+O\left(b^7\right)$$
Taking an example using $a=-0.1$, $b=0.1$, $m=1$, this would lead to
$$A=\frac{1}{2} \left(-1+5 \log \left(\frac{11}{10}\right)+5 \log
   \left(\frac{10}{9}\right)\right)\approx 0.00167674$$
$$B=-150 (1+10 \log (3)-5 \log (11))\approx 0.503022$$
A: I will present an elementary calculus solution. I will first rename your $g(x)$ into $f(x)$ because I can and because it looks nicer to the eye. It is apparent that the best linear approximation to any function at $x=0$ must be the tangent line to the function at $x=0$. In our case, since $f(0)=0$, our tangent line must pass through $(0,0)$ too and so it must be of the form $y=mx$. Since it is the tangent line, the gradient $m$ of this line must have the same "gradient" (i.e. rate of change) as $f$ at $0$. To find the gradient, thus, we set
$$ m = f'(0) $$
Where the RHS is the derivative at $0$; so the rate of change as previously mentioned. So now we just have to take the derivative of $f$ and evaluate it at $0$ to find $m$. I'm sure you know how to do this, but I'll put it here for completeness's sake:
$$\frac{d}{dx} \frac{x}{1-x} = \frac{(1-x)-(-1)x}{(1-x)^2}$$
Evaluated at $0$ this is $1$. So $m=1$ and the closest linear approximation would be $y=mx=x$.
A: $$\frac d{dx} \frac1 {1-x} = \frac1{(1-x)^2}$$ Additionally,
since $\frac1{1-x} = \sum_{k=0}^{\infty}x^k$ clearly $$\frac1{(1-x)^2} = \sum_{k=1}^{\infty}kx^{k-1}$$Unfortunately, it does not really make sense to approximate linearly something like this in any global sense. Locally, evaluating $m = \frac1{(1-x_0)^2}$ would give you the slope of a line tangent to $\frac1{(1-x)}$ at point $x_0$ which locally approximates it linearly. I believe this is what you're asking.
