How do I transform $\frac{1-\exp(-Vx^m)}{1-(1-x^m)^V}-1$ so I can evaluate numerically for $V$ and $m$ between $1$ and $30$, and $0\leq x\leq 1$?

Question

I need to evaluate and plot the following expression:

$$ f(x,V,m)=\frac{1-\exp(-Vx^m)}{1-(1-x^m)^V}-1 $$

$V$ and $m$ are numbers from $1$ to $\sim30$ and $0\leq x\leq1$.

The issue is that for $x$ close to zero, the exponential, and the power term in the denominator take values very close to 1, making it difficult to evaluate $f$ numerically. I've tried MATLAB symbolic toolbox which, just to give one case, gives $f(10^{-3},1,6)=-1$. Mathematica also runs into similar problems. I know this to be incorrect because:

$$ \lim_{x\to 0} f(x,V,m)=0 $$

Is there a manipulation of the equation (the first term is the problem) that I am missing that would make it more amenable to evaluate numerically or simplify it? I want to know the behavior of the function for small values of $x$ but short of plotting I am struggling to conclude anything, because even though both numerator and denominator are monotonic, I cannot say the same about their ratio.

Answer 1

Using Taylor expansion of $$g(y)= \frac{1-\exp(-Vy)}{(1-(1-y)^V)}-1$$ at order one, you find

$$g(y) = -\frac{1}{2}y + O(y^2)$$ and therefore

$$f(x,V,m)=-\frac{1}{2}x^m + O(x^{2m})$$ around zero. Which provides a practical way to evaluate $f$ for $x$ near zero. Interesting is that this doesn't depend (at order one) on $V$.

Answer 2

Note that with $t=vx^m$, the expression is

$$\frac{1-e^{-t}}{1-\left(1-\dfrac tv\right)^v}-1=\frac{\left(1-\dfrac tv\right)^v-e^{-t}}{1-\left(1-\dfrac tv\right)^v}.$$

We know that $\left(1-\dfrac tv\right)^v$ is an approximation to $e^{-t}$, with the development

$$1-t+\left(1-\frac1v\right)\frac{t^2}2-\left(1-\frac3v+\frac2{v^2}\right)\frac{t^3}{3!}+\cdots$$

Your function is approximated by

$$\frac tv\,\frac{-\dfrac{1}2-\left(-3+\dfrac2{v}\right)\dfrac{t}{3!}+\cdots}{-1+\left(1-\dfrac1v\right)\dfrac{t}2-\left(1-\dfrac3v+\dfrac2{v^2}\right)\dfrac{t^2}{3!}+\cdots}$$