I'm looking for an accurate but as simple as possible approximation of
$S(x,\lambda) = \frac{1}{(1-x) [x-(1-\lambda )]}\left((1+\lambda ) \left(\frac{x(1+\lambda)}{1-\lambda }\right)^{\frac{x(1-\lambda )-(1-\lambda )}{x(1+\lambda )-(1-\lambda )}}-\lambda -x\right)$
for $x \in (0,1), \lambda \in (0,1)$.
Does such a thing exist, and how can one find it? From what I've tried, Taylor expansion does not seem very effective in approximating the function in question... any other ideas? Many thanks!
The graph of $S$ as a function of $x$ looks very much like a hyperbola, which suggests using the approximation $S(x,\lambda)\approx a+\frac{b}{x+c}$ where $a,b,c$ depend on $\lambda$ in some way.
I first tried to fit a hyperbola to the data $$S(0)=\frac{\lambda}{1-\lambda},\quad S'(0)=-\frac{\lambda(1+2\lambda)}{(1-\lambda)^2}, \quad S(1)=\frac{1}{2\lambda^2}\left(2\lambda-(1-\lambda^2)\ln\frac{1+\lambda}{1-\lambda}\right)$$ by solving the system $a+b/c=S(0)$, $-b/c^2=S'(0)$, and $a+b/(1+c)=S(1)$. Unfortunately this results in a substantial underestimate. The reason is that $S'(0)$ is a poor indicator of the behavior of the curve: $S'$ changes rapidly near $0$.
Then I fit the model $a+b/(x+c)$ to three points $S(0), S(1/3), S(1)$, and this worked much better.
Here is the comparison for $\lambda=0.17$ ($S$ in red, approximation in blue):
For $\lambda=0.37$
For $\lambda=0.87$
The coefficients $a,b,c$ are ugly, so I did not reproduce them here. The ugliness does not matter much if you work with specific $\lambda$. If you also need reasonably simple dependency on $\lambda$, let us know: maybe someone will come up with a better way to get $a,b,c$.
(Added)
Here is an approximation with simple dependency on $\lambda$. I found it more convenient to work with $\tilde S=1-\frac{1-\lambda}{\lambda}S$, since this function is increasing from $0$ to something in the range 0.3..1, which is to say it has a milder dependency on $\lambda$ than the original one. I picked the model $\tilde S\approx \alpha x/(\beta x+\gamma)$ where $\alpha,\beta,\gamma$ are linear functions of $\lambda$. Here is one possibility: $$ \tilde S\approx \frac{(1+2\lambda)x}{(2+\lambda)x+1-\lambda} $$ This approximation is not great, but you can tweak the six (well, actually five) numeric parameters to make it look better.