Expressing an formula in term of another one

Question

I have this formula

$$-\frac 1\lambda\left[\lambda D+1+W_{-1}\left(-r\exp(-\lambda D-1)\right)\right]$$

with $r$ , $\lambda$ and $D$ >0.

Where $W$ is the Lambert W function http://en.wikipedia.org/wiki/Lambert_W_function

I have some reason to believe that this expression is somehow equivalent to a simple:

$$\gamma+\alpha x^\beta $$ where $\beta<0$, $\alpha$ and $\gamma >0 $.

I believe this just because from numerical simulation I can almost perfectly fit the second equation starting from the first one, treating $\lambda$ as $x$, as shown in the figure The blue circles represent the first equation, the solid black line represent the second one, fitted over the first one. The fit is not perfect, but it is pretty good, and it is still good when I play around with the different parameters. So, I am wondering: are these two equations the same? If so, how can I prove that?

Thank you very much.

Answer 1

Like Thomas says, it would be nice to have an approximation of $W_{-1}(z)$ to replace it in the final function and have it over with.

Unfortunately the plain Taylor series expansion of the -1 branch of the Lambert is unsuited for this case, because the domain of the OP's function exceeds its radius of convergence.

If however, we use the compound expansion of $W_{-1}(-\exp(-5x-1))$ around approximately half the range of 0..8, we are in business (note that $-\exp(-5x-1)$, for $x\in(0,8)$, falls inside the domain $(-1/e,0)$ of the branch $k=-1$ of the Lambert, so such an expansion is valid).

The original function:

W:=LambertW;

f := proc (lambda) options operator, arrow; -(5*lambda+1+W(-1, -exp(-5*lambda-1)))/lambda end proc;

Using Maple and expanding the Lambert term around x=3, using $\mathit{only}$ 2 terms:

s := series(W(-1, -exp(-5*x-1)), x = 3.0, 2);

s := -18.94134722-5.278685871*(x-3.0)+O((x-3.0)^2)

p := convert(s, polynom);

sp2 := unapply(p, x):

Now substitute the series found in place of the Lambert term:

g2 := proc (x) options operator, arrow; -(5*x+1+sp2(x))/x end proc;

Let's try it now:

p1 := plot(f(x), x = 0 .. 8, 0 .. 6, color = red):

p2 := plot(g2(x), x = 0 .. 8, 0 .. 6, color = green):

display(p1, p2);

The approximation is now revealed as:

simplify(g2(x));

expand(%);

.2786858710+2.105289610/x,

which can easily be seen as being of the form $\gamma+\alpha\cdot x^{\beta}$, with $\beta=-1<0$ and $\gamma,\alpha>0$.

Tweeking these constants a bit may make the match even better.