The Lambert W function is defined as follows:
$$z = W(z)e^{W(z)}$$
for any complex number z.
Many equations involving exponentials can be solved using the W function. For example:
$$ Y = X e ^ X \; \Longleftrightarrow \; X = W(Y)\ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
My question is the following. If we know the value assumed by the function in $x=a$ (for simplicity we assume real values), i.e. we know $W(a)$, can we deduce a new value $W(b), b\neq a$ without applying the (1)?
Assuming that I properly understood the question, you can perform a Taylor expansion of the Lambert function at $x=a$. What is beautiful with this function is that all its derivatives express just involve the Lambert function itself.
As a result,$$W(x)=W(a)+\frac{W(a) }{a (W(a)+1)}(x-a)-\frac{\left(W(a)^3+2 W(a)^2\right) }{2 a^2 (W(a)+1)^3}(x-a)^2+\frac{W(a)^3 \left(2 W(a)^2+8 W(a)+9\right) }{6 a^3 (W(a)+1)^5}(x-a)^3+O\left((x-a)^4\right)$$ In order to check, let us use $a=2$ and $b=3$ (which is a big move); the first order will give $1.08272$, the second order $1.04195$, the third order $1.05217$, the fourth order $1.04921$, the fifth order $1.05014$ while the exact value is $W(3)=1.04991$