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Taking $\alpha$ as given, I am trying to understand/know under which restrictions on $n$, the following holds: $$ 0 \leq n+\alpha W\left(-e^{\frac{-n}{\alpha}} \right) \leq 1.$$ i.e I want to solve for $n$ the following two inequalities: a) $0 \leq n+\alpha W\left(-e^{\frac{-n}{\alpha}} \right)$, and b) $n+\alpha W\left(-e^{\frac{-n}{\alpha}} \right) \leq 1$. In addition, I know $\alpha >0$, that both $n$ and $\alpha$ are reals.

  • I manage to get that a) is satisfied for $n\geq \alpha$, using: Reduce[n+a*ProductLog[-E^(-n/a)]>=0, n, Reals] which returns (a<0&&n==a)||(a>0&&n≥a).
  • However I struggle to find any solution for the inequality b). I have tried all of the following:
(* for convenience I first define the function *)
f[n_, a_] =n+a*ProductLog[-E^(-n/a)]

Reduce[f[n, a]<=1, n, Reals]
Reduce[f[n, a]<=1 && a>0, n, Reals]
Solve[f[n, a]<=1, n, Reals]
Solve[f[n, a]<=1 && a>0, n, Reals]

Which all return:

"This system cannot be solved with the methods available to Solve/Reduce."

Even though I specified the domain as suggested here.

Is the inequality in b) really not solvable for $n$? Could Anyone think of an alternative command or commands that would do the trick?

Thank you,

EDIT:

In my context, f[n_, a_]being strictly increasing in n, and also because I know that (due to the domain of the Lambert W function) we must have n ≥ a (on the graph of f[n, a]below, we see clearly that it is not defined otherwise), even the solution for the equality to 1 would suffice me.

Manipulate[Plot[f[n, a], {n, -1, 10}], {a, 0, 5, 0.05, Appearance -> "Labeled"}]

increasing in n

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    $\begingroup$ Maybe: Solve[f[n, a] == 1, n] ? $\endgroup$ Aug 1 at 10:08
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We can change variable.

f[n_, a_] = n + a*ProductLog[-E^(-n/a)];
f[n, a] <= 1 /. n -> t*a // Simplify

a (t + ProductLog[-E^-t]) <= 1

so for a>0 we just need to solve

t + ProductLog[-E^-t]<=1/a

Since

FunctionRange[D[t + ProductLog[-E^-t], t], t, y]

y > 1

so t + ProductLog[-E^-t] is increasing and we find its inverse.

InverseFunction[Function[t, t + ProductLog[-E^-t]]][1/a]

t<=1/a + E^(-1/a)

n/a<=1/a + E^(-1/a)

n<=1+a* E^(-1/a)

Edit

For a fixed a>0,both

f[n_, a_] = n + a*ProductLog[-E^(-n/a)];
Solve[f[n, a] == c, n]
InverseFunction[Function[n, f[n, a]]][c, a]

get

E^(-(c/a)) (a + c E^(c/a)

means that f[n,a]<=c is solvatable.

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