then
f(x)=1a(x)W(a(x)c(x)eb(x)+a(x)d(x)c(x))−d(x)c(x)
where W is the Lambert W Function. You can click through to Wikipedia for a proof. Here I merely list some special cases which every good citizen should instantly recognize (and it can be mildly tedious to reproduce them).
Recovering the definition of W(x)
Just to check we are on the right planet set a=1, c=1, b=ln(x) and d=0 to get xe−f(x)=f(x)
or equivalently f(x)ef(x)=x
which is the definition of the Lambert W function. Sure enough those substitutions recover
f(x)=W(x)
as expected. A minor modification uses a(x)=1, c(x)=1, b(x)=0 but leaves d(x) general.
The solution to e−f(x)−f(x)=d(x)
is f(x)=W(ed(x))
Only slightly more generally, set a(x)=a, b=ln(g(x)),c=1,d=1 for those cases where g(x) is on the wrong side, as it were.
The solution to e−af(x)=f(x)g(x)
or equivalently f(x)eaf(x)g(x)=1 is f(x)=1aW(ag(x))
In particular,
The solution to xke−af(x)f(x)=1
which seems to crop up a fair bit for your author is f(x)=1aW(ax−k)
Similarly setting a=−1 ...
The solution to xef(x)f(x)=1
(i.e. where x is on the wrong side but we otherwise have the Lambert W definition) must be f(x)=−W(−1x)
We might also take b=ln(g(x)) and thus
The solution to g(x)e−af(x)=f(x)
or equivalently f(x)eaf(x)=g(x) is f(x)=1aW(ag(x))
which reduces to the Lambert W function for a=1, as we expect. It is also pretty obvious from first principles, because if we multiply both sides by a we have
(af)e(af)=ag
and thus af=W(ag). Next suppose we want a power of f to appear. Let b=β(x)/k, c=γ1/k, a=k/α and d=0. And then raise both sides to the power k. It follows that...
The solution to eαf(x)+β(x)=γf(x)k
is f(x)=αkW(ke1kβ(x)αγ1/k)
and if, in particular, β(x)=−ln(g(x)) then
The solution to eαf(x)=g(x)γf(x)k
is f(x)=αkW(kαg(x)1/kγ1/k)
and if we take c=1γ and g(x)=x and α=s2 and k=2 then
p>
The solution to e−s2f(x)xf(x)2=c
is f(x)=s2kW(2ks√cx)
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