2

I would like to find the range of the function below, but only for values of $a$ strictly greater than zero.

v[a_]=-a*Log[1-(a - a E^(-1 - 1/a + ProductLog[E^(1 + 1/a)]))/a]

In other words: what is the range of $v(a)$, for/conditional on $a>0$ (reals).

  • I have tried this: (ie without restricting) FunctionRange[v[a], a, y], but the computation time is too long, I can't manage to get an output.

  • I also have tried to define the $v(a)$ function over the domain that interests me using this:

    v[a_ /; 0 < a] := -a*Log[1-(a - a E^(-1 - 1/a + ProductLog[E^(1 + 1/a)]))/a] (solution found here), and then FunctionRange[v[a], a, y], but the computation time seems to bee stil too long. Probably due to the fact that my restricted definition doesn't seem to have worked: even in that case, FunctionDomain[v[a],a]returns $a≠0$ (instead of $a>0$ as I would have expected since it is what I defined in the first place).

EDIT

  • This doesn't seem to work either: Assuming[a>0,FunctionRange[v[a], a, y]].
Banalaude
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2 Answers2

5
FunctionRange[{-a*
   Log[1 - (a - a E^(-1 - 1/a + ProductLog[E^(1 + 1/a)]))/a], 
  a > 0}, a, y]

5.14988*10^-9 <= y <= 0.5

and warming message:

Unable to find the exact range. Returning bounds on the range
computed using numeric optimization methods

So we have to consider Derivative

FunctionRange[{D[-a*
     Log[1 - (a - a E^(-1 - 1/a + ProductLog[E^(1 + 1/a)]))/a], a] // 
   Evaluate, a > 0}, a, y]

y > 0

It means that v[a] is a monotonic function ! So we just need to calculate the Limit just as @Akku14 have done.

Limit[-a*Log[1 - (a - a E^(-1 - 1/a + ProductLog[E^(1 + 1/a)]))/a], 
 a -> 0, Direction -> "FromAbove"]
Limit[-a*Log[1 - (a - a E^(-1 - 1/a + ProductLog[E^(1 + 1/a)]))/a], 
 a -> ∞]

0

1/2

cvgmt
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3
Limit[v[a], a -> 0, Direction -> -1]

(*   0   *)

Limit[v[a], a -> Infinity]

(*   1/2   *)

NMaximize[{v[a], a > 0}, a, WorkingPrecision -> 100]

(*   {0.4999999999999999999999999999999996695802781800686528047791793409305\
917941358131704720756957033359931, {a -> 
 1.89153358206810035345892896405454026939268176767953621670767804301\
 9662196519029316421185286618066641*10^32}}   *)

NMinimize[{v[a], a > 0}, a, WorkingPrecision -> 100]

(*   {9.2396091222066804246261095335668388330820970607385777570018055570761\
38791443783933838210621794192294*10^-7, {a -> 
 5.52917699540862549895289484183558798401702920323792264099048675666\
3524280963344309794142794231985742*10^-8}}   *)
Akku14
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  • Series[-a* Log[1 - (a - a E^(-1 - 1/a + ProductLog[E^(1 + 1/a)]))/a], {a, 0, 2}, Assumptions -> a > 0] results in $-a \log (a)+a^2 (\log (a)+1)+O\left(a^3\right)$, confirming positive values of v[a] and Series[-a*Log[ 1 - (a - a E^(-1 - 1/a + ProductLog[E^(1 + 1/a)]))/a], {a, Infinity, 2}] results in $\frac{1}{2}-\frac{1}{16 a}+\frac{1}{192 a^2}+O\left(\left(\frac{1}{a}\right)^3\right)$, confirming the inequality v[a]<1/2 at Infinity. – user64494 Jul 20 '21 at 06:50