MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Define g(x) = (1+x) ln(1+x) - x. One can check that g is strictly monotonically increasing for x>=0 by checking its derivative is ln(1+x). So g is invertible and its inverse is also strictly monotically increasing.

Is there an explicit closed form for its inverse?

With a page of calculations I can prove that (1/2) x/ln(1+sqrt(x)) <= g^{-1}(x) <= 2 x/ln(1+sqrt(x)) for all x>=0. Is this obvious? Can estimates like this be found in the literature?

share|cite|improve this question
I don't think there's any reason to expect one. You could try writing it in terms of the Lambert W-function, although I don't see how this could possibly help anybody. You might also try Lagrange inversion: – Qiaochu Yuan Jul 8 '10 at 16:52
Why do you say that writing it in terms of Lambert-W couldn't help anyone? Lambert-W hasn't made it into the high school curriculum yet, but it's incorporated in the major symbolic math engines, which know how to evaluate it, manipulate it, etc. In particular, if what OP wants is estimates, well, it shouldn't be too hard to find useful info on Lambert-W on the web. – Gerry Myerson Jul 9 '10 at 7:14
up vote 6 down vote accepted

Just to expand on Qiaochu's comment: let $W$ stand for the Lambert W-function, then if $g(x)=z$, we readily find that $$ x=\exp\big(W((z-1)/e)+1\big)-1. $$

share|cite|improve this answer
Thanks very much for this information, I was not familiar with Lambert-W. – Nick Harvey Jul 9 '10 at 13:31
Conversely, my estimates to g^{-1} imply estimates for W. Specifically, for all z>-1/e, I can prove W(z) <= ln( 2*(ez+1)/ln(1+sqrt(ez+1)) + 1)-1 and W(z) >= ln( 0.5*(ez+1)/ln(1+sqrt(ez+1)) + 1)-1 Is this obvious or well-known? The Taylor's series for W(z) around z=0 given at Wikipedia is only useful for z close to 0. My estimates are best for z close to -1/e, and have a constant additive error for z -> infinity. – Nick Harvey Jul 9 '10 at 19:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.