MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

I am trying to prove a result for which I need the nth term of the Baker-Campbell-Hausdorff formula. I came at this particular result (which is not of significance for the question, but mentioning for context) by hypothesizing and using the first few terms of the Baker formula to verify. In order to prove my result rigorously, I think I need the nth term of the Baker formula. Is there an expression for that - I could not find it through online research. ? It would also help if I could find a proof of the BCH formula which is based on recursion i.e. if I could see how the nth term relates recursively to the (n-1)th term.

share|cite|improve this question
Useful reference: Jean-Louis Loday, Série de Hausdorff, idempotents Eulériens et algèbres de Hopf, Expo. Math. 12 (1994), pp. 165--178. IIRC the paper used to be on Loday's website which got taken down after his death (why?). Here is a link, with no guarantees of permanence:… . Also, Emily Burgunder's papers might be relevant. – darij grinberg Jul 22 '14 at 20:10
The paper is now back on Loday's page: (the website itself is ). Thanks are due to Alexis Palaticky from the IRMA! – darij grinberg Jul 23 '14 at 14:10
up vote 8 down vote accepted

The Dynkin formula is somewhat cumbersome. Maybe a better choice is Goldberg's version In the commutator form Goldberg's result is reformulated in (Cyclic relations and the Goldberg coefficients in the Campbell-Baker-Hausdorff formula, by Robert C. Thompson).

By the way an interesting early history of the Baker-Campbell-Hausdorff-Dynkin formula can be found in (The early proofs of the theorem of Campbell, Baker, Hausdorff, and Dynkin, by R. Achilles and A. Bonfiglioli).

share|cite|improve this answer
Zurab, Would it be possible for you to share the first paper on the formal power series ? I do not have access to this journal. Thanks. – cleanplay Jul 22 '14 at 15:29
@cleanplay: I'm trying to get a copy. I also do not have a direct access. – Zurab Silagadze Jul 24 '14 at 11:23
@cleanplay: you can get the paper from here: – Zurab Silagadze Jul 28 '14 at 3:12
Zurab, thanks a lot :-) – cleanplay Jul 28 '14 at 3:40

There is a formula due to Dynkin (1947). It can for instance be found as Theorem 2.2 in these lecture notes. You may also google Baker-Campbell-Hausdorff-Dynkin formula.

share|cite|improve this answer
For the paper of Dynkin, see Dynkin, Eugene Borisovich (1947). "Вычисление коэффициентов в формуле Campbell–Hausdorff" [Calculation of the coefficients in the Campbell–Hausdorff formula]. Doklady Akademii Nauk SSSR (in Russian) 57: 323–326. – Dietrich Burde Jul 22 '14 at 8:12
Probably a more accessible source is "Selected Papers of E.B. Dynkin With Commentary", AMS, 2000, pp. 31-35 – Pasha Zusmanovich Jul 22 '14 at 20:11
There is also another Dynkin paper on this: ("О представлении ряда $\log(e^xe^y)$ от некоммутирующих $x$ и $y$ через коммутаторы"). It claims that the proof of the formula in the 1947 paper was indirect, in the sense of relying on the existence of such a formula, while the one in the new paper would work from scratch. – darij grinberg Jul 22 '14 at 22:14

A completely explicit formula is obtained in a nice way in Bourbaki, Lie Groups and Lie Algebras, ch. 2, §6.

share|cite|improve this answer
Comment to the answer (v1): You mean p.90? It seems that Bourbaki there only calculates the BCH-terms linear in one of the two inputs, i.e. the terms whose coefficients are given by the Bernoulli numbers. Not the full Dynkin formula. – Qmechanic Jul 26 '14 at 21:30

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.