# What was Lambert's solution to $x^m+x=q$?

I've been thinking about Lambert's Trinomial Equation quite a bit, and I want to see his solution. The only solution I could find was in Euler's form, and I still don't quite understand how he got from one equation to the other. Lambert's solution is probably somewhere in here, since this is the paper that was cited in the first link, but this text is absolutely foreign to me; I can't make heads or tails of it. He's using the long "s" and I don't speak latin, and algebra was pretty different back then. I asked this on MSE but I didn't get any answers yet. Is this the sort of question that's appropriate on this site, since it has to do with research? Thanks!

• Thanks for mentioning that you had asked the question on math.stackexchange.com . Here is a link to the question: math.stackexchange.com/questions/274853/… . I think your main question is appropriate here, but it would have been nice if you had waited a little longer to make sure you didn't get any answers on MSE.
– HJRW
Jan 10, 2013 at 13:40
• A power series expansion $x=x(q)$ around $q=0$, that is, an expansion of the local inverse at $0$ of $f(x):=x+x^m$, is given by the Lagrange inversion formula: en.wikipedia.org/wiki/… . The method works well because all powers of $x+x^m$ have simple power series expansions (via the Newton series). You can easily write it by yourself. Enjoy the computation! Jan 10, 2013 at 14:14
• Also, check the answer to this related question mathoverflow.net/questions/60943/root-estimation/60997#60997 Jan 10, 2013 at 17:45

I understand your question as a historical one: how did Lambert solve the trinomial equation? Let me try to walk you through his derivation in his 1758 paper "Observationes variae in mathesin puram".

In paragraph 35 (see here) Lambert first explains his method of successive inequalities for the simplest case $m=1$, so to solve $x+px=q$ for $x$. He considers separately the cases $p<1$ (left column of inequalities) and $p>1$ (right column). He thus arrives at the series

$x=q-pq+p^2q-p^3q+p^4q+\cdots$ for $p<1$ (notice the typo in the sign of the fourth term)

$x=q/p -q/p^2+q/p^3-q/p^4+\cdots$ for $p>1$.

Both series converge to $q/(1+p)$, which is indeed the solution.

He then goes on in paragraph 36 to apply the same method to the case $m=2$, solving $x^2+px=q$, concluding in paragraph 39 with the general case $x^m+px=q$:

$$x=q/p-q^m/p^{m+1}+mq^{2m-1}/p^{2m+1}-\tfrac{1}{2}m(3m-2)q^{3m-2}/p^{3m+1}+\cdots$$

(The paragraphs where this general solution is given are missing from the scanned version of his paper that you found online.)

In paragraph 40 he gives the convergence criterium for the series as $(m-1)^{m-1}p^m>m^m q^{m-1}$. [note]

For historical background on Lambert and the discovery of the method of Lagrange inversion, see "A Historian Looks Back" by Judith Grabiner, page 52 and following.

[note added October 2014] Grabiner gives a weaker convergence criterion, $p^m>q^{m-1}$, but as Tom Copeland noted, the original Lambert criterion is consistent with the answer to this MSE question (for $m=4$, $p=1$).

• Grabiner? Gerhard "Proof-reading Minds Want To Know" Paseman, 2013.01.12 Jan 13, 2013 at 1:02
• typo corrected, thank you, Gerhard. Jan 13, 2013 at 8:20
• @Carlo Beenakker, how does your convergence criterion square with math.stackexchange.com/questions/683372/…, which seems consistent with Lambert's criterion? Oct 15, 2014 at 19:09
• @TomCopeland --- good point, I have edited the answer accordingly Oct 15, 2014 at 20:04
• @Carlo Beenakker: Based on the discriminant, Lambert's criterion is correct. How did he determine it? For those who may not know, the coefficients to these series give the Fuss-Catalan numbers whose combinatorial interpretations were explored by Euler and Fuss, among others. Oct 15, 2014 at 22:09

Probably it is too late but note that lambert made a mistake(or typo) in his paper, that fourth term should be -1/2m(3m - 1) q^{3m-2}/p^{3m+1}. O.F