Timeline for How do i solve this : $\displaystyle \ f'=e^{{f}^{-1}}$?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2017 at 23:00 | comment | added | Tom Copeland | Link above should read oeis.org/A145271. | |
| Jan 17, 2017 at 22:42 | comment | added | Tom Copeland | @PietroMajer, I pay tribute to Lagrange by calling pretty much any formula an LlF, and there are very many, that give a series expansion equivalent to that of your LIF. Think Lagrange inversion = compostitional inversion via series whether o.g.f.s, e.g.f.s, or other series reps. For non-series inversion, I might use directly $g(g^{-1}(x) )= x$, or a Laplace-like transform with a change of variables--any way to skin the cat analytically--but we don't have a well-defined analytic function, forward or inverse, to begin with here though, so bootstrap methods only come to mind. | |
| Jan 17, 2017 at 21:26 | comment | added | zeraoulia rafik | @PietroMajer, i think he meant by LIF the lagrange inverse formula theorem | |
| Jan 17, 2017 at 1:56 | comment | added | Pietro Majer | To me,"LIF" is the explicit (non-recursive) formula I mentioned above, in terms of the coefficients of negative powers of $f$. So my question was, if you can exploit this more advanced formula, and get more information, of course. I mean, like a non-recursive formula for these sequences, or recursive but not as complicated as those one obtains equating the coefficients in $f'(x)=\exp(f^{(-1)}(x))$ or in $g'(x)=\exp(g(g(x))$. | |
| Jan 17, 2017 at 1:56 | comment | added | Pietro Majer | Of course, in the sense of formal power series, for $f:=\sum_{j=0}^{+\infty}f_jx^j$, say with complex coefficients, we have an elementary inverse function theorem, as you said (if $f_0=0$, and $f_1\neq0$ then $f$ is invertible) and the coefficients of the inverse are determined inductively; we also have a formal derivative and the rule of derivative of the inverse function holds, and all that is sufficient to determine recursively the coefficients of $f$ and its inverse here. It was not clear to me what you mean by "LIF", | |
| Jan 17, 2017 at 1:23 | comment | added | Tom Copeland | A divergent series doesn't really have any derivatives, so one can't invoke the inverse function theorem in its clearly geometric interpretation but only as a relation beween the coefficients of a formal series and its 'formal inverse' as defined by the series involutions denoted as Lagrange inversion formulas in the OEIS, which apply to convergent analytic series at the origin with f(0)=0 and a nonzero first derivation. These involutions define relations among the coefficients of the pair of series independent of convergence, that are consistent with composition and reciprocation. | |
| Jan 16, 2017 at 21:33 | comment | added | Tom Copeland | @PietroMajer, both. See, for example, tcjpn.wordpress.com/2016/11/01/… and the July 2015 formula in oeis.org/A133437. Lagrange a la Lah has different series reps of the Lagrange inversion formula, all available on the OEIS (see cross-references in oeis.org/A145257). | |
| Jan 16, 2017 at 18:43 | comment | added | Pietro Majer | When you talk of the Lagrange inversion theorem, do you mean the plain derivation of coefficients of the series for $f^{(-1)}$, or also the residue formula $[x^n] f^{(-1)}={1\over n}\mathrm{Res}(f^{-n})$ ? (which could be of use here to get some explicit formula for the coefficients, although I don't see it) | |
| Jan 4, 2017 at 18:29 | comment | added | Tom Copeland | The inverse function theorem here might be more aptly called the inverse formal series theorem. As you can see, the differential equations and inverses here in analytic guise are concise statements of relations among the coefficients of formal series (e.g.f.s or o.g.f.s). | |
| Jan 4, 2017 at 5:22 | history | edited | Tom Copeland | CC BY-SA 3.0 |
Corrected an index
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| Jan 3, 2017 at 19:21 | history | answered | Tom Copeland | CC BY-SA 3.0 |