5 deleted 1 characters in body

Consider $g(x)=e^x-1$. Then $g^n(x)= x+\frac{1}{2!}n x^2+\frac{1}{3!} \left(\frac{3 n^2}{2}-\frac{n}{2}\right) x^3+\frac{1}{4!} \left(3 n^3-\frac{5 n^2}{2}+\frac{n}{2}\right) x^4$ $+\frac{1}{5!} \left(\frac{15 n^4}{2}-\frac{65 n^3}{6}+5 n^2-\frac{2 n}{3}\right) x^5$

$+\frac{1}{6!} \left(\frac{45 n^5}{2}-\frac{385 n^4}{8}+\frac{445 n^3}{12}-\frac{91 n^2}{8}+\frac{11 n}{12}\right) x^6$

$+\frac{1}{7!}\left(\frac{315 n^6}{4}-\frac{1827 n^5}{8}+\frac{6125 n^4}{24}-\frac{1043 n^3}{8}+\frac{637 n^2}{24}-\frac{3 n}{4}\right) x^7 + \cdots$

Note that $g^0(x)=x, g^1(x)=e^x-1$ and that

$g^\frac{1}{2}(x)=x+\frac{x ^2}{4}+ +\frac{x^3}{48} \frac{x^3}{48} +\frac{x^5}{3840}-\frac{7 x^6}{92160} +\frac{x^7}{645120}$ which is consistent with what Greg Kuperburg obtained. A symbolic mathematical program will also confirm that $g^m(g^n(x))=g^{m+n}(x) +O(x^8)$

4 added 2 characters in body

Consider $g(x)=e^x-1$. Then $g^n(x)= x+\frac{1}{2!}n x^2+\frac{1}{3!} \left(\frac{3 n^2}{2}-\frac{n}{2}\right) x^3+\frac{1}{4!} \left(3 n^3-\frac{5 n^2}{2}+\frac{n}{2}\right) x^4$ $+\frac{1}{5!} \left(\frac{15 n^4}{2}-\frac{65 n^3}{6}+5 n^2-\frac{2 n}{3}\right) x^5$

$+\frac{1}{6!} \left(\frac{45 n^5}{2}-\frac{385 n^4}{8}+\frac{445 n^3}{12}-\frac{91 n^2}{8}+\frac{11 n}{12}\right) x^6$

$+\frac{1}{7!}\left(\frac{315 n^6}{4}-\frac{1827 n^5}{8}+\frac{6125 n^4}{24}-\frac{1043 n^3}{8}+\frac{637 n^2}{24}-\frac{3 n}{4}\right) x^7 + \cdots$

Note that $g^0(x)=x, g^1(x)=e^x-1$ and that

$g^\frac{1}{2}(x)=x+\frac{x ^2}{4}+ +\frac{x^3}{48} +\frac{x^5}{3840}-\frac{7 x^6}{92160} +\frac{x^7}{645120}$ which is consistent with what Greg Kuperburg obtained. A symbolic mathematical program will also confirm that $g^m(g^n(x))=g^{m+n}(x) +O(8)$O(x^8)$See The Euler-Arnold equation for more information. 3 Added referrence link Consider$g(x)=e^x-1$. Then$g^n(x)= x+\frac{1}{2!}n x^2+\frac{1}{3!} \left(\frac{3 n^2}{2}-\frac{n}{2}\right) x^3+\frac{1}{4!} \left(3 n^3-\frac{5 n^2}{2}+\frac{n}{2}\right) x^4 +\frac{1}{5!} \left(\frac{15 n^4}{2}-\frac{65 n^3}{6}+5 n^2-\frac{2 n}{3}\right) x^5  +\frac{1}{6!} \left(\frac{45 n^5}{2}-\frac{385 n^4}{8}+\frac{445 n^3}{12}-\frac{91 n^2}{8}+\frac{11 n}{12}\right) x^6  +\frac{1}{7!}\left(\frac{315 n^6}{4}-\frac{1827 n^5}{8}+\frac{6125 n^4}{24}-\frac{1043 n^3}{8}+\frac{637 n^2}{24}-\frac{3 n}{4}\right) x^7 + \cdots$Note that$g^0(x)=x, g^1(x)=e^x-1$and that$g^\frac{1}{2}(x)=x+\frac{x ^2}{4}+ +\frac{x^3}{48} +\frac{x^5}{3840}-\frac{7 x^6}{92160} +\frac{x^7}{645120}$which is consistent with what Greg Kuperburg obtained. A symbolic mathematical program will also confirm that$g^m(g^n(x))=g^{m+n}(x) +O(8)\$