Let $f,g$ be smooth functions from $\mathbb R$ to $\mathbb R$. Then $$ \frac{(g\circ f)^{(n)}}{n!}=\sum_{1\le r\le n}\frac{(g^{(r)}\circ f)}{r!} \sum_{\substack{(n_1,\dots, n_r)\in {\mathbb N^*}^r\\n_1+\dots+n_r=n}}\prod_{1\le j\le r} \frac{f^{(n_j)}}{n_j!}, \tag{$\ast$}$$ is a way to write the so-called Faà di Bruno Formula. However that formula is usually written in a more complicated (equivalent) way which is $$ \frac{(g\circ f)^{(n)}}{n!}=\sum_{\substack{l_1+2l_2+\dots +nl_n=n\\r=l_1+\dots+l_n}}\frac{g^{(r)}\circ f}{l_1!\dots l_n! }\prod_{1\le j\le n} \left(\frac{f^{(j)}}{j!}\right)^{l_j}. \tag{$\ast\ast$}$$ My question: it seems that $(\ast)$ is quite easy to memorize whereas $(\ast\ast)$ is hard to recall and also that $(\ast)$ is handy to prove that Gevrey functions ($s\ge 1$, Gevrey of index 1 is analytic functions) make an algebra for composition. Why is $(\ast\ast)$ so popular and $(\ast)$ is absent from the literature?
$\begingroup$
$\endgroup$
6
-
$\begingroup$ See hal.archives-ouvertes.fr/hal-00950525/document $\endgroup$– user21574Commented Jan 22, 2016 at 13:15
-
6$\begingroup$ Because it is usual to collect proportional terms. Say, formula $(x+y)^2=x^2+xy+yx+y^2$ is less popular than $x^2+2xy+y^2$. $\endgroup$– Fedor PetrovCommented Jan 22, 2016 at 14:01
-
$\begingroup$ @Fedor Petrov Yes, but if you want to prove that $G^{s}$ is a composition algebra for $s\ge 1$, $(\ast)$ is easier to use than $(\ast\ast)$. $\endgroup$– BazinCommented Jan 24, 2016 at 19:24
-
$\begingroup$ @Bazin I agree that the first formula may be more appropriate in some situations (moreover, I personally always prefer it myself). But the explanation why it is less popular looks to be just as said. $\endgroup$– Fedor PetrovCommented Jan 24, 2016 at 20:16
-
$\begingroup$ Also, I don't quite understand the need to memorise. Either way, this formula is immediate from the Taylor formula and the multinomial theorem (for which reason the comment of Fedor Petrov is even more to the point than it may appear at a first glance), and attempting to memorise is the easiest way to get some bits wrong. $\endgroup$– Vladimir DotsenkoCommented Jan 25, 2016 at 2:21
|
Show 1 more comment