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I have noticed there is an identity for Bell polynomials that can apply of Faà di Bruno's formula. This is a convolution identity that states: $$ (x \ast y)_n = \sum_{j=1}^{n-1} {n \choose j} x_j y_{n-j}, $$

which is applied to Bell polynomials which an example is shown on this wiki page. But I have yet to see a proof of this formula using this identity. Since Faà di Bruno's formula can be expressed as:

$$ \frac{d^n}{dx^n}[f(g(x))] = \sum_{k=1}^{n}f^{(k)}(g(x)) B_{n,k}(g'(x),g''(x),\dots,g^{(n-k+1)}(x)). $$ Is there a proof of this formula using the convolution identity for Bell polynomials, or would this be unnecessarily rigorous? Either way it would be interesting to see.

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For a sequence $x=\left(x_{1},x_{2},...,x_{N}\right)$, define the convolution with: $$(x\ast x)_{n}=\sum_{j=1}^{n-1}{n \choose j}x_{j}x_{n-j}.$$ Multiple application of this convolution yields: $$\underbrace{(x\ast x\ast ...\ast x)}_{k\ factors}=\sum_{i_{1}=1}^{N}\sum_{i_{2}=1}^{i_{1}-1}\sum_{i_{3}=1}^{i_{2}-1}...\sum_{i_{k}=1}^{i_{k-1}-1}{i_{1}\choose i_{2}}{i_{2}\choose i_{3}}...{i_{k-1}\choose i_{k}}x_{i_{1}-i_{2}}x_{i_{2}-i_{3}}...x_{i_{k-1}-{i_k}}x_{i_{k}}$$ $$=k!\sum_{n=1}^{N}\sum_{\pi_{n,k}}{n!\over j_1!j_2!...j_{n-k+1}!}\left({x_1\over 1!}\right)^{j_1}\left({x_2\over 2!}\right)^{j_2}...\left({x_{n-k+1}\over(n-k+1)!}\right)^{j_{n-k+1}}$$ $$=k!\sum_{n=1}^{N}B_{n,k}(x_1,x_2,...,x_{n-k+1}),$$ $\pi_{n,k}=\pi_{n,k}(j_1,j_2,...,j_{n-k+1})$ are the partitions of integer $n$ into exactly $k$ parts.

This is the relation between Bell polynomials and your convolution. For proving Faà di Bruno's formula, it is not necessary to use the convolution.

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It is not full answer for your question but the method of proof might be such way

Taylor expansion of $f(g(x+h))$ can be written as

$$ f(g(x+h))=f(g(x))+h\frac{d}{dx} \left( f(g(x)) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left(f(g(x)) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(g(x) \right)+.... \tag 1 $$

$$ f(g(x+h))=f \left( g(x)+hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....\right) $$

$$ f(g(x+h))=f(g(x))+(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)f'(g(x))+\frac{(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^2}{2!}f''(g(x))+\frac{(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^3}{3!}f'''(g(x))+..... $$

$$ f(g(x+h))=\sum_{k=0}^\infty \frac{(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^k}{k!}f^{(k)}(g(x)) $$

$$ \frac{d^n }{dh^{n}}f(g(x+h))=\frac{d^n }{dh^{n}}\left(\sum_{k=0}^\infty \frac{(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^k}{k!}f^{(k)}(g(x))\right) $$

$$ \frac{d^n }{dh^{n}}f(g(x+h))|_{h=0}=\frac{d^n }{dh^{n}}\left(\sum_{k=0}^\infty \frac{(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^k}{k!}f^{(k)}(g(x))\right)|_{h=0} $$ If we use Equation 1 and we can get

$$ \frac{d^n }{dx^{n}}\left( f(g(x)) \right)=\sum_{k=0}^\infty \frac{d^n }{dh^{n}} \left((hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^k \right)|_{h=0} \frac{ f^{(k)}(g(x))}{k!} $$

You need to use binominal expansion formula after this.

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  • $\begingroup$ This is not what I was looking for, But it is another perspective. I like the approach but I would like it very much if you could relate this to bell polynomials or convolution identity as it is mostly what I am looking for. $\endgroup$
    – Eric L
    Nov 21 '14 at 22:07

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