# Proof of Faà di Bruno's formula using a convolution identity for Bell polynomials?

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.

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.

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.

• 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. Nov 21 '14 at 22:07