- Faà di Bruno's formula
**Faà di Bruno's formula**is an identity inmathematics generalizing thechain rule to higher derivatives, named in honor ofFrancesco Faà di Bruno (1825–1888), who was (in chronological order) a military officer, a mathematician, and a priest, and was beatified by the Pope a century after his death. Perhaps the most well-known form of Faà di Bruno's formula says that:$\{d^n\; over\; dx^n\}\; f(g(x))=sum\; frac\{n!\}\{m\_1!,1!^\{m\_1\},m\_2!,2!^\{m\_2\},cdots,m\_n!,n!^\{m\_n\; f^\{(m\_1+cdots+m\_n)\}(g(x))\; prod\_\{j=1\}^nleft(g^\{(j)\}(x)\; ight)^\{m\_j\},$

where the sum is over all "n"-tuples ("m"

_{1}, ..., "m"_{"n"}) satisfying the constraint:$1m\_1+2m\_2+3m\_3+cdots+nm\_n=n.,$

Sometimes, to give it a pleasing and memorable pattern, it is written in a way in which the coefficients that have the combinatorial interpretation discussed below are less explicit:

:$\{d^n\; over\; dx^n\}\; f(g(x))=sum\; frac\{n!\}\{m\_1!,m\_2!,cdots,m\_n!\}f^\{(m\_1+cdots+m\_n)\}(g(x))prod\_\{j=1\}^nleft(frac\{g^\{(j)\}(x)\}\{j!\}\; ight)^\{m\_j\}.$

Combining the terms with the same value of $m\_1+m\_2+cdots+m\_n=k$ leads to another somewhat simpler formula expressed in terms of

Bell polynomial s $B\_\{n,k\}(x\_1,dots,x\_\{n-k+1\})$::$\{d^n\; over\; dx^n\}\; f(g(x))\; =\; sum\_\{k=0\}^n\; f^\{(k)\}(g(x))\; B\_\{n,k\}left(g\text{\'}(x),g"(x),dots,g^\{(n-k+1)\}(x)\; ight).$

**Combinatorial form**The formula has a "combinatorial" form:

:$\{d^n\; over\; dx^n\}\; f(g(x))=(fcirc\; g)^\{(n)\}(x)=sum\_\{piinPi\}\; f^\{(left|pi\; ight|)\}(g(x))cdotprod\_\{Binpi\}g^\{(left|B\; ight|)\}(x)$

where

*π runs through the set Π of all partitions of the set { 1, ..., "n" },

*"B" ∈ π" means the variable "B" runs through the list of all of the "blocks" of the partition π, and

*|"A"| denotes the cardinality of the set "A" (so that |π| is the number of blocks in the partition π and |"B"| is the size of the block "B").

**Explication via an example**The combinatorial form may initially seem forbidding, so let us examine a concrete case, and see what the pattern is:

:$egin\{align\}(fcirc\; g)""(x)\; =\; f""(g(x))g\text{\'}(x)^4\; +\; 6f"\text{\'}(g(x))g"(x)g\text{\'}(x)^2\; \backslash \; \{\}\; quad+;\; 3f"(g(x))g"(x)^2+\; 4f"(g(x))g"\text{\'}(x)g\text{\'}(x)\; \backslash \; \{\}\; quad+;\; f\text{\'}(g(x))g""(x).end\{align\}$

What is the pattern?

:$egin\{align\}\; g\text{\'}(x)^4\; leftrightarrow\; 1+1+1+1\; leftrightarrow\; f""(g(x))\; leftrightarrow\; 1\; \backslash \; \backslash \; g"(x)g\text{\'}(x)^2\; leftrightarrow\; 2+1+1\; leftrightarrow\; f"\text{\'}(g(x))\; leftrightarrow\; 6\; \backslash \; \backslash g"(x)^2\; leftrightarrow\; 2+2\; leftrightarrow\; f"(g(x))\; leftrightarrow\; 3\; \backslash \; \backslash g"\text{\'}(x)g\text{\'}(x)\; leftrightarrow\; 3+1\; leftrightarrow\; f"(g(x))\; leftrightarrow\; 4\; \backslash \; \backslash g""(x)\; leftrightarrow\; 4\; leftrightarrow\; f\text{\'}(g(x))\; leftrightarrow\; 1end\{align\}$

The factor $scriptstyle\; g"(x)g\text{\'}(x)^2\; ;$ corresponds to the partition 2 + 1 + 1 of the integer 4, in the obvious way. The factor $scriptstyle\; f"\text{\'}(g(x));$ that goes with it corresponds to the fact that there are "three" summands in that partition. The coefficient 6 that goes with those factors corresponds to the fact that there are exactly six partitions of a set of four members that break it into one part of size 2 and two parts of size 1.

Similarly, the factor $scriptstyle\; g"(x)^2\; ;$ in the third line corresponds to the partition 2 + 2 of the integer 4, (4, because we are finding the fourth derivative), while $scriptstyle\; f"(g(x))\; ,!$ corresponds to the fact that there are "two" summands (2 + 2) in that partition. The coefficient 3 corresponds to the fact that there are 3 ways of partitioning 4 objects into groups of 2 (

^{4}C_{2}÷ 2). The same concept applies to the others.**Combinatorics of the Faà di Bruno coefficients**These partition-counting

**Faà di Bruno coefficients**have a "closed-form" expression. The number of partitions of a set of size "n" corresponding to theinteger partition :$displaystyle\; n=underbrace\{1+cdots+1\}\_\{m\_1\},+,\; underbrace\{2+cdots+2\}\_\{m\_2\}\; ,+,\; underbrace\{3+cdots+3\}\_\{m\_3\}+cdots$

of the integer "n" is equal to

:$frac\{n!\}\{m\_1!,m\_2!,m\_3!,cdots\; 1!^\{m\_1\},2!^\{m\_2\},3!^\{m\_3\},cdots\}.$

These coefficients also arise in the

Bell polynomials , which are relevant to the study ofcumulant s.**A multivariable version**Let "y" = "g"("x"

_{1}, ..., "x"_{"n"}).Then the following identity holds regardless of whether the "n" variables are all distinct, or all identical, or partitioned into several distinguishable classes of indistinguishable variables (if it seems opaque, see the very concrete example below)::$\{partial^n\; over\; partial\; x\_1\; cdots\; partial\; x\_n\}f(y)=\; sum\_\{piinPi\}\; f^\{(left|pi\; ight|)\}(y)cdotprod\_\{Binpi\}\{partial^\{left|B\; ighty\; over\; prod\_\{jin\; B\}\; partial\; x\_j\}$

where (as above)

*π runs through the set Π of all partitions of the set { 1, ..., "n" },

*"B" ∈ π" means the variable "B" runs through the list of all of the "blocks" of the partition π, and

*|"A"| denotes the cardinality of the set "A" (so that |π| is the number of blocks in the partition π and |"B"| is the size of the block "B").

See "Hardy, Michael, " [

*http://www.combinatorics.org/Volume_13/PDF/v13i1r1.pdf Combinatorics of Partial Derivatives*] ", [*http://www.combinatorics.org Electronic Journal of Combinatorics*] ,**13**(2006), #R1.**Example**The five terms in the following expression correspond in the obvious way to the five partitions of the set { 1, 2, 3 }, and in each case the order of the derivative of "f" is the number of parts in the partition:

:$\{partial^3\; over\; partial\; x\_1,\; partial\; x\_2,\; partial\; x\_3\}f(y)=\; f\text{'}(y)\{partial^3\; y\; over\; partial\; x\_1,\; partial\; x\_2,\; partial\; x\_3\}$

::::$\{\}\; +\; f"(y)\; left(\; \{partial\; y\; over\; partial\; x\_1\}cdot\{partial^2\; y\; over\; partial\; x\_2,\; partial\; x\_3\}+\{partial\; y\; over\; partial\; x\_2\}cdot\{partial^2\; y\; over\; partial\; x\_1,\; partial\; x\_3\}+\; \{partial\; y\; over\; partial\; x\_3\}cdot\{partial^2\; y\; over\; partial\; x\_1,\; partial\; x\_2\}\; ight)$

:::::$\{\}\; +\; f"\text{\'}(y)\; \{partial\; y\; over\; partial\; x\_1\}cdot\{partial\; y\; over\; partial\; x\_2\}cdot\{partial\; y\; over\; partial\; x\_3\}.$

If the three variables are indistinguishable from each other, then three of the five terms above are also indistinguishable from each other, and then we have the classic one-variable formula.

**Formal power series version**In the

formal power series :$f(x)=sum\_n\; \{a\_n\; over\; n!\}x^n,$

we have the "n"th derivative at 0:

:$f^\{(n)\}(0)=a\_n.\; ;$

This should not be construed as the value of a function, since these series are purely formal; there is no such thing as convergence or divergence in this context.

If

:$g(x)=sum\_\{n=1\}^infty\; \{b\_n\; over\; n!\}\; x^n$

and

:$f(x)=sum\_\{n=1\}^infty\; \{a\_n\; over\; n!\}\; x^n$

and

:$g(f(x))=h(x)=sum\_\{n=1\}^infty\{c\_n\; over\; n!\}x^n,$

then the coefficient "c"

_{"n"}(which would be the "n"th derivative of "h" evaluated at 0 if we were dealing with convergent series rather than formal power series) is given by:$c\_n=sum\_\{pi=left\{,B\_1,,dots,,B\_k,\; ight\; a\_\{left|B\_1\; ightcdots\; a\_\{left|B\_k\; ight\; b\_k$

where π runs through the set of all partitions of the set { 1, ..., "n" } and "B"

_{1}, ..., "B"_{"k"}are the blocks of the partition π, and | "B"_{"j"}| is the number of members of the "j"th block, for "j" = 1, ..., "k".This version of the formula is particularly well suited to the purposes of

combinatorics . See the "compositional formula" in Chapter 5 of " [*http://www-math.mit.edu/~rstan/ec/ Enumerative Combinatorics, Volumes 1 and 2*] ", Richard P. Stanley, Cambridge University Press, 1997 and 1999, ISBN 0-521-55309-1N.We can also write

:$g(f(x))\; =\; sum\_\{n=1\}^infty\{sum\_\{k=1\}^\{n\}\; b\_k\; B\_\{n,k\}(a\_1,dots,a\_\{n-k+1\})\; over\; n!\}\; x^n.$

where the expressions

:$B\_\{n,k\}(a\_1,dots,a\_\{n-k+1\})$

are

Bell polynomials .**A special case**If "f"("x") = e

^{"x"}then all of the derivatives of "f" are the same, and are a factor common to every term. In case "g"("x") is acumulant -generating function, then "f"("g"("x")) is a moment-generating function, and the polynomial in various derivatives of "g" is the polynomial that expresses the moments as functions of the cumulants.**External links*** W.P. Johnson, "The Curious History of Faà di Bruno's Formula", "

American Mathematical Monthly ", Vol. 109, March 2002, 217-234, [*http://www.maa.org/news/monthly217-234.pdf online*]

* [*http://mathworld.wolfram.com/FaadiBrunosFormula.html Faà di Bruno's Formula on Mathworld*]

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