Faà di Bruno's formula is an identity in mathematics generalizing the chain rule to higher derivatives. It is named after, although he was not the first to state or prove the formula. In 1800, more than 50 years before Faà di Bruno, the French mathematician Louis François Antoine Arbogast had stated the formula in a calculus textbook,[1] which is considered to be the first published reference on the subject.[2]
Perhaps the most well-known form of Faà di Bruno's formula says that
where the sum is over all
n
(m1,\ldots,mn)
Sometimes, to give it a memorable pattern, it is written in a way in which the coefficients that have the combinatorial interpretation discussed below are less explicit:
{dn\overdxn}f(g(x)) =\sum
n! | |
m1!m2! … mn! |
(m1+ … +mn) | |
⋅ f |
| ||||
(g(x)) ⋅ \prod | ||||
j=1 |
mj | |
\right) |
.
Combining the terms with the same value of
m1+m2+ … +mn=k
mj
j>n-k+1
Bn,k(x1,\ldots,xn-k+1)
{dn\overdxn}f(g(x))=
n | |
\sum | |
k=0 |
f(k)(g(x)) ⋅ Bn,k\left(g'(x),g''(x),...,g(n-k+1)(x)\right).
The formula has a "combinatorial" form:
{dn\overdxn}f(g(x))=(f\circg)(n)(x)=\sum\pi\in\Pif(\left|\pi\right|)(g(x)) ⋅ \prodB\in\pig(\left|B\right|)(x)
where
\pi
\Pi
\{1,\ldots,n\}
B\in\pi
B
\pi
|A|
A
|\pi|
\pi
|B|
B
The following is a concrete explanation of the combinatorial form for the
n=4
\begin{align} (f\circg)''''(x) ={}&f''''(g(x))g'(x)4 +6f'''(g(x))g''(x)g'(x)2\\[8pt] &{}+ 3f''(g(x))g''(x)2 +4f''(g(x))g'''(x)g'(x)\\[8pt] &{}+ f'(g(x))g''''(x). \end{align}
The pattern is:
\begin{array}{cccccc} g'(x)4&&\leftrightarrow&&1+1+1+1&&\leftrightarrow&&f''''(g(x))&&\leftrightarrow&&1\\[12pt] g''(x)g'(x)2&&\leftrightarrow&&2+1+1&&\leftrightarrow&&f'''(g(x))&&\leftrightarrow&&6\\[12pt] g''(x)2&&\leftrightarrow&&2+2&&\leftrightarrow&&f''(g(x))&&\leftrightarrow&&3\\[12pt] g'''(x)g'(x)&&\leftrightarrow&&3+1&&\leftrightarrow&&f''(g(x))&&\leftrightarrow&&4\\[12pt] g''''(x)&&\leftrightarrow&&4&&\leftrightarrow&&f'(g(x))&&\leftrightarrow&&1 \end{array}
The factor
g''(x)g'(x)2
f'''(g(x))
Similarly, the factor
g''(x)2
f''(g(x))
\tfrac{1}{2}\tbinom{4}{2}=3
A memorizable scheme is as follows:
\begin{align}&
D1(f\circ{ | |
g)}{1!} |
&=\left(f(1)\circ{}g\right)
| |||||
1! |
\\[8pt] &
D2(f\circg) | |
2! |
&=\left(f(1)\circ{}g\right)
| |||||
1! |
&{}+\left(f(2)\circ{}g\right)
| ||||||||
2! |
\\[8pt] &
D3(f\circg) | |
3! |
&=\left(f(1)\circ{}g\right)
| |||||
1! |
&{}+\left(f(2)\circ{}g\right)
| |||||
1! |
| |||||
1! |
&{}+\left(f(3)\circ{}g\right)
| |||||||||||
3! |
\\[8pt] &
D4(f\circg) | |
4! |
&=\left(f(1)\circ{}g\right)
| |||||
1! |
&{}+\left(f(2)\circ{}g\right)\left(
| |||||
1! |
| + | |||||
1! |
| ||||||||
2! |
\right)&{}+\left(f(3)\circ{}g\right)
| ||||||||
2! |
| |||||
1! |
&{}+\left(f(4)\circ{}g\right)
| ||||||||||||||
4! |
\end{align}
Let
y=g(x1,...,xn)
n
{\partialn\over\partialx1 … \partialxn}f(y) =\sum\pi\in\Pif(\left|\pi\right|)(y) ⋅ \prodB\in\pi{\partial\left|B\right|y\over\prodj\in\partialxj}
where (as above)
\pi
\Pi
\{1,\ldots,n\}
B\in\pi
B
\pi
|A|
A
|\pi|
\pi
|B|
B
More general versions hold for cases where the all functions are vector- and even Banach-space-valued. In this case one needs to consider the Fréchet derivative or Gateaux derivative.
The five terms in the following expression correspond in the obvious way to the five partitions of the set
\{1,2,3\}
f
\begin{align} {\partial3\over\partialx1\partialx2\partialx3}f(y) ={}&f'(y){\partial3y\over\partialx1\partialx2\partialx3}\\[10pt] &{}+f''(y)\left({\partialy\over\partial
2 | |
x | |
1} ⋅ {\partial |
y\over\partialx2\partialx3} +{\partialy\over\partial
2 | |
x | |
2} ⋅ {\partial |
y\over\partialx1\partialx3} +{\partialy\over\partial
2 | |
x | |
3} ⋅ {\partial |
y\over\partialx1\partialx2}\right)\\[10pt] &{}+f'''(y){\partialy\over\partialx1} ⋅ {\partialy\over\partialx2} ⋅ {\partialy\over\partialx3}. \end{align}
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.
Suppose
infty | |
f(x)=\sum | |
n=0 |
{an}xn
infty | |
g(x)=\sum | |
n=0 |
{bn}xn
b0=0
Then the composition
f\circg
n, | |
f(g(x))=\sum | |
n}x |
c0=a0
cn
n\geq1
n
n
cn=\sumi\inn
l{C}n=\{(i1,i2,...,ik): 1\lek\len, i1+i2+ … +ik=n\}
n
k
or
cn=
n | |
\sum | |
k=1 |
ak\sum\pi\inn,k
l{P}n,k=\{(\pi1,\pi2,...,\pin): \pi1+\pi2+ … +\pin=k, \pi1 ⋅ 1+\pi2 ⋅ 2+ … +\pin ⋅ n=n\}
n
k
The first form is obtained by picking out the coefficient of
xn
(b1x+b2x2+ … )k
The special case
f(x)=ex
g(x)=\sumn\geq
1 | |
n! |
anxn
f(x)=1/(1-x)
g(x)=\sumn\geq(-an)xn
\sumn\geqanxn
a0=1
Stanley [4] gives a version for exponential power series.In the formal power series
f(x)=\sumn{
an | |
n! |
we have the
n
f(n)(0)=an.
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
infty | ||
g(x)=\sum | { | |
n=0 |
bn | |
n! |
and
infty | ||
f(x)=\sum | { | |
n=1 |
an | |
n! |
and
| ||||
g(f(x))=h(x)=\sum | ||||
n=0 |
then the coefficient
cn
n
h
cn=\sum
\pi=\left\{B1,\ldots,Bk\right\ |
where
\pi
\{1,\ldots,n\}
B1,\ldots,Bk
\pi
|Bj|
j
j=1,\ldots,k
This version of the formula is particularly well suited to the purposes of combinatorics.
We can also write with respect to the notation above
g(f(x))=b0+
infty | |
\sum | |
n=1 |
| ||||||||||
n! |
xn,
where
Bn,k(a1,\ldots,an-k+1)
If
f(x)=ex
f
{dn\overdxn}eg(x)=eg(x)Bn\left(g'(x),g''(x),...,g(n)(x)\right),
where
Bn(x)
In case
g(x)
f(g(x))
g