Generalization of the product rule in calculus
In calculus , the general Leibniz rule ,[1] named after Gottfried Wilhelm Leibniz , generalizes the product rule (which is also known as "Leibniz's rule"). It states that if f {\displaystyle f} and g {\displaystyle g} are n -times differentiable functions , then the product f g {\displaystyle fg} is also n -times differentiable and its n -th derivative is given by
( f g ) ( n ) = ∑ k = 0 n ( n k ) f ( n − k ) g ( k ) , {\displaystyle (fg)^{(n)}=\sum _{k=0}^{n}{n \choose k}f^{(n-k)}g^{(k)},} where
( n k ) = n ! k ! ( n − k ) ! {\displaystyle {n \choose k}={n! \over k!(n-k)!}} is the
binomial coefficient and
f ( j ) {\displaystyle f^{(j)}} denotes the
j th derivative of
f (and in particular
f ( 0 ) = f {\displaystyle f^{(0)}=f} ).
The rule can be proven by using the product rule and mathematical induction .
Second derivative [ edit ] If, for example, n = 2 , the rule gives an expression for the second derivative of a product of two functions:
( f g ) ″ ( x ) = ∑ k = 0 2 ( 2 k ) f ( 2 − k ) ( x ) g ( k ) ( x ) = f ″ ( x ) g ( x ) + 2 f ′ ( x ) g ′ ( x ) + f ( x ) g ″ ( x ) . {\displaystyle (fg)''(x)=\sum \limits _{k=0}^{2}{{\binom {2}{k}}f^{(2-k)}(x)g^{(k)}(x)}=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x).} More than two factors [ edit ] The formula can be generalized to the product of m differentiable functions f 1 ,...,f m .
( f 1 f 2 ⋯ f m ) ( n ) = ∑ k 1 + k 2 + ⋯ + k m = n ( n k 1 , k 2 , … , k m ) ∏ 1 ≤ t ≤ m f t ( k t ) , {\displaystyle \left(f_{1}f_{2}\cdots f_{m}\right)^{(n)}=\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}\prod _{1\leq t\leq m}f_{t}^{(k_{t})}\,,} where the sum extends over all
m -tuples (
k 1 ,...,
k m ) of non-negative integers with
∑ t = 1 m k t = n , {\textstyle \sum _{t=1}^{m}k_{t}=n,} and
( n k 1 , k 2 , … , k m ) = n ! k 1 ! k 2 ! ⋯ k m ! {\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}} are the
multinomial coefficients . This is akin to the
multinomial formula from algebra.
The proof of the general Leibniz rule proceeds by induction. Let f {\displaystyle f} and g {\displaystyle g} be n {\displaystyle n} -times differentiable functions. The base case when n = 1 {\displaystyle n=1} claims that:
( f g ) ′ = f ′ g + f g ′ , {\displaystyle (fg)'=f'g+fg',} which is the usual
product rule and is known to be true. Next, assume that the statement holds for a fixed
n ≥ 1 , {\displaystyle n\geq 1,} that is, that
( f g ) ( n ) = ∑ k = 0 n ( n k ) f ( n − k ) g ( k ) . {\displaystyle (fg)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}f^{(n-k)}g^{(k)}.} Then,
( f g ) ( n + 1 ) = [ ∑ k = 0 n ( n k ) f ( n − k ) g ( k ) ] ′ = ∑ k = 0 n ( n k ) f ( n + 1 − k ) g ( k ) + ∑ k = 0 n ( n k ) f ( n − k ) g ( k + 1 ) = ∑ k = 0 n ( n k ) f ( n + 1 − k ) g ( k ) + ∑ k = 1 n + 1 ( n k − 1 ) f ( n + 1 − k ) g ( k ) = ( n 0 ) f ( n + 1 ) g ( 0 ) + ∑ k = 1 n ( n k ) f ( n + 1 − k ) g ( k ) + ∑ k = 1 n ( n k − 1 ) f ( n + 1 − k ) g ( k ) + ( n n ) f ( 0 ) g ( n + 1 ) = ( n + 1 0 ) f ( n + 1 ) g ( 0 ) + ( ∑ k = 1 n [ ( n k − 1 ) + ( n k ) ] f ( n + 1 − k ) g ( k ) ) + ( n + 1 n + 1 ) f ( 0 ) g ( n + 1 ) = ( n + 1 0 ) f ( n + 1 ) g ( 0 ) + ∑ k = 1 n ( n + 1 k ) f ( n + 1 − k ) g ( k ) + ( n + 1 n + 1 ) f ( 0 ) g ( n + 1 ) = ∑ k = 0 n + 1 ( n + 1 k ) f ( n + 1 − k ) g ( k ) . {\displaystyle {\begin{aligned}(fg)^{(n+1)}&=\left[\sum _{k=0}^{n}{\binom {n}{k}}f^{(n-k)}g^{(k)}\right]'\\&=\sum _{k=0}^{n}{\binom {n}{k}}f^{(n+1-k)}g^{(k)}+\sum _{k=0}^{n}{\binom {n}{k}}f^{(n-k)}g^{(k+1)}\\&=\sum _{k=0}^{n}{\binom {n}{k}}f^{(n+1-k)}g^{(k)}+\sum _{k=1}^{n+1}{\binom {n}{k-1}}f^{(n+1-k)}g^{(k)}\\&={\binom {n}{0}}f^{(n+1)}g^{(0)}+\sum _{k=1}^{n}{\binom {n}{k}}f^{(n+1-k)}g^{(k)}+\sum _{k=1}^{n}{\binom {n}{k-1}}f^{(n+1-k)}g^{(k)}+{\binom {n}{n}}f^{(0)}g^{(n+1)}\\&={\binom {n+1}{0}}f^{(n+1)}g^{(0)}+\left(\sum _{k=1}^{n}\left[{\binom {n}{k-1}}+{\binom {n}{k}}\right]f^{(n+1-k)}g^{(k)}\right)+{\binom {n+1}{n+1}}f^{(0)}g^{(n+1)}\\&={\binom {n+1}{0}}f^{(n+1)}g^{(0)}+\sum _{k=1}^{n}{\binom {n+1}{k}}f^{(n+1-k)}g^{(k)}+{\binom {n+1}{n+1}}f^{(0)}g^{(n+1)}\\&=\sum _{k=0}^{n+1}{\binom {n+1}{k}}f^{(n+1-k)}g^{(k)}.\end{aligned}}} And so the statement holds for
n + 1 {\displaystyle n+1} , and the proof is complete.
Multivariable calculus [ edit ] With the multi-index notation for partial derivatives of functions of several variables, the Leibniz rule states more generally:
∂ α ( f g ) = ∑ β : β ≤ α ( α β ) ( ∂ β f ) ( ∂ α − β g ) . {\displaystyle \partial ^{\alpha }(fg)=\sum _{\beta \,:\,\beta \leq \alpha }{\alpha \choose \beta }(\partial ^{\beta }f)(\partial ^{\alpha -\beta }g).} This formula can be used to derive a formula that computes the symbol of the composition of differential operators. In fact, let P and Q be differential operators (with coefficients that are differentiable sufficiently many times) and R = P ∘ Q . {\displaystyle R=P\circ Q.} Since R is also a differential operator, the symbol of R is given by:
R ( x , ξ ) = e − ⟨ x , ξ ⟩ R ( e ⟨ x , ξ ⟩ ) . {\displaystyle R(x,\xi )=e^{-{\langle x,\xi \rangle }}R(e^{\langle x,\xi \rangle }).} A direct computation now gives:
R ( x , ξ ) = ∑ α 1 α ! ( ∂ ∂ ξ ) α P ( x , ξ ) ( ∂ ∂ x ) α Q ( x , ξ ) . {\displaystyle R(x,\xi )=\sum _{\alpha }{1 \over \alpha !}\left({\partial \over \partial \xi }\right)^{\alpha }P(x,\xi )\left({\partial \over \partial x}\right)^{\alpha }Q(x,\xi ).} This formula is usually known as the Leibniz formula. It is used to define the composition in the space of symbols, thereby inducing the ring structure.
See also [ edit ] References [ edit ]