General Leibniz rule

Generalization of the product rule in calculus
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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 jth 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

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

The formula can be generalized to the product of m differentiable functions f1,...,fm.

( 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 (k1,...,km) 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.

Proof

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

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

  • Binomial theorem – Algebraic expansion of powers of a binomial
  • Derivation (differential algebra) – Algebraic generalization of the derivative
  • Derivative – Instantaneous rate of change (mathematics)
  • Differential algebra – Algebraic study of differential equations
  • Pascal's triangle – Triangular array of the binomial coefficients in mathematics
  • Product rule – Formula for the derivative of a product
  • Quotient rule – Formula for the derivative of a ratio of functions

References

  1. ^ Olver, Peter J. (2000). Applications of Lie Groups to Differential Equations. Springer. pp. 318–319. ISBN 9780387950006.
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