Binomial formula for powers of a derivation

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Suppose R is a Non-associative ring (?) (i.e., a not necessarily associative ring -- this includes the associative ring, Lie ring, and other cases) and d is a derivation of R. For any nonnegative integer n, if we denote by d^n the composition of d with itself n times, with d^0 defined as the identity map.

We then have:

\! d^n(x * y) = \sum_{i=0}^n \binom{n}{i} \left(d^i(x) * d^{n-i}(y)\right)

Note that this in particular applies to the case of a Derivation of a Lie ring (?) and a Derivation of an associative ring (?).

Related facts



Proof idea

The idea is to use the Leibniz rule (which establishes the case n = 1; n = 0 is vacuous) and prove by induction. The proof is almost exactly like the proof of the binomial formula for (x + y)^n for a commutative associative algebra.

Proof details: case n = 0

In the case n = 0, both sides are x * y, so the equality holds vacuously.

Proof details: case n = 1

In this case, the left side is d(x * y). The right side is:

\! \binom{1}{0} (d^0(x) * d^1(y)) + \binom{1}{1}(d^{1}(x) * d^{0}(y)) = (x * dy) + (dx * y)

By the Leibniz rule, the left side equals the right side.

Proof details: inductive step

Inductive hypothesis: \! d^{n-1}(x * y) = \sum_{i=0}^{n-1} \binom{n - 1}{i} d^i(x) * d^{n-1-i}(y)

To prove: \! d^{n}(x * y) = \sum_{i=0}^{n} \binom{n}{i} d^i(x) * d^{n-1-i}(y)

Proof: We have by definition:

\! d^n(x * y) = d\left(d^{n-1}(x * y)\right)

By the inductive hypothesis, we can expand the inside and we get:

\! d^n(x * y) = d\left\{\sum_{i=0}^{n-1} \binom{n - 1}{i} d^i(x) * d^{n-1-i}(y)\right\}

Since d is additive, we can pull it inside the summation on the right side and get:

\! d^n(x * y) = \sum_{i=0}^{n-1}\binom{n - 1}{i}d\left\{d^i(x) * d^{n-1-i}(y)\right\}

Now using the Leibniz rule on the inside, we get:

\! d^n(x * y) = \sum_{i=0}^{n-1}\binom{n - 1}{i}\left\{(d^{i+1}(x) * d^{n-(i+1)}(y)) + (d^i(x) * d^{n-i}(y))\right\}

We now rearrange the summation and obtain:

\! d^n(x * y) = \sum_{i=0}^n \left(\binom{n - 1}{i} + \binom{n - 1}{i-1}\right)(d^i(x) * d^{n-i}(y))

We use Pascal's identity on the sum of binomial coefficients and obtain what we want:

\! d^{n}(x * y) = \sum_{i=0}^{n} \binom{n}{i} d^i(x) * d^{n-1-i}(y)