Cauchy product


In mathematics, more specifically in mathematical analysis, the Cauchy product is the discrete convolution of two infinite series. It is named after the French mathematician Augustin-Louis Cauchy.

Definitions

The Cauchy product may apply to infinite series or power series. When people apply it to finite sequences or finite series, that can be seen merely as a particular case of a product of series with a finite number of non-zero coefficients.
Convergence issues are discussed in the [|next section].

Cauchy product of two infinite series

Let and be two infinite series with complex terms. The Cauchy product of these two infinite series is defined by a discrete convolution as follows:

Cauchy product of two power series

Consider the following two power series
with complex coefficients and. The Cauchy product of these two power series is defined by a discrete convolution as follows:

Convergence and Mertens' theorem

Let and be real or complex sequences. It was proved by Franz Mertens that, if the series converges to and converges to, and at least one of them converges absolutely, then their Cauchy product converges to. The theorem is still valid in a Banach algebra.
It is not sufficient for both series to be convergent; if both sequences are conditionally convergent, the Cauchy product does not have to converge towards the product of the two series, as the following example shows:

Example

Consider the two alternating series with
which are only conditionally convergent. The terms of their Cauchy product are given by
for every integer. Since for every we have the inequalities and, it follows for the square root in the denominator that, hence, because there are summands,
for every integer. Therefore, does not converge to zero as, hence the series of the diverges by the term test.

Proof of Mertens' theorem

For simplicity, we will prove it for complex numbers. However, the proof we are about to give is formally identical for an arbitrary Banach algebra.
Assume without loss of generality that the series converges absolutely.
Define the partial sums
with
Then
by rearrangement, hence
Fix. Since by absolute convergence, and since converges to as, there exists an integer such that, for all integers,
. Since the series of the converges, the individual must converge to 0 by the term test. Hence there exists an integer such that, for all integers,
Also, since converges to as, there exists an integer such that, for all integers,
Then, for all integers, use the representation for, split the sum in two parts, use the triangle inequality for the absolute value, and finally use the three estimates, and to show that
By the definition of convergence of a series, as required.

Cesàro's theorem

In cases where the two sequences are convergent but not absolutely convergent, the Cauchy product is still Cesàro summable. Specifically:
If, are real sequences with and then
This can be generalised to the case where the two sequences are not convergent but just Cesàro summable:

Theorem

For and, suppose the sequence is summable with sum A and is summable with sum B. Then their Cauchy product is summable with sum AB.

Examples

  • For some, let and. Then by definition and the binomial formula. Since, formally, and, we have shown that. Since the limit of the Cauchy product of two absolutely convergent series is equal to the product of the limits of those series, we have proven the formula for all.
  • As a second example, let for all. Then for all so the Cauchy product does not converge.

    Generalizations

All of the foregoing applies to sequences in . The Cauchy product can be defined for series in the spaces where multiplication is the inner product. In this case, we have the result that if two series converge absolutely then their Cauchy product converges absolutely to the inner product of the limits.

Products of finitely many infinite series

Let such that and let be infinite series with complex coefficients, from which all except the th one converge absolutely, and the th one converges. Then the limit
exists and we have:

Proof

Because
the statement can be proven by induction over : The case for is identical to the claim about the Cauchy product. This is our induction base.
The induction step goes as follows: Let the claim be true for an such that, and let be infinite series with complex coefficients, from which all except the th one converge absolutely, and the -th one converges. We first apply the induction hypothesis to the series. We obtain that the series
converges, and hence, by the triangle inequality and the sandwich criterion, the series
converges, and hence the series
converges absolutely. Therefore, by the induction hypothesis, by what Mertens proved, and by renaming of variables, we have:
Therefore, the formula also holds for.

Relation to convolution of functions

A finite sequence can be viewed as an infinite sequence with only finitely many nonzero terms, or in other words as a function with finite support. For any complex-valued functions f, g on with finite support, one can take their convolution:
Then is the same thing as the Cauchy product of and.
More generally, given a monoid S, one can form the semigroup algebra of S, with the multiplication given by convolution. If one takes, for example,, then the multiplication on is a generalization of the Cauchy product to higher dimension.