Cauchy product

Definitions

The Cauchy product may apply to infinite series or power series. When people apply it to finite sequences or finite series, it is by abuse of language: they actually refer to discrete convolution.

Convergence issues are discussed in the next section.

Cauchy product of two infinite series

Let ∑ i = 0 ∞ a i and ∑ j = 0 ∞ b j 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 { a i } and { b j } . The Cauchy product of these two power series is defined by a discrete convolution as follows:

Convergence and Mertens' theorem

Let (a_n)_n≥0 and (b_n)_n≥0 be real or complex sequences. It was proved by Franz Mertens that, if the series ∑ n = 0 ∞ a n converges to A and ∑ n = 0 ∞ b n converges to B, and at least one of them converges absolutely, then their Cauchy product converges to AB.

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 divergence of the series of the absolute values follows from the direct comparison test and the divergence of the harmonic series). The terms of their Cauchy product are given by

for every integer n ≥ 0. Since for every k ∈ {0, 1, ..., n} we have the inequalities k + 1 ≤ n + 1 and n – k + 1 ≤ n + 1, it follows for the square root in the denominator that √(k + 1)(n − k + 1) ≤ n +1, hence, because there are n + 1 summands,

for every integer n ≥ 0. Therefore, c_n does not converge to zero as n → ∞, hence the series of the (c_n)_n≥0 diverges by the term test.

Proof of Mertens' theorem

Assume without loss of generality that the series ∑ n = 0 ∞ a n converges absolutely. Define the partial sums

with

Then

by rearrangement, hence

Fix ε > 0. Since ∑ k ∈ N | a k | < ∞ by absolute convergence, and since B_n converges to B as n → ∞, there exists an integer N such that, for all integers n ≥ N,

(this is the only place where the absolute convergence is used). Since the series of the (a_n)_n≥0 converges, the individual a_n must converge to 0 by the term test. Hence there exists an integer M such that, for all integers n ≥ M,

Also, since A_n converges to A as n → ∞, there exists an integer L such that, for all integers n ≥ L,

Then, for all integers n ≥ max{L, M + N}, use the representation (1) for C_n, split the sum in two parts, use the triangle inequality for the absolute value, and finally use the three estimates (2), (3) and (4) to show that

By the definition of convergence of a series, C_n → AB 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 ( a n ) n ≥ 0 , ( b n ) n ≥ 0 are real sequences with ∑ a n → A and ∑ b n → B then

This can be generalised to the case where the two sequences are not convergent but just Cesàro summable:

Theorem

For r > − 1 and s > − 1 , suppose the sequence ( a n ) n ≥ 0 is ( C , r ) summable with sum A and ( b n ) n ≥ 0 is ( C , s ) summable with sum B. Then their Cauchy product is ( C , r + s + 1 ) summable with sum AB.

Examples

by definition and the binomial formula. Since, formally, exp ⁡ ( x ) = ∑ a n and exp ⁡ ( y ) = ∑ b n , we have shown that exp ⁡ ( x + y ) = ∑ c n . 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 exp ⁡ ( x + y ) = exp ⁡ ( x ) exp ⁡ ( y ) for all x , y ∈ R .

Generalizations

All of the foregoing applies to sequences in C (complex numbers). The Cauchy product can be defined for series in the R n spaces (Euclidean 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 n ∈ N such that n ≥ 2 (actually the following is also true for n = 1 but the statement becomes trivial in that case) and let ∑ k 1 = 0 ∞ a 1 , k 1 , … , ∑ k n = 0 ∞ a n , k n be infinite series with complex coefficients, from which all except the n th one converge absolutely, and the n th one converges. Then the series

converges and we have:

This statement can be proven by induction over n : The case for n = 2 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 n ∈ N such that n ≥ 2 , and let ∑ k 1 = 0 ∞ a 1 , k 1 , … , ∑ k n + 1 = 0 ∞ a n + 1 , k n + 1 be infinite series with complex coefficients, from which all except the n + 1 th one converge absolutely, and the n + 1 th one converges. We first apply the induction hypothesis to the series ∑ k 1 = 0 ∞ | a 1 , k 1 | , … , ∑ k n = 0 ∞ | a n , k n | . 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 n + 1 .

Relation to convolution of functions

A finite sequence can be viewed as an infinite sequence with only finitely many nonzero terms. A finite sequence can be viewed as a function f : N → C with finite support. For any complex-valued functions f, g on N with finite support, one can take the convolution of them:

Then ∑ ( f ∗ g ) ( n ) is the same thing as the Cauchy product of ∑ f ( n ) and ∑ g ( n ) .

More generally, given a unital semigroup S, one can form the semigroup algebra C [ S ] of S, with, as usual, the multiplication given by convolution. If one takes, for example, S = N d , then the multiplication on C [ S ] is a sort of a generalization of the Cauchy product to higher dimension.