Cauchy product

In mathematics, the Cauchy product, named after Augustin Louis Cauchy, of two sequences $\textstyle (a_n)_{n\geq0}$, $\textstyle (b_n)_{n\geq0}$, is the discrete convolution of the two sequences, the sequence $\textstyle (c_n)_{n\geq0}$ whose general term is given by

$c_n=\sum_{k=0}^n a_k b_{n-k}.$

In other words, it is the sequence whose associated formal power series $\textstyle \sum_{n=0}^\infty c_nX^n$ is the product of the two series similarly associated to $(a_n)_{n\geq0}$ and $(b_n)_{n\geq0}$.

Series

A particularly important example is to consider the sequences $\textstyle a_n, b_n$ to be terms of two strictly formal (not necessarily convergent) series

$\sum_{n=0}^\infty a_n,\qquad \sum_{n=0}^\infty b_n,$

usually, of real or complex numbers. Then the Cauchy product is defined by a discrete convolution as follows.

$\left(\sum_{n=0}^\infty a_n\right) \cdot \left(\sum_{n=0}^\infty b_n\right) = \sum_{n=0}^\infty c_n,\qquad\mathrm{where}\ c_n=\sum_{k=0}^n a_k b_{n-k}$

for n = 0, 1, 2, ...

"Formal" means we are manipulating series in disregard of any questions of convergence. These need not be convergent series. See in particular formal power series.

One hopes, by analogy with finite sums, that in cases in which the two series do actually converge, the sum of the infinite series

$\sum_{n=0}^\infty c_n$

is equal to the product

$\left(\sum_{n=0}^\infty a_n\right) \left(\sum_{n=0}^\infty b_n\right)$

just as would work when each of the two sums being multiplied has only finitely many terms. This is not true in general, but see Mertens' Theorem and Cesàro's theorem below for some special cases.

Convergence and Mertens' theorem

This section is not about Mertens' theorems concerning distribution of prime numbers.

Let $\textstyle (a_n)_{n\geq0}$ and $\textstyle (b_n)_{n\geq0}$ be real sequences. It was proved by Franz Mertens that if the series $\textstyle (b_n)_{n\geq0}$ converges to B and the series $\textstyle (a_n)_{n\geq0}$ converges absolutely to A then their Cauchy product converges to AB. It is not sufficient for both series to be conditionally convergent. For example, the sequences $\textstyle a_n = b_n = (-1)^n / \sqrt{n+1}$ are conditionally convergent but their Cauchy product does not converge.

Proof of Mertens' theorem

Let $\textstyle A_n$, $\textstyle B_n$ and $\textstyle C_n$ denote the partial sums

$A_n = \sum_{i=0}^n a_i,\; B_n = \sum_{i=0}^n b_i \text{ and }C_n = \sum_{i=0}^n\sum_{k=0}^ia_kb_{i-k}.$

Then

$C_n = \sum_{i=0}^n \sum_{k=0}^i a_k b_{i-k} = \sum_{i=0}^n B_i a_{n-i}$

by rearrangement. So

$C_n = \sum_{i=0}^n(B_i-B)a_{n-i}+BA_n.$

Fix ε > 0. Since $\textstyle \sum a_n$ is absolutely convergent and $\textstyle \sum b_n$ is convergent then there exists an integer N such that for all $\textstyle n\geq N$ we have

$|B_n-B|<\frac{\varepsilon/4}{\sum_{n=0}^\infty |a_n|+1},$

and an integer M such that for all $\textstyle n\geq M$ it holds that

$|a_{n-N}|<\frac{\varepsilon}{4N\sup |B_n-B|+1}$

(since the series converges, the sequence must converge to 0). Also, there exists an integer L such that if $\textstyle n\geq L$ then

$|A_n-A|<\frac{\varepsilon/2}{|B|+1}$.

Therefore for $\textstyle n\geq\max\{L,M,N\}$ we have

$|C_n - AB| = |\sum_{i=0}^n (B_i-B)a_{n-i}+B(A_n-A)| \leq \sum_{i=0}^{N-1} |B_i-B||a_{n-i}|+\sum_{i=N}^n |B_i-B||a_{n-i}|+|B||A_n-A|<\varepsilon.$

By the definition of convergence of a series $\textstyle C_n\to AB$ as required.

Examples

Finite series

Suppose $\textstyle a_i = 0$ for all i > n and $\textstyle b_i = 0$ for all $\textstyle i><span class=$m" border="0">. Here the Cauchy product of $\textstyle \sum a_n$ and $\textstyle \sum b_n$ is readily verified to be $\textstyle (a_0+\cdots + a_n)(b_0+\cdots+b_m)$. Therefore, for finite series (which are finite sums), Cauchy multiplication is direct multiplication of those series.

Infinite series

• For some $\textstyle x,y\in\mathbb{R}$, let $\textstyle a_n = x^n/n!\,$ and $\textstyle b_n = y^n/n!\,$. Then

$c_n = \sum_{i=0}^n\frac{x^i}{i!}\frac{y^{n-i}}{(n-i)!} = \frac{1}{n!}\sum_{i=0}^n\binom{n}{i}x^i y^{n-i} = \frac{(x+y)^n}{n!}$

by definition and the binomial formula. Since, formally, $\textstyle \exp(x) = \sum a_n$ and $\textstyle \exp(y) = \sum b_n$, we have shown that $\textstyle \exp(x+y) = \sum c_n$. 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 $\textstyle \exp(x+y) = \exp(x)\exp(y)$ for all $\textstyle x,y\in\mathbb{R}$.

• As a second example, let $\textstyle a_n=b_n = 1$ for all $\textstyle n\in\mathbb{N}$. Then $\textstyle c_n = n+1$ for all $n\in\mathbb{N}$ so the Cauchy product $\textstyle \sum c_n = (1,1+2,1+2+3,1+2+3+4,\dots)$ does not converge.

Cesàro's theorem

In cases where the two sequences are convergent but not absolutely convergent, the Cauchy product is still Cesàro summable. Specifically:

If $\textstyle (a_n)_{n\geq0}$, $\textstyle (b_n)_{n\geq0}$ are real sequences with $\textstyle \sum a_n\to A$ and $\textstyle \sum b_n\to B$ then

$\frac{1}{n}\left(\sum_{i=1}^n\sum_{k=0}^i a_k b_{n-k}\right)\to AB.$

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

Theorem

For $\textstyle r>-<span class=$1" border="0"> and $\textstyle s>-<span class=$1" border="0">, suppose the sequence $\textstyle (a_n)_{n\geq0}$ is $\textstyle (C,\; r)$ summable with sum A and $\textstyle (b_n)_{n\geq0}$ is $\textstyle (C,\; s)$ summable with sum B. Then their Cauchy product is $\textstyle (C,\; r+s+1)$ summable with sum AB.

Generalizations

All of the foregoing applies to sequences in $\textstyle \mathbb{C}$ (complex numbers). The Cauchy product can be defined for series in the $\textstyle \mathbb{R}^n$ 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.

Relation to convolution of functions

One can also define the Cauchy product of doubly infinite sequences, thought of as functions on $\textstyle \Z$. In this case the Cauchy product is not always defined: for instance, the Cauchy product of the constant sequence 1 with itself, $\textstyle (\dots,1,\dots)$ is not defined. This doesn't arise for singly infinite sequences, as these have only finite sums.

One has some pairings, for instance the product of a finite sequence with any sequence, and the product $\textstyle \ell^1 \times \ell^\infty$. This is related to duality of L^p spaces.

References

• Apostol, Tom M. (1974), Mathematical Analysis (2nd ed.), Addison Wesley, p. 204, ISBN 978-0201002881 
• Hardy, G. H. (1949), Divergent Series, Oxford University Press, p. 227–229