Cauchy product

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In mathematics, the Cauchy product, named in honor of Augustin Louis Cauchy, of two sequences $a n, b n$ , is the discrete convolution

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

That is, it is the product of the sequences, considered as elements of the semigroup ring of the natural numbers $R[\N]$ .

1 Series
2 Examples
- 2.1 Finite series
- 2.2 Infinite series
3 Convergence and Mertens' theorem
- 3.1 Proof of Mertens' theorem
4 Cesàro's theorem
5 Generalizations
6 Relation to convolution of functions

[edit] Series

A particularly important example is to consider the sequences $a n, b n$ 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, is defined by a discrete convolution as follows. Then the Cauchy product is

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

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.

In sufficiently well-behaved cases, this works. But—and this is an important point—the Cauchy product of two sequences exists even when either or both of the corresponding infinite series fails to converge.

[edit] Examples

[edit] Finite series

$x i = 0$ for all $i > n$ and $y i = 0$ for all $i > m$ . Here the Cauchy product of $\sum x$ and $\sum y$ is readily verified to be $(x_0+\cdots + x_n)(y_0+\dots+y_m)$ . Therefore, for finite series (which are finite sums), Cauchy multiplication is direct multiplication of those series.

[edit] Infinite series

For some $a,b\in\mathbb{R}$ , let $x_n = a^n/n!\,$ and $y_n = b^n/n!\,$ . Then

$C(x,y)(n) = \sum_{i=0}^n\frac{a^i}{i!}\frac{b^{n-i}}{(n-i)!} = \frac{(a+b)^n}{n!}$

by definition and the binomial formula. Since, formally, $\exp(a) = \sum x$ and $\exp(b) = \sum y$ , we have shown that $\exp(a+b) = \sum C(x,y)$ . Since the limit of the Cauchy product of two absolutely convergent series is equal to the product of the limits of those series (see below), we have proven the formula $exp(a + b) = exp(a)exp(b)$ for all $a,b\in\mathbb{R}$ .

As a second example, let $x (n) = 1$ for all $n\in\mathbb{N}$ . Then $C (x, x)(n) = n + 1$ for all $n\in\mathbb{N}$ so the Cauchy product $\sum C(x,x) = (1,1+2,1+2+3,1+2+3+4,\dots)$ and it does not converge.

[edit] Convergence and Mertens' theorem

Let x, y be real sequences. It was proved by Franz Mertens that if the series $\sum y$ converges to Y and the series $\sum x$ converges absolutely to X then their Cauchy product $\sum C(x,y)$ converges to XY. It is not sufficient for both series to be conditionally convergent. For example, the sequence $x n = ( - 1) n / n$ generates a conditionally convergent series but the sequence $C (x, x)$ does not converge to 0. Here is a proof.

[edit] Proof of Mertens' theorem

Let $X_n = \sum_{i=0}^n x_i$ , $Y_n = \sum_{i=0}^n y_i$ and $C_n = \sum_{i=0}^n C(x,y)(i)$ . Then $C_n = \sum_{i=0}^n \sum_{k=0}^i x_k y_{i-k} = \sum_{i=0}^n Y_i x_{n-i}$ by rearrangement. So $C_n = \sum_{i=0}^n(Y_i-Y)x_{n-i}+YX_n$ . Fix ε > 0. Since $\sum x$ is absolutely convergent and $\sum y$ is convergent then there exists an integer N such that for all n ≥ N $|Y_n-Y|<\frac{\varepsilon/4}{\sum_{n=0}^\infty |x_n|+1}$ and an integer M such that for all $n\geq M$ $|x_{n-N}|<\frac{\varepsilon}{4N\sup |Y_n-Y|+1}$ (since the series converges, the sequence must converge to 0). Also, there exists an integer L such that if $n\geq L$ then $|X_n-X|<\frac{\varepsilon/2}{|Y|+1}$ . Therefore,

$|C_n - XY| = |\sum_{i=0}^n (Y_i-Y)x_{n-i}+Y(X_n-X)| \leq \sum_{i=0}^{N-1} |Y_i-Y||x_{n-i}|+\sum_{i=N}^n |Y_i-Y||x_{n-i}|+|Y||X_n-X|<\varepsilon$

for all integers n larger than N, M, and L. By the definition of convergence of a series $\sum C(x,y)\to XY.$

[edit] Cesàro's theorem

If x and y are real sequences and $\sum x\to A$ and $\sum y\to B$ then $\frac{1}{n}\left(\sum_{i=0}^n C(x,y)_n\right)\to AB.$

[edit] Generalizations

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

[edit] Relation to convolution of functions

One can also define the Cauchy product of doubly-infinite sequences, thought of as functions on $\Z$ . In this case the Cauchy product is not always defined: for instance, the Cauchy product of the constant sequence 1 with itself, $(\dots,1,\dots)$ 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 $\ell^1 \times \ell^\infty$ . This is related to duality of L^p spaces.