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^[1]^[2]^[3]^[4]^[5]^[6]^[7]^[8]^[9]^[10]^[11] or power series.^[12]^[13] When people apply it to finite sequences^[14] 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 $\textstyle \sum_{i=0}^\infty a_i$ and $\textstyle \sum_{j=0}^\infty b_j$ be two infinite series with complex terms. The Cauchy product of these two infinite series is defined by a discrete convolution as follows:

\left(\sum_{i=0}^\infty a_i\right) \cdot \left(\sum_{j=0}^\infty b_j\right) = \sum_{k=0}^\infty c_k

where

c_k=\sum_{l=0}^k a_l b_{k-l}

Cauchy product of two power series

Consider the following two power series

\sum_{i=0}^\infty a_i x^i

and

\sum_{j=0}^\infty b_j x^j

with complex coefficients $\{a_i\}$ and $\{b_j\}$ . The Cauchy product of these two power series is defined by a discrete convolution as follows:

\left(\sum_{i=0}^\infty a_i x^i\right) \cdot \left(\sum_{j=0}^\infty b_j x^j\right) = \sum_{k=0}^\infty c_k x^k

where

c_k=\sum_{l=0}^k a_l b_{k-l}

Convergence and Mertens' theorem

Not to be confused with Mertens' theorems concerning distribution of prime numbers.

Let $(a n) n \geq0$ and $(b n) n \geq0$ be real or complex sequences. It was proved by Franz Mertens that, if the series $\textstyle \sum_{n=0}^\infty a_n$ converges to $A$ and $\textstyle \sum_{n=0}^\infty b_n$ 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

a_n = b_n = \frac{(-1)^n}{\sqrt{n+1}}\,,

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

c_n = \sum_{k=0}^n \frac{(-1)^k}{\sqrt{k+1}} \cdot \frac{ (-1)^{n-k} }{ \sqrt{n-k+1} } = (-1)^n \sum_{k=0}^n \frac{1}{ \sqrt{(k+1)(n-k+1)} }

for every integer $n \geq 0$ . Since for every $k \in {0, 1, ..., n}$ we have the inequalities $k + 1 \leq n + 1$ and $n - k + 1 \leq n + 1$ , it follows for the square root in the denominator that $\sqrt (k + 1)(n - k + 1) \leq n +1$ , hence, because there are $n + 1$ summands,

|c_n| \ge \sum_{k=0}^n \frac{1}{n+1} \ge 1

for every integer $n \geq 0$ . Therefore, $c n$ does not converge to zero as $n \to \infty$ , hence the series of the $(c n) n \geq0$ diverges by the term test.

Proof of Mertens' theorem

Assume without loss of generality that the series $\textstyle \sum_{n=0}^\infty a_n$ converges absolutely. Define the partial sums

A_n = \sum_{i=0}^n a_i,\quad B_n = \sum_{i=0}^n b_i\quad\text{and}\quad C_n = \sum_{i=0}^n c_i

with

c_i=\sum_{k=0}^ia_kb_{i-k}\,.

Then

C_n = \sum_{i=0}^n a_{n-i}B_i

by rearrangement, hence

$C_n = \sum_{i=0}^na_{n-i}(B_i-B)+A_nB\,.$

(1)

Fix $ε > 0$ . Since $\textstyle \sum_{k\in{\mathbb N}} |a_k|<\infty$ by absolute convergence, and since $B n$ converges to $B$ as $n \to \infty$ , there exists an integer $N$ such that, for all integers $n \geq N$ ,

$|B_n-B|\le\frac{\varepsilon/3}{\sum_{ k\in{\mathbb N} } |a_k|+1}$

(2)

(this is the only place where the absolute convergence is used). Since the series of the $(a n) n \geq0$ converges, the individual $a n$ must converge to 0 by the term test. Hence there exists an integer $M$ such that, for all integers $n \geq M$ ,

$|a_n|\le\frac{\varepsilon}{3N(\sup_{ i\in\{0,\dots,N-1\} } |B_i-B|+1)}\,.$

(3)

Also, since $A n$ converges to $A$ as $n \to \infty$ , there exists an integer $L$ such that, for all integers $n \geq L$ ,

$|A_n-A|\le\frac{\varepsilon/3}{|B|+1}\,.$

(4)

Then, for all integers $n \geq max{L, M + N}$ , 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

\begin{align} |C_n - AB| &= \biggl|\sum_{i=0}^n a_{n-i}(B_i-B)+(A_n-A)B\biggr| \\ &\le \sum_{i=0}^{N-1}\underbrace{|a_{\underbrace{\scriptstyle n-i}_{\scriptscriptstyle \ge M}}|\,|B_i-B|}_{\le\,\varepsilon/(3N)\text{ by (3)}}+{}\underbrace{\sum_{i=N}^n |a_{n-i}|\,|B_i-B|}_{\le\,\varepsilon/3\text{ by (2)}}+{}\underbrace{|A_n-A|\,|B|}_{\le\,\varepsilon/3\text{ by (4)}}\le\varepsilon\,. \end{align}

By the definition of convergence of a series, $C n \to AB$ 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 $\textstyle (a_n)_{n\geq0}$ , $\textstyle (b_n)_{n\geq0}$ are real sequences with $\textstyle \sum a_n\to A$ and $\textstyle \sum b_n\to B$ then

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

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

Theorem

For $\textstyle r>-1$ and $\textstyle s>-1$ , 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.

Examples

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.

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.

Products of finitely many infinite series

Let $n \in \mathbb N$ such that $n \ge 2$ (actually the following is also true for $n=1$ but the statement becomes trivial in that case) and let $\sum_{k_1 = 0}^\infty a_{1, k_1}, \ldots, \sum_{k_n = 0}^\infty a_{n, k_n}$ 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

\sum_{k_1 = 0}^\infty \sum_{k_2 = 0}^{k_1} \cdots \sum_{k_n = 0}^{k_{n-1}} a_{1, k_n} a_{2, k_{n-1} - k_n} \cdots a_{n, k_1 - k_2}

converges and we have:

\sum_{k_1 = 0}^\infty \sum_{k_2 = 0}^{k_1} \cdots \sum_{k_n = 0}^{k_{n-1}} a_{1, k_n} a_{2, k_{n-1} - k_n} \cdots a_{n, k_1 - k_2} = \prod_{j=1}^n \left( \sum_{k_j = 0}^\infty a_{j, k_j} \right)

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 $n \in \mathbb N$ such that $n \ge 2$ , and let $\sum_{k_1 = 0}^\infty a_{1, k_1}, \ldots, \sum_{k_{n+1} = 0}^\infty a_{n+1, k_{n+1}}$ 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 $\sum_{k_1 = 0}^\infty |a_{1, k_1}|, \ldots, \sum_{k_n = 0}^\infty |a_{n, k_n}|$ . We obtain that the series

\sum_{k_1 = 0}^\infty \sum_{k_2 = 0}^{k_1} \cdots \sum_{k_n = 0}^{k_{n-1}} |a_{1, k_n} a_{2, k_{n-1} - k_n} \cdots a_{n, k_1 - k_2}|

converges, and hence, by the triangle inequality and the sandwich criterion, the series

\sum_{k_1 = 0}^\infty \left| \sum_{k_2 = 0}^{k_1} \cdots \sum_{k_n = 0}^{k_{n-1}} a_{1, k_n} a_{2, k_{n-1} - k_n} \cdots a_{n, k_1 - k_2} \right|

converges, and hence the series

\sum_{k_1 = 0}^\infty \sum_{k_2 = 0}^{k_1} \cdots \sum_{k_n = 0}^{k_{n-1}} a_{1, k_n} a_{2, k_{n-1} - k_n} \cdots a_{n, k_1 - k_2}

converges absolutely. Therefore, by the induction hypothesis, by what Mertens proved, and by renaming of variables, we have:

\begin{align} \prod_{j=1}^{n+1} \left( \sum_{k_j = 0}^\infty a_{j, k_j} \right) & = \left( \sum_{k_{n+1} = 0}^\infty \overbrace{a_{n+1, k_{n+1}}}^{=:a_{k_{n+1}}} \right) \left( \sum_{k_1 = 0}^\infty \overbrace{\sum_{k_2 = 0}^{k_1} \cdots \sum_{k_n = 0}^{k_{n-1}} a_{1, k_n} a_{2, k_{n-1} - k_n} \cdots a_{n, k_1 - k_2}}^{=:b_{k_1}} \right) \\ & = \sum_{k_1 = 0}^\infty \sum_{k_2 = 0}^{k_1} a_{n+1, k_1 - k_2} \sum_{k_3 = 0}^{k_2} \cdots \sum_{k_{n+1} = 0}^{k_n} a_{1, k_{n+1}} a_{2, k_n - k_{n+1}} \cdots a_{n, k_2 - k_3} \end{align}

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 $f: \mathbb{N} \to \mathbb{C}$ with finite support. For any complex-valued functions f, g on $\mathbb{N}$ with finite support, one can take the convolution of them:

(f * g)(n) = \sum_{i + j = n} f(i) g(j)

Then $\sum (f *g)(n)$ is the same thing as the Cauchy product of $\sum f(n)$ and $\sum g(n)$ .

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

Notes

↑ Canuto & Tabacco 2015, p. 20.
↑ Bloch 2011, p. 463.
↑ Friedman & Kandel 2011, p. 204.
↑ Ghorpade & Limaye 2006, p. 416.
↑ Hijab 2011, p. 43.
↑ Montesinos, Zizler & Zizler 2015, p. 98.
↑ Oberguggenberger & Ostermann 2011, p. 322.
↑ Pedersen 2015, p. 210.
↑ Ponnusamy 2012, p. 200.
↑ Pugh 2015, p. 210.
↑ Sohrab 2014, p. 73.
↑ Canuto & Tabacco 2015, p. 53.
↑ Mathonline, Cauchy Product of Power Series.
↑ Weisstein, Cauchy Product.

References

Apostol, Tom M. (1974), Mathematical Analysis (2nd ed.), Addison Wesley, p. 204, ISBN 978-0-201-00288-1 .

Bloch, Ethan D. (2011), The Real Numbers and Real Analysis, Springer .

Canuto, Claudio; Tabacco, Anita (2015), Mathematical Analysis II (2nd ed.), Springer .

Friedman, Menahem; Kandel, Abraham (2011), Calculus Light, Springer .

Ghorpade, Sudhir R.; Limaye, Balmohan V. (2006), A Course in Calculus and Real Analysis, Springer .

Hardy, G. H. (1949), Divergent Series, Oxford University Press, p. 227–229 .

Hijab, Omar (2011), Introduction to Calculus and Classical Analysis (3rd ed.), Springer .

Mathonline, Cauchy Product of Power Series .

Montesinos, Vicente; Zizler, Peter; Zizler, Václav (2015), An Introduction to Modern Analysis, Springer .

Oberguggenberger, Michael; Ostermann, Alexander (2011), Analysis for Computer Scientists, Springer .

Pedersen, Steen (2015), From Calculus to Analysis, Springer .

Ponnusamy, S. (2012), Foundations of Mathematical Analysis, Birkhäuser .

Pugh, Charles C. (2015), Real Mathematical Analysis (2nd ed.), Springer .

Sohrab, Houshang H. (2014), Basic Real Analysis (2nd ed.), Birkhäuser .

Weisstein, Eric W., "Cauchy Product", From MathWorld--A Wolfram Web Resource .

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