Cauchy product

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

Definitions[edit]

The Cauchy product may apply to finite sequences,[1][2] infinite sequences, finite series,[3] infinite series,[4][5][6][7][8][9][10][11][12][13][14] power series,[15][16] etc. Convergence issues are discussed further down in the sections on Mertens' theorem and Cesàro's theorem.

Cauchy product of two finite sequences[edit]

Let \{a_i\} and \{b_j\} be two finite sequences of complex numbers with the same length n. The Cauchy product of these two finite sequences is equal to the Cauchy product of the finite series \textstyle \sum_{i=0}^{n} a_i and \textstyle \sum_{j=0}^{n} b_j.

Cauchy product of two infinite sequences[edit]

Let \{a_i\} and \{b_j\} be two infinite sequences of complex numbers. The Cauchy product of these two infinite sequences is equal to the Cauchy product of the infinite series \textstyle \sum_{i=0}^\infty a_i and \textstyle \sum_{j=0}^\infty b_j.

Cauchy product of two finite series[edit]

Let \textstyle \sum_{i=0}^{n} a_i and \textstyle \sum_{j=0}^{n} b_j be two finite series with complex terms. The Cauchy product of these two finite series is defined by a discrete convolution as follows:

\left(\sum_{i=0}^{n} a_i\right) \cdot \left(\sum_{j=0}^{n} b_j\right) = \sum_{k=0}^{n} c_k\qquad\mathrm{where}\ c_k=\sum_{l=0}^k a_l b_{k-l}.

Cauchy product of two infinite series[edit]

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\qquad\mathrm{where}\ c_k=\sum_{l=0}^k a_l b_{k-l}.

Cauchy product of two power series[edit]

Consider the following two power series with complex coefficients \{a_i\} and \{b_j\}:

\sum_{i=0}^\infty a_i x^i \qquad \mathrm{and} \qquad \sum_{j=0}^\infty b_j x^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\qquad\mathrm{where}\ c_k=\sum_{l=0}^k a_l b_{k-l}.

If these power series are formal power series, then we are manipulating series in disregard of any question of convergence: they need not be convergent series. Otherwise, see Mertens' theorem and Cesàro's theorem below for convergence criteria.

Property[edit]

Let \textstyle \sum_{i=0}^{n} a_i and \textstyle \sum_{j=0}^{n} b_j be two finite series with complex terms. The product of these two finite series satisfies the equation:

\left(\sum_{k=0}^{n} a_k\right) \cdot \left(\sum_{k=0}^{n} b_k\right)=\sum_{k=0}^{2n} \sum_{i=0}^k a_ib_{k-i} - \sum_{k=0}^{n-1} \left(a_k \sum_{i=n+1}^{2n-k}b_i +b_k \sum_{i=n+1}^{2n-k} a_i\right).

Convergence and Mertens' theorem[edit]

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

Let (an)n≥0 and (bn)n≥0 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[edit]

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 ≥ 0. Since for every k ∈ {0, 1, ..., n} we have the inequalities k + 1 ≤ n + 1 and nk + 1 ≤ n + 1, it follows for the square root in the denominator that (k + 1)(nk + 1)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 ≥ 0. Therefore, cn does not converge to zero as n → ∞, hence the series of the (cn)n≥0 diverges by the term test.

Proof of Mertens' theorem[edit]

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 Bn converges to B as n → ∞, there exists an integer N such that, for all integers nN,

|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 (an)n≥0 converges, the individual an must converge to 0 by the term test. Hence there exists an integer M such that, for all integers nM,

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

 

 

 

 

(3)

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

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

 

 

 

 

(4)

Then, for all integers n ≥ max{L, M + N}, use the representation (1) for Cn, 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, CnAB as required.

Examples[edit]

Finite series[edit]

Suppose \textstyle a_i = 0 for all i>n and \textstyle b_i = 0 for all \textstyle i>m. 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[edit]

  • 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[edit]

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[edit]

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.

Generalizations[edit]

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[edit]

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 nth one converge absolutely, and the nth 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+1th one converge absolutely, and the n+1th 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[edit]

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 Lp spaces.

Notes[edit]

  1. ^ Dyer & Edmunds 2014, p. 190.
  2. ^ Weisstein, Cauchy Product.
  3. ^ Oberguggenberger & Ostermann 2011, p. 322.
  4. ^ Canuto & Tabacco 2015, p. 20.
  5. ^ Bloch 2011, p. 463.
  6. ^ Friedman & Kandel 2011, p. 204.
  7. ^ Ghorpade & Limaye 2006, p. 416.
  8. ^ Hijab 2011, p. 43.
  9. ^ Montesinos, Zizler & Zizler 2015, p. 98.
  10. ^ Oberguggenberger & Ostermann 2011, p. 322.
  11. ^ Pedersen 2015, p. 210.
  12. ^ Ponnusamy 2012, p. 200.
  13. ^ Pugh 2015, p. 210.
  14. ^ Sohrab 2014, p. 73.
  15. ^ Canuto & Tabacco 2015, p. 53.
  16. ^ Mathonline, Cauchy Product of Power Series.

References[edit]

  • Bloch, Ethan D. (2011), The Real Numbers and Real Analysis, Springer .
  • Canuto, Claudio; Tabacco, Anita (2015), Mathematical Analysis II (2nd ed.), Springer .
  • Dyer, R.H.; Edmunds, D.E. (2014), From Real to Complex Analysis, Springer .
  • Friedman, Menahem; Kandel, Abraham (2011), Calculus Light, Springer .
  • Ghorpade, Sudhir R.; Limaye, Balmohan V. (2006), A Course in Calculus and Real Analysis, Springer .
  • Hijab, Omar (2011), Introduction to Calculus and Classical Analysis (3rd ed.), Springer .
  • 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 .