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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.

Contents

DefinitionsEdit

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 seriesEdit

Let   and   be two infinite series with complex terms. The Cauchy product of these two infinite series is defined by a discrete convolution as follows:

      where      .

Cauchy product of two power seriesEdit

Consider the following two power series

      and      

with complex coefficients   and  . The Cauchy product of these two power series is defined by a discrete convolution as follows:

      where      .

Convergence and Mertens' theoremEdit

Let (an)n≥0 and (bn)n≥0 be real sequences. If   and  , and at least one of them converges absolutely, then their Cauchy product converges to AB. (Rudin, Principles of Mathematical Analysis, Theorem 3.50)


Let (an)n≥0 and (bn)n≥0 be real or complex sequences. It was proved by Franz Mertens[citation needed] that, if the series   converges to A and   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:

ExampleEdit

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 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,

 

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' theoremEdit

Assume without loss of generality that the series   converges absolutely. Define the partial sums

 

with

 

Then

 

by rearrangement, hence

 

 

 

 

 

(1)

Fix ε > 0. Since   by absolute convergence, and since Bn converges to B as n → ∞, there exists an integer N such that, for all integers nN,

 

 

 

 

 

(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,

 

 

 

 

 

(3)

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

 

 

 

 

 

(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

 

By the definition of convergence of a series, CnAB as required.

Cesàro's theoremEdit

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

If  ,   are real sequences with   and   then

 

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

TheoremEdit

For   and  , suppose the sequence   is   summable with sum A and   is   summable with sum B. Then their Cauchy product is   summable with sum AB.

ExamplesEdit

  • For some  , let   and  . Then
 
by definition and the binomial formula. Since, formally,   and  , we have shown that  . 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   for all  .
  • As a second example, let   for all  . Then   for all   so the Cauchy product   does not converge.

GeneralizationsEdit

All of the foregoing applies to sequences in   (complex numbers). The Cauchy product can be defined for series in the   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 seriesEdit

Let   such that   (actually the following is also true for   but the statement becomes trivial in that case) and let   be infinite series with complex coefficients, from which all except the  th one converge absolutely, and the  th one converges. Then the series

 

converges and we have:

 

This statement can be proven by induction over  : The case for   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   such that  , and let   be infinite series with complex coefficients, from which all except the  th one converge absolutely, and the  th one converges. We first apply the induction hypothesis to the series  . 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  .

Relation to convolution of functionsEdit

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   with finite support. For any complex-valued functions f, g on   with finite support, one can take their convolution:

 

Then   is the same thing as the Cauchy product of   and  .

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

NotesEdit

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

ReferencesEdit

  • Apostol, Tom M. (1974), Mathematical Analysis (2nd ed.), Addison Wesley, p. 204, ISBN 978-0-201-00288-1.
  • Canuto, Claudio; Tabacco, Anita (2015), Mathematical Analysis II (2nd ed.), 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.
  • Pugh, Charles C. (2015), Real Mathematical Analysis (2nd ed.), Springer.
  • Sohrab, Houshang H. (2014), Basic Real Analysis (2nd ed.), Birkhäuser.