en>Yobot: WP:CHECKWIKI error fixes using AWB (10093)

2014-05-05T12:47:35Z

WP:CHECKWIKI error fixes using AWB (10093)

en>Tsirel: clarification

2013-12-20T06:59:51Z

clarification

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{{Expert-subject|Mathematics|talk=Notation and references|date=May 2011}}
In [[mathematics]], the '''Cauchy product''', named after [[Augustin Louis Cauchy]], of two [[sequence]]s <math>\textstyle (a_n)_{n\geq0}</math>, <math>\textstyle (b_n)_{n\geq0}</math>, is the discrete [[convolution]] of the two sequences, the sequence <math>\textstyle (c_n)_{n\geq0}</math> whose general term is given by

:<math>c_n=\sum_{k=0}^n a_k b_{n-k}.</math>

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

==Series==
A particularly important example is to consider the sequences <math>\textstyle a_n, b_n</math> to be terms of two strictly formal (not necessarily convergent) [[series (mathematics)|series]]

:<math>\sum_{n=0}^\infty a_n,\qquad \sum_{n=0}^\infty b_n,</math>

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

:<math>\left(\sum_{n=0}^\infty a_n\right) \cdot \left(\sum_{m=0}^\infty b_m\right) = \sum_{j=0}^\infty c_j,\qquad\mathrm{where}\ c_j=\sum_{k=0}^j a_k b_{j-k}</math>
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]]

:<math>\sum_{j=0}^\infty c_j</math>

is equal to the product

:<math>\left(\sum_{n=0}^\infty a_n\right) \left(\sum_{m=0}^\infty b_m\right)</math>

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.

==Finite summations==
The product of two finite series ''ak'' and ''bk'' with ''k'' between 0 and 2''n'' satisfies the equation:

:<math>\left(\sum_{k=0}^{2n} a_k\right) \cdot \left(\sum_{k=0}^{2n} 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)</math>

==Convergence and Mertens' theorem==
{{distinguish2|[[Mertens' theorems]] concerning distribution of prime numbers}}

Let {{math|(''an'')''n''≥0}} and {{math|(''bn'')''n''≥0}} be real or complex sequences. It was proved by [[Franz Mertens]] that, if the series <math>\textstyle \sum_{n=0}^\infty a_n</math> [[Convergent series|converges]] to {{math|''A''}} and <math>\textstyle \sum_{n=0}^\infty b_n</math> converges to {{math|''B''}}, and at least one of them [[Absolute convergence|converges absolutely]], then their Cauchy product converges to {{math|''AB''}}.

It is not sufficient for both series to be [[conditional convergence|conditionally convergent]], as the following example shows.

===Example===
Consider the two [[alternating series]] with

:<math>a_n = b_n = \frac{(-1)^n}{\sqrt{n+1}}\,,</math>

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 (mathematics)|harmonic series]]). The terms of their Cauchy product are given by

:<math>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)}}</math>

for every integer {{math|''n'' ≥ 0}}. Since for every {{math|''k'' ∈ {0, 1, ..., ''n''<nowiki>}</nowiki>}} we have the inequalities {{math|''k'' + 1 ≤ ''n'' + 1}} and {{math|''n'' – ''k'' + 1 ≤ ''n'' + 1}}, it follows for the square root in the denominator that {{math|{{sqrt|(''k'' + 1)(''n'' − ''k'' + 1)}} ≤ ''n'' +1}}, hence, because there are {{math|''n'' + 1}} summands,

:<math>|c_n| \ge \sum_{k=0}^n \frac{1}{n+1} = 1</math>

for every integer {{math|''n'' ≥ 0}}. Therefore, {{math|''cn''}} does not converge to zero as {{math|''n'' → ∞}}, hence the series of the {{math|(''cn'')''n''≥0}} diverges by the [[term test]].

===Proof of Mertens' theorem===
We assume [[without loss of generality]] that the series of the {{math|(''an'')''n''≥0}} converges absolutely.
Define the [[partial sums]]

:<math>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</math>

with

:<math>c_i=\sum_{k=0}^ia_kb_{i-k}\,.</math>

Then

:<math>C_n = \sum_{i=0}^n a_{n-i}B_i</math>

by rearrangement, hence

{{NumBlk|:|<math>C_n = \sum_{i=0}^na_{n-i}(B_i-B)+A_nB\,.</math>|{{EquationRef|1}}}}

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

{{NumBlk|:|<math>|B_n-B|\le\frac{\varepsilon/3}{\sum_{k\in{\mathbb N}} |a_k|+1}</math>|{{EquationRef|2}}}}

(this is the only place where the absolute convergence is used). Since the series of the {{math|(''an'')''n''≥0}} converges, the individual {{math|''an''}} must converge to 0 by the [[term test]]. Hence there exists an integer {{math|''M''}} such that, for all integers {{math|''n'' ≥ ''M''}},

{{NumBlk|:|<math>|a_n|\le\frac{\varepsilon}{3N(\sup_{i\in\{0,\dots,N-1\}} |B_i-B|+1)}\,. </math>|{{EquationRef|3}}}}

Also, since {{math|''An''}} converges to {{math|''A''}} as {{math|''n'' → ∞}}, there exists an integer {{math|''L''}} such that, for all integers {{math|''n'' ≥ ''L''}},

{{NumBlk|:|<math>|A_n-A|\le\frac{\varepsilon/3}{|B|+1}\,.</math>|{{EquationRef|4}}}}

Then, for all integers {{math|''n'' ≥ max{''L'', ''M'' + ''N''}}}, we use the represention ({{EquationNote|1}}) for {{math|''Cn''}}, split the sum in two parts, use the [[triangle inequality]] for the [[absolute value]], and finally use the three estimates ({{EquationNote|2}}), ({{EquationNote|3}}) and ({{EquationNote|4}}) to show that

:<math>\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}</math>

By the [[Convergent series|definition of convergence of a series]], {{math|''Cn'' → ''AB''}} as required.

==Examples==

===Finite series===

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

===Infinite series===

* For some <math>\textstyle x,y\in\mathbb{R}</math>, let <math>\textstyle a_n = x^n/n!\,</math> and <math>\textstyle b_n = y^n/n!\,</math>. Then

:<math> 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!}</math>

by definition and the [[binomial formula]]. Since, [[formal series|formally]], <math>\textstyle \exp(x) = \sum a_n</math> and <math>\textstyle \exp(y) = \sum b_n</math>, we have shown that <math>\textstyle \exp(x+y) = \sum c_n</math>. Since the limit of the Cauchy product of two [[absolute convergence|absolutely convergent]] series is equal to the product of the limits of those series, we have proven the formula <math>\textstyle \exp(x+y) = \exp(x)\exp(y)</math> for all <math>\textstyle x,y\in\mathbb{R}</math>.

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

==Cesàro's theorem==


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

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

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

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

===Theorem===
For <math>\textstyle r>-1</math> and <math>\textstyle s>-1</math>, suppose the sequence <math>\textstyle (a_n)_{n\geq0}</math> is <math>\textstyle (C,\; r)</math> summable with sum ''A'' and <math>\textstyle (b_n)_{n\geq0}</math> is <math>\textstyle (C,\; s)</math> summable with sum ''B''. Then their Cauchy product is <math>\textstyle (C,\; r+s+1)</math> summable with sum ''AB''.

==Generalizations==

All of the foregoing applies to sequences in <math>\textstyle \mathbb{C}</math> ([[complex number]]s). The '''Cauchy product''' can be defined for series in the <math>\textstyle \mathbb{R}^n</math> 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 <math>\textstyle \Z</math>. In this case the Cauchy product is not always defined: for instance, the Cauchy product of the constant sequence 1 with itself, <math>\textstyle (\dots,1,\dots)</math> 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 <math>\textstyle \ell^1 \times \ell^\infty</math>.
This is related to [[Lp space#Dual spaces|duality of Lp spaces]].

==References==
*{{citation|first=Tom M.|last=Apostol|authorlink=Tom M. Apostol|title=Mathematical Analysis|edition=2nd|year=1974|publisher=Addison Wesley|isbn=978-0-201-00288-1|page=204}}
*{{citation|first=G. H.|last=Hardy|authorlink=G. H. Hardy|title=Divergent Series|year=1949|publisher=Oxford University Press|page=227–229}}

[[Category:Real analysis]]
[[Category:Complex analysis]]
[[Category:Sequences and series]]
[[Category:Articles containing proofs]]

Separable sigma algebra - Revision history

en>Yobot: WP:CHECKWIKI error fixes using AWB (10093)

en>Tsirel: clarification