Английская Википедия:Cauchy product
Шаблон:Short description 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]Шаблон:Excessive citations inline or power series.[12][13] When people apply it to finite sequences[14] or finite series, that can be seen merely as a particular case of a product of series with a finite number of non-zero coefficients (see discrete convolution).
Convergence issues are discussed in the next section.
Cauchy product of two infinite series
Let <math display="inline"> \sum_{i=0}^\infty a_i</math> and <math display="inline"> \sum_{j=0}^\infty b_j</math> be two infinite series with complex terms. The Cauchy product of these two infinite series is defined by a discrete convolution as follows:
- <math>\left(\sum_{i=0}^\infty a_i\right) \cdot \left(\sum_{j=0}^\infty b_j\right) = \sum_{k=0}^\infty c_k</math> where <math>c_k=\sum_{l=0}^k a_l b_{k-l}</math>.
Cauchy product of two power series
Consider the following two power series
- <math>\sum_{i=0}^\infty a_i x^i</math> and <math>\sum_{j=0}^\infty b_j x^j</math>
with complex coefficients <math>\{a_i\}</math> and <math>\{b_j\}</math>. The Cauchy product of these two power series is defined by a discrete convolution as follows:
- <math>\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</math> where <math>c_k=\sum_{l=0}^k a_l b_{k-l}</math>.
Convergence and Mertens' theorem
Let Шаблон:Math and Шаблон:Math be real or complex sequences. It was proved by Franz Mertens that, if the series <math display="inline"> \sum_{n=0}^\infty a_n</math> converges to Шаблон:Math and <math display="inline"> \sum_{n=0}^\infty b_n</math> converges to Шаблон:Math, and at least one of them converges absolutely, then their Cauchy product converges to Шаблон:Math.[15] The theorem is still valid in a Banach algebra (see first line of the following proof).
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
<math display="block">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). The terms of their Cauchy product are given by
<math display="block">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 Шаблон:Math. Since for every Шаблон:Math we have the inequalities Шаблон:Math and Шаблон:Math, it follows for the square root in the denominator that Шаблон:Math, hence, because there are Шаблон:Math summands,
<math display="block">|c_n| \ge \sum_{k=0}^n \frac{1}{n+1} = 1</math>
for every integer Шаблон:Math. Therefore, Шаблон:Math does not converge to zero as Шаблон:Math, hence the series of the Шаблон:Math diverges by the term test.
Proof of Mertens' theorem
For simplicity, we will prove it for complex numbers. However, the proof we are about to give is formally identical for an arbitrary Banach algebra (not even commutativity or associativity is required).
Assume without loss of generality that the series <math display="inline"> \sum_{n=0}^\infty a_n</math> converges absolutely. Define the partial sums
<math display="block">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 display="block">c_i=\sum_{k=0}^ia_kb_{i-k}\,.</math>
Then
<math display="block">C_n = \sum_{i=0}^n a_{n-i}B_i</math>
by rearrangement, hence
Fix Шаблон:Math. Since <math display="inline"> \sum_{k \in \N} |a_k| < \infty</math> by absolute convergence, and since Шаблон:Math converges to Шаблон:Math as Шаблон:Math, there exists an integer Шаблон:Math such that, for all integers Шаблон:Math,
(this is the only place where the absolute convergence is used). Since the series of the Шаблон:Math converges, the individual Шаблон:Math must converge to 0 by the term test. Hence there exists an integer Шаблон:Math such that, for all integers Шаблон:Math,
Also, since Шаблон:Math converges to Шаблон:Math as Шаблон:Math, there exists an integer Шаблон:Math such that, for all integers Шаблон:Math,
Then, for all integers Шаблон:Math, use the representation (Шаблон:EquationNote) for Шаблон:Math, split the sum in two parts, use the triangle inequality for the absolute value, and finally use the three estimates (Шаблон:EquationNote), (Шаблон:EquationNote) and (Шаблон:EquationNote) to show that
<math display="block">\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/3\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 definition of convergence of a series, Шаблон:Math 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 <math display="inline"> (a_n)_{n \geq 0}</math>, <math display="inline"> (b_n)_{n \geq 0}</math> are real sequences with <math display="inline"> \sum a_n\to A</math> and <math display="inline"> \sum b_n\to B</math> then
<math display="block">\frac{1}{N}\left(\sum_{n=1}^N\sum_{i=1}^n\sum_{k=0}^i a_k b_{i-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 display="inline"> r>-1</math> and <math display="inline"> s>-1</math>, suppose the sequence <math display="inline"> (a_n)_{n \geq 0}</math> is <math display="inline"> (C,\; r)</math> summable with sum A and <math display="inline"> (b_n)_{n \geq 0}</math> is <math display="inline"> (C,\; s)</math> summable with sum B. Then their Cauchy product is <math display="inline"> (C,\; r+s+1)</math> summable with sum AB.
Examples
- For some <math display="inline"> x,y \in \Reals</math>, let <math display="inline"> a_n = x^n/n!</math> and <math display="inline"> b_n = y^n/n!</math>. Then <math display="block"> 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, formally, <math display="inline"> \exp(x) = \sum a_n</math> and <math display="inline"> \exp(y) = \sum b_n</math>, we have shown that <math display="inline"> \exp(x+y) = \sum c_n</math>. 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 <math display="inline"> \exp(x+y) = \exp(x)\exp(y)</math> for all <math display="inline"> x,y \in \Reals</math>.
- As a second example, let <math display="inline"> a_n = b_n = 1</math> for all <math display="inline"> n \in \N</math>. Then <math display="inline"> c_n = n+1</math> for all <math>n \in \N</math> so the Cauchy product <math display="block"> \sum c_n = (1,1+2,1+2+3,1+2+3+4,\dots)</math> does not converge.
Generalizations
All of the foregoing applies to sequences in <math display="inline"> \Complex</math> (complex numbers). The Cauchy product can be defined for series in the <math display="inline"> \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.
Products of finitely many infinite series
Let <math>n \in \N</math> such that <math>n \ge 2</math> (actually the following is also true for <math>n=1</math> but the statement becomes trivial in that case) and let <math display="inline">\sum_{k_1 = 0}^\infty a_{1, k_1}, \ldots, \sum_{k_n = 0}^\infty a_{n, k_n}</math> be infinite series with complex coefficients, from which all except the <math>n</math>th one converge absolutely, and the <math>n</math>th one converges. Then the limit <math display="block">\lim_{N\to\infty}\sum_{k_1+\ldots+k_n\leq N} a_{1,k_1}\cdots a_{n,k_n}</math> exists and we have: <math display="block">\prod_{j=1}^n \left( \sum_{k_j = 0}^\infty a_{j, k_j} \right)=\lim_{N\to\infty}\sum_{k_1+\ldots+k_n\leq N} a_{1,k_1}\cdots a_{n,k_n}</math>
Proof
Because <math display="block">\forall N\in\mathbb N:\sum_{k_1+\ldots+k_n\leq N}a_{1,k_1}\cdots a_{n,k_n}=\sum_{k_1 = 0}^N \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}</math> the statement can be proven by induction over <math>n</math>: The case for <math>n = 2</math> 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 <math>n \in \N</math> such that <math>n \ge 2</math>, and let <math display="inline">\sum_{k_1 = 0}^\infty a_{1, k_1}, \ldots, \sum_{k_{n+1} = 0}^\infty a_{n+1, k_{n+1}}</math> be infinite series with complex coefficients, from which all except the <math>n+1</math>th one converge absolutely, and the <math>n+1</math>-th one converges. We first apply the induction hypothesis to the series <math display="inline">\sum_{k_1 = 0}^\infty |a_{1, k_1}|, \ldots, \sum_{k_n = 0}^\infty |a_{n, k_n}|</math>. We obtain that the series <math display="block">\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}|</math> converges, and hence, by the triangle inequality and the sandwich criterion, the series <math display="block">\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|</math> converges, and hence the series <math display="block">\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}</math> converges absolutely. Therefore, by the induction hypothesis, by what Mertens proved, and by renaming of variables, we have: <math display="block">\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) \\
& = \left( \sum_{k_1 = 0}^\infty \overbrace{\sum_{k_2 = 0}^{k_1} \sum_{k_3 = 0}^{k_2} \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}}^{=:a_{k_1}} \right) \left ( \sum_{k_{n+1} = 0}^\infty \overbrace{a_{n+1, k_{n+1}}}^{=:b_{k_{n+1}}} \right) \\
& = \left( \sum_{k_1 = 0}^\infty \overbrace{\sum_{k_3 = 0}^{k_1} \sum_{k_4 = 0}^{k_3} \cdots \sum_{k_n+1 = 0}^{k_{n}} a_{1, k_{n+1}} a_{2, k_{n} - k_{n+1}} \cdots a_{n, k_1 - k_3}}^{=:a_{k_1}} \right) \left ( \sum_{k_{2} = 0}^\infty \overbrace{a_{n+1, k_{2}}}^{=:b_{n+1,k_{2}}=:b_{k_{2}}} \right) \\
& = \left( \sum_{k_1 = 0}^\infty a_{k_1} \right) \left ( \sum_{k_{2} = 0}^\infty b_{k_2} \right) \\
& = \left( \sum_{k_1 = 0}^\infty \sum_{k_{2} = 0}^{k_1} a_{k_2}b_{k_1 - k_2} \right) \\
& = \left( \sum_{k_1 = 0}^\infty \sum_{k_{2} = 0}^{k_1} \left ( \overbrace{\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}}^{=:a_{k_2}} \right) \left ( \overbrace{a_{n+1, k_1 - k_2}}^{=:b_{k_1 - k_2}} \right) \right) \\
& = \left( \sum_{k_1 = 0}^\infty \sum_{k_{2} = 0}^{k_1} \overbrace{\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}}^{=:a_{k_2}} \overbrace{a_{n+1, k_1 - k_2}}^{=:b_{k_1 - k_2}} \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}</math>
Therefore, the formula also holds for <math>n+1</math>.
Relation to convolution of functions
A finite sequence can be viewed as an infinite sequence with only finitely many nonzero terms, or in other words as a function <math>f: \N \to \Complex</math> with finite support. For any complex-valued functions f, g on <math>\N</math> with finite support, one can take their convolution: <math display="block">(f * g)(n) = \sum_{i + j = n} f(i) g(j).</math> Then <math display="inline">\sum (f *g)(n)</math> is the same thing as the Cauchy product of <math display="inline">\sum f(n)</math> and <math display="inline">\sum g(n)</math>.
More generally, given a monoid S, one can form the semigroup algebra <math>\Complex[S]</math> of S, with the multiplication given by convolution. If one takes, for example, <math>S = \N^d</math>, then the multiplication on <math>\Complex[S]</math> is a generalization of the Cauchy product to higher dimension.
Notes
References
External links
