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.
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, 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.
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:
infty | |
\left(\sum | |
i=0 |
ai\right) ⋅
infty | |
\left(\sum | |
j=0 |
bj\right)=
infty | |
\sum | |
k=0 |
ck
ck=\sum
k | |
l=0 |
albk-l
Consider the following two power series
infty | |
\sum | |
i=0 |
aixi
infty | |
\sum | |
j=0 |
bjxj
with complex coefficients
\{ai\}
\{bj\}
infty | |
\left(\sum | |
i=0 |
aixi\right) ⋅
infty | |
\left(\sum | |
j=0 |
bjxj\right)=
infty | |
\sum | |
k=0 |
ckxk
ck=\sum
k | |
l=0 |
albk-l
Let and be real or complex sequences. It was proved by Franz Mertens that, if the series converges to and converges to, and at least one of them converges absolutely, then their Cauchy product converges to .[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:
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 . Since for every we have the inequalities and, it follows for the square root in the denominator that, hence, because there are summands,
for every integer . Therefore, does not converge to zero as, hence the series of the diverges by the term test.
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 converges absolutely.Define the partial sums
with
Then
by rearrangement, hence
Fix . Since by absolute convergence, and since converges to as, there exists an integer such that, for all integers,
(this is the only place where the absolute convergence is used). Since the series of the converges, the individual must converge to 0 by the term test. Hence there exists an integer such that, for all integers,
Also, since converges to as, there exists an integer such that, for all integers,
Then, for all integers, use the representation for, split the sum in two parts, use the triangle inequality for the absolute value, and finally use the three estimates, and to show that
\, | B_i-B |
\, | B |
By the definition of convergence of a series, as required.
In cases where the two sequences are convergent but not absolutely convergent, the Cauchy product is still Cesàro summable.[16] 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:
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.
n\in\N
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.
Let
n\in\N
n\ge2
n=1
n
n
Becausethe statement can be proven by induction over
n
n=2
The induction step goes as follows: Let the claim be true for an
n\in\N
n\ge2
n+1
n+1
n+1
A finite sequence can be viewed as an infinite sequence with only finitely many nonzero terms, or in other words as a function
f:\N\to\Complex
\N
\Complex[S]
S=\Nd
\Complex[S]