Product of Cauchy

In analyzes, the produced of Cauchy is an operation carrying on some series. It makes it possible to generalize the property of Distributivité. Its name is a homage to the French analyst Augustin Louis Cauchy. It is about a discrete product of Convolution.

Preliminary: a writing of the product of polynomials

A particular writing of the coefficients of the product of polynomials makes it possible to include/understand the introduction of the formula of the product of Cauchy. Is two Polynôme S with complex coefficients P and Q given by their decomposition in the canonical base

P= \ sum_ {i=0} ^ {+ \ infty} a_i X^i \ qquad Q= \ sum_ {j=0} ^ {+ \ infty} b_j X^j

where the coefficients of P and Q are null starting from a certain row. Then their product breaks up like

PQ= \ sum_ {I \ in \ NR, J \ in \ NR} a_i b_j X^ {i+j} = \ sum_ {s=0} ^ {+ \ infty} \ sum_ {k=0} ^ S. has. _k b_ {s-k} X^s

The reindexation necessary does not raise a difficulty since the sum in fact is finished.

Product of Cauchy of complex series

The produces of Cauchy series $\ sum a_n$ and $\ sum b_n$ of complex numbers is the series of general term

c_n= \ sum_ {k=0} ^n a_k b_ {n-k}

Under suitable assumptions, this series will converge, and one will be able to write the formula of generalized distributivity

\ sum_ {n=0} ^ {+ \ infty} c_n= \ left (\ sum_ {i=0} ^ {+ \ infty} a_i \ right) \ left (\ sum_ {j=0} ^ {+ \ infty} b_j \ right)

If the series are both with null terms starting from a certain row, it is enough to use the result of the preceding paragraph in the case X=1 . But in general, it is not possible to affirm that the property is true since one arbitrarily cannot réindexer of the sums of series (see summable Famille for a justification).

Case of two absolutely convergent series

When the series $\ sum a_n$ and $\ sum b_n$ are both absolutely convergent, their product of Cauchy converges and the formula of generalized distributivity is checked. It is indeed enough to use the property of commutation and associativeness of the summable families.

In particular, for two complexes has and B , one can make the product of Cauchy of the series defining the Exponentielle

e^a. e^b = \ sum_ {n=0} ^ {+ \ infty} \ left (\ sum_ {k=0} ^n \ frac {a^kb^ {n-k}} {K! (n-k)!}\ right)

\ sum_ {n0} ^ {+ \ infty} \ frac1 {N!} (a+b)^n e^ {a+b}

Starting from this property, it is possible also to define the product of Cauchy of two whole series, whose properties are studied hereafter.

Theorem of Mertens

Theorems of convergence

If two convergent series there are however positive results of convergence for their product of Cauchy. By taking again the notations $a_n, \, b_n, \, c_n$ for the general terms of the two series and the series produces of Cauchy, and by noting has and B the sums of the first two series

if the series produced $\ sum c_n$ converges, then it can be only towards the product $AB$ , it is a consequence of the Théorème of Abel

there is in any case always a convergence in a weaker direction, within the meaning of the Procédé of summation of Cesàro. I.e.

\ lim \ limits_ {N \ to + \ infty} \ frac1 {n+1} \ sum_ {k=0} ^n \ left (\ sum_ {j=0} ^k c_j \ right) = AB

Product of Cauchy of whole series

Two whole series $\ sum a_n x^n$ and $\ sum b_n x^n$ being given, their product of Cauchy is also a whole series, since the general term is worth

\ sum_ {k=0} ^n a_k x^k b_ {n-k} x^ {n-k} = \ left (\ sum_ {k=0} ^n a_k b_ {n-k} \ right) x^n

By noting $R_ 1, R _2$ the respective rays of convergence of the two whole series, the ray of convergence $R_ \ Pi$ of the series produced checks the inequality

R_ \ pi \ geq \ mathrm {Min} (R_ 1,R _2)

Indeed, if one considers a complex of module strictly lower than this minimum, the two whole series converge absolutely, the series also produced, and its function sum is the product of the functions sums of the two series. One from of deduced that the product of two developable functions in whole series on open is him also developable in whole series.

The preceding inequality can be strict. It is the case for example if one takes for the two series $\ sum x^n$ (ray 1) on the one hand and $1-x$ on the other hand (infinite ray). The series produced is reduced to 1 and has an infinite ray of convergence.

More surprising, the ray of the series produced can be infinite while at the same time the two rays of the initial series are finished. For example if one considers the development of $\ sqrt {1-x}$ in whole series, the ray of convergence is 1. But when one makes the product of Cauchy of this series with itself, one obtains the series $1-x$ , therefore a polynomial of infinite ray.

Generalization with the algebras of Banach

It is supposed that has is a Algèbre of Banach. Then it is possible to define the concept of product of Cauchy of two series in values in has . Moreover, the product of Cauchy of two absolutely convergent series converges, and formulates it generalized distributivity always holds.

For example, it is possible to take again the calculation of the product of two exponential S carried out in the complex case. The only property which misses to be able to write the formula is the possibility of applying the formula of the Binomial theorem, which requires to suppose for example that has and B commutates. Under this assumption

e^ {a+b} =e^a \ times e^b

For example, if T, U are scalars one always has

e^ {(t+u) has} =e^ {your} \ times e^ {ua}

Another important formula: if b=-a ,

e^a \ times e^ {- has} =e^0=1

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