Cauchy product formula

The Cauchy product formula , also Cauchy product or Cauchy convolution , named after the French mathematician Augustin Louis Cauchy, allows the multiplication of infinite series . This is a discrete convolution .

definition

If and are two absolutely convergent series, then it is the series ${\displaystyle \sum _{n=0}^{\infty }a_{n}}$ ${\displaystyle \sum _{n=0}^{\infty }b_{n}}$

${\displaystyle \sum _{n=0}^{\infty }c_{n}}$ With ${\displaystyle c_{n}=\sum _{k=0}^{n}{a_{k}b_{n-k}}=\sum _{i+j=n}a_{i}b_{j}}$

also an absolutely convergent series and it holds

${\displaystyle \left(\sum _{n=0}^{\infty }a_{n}\right)\cdot \left(\sum _{n=0}^{\infty }b_{n}\right)=\sum _{n=0}^{\infty }c_{n}.}$

The series is called Cauchy Product of the series and . The coefficients can be understood as the discrete convolution of the vectors and . ${\displaystyle \sum _{n=0}^{\infty }c_{n}}$ ${\displaystyle \sum _{n=0}^{\infty }a_{n}}$ ${\displaystyle \sum _{n=0}^{\infty }b_{n}}$ ${\displaystyle c_{n}}$ ${\displaystyle (a_{0},a_{1},\dots ,a_{n})}$ ${\displaystyle (b_{0},b_{1},\dots ,b_{n})}$

If you write out this formula, you get:

${\displaystyle \left(\sum _{n=0}^{\infty }a_{n}\right)\cdot \left(\sum _{n=0}^{\infty }b_{n}\right)=\underbrace {(a_{0}b_{0})} _{c_{0}}+\underbrace {(a_{0}b_{1}+a_{1}b_{0})} _{c_{1}}+\underbrace {(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0})} _{c_{2}}+...+\underbrace {(a_{0}b_{n}+a_{1}b_{n-1}+...+a_{k}b_{n-k}+...+a_{n}b_{0})} _{c_{n}}+...}$

If you break off this series at a certain value of , you get an approximation for the product you are looking for. ${\displaystyle n}$

The following applies in particular to the multiplication of power series

${\displaystyle \left(\sum _{n=0}^{\infty }\alpha _{n}{(x-x_{0})}^{n}\right)\cdot \left(\sum _{n=0}^{\infty }\beta _{n}{(x-x_{0})}^{n}\right)=\sum _{n=0}^{\infty }\left(\sum _{k=0}^{n}{\alpha _{k}\beta _{n-k}}\right)(x-x_{0})^{n}.}$

Examples

Application to the exponential function

As an application example, it should be shown how the functional equation of the exponential function can be derived from the Cauchy product formula. As is known, the exponential function converges absolutely. Hence, the product can be calculated and obtained using the Cauchy product ${\displaystyle \textstyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}}$ ${\displaystyle e^{x}e^{y}}$

${\displaystyle e^{x}e^{y}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\cdot \sum _{n=0}^{\infty }{\frac {y^{n}}{n!}}=\sum _{n=0}^{\infty }\sum _{k=0}^{n}{\frac {1}{k!}}{\frac {1}{(n-k)!}}x^{k}y^{n-k}}$

After defining the binomial coefficient , this can be further transformed as ${\displaystyle \textstyle {n \choose k}={\frac {n!}{k!(n-k)!}}}$

${\displaystyle =\sum _{n=0}^{\infty }{\frac {1}{n!}}\sum _{k=0}^{n}{\frac {n!}{k!(n-k)!}}x^{k}y^{n-k}=\sum _{n=0}^{\infty }{\frac {1}{n!}}\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}=\sum _{n=0}^{\infty }{\frac {1}{n!}}(x+y)^{n}=e^{x+y}}$

where the penultimate equal sign is justified by the binomial theorem .

A divergent series

It's supposed to be the Cauchy product

${\displaystyle \left(\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{\sqrt {n+1}}}\right)\cdot \left(\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{\sqrt {n+1}}}\right)}$

a series that is only partially convergent with itself.

The following applies here

${\displaystyle c_{n}=\sum _{k=0}^{n}{\frac {(-1)^{k}}{\sqrt {k+1}}}\cdot {\frac {(-1)^{n-k}}{\sqrt {n-k+1}}}=(-1)^{n}\sum _{k=0}^{n}{\frac {1}{\sqrt {(k+1)(n-k+1)}}}\ .}$

With the inequality of the arithmetic and geometric mean applied to the root in the denominator, it follows ${\displaystyle {\sqrt {ab}}\leq {\tfrac {1}{2}}(a+b)}$

${\displaystyle |c_{n}|\geq \sum _{k=0}^{n}{\frac {2}{n+2}}={\frac {2(n+1)}{n+2}}\geq 1\ .}$

Since they do not form a zero sequence , the series diverges ${\displaystyle c_{n}}$ ${\displaystyle \sum _{n=0}^{\infty }c_{n}.}$

Calculation of the inverse power series

With the help of the Cauchy product formula, the inverse of a power series with real or complex coefficients can be calculated. We use and . We calculate the coefficients using: ${\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}}$ ${\displaystyle {\frac {1}{f(z)}}=\sum _{m=0}^{\infty }b_{m}z^{m}}$ ${\displaystyle b_{m}}$

${\displaystyle 1=f(z)\cdot {\frac {1}{f(z)}}=\sum _{n=0}^{\infty }a_{n}z^{n}\sum _{m=0}^{\infty }b_{m}z^{m}=\sum _{r=0}^{\infty }\left(\sum _{l=0}^{r}a_{l}b_{r-l}\right)\cdot z^{r}\ }$,

where we used the Cauchy product formula in the last step. With a coefficient comparison it follows:

${\displaystyle r=0:\ a_{0}b_{0}=1\Rightarrow b_{0}={\frac {1}{a_{0}}}\ .}$

${\displaystyle r=1:\ a_{0}b_{1}+a_{1}b_{0}=0\Rightarrow b_{1}=-{\frac {a_{1}b_{0}}{a_{0}}}=-{\frac {a_{1}}{a_{0}^{2}}}.}$

${\displaystyle r=2:\ a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0}=0\Rightarrow b_{2}=-{\frac {a_{1}b_{1}}{a_{0}}}-{\frac {a_{2}b_{0}}{a_{0}}}={\frac {a_{1}^{2}}{a_{0}^{3}}}-{\frac {a_{2}}{a_{0}^{2}}}.}$

${\displaystyle r=3:\ a_{0}b_{3}+a_{1}b_{2}+a_{2}b_{1}+a_{3}b_{0}=0\Rightarrow b_{3}=-{\frac {a_{1}b_{2}}{a_{0}}}-{\frac {a_{2}b_{1}}{a_{0}}}-{\frac {a_{3}b_{0}}{a_{0}}}=-{\frac {a_{1}^{3}}{a_{0}^{4}}}+{\frac {2a_{2}a_{1}}{a_{0}^{3}}}-{\frac {a_{3}}{a_{0}^{2}}}.}$

${\displaystyle \dots }$

To simplify and o. B. d. A. we set and find . ${\displaystyle a_{0}=1}$ ${\displaystyle {\frac {1}{f(z)}}=1-a_{1}z+(a_{1}^{2}-a_{2})z^{2}+(-a_{1}^{3}+2a_{1}a_{2}-a_{3})z^{3}+\dots =\sum _{i=0}^{\infty }(-1)^{i}\cdot \left(\sum _{n=1}^{\infty }a_{n}z^{n}\right)^{i}}$

Generalizations

According to Mertens' theorem , it is sufficient to require that at least one of the two convergent series converges absolutely so that its Cauchy product converges (not necessarily absolute) and its value is the product of the given series values.

If both series only converge to a limited extent, it is possible that their Cauchy product does not converge, as the above example shows. In this case, however, if the Cauchy product converges, then, according to Abel's theorem , its value coincides with the product of the two series values.

literature

• Konrad Königsberger : Analysis 1 . Springer, Berlin 2004, ISBN 3-540-41282-4