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Wikipedia claims that Laurent series cannot in general be multiplied. I am wondering why not? Suppose we have $f(z),g(z)$ analytic in the annulus: $r<|z-a|<R (0\le r<R\le\infty)$, then $$ f(z)=\sum_{n=-\infty}^\infty c_n(z-a)^n $$ is the Laurent series of $f(z)$ in the annulus, and $$ g(z)=\sum_{n=-\infty}^\infty d_n(z-a)^n $$ is the Laurent series of $g(z)$ in the annulus. We also know that the Laurent series absoltely and uniformly converges to funtions, then why can't we take the convolution of $f(z)$ and $g(z)$ which should be the Laurent series of $f(z)g(z)$ ?

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If you know $f,g$ converges together on some nonempty annulus, then yes you can multiply them (because of local absolute uniform convergence).

The wikipedia statement is really referring to the case for general Laurent series $f,g$ which you don't know whether they converge at all. Then you may not be able to multiply, e.g., $f(z)=\sum_{n\geq 0}z^n$, $g(z)=\sum_{n\leq 0}z^n$.

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  • $\begingroup$ Thank you! Crystal clear. $\endgroup$ – Bach May 18 at 15:16
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If you have two Laurent series, both convergent at least for $a<|z|<b$, then you can form the product series, which then represents the product of the two individual functions (an example is treated here: Products of Laurent Series ). If, however, the two series have no common annulus of convergence you cannot hope for much.

But there is a difference to the convolution multiplication of two power series $\sum_{j\geq0}a_j z^j$ and $\sum_{k\geq0} b_kz^k$: Each individual coefficient $c_r$ of the product series $\sum_{r\geq0}c_r z^r$ can be obtained by a finite number of operations, while with the product of two Laurent series (with infinite ends on both sides) each coefficient of the product series is the sum of an infinite series of complex products.

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