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vinovinovino
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I wonder why z^(-1/2) cannot be expanded in Laurent series with center z=0. Anyone knows?
micromass said:That function is not holomorphic on [itex]\mathbb{C}[/itex]. Indeed, there is a line through the origin on which the function is not continuous.
To see this, we use the definition of complex exponentiation
[tex]z^{-1/2}=e^{-\frac{1}{2}Log(z)}[/tex]
but the complex logarithm isn't holomorphic, so the composition isn't either.
The function z^(-1/2) is not analytic at z=0, which means it is not continuous and does not have a well-defined derivative at this point. Therefore, it cannot be expanded in a Laurent series, which is a representation of an analytic function.
No, z^(-1/2) also cannot be expanded in a Taylor series because it is not analytic at z=0. A Taylor series is a special case of a Laurent series, so if a function cannot be expanded in a Laurent series, it also cannot be expanded in a Taylor series.
Yes, although z^(-1/2) cannot be expanded in a Laurent series with center z=0, it can still be approximated using other methods, such as numerical methods or by using a series expansion at a different point.
Understanding the limitations of Laurent series expansions is important because it allows us to accurately represent and approximate functions. If we try to expand a function in a series when it is not analytic, we can end up with incorrect or nonsensical results.
No, only functions that are analytic can be expanded in a Laurent series. Functions that are not analytic, such as z^(-1/2), cannot be expanded in a series with a finite number of terms.