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当前位置:首页 > 高等教育 > 理学 > 北大数学物理方法(A)-复变函数教案08解析函数的Laurent展开
Outline1lù)Û¼êLaurentÐmÔnÆêÆÔn{§|2007cSC.S.Wu1lù)Û¼êLaurentÐmOutlineùÇ:1LaurentÐmÐm½nÞ~õ¼êLaurentÐm2ü¼êáÛ:áÛ:áÛ:©a¼ê3á?ÛÉ53)Ûòÿ~f)ÛòÿVgC.S.Wu1lù)Û¼êLaurentÐmOutlineùÇ:1LaurentÐmÐm½nÞ~õ¼êLaurentÐm2ü¼êáÛ:áÛ:áÛ:©a¼ê3á?ÛÉ53)Ûòÿ~f)ÛòÿVgC.S.Wu1lù)Û¼êLaurentÐmOutlineùÇ:1LaurentÐmÐm½nÞ~õ¼êLaurentÐm2ü¼êáÛ:áÛ:áÛ:©a¼ê3á?ÛÉ53)Ûòÿ~f)ÛòÿVgC.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationReferencesÇÂÁ§5êÆÔn{6§§5.4—5.7ù&§5êÆÔn{6§§3.5,3.4nÎ!X1Á§5êÆÔn{6§§3.4,3.5C.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationReferencesÇÂÁ§5êÆÔn{6§§5.4—5.7ù&§5êÆÔn{6§§3.5,3.4nÎ!X1Á§5êÆÔn{6§§3.4,3.5C.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationReferencesÇÂÁ§5êÆÔn{6§§5.4—5.7ù&§5êÆÔn{6§§3.5,3.4nÎ!X1Á§5êÆÔn{6§§3.4,3.5C.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuation)Û¼êLaurentÐm¼êØ3)Û:TaylorÐm §kIò§3Û:NCÐm¤?êùÒ´LaurentÐmC.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuation)Û¼êLaurentÐm¼êØ3)Û:TaylorÐm §kIò§3Û:NCÐm¤?êùÒ´LaurentÐmC.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuation)Û¼êLaurentÐm¼êØ3)Û:TaylorÐm §kIò§3Û:NCÐm¤?êùÒ´LaurentÐmC.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsùÇ:1LaurentÐmÐm½nÞ~õ¼êLaurentÐm2ü¼êáÛ:áÛ:áÛ:©a¼ê3á?ÛÉ53)Ûòÿ~f)ÛòÿVgC.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsÐm½n(Laurent)¼êf(z)3±b%/«R1≤|z−b|≤R2þü)Û§KéuS?Ûz:§f(z)±ÐmLaurent?êf(z)=∞Xn=−∞an(z−b)nR1|z−b|R2an=12πiICf(ζ)(ζ−b)n+1dζC´S7S±?¿^4ÜC.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsÐm½n(Laurent)(:)f(z)=∞Xn=−∞an(z−b)nR1|z−b|R2an=12πiICf(ζ)(ζ−b)n+1dζòS .©OPC1ÚC2§âEëÏ«CauchyÈ©úª§éu/«S?¿:z§kf(z)=12πiIC2f(ζ)ζ−zdζ−12πiIC1f(ζ)ζ−zdζe¡©OO÷C1ÚC2È©C.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsÐm½n(Laurent)(:)f(z)=∞Xn=−∞an(z−b)nR1|z−b|R2an=12πiICf(ζ)(ζ−b)n+1dζòS .©OPC1ÚC2§âEëÏ«CauchyÈ©úª§éu/«S?¿:z§kf(z)=12πiIC2f(ζ)ζ−zdζ−12πiIC1f(ζ)ζ−zdζe¡©OO÷C1ÚC2È©C.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsÐm½n(Laurent)(:)f(z)=∞Xn=−∞an(z−b)nR1|z−b|R2an=12πiICf(ζ)(ζ−b)n+1dζòS .©OPC1ÚC2§âEëÏ«CauchyÈ©úª§éu/«S?¿:z§kf(z)=12πiIC2f(ζ)ζ−zdζ−12πiIC1f(ζ)ζ−zdζe¡©OO÷C1ÚC2È©C.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsÐm½n(Laurent)(:)f(z)=∞Xn=−∞an(z−b)nR1|z−b|R2an=12πiICf(ζ)(ζ−b)n+1dζéu÷C1È©−12πiIC1f(ζ)ζ−zdζ=12πiIC1f(ζ)(z−b)−(ζ−b)dζC.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsÐm½n(Laurent)(:)f(z)=∞Xn=−∞an(z−b)nR1|z−b|R2an=12πiICf(ζ)(ζ−b)n+1dζéu÷C1È©−12πiIC1f(ζ)ζ−zdζ=12πiIC1f(ζ)z−b∞Xk=0ζ−bz−bkdζC.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsÐm½n(Laurent)(:)f(z)=∞Xn=−∞an(z−b)nR1|z−b|R2an=12πiICf(ζ)(ζ−b)n+1dζéu÷C1È©−12πiIC1f(ζ)ζ−zdζ=12πiIC1f(ζ)z−b∞Xk=0ζ−bz−bkdζ=∞Xk=012πiIC1f(ζ)(ζ−b)kdζ(z−b)−k−1(|z−b|R1)C.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsÐm½n(Laurent)(:)f(z)=∞Xn=−∞an(z−b)nR1|z−b|R2an=12πiICf(ζ)(ζ−b)n+1dζéu÷C1È©−12πiIC1f(ζ)ζ−zdζ=12πiIC1f(ζ)z−b∞Xk=0ζ−bz−bkdζ=−∞Xn=−112πiIC1f(ζ)(ζ−b)n+1dζ(z−b)−n(|z−b|R1)C.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsÐm½n(Laurent)(:)f(z)=∞Xn=−∞an(z−b)nR1|z−b|R2an=12πiICf(ζ)(ζ−b)n+1dζéu÷C2È©§Ú^TaylorÐm(J12πiIC2f(ζ)ζ−zdζ=12πiIC2f(ζ)ζ−b∞Xn=0z−bζ−bndζ=∞Xn=012πiIC2f(ζ)(ζ−b)n+1dζ(z−b)n(|z−b|R2)C.S.Wu1lù)Û¼êLaurentÐmExpansioninLaurentSeriesIsolatedSingularitiesofUniformFunctionAnalyticContinuationTheorem(Laurent)IllustrativeExamplesLaurentExpansion:MultivaluedFunctionsÐm½n(Laurent)(:)f(z)=∞Xn=−∞an(z−b)nR1|z−b|R2an=12πiICf(ζ)(ζ−b)n+1dζò
本文标题:北大数学物理方法(A)-复变函数教案08解析函数的Laurent展开
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