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  • 莫比乌斯反演推导

    Proof : (g(m)=sumlimits_{d|m}f(d) iff f(m)=sumlimits_{d|m}mu(d)g(frac{m}{d}))
    ( ightarrow\sumlimits_{d|m}mu(d)g(frac{m}{d})=sumlimits_{d|m}mu(frac{m}{d})g(d)=sumlimits_{d|m}mu(frac{m}{d})sumlimits_{k|d}f(k)=sumlimits_{k|m}f(k)sumlimits_{d|frac{m}{k}}mu(frac{m}{kd})=sumlimits_{k|m}f(k)sumlimits_{d|frac{m}{k}}mu(d)=sumlimits_{k|m}f(k)[frac{m}{k}=1]=f(m))
    (leftarrow\ sumlimits_{d|m}f(d)=sumlimits_{d|m}sumlimits_{k|d}mu(k)g(frac{d}{k})=sumlimits_{d|m}sumlimits_{k|d}mu(frac{d}{k})g(k)=sumlimits_{k|m}g(k)sumlimits_{d|frac{m}{k}}mu(d)=sumlimits_{k|m}g(k)[frac{m}{k}=1]=g(m))
    Proof : (g(m)=sumlimits_{dgeq1}f(frac{m}{d}) iff f(m)=sumlimits_{dgeq1}mu(d)g(frac{m}{d}))
    ( ightarrow\sumlimits_{dgeq1}mu(d)g(frac{m}{d})=sumlimits_{dgeq1}mu(d)sumlimits_{kgeq1}f(frac{m}{kd})=sumlimits_{ngeq1}f(frac{m}{n})sumlimits_{k,dgeq1}mu(d)[n=kd]=sumlimits_{ngeq1}f(frac{m}{n})sumlimits_{d|n}mu(d)=sumlimits_{ngeq1}f(frac{m}{n})[n=1]=f(m))
    (leftarrow \ sumlimits_{dgeq1}f(frac{m}{d})=sumlimits_{dgeq1}sumlimits_{kgeq1}mu(k)g(frac{m}{kd})=sumlimits_{ngeq1}g(frac{m}{n})sumlimits_{k,dgeq1}mu(k)[n=kd]=sumlimits_{ngeq1}g(frac{m}{n})sumlimits_{k|n}mu(k)=sumlimits_{ngeq1}g(frac{m}{n})[n=1]=g(m))

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  • 原文地址:https://www.cnblogs.com/lemon-jade/p/13881395.html
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