莫比乌斯反演的非卷积形式证明

\[f(n)=\sum_{i=1}^n t(i) g(\lfloor\frac ni\rfloor)\\ 求g\\ 推导(*为狄利克雷卷积\ ·为点乘):\\ f(n)=\sum_{i=1}^n t(i)\sum_{j=1}^{\lfloor\frac ni\rfloor}\Delta g(j)\\ =\sum_{i*j\leq n}^n t(i)\Delta g(j)\\ \therefore \Delta f(n)=\sum_{i|n}t(\frac ni)*\Delta g(i)=(t*\Delta g)(n)\\ \Delta g(n)=(\Delta f*t^{-1})(n)\\ g(n)=\sum_{i=1}^n(\Delta f*t^{-1})(i)\\ =\sum_{i=1}^n\sum_{j|i}\Delta f(\frac ij)t^{-1}(j)\\ =\sum_{i=1}^nt^{-1}(i)\sum_{k}^{\lfloor\frac ni\rfloor}\Delta f(k)\\ =\sum_{i=1}^nt^{-1}(i)f(\lfloor\frac ni\rfloor)\\ 若t为完全积性函数则t^{-1}为t·\mu \]