Question Number 216841 by hardmath last updated on 22/Feb/25
$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\:\:\delta\left(\mathrm{n}\right)\:=\:\underset{\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{n}}}} {\sum}\:\boldsymbol{\varphi}\left(\mathrm{d}\right)\:\boldsymbol{\tau}\left(\frac{\mathrm{n}}{\mathrm{d}}\right) \\ $$$$\boldsymbol{\delta}\left(\mathrm{n}\right)\:=\:\underset{\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{n}}}} {\sum}\:\mathrm{d}\:\:\:,\:\:\:\boldsymbol{\tau}\left(\mathrm{n}\right)\:=\:\underset{\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{n}}}} {\sum}\:{l}\:\:\:\mathrm{and}\:\:\:\varphi-\mathrm{Eyler}.\mathrm{f} \\ $$
Answered by MrGaster last updated on 23/Feb/25
$$\mathrm{Let}\:{f}\left({n}\right)=\underset{{d}\mid{n}} {\sum}\varphi\left({d}\right)\tau\left(\frac{{n}}{{d}}\right) \\ $$$$\Rightarrow,{f}\left({n}\right)=\underset{{d}\mid{n}} {\sum}\varphi\left({d}\right)\underset{{k}\mid\frac{{n}}{{d}}} {\sum}\mathrm{1}=\underset{{d}\mid{n}} {\sum}\varphi\left({d}\right)\underset{{k}\mid\frac{{n}}{{d}}} {\sum}\mathrm{1}=\underset{{d}\mid{n}} {\sum}\varphi\left({d}\right)\left(\frac{{n}}{{d}}\right) \\ $$$$\underset{{d}\mid{n}} {\sum}\varphi\left({d}\right)={n} \\ $$$$\mathrm{have}:{f}\left({n}\right)=\underset{{d}\mid{n}} {\sum}\varphi\left({d}\right)\left(\frac{{n}}{{d}}\right)={n}\underset{{d}\mid{n}} {\sum}\frac{\varphi\left({d}\right)}{{d}} \\ $$$$\mathrm{But}\:\underset{{d}\mid{n}} {\sum}=\frac{\varphi\left({d}\right)}{{d}}=\mathrm{1} \\ $$$$\Rightarrow{f}\left({n}\right)={n}×\mathrm{1}={n} \\ $$$$\therefore\delta\left({n}\right)={n} \\ $$$$ \\ $$
Commented by hardmath last updated on 23/Feb/25
$$ \\ $$Excellent solution, thank you very much dear professor