I am unsure of the proof of the identity involving the identity arithmetic function, $e$. That is, the identity $f\ast e=f$. My proof so far is: $$[f\ast e](n)=\sum\limits_{d\mid n}f(d)e(\frac{n}{d})$$ Letting $d=n$ to make $e\neq0$: $$\sum\limits_{d\mid n}f(d)e(\frac{n}{d})=f(n)$$ The step I don't understand is the part where $e$ CAN'T equal 0. I know it's nice when we say is doesn't equal zero, but what lets us do this?
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By definition, $e(n)=0$ if $n>1$ and $e(1)=1$.
So, in the sum $\sum\limits_{d\mid n}f(d)e(\frac{n}{d})$, the only non-zero term is when $\frac{n}{d}=1$.
lhf
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Oh, thanks, I forgot about the sum part. – ceols Mar 09 '15 at 01:08