Additivity and Multiplicativity of Number-Theoretic Functions - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-06-19T07:53:15Z http://mathoverflow.net/feeds/question/48898 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48898/additivity-and-multiplicativity-of-number-theoretic-functions Additivity and Multiplicativity of Number-Theoretic Functions Jose Arnaldo Dris 2010-12-10T08:44:50Z 2010-12-11T17:38:07Z <p>An additive function $f(n)$ is said to be totally/completely additive if</p> <p>$$f(ab) = f(a) + f(b)$$ </p> <p>holds for all positive integers $a$ and $b$.</p> <p>Additionally, if $f$ is a totally/completely additive function, then $f(1) = 0$.</p> <p>On the other hand, a multiplicative function $g(n)$ is said to be totally/completely multiplicative if </p> <p>$$g(ab) = g(a)g(b)$$ </p> <p>holds for all positive integers $a$ and $b$. </p> <p>Additionally, if g is a totally/completely multiplicative function, then $g(1) = 1$.</p> <p>So now suppose there is a function $h(n)$ such that </p> <p>$$h(n) = \frac{1}{2}[f(n) + g(n)]$$</p> <p>where $f$ is totally additive and $g$ is totally multiplicative. (I removed the erroneous assumption on the function $h(n)$, thank you Gerry for pointing that out!)</p> <p>Then </p> <p>$$h(1) = \frac{1}{2}(f(1) + g(1)) = \frac{1}{2}$$</p> <p>Now, let </p> <p>$$i(n) = h(n + 1)$$</p> <p>so that </p> <p>$$i(0) = h(1) = \frac{1}{2}$$</p> <p>My question is:</p> <p>How do we then characterize such number-theoretic functions $i(n)$?</p>