"inversion" of a convolution - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:12:32Z http://mathoverflow.net/feeds/question/34826 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34826/inversion-of-a-convolution "inversion" of a convolution Martin Rubey 2010-08-07T10:11:21Z 2010-08-09T12:03:45Z <p>I have the following relation:</p> <p>$$\sum_{d|n} (1+1/x)^{d-1} F_{n/d}(x^d)=L_n(x)$$</p> <p>where the right hand side is (for every $n$) a polynomial in $x$, which I have an expression for, but it's not extremely beautiful. The family of polynomials $F_k(x)$ is unknown, and is what I'm looking for.</p> <p>Since this is close to Dirichlet convolution, I have not quite given up hope that there is something similar to Möbius inversion, that would give me $F_k(x)$ explicitely. Is this possible? Related instances of such a problem may also be interesting.</p> <p>A possibly weaker, but still sufficient solution would be an expression in terms of $L_k$ and $R_k$ of the expression</p> <p>$$\sum_{d|n} R_d(x) F_{n/d}(x^d)$$</p> <p>where $R_k(x)$ is another family of polynomials, which is also unknown.</p> http://mathoverflow.net/questions/34826/inversion-of-a-convolution/34979#34979 Answer by Martin Rubey for "inversion" of a convolution Martin Rubey 2010-08-09T08:12:57Z 2010-08-09T12:03:45Z <p>At least computing $F_k(x)$ turned out not to be that hard after all. Slightly more generally, consider</p> <p>$$\sum_{d|k} Z_d(x) F_{n/d}(x^d) = L_n(x),$$ with $Z_1(x)=1$.</p> <p>Then we have $$F_n(x)=\sum_{1=d_0|d_1|\dots|d_k|n}L_{n/d_k}(x^{d_k}) (-1)^k\prod_{i=0}^{k-1} Z_{d_{i+1}/d_i}(x^{d_i}),$$</p> <p>where in the sum $d_0 &lt; d_1 &lt; \dots &lt; d_k \leq n$. In other words, we are summing over all chains starting at $1$, below $n$. The formula is easily shown by induction.</p>