Timeline for uniform approximation by a particular set of functions
Current License: CC BY-SA 4.0
15 events
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| Apr 8, 2019 at 13:31 | comment | added | Pietro Majer | I don't know. The $f_k$ would then have a different form. Maybe one may start from special $C(a)$ for which the expression of the $f_k$ are still simple, e.g. $C(a):={\exp(pa)\over1-a}$ and see what happens | |
| Apr 8, 2019 at 13:14 | comment | added | Ali | Right, but don’t you think the density should still hild if say these constants are all positive? | |
| Apr 8, 2019 at 13:11 | comment | added | Pietro Majer | Say now the $f_k$ are given by $$F(a):=\sum_{k=0}^\infty f_k a^k=\prod_{j=1}^nC(a\mu_j)$$ with $$C(a):=1+\sum_{k=1}^\infty c_k a^k$$ and you would like to have $f_k$ in the form $$f_k=\sum_{j=1}^n\lambda_j\mu_j^k$$ as before? The latter implies $F(a)=\sum_{j=1}^n{\lambda_j\over 1-\mu_ja}={P(a)\over\prod_{1\le j\le n}(1-\mu_ja)}$ with $\text{deg}(P)<n$, and if this has to factor as $\prod_{j=1}^nC(a\mu_j)$ the polynomial $P$ must be a constant, hence $1$. | |
| Apr 8, 2019 at 13:08 | comment | added | Pietro Majer | Indeed, I think the form $f_k=\sum_{j=1}^n\lambda_j\mu_j^k$ forces all $c_k's$ to be $1$ | |
| Apr 8, 2019 at 8:45 | comment | added | Ali | you made a very nice trick to symmetrize all the expressions again and therefore the same approach works. I just wonder if the same result must always be true. If I randomly put some positive coefficients $c_k's$ as follows $(1+c_1a\mu_1+c_2a^2\mu_1^2+\ldots)\ldots(1+c_1a\mu_n+c_2a^2\mu_n^2+\ldots)$ then the same approach can not work right? | |
| Apr 7, 2019 at 20:16 | comment | added | Pietro Majer | Note: dividing by $\lambda_n$ requires it has constant sign, that is, $\mu_n$ has constant sign. I guess this assumption is not necessary, though. | |
| Apr 7, 2019 at 20:12 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Apr 7, 2019 at 20:09 | vote | accept | Ali | ||
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| Apr 7, 2019 at 20:00 | comment | added | Pietro Majer | (details added) | |
| Apr 7, 2019 at 20:00 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Apr 7, 2019 at 19:34 | comment | added | Pietro Majer | I'd say you can repeat the argument with minor modifications. In this case, you get some coefficients functions $\lambda_j(t)$ (depending on the $\mu_j$'s) in front of $f\circ\mu_j$ in the expression of the elements in the closure. The $\lambda_j$ have constant sign, and in particular, $\lambda_n>0$. So you only have to modify conveniently the of the iteration formula for $f$ adding these coefficients. | |
| Apr 7, 2019 at 18:50 | comment | added | Ali | Thanks for the responses. Do you think any of these proofs generalizes if I were to change the definition of $f_k's$ to for example the coefficient of $a^k$ in $(1+a\mu_1+a^2\mu_1^2+\ldots)\ldots(1+a\mu_n+a^2\mu_n^2+\ldots)$? | |
| Apr 6, 2019 at 23:16 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Apr 6, 2019 at 22:50 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Apr 6, 2019 at 17:31 | history | answered | Pietro Majer | CC BY-SA 4.0 |