Timeline for The functional equation $f(x) = qx + qxf(x) - f(x^2)$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2016 at 20:13 | comment | added | D. Ror. | $a_n = a_n^{(q)}$ satisfies the following recursion: \begin{eqnarray*} a_0 & = & 0;\\ a_1 & = & q;\\ a_{2k} & = & qa_{2k-1} - a_k;\\ a_{2k+1} & = & qa_{2k}. \end{eqnarray*} | |
| Oct 14, 2016 at 20:06 | comment | added | Timothy Chow | It might be nice for you to provide a table of values of $a_n^{(q)}$ for small values of $n$ and $q$. | |
| Oct 14, 2016 at 18:41 | comment | added | user44191 | Ah, you're right, I made a mistake when trying to deal with $f(x^2)$. | |
| Oct 14, 2016 at 14:52 | comment | added | D. Ror. | @user44191 The numerator in the summation has $q\cdot x^{2^j}$ not $(qx)^{2^j}$. | |
| Oct 14, 2016 at 5:30 | comment | added | user44191 | Also, as such, any information about $f_q(q^{-2}) = f(\frac{1}{q})$, so any information about that probability is equivalent to information about your original function itself. | |
| Oct 14, 2016 at 5:22 | comment | added | user44191 | Something interesting I noticed: there's an easy way to get $f_q(x)$ from any other $f_q(x)$, by scaling $x$ by $q$. As such, you only need to care about $f_1(x)$. | |
| Oct 13, 2016 at 20:04 | history | edited | D. Ror. | CC BY-SA 3.0 |
string -> word (one instance); example added
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| Oct 13, 2016 at 18:51 | history | asked | D. Ror. | CC BY-SA 3.0 |