Timeline for Solving recursion of a complex function
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 21, 2020 at 4:24 | vote | accept | Koko Nanahji | ||
| May 21, 2020 at 3:05 | comment | added | Max Alekseyev | @მამუკაჯიბლაძე: Good point, thanks! | |
| May 21, 2020 at 0:57 | comment | added | Koko Nanahji | Ok, sounds good. Thank you very much for your help | |
| May 20, 2020 at 23:53 | comment | added | Max Alekseyev | @KokoNanahji: This is just an application of the formula for the product of two exponential generating functions. | |
| May 20, 2020 at 19:50 | comment | added | მამუკა ჯიბლაძე | Not sure if this is usable, but for the generating function ${\mathcal G}(x,z):=\sum_hG_h(x)z^h$ from a formula for Hadamard product your relations give$$\frac\partial{\partial x}{\mathcal G}(x,z)=z\int_0^1{\mathcal G}(\frac x2,\sqrt ze^{2\pi it}){\mathcal G}(\frac x2,\sqrt ze^{-2\pi it})dt$$ | |
| May 20, 2020 at 18:16 | comment | added | Koko Nanahji | Thank you for your time and clear explanation. I just have a question regarding $$G_h'(x) - G_{h-1}'(x) = G_{h-1}(x/2)^2 - G_{h-2}(x/2)^2.$$ From the lhs I get $$ \sum_{n\geq 1} (g_n(h)-g_n(h-1) ) \frac{x^{n-2}}{(n-2)!} $$ assuming $$ G_h'(x) = \frac{d(G_h(x))}{dx}$$ and I am not too sure how this equals to $$G_{h-1}(x/2)^2 - G_{h-2}(x/2)^2$$. | |
| May 20, 2020 at 15:14 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added 149 characters in body
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| May 20, 2020 at 14:56 | history | answered | Max Alekseyev | CC BY-SA 4.0 |