Timeline for A new simple formula is needed
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 22, 2021 at 19:03 | vote | accept | Maksym Voznyy | ||
| Feb 22, 2021 at 19:02 | vote | accept | Maksym Voznyy | ||
| Feb 22, 2021 at 19:03 | |||||
| Feb 4, 2021 at 12:01 | comment | added | Gerry Myerson | We aim to please, Maksym. | |
| Feb 4, 2021 at 5:14 | comment | added | Maksym Voznyy | @Gerry Myerson: Thank you so much for the idea to convert to a hyperbola! I posted a much more productive code based on it in the answer. | |
| Feb 4, 2021 at 5:10 | vote | accept | Maksym Voznyy | ||
| Feb 22, 2021 at 19:02 | |||||
| Feb 4, 2021 at 5:10 | vote | accept | Maksym Voznyy | ||
| Feb 4, 2021 at 5:10 | |||||
| Feb 4, 2021 at 5:10 | answer | added | Maksym Voznyy | timeline score: 0 | |
| Feb 4, 2021 at 4:59 | history | edited | Maksym Voznyy | CC BY-SA 4.0 |
added print(len(listA)) to the code for future comparison
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| Jan 25, 2021 at 3:09 | vote | accept | Maksym Voznyy | ||
| Feb 4, 2021 at 5:10 | |||||
| Jan 24, 2021 at 15:51 | history | edited | Maksym Voznyy | CC BY-SA 4.0 |
added 9 characters in body
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| Jan 24, 2021 at 14:50 | comment | added | LSpice |
I converted some \fracs into \dfracs, which I think enhances readability. However, you say "Applying (2) and (2) consecutively", which probably is not what you mean.
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| Jan 24, 2021 at 14:50 | history | edited | LSpice | CC BY-SA 4.0 |
\frac -> \dfrac
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| Jan 24, 2021 at 11:39 | comment | added | Gerry Myerson | From one value of $a$, your five formulas give you five new values of $a$. From one value of $a$, my construction gives you an exhaustive infinity of new values of $a$. | |
| Jan 24, 2021 at 10:54 | answer | added | Nulhomologous | timeline score: 1 | |
| Jan 24, 2021 at 10:47 | comment | added | Xarles | I am not sure what you are really asking. What it seems to me, you are asking is for rational functions $b=b(a)$ such that, if $a(a+1)=2e^2$ for some $e$, then $b(b+1)=2d^2$ for some $d$ (in terms of $a$ and $e$). It is not explained at all what it is the relation with elliptic curves, and it is probably misleading. About the code, you are only searching solutions of a "Pell equation", easily parametrizable (as explained by @GerryMyerson). | |
| Jan 24, 2021 at 4:42 | history | edited | Maksym Voznyy | CC BY-SA 4.0 |
added 2 characters in body
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| Jan 24, 2021 at 1:35 | comment | added | Maksym Voznyy | @Gerry Myerson: $b$ is a different value in the same list. E.g., if $r=2$, $top=289$, $a=-2$, then formulas $(1)-(5)$ produce $b_1=\frac{1}{7}$, $b_2=\frac{2}{7}$, $b_3=-\frac{81}{49}$, $b_4=\frac{18}{7}$, $b_5=-289$, all of which are in the same list. | |
| Jan 23, 2021 at 23:20 | comment | added | Gerry Myerson | You use the symbol $b$, but never define it. Anyway, you want $\sqrt{a(a+1)}=c\sqrt2$, This equation can be rewritten in the form $u^2-2d^2=1$, where $u=2a+1$, and this gives you $\sqrt{a(a+1)}=c\sqrt2$ with $d=2c$. Taking a point on $u^2-2d^2=1$, such as $u=1$, $d=0$, drawing a line through it with rational slope $t$, and finding the other intersection of this line with the hyperbola, should give all rational solutions, parametrized by $t$. | |
| Jan 23, 2021 at 23:04 | history | asked | Maksym Voznyy | CC BY-SA 4.0 |