Timeline for How do you determine which representation of a function to use for Newton's method?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 23, 2019 at 7:52 | audit | First posts | |||
| Sep 23, 2019 at 7:54 | |||||
| Sep 21, 2019 at 22:59 | audit | First posts | |||
| Sep 21, 2019 at 23:23 | |||||
| Sep 18, 2019 at 4:20 | comment | added | ruakh | @ClaudeLeibovici: Oh, wait, I see. You never mention what equation you were solving, and I made a bad assumption: I thought you were solving the equation $f(x) = y_0$ for some (arbitrary) $y_0$, and I didn't see what was gained by "transforming" that to $g(x) = z_0$ unless there was some relationship between $g$ and $f$ (such as $g(x) = \log f(x)$). I now realize that you were solving the equation $f(x) = 0$, so your "transformation" involves converting that to $f_A(x) = f_B(x)$, then to $g_A(x) = g_B(x)$, then to $g(x) = 0$, where $g_A(x) = \log f_A(x)$ and $g_B(x) = \log f_B(x)$. | |
| Sep 18, 2019 at 3:23 | comment | added | Claude Leibovici | @ruakh. Not at all. What you suggest is better that the original function but is almost hyperbolic while $g(x)$ is almost linear. | |
| Sep 17, 2019 at 18:49 | comment | added | ruakh | I think you meant to write $$f(x)=\frac{\sum_{i=1}^n a_i^x}{b^x}$$ (with division rather than subtraction). Otherwise your "transform" to $g(x)$ gives a more-or-less unrelated function; no?) | |
| Sep 17, 2019 at 12:43 | history | edited | user21820 | CC BY-SA 4.0 | deleted 5 characters in body |
| Sep 17, 2019 at 9:52 | history | edited | Claude Leibovici | CC BY-SA 4.0 | added 5 characters in body |
| Sep 17, 2019 at 1:43 | history | edited | Claude Leibovici | CC BY-SA 4.0 | added 42 characters in body |
| Sep 17, 2019 at 1:40 | vote | accept | Sunden | ||
| Sep 16, 2019 at 17:18 | comment | added | Claude Leibovici | @mathmandan. This is correct. Thanks & cheers | |
| Sep 16, 2019 at 16:52 | comment | added | mathmandan | Nice (+1)! In your example, I think the notation $p_i$ means the $i$th prime? So for $n=6$, you'd have $(a_1, a_2, \ldots, a_6) = (2, 3, \ldots, 13)$, and $b=17$? | |
| Sep 16, 2019 at 5:53 | history | answered | Claude Leibovici | CC BY-SA 4.0 |