Timeline for Inverse of a complicated function
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 22, 2012 at 4:07 | answer | added | Jens | timeline score: 3 | |
| Jun 22, 2012 at 3:37 | comment | added | whuber |
You have to track the root crossings carefully: interesting things happen at $\pi/2$ and $3\pi/2$! E.g., g = Solve[ Cos[s] == (-1 + 4.20278 r (0.008712/r^2 + 0.475876/r - 1/(1 + r)))/ Sqrt[1 - 10.598 r^2 (0.008712/r^2 + 0.475876/r - 1/(1 + r))], r]; Show[Plot[r /. g[[3]], {s, 0, \[Pi]/2}, PlotRange -> {{0, \[Pi]}, {0.15, 0.65}}], Plot[r /. g[[4]], {s, \[Pi]/2, \[Pi]}]]
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| Jun 22, 2012 at 0:46 | answer | added | Vitaliy Kaurov | timeline score: 13 | |
| Jun 22, 2012 at 0:06 | comment | added | Tuckerman | When I plot f[s], I see that the inverse of the function should increase from fmin~.19 to fmax~.602 as s varies from 0 to Pi. When plotting any of the Root[] functions over the interval (0,Pi) I find that they all have the same value at 0 and Pi. | |
| Jun 21, 2012 at 23:46 | comment | added | J. M.'s missing motivation♦ |
You could help things a bit by giving Solve[Cos[s] == (* stuff inside the ArcCos[] *), r], but that yields a bunch of Root[] objects, and you need to pick which solution is appropriate...
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| Jun 21, 2012 at 23:37 | history | asked | Tuckerman | CC BY-SA 3.0 |