Timeline for How would I use mathematica to solve this equation?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 10, 2018 at 15:45 | comment | added | Alexei Boulbitch | @ Carl Woll That was also my concern. On the other hand, taking a derivative is a rather standard trick for integral equations, though, of course, all information about A is lost. | |
| Jan 10, 2018 at 15:10 | comment | added | Carl Woll | @AlexeiBoulbitch You're making the same error I originally made. Consider the equation $x^2=1$. Taking a derivative with respect to $x$ does not yield the same roots. | |
| Jan 10, 2018 at 15:09 | history | edited | Carl Woll | CC BY-SA 3.0 |
Fix a few bugs
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| Jan 10, 2018 at 11:33 | comment | added | Alexei Boulbitch |
I am not quite sure, but it seems that the by calculating the derivative of the initial equation with respect to u one immediately finds f^(-1)(u)==0, is not it?
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| Jan 9, 2018 at 22:34 | history | edited | Carl Woll | CC BY-SA 3.0 |
Remove unnecessary code
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| Jan 9, 2018 at 22:13 | history | undeleted | Carl Woll | ||
| Jan 9, 2018 at 22:13 | history | edited | Carl Woll | CC BY-SA 3.0 |
Fix error
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| Jan 9, 2018 at 18:01 | history | deleted | Carl Woll | via Vote | |
| Jan 9, 2018 at 17:39 | comment | added | user56834 | Where did you get $0=f^{-1}(u)$? | |
| Jan 9, 2018 at 14:57 | comment | added | user56834 | I don't understand why those two constraints imply $a=0$? | |
| Jan 9, 2018 at 14:36 | history | answered | Carl Woll | CC BY-SA 3.0 |