Timeline for Solving equation
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 18, 2016 at 22:05 | vote | accept | John | ||
| Apr 18, 2016 at 22:05 | vote | accept | John | ||
| Apr 18, 2016 at 22:05 | |||||
| Apr 18, 2016 at 22:04 | comment | added | John | @MichaelE2, thank you very much. You solved my problem. Comments from @rewi are also useful. Again, thank you! | |
| Apr 18, 2016 at 18:33 | comment | added | Michael E2 |
@rewi Don't you just need to use a negative value for c and then they'll look the same? DSolve probably just did something like that, since the c disappears from my DE.
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| Apr 18, 2016 at 13:02 | comment | added | user36273 | @Michael E2 Thanks Michael E2. Is my thinking wrong when I build the inverse of f in my solution? Should I delete my answer? | |
| Apr 18, 2016 at 12:46 | comment | added | Michael E2 |
@rewi Do you mean Log[x] should be an issue? In Mathematica Log is defined for negative real numbers. It's complex-valued, but the logarithms in the inverse function cancel out the imaginary part of Log[x]. E.g., Solve[Log[-1] == Log[x]] and FindRoot[Log[-1] == Log[x], {x, -0.1}]. Is that what you meant? (Otherwise it is clear from the equation that the solution should be symmetric in this way.)
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| Apr 18, 2016 at 12:32 | comment | added | user36273 | @Michael E2 Your function is mirrored at the x axis | |
| Apr 18, 2016 at 10:13 | comment | added | Michael E2 |
@rewi I'm not sure what you mean by "course." The second formula, which I got from manipulating the DSolve result, has for its argument to InverseFunction the left-hand side of the OP's equation up to a constant factor, with #1 replacing a. I thought that clarified the relationship of the DSolve solution to the equation, especially the relationship between the constants. But perhaps you meant something else?
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| Apr 18, 2016 at 9:59 | comment | added | Michael E2 |
@JohnM. Over a finite interval, one can construct an interpolating function with NDSolve or FunctionInterpolation, which would evaluate faster than InverseFunction. Or one could use another approximation method, such as a Chebyshev series expansion I used here. One might try fitting a formula (see FindFit, LinearModelFit, and the whole *Fit family).
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| Apr 18, 2016 at 9:37 | comment | added | user36273 | @Michael E2 Why you have another functional course? | |
| Apr 18, 2016 at 0:38 | comment | added | John |
Thank you again. This was useful. I noticed that I can not represent dsol with Tanh[x], they are similar but not the same. However, is there a way to represent this curve (sigmoid function) with some function (approximate solution) in term of x?
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| Apr 17, 2016 at 23:23 | comment | added | Michael E2 |
@JohnM. You're welcome. I think you can see from this image that, while dsol produces a sigmoid-like plot similar to Tanh, it is not in fact the same.
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| Apr 17, 2016 at 23:23 | history | edited | Michael E2 | CC BY-SA 3.0 |
Explanation
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| Apr 17, 2016 at 22:53 | comment | added | John |
thank you for your answer. Your graph (dsol) presents solution to this equation and it looks like Tanh[Constant*x], where Constant is related to c. For different values of c the shape of Tanh[] will be different. So, how can I find this Constant? For example, to have solution in form of Tanh[x].
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| Apr 17, 2016 at 14:50 | history | answered | Michael E2 | CC BY-SA 3.0 |
