Timeline for Solving symbolically a trascendental equation containing an exponential
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| May 27, 2022 at 8:03 | vote | accept | umby | ||
| May 24, 2022 at 12:58 | history | edited | rhermans | CC BY-SA 4.0 |
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| May 24, 2022 at 11:41 | history | edited | rhermans | CC BY-SA 4.0 |
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| May 24, 2022 at 11:08 | vote | accept | umby | ||
| May 24, 2022 at 11:13 | |||||
| May 24, 2022 at 11:07 | comment | added | umby | @rhermans, thank you, I see that it is probably impossible to get a simple expression. Your approach works quite fine however. | |
| May 24, 2022 at 10:54 | history | edited | rhermans | CC BY-SA 4.0 |
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| May 24, 2022 at 10:46 | comment | added | rhermans | @Nasser see the second figure, where I plot RHS too for those arbitrary values. It means the LHS and RHS cross in two places. | |
| May 24, 2022 at 10:45 | history | edited | rhermans | CC BY-SA 4.0 |
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| May 24, 2022 at 10:43 | comment | added | Nasser | Your answer is good. I was just wondering what the second value mean, that is all. | |
| May 24, 2022 at 10:41 | comment | added | Nasser |
i.e. Plot[{Exp[-x^2], c1/Sqrt[c2 - c3* x]} /. {c1 -> 1, c2 -> 2, c3 -> 3}, {x, -2, 0}] shows one root at x->-0.874565 which your method gives correctly when using x=-1 for approximation. But second value it also gives, which not sure what it means.
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| May 24, 2022 at 10:39 | comment | added | Nasser |
When using approximation at -1, now your method gives good agreement. -0.874565 But it also gives second one -0.216816, which does not show in the plot as being a root?
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| May 24, 2022 at 10:33 | comment | added | Nasser |
This does not match the root for x<0 which is x -> -0.875079 when using {c1 -> 1, c2 -> 2, c3 -> 3} ? Your gives -1.78078 for these c values.
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| May 24, 2022 at 10:32 | history | edited | rhermans | CC BY-SA 4.0 |
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| May 24, 2022 at 10:26 | history | answered | rhermans | CC BY-SA 4.0 |