Timeline for How do I improve this code for hyperbolic numbers so to allow equation-solving?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Dec 12, 2021 at 10:12 | answer | added | Anixx | timeline score: 0 | |
| Nov 3, 2021 at 9:49 | answer | added | user3257842 | timeline score: 1 | |
| Nov 3, 2021 at 7:51 | comment | added | Anixx | @user3257842 real. | |
| Nov 2, 2021 at 17:06 | comment | added | user3257842 | Are the $a$ and $b$ in your equations supposed to be real, or split-complex numbers themselves? | |
| Nov 2, 2021 at 16:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 3, 2021 at 15:45 | answer | added | Daniel Lichtblau | timeline score: 0 | |
| Oct 3, 2021 at 14:41 | comment | added | Anixx | @DanielHuber look, in split-complex numbers there is a formula: $f(a+bj)=\frac{1}{2} (f(a-b)+f(a+b))+\frac{j}{2} (f(a+b)-f(a-b))$, take $f(z)=\ln z, a=0,b=1$ and you will see the result. $i \pi$ is logarithm of $-1$. $e^{i\pi}=-1$. | |
| Oct 3, 2021 at 14:35 | comment | added | Daniel Huber | You are right. But what about defining Ln[J]= n Pi I , where n is an Integer? | |
| Oct 3, 2021 at 13:57 | comment | added | Anixx | @DanielHuber of course, not. This equality does not work even for -1. $\log j=\frac{i \pi }{2}-\frac{i j \pi }{2}$ You can see it using my code. | |
| Oct 3, 2021 at 13:51 | comment | added | Daniel Huber | As J^2=1 it follows 2 Ln[J]= 0. Therefore Ln[J]=0. | |
| Oct 3, 2021 at 13:36 | comment | added | Anixx |
@DanielHuber in that case you cannot evaluate expressions of split-complex numbers, such as Log[J], etc. That's why I need improvement of my code. I need solving logarithmic equations, exponential equations, trigonometric equations, etc.
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| Oct 3, 2021 at 13:34 | comment | added | Daniel Huber | Note, I did not use your function, I simply set J^2=1 and got he above result. And try a new kernel in case you have old definitions. | |
| Oct 3, 2021 at 13:24 | comment | added | Anixx |
@DanielHuber and for Solve[(a+b J)^2==1,{a,b}] I do not get anything.
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| Oct 3, 2021 at 13:23 | history | edited | Anixx | CC BY-SA 4.0 |
added 37 characters in body
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| Oct 3, 2021 at 13:20 | comment | added | Anixx |
@DanielHuber I get nonsciential MatrixFunction[ Function[J, {{b -> -Sqrt[J] - a J}, {b -> Sqrt[J] - a J}}], J] Is it what you get as well?
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| Oct 3, 2021 at 13:15 | comment | added | Daniel Huber |
I have MMA 12.3. what version do you have? For your first equation I get the result: : {{b -> -Sqrt[J] - a J}, {b -> Sqrt[J] - a J}}
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| Oct 3, 2021 at 10:12 | history | edited | Anixx | CC BY-SA 4.0 |
added 111 characters in body
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| Oct 3, 2021 at 10:09 | comment | added | Anixx | @DanielHuber does not work for me, makes things worse, even expressions after this are not evaluated. | |
| Oct 3, 2021 at 10:01 | comment | added | Daniel Huber |
Why not simply define J^2==1 ? (after unprotecting Power)
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| Oct 3, 2021 at 8:32 | history | asked | Anixx | CC BY-SA 4.0 |