Timeline for How does Mathematica Find Integer Roots For Certain Transcendental Functions?
Current License: CC BY-SA 4.0
14 events
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| Jul 23, 2019 at 22:14 | comment | added | DUO Labs | @yarchik Then how would you solve these types of polynomial equations? | |
| Jul 23, 2019 at 20:45 | comment | added | yarchik | Your equations are also algebraic for $n\in\mathbb{Z}$. Thus, methods of solving polynomial equations apply. I believe this is how MA does it. | |
| Jul 23, 2019 at 18:40 | comment | added | user64494 | Neither Reduce[Cos[nPi/x]*Cos[xPi] == 1, x, Reals] nor Reduce[Cos[5*Pi/x]*Cos[xPi] == 1, x, Reals] do the job in version 12.0. Could you support your claim by a working code? The FindInstance[Cos[5*Pi/x]*Cos[xPi] == 1, x, Reals, 2] command fails too. | |
| Jul 23, 2019 at 16:46 | comment | added | N.J.Evans | @QuoteDave Can you provide the actual code you used to find the roots? | |
| Jul 23, 2019 at 16:35 | comment | added | DUO Labs | @N.J.Evans Well, that only divides any given range (where only one root must lie in) by around 50% if you know that one x is odd. That still doesn't answer how can they do it. | |
| Jul 23, 2019 at 16:32 | comment | added | N.J.Evans | @QuoteDave For f(x), Cos has a range [-1,1], so Cos(a)*Cos(b) = 1 iff |Cos(a)|=|Cos(b)|=1 and the signs are the same. Since Cos(a)=+/-1 iff a = j*Pi where j is an integer whose parity determines the sign, both x and n/x must be integers of the same parity. | |
| Jul 23, 2019 at 15:59 | comment | added | DUO Labs | @N.J.Evans You said something about "guessing the solutions" using "some logical arguments". What do you mean by that? | |
| Jul 23, 2019 at 15:55 | comment | added | DUO Labs | @N.J.Evans Maybe, I know the roots to $f(x)$ are maxima and for $g(x)$ it's minima-- so maybe they calculate the deriative, find all the zeroes in the range (somehow), then check all of them. However, I do't think that's true as for big $n$, there are a lot of extrema very close together. | |
| Jul 23, 2019 at 15:52 | comment | added | N.J.Evans | It's pretty straightforward to guess the solutions to each using some logical arguments - not strictly symbol manipulation. I'm not sure how much of that MMA can do though. Maybe it finds numerical solutions and has some provision to test integers if the numerical solution is converging to the integer within some precision? Some of the folks here have a pretty deep knowledge of how these things work, maybe one of them will chime in. | |
| Jul 23, 2019 at 15:41 | comment | added | DUO Labs | Hmmm, maybe they check where $n/x$ is an integer and see whether $n$ is an integer as well? | |
| Jul 23, 2019 at 15:26 | history | edited | DUO Labs | CC BY-SA 4.0 |
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| Jul 23, 2019 at 14:47 | comment | added | bill s | I doubt its possible to know exactly. But notice the simple case: Solve[Cos[x Pi] == 1, x] is solved easily. One might guess that more complex versions (such as your f(x) and g(x) build on this kind of simple (and likely "built in") answer. | |
| Jul 23, 2019 at 14:43 | history | edited | DUO Labs | CC BY-SA 4.0 |
added 30 characters in body
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| Jul 23, 2019 at 13:42 | history | asked | DUO Labs | CC BY-SA 4.0 |