Timeline for Software to decide equality between expressions involving powers and trigonometry
Current License: CC BY-SA 3.0
8 events
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| Jun 20, 2012 at 20:11 | history | edited | Robert Israel | CC BY-SA 3.0 |
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| Jun 20, 2012 at 19:03 | comment | added | Manu | Thanks! This seems to solve basically what I asked, except I guess you can only handle integer exponents this way? It is also not hard to implement. If you make it an answer I'll accept it. | |
| Jun 19, 2012 at 15:27 | comment | added | Robert Israel | For trigonometric polynomials in rational multiples of $\pi$ with rational coefficients, equality can be checked algebraically. Let $x = \sum_{j=1}^n a_j \cos(b_j \pi/m)$ where $m$, $a_j$ and $b_j$ are integers, $0 \le b_j < 2 m$. Then if $\omega = e^{i\pi/m}$ we have $x = \sum_{j=1}^n a_j (\omega^{b_j} + \omega^{-b_j})/2$. So $x = 0$ iff the minimal polynomial of $\omega$, which is the cyclotomic polynomial $\Phi_m(z)$, divides $\sum_{j=1}^n a_j (z^{2m+b_j}+z^{2m-b_j})$. | |
| Jun 19, 2012 at 14:22 | comment | added | Manu | Thanks, but my question still stands. One can obtain a zero-error "testeq" by trying all random choices. Is this available somewhere? Also, I think this all boils down to bounding the distance from 0 of non-zero expressions -- are there bounds available for expressions involving cos/sin and powers? I am aware of Baker's work on logarithmic forms. | |
| Jun 18, 2012 at 15:40 | comment | added | Jacques Carette | testeq is by design probabilistic. If it returns false, it is always correct, if it returns true, there is a very small chance it is wrong. User beware. For certainty, see my comments on my answer. | |
| Jun 18, 2012 at 13:41 | comment | added | Manu | Thanks, this looks useful. Is there an implementation of testeq that is always correct but may take a long time? | |
| Jun 18, 2012 at 3:46 | history | edited | Robert Israel | CC BY-SA 3.0 |
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| Jun 18, 2012 at 3:38 | history | answered | Robert Israel | CC BY-SA 3.0 |