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Seth says | |||
| ok so i was doing my math hw and i stared getting really annoyed with finding roots of cubics because i had no therums that didnt require trail-and-error to some degree so i started googleing around and i think this formula might work the best of any other that i found. can anyone better at math than me verify that this does indeed work? http://planetmath.org/encyclopedia/CubicFormula.html |
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| Total Topic Karma: 19 | - More by this Author | |||
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Seth says |
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| this is i mean does it work for ALL cases? | ||||||
| - 03 February, 2007 | ||||||
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Seth says |
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| *that (not "this") | ||||||
| - 03 February, 2007 | ||||||
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jared.nance says |
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| seth, that formula will have problems with certain coefficients where you will end up dividing by zero... not so good. check out mathworld for an excellent treatment of this problem, and many others as well... http://mathworld.wolfram.com/CubicFormula.html |
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| - 03 February, 2007 | ||||||
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Seth says |
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| dang, i was hoping that it would be all in one formula | ||||||
| - 04 February, 2007 | ||||||
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Seth says |
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| im confused. everything in the mathworld eqution seems to come down to lines (53-55) which seems to be the same as the equation from PlanetMath, just with S and T so that its easier to read | ||||||
| - 05 February, 2007 | ||||||
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Joe says |
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| I haven't worked it out, but you can just substitute S and T, and then Q and R into that final equation. That will most likely give you that insane equation listed on the other site. If so, that's pretty awesome. Now all you have to do is prove mathematicians of the past few hundred years wrong and generalize up to polynomials of order n. | ||||||
| - 06 February, 2007 | ||||||
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Seth says |
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| but thats what im saying jared told me the insane one on the other site didnt work because of divid by zero problems. btw, jared are those div by 0 problems in any specific equations? cause if i can solve for r1, then it can just divide the cubic by (x-r1) and get a quadratic |
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| - 06 February, 2007 | ||||||
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jared.nance says |
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| yeah, it seems that the division by zero occurs for certain values of the coefficients. not so good. the best thing to do to solve cubics is just follow cordano's procedure... and believe me, you don't even want to see the quartic formula. and the quintic, luckily, doesn't exist. it has been proven that you cannot analytically get the roots. | ||||||
| - 06 February, 2007 | ||||||
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Joe says |
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| I'm still not seeing where there is a problem dividing by zero. All of the coefficients go in the numerator and it's only constants in the denominator. Unless I'm missing something... | ||||||
| - 06 February, 2007 | ||||||
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(Guest) Seth ( forgot to sign in) says |
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| which coefficiants would that be for? |
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| - 06 February, 2007 | ||||||
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Seth says |
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| http://www.1728.com/cubic2.htm thats a another thing i found. wyhat do u think of that one? |
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| - 06 February, 2007 | ||||||
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Joe says |
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| Here's something interesting. The "cubic formula" as given above only works if all the roots are real. It does not work if two of them are complex. I'm not sure why, but I'll try to find out. | ||||||
| - 07 February, 2007 | ||||||
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Seth says |
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| which one? the 1728.com one? |
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| - 08 February, 2007 | ||||||
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Joe says |
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| No, the first planet math one. | ||||||
| - 09 February, 2007 | ||||||
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Viczy says |
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| Don't know if this helps or not, but you can find the values of x for the critical points of f(x) as follows: ± I think that checks out. Hope it helps! |
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| - 21 February, 2007 | ||||||
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Viczy says |
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| Gah, ignore tha A^ thingy. Damned no edits... | ||||||
| - 21 February, 2007 | ||||||
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Seth says |
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| ur math chacks out but im not sure it helps all that much cuz as u said, thats finding the critical points. i dont think that really helps in finding the roots though. | ||||||
| - 22 February, 2007 | ||||||
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Viczy says |
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| Ah, alright. My bad. | ||||||
| - 23 February, 2007 | ||||||
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