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Old 04-19-2008, 07:16 AM
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Default Derive the Cubic formula

Hi, well, I tried deriving the cubic formula using solving the cube method but the cx and bx^2 seems to be a nuisance.

Ice Sync
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Old 04-19-2008, 07:50 AM
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Hi, well, I tried deriving the cubic formula using solving the cube method but the cx and bx^2 seems to be a nuisance.

Ice Sync
If you are serious about this, try Cardano's method.

-Dan
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Old 04-20-2008, 06:28 AM
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If you are serious about this, try Cardano's method.

-Dan
Yeah well, I saw read that article before I posted it,
anyways I did derive a formula by "completing the cube" method, just took 5 minutes, sadly it only works for perfect cubes or cubic equations with one root.

It is [-b+ CUBEROOT(-27a^2d+b^3)]/3a

-27a^2d+b^3 is the discriminant, funny part is, the discriminant is always 0 so the actual formula is -b/3a.


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Old 04-20-2008, 06:38 AM
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Originally Posted by ice_syncer View Post
Yeah well, I saw read that article before I posted it,
anyways I did derive a formula by "completing the cube" method, just took 5 minutes, sadly it only works for perfect cubes or cubic equations with one root.

It is [-b+ CUBEROOT(-27a^2d+b^3)]/3a

-27a^2d+b^3 is the discriminant, funny part is, the discriminant is always 0 so the actual formula is -b/3a.


Ice SYnc
There are occasionally neat formulas that work for specific kinds of cubics, but solving one in general is rather difficult. And if you think cubics are bad, you should try quartics!

-Dan
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Old 04-20-2008, 06:43 AM
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There are occasionally neat formulas that work for specific kinds of cubics, but solving one in general is rather difficult. And if you think cubics are bad, you should try quartics!

-Dan
And if you want to contemplate the impossible, try quintics
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Old 04-20-2008, 06:46 AM
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Hi, well, I tried deriving the cubic formula using solving the cube method but the cx and bx^2 seems to be a nuisance.

Ice Sync
Cubic function - Wikipedia, the free encyclopedia

That is the general case....that would be nearly impossible to solve for...Wow

just one root generally would be x=---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- that long basically
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\int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1)-\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}
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Old 04-20-2008, 08:34 AM
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There are occasionally neat formulas that work for specific kinds of cubics, but solving one in general is rather difficult. And if you think cubics are bad, you should try quartics!

-Dan
OK, forget the formulas, here's the best, use remainder theorem and reduce it to a quadratic equation , the use the quadratic formula!!


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Old 04-20-2008, 08:37 AM
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And if you want to contemplate the impossible, try quintics
How about n-thics ( degree as n )

wow , then we'd get a universal formula for solving any equation, forget it, use the remainder theorem
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Old 04-20-2008, 09:16 AM
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How about n-thics ( degree as n )

wow , then we'd get a universal formula for solving any equation, forget it, use the remainder theorem
Well, no "n-thics" can yield a formula in terms of its coefficients for n > 4 by Abels Impossibility theorem. Try this link Abel's Impossibility Theorem -- from Wolfram MathWorld
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Old 04-20-2008, 09:37 AM
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Well, no "n-thics" can yield a formula in terms of its coefficients for n > 4 by Abels Impossibility theorem. Try this link Abel's Impossibility Theorem -- from Wolfram MathWorld
Wait isnt due to Galois Theory that we know there cant be a quintic equation?
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\int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1)-\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}
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