Redpill me on cardano's method for solving cubic's Veeky Forums...

redpill me on cardano's method for solving cubic's Veeky Forums. all youtube gives me are fucking pajeets mumbling on in some gutterspeak with quantitative coefficients and crazy pajeet calculating and its just nonsensical habberdash.

pic related its what im trying to get from

ax^3 + bx^2 + cx + d

take your numbers elsewhere i want valid general proof for the general not 133 = b pajeets i want rigor godfuckingdammit

Other urls found in this thread:

youtube.com/watch?v=yXRdwvxxe2I
warosu.org/sci/thread/7529602
youtube.com/watch?v=HZYgPx031pA
twitter.com/NSFWRedditImage

Divide by the dominant coefficient to get a unitary polynomial.
[eqn]X^3 \,+\, a\,X^2 \,+\, b\,X \,+\, c[/eqn]

Substitute [math]X \,=\, Y \,-\, \frac{a}3[/math] to suppress the square term. You get something like:
[eqn]Y^3 \,+\, p\,Y \,+\, q[/eqn]

Suppose the solution is under the form [math]\sqrt[3]{u} \,+\, \sqrt[3]{v}[/math] and solve for these. You should come across the easy problem "find two numbers knowing their sum and product."

you must be a pajeet if youre teasing me that hard

yes its just algebra and that y-a/3 is so fucking wet

anyway i found the real fucking shill for us god damn is this what im in for if commit debt for maths? this is so lit

youtube.com/watch?v=yXRdwvxxe2I

>redpill me
gtfo pill-popping /pol/esmoker

I'm bumping this thread first to keep it alive, then I'll reply.

Now that the thread is alive, hello OP, you or anyone else may take interest in this (ultimately autistic high-school tier) problem.

Still, it's of interest to me. I groped towards my own derivation by following wiki about two years ago. The results are archived here:

warosu.org/sci/thread/7529602

I did not literally follow Cardano's method, but instead used third roots of unity and a trick which I attribute to John Derbyshire (though it almost certainly was used by someone else), to get a sort of eight/nine step process which made sense to me personally at the time.

Pic related depicts the three roots of the general cubic equation

[math] ax^3 + bx^2 + cx + d = 0 [/math]

, assuming real coefficients a,b,c,d. These expressly depicted forms are what you end up with once you "un-simplify" your whole process.

An excuse to repost these is always fun. Compare with the picture I meant to post just now.

[math] \displaystyle x = \bigg( - \frac{1}{3a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 + \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } + \bigg( - \frac{1}{3a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 - \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } - \frac{b}{3a} [/math]

[math] \displaystyle x = \bigg( \frac{ 1 + \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 + \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } + \bigg( \frac{ 1 - \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 - \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } - \frac{b}{3a} [/math]

[math] \displaystyle x = \bigg( \frac{ 1 - \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 + \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } + \bigg( \frac{ 1 + \sqrt{3} i }{6a} \bigg) \sqrt[3]{ \frac{1}{2} \bigg( 27a^2 d - 9abc + 2b^3 - \sqrt{ 729a^4 d^2 - 486a^3 bcd + 108a^3 c^3 + 108a^2 b^3 d - 27a^2 b^2 c^2 } \bigg) } - \frac{b}{3a} [/math]

These forms are actually /relatively attractive/ in that they don't involve any nasty denominator shit, however wiki suggests that they don't cover the most general complex case. I've never closed the loop on this, honestly.

So anyway real talk OP, this is generally what you want to do.

First of all, get your terminology straight. you wrote "ax^3 + bx^2 + cx + d", which is a polynomial. But it's not a polynomial /equation/. For that, set what you wrote equal to zero.

Also recognize that your polynomial involves /one, single variable/. Thus it is a /univariate/ polynomial, as opposed to a /multivariate/ polynomial (where the latter adjective describes more than one variable).

Next, divide through by the leading (manifestly non-zero, so you can divide by it) coefficient a. This leads to a new, /monic/ polynomial, or monic univariate polynomial equation.

Now comes the first real step where you get to do your own work to convince yourself. you make the substitution x = t - b/(3a), which ultimately leads to a new polynomial equation where its next-highest degree term has a zero coefficient, and therefore vanishes. Such a polynomial equation, with a zero-coefficient for its second highest degree, is called a /depressed/ polynomial equation.

When you put all of these terms together, what do you end up with? /A monic, depressed, univariate, polynomial, equation/. I refer to it as an "MDUPE". This is partly autistic, but partly-not. This whole problem is bookkeeping, and you need to keep your books straight and understand clearly exactly what it is that you're working with. that's why I've stressed the above terms in this order.

Your situation is the analogue of the quadratic equation, leading to the quadratic formula. It's just two/three-ish orders of magnitude more complex.

>>>/lgtb/

I'm not even "white"

So now we can talk turkey about the rest of (my version of) the proof.

1) divide through by the leading coefficient a. (get a monic univariate polynomial equation)

2) substitute x = (t - b/(3a)) and rearrange to get a new cubic equation in terms of t, which manifestly has no quadratic term (that is, the coefficient vanishes - get an MDUPE. This type of substitution is called a /Tschirnhausen substitution/. Protip: it is equally valid to derive the quadratic formula. Try it!)

3) give simple names to the coefficients of this new cubic in t, which are given in terms of the original a,b,c,d. Call them p and q, say. Bookeeping. Just never forget that you want to get back to those a,b,c,d when you're all done.

4) set t = w - p/(3w). This is a type of substitution, a further trick, which may be called a /Viete substitution/. You must convince yourself that there really exists such a w! This is an important exercise which I leave for you (use the quadratic formula).

5) If you're good that this "w" really exists, then you can yet-agian substitute in the above, in place of t.

6) Rearrange the above. You get a polynomial of degree six in terms of w, with just one other term of degree three in terms of w, and a constant term (with respect to w). note that the above p and q terms act in the role of coefficients just now. In other words... you have a quadratic in terms of w^3, so we may as well call w^3 = z.

7) Solve for z using the quadratic formula. Since z = w^3, simply (and brashly) solve for w, by just... taking the third root!

8) At this point, I found it helpful to identify and specify, for my personal use, a number of "discriminant" forms, and "conjugate" forms. It is left to you as an exercise to know exactly what those things are, to keep your bookkeeping straight.

9) This is the big non-obvious trick which Derbyshire invoked, and which I used to complete my own proof:

[math] (w + \bar{w})^3 + p(w + \bar{w}) + q = 0 [/math]

cont.

10) The next part is to use /all three third roots of unity in general/.

If you really want to figure this out for yourself OP, I've given you almost all the tools to do the autistic paper chase. It's on you whether you really want to do the work to make your own understanding.

>I'm not even a /pol/esmoker
>I'm not, I'm not, I'm not
Yes, you are.

>we can talk turkey
people don't talk turkey any more, Grandpa

OP here bumping ill type a response once i have my head around this.

Honestly just use my write-up and follow wiki. I use slightly different notation than wiki because that's my preference, but it comes to much the same thing..

It will probably take you a day or two of grinding to really start getting it, but it's all just one big grind, but satisfying when you've been through it. If you get stuck, bump this thread and I'll be happy to help once you have a detailed, substantive question of some kind.

>redpill me
>i want rigor
gtfo

>im just going to post my thoughts for the night after chipping away at the cheese

im having trouble with the jsMath script on the warosu page, after the t = w + wbar sub it becomes indiscernible

so getting to the cube root invocation was straight forward enough once i got my signs from the binomial expansions straight and was able to successfully combine like terms for a greatly simplified equation.

i find it very interesting we "complete the cube" of the original cubic equation to depress the square term (your tschirnhausen sub), and then complete another cube in the resulting associated cubic (your viete sub) to find ourselves with a quadratic equation.

perhaps a platonic (trite ) way of stating this could be that a /particular/ quadratic equation is emanating from a /general/ cubic function.

an interesting thought occurs making me wonder if we cannot from the general quadratic find ourselves with a particular cubic.

calculus was long ago but this seems to be ringing some
derivative and integral bells. fx = x^3 ; f'x = 3x^2 ; f"x = 6x.
but "condensing" a cubic into a quadratic is, im guessing,
functionally different from "expanding" a quadratic into a cubic?

>does the resulting equation for z involving the initial cube root on w^3 enclosing the quadratic term z = (-q/2 +/- (q^2/4 + p^3/27)^1/2)^1/3
>provide us with the box we plug our real values into?

work backwards with given abcd? ill plug and chug it.

the derbyshire trick is another beast that ill work tomorrow im assuming its where the trig identities enter the process.

cube root invoked. so am i understanding correctly we have increased the number of roots.

> ((1)^1/3)
> (((1)^2)1/3)
> (((1)^3)1/3)

You are halfway home. your p, q and w look right next to my notes. Keep going.

Unfortunately, warosu does not perserve tex but at least the code is there.

Now that I check, my "w-bar" is just a notation I made up in order to specify both of the w's depending on whether the + or - sign is taken in the quadratic formula form.

Juust one little bump to give the OP a chance to report back.

OP here, back from the depths of this algebraic forest, though i havent completed the proof, i feel goddamn close.

as a quick aside this type of thinking is awfully similar to the thinking that goes into composing the contrapuntal vocal polyphony the italians were loving during the same period cardano was jerking his pencil around this cubic. totally unsurprisingly gombert and cardano were contemporary comrades in italy during the 16th.

>in that you must know where youre going before you can know what direction in which to begin
youtube.com/watch?v=HZYgPx031pA

anyway, ive worked now to what i believe to be some type of bottom to the original equation in terms of w wbar, which as we recall is just the representation of two conjugate quadratic terms (+/-) u and v which are themselves in terms of p and q which turn to our original abcd coefficients. i still have to walk the whole thing back up, but it felt good to hit what i think is the bottom.

i was particularly surprised to find w and wbar cancelling in such a way as to form the constants -q and -p/3 to provide the requirements of cardanos conditions {u^3 + v^3 = -q} and {uv = -p/3}. it seems like sorcery that these terms cancel in just this way. there are a lot of freedoms being taken with the radicals but this doesnt detract from the beauty im encountering here.

its really quite fascinating to me the way this whole thing ends up looping back to itself.

but there seem to be multiple methods of finding solutions to the depressed cubic, depending on whether you start with real coefficients (i think your post was to find if cardanos method held up for complex coefficients) or not.

your detailed post and particularly the wikis are turning out to be invaluable.

this time around i dont have clean sheets to scan as there was substantially more trial and error in figuring out what went where when. but i will.

see whats happened is ive now only proven the existence of such values that when given will zero the equation. and is this the zinger?

>how do we find such values

sure there are infinitely many of them, but they are very hard to find arent they?

we reduced the equation by a substitution method. and then we reduce the equation again by another substitution method. and then through another similar transformation (we just use the quadratic equation, but the proof of such involves using an identical substitution technique). we can increase the number of potential roots, and through a (im reaching here) matrix evaluation, by knowing that sqrt1 and sqrt1^2 share at least one root with cbrt1 and cbrt1^2 and cbrt1^3. and when given two variables to play with we can find sets of related root patterns who share with other sets of root patterns and by a process of elimination determine that such values exist who can zero the function.

but how do we find the values? because it doesnt just work with any number you inject into abcd, and working from the bottom isnt any more fruitful.

Hello OP, other guy here. Since you're putting in good effort let me give context for my reliance on Derbyshire. I fail to find exact citation at-a-glanc but he gives the history and actually crunches the letters around pages 30-100-ish or so in his /Unknown Quantity/, which impressed itself on me when I was younger. I may have just gleaned something from him without him actually doing it thereabouts, but there you go.

Wiki seems to have been edited significantly since I used it for this exercise. The latter bits go like this, if you haven't done already.

You want to pass from

[eqn] (w + \bar{w}) ^3 + p (w + \bar{w}) + q = 0 [/eqn]

to

[eqn] (uw + \bar{u} \bar{w}) ^3 + p (uw + \bar{u} \bar{w}) + q = 0 [/eqn]

where the u-u-bars are the third roots of unity as needed. (check it out, it works). This entails assuming knowledge of things that Cardano didn't have available (just like FTAlgebra), but that works in my modern version of things, to use all available tools at-will. Then you can wrap things up.

NOW, the big sticking points which I've never personally dealt with as of yet, but which are closely addressed in wiki etc: algebraic incompleteness of the reals. these operations assume real coefficients throughout, branching via root extraction, and other hairy stuff. This is what has spurred me to get back my old notes.

bump