Talk:Quartic equation

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1 It is too long: may be is adapt for a math book not an encyclopedia
2 solved issues
3 Rational Roots therom
4 Ferrari's solution
5 Substitution x→u

[edit] It is too long: may be is adapt for a math book not an encyclopedia

Perhaps this article would be better in a mathematical book (wikibook?). It is long, complex and difficult to read. It would be better if the article was shorter. In my opinion, all the demostration should be summarize up. A link to the full explanation of the methods could be given instead. Historic and modern view of the problem could be expand instead. AnyFile 16:57, 24 Oct 2004 (UTC)

How to solve a polynomial equation is a good example of a mathematical topic of general interest for which good information is hard to find. It's not just mathematicians who might want to know, and anyone who actually does want to know is going to want a working explanation. That being said, I think it could be made more concise, but should it be? The long part is "The General Case, along Ferrari's lines" which spends a lot of time bludgeoning the the obvious. user: Gene Ward Smith

But this is an encyclopedia not a tutorial for everything. Should I insert how to insert a PCI card in a computer or how to install a Range Hood or quote the complet text of a novel in case sombody needs it? There exist en.wikibook.org WikiBook, I suppose it could be a more appropriate place for a complete dissertation. AnyFile 12:53, 12 Nov 2004 (UTC)

I think AnyFile has a point in that long proofs and calculations do not belong in an encyclopedic article. However, I think that the final formula for the roots (however horrible it may be) should be in the article, and given the historical importance, Ferrari's approach should also be explained. Perhaps the best solution would be if "The General Case, along Ferrari's line" was made a bit more compact, if possible; a full discussion with all the gritty details minutiously explained could be given somewhere else (Wikibooks might indeed be an appropriate place for this). However, I don't think that the article is a "tutorial" or a "dissertation"; I think it is too detailed. -- Jitse Niesen 22:20, 13 Nov 2004 (UTC)

Nobody said this article is a tutorial. The article wants to describe an old, rather unique procedure for solving a quartic equation, while inserting something into something else, is quite easy (maybe you should add it to the article Greece; do not forget to mention, what happens if the thing is too long or too thick (note my subtile sense of humor. Thx)). Cheers. --83.129.175.32 02:47, 14 Nov 2004 (UTC)

I've just put in a slick solution to the quartic, since I have an actual application for solving the quartic and came here for the answer. A good first step for clearing up this article would be to eliminate all quadratic formulae from it. They are well known and can be referred elsewhere. Goatchurch 00:11, 6 November 2006 (UTC)

Very slick indeed. An upcoming article thanks you in advance. mdf 22:35, 21 November 2006 (UTC)

[edit] solved issues

Can somebody explain me, how in Quartic_equation#Reduction_to_a_biquadradic the formula $y 4 ...$ is created, please? Somehow I do not see, if it is substitution, and if yes, what by what. Thx. --Riddick 13:11, 6 Oct 2004 (UTC)

You can compute it by using a resultant, and eliminating x between

x 4 + c x 2 + d x + e = 0

and y - x² - px - q = 0, where p and q are as given. User:Gene Ward Smith

Is it possible to predict which combination of solution for the long and short y-equations leads to a valid solution for equation (1) (in Quartic_equation#Reduction_to_a_biquadradic)? Thx. --Riddick 16:14, 6 Oct 2004 (UTC)

It would be if instead we used y = (ax+b)/(cx+d) but I couldn't get that to work, though it seems it should. User:Gene Ward Smith

I was just wondering how in the summary, x (the solution) gives only two values when we are solving a quartic which should have four solutions. Thanks! (p.voges@qut.edu.au)

The signs were a little bit tricky, but I think, I fixed it... --Riddick 13:11, 6 Oct 2004 (UTC)

A certain Arne noted on Wikipedia:Reference desk that there seems to be a mistake in the solution given in the article. I had a short look and I think he is right. The formula for v, which solves equation (5), seems suspect. -- Jitse Niesen 16:08, 28 Sep 2004 (UTC)

It was me (Arne) :-)) --Riddick 16:42, 28 Sep 2004 (UTC)

Now it looks to me like the formula for v (after equation (5)) is correct for real v (I tried it with some example numbers).

But: If v is a complex number, then the formula for v does not bring values that solve (5). --Riddick 18:25, 28 Sep 2004 (UTC)

With the coefficients A=1, B=0, C=6, D=-60 and E=36 the formula for x delivers just two solutions x1=~3.100 and x2=~0.6444 (if I toggle the +/- sign; but indeed toggling the sign of the first square root (where no +/- sign can be found), results in invalid values for x) with v=9.0097912 (which is approximately a real root of (5) but not the result of the given formula for v). The remaining two solutions can be found by the procedure in the last section. But that does not look so smooth... (I strongly believe, that is because v is a real number in this example; maybe I use the wrong calculator). When I use the formulas from the summary I get (with + instead of +/-) x=~2.275-~1.347i and v=~6.304-~1.562i (both values are no roots of their corresponding equations)

on Quartic Equation -- from MathWorld I have found another description of Ferrari's solution (I just tested it, and its works fine (with coefficients from Ferrari's first one): x1=~3.100, x2=~0.6444, x3=~-1.872+~3.810i and x4=~-1.872-~3.810i=x3* with y1=~20.019582457; it sounds a little bit different at some lines (e. g. they demand a real root of the cubic equation)). --Riddick 21:47, 28 Sep 2004 (UTC)

Should I change something, although I do not know, where the mistake is? --Riddick 21:47, 28 Sep 2004 (UTC)

Arne says: Yes. --Riddick 20:38, 1 Oct 2004 (UTC)

I took two formulae from Cubic equation: $u=\sqrt[3]{{q\over 2}\pm \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}$ and $v={p\over 3u}$

and inserted them at the right(?) place in this article. --Riddick 20:00, 1 Oct 2004 (UTC)

experimental results: the sign patterns +++ = x1, -+- = x2, --- = x3, +-+ = x4 are good sign patterns (the others are not)

Hereby I recommend further tests and/or proves. Thx. --Riddick 20:38, 1 Oct 2004 (UTC)

A=1, B=0, C=6, D=-60, E=36: check
A=1, B=10, C=-6, D=60, E=36: check
A=1, B=-28, C=294, D=-1372, E=2401: check (remark: (x-7)^4 (it is a little bit tricky due to many zeros as inverse (mult)))

[edit] Why should S be 1?

As far as I understood mathematics $S$ can be -1, too, because: $\beta\in\mathbb{R}^-$ can be true and by convention $\sqrt{x}\in\mathbb{R}^+$ with $x\in\mathbb{R}^+$ is true (that's why we use a $\pm$ -sign in front of a $\sqrt.$ -sign sometimes)...

example:

A=1
B=-2
C=3
D=-4
E=-5
a = -3*B*B/(8*A*A) + C/A;
b = B*B*B/(8*A*A*A) - B*C/(2*A*A) + D/A;
c = -3*B*B*B*B/(256*A*A*A*A) + C*B*B/(16*A*A*A) - B*D/(4*A*A) + E/A;
P = -a*a/12 - c;
Q = -a*a*a/108 + a*c/3 - b*b/8;
R = Q/2 + sqrt(Q*Q/4+P*P*P/27);
U = exp(ln(R)/3);
y = -5*a/6 + P/(3*U) - U;
S = b/(2*sqrt(a+2*y)*sqrt(y^2+2*y*a+a^2-c))
a
       1.5
b
       -2
c
       -6.4375
P
       6.25
Q
       -3.75
R
       1.66870856171474269987
U
       1.18611509075441016287
y
       ~-.67968070291750861840
S
       ~-1.00000000000000000003
sic!

I recommend a revert asap. --83.129.237.17 02:59, 13 Nov 2004 (UTC)

It's correct it can be both 1 and -1, I was thinking (not thinking clearly) that people could select the sign of root for it to be correct. In any case, I changed it to ±1, and later rewrote some of the section without the S, since I did't like S much anyway... Κσυπ Cyp 2004年11月13日 (土) 18:45 (UTC)

I did not cross check your changes closely; I hope you will do or have done at least some tests with constructed examples, that have well known solutions (hint: keep in mind, that two polynoms of degree 4 are equal, if 5 points are equal, which helps to cross check, if the coefficients A-E are correct). --83.129.175.32 02:22, 14 Nov 2004 (UTC)

Apart from removing the S (and correcting a missing A² earlier in the article, my changes shouldn't have changed the math, only the layout of the math. I just checked with A to E being pseudorandom numbers between -2 and 2, and it found all 4 roots. Κσυπ Cyp 2004年11月14日 (日) 04:42 (UTC)

[edit] Why should we always have 4 pairwise different solutions?

With $A = 1, B = - 1, C = - 1, D = 1, E = 0$ we just have 3 solutions, so that the statement "Each sign pattern gives a different root" seems to be false or misleading (because: it rises the question "different from what?"). I recommend a revert or a completely different phrase (maybe by introduction of $s,r\in\{-1,1\}$ ). --83.129.237.17 03:26, 13 Nov 2004 (UTC)

Hopefully it's clearer now. Κσυπ Cyp 2004年11月13日 (土) 18:45 (UTC)

I do not see that "(although one would then have gotten a biquadratic equation earlier and therefore not be using Ferrari's solution)" is true (there is at least one case where $B\ne0 \wedge D\ne0$ (e. g.

(x - 1) 4 = x 4 - 4 x 3 + 6 x 2 - 4 x + 1

) where we have just one solution), but I do not care so much anymore. Maybe the statement "The set of values for x (4 +/- possibilities) denotes all solutions of (1) with $x\in\mathbb{C}$ , which can be less than 4 for various reasons." would be more professional.

Keep in mind, please, that in Mathematics precise and true statements are quite important, because else it is not an exact science but just the common nonsense like medicine science or engineer science...

Keep in mind, please, that I did not want to push you by the phrase "asap"; maybe I was just a little bit annoyed, because somebody made not so urgent changes, that have to be cross checked, which is quite time consuming, so that I will not do any cross checking anymore, because the general quality of Wikipedia articles is not what I would expect anyway, so that I do not work here anymore. Thx. --83.129.175.32 02:22, 14 Nov 2004 (UTC)

(x - 1) 4 = x 4 - 4 x 3 + 6 x 3 - 4 x + 1 = u 4 + 0 u 2 + 0 u

, which is a biquadratic equation, a case mentioned just before the start of the "Ferarri's solution" paragraph. Of course, it should be clearer, and say it's the β=0 case, instead of mentioning that it transforms into a biquadratic equation in a different part of the article. Κσυπ Cyp 2004年11月14日 (日) 04:42 (UTC)

[edit] Why should there be a +/- sign in the formula for R?

Since we just need one $R$ , the + suffices. I recommend a revert asap. --83.129.237.17 02:59, 13 Nov 2004 (UTC)

Think it's clearer with a ± and comment saying both roots work, since then people won't worry that they might be taking the "wrong" complex root. Κσυπ Cyp 2004年11月13日 (土) 18:45 (UTC)

Since there was no further specification for the value of $\sqrt.$ it means by convention, that we look for a value $n=\sqrt{m}$ that fulfills

n 2 = m

(there r special cases for $m\in\mathbb{R}^+$ ). Your notation might be misleading, because people might think, that square roots have a sign nowadays, or that there would be more than 4 solutions (maybe 8 or even 24... huh!? :-) )... But of course I don't care for this misleading $\pm$ ... --83.129.175.32 02:22, 14 Nov 2004 (UTC)

I don't mind whether it's there or not. Think it's slightly better with the ±, but it's probably clear without, too. Κσυπ Cyp 2004年11月14日 (日) 04:42 (UTC)

[edit] Why should we compactify the formula for x1-4?

Compactification is not necessarily simpler... I would like to see an explanation. --83.129.237.17 02:59, 13 Nov 2004 (UTC)

I think "Divide both sides by −4, and move the −β2/4 to the right," is simpler than "Multiply both sides by −1, then add β2 to both sides, divide both sides by 4, then subtract β2/4 from both sides," in this case, since it's one division versus one division, one multiplication, one addition and one subtraction. Perhaps it would be simpler just to say "Divide both sides by −4,". Κσυπ Cyp 2004年11月13日 (土) 18:45 (UTC)

I disagree, because: Many simple steps can be better than just a few more complicated steps.

Anyway I was talking of the formula for

x 1,2,3,4

. But I don't care anymore... --83.129.175.32 02:22, 14 Nov 2004 (UTC)

Sorry, didn't read the header, just assumed you were referring to something I did... My reply was a bit irrelevant... Think it would be better if the formula was there like you said. Κσυπ Cyp 2004年11月14日 (日) 04:42 (UTC)

[edit] Rational Roots therom

Why is the Rational Roots therom left out? I would include it before the Galois soltion, because its essentially Proved by Galois, and its Galois simplest lemma, or at the very start when you start looking for real roots, vs Imaginary roots. Actally The article is good. Very good infact, and shows a lot of quality mathamatics. Artoftransformation 10:14, 10 November 2005 (UTC)

[edit] Ferrari's solution

I'm currently working on an applet to calculate ferrari's solution to an equation. The funny thing is that the sanity check fails.(putting the found values in the equation).

I've been looking for this one whole day and I'm not sure if it is my mistake or a mistake in the formula. Could you please post the values you get for alpha,beta, etc... on the x^4 + 6x^2 - 60x +36 = 0 equation? This would really help me out.

You test root finders by constructing polynomials with known answers. Simply expand (x+a)(x+b)(x+c)(x+d) by hand or do it in the computer and feed the result into your solver. mdf 16:46, 23 November 2006 (UTC)

[edit] Substitution x→u

I was wondering, anyone know what name the substitution of

$x = u - {B \over 4 A}$

is called? And how did Ferrari resolve to this method to depress the $x 3$ term?

The general class of substitutions are called Tschirnhaus transformations. As for how to arrive at these, I imagine that Ferrari probably more or less just guessed. There is a general theory of these sorts of tricks due to Lagrange, however. It is based on analyzing the properties of symmetric functions of the roots. For a bit of the flavor of this, you can look at chapter 1 in Galois Theory by Stewart (its a reference at the bottom of this page). Grokmoo 19:06, 10 March 2007 (UTC)

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