Wikipedia:Reference desk/Archives/Mathematics/2013 November 29

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November 29

Quartic formula

The image in the quartic function article displays a formula for quartics whose leading coefficient is 1. We need to substitute an image that displays a formula for all quartics. Go to:

http://www.josechu.com/ecuaciones_polinomicas/cuartica_solucion.htm

for such a formula. Any thoughts on which is better?? Georgia guy (talk) 01:44, 29 November 2013 (UTC)

Surely the general formula will only differ from the special case by a multiplicative factor applied to the coefficients ? In other words, if

$f(p,q,r,s)$

is a root of

$x^{4}+px^{3}+qx^{2}+rx+s$

then

$f\left({\frac {b}{a}},{\frac {c}{a}},{\frac {d}{a}},{\frac {e}{a}}\right)$

or, alternatively,

${\frac {1}{a}}f(b,ac,a^{2}d,a^{3}e)$

is a root of

$ax^{4}+bx^{3}+cx^{2}+dx+e$

Since the full quartic formula is only shown as a curiousity, I would say it doesn't matter which image is used. Gandalf61 (talk) 09:16, 29 November 2013 (UTC)

Certainly not. Consider two polynomials:
${\begin{aligned}x^{4}-4x^{3}+6x^{2}-4x+1&=(x-1)^{4}\\2x^{4}-4x^{3}+6x^{2}-4x+1&=(x-1)^{4}+x^{4}\end{aligned}}$
They differ only in their fist coefficient, just like you wrote, however the first one has a root x=1 while the second has no real roots.
Please use distinctive symbols for b, c, d and e in both polynomials. --CiaPan (talk) 11:50, 29 November 2013 (UTC)

Well, I thought it was obvious from context that my b, c, d and e were variable coefficients and so did not take the same values in the two quartics. But I have changed them in my post above to make this clear. And, for further clarity, I am not claiming that

$2x^{4}-4x^{3}+6x^{2}-4x+1$

has the same roots as

$x^{4}-4x^{3}+6x^{2}-4x+1$

... but it does have the same roots as

$x^{4}-2x^{3}+3x^{2}-2x+{\frac {1}{2}}$

and its roots are 1/2 times the roots of

$x^{4}-4x^{3}+12x^{2}-16x+8$

My point is that once we have a formula for finding the roots of a monic quartic, then generalising it to a formula for the roots of a non-monic quartic is trivial. Gandalf61 (talk) 12:14, 29 November 2013 (UTC)

Agreed. Sticking in the leading coefficient everywhere is like writing zero cents when talking about the budget. It just clutters things up for no real gain. Dmcq (talk) 12:27, 29 November 2013 (UTC)

Disagreed. Passing from the general formula to the particular case is trivial to an extent that passing from the particular case to the general case is not. YohanN7 (talk) 17:02, 29 November 2013 (UTC)

What's there is how reliable sources do it. They don't stick in big machine generated formulae, they explain what they're up to. This is an encyclopaedia. Dmcq (talk) 17:31, 29 November 2013 (UTC)

Well I guess RS doesn't apply so much to illustrations and that's is just there as a curiosity. Personally I think it should be removed altogether if people are thinking of it as actual content. Dmcq (talk) 17:37, 29 November 2013 (UTC)

Perhaps you misunderstand me, and perhaps I haven't read all this (and the article) thoroughly enough to understand the finer points. They seemed irrelevant to me at the moment. You should look in an encyclopedia to find the general answer, then look up the details (special cases) yourself. You seem to argue that you should go to an encyclopedia to find a trimmed (specific) answer, and then use references (which usually cost $20+) to find the general answer.

Dividing all 5 coefficients through the first one does not cost $20. It "costs" perhaps 20 seconds. Either way, we need " reliable non-copyrighted sources", and yours isn't. (Don't get me wrong, it's "true", but unfortunately this is not enough, according to Wikipedia:Standards). — 79.118.173.2 (talk) 22:05, 29 November 2013 (UTC)

References do cost, but people as brilliant as you do not have to pay a fee. It's just 20 seconds for you. (Took me 3 seconds.) From where do you get that I'm referring to unreliable sources?

You mean an idiot savant ? :-) No, you don't have to be Rain Man to figure that one out. :-) — 79.113.209.133 (talk) 22:36, 30 November 2013 (UTC)

I apologize for my unbalanced comment. It will not happen again. I removed the most offending part. YohanN7 (talk) 21:29, 1 December 2013 (UTC)

Less important, but still illustrating my point, is that I have recently implemented the general solution of forth-degree polynomial equations in a computer program. The extra cost for this to me (in time) was a faulty (and still special) formula (picked from a "reliable source", i.e. an old copy of Spiegels "Mathematical Handbook"). Then, of course, there was an added bonus: I learned a little about Galois field extensions in the process of finding the bug in my program. That was well worth the extra time!!! YohanN7 (talk) 19:14, 29 November 2013 (UTC)

The point is that most sources discussing the quadric formula only do the monic case. The transformation to go from the monic to the general case is completely trivial, but including the general formula adds needless complexity to a formula that is already quite complex. Sławomir Biały (talk) 19:50, 29 November 2013 (UTC)

Some reliable sources (e.g GMT 242) display the general case. While passing from the particular case to the general case is trivial to the mathematician, it is not as trivial as it is to pass from the general case to the specific case for the general audience. It happens to be nearly trivial, but my refs devote a sentence to it (making it not entirely nontrivial). But I don't want to argue this point further. YohanN7 (talk) 20:38, 29 November 2013 (UTC)

p=b/a, etc, is completely trivial to any member of the "general audience" for whom the quartic formula would be in the least bit meaningful. Sławomir Biały (talk) 22:54, 29 November 2013 (UTC)

I'd be rather surprised if you can use those formulae directly. You have to choose the appropriate root in some cases and keep the implications consistent elsewhere plus I'd have though even when the roots are real you'd sometimes have complex roots in the intermediate parts. I'd have thought there was a deal more to it than just applying those, even getting a good algorithm for the roots of a quadratic is non-trivial. Have you actually checked your results? You'd be far better off just using a good package. Dmcq (talk) 23:28, 29 November 2013 (UTC)

Plus what is GMT 242?, you should make things plainer. Dmcq (talk) 23:37, 29 November 2013 (UTC)

Sorry, GMT 242 = Graduate Texts in Mathematics 242 (the Springer series). Should perhaps have been GTM. "You have to choose the appropriate root in some cases and keep the implications consistent elsewhere plus I'd have though even when the roots are real you'd sometimes have complex roots in the intermediate parts." Right, and I am aware of this, and the thing works. YohanN7 (talk) 08:29, 30 November 2013 (UTC)

Abstact Algebra by Grillet? There's no great big expression in that book. It is a good book and doesn't do silly thing like that. And even if it did being in a volume of work does not of itself confer enough interest to include in Wikipedia. As it says in WP:NOT Wikipedia is not a manual or textbook or scientific journal not is it an indiscriminate collection of information. Dmcq (talk) 13:11, 30 November 2013 (UTC)~

The book by Grillet decidedly displays the general formula for the cubic equation. I don't remember if the quartic formula is general, and I don't have the book available a t m. But we shouldn't argue over this, it's not really important whether the article displays the general formula or not. YohanN7 (talk) 21:29, 1 December 2013 (UTC)

A+B+C = X+Y+Z and A*B*C= X*Y*Z

Are there any other solutions to this apart from the trivial case {A,B,C} = {X,Y,Z} ? Either way, how would we actually prove it ? This is related to an interesting geometry question. — 79.118.173.2 (talk) 23:21, 29 November 2013 (UTC)

-5 - 5 - 4 = 1 + 5 - 20, (-5)*(-5)*(-4) = (1)*(5)*(-20). DTLHS (talk) 23:32, 29 November 2013 (UTC)

The six unknowns have to be positive, since they represent the sides of two triangles. Also, this would imply A+B > C and X+Y > Z. (Either way, thank you, I was interested in the more general case as well). — 79.118.173.2 (talk) 23:42, 29 November 2013 (UTC)

A=B=10, C=18, X= 12, Y = 13+√19 ≈ 17.36, Z = 13-√19 ≈ 8.64. --RDBury (talk) 00:15, 30 November 2013 (UTC)

Here are integer cases from a brute force search with numbers up to 30 which can represent the sides of two triangles: 5+8+9 = 6+6+10, 7+12+12 = 8+9+14, 8+12+15 = 9+10+16, 8+15+15 = 10+10+18, 8+15+18 = 9+12+20, 9+14+20 = 10+12+21, 9+15+16 = 10+12+18, 9+20+20 = 10+15+24, 10+16+18 = 12+12+20, 10+18+22 = 11+15+24, 10+20+21 = 12+14+25, 10+20+22 = 11+16+25, 11+16+21 = 12+14+22, 11+18+20 = 12+15+22, 12+18+26 = 13+16+27, 12+20+21 = 14+15+24, 12+21+24 = 14+16+27, 12+21+25 = 15+15+28, 12+21+26 = 13+18+28, 12+24+25 = 15+16+30, 12+25+26 = 13+20+30, 13+18+25 = 15+15+26, 13+20+24 = 15+16+26, 13+21+24 = 14+18+26, 14+18+20 = 15+16+21, 14+20+27 = 15+18+28, 14+24+24 = 16+18+28, 14+24+25 = 15+20+28, 14+25+27 = 15+21+30, 15+20+24 = 16+18+25, 15+24+24 = 16+20+27, 15+24+27 = 18+18+30, 15+24+28 = 16+21+30, 16+21+27 = 18+18+28, 16+25+27 = 18+20+30, 18+24+25 = 20+20+27, 18+25+28 = 20+21+30, 20+27+28 = 21+24+30. PrimeHunter (talk) 00:22, 30 November 2013 (UTC)

Here are triples with integers up to 50: 18+33+40 = 20+27+44 = 22+24+45, 20+33+42 = 21+30+44 = 22+28+45, 20+35+36 = 21+30+40 = 24+25+42, 24+40+45 = 25+36+48 = 27+32+50. PrimeHunter (talk) 00:40, 30 November 2013 (UTC)

Resolved

— Thanks ! :-) — 79.118.173.2 (talk) 00:25, 30 November 2013 (UTC)

Now let's add one more condition: (s-a)(s-b)(s-c) = (s-x)(s-y)(s-z), where s = ⁠a+b+c/2⁠ = ⁠x+y+z/2⁠ . Could someone find such a non-trivial example, or prove that it does not exist ? — 79.118.173.2 (talk) 01:15, 30 November 2013 (UTC)

But this is equivalent to ab + bc + ca = xy + yz + zx. But then, according to Vieta's formulas and to the fundamental theorem of algebra, {a,b,c} and {x,y,z} would have to be the (three) roots of a cubic polynomial. Hence the two pairs have to be equal. Thanks again! — 79.118.173.2 (talk) 01:51, 30 November 2013 (UTC)

There are three unknowns: X,Y,Z, but only two equations: A+B+C = X+Y+Z and ABC = XYZ. Chose P freely and solve the cubic equation x³−(A+B+C)x²+Px−ABC=0. There is essentially one solution to the original problem for each value of P. (Permutations of the same numbers (X,Y,Z) are not considered different solutions). Bo Jacoby (talk) 10:29, 3 December 2013 (UTC).