Discovering a formula for the cubic | Gowers's Weblog

]]>Discovering a formula for the cubic | Gowers's Weblog

]]>I was reading the discussion at your web page where you said that you haven’t a plausible explanation for the substitution y=w+p/3w.

I think that this is natural because the expression x^3-3x reminds you of the triple angle cosine. And because 2cos t = z + 1/z, where z=exp(it), one can adapt this substitution in order to arrive at y=w+p/3w.

]]>*Thanks — I’ve corrected the mistake. (I hope that I made all the necessary follow-on changes and got them right.)*

Before we try to solve the cubic,Let’s investigate how to solve the

quadratic.Let the roots of the quadratic be .According to the factor theorem,we have In order to figure out and ,we need another linear equation ,where are constants and both are nonzero,and ,so that we obtain a system of linear

equations then by solving this system of linear equations we could get .Obviously, is not a symmetric polynomial of ,but is a symmetric polynomial of .By the fundamental theorem of symmetric polynomials, can be expressed as a polynomial of $b$ and $c$,more specifically,

We HOPE that there is a relationship between and ,so that when we combine this relationship with the equation (1),we could obtain a value for .Suppose

then we get that ,so .When ,,this is not what we want.So let ,then .Then the equation (1) becomes By solving it we obtain that Then we managed to get a system of linear equations Solve it ,we can get .By using the same method,we try to solve the cubic.Let the roots of the cubic be ,then according to the factor theorem,we have We need another two linear equations together with the linear equation ,then by solving this sysmtem of linear equations we can get .Obviously,the polynomial is not a symmetric polynomial of ,because if it is symmetric,then ,then the system of linear equations do not have a unique solution,this is not what we want.But must be a symmetric polynomial of .By using the fundamental theorem of symmetric polynomials,this symmetric polynomial can be expressed as a polynomial of ,we denote it as .Now let’s see two sets of linear polynomials and Each of the

polynomial is a factor of and the polynomial in each set has the cyclic symmetric relation.We HOPE that there is a relationship among the polynomials in each sets.Assume that Then we get that ,.When ,,this is not what we want,when ,where ,at this

time,.Similary,when ,.According to the symmetry,let ,.Then we have Let ,,then we have It is also easy to

see that is a symmetric polynomial of ,so can be expressed as a polynomial of ,denoted as .So So and

can be easily solved.Finally,we managed to get a system of linear equations Then can be solved!(Note that finally will disappear.)

It should be $3u^2-3vw=c$, I think.

I don’t see how the substitution $v = s+t$, $w = s-t$ helps; $v^3 = s+t$, $w^3 = s-t$ is what we need; I don’t see why this would be preferred on symmetry grounds.

Dear Prof.Tao,

I give up your 245a real analysis blog series for a moment and decide to turn to complex analysis .Ancient mathematicians first encounter imaginary numbers while they were studying the root of cubic equations.Then I am troubled by the cubic equation,because I think that the method of solving the cubic equation is too unacceptable by me,only genius will discover them and accept them without any uncomfortable!

Luckily I see your comment.I think that the method of using discrete fourier transform is very good,although I think it is also not easy to come up with this method because I think there is no direct hint on why we should use fourier transform…

]]>This is further simplified by changing coordinates, i.e. by using a different basis than $n^j$ for the vector space of polynomials. A better basis is $n, n(n-1), n(n-1)(n-2), \dots, n(n-1)\cdots(n-k)$ since these functions have nice sums. In other words, try to find an explicit formula for $\sum_{a=0}^n a(a-1)(a-2)$ and be delighted ðŸ™‚

]]> JÃ¶rg Bewersdorff. Galois Theory for Beginners: A Historical Perspective

http://tinyurl.com/2u6k52

Lagrange came up with the sum-of-roots-of-unity ansatz while trying to understand what’s going on with the quintic.

In fact, I find the usual “modern” presentation of Galois theory utterly incomprehensible. I mean, it’s not so bad and quite simple after knowing the historical approach of permuting the roots. But why, why is the easy way pretty much forgotten? Why struggle months with something abstract-nonsensical and void of motivation when there’s a dead simple and even more illuminating approach to understand it in hours? As Feynman (?) put it: “If you can’t explain it to a six year old, you don’t really understand it.” But I’m preaching to the choir ðŸ™‚

]]>Let s = a + b + c, t = a + wb +w^2c, u = a + w^2b + wc

Then express t^3 + u^3 and t^3 * u^3 in terms of the symmetric terms.

There is a nice story to that solution you can find it in Mark Kac’s autobiography “Enigmas of Chance”.

It seems to me that this solution is related to Terence Tao’s solution as a sort of inverse transformation.

Since someone cited sums of powers I remembered finding a nice “geometrical” solution to the sum of squares 1^2 + 2^2 + 3^2 +…+ n^2. Organize one 1, two 2’s, three 3’s, …., and n n’s into an equilateral triangle in the natural way, then add the corresponding elements of the three versions of this triangle (the original and the rotated by 120 and 240 degrees) it nice to see that the sum is constant and equal to 2n+1. Since there are n(n+1)/2 elements in the triangle the total sum is n(n+1)/2*(2n+1) and the sum of the elements in the original triangle n(n+1)(2n+1)/6 which is the sum of the squares.

I would like to present this in a motivated way but I am afraid that is somewhat longer. I also tried to generalize this idea to get the sum of cubes but failed. Since it has such a nice sum I suppose that it could also have a nice geometrical method to it. Anyone has an idea?

]]>On an unrelated note, when you say “we end up needing to solve quadratics and take cube roots, both of which we are allowed to assume that we can do”, I wonder how many people would actually consider cube roots of a complex number something basic they “can calculate”, as, unlike in the square root case, you can’t find the real and imaginary parts in terms of roots of only real numbers, so presumably would resort to using transcendental functions (in which case you could of course solve the quintic too.) Utlimately you have to make a choice of “numbers I can deal with” – something you’ve talked about before in other places.

]]>At first I tried ‘completing the cube’, which seemed a natural thing to try but I was heading in the wrong direction.. ]]>

There is a discussion in section 6.5 of Concrete Mathematics, by Graham, Knuth and Patashnik, of how one might discover and prove the formula for this sum by nothing but sheer obstinancy. Later, in section 7.3, they describe how the problem becomes easier when armed with the tool of generating functions. I’ll sketch the latter attack here, because generating functions are a great tool. Set S(m,n)=1+2^m+3^m+..+n^m. There are four approaches you might try in a generating function attack: you could define any of the four functions

A_m(w)=sum_n S(m,n) w^n, B_m(w)=sum_n S(m,n) w^n/n!, C_n(z)=sum_m S(m,n) z^m, or D_n(z)=sum_m S(m,n) z^m/m!

and try to get this function into a simple form. With A, B and C, we strike out. But, if we come back for a fourth swing, D_n has the nice closed form

(e^{nz}-1)/(e^z-1)=(sum_{k>=1} n^k z^k/k! ) (1/z-1/2+z/6-z^3/30+…)

and we get the closed formula immediately. So one approach to this sum comes down to knowing about generating functions, being willing to try again when the first attack fails, and some basic comfort manipulating series.

]]>It was stated on your home page that any reasonable person wouldn’t be expected to solve the cubic in a few hours or so. However, how long would you expect say a cambridge undergraduate with no knowledge of solving the cubic to discover the solution?

And I wondered if it would be a good idea if you were to write a similar article on finding the closed form sum of 1+(2^m)+(3^m)…+n^m where n and m are positive integers, to offer some insight into how one might discover the formula for that?

Thanks,

John

http://rigtriv.wordpress.com/2007/08/29/invariants-and-solving-polynomials/

Basically, the philosophy here is to only permit yourself to write down expressions (such as the discriminant) which transform nicely under projective changes of coordinates. There are relatively few of these expressions, and so you have a smaller “search space” in which to find the invariants that factorise the original polynomial into linear factors.

]]>I think you have a typo where you have written x3 – dx – c 3 instead of x2 – dx – c 3 ]]>

For matrices in wordpress, I find that \begin{pmatrix} … \\ … \end{pmatrix} works nicely, e.g. . (If you “edit” this comment you will be able to steal the latex code.)

It is amusing to recast your above discussion through the lens of Fourier analysis. One can view solving a polynomial as solving a system of symmetrised equations. For instance, by the factor theorem, the task of finding the three roots x,y,z of the cubic is equivalent to solving the system of equations

.

Now these equations are invariant under cyclic shift of the x,y,z. One of the key insights of Fourier analysis is that any mathematical problem which enjoys a translation invariance symmetry is likely to be clarified by use of the Fourier transform. This would motivate the Fourier substitution , , , which leads to

.

This lets one solve for u; to solve for v and w one can then observe a residual cyclic symmetry between v and w, prompting another Fourier transform , , which soon lets one solve for everything.

Presumably this can all be interpreted nicely in terms of the Galois group , and in particular to the solvability of that group, especially given that solvable groups can be “built up” from abelian groups such as and , which are precisely the groups which enjoy nice Fourier transforms. My Galois theory is incredibly rusty, though, so I don’t see the connection clearly.

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