# Is there a lagrangian resolvent for a quadratic polynomial?

i found this brilliant.org article mentioning a technique called lagrangian resolvents to simplify a polynomial, making it easier to solve:

Using Lagrange's resolvents, to solve the cubic, one has to first solve a quadratic. Given the general cubic,
$$x^3 + ax^2 + bx + c = 0$$
It's resolvent equation is given by: $$z^2 + z( 2a^3 -9ab +27c )+ (a^2-3b)^3 =0$$
such that that the solution is given by: $$x= \frac{ -a+z_{1}^{\frac{1}{3}} + z_{2}^{\frac{1}{3}} }{3}$$
where $$z_{1,2}$$ are roots of the resolvent

it is not explained how the lagrangian resolvent of a given polynomial is derived either, if there was a general method described I could have attempted it on the quadratic but I have next to nothing to go on. googling was not helpful at all and I think the wikipedia article was written for the benefit of people who already knows how to derive it.

there are some questions on this site that give clues as to the process of determining the Lagrangian resolvent of a given polynomial.

This question says you multiply the sum of the roots of the polynomial by a variable, in increasing powers starting from zero. I don't know how to go from that to what is produced in the article i linked. probably something to do with the definition of the variable. if only I knew it, that'd be great.

the main point of this question though, is can it be done to a quadratic.

• – dxiv
Jun 6, 2023 at 20:28

The resolvent for the quadratic is the discriminant itself.

The Lagrangian method of resolvents is not fully algorithmic - the actual expression for resolvent is basically a guess, and if the guess is correct, everything works out.

Here's an abbreviated description:

Given an original polynomial

OP( x ) = 0 = pN_aN x^n + pN_a(N-1) x^( n - 1 ) + ... pN_a0

[1] First, transform the general polynomial into a monic polynomial by dividing all the coefficients by the top coefficient, which cannot be 0 since it would then not be a polynomial of that order.

MOP( x ) = OP( x ) / pN_aN -> mN_aN = 1

[2] Next, set up an offset substitution variable that is simply the original variable plus a free-term offset amount such that when the original variable is substituted with its equivalent in the substitution variable (and thus would be minus that offset), the resulting coefficient for the ( N - 1 ) power goes to zero, yielding a depressed polynomial (this offset seems to be referred to as a Tchirnhaus reduction). For the quadratic formula, this depressed offset (DO) is that [ - B / ( 2 A ) ] term.

t -> depressed substitution variable

DO = - ( 1 / N ) mN_a(N-1) = ( 1 / N ) ( aN_(N-1) / aN_N )

t = x - DO -> x = t + DO

DP( t ) = MOP( x( t ) ) = MOP( t + DO ) -> dN_a(N-1) = 0

= t^N + dN_a(N-2) t^( N - 2 ) ... + dN_a0

E.G., for a quadratic (doing all the algebraic manipulation):

d2_a1 = 0

d2_a0 = [ m2_a0 - (1/4) m2_a1^2 ]

AND SO IF A = m2_a2 = 1 , B = m2_a1 = o2_a1 , C = m2_a0 = p2_a0, THEN

d2_a0 = (1/4) [ - { B^2 - 4 A C } ]

DOES {THIS} LOOK FAMILIAR? :) It's the famous quadratic discriminant Δ!

d2_a0 = - (1/4) Δ

At this point, there are now ( n - 1 ) coefficients (d_a_#) , so it might be possible to squinch everything together into a resolvent polynomial whose degree is 1 less than the original.

The depressed polynomial DP( t ) has the # of solutions (or "roots") that are equal to the degree (Fundamental Theorem of Algebra), and thus the coefficients are equal to the corresponding elementary symmetric polynomials in terms of the (depressed) solutions. Of course, the offset is selected so that the t^1 is 0, so there is the EQ that the sum of the solutions is 0 (i.e., for any degree polynomial). So at this point there are N # of EQ that relate the depressed solutions to the depressed coefficients

EDP( t ) -> polynomial in terms of elementary symmetric polynomials of the depressed solutions

and of course: EDP( t ) = DP( t ) = 0

E.G., for a cubic:

EDP3( t ) = t^3 + [ - ( T30 + T31 + T32 ) ] t^2 + [ T30 T31 + T31 T32 + T32 T30 ] t + [ - ( T30 T31 T32 ) ]

and since

0 = EDP3( t ) = DP3( t )

0 = d3_a2 = T30 + T31 + T32 <- THIS (N-1) COEFFICIENT IS ALWAYS THE SUM OF THE DEPRESSED COEFFICIENTS AND IS 0 FOR ANY DEGREE OF THE ORIGINAL POLYNOMIAL

d3_a1 = T30 T31 + T31 T32 + T32 T30

d3_a0 = T30 T31 T32

EDP2( t ) = t^2 + [ - ( T20 + T21 ) ] t + [ T20 T21 ]

0 = d2_a1 = T20 + T21

d2_a0 = T20 T21

RECALL that the values for d_a_# had been computed from the original a_# coefficients, and thus are "known" values.

T# -> solutions in t for DP( t ) -> DP( T# ) = 0

And here is where the resolvents come in.

[3] Propose a set of expressions (functions) for a set - of size ( N - 1 ) - of resolvents R# that are in terms of the reduced polynomial solutions T#, such that these expressions are symmetric within a set (symmetricity means the input arguments can be swapped around), and such that this symmetric set is of size ( N - 1 ).

E.G., for an original cubic, there is the proposed expression in all 6 permutations of j,k,l

ω3 = e^[ i (120°) ] <- deMoivre root of unity

ω3^3 = 1

R3jkl = R3012( T3j , T3k , T3l ) = [ (1/3) ( T3j + ω3 T3k + ω3^2 T3l ) ]^3

IT TURNS OUT THAT OF THE 6 PERMUTATIONS, SETS OF 3 OF THE 6 ARE THE SAME VALUE, SO THERE ARE ONLY 2 EXPRESSIONS

R30 = R3012 = R3120 = R3210

R31 = R3210 = R3102 = R3021

E.G., for an original quadratic, the proposed expression is

R20 = (1/4) ( T20 - T21 )^2 = (1/4) ( T21 - T20 )^2 <- symmetric, both are the same expression

E.G., for a quartic, the proposed expression in all 24 permutations could be

R3jklm = ( T4j + T4k ) ( T4l + T4m )

IT TURNS OUT THAT OF THE 24 PERMUTATIONS, SETS OF 8 ARE THE SAME VALUE, SO THERE ARE ONLY 3 EXPRESSIONS - E.G;, WHICH MATCHES THE 3 DIFFERENT WAYS THAT 4 CARDS CAN BE DIVIDED INTO SETS OF 2 CARDS, OR 1/2 OF THE NUMBER OF COMBINATIONS OF 4 TAKEN 2 AT A TIME (I.E., AFTER TAKING THIS COMBINATION, THE REMAINDER WILL BE DEFINED AS A COMBINATION ITSELF). (NOTE: THERE ARE OTHER EXPRESSION FORMS THAT WORK AS WELL.)

[4] With the ( N - 1 ) resolvent EQs and the EQ of the sum of the depressed solutions T# being 0 - i.e., a total of N EQs - come up expressions of the depressed solutions T# in terms of the resolvents R# (i.e., inverse function set of the resolvents). These expressions will necessarily contain root operations and the complex angle for DeMoivre roots of unity (which is simply -1 for a quadratic original polynomial) of whatever root operation needs to be done.

E.G., for a cubic, this inversion yields the 3 functions of the depressed solutions in terms of the resolvent parameters

T30 = T30( R30 , R31 )

T31 = T31( R30 , R31 )

T32 = T32( R30 , R31 )

E.G., for a quadratic, there is

T20 = T20( R20 )

T21 = T21( R20 )

but with the good guess from above, it turns out that these expressions are the square root of the resolvent

T20 = R20^(1/2)

T21 = - TR0 = - R20^(1/2)

[5] The resolvent polynomial ER( r ) will have its coefficients be the elementary symmetric polynomials for the resolvents R#, and which can be expressed in terms of the depressed solutions T#, and - THIS IS WHERE THE MAGIC HAPPENS - the expressions in terms of the depressed solutions can be expressed in terms of the coefficients for the depressed polynomial, which means the resolvent polynomial has coefficients that are functions of the depressed coefficients (which are "KNOWN VALUES"), and thus there is a new polynomial EQ of degree -1 from the original.

E.G., for a cubic, there is

E3R( r ) = r^2 + [ - ( R30 + R31 ) ] r + [ R30 R31 ]

e3r_2 = 1

e3r_1 = - ( R30 + R31 )

e3r_0 = R30 R31

THE MAGIC IS THAT IT END UP such THAT

e3r_1 = E3R_1( d_a1 , d_a0 )

e3r_0 = E3R_0( d_a1 , d_a0 )

and so the values for the resolvents for the cubic are solutions of the quadratic EQ of the resolvents:

R3# = (1/2) [ - d2_a1 +/- ( d2_a1^2 - 4 d2_a2 )^(1/2) ]

E.G., for the quadratic, there is simply

E2R( r ) = r + [ - R20 ] => r = R30

e2r_0 = - R30 = - r

but recall that

d2_a0 = - (1/4) Δ

and so

R20 = - e2r_0 = - d2_a0 = (1/4) Δ

and finally

T20 = R20^(1/2) = [ (1/4) Δ ]^(1/2) = (1/2) Δ^(1/2)

T21 = - T20 = (1/2) Δ^(1/2)

AND IF A = 1, ETC. AS BEFORE

T20,T21 = +/- ( B^2 - 4 A C )^(1/2) / ( 2 A )

LOOKING EVEN MORE FAMILIAR? :)

[6] Use the already known formulae for a polynomial of the degree for the resolvent polynomial, and then plug them into the expressions for the reduced solutions, and then add the offset term to get solutions for the original polynomial.

XN# -> solutions for the original EQ

E.G., for the cubic there is

X3# = T3# + DO3

E.G., for the quadartic, there is

X2# = T2# + DO2 = T2# + [ - a_1 / ( 2 a_0 ) ]

AND IF A = 1, ETC. AS BEFORE

X20,X21 = [ - B +/- ( B^2 - 4 A C )^(1/2) ] / ( 2 A )

FINALLY! :)

Something to keep in mind is that the root operations (that I had said were a necessary part of the expressions for the solutions in terms of the resolvents) are as per all possible DeMoivre root indices, and thus every time there is such a root operation, there is a cardinality dimension of possible values. Also, since the solutions must have underlying symmetry, they will all have the same expression, just with different combinations of those DeMoivre roots indices.

The quadratic formula is the same expression for both solutions, just with a plus/minus for the square root of the discriminant (which is also the resolvent); this plus/minus represents the 2 indices for the DeMoivre roots.

For the cubic, the solution expression has the cube roots of the 2 resolvents for a total of 9 solutions, while for the quartic, the solution expression has square roots for the 3 resolvents for a total of 8 solutions. In both cases, there are extraneous solutions, but there is a constraint due to the underlying symmetry (it can be calculated as part of the total calculation effort) such that there is an EQ relating all the resolvents. In both cases, the product of all the resolvents is constrained to be the coefficient for the t^1 term (which for a cubic or quartic is not 0, unlike for the quadratic), and indeed this coefficient cannot be 0, or else there is a division-by-zero degeneracy. However, if this coefficient happens to be zero, then the cubic becomes a simply free-term polynomial ( t^3 + z ), and the quartic becomes a biquadratic ( t^4 + y t^2 + z ), which are easily solved.

Wikipedia has articles on the cubic & quartic polynomials, with a derivation of the solution formulae. For the cubic, the parameters u & v are cubic roots of the resolvents, while for the quartic, the parameters s#^2 (in Lagrange's solution) or α, β, γ (in Euler's solution). In both cases, the (reduced) solutions are simply the sum of roots of the resolvents.