How to solve a Cubic Equation Part 3 General Depression and a New Covariant

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1 ow to Solve a ubic Equation Part 3 ow to solve a ubic Equation Part 3 General Deression and a New ovariant James F. Blinn Microsoft Research blinn@microsoft.com Originally ublished in IEEE omuter Grahics and Alications Nov/Dec 006, ages 9 0 V is good for you. hat s one of the conclusions of this month s book recommendation: Everything Bad Is Good for You: ow oday's Poular ulture Is Actually Making Us Smarter by Steven Johnson. he basic remise is that television series (his rimary examle) have gotten so comlex in their seed of exosition and involved interersonal relationshis that it really exercises your brain just to follow them. And each eisode is not an individual story any more; a given rogram is one continuing mini-series that has to be seen in order without gas for it to make sense. For examle I ve been totally sucked into the series ost, and the whole fun of the series is watching the mystery unfold by seeing the eisodes in sequence. But I missed the first few eisodes of the second season, so I didn t dare look at any of the later eisodes. I ve been forced to wait for the second season to come out on DVD and not look at any fan sites until then. Given that V shows are now all miniseries I don t feel so bad about the fact that this column has evolved into various multi-art series. I had eight articles on 3D lines, two articles on solving quadratic equations, and this is the third one on cubic equations. (It would be nice to make it three articles on cubics, and four on quartics, but there is more than three articles worth say on cubics so it will robably be four). I do try to make each article have a theme so they are roughly self-contained, but I still must start out with the same hrase as all the serialized V shows Previously on Jim Blinn s orner We are looking for the roots [x,w] of the equation 3 3 f x, w Ax 3Bx w 3xw Dw 0 he essian matrix is a useful intermediate result and is calculated as 3 A B AD B BD 3 3 he discriminant of the cubic is the determinant of this matrix det 4 A D 6ABD 4A 4B D 3B 3 3 (0.) (0.)

2 ow to Solve a ubic Equation Part 3 If 0 there are three real roots, if 0 there is a double or trile root, and if 0 there is a single real root. he first ste in solving the cubic is to deress it by transforming in arameter sace to get a cubic with a second coefficient of zero. We transform by the substitution: his gives a transformation in coefficient sace of t u x w x w s v (0.3) 3 3 A t 3t u 3tu u A t s tus t v u s tuv u v B B ts us tsv usv tv uv D 3 3 s 3s v 3sv v D (0.4) We found two transformation matrices that result in a cubic with B 0 : t u 0 0 or s v B A D (0.5) he two ossibilities generate two variants of the algorithm (which I called A and D after the determinant of the transformation.) Each of these generate a deressed olynomial (though some authors use the more rosaic term reduced olynomial ) with A D A 0 0 B or A A B D BD D 3 3 AB 3AB A D D D A 3DB Each of these has a common homogeneous factor which we can toss out giving (0.6) A 0 B 0 or A B BD D 3 3 B 3AB A D D A 3DB So both generate a canonical olynomial to solve: 3 x x D 3 0 Solve this (see [] for details) and transform by the original matrix to get the answer 0 0 B A D x w x or x Both these variants work fine algebraically, but each is numerically good for regions in arameter sace that is different from the other. So far, I have only finished the calculations for the case 0 where there is just one real root, a, and a comlex conjugate air b ic. We found that the best (so far) numerical solution to be:

3 ow to Solve a ubic Equation Part 3 if 3 3 B D A {equivalent to} a b c use algorithm A to calculate a, use algorithm D to calculate b and c otherwise use algorithm D to calculate a, use algorithm A to calculate b and c. Generalizing the deressing transformation Given the two choices of deressing transformations, Equation (0.5), it is natural to ask if there are any others. Of course there are. So let s find all ossible transformations (values of s,,v) that make B 0. From Equation (0.4) we have (0.7) B t sa tus t v B u s tuv u vd Proceeding as we did with quadratic olynomials we will arameterize the desired set of transformations in terms of the first row []. So given we find the roer that makes B 0 by rewriting equation (0.7) as An that satisfies this is : B s t A tub u v t B tu u D 0 s t B tu u D v t A tub u (0.8) Actually any nonzero multile of these values will also work, but for now I ll stick with just the values in Equation (0.8). he transformation this generates is singular if the following is zero. tv su t t A tub u t B tu u D u 3 3 t A 3t ub 3tu u D f t, u In other words, this transformation will be singular if the () we chose was a root of the original cubic. his situation will turn out to be not all that bad, but I ll defer a comlete discussion of this until Part 5. Now that we ve transformed to get B 0 let s see what the other coefficients are as functions of (). Evaluating A, and D We first exand out the matrix formulation of equation (0.4) to get the formulas for A,, D. 3 3 A t A 3t ub 3tu u D ts A us tsv B usv tv uv D 3 3 D s A 3s vb 3sv v D (0.9) 3

4 ow to Solve a ubic Equation Part 3 We recognize A as just f() which is also the determinant of the transformation. hinking of A as now a A t, u f t, u. he other two values and D can be had in terms of function of () we have simly just t and u by simly lugging the values from equation (0.8) into equation (0.9). his will result in a exression for that is fifth order in () and for D that is sixth order in (). Needless to say, doing this exlicitly will turn into an algebraic quagmire. And this quagmire obscures the very nice algebraic fact that,, A t, u as a factor. We saw a secial the exression for t u and Dt u both contain the olynomial case of this in equation (0.6) of the original Algorithm A (where []=[,0]) and Algorithm D (where []=[0,]). o see the general case more clearly I will drag in my favorite algebraic simlification tool, the tensor diagram. ensor Diagram Essentials I ve described this notation scheme in various revious articles []. But there are a few new constructions that we will need here so let s start out with a brief review. Polynomials as ensors We are interested in linear, quadratic and cubic homogeneous olynomials l x, w Ax Bw q x, w Ax Bxw w f x, w Ax 3Bx w 3xw Dw 3 3 We can write these as vector-matrix roducts (though the cubic olynomial is a bit clunky in this notation). A l x, w x w B A B x q x, w x w B w A B B x x f x, w x w B D w w he coefficients form tensors of rank (a vector for the linear olynomial), rank (a matrix for the quadratic) and rank 3 (for the cubic). Next, we will give names to the tensors and give alternate, indexed, names to their elements. A 0 B A B 00 0 B 0 A B B B D he vector will be dot-roducted with these. As a bookkeeing convention we will write its elements indices as suerscrits. 4

5 ow to Solve a ubic Equation Part 3 0 x w hen we can write the olynomial functions as nested sums. l q f i i, j i, j, k i i i j i, j i j k ijk (0.0) Note that I m kinda running out of letters here so I ve overloaded the notation a bit. I use boldface to reresent the tensor and italic with subscrits to reresent its elements, and I use italic without subscrits as one of the elements of. We will next disense with exlicit indices anyway so this is just temorary scaffolding. o disense with exlicit indices we write equation (0.0) in diagram form as follows. l q f Each vector/matrix/tensor is a node. he number of connection oints equals the number of indices. Each summed-over index is an arc connecting the two nodes containing that index. he arrow on the arc oints from a connection oint on a arameter vector (reresenting a suerscrit or contravariant index) to a connection oint on a coefficient tensor (reresenting a subscrited or covariant index) I want to oint out something here that is retty obvious, but that we will build on later. Given a tensor, we may extract out the coefficients by lugging in articular values for the arameter vector. For examle we can lug in the arameter vectors [, 0] and [0, ] to get the A and D coefficients.,0 0, A,0 D 0,,0 Slightly less obvious is the following. We can extract out the other coefficients by lugging different arameter vectors into the three inuts. I have colored the nodes here simly to emhasize the attern. 0, (0.),0,0 B 0, 0,,0 0, (0.) 5

6 ow to Solve a ubic Equation Part 3 ransformations If we transform the arameter vector via a matrix he diagram version is t u x w x w s v = Note that the matrix has one arrow in and one out. his is called a mixed tensor in distinction to the ure covariant tensor that reresents a quadratic olynomial. o see the effect of on the tensor we form = So the transformed tensor is just the inside ortion = (0.3) Given the tensor we can extract the coefficients A, B,, D by lugging in the arameter vectors [,0] and [0,] in the atterns from equation (0.) and (0.). Given the internal structure of from equation (0.3) these will go straight into the transformation matrix. We observe that,0 = 0, = his means that the coefficients A, B,, D in terms of elements of the original tensor are: 6

7 ow to Solve a ubic Equation Part 3 A B D his is the diagram version of equation (0.4) and is called blossoming. (0.4) he Esilon ensor wo vectors and 0 0 are homogeneously equivalent (differ by only a nonzero homogeneous scale factor) if the following is true or We can write this exression as a matrix roduct We give the name esilon to the x anti-symmetric matrix. In diagram notation it would look like a node with two connecting arcs joined to the and nodes. But we have to be careful. Since esilon is antisymmetric reversing the ositions of and will fli the sign: (0.5) I want to make this visually aarent with the node diagram so I make the node shae itself asymmetric. he diagram version of equation (0.5) is = - A rotation of a diagram or other connectivity-reserving rearrangement of its nodes doesn t affect its algebraic meaning. But a mirror reflection of an esilon node will fli the sign. = - It s imortant to note that, in this diagram, and don t have to be just single vectors. hey can reresent any other more comlex collections of nodes and arcs; if you fli an esilon anywhere inside a comlex diagram you fli its algebraic sign. he esilon tensor is also good for reresenting determinants of matrices. onsider the following A B 0 B A B 0 B 7

8 ow to Solve a ubic Equation Part 3 Now ut two of these end to end B A B A B A 0 B B 0 B A his is (minus) the determinant of, times an identity matrix. Diagrammatically: (0.6) = -det You can write the determinant as a simle scalar by connecting the dangling arcs, which is the same as taking the trace of the matrix. det (0.7) (0.8) he factor of on the right hand side comes from taking the trace of the right hand side of equation (0.6). Another way to look at this is to take the trace of the version in equation (0.7) and note that a single arc reresents a x identity matrix whose trace is. Why ensor Diagrams are ool here are several things I like about tensor diagram notation. First of all, they are a good visualization tool since they reresent certain algebraic roerties visually: Symmetry of the tensors is reflected in the roundness of the nodes, so that exchanging any two inut arcs leaves the algebraic values unchanged. And anti-symmetry of the esilon is reflected in the shae of its node. But there are two main things that are articularly nice: tensor diagrams automatically reresent transformationally invariant quantities, and they allow easy algebraic maniulations. et s look at the first of these. Invariants onsider the following construction t s 0 t u 0 tv su u v 0 s v tv su 0 he right hand side is just the determinant of times the esilon matrix. hat is: = det Now look at what haens when we form the determinant of a transformed matrix (0.9) 8

9 ow to Solve a ubic Equation Part 3 = det Each air of s on either side of an esilon extracts out into a factor of det for the resulting exression. Since there are two such factors here, the sign of the determinant of is unchanged by any arameter transformation. he mere writing of an exression as a tensor exression (in either diagram or index form) constitutes a roof that it reresents a transformationally invariant quantity. Each esilon in the diagram will turn into a factor of det when a transformation is alied. So if the diagram contains an even number of esilons its sign is invariant under transformation. If it contains an odd number just the zeroness is invariant. he number of esilon nodes in a diagram is called the weight. An Invariant of the ubic As we stated in art of this series [3], the essian of the cubic is a quadratic olynomial formed as the determinant of the matrix of second derivatives. What does this look like in diagram notation? he first derivative of f(x,w) is a vector comosed of the artial derivatives fx, f w which can be written in diagram notation by simly loing off one inut node: f f 3 Ax Bxw w Bx xw Dw x w x,w = 3 he second derivative matrix comes from loing of two nodes x,w (0.0) fxx fxw Ax Bw Bx w 6 6 fxw f ww Bx w x Dw x,w he determinant of this, as rescribed by equation (0.8), will be fxx fxw det 8 fxw f ww x,w x,w I decided, for convenience, to define the essian matrix (equation (0.)) without the constant factor of 8 giving us Next we look at the cubic discriminant, = - det, which in diagram form is (0.) 9

10 ow to Solve a ubic Equation Part 3 det Plug in equation (0.) twice, once for each, and the minus s from that equation go away leaving us with the cubic invariant he existence of six esilons in this diagram immediately verifies the statement made in art that the discriminant of a transformed cubic differs from the discriminant of the original by a factor of 6 det ovariants Now let s look at what haens to the essian when we transform the tensor. he essian of the transformed tensor is = - Alying equation (0.3) and then equation (0.9) gives us = - = - det = det So the essian transforms, as a quadratic, by the same transform that was alied to the cubic; the transform of the essian is the essian of the transform. his not-entirely-surrising fact is still imortant. It means that the essian is geometrically locked to the cubic and reresents a transformationally invariant roerty. his tye of object is called a covariant of the cubic. In fact, the main difference between an invariant and a covariant is that an invariant is a scalar and a covariant is a tensor it still has some dangling arcs. Much of the early work on invariants and covariants of homogeneous olynomials (also called forms or quantics) was due to David ilbert. A good resource is [4] which is a recent rerint of a series of lectures originally given in 897 and which gave me a lot of the insiration for this and later derivations. ilbert did not use diagrams, but rather develoed what amounts to a set of lexicograhic rules for testing whether olynomials in A,B,,D like equation (0.) were invariants. For examle if you rename the coefficients 0

11 ow to Solve a ubic Equation Part 3 with indices A a3, B a, a, D a0 and consider owers to be just relications of a quantity (so that A D a3a3a0a ) then a necessary (but not sufficient) condition for a olynomial in the a 0 i s to be an invariant is that the sum of the subscrits in each term is the same. For equation (0.) you can verify for yourself that that index sum for each term is 6 (which equals the weight of the exression, or diagramwise, the number of esilons). Deression in diagram form o make B 0 we had to choose the second row of the transformation according to Equation (0.8). We can write this in terms of esilon with s v t B tu u D t A tub u 0 t A tub u t B tu u D 0 omaring this to the first derivative diagram of equation (0.0) shows us the diagram form of [] = Plugging this into the diagrams of equation (0.4) gives us the following. For B we have (0.) B = his is of form Which is identically zero, as we exect, since we designed () to make this haen. Next, for we have = and for D we have

12 ow to Solve a ubic Equation Part 3 D = Now let s see how to maniulate these diagrams into something simler. Doing hings with Diagrams So far, I ve used tensors only as a reresentation scheme for olynomial exressions. he real ower is in using them to maniulate exressions. he technique is based on the observation that there can be only two linearly indeendent -vectors. Any third vector must be a linear combination of the first two. A little fiddling can show that the scale factors in the linear combination are given by the following identity: In diagram notation this looks like M M M M M M = + M M Rearranging this a bit to ut, and M in a fixed osition, and lugging in another tensor as a laceholder, we get M = + M M (0.3) Now the cool thing is that this works even if,,, and M are not simly vectors. hey can be more general tensors with any number of other arrows ointing into and out of them that don t directly articiate in the identity, but just go along for the ride. his basic identity then allows us to reconnect the arrows of any diagram to construct two new diagrams whose sum is equal to the original one. As a useful examle, suose the nodes and M were actually quadratics R and with an esilon between them. Our diagram equation would look like R = + R R lean this u with a coule of esilon mirrors (and associated sign flis) for neatness and get the following identity:

13 ow to Solve a ubic Equation Part 3 R = - R As a secial case of this identity, consider what haens if you lug in R=. he identity turns into something we ve seen before, a combination of equations (0.7) and (0.8). = + R (0.4) (0.5) So whenever we see a chain of two identical quadratics, we can aly this identity and slit the diagram into two disjoint ieces. his trick alies immediately to our diagram for Simlifying ets start with the diagram and bend it around a bit (with a sign fli) to get = - We can now aly identity (0.5), treating the two diagram ieces inside the gray ellises as the quadratic. We get he diagram has fallen aart into two disjoint ieces. Each iece is an indeendent factor of the net exression. And one of the ieces is, as advertised, the value of A. Simlifying D (0.6) o simlify the diagram for D we again start by bending the diagram around a bit and fliing an esilon (and the sign). 3

14 ow to Solve a ubic Equation Part 3 = - his time we aly the more general identity of Equation (0.4) to the to gray regions and get D - Now look at the first term on the right hand side. What do you see? et me give you a hint by the following grouing. Remarkably, this is two identical things (indicated by the blue areas) on each side of an esilon (the middle one of the three). So it s ZERO. his is great. We like things that are zero. his makes D equal to just the second term. D Again notice the factor of A. (0.7) 4

15 ow to Solve a ubic Equation Part 3 General Deression et s ut all this together. Using the general transformation of equation (0.3) with the second row, [], 3 3 defined by equation (0.) we have deressed our olynomial into the form Ax 3xw Dw. hen A f t, u from looking at equations (0.6) and (0.7) we see that we can remove the constant factor of this exression giving the net equation we need to solve as: Where 3 3 x 3xw Dw 0 D (0.8) Algorithms A and D are simly secial cases of this with []=[,0] and []=[0,] resectively. But now we have a whole continuum of and D values arameterized by []. We now think of and D as olynomial functions of (). he exression for is something we ve seen already; it s just the half of the essian evaluated at (). But what about D? What is D? D is now a homogeneous cubic olynomial in (). And, since equation (0.8) shows it written as a tensor diagram, it is also a covariant of the original cubic tensor. (ilbert calls this the skew covariant because the weight is odd.) So you can also think of D as a maing of one cubic olynomial to another one, D. For concreteness, let s exlicitly find this maing. Equation (0.8), written as a matrix roduct looks like s Dt, u t u v Plugging in equation (0.8) for [], exanding out and collecting like terms for t 3 etc gives with 3 3, 3 3 D t u A t B t u tu D u D D D D 3 A A D 3AB B D B A ABD B D AD B D B D 3 D AD 3BD et s see what else we can find out about this maing. D (0.9) 5

16 ow to Solve a ubic Equation Part 3 D is a transformation of I will now show that the maing D is just a arameter-sace transformation of. Of course any cubic of a articular tye is just a arameter transformation of other cubics of the same tye. But this transformation is secial; it s just the roduct of with esilon,. In diagram form this is = Note that lugging in an esilon has changed the two-arrows-in matrix to a one-in-and-one-out matrix that tyifies a transformation matrix. he transformed cubic is then ˆ o see why Ĉ is homogeneously equivalent to D we again do a derivation using tensor diagrams. I m not going to do this in detail here, but the rocess is very similar to what we ve seen before. he net result is We can see that this is at least reasonable since the number of nodes, esilons and arcs is the same; they are just connected together differently. he only slightly tricky art in the roof is that you will encounter the following diagram fragment = - his fragment is identically zero, which you can rove by alying the basic identity of equation (0.3) to the arcs marked in red. Now what about the essian of the cubic D? Since the essian oeration is a covariant and since we transform by D we can simly transform the essian of by to find the essian of D. We get to get 6

17 ow to Solve a ubic Equation Part 3 = So we get almost trivially that the essians of and D are the same u to a homogeneous scalar. In art we talked about the sace of ossible cubics as being foliated by a network of lines of constant essian. We have just shown that and D lie on the same line in this sace. Finally let s think about what haens if we aly the D oeration twice; what is DD? Well, Equation (0.7) basically tells us that the matrix is its own inverse (u to a homogeneous scale factor). In other words alying the D oeration twice gets us back to the original cubic. D is a function of other covariants At first it might seem that the discovery of D gives us something new about the cubic olynomial. But in fact it turns out that D is a function of the other three in/covariants that we already know about. Just looking at equation (0.8) it s not immediately obvious how this can be. Any attemt to rocess the diagram for D into something else, using the basic identity in Equation (0.3), doesn t get us anywhere. But if we instead start laying with the square of D (formed by just lacing two coies side by side) we make some rogress. he first ste uses Equation (0.3) on this air as follows = + Figure shows the ultimate result (which requires a few more alications of Equation (0.3)). Again, as a visual sanity check note that each term in figure has the same number of s, s and esilons; they are just connected together in different ways. 7

18 ow to Solve a ubic Equation Part 3 = + Figure Syzygy relating the in/covariants of We can write the diagram of figure algebraically as Which simlifies to Dx, w Ax, w x, w 3, 4,, 3 D x w x w A x w his is the generalization of equation 3 of art to a general [] instead of [,0]. his sort of relation is called a syzygy. As a general rincile, any diagram that contains an odd number of esilons can usually be related to simler in/covariants by rocessing its square in this fashion. Examles et s round out our intuition about the essian and the D oeration by looking at what they do to each of the four tyes of cubics. I ll start with a secific examle of each tye and aly equation (0.) to get the essian, both in the form of the matrix and of its associated quadratic olynomial which I ll write as (f), and then aly equation (0.9) to calculate D f. hen I ll discuss how this looks for a more general cubic of that tye. ye 3 3 he simlest exemlar of trile-root cubics is f x, w x. Plugging this into equation (0.) shows that the essian is identically zero (this is the defining characteristic of a trile-root cubic). hen equation (0.9) shows that D is also identically zero. We ll catalog these as ye his is tyified by the function x D x f x, w 3x w which leads to

19 ow to Solve a ubic Equation Part x w x D 3x w x 3 he transformation that mas f to D is singular, so it is able to change the tye of the cubic from ye to ye 3 that is, to the cubic that has a trile root where the original f has a double root. here s a nice geometric interretation of this henomenon. Recall that in art we found that the discriminant surface 0, where all tye cubics reside, is a ruled surface. A articular ruling line is the set of all tye cubics that share the same essian. he set of all tye 3 cubics lies on a crease in the discriminant surface and each ruling line crosses this crease exactly once. he oeration of forming the D of a tye cubic then consists of following along its ruling line to the crease. Generalizing the single root at w=0 to instead be at Ax 3Bw 0 gives us, after the aroriate comutation: 0 0 B 0, 3 3 Ax Bx w B x 0 B, D Ax 3Bx w B x he coefficient A does not aear in the formulas for and D ; only the location of the double root matters. his means that a ruling line (a line of constant essian) is also a line of constant double root. Sliding u and down along this line just changes the single root until, at the crease, the single root momentarily coincides with the double root. ye 3 A nice examle of a tye cubic is f x, w xx 3wx 3w x 3xw. If you lot the three roots as lines through the origin in the x,w lane you find that they are equally saced 60 degrees aart, Figure. Some comutation gives: 0, x 3 xw x w D x 3xw, 0 3x w 3x w w he essian has no real roots. he matrix that transforms f to D is a 90 degree rotation so D has three roots interleaved between the three roots of f, again see figure. 9

20 ow to Solve a ubic Equation Part 3 w Roots of f x Roots of D f Figure. Roots of tye cubic and its covariants For a general tye cubic, the essian will always have no real roots and be negative definite. Now recall that the roots of corresond to the eigenvectors of, so the transformation from f to D has no real eigenvectors. his means that it is always a generalized rotation that is, one with some ossible additional shearing. So we have the situation that sliding a tye cubic along a line of constant essian basically rotates its roots until it gets to D. But there s something else that s interesting. In figure note that a rotation of 90 degrees is not the only rotation that can change f into D. A rotation of 30 degrees and - 30 degrees will also work. In the general case the matrix is not the only transformation that does the job. Another matrix that works is one that is effectively the cube root of. ye f x, w x x 3w x 3xw ere our concrete examle will be omutation reveals 3 so it has one real root at [x,w]=[0,]. 0 3, 3 0 x xw x w 0, D x 3 xw 3 x w w he essian has two real roots, [x,w]=[,] and [x,w]=[-,]. he transformation matrix simly exchanges x and w; it is a mirror about the line x=w. So D is a cubic with one real root at [x,w]=[,0]. See figure 3. 0

21 ow to Solve a ubic Equation Part 3 w Root of f x Root of D f Roots of f Figure 3. Roots of tye cubic and its covariants In the general case will always have two real eigenvectors and a negative determinant (since 0). D f. But here s the really interesting So it will mirror the single real root of f into the single real root of thing: Recall that in art I showed that lines of constant essian for tye cubics intersect the trileroot curve at two oints. hese two trile-root cubics are the cubes of the roots of the essian. In our secial case these are x w 3 and x w 3 D f, that is also on this constant. Now we have a new cubic, essian line. his means that these four cubic olynomials are linearly deendent; in articular in our secial case we can easily see that: f D f x w f D f x w So traveling along the constant-essian line we encounter, in order, f, x w, D f, x w get back to f. What can we do with this? and then Our ultimate goal here is twofold. We want to get insight into how cubic olynomials work: what are the in/covariants and their geometric interretations, how does a transformation in arameter sace relate to the transformation in coefficient sace, etc. But, more ractically, we are also interested in finding numerically sound ways to calculate the roots of a cubic. o do this it is useful to have more than one way to calculate the roots. For examle, when we were solving quadratics we found numerical roblems in the exression B B A / A when B is ositive and A B. But knowing about the roblem was only useful because there was another way to calculate the same value: / B B A, that didn t have numerical roblems. Similarly, last time we found two algorithms that comlement each other in solving tye cubics so that at least one of them (retty much) will be numerically nice. In art 4 we will show a reliminary solution to tye cubics, and in art 5, with the insights gained in this article, we will work on finding numerically nice solutions to all cubics. Meanwhile, I ll get back to my ost DVD s. I wonder if Walt will get rescued. (I ll bet he does.)

22 ow to Solve a ubic Equation Part 3 References [] J. F. Blinn, ow to Solve a ubic Equation, Part he case, IEEE G&A, Vol 6, No 4, Jul/Aug 006, ages [] J. F. Blinn, Polynomial Discriminants Part ensor Diagrams IEEE G&A, Jan/Feb 00, and J. F. Blinn, uartic Discriminants and ensor Invariants, IEEE G&A, Mar/Ar 00, both rerinted in Jim Blinn s orner: Notation, Notation, Notation, hater 0, Morgan auffman, 003. [3] J. F. Blinn, ow to Solve a ubic Equation, Part he Shae of the Discriminant, IEEE G&A, Vol 6, No 3, 006, ages [4] D. ilbert, heory of Algebraic Invariants, ambridge University Press, 993.