^{4 }+ Ax

^{3 }+ Bx

^{2}+

^{ }Cx + D but which is generally "normalized" to yield a reduced quartic in the form: x

^{4}

^{ }- Ax

^{2 }-

^{ }Bx -

^{ }C = 0.

The solution is then obtained from what is called the "resolvent cubic" and using those roots to generate the roots of the quartic, call them x1, x2, x3, and x4. But what if everything is done according to the letter, but the roots obtained do not satisfy the (reduced) quartic? We then must deal with the emergence of errors somewhere in the approach.

While the typical American often views errors as a nuisance, or even demonizes them (who can blame them after a Mars lander missed Mars by over 3,000 miles some ten years ago because the wrong units were used) there is another take which more aligns with the Italian verb errare, i.e. "to wander" (a rough equivalent in English being the Knight Errant -send wandering over the countryside to right wrongs.). And when one errs, as in takes a "wrong turn" it is exactly the task at hand to find where the wrong turn was made.

In their article 'Right or Wrong? That is the Question', authors M. Dedo and L. Sferch (Notices of the American Mathematical Society, p. 924) invite the reader to look with a different perspective on errors. As they observe:

"If we want to free ourselves of the concept of error as 'fault' and rehabilitate it as an ally in the process of learning, work on error must be carried out according to very precise practices that are shared with students and have a real effect on evaluation. First of all, one crucial premise is that working on errors - knowing how to recognize, interpret and transform them - is not a natural ability but rather one that must be discovered, cultivated and transformed".

This is what I hope to demonstrate in this blog for any math maven readers (or merely the intellectually curious, say interested in applications of data streams and how their correct application often invites errors). In other words, what I am about to present can be regarded as an excellent exercise in error detection for anyone who works with volumes of data, and applies it to real world situations.

The gist of it is quite simple: I will take readers through a textbook solution of a (reduced) quartic equation, where each major, intermediary criterion for proper solution is met - but for which the final roots yielded happen not to be the correct ones, i.e. that actually solve the quartic equation.

The reduced quartic equation is: x

^{4}

^{ }+ 3x

^{2 }+

^{ }4x +

^{ }2 = 0

The resolvent cubic to solve is: z

^{3 }+ (3/2)z

^{2}+

^{ }(1/16)z

^{ }- 1/4 = 0

The coefficients of the terms (other than the first) are found by taking, in turn: (A/2), (A

^{2}/16 - C/4) and B

^{2}/64.

Using this, and applying the procedure shown in the earlier link for the solution of cubics- for which we need to solve the equation of the form: x

^{3}– px

^{2 }+ qx – r = 0, or in our specific case: x

^{3}+ (3/2)x

^{2}+ (1/16)x - 1/4 = 0

Based on the preceding, we have for the cubic parameters: p = -3/2, q = 1/16 and r = ¼

The following computations then follow the earlier (cubic) template and are as follows:

DELTA = 1.128

Epsilon1 = 1.277 - 0.658i

Epsilon2 = 1.277 + 0.658i

The cube roots of unity, e.g for (1)^1/3 are 1, rho and rho2:

where rho = -½ + [-3]^½/ 2 and rho2 = -½ - [-3]^½/ 2

Then the roots of the resolvent cubic are defined from:

φ1 = 1/3 (p + Epsilon1 + Epsilon2) = 0.351

φ2 = 1/3 (p + rho2 * Epsilon1 + rho *Epsilon2) = -1.305

φ3 = 1/3(p + rho* Epsilon1 + rho2 *Epsilon2) = -0.546

To ensure these are actually the roots of the resolvent cubic, three conditions must be met:

1) φ1 + φ2 +φ3 = p

check: 0.351 + (-1.305) + (-0.546) = -1.5 = p

2) φ1 φ2 + φ2 φ3 + φ3 φ1 = 0.063 = 1/16

check: (0.351)(-1.305) + (-1.305)(-0.546) + (-0.546)(0.351) = 0.63

3) φ1 φ2 φ3 = r = 0.25

check: (0.351)(-1.305)(-0.546) = 0.25

Finally, the solution of the reduced quartic is validated provided we have:

[φ1]^½ [φ2] ^½[φ3]^½ = B / 8 = 4/8 = 0.5

This is also fulfilled as the reader can check.

The roots of the quartic are therefore given, e.g. by Burington Mathematical Tables and Formulas (2nd Ed., p. 9) as:

x1 = [φ1]^½ + [φ2] ^½+ [φ3]^½

x2 = [φ1]^½ - [φ2] ^½ - [φ3]^½

x3 = - [φ1]^½ + [φ2] ^½ - [φ3]^½

x4 = - [φ1]^½ - [φ2] ^½+ [φ3]^½

The problem is that NONE of the above roots, e.g. x1 = 0.592 - 0.404i, deliver the solutions if substituted into the reduced quartic. When so substituted, we don't get the expected 0 but rather: 4.74 - 3.231 i

So, WHERE did the error occur? (And be sure to confirm by providing the correct roots!!) First reader to deliver the solution, via comment, receives a free copy of my recent book on the JFK Assassination, and also 'Astronomy and Astrophysics: Notes, Problems and Solutions'. Good luck!

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