IFPUG-FPA (Function Point Analisys)

We begin by reviewing the state of the art regarding solutions of polynomial equations, as it was just before the time of Galois. We consider linear, quadratic, cu-bic, quartic, and quintic equations in turn. In the case of the quintic, we also describe some ideas that were discovered after Galois. Throughout, we make the default as-sumption of the period: the coefficients of the equation are complex numbers.

Linear Equations

Let a, b ∈ C with a 6= 0. The general linear equation is at+ b = 0

and the solution is clearly

t= −b a Quadratic Equations

Let a, b, c ∈ C with a 6= 0. The general quadratic equation is at²+ bt + c = 0

Dividing by a and renaming the coefficients, we can consider the equivalent equation t²+ at + b = 0

The standard way to solve this equation is to rewrite it in the form

 t+a

2

2

=a² 4 − b Taking square roots,

t+a 2 = ±

ra² 4 − b

Solution by Radicals 25

which is the usual quadratic formula except for a change of notation. The process used here is called completing the square; as remarked in the Historical Introduction, it goes back to the Babylonians 3600 years ago.

Cubic Equations

Let a, b, c ∈ C with a 6= 0. The general cubic equation can be written in the form t³+ at²+ bt + c = 0

where again we have divided by the leading coefficient to avoid unnecessary compli-cations in the formulas.

The first step is to change the variable to make a = 0. This is achieved by setting y= t +^a₃, so that t = y −^a₃. Such a move is called a Tschirnhaus transformation, after the person who first made explicit and systematic use of it. The equation becomes

y³+ py + q = 0 (1.3)

To find the solution(s) we try (rabbit out of hat) the substitution y=√³ We now choose u and v to make both terms vanish:

u+ v + q = 0 (1.4)

Multiply (1.6) by u and subtract (1.7) to get

u(u + v) − uv = −qu +p³ 27 which can be rearranged to give

u²+ qu −p³ 27= 0 which is a quadratic.

The solution of quadratics now tells us that u= −q

Changing the sign of the square root just permutes u and v, so we can set the sign to +. Thus we find that

which (by virtue of publication, not discovery) is usually called Cardano’s formula.

(This version differs from the formula in the Historical Introduction because Cardano worked with x²+ px = q, so q changes sign.) Finally, remember that the solution t of the original equation is equal to y − a/3.

Peculiarities of Cardano’s Formula

An old Chinese proverb says ‘Be careful what you wish for: you might get it’. We have wished for a formula for the solution, and we’ve got one. It has its peculiarities.

First: recall that over C every nonzero complex number z has three cube roots. If one of them is α, then the other two are ωα and ω²α , where is a primitive cube root of 1. Then

ω²= −1 2− i

√3 2

The expression for y therefore appears to lead to nine solutions, of the form α + β α + ω β α + ω²β

ω α + β ω α + ω β ω α + ω²β ω²α + β ω²α + ω β ω²α + ω²β

Solution by Radicals 27 where α, β are specific choices of the cube roots.

However, not all of these expressions are zeros. Equation (1.5) implies (1.7), but (1.7) implies (1.5) only when we make the correct choices of cube roots. If we choose α , β so that 3α β + p = 0, then the solutions are

α + β ω α + ω²β ω²α + ω β

Another peculiarity emerges when we try to solve equations whose solutions we already know. For example,

which seems a far cry from 3. However, further algebra converts it to 3: see Exercise 1.4.

As Cardano observed in his book, it gets worse: if his formula is applied to

t³− 15t − 4 = 0 (1.9)

in contrast to the obvious solution t = 4. This is very curious even today, and must have seemed even more so in the Renaissance period.

Cardano had already encountered such baffling expressions when trying to solve the quadratic t(10 −t) = 40, with the apparently nonsensical solutions 5 +√

−15 and 5 −√

−15, but there it was possible to see the puzzling form of the ‘solution’ as ex-pressing the fact that no solution exists. However, Cardano was bright enough to spot that if you ignore the question of what such expressions mean, and just manipulate them as if they are ordinary numbers, then they do indeed satisfy the equation. ‘So,’

Cardano commented, ‘progresses arithmetic subtlety, the end of which is as refined as it is useless.’

However, this shed no light on why a cubic could possess a perfectly reasonable solution, but the formula (more properly, the equivalent numerical procedure) could not find it. Around 1560 Raphael Bombelli observed that (2 ±√

−1)³= 2 ±√

−121, and recovered (see Exercise 1.7) the solution t = 4 of (1.9) from the formula (1.10), again assuming that such expressions can be manipulated just like ordinary num-bers. But Bombelli, too, expressed scepticism that such manoeuvres had any sensi-ble meaning. In 1629 Albert Girard argued that such expressions are valid as formal solutions of the equations, and should be included ‘for the certitude of the general rules’. Girard was influential in making negative numbers acceptable, but he was way ahead of his time when it came to their square roots.

In fact, Cardano’s formula is pretty much useless whenever the cubic has three realroots. This is called the ‘irreducible case’ of the cubic, and the traditional escape

route is to use trigonometric functions, Exercise 1.8. All this rather baffled the Re-naissance mathematicians, who did not even have effective algebraic notation, and were wary of negative numbers, let alone imaginary ones.

Using Galois theory, it is possible to prove that the cube roots of complex num-bers that arise in the irreducible case of the cubic equation cannot be avoided. That is, there are no formulas in real radicals for the real and imaginary parts. See Van der Waerden (1953) volume 1 page 180, and Isaacs (1985).

Quartic Equations

An equation of the fourth degree

t⁴+ at³+ bt²+ ct + d = 0

is called a quartic equation (an older term is biquadratic). To solve it, start by making the Tschirnhaus transformation y = t + a/4, to get

y⁴+ py²+ qy + r = 0 (1.11) Introduce a new term u, and observe that



We choose u to make the right hand side a perfect square. If it is, it must be the square of√

2uy − ^q

2√

2u, and then we require

−r + p²

4 + pu + u²=q² 8u Provided u 6= 0, this becomes

8u³+ 8pu²+ (2p − 8r)u − q²= 0 (1.12)

Solution by Radicals 29 which is a cubic in u. Solving by Cardano’s method, we can find u. Now



Finally, we can solve the above two quadratics to find y.

If u = 0 we do not obtain (1.12), but if u = 0 then q = 0, so the quartic (1.11) is a quadratic in y², and can be solved using only square roots.

Equation (1.12) is called the resolvent cubic of (1.11). Explicit formulas for the roots can be obtained if required. Since they are complicated, we shall not give them here.

An alternative approach to the resolvent cubic, not requiring a preliminary Tschirnhaus transformation, is described in Exercise 1.13.

Quintic Equations

So far, we have a series of special tricks, different in each case. We can start to solve the general quintic equation

t⁵+ at⁴+ bt³+ ct²+ dt + e = 0

in a similar way. A Tschirnhaus transformation y = t + a/5 reduces it to y⁵+ py³+ qy²+ ry + s = 0

However, all variations on the tricks that we used for the quadratic, cubic, and quartic equations grind to a halt.

In 1770–1771 Lagrange analysed all of the above special tricks, showing that they can all be ‘explained’ using general principles about symmetric functions of the roots. When he applied this method to the quintic, however, he found that it

‘reduced’ the problem to a sextic—an equation of degree 6. Instead of helping, the method made the problem worse. A fascinating description of these ideas, together with a method for solving quintics whenever they are soluble by radicals, can be found in a lecture by George Neville Watson, rescued from his unpublished papers and written up by Berndt, Spearman and Williams (2002). The same article contains a wealth of other information about the quintic, including a long list of historical and recent references. Because the formulas are messy and the story is lengthy, the most we can do here is give some flavour of what is involved.

Lagrange observed that all methods for solving polynomial equations by radicals involve constructing rational functions of the roots that take a small number of values when the roots αjare permuted. Prominent among these is the expression

δ =

∏

1≤ j<k≤n

(α_j− αk) (1.13)

where n is the degree. This takes just two values, ±δ : plus for even permutations and minus for odd ones. Therefore ∆ = δ²(known as the discriminant because it is nonzero precisely when the roots are distinct, so it ‘discriminates’ among the roots) is a rational function of the coefficients. This gets us started, and it yields a complete solution for the quadratic, but for cubics upwards it does not help much unless we can find other expressions in the roots with similar properties under permutation.

Lagrange worked out what these expressions look like for the cubic and the quar-tic, and noticed a pattern. For example, if a cubic polynomial has roots α1, α2, α3and ω is a primitive cube root of unity, then the expression

u= (α1+ ωα2+ ω²α3)³

takes exactly two distinct values. In fact, even permutations leave it unchanged, while odd permutations transform it to

v= (α1+ ω²α2+ ωα3)³

It follows that u + v and uv are fixed by all permutations of the roots, and must there-fore be expressible as rational functions of the coefficients. So u + v = a, uv = b where a, b are rational functions of the coefficients. Therefore u and v are the solu-tions of the quadratic equation t²− at + b = 0, so they can be expressed using square roots. But now the further use of cube roots expresses α1+ ωα2+ ω²α3=√³

uand α1+ ω²α2+ ωα3=√³

vby radicals. Since we also know that α1+ α2+ α3is minus the coefficient of t², we have three independent linear equations in the roots, which are easily solved.

Something very similar works for the quartic, with expressions like (α1− α2+ α3− α4)²

But when we try the same idea on the quintic, an obstacle appears. Suppose that the roots of the quintic are α1, α2, α3, α4, α5. Let ζ be a primitive fifth root of unity.

Following Lagrange’s lead, it is natural to consider

w= (α1+ ζ α2+ ζ²α₃+ ζ³α₄+ ζ⁴α₅)⁵

There are 120 permutations of 5 roots, and they transform w into 24 distinct expres-sions. Therefore w is a root of a polynomial of degree 24—a big step in the wrong direction, since we started with a mere quintic.

The best that can be done is to use an expression derived by Arthur Cayley in 1861, based on an idea of Robert Harley in 1859. This expression is

x= (α1α2+ α2α3+ α3α4+ α4α5+ α5α1− α1α3− α2α4− α3α5− α4α1− α5α2)² It turns out that x takes precisely 6 values when the variables are permuted in all 120 possible ways. Therefore x is a root of a sextic equation. The equation is very complicated and has no obvious roots; it is, perhaps, better than an equation of degree 24, but it is still no improvement on the original quintic. Except when the sextic

Exercises 31 happens, by accident, to have a root whose square is rational, in which case the quintic is soluble by radicals. Indeed, this is a necessary and sufficient condition for a quintic to be soluble by radicals, see Berndt, Spearman and Williams (2002). For instance, as they explain in detail, the equation

t⁵+ 15t + 12 = 0 with similar expressions for the other four roots.

Lagrange’s general method, then, fails for the quintic. This does not prove that the general quintic is not soluble by radicals, because for all Lagrange or anyone else knew, there might be other methods that do not make the problem worse. But it does suggest that there is something very different about the quintic. Suspicion began to grow that no method would solve the quintic by radicals. Mathematicians stopped looking for such a solution, and started looking for an impossibility proof instead.

EXERCISES

1.1 Use (1.1) to prove that multiplication of complex numbers is commutative and associative. That is, if u, v, w are complex numbers, then uv = vu and (uv)w = u(vw).

1.2 Prove that√

2 is irrational, as follows. Assume for a contradiction that there exist integers a, b, with b 6= 0, such that (a/b)²= 2.

1. Show that we may assume a, b > 0.

2. Observe that if such an expression exists, then there must be one in which bis as small as possible.

3. Show that

qis rational if and only if q is a perfect square; that is, it can be written in the form q = p^a₁¹· · · p^a_nⁿ where the integers a_j, which may be positive or negative, are all even.

1.4* Prove without using Cardano’s formula that

1.6 Let ω be a primitive cube root of unity in C. With the notation of Exercise 1.5, show that the map

p+ qα + rα²7→ p + qωα + rω²α² is a monomorphism onto its image, but not an automorphism.

1.7 Use Bombelli’s observation that (2 ±√

−1)³= 2 ±√

−121 to show that (with one choice of values of the cube roots)

q3

1.10 When 27q²+ 4p³< 0 it is possible to try to make sense of Cardano’s formula by generalising Bombelli’s observation; that is, to seek α, β such that



1.11* Let P(n) be the number of ways to arrange n zeros and ones in a row, given that ones occur in groups of three or more. Show that

P(n) = 2P(n − 1) − P(n − 2) + P(n − 4)

and deduce that as n → ∞ the ratio ^P(n+1)_P(n) → x, where x > 0 is real and x⁴− 2x³+ x²− 1 = 0. Factorise this quartic as a product of two quadratics, and hence find x.

1.12* The largest square that fits inside an equilateral triangle can be placed in any of three symmetrically related positions. Eugenio Calabi noticed that there is ex-actly one other shape of triangle in which there are three equal largest squares, Figure 7. Prove that in this triangle the ratio x of the longest side the other two is a solution of the cubic equation 2x³− 2x²− 3x + 2 = 0, and find an approximate value of x to three decimal places.

Exercises 33

FIGURE 7: Calabi’s triangle.

1.13 Investigate writing the general quartic t⁴+ at³+ bt²+ ct + d in the form (t²+ pt + q)²− (rt + s)²

which, being a difference of two squares, factorises into two quadratics (t²+ pt + q + rt + s)(t²+ pt + q − rt − s)

and can thus be solved in radicals if p, q, r, s can be expressed in terms of the original coefficients a, b, c, d.

Show that doing this leads to a cubic equation.

1.14 Mark the following true or false.

(a) -1 has no square root.

(b) -1 has no real square root.

(c) -1 has two distinct square roots in C.

(d) Every subring of C is a subfield of C.

(e) Every subfield of C is a subring of C.

(f) The set of all numbers p + q√⁷

5 for p, q ∈ Q is a subring of C.

(g) The set of all numbers p + q√⁷

5 for p, q ∈ C is a subring of C.

(h) Cardano’s formula always gives a correct answer.

(i) Cardano’s formula always gives a sensible answer.

(j) A quintic equation over Q can never be solved by radicals.

Chapter 2

In document Analisis de Metodologias para la estimacion de tamanos de productos software. (página 47-55)