1. This question is concerned with solutions of the equation C (z) = 0, where

C (z) = z3 + pz + q,

for real parameters p and q . The discriminant of C is given by

∆ = q2  + p3

There are three real roots if ∆ ≤ 0 (including repeated roots if ∆ = 0). Otherwise there is one real root and a complex conjugate pair. In your analysis, you can assume that p  0 and q  0. When taking a complex cube root, keep in mind that there are three possible values (unless z = 0).

Thus, if z = reiθ  with r > 0 and −π < θ  ≤ π, then

z1/3  = ^3rei(2kπ+θ)/3,    k = 0, 1, 2.

You need not prove that taking the cube root of a complex number is a safe operation, and you

can assume that ∆ itself is safely calculated.

(a) Verify that the three roots of C are given by

zj  = S+ + S−     where   S±  = (−  ± ∆ 1/2 )1/3 ,

(*)

roots can be taken in the‘obvious’way, because using the alternative values just interchanges S+  and S− .)

Avoid messy algebra: you may find that it helps to calculate S + S and S+ S−  first.

(b)  Explain how both S+  and S−  can always be safely calculated (i.e. without any danger of

catastrophic cancellation).

(c) Show that if ∆ ≤ 0 then S−  = S+  (the complex conjugate).  Hence explain how all three roots can be safely calculated in this case.

(d) Show that if p < 0 but ∆ ≥ 0 then the real root can be safely calculated.  The complex roots can then be calculated through their relationships to the coefficients of C . Can this be achieved without any danger of catastrophic cancellation? Why (or why not)?

(e) Show that the imaginary parts of the complex roots can always be safely calculated if p > 0. Can the other information about the roots (the real root and the real part of the complex roots) be safely obtained in this case? Why (or why not)?

Hint:  find and simplify an expression for the product of the complex roots.

(f)   (i) Write a Maple procedure that takes as its arguments real numbers p and q and returns accurately calculated roots for C (z) as its results. The procedure should work by first calculating S+  and S− , and then using one or both of these to compute the roots, taking advantage of the above analysis. Your procedure should handle p = 0 and/or q = 0 as special cases.

Be careful taking cube roots.  In Maple, z^(1/3) gives the principle value (i.e. with argument divided by 3) .  In contrast, surd(  x  ,  3  ) returns a real cube root for any real x .

(ii) Test your procedure with some different values for p and q, and check that the results are accurate (no more than one or two significant figures should be lost). Make sure all three cases (p > 0, p < 0 with ∆ > 0 and ∆ < 0) are tested, and include some cases where one might expect to see cancellation were Cardano’s formula to be applied naively.

2.  Let f (x) be a quartic polynomial with distinct, real roots α 1 , . . . , α4 .

Show that

S (x) =  =  +  +  +  ,

and find a similar expression for T (x) =  −   .

(b)  Let x0  and λ be real numbers such that x0   αj  for j = 1, 2, 3, 4 and λ  x0 , and consider

the quadratic

Q(x) = K (x − x0 )2  − (x − λ)2 ,

where

(i) What can be said about roots of Q(x) between x0  and αj ?

(ii)  By writing αj  − λ = (αj  − x0 ) + (x0 − λ), prove that

K = 4 − 2(x0 − λ)S0 + (x0 − λ)2T0 ,

where S0  = S (x0 ) and T0  = T (x0 ).

(c)   (i)  Make the change of variables y = x − x0  and µ = x0  − λ to obtain an expression for

roots of Q as close as possible to the roots of f — occurs at the point where ∂Q/∂µ = 0.

Denote this value by µ0 ; find its value, and then show that Q(y, µ0 ) = 0 if

y2 [3T0 − S0(2)] − 2S0y − 4 = 0.

In your answer, you may assume that T0y2    1.

(ii)  Is it possible to have T0y2  = 1 and ∂Q/∂µ = 0 at the root? Why (or why not)?

(d)  By solving the quadratic from part (c) and writing f0  = f (x0 ), f = f\ (x0 ) etc. show that

x1  = ϕ(x0 ) = x0 −                4f0                        

Hint: move the square root into the denominator before eliminating S0  and T0 .