The History of Mathematics: A Brief Course

QUESTIONS AND PROBLEMS^435

14.9. Consider the cubic equation of Sharaf al-Tusi's third type, which we write
as -x^3 - ax^2 + bx - c = 0. Using Horner's method, as described in Section 2, show
that if the first approximation is .ô = m, where m satisfies 3m^2 + 2am — 6 = 0, then
the equation to be satisfied at the second approximation is y^2 — (3m + a)y^2 — (m^3 +
am^2 — bm + c) = 0. That is, carry out the algorithm for reduction and show that
the process is
—c — m^3 — am^2 + bm. — c
b -3m^2 - 2am + b ( = 0)
—a —3m — a
-1 -1
14.10. Consider Problem 27 of Book 1 of De numeris datis: Two numbers are
given whose sum is 10. If one is divided by 4 and the other by 2, the product of the
quotients is 2. What are the two numbers? Solve this problem in your own way,
then solve it following Jordanus' recipe, which we paraphrase as follows. Let the
two numbers be ÷ and y, and let the quotients be e and / when ÷ and y are divided
by c and d respectively; let the product of the quotients be ef = b. Let be — h,
which is the same as fee or fx. Then multiply d by h to produce j, which is the
same as xdf or xy. Since we now know both ÷ + y and xy, we can find ÷ and y.

14.11. Solve the equation x^3 + 60x = 992 using the recipe given by Tartaglia.

14.12. How can you prove that vVfi)8 + 10 - íË/Ú08- 10 = 2?

14.13. If you know the polar form of complex numbers æ = r cos è + ir sin È, show

that the problem of taking the cube root of a complex number is equivalent to

solving two of the classical problems of antiquity simultaneously, just as Viete

claimed: the problem of two mean proportionals and the problem of trisecting the

angle.

14.14. Consider Viete's problem of finding three numbers in direct proportion given

the middle number and the difference between the largest and smallest. Show that

this problem amounts to finding ÷ and y given ^/xy and y — x. How do you solve

such a problem?

14.15. Show that the equation x^3 = px + q, where ñ > 0 and q > 0, has the

solution ÷ = ã/4ñ/3 cosfl, where è — \ arccos ((9v/27)/(2v/P^3 ))- In order for this

inverse cosine to exist it is necessary and sufficient that g^2 /4 — p^3 /27 < 0, which is

precisely the condition under which the Cardano formula requires the cube root of

a complex number. [Hint:Use the formula 4 cos^3 è — 3cos# = cos(30).]

Observe that

1 fl 1

where a = (q\/27)/(2y/jft). Thus, the solution of the cubic equation has a connec-

tion with the integral of an algebraic function 1/y, where y satisfies the quadratic

equation y^2 = 1 — x^2. This kind of connection turned out to be the key to the solu-

tion of higher-degree algebraic equations. As remarked in the text, Viete's solution

of the cubic uses a transcendental method, even though an algebraic method exists.