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166 ALGEBRA

5 . Budan–Fourier method.Let N(x) be the number of sign alterations in the sequence
◦
P n (x), ... , P n (n) (x) consisting of the values of the polynomial (5.1.5.2) and its derivatives.
Then the number of real roots of equation (5.1.5.1) on the interval [a, b] with P n (a) ≠ 0,
P n (b) ≠ 0 is either equal to N(a)– N(b) or is less by an even number. When calculating
N(a), zero terms of the sequence are dropped. When calculating N(b), it may happen that
(i)
(i)
P (b)= 0 for k ≤ i ≤ m and P n (k–1) (b) ≠ 0, P n (m+1) (b) ≠ 0;then P (b) should be replaced
n
n
by (–1) m+1–i sign P n (m+1) (b).
6 . Sturm method for finding the number of real roots. Consider a polynomial P n (x) with
◦
no multiple roots and denote by N(x) the number of sign alterations in the sequence of
values of the polynomials (zero terms of the sequence are not taken into account):
f 0 (x)= g 0 (x)f 1 (x)– f 2 (x),
f 1 (x)= g 1 (x)f 2 (x)– f 3 (x),
........ ......... ....... ,

where f 0 (x)= P n (x), f 1 (x)= P (x); for k > 1, every polynomial –f k (x) is the residue after

n
dividing the polynomial f k–2 (x)by f k–1 (x); the last polynomial f n (x) is a nonzero constant.
Then the number of all real roots of equation (5.1.5.1) on the segment [a, b]for P n (a) ≠ 0,
P n (b) ≠ 0 is equal to N(a)– N(b).
Remark 1. Taking a =–L and b = L and passing to the limit as L →∞, we obtain the overall number
of real roots of the algebraic equation.

Example 9. Consider the following cubic equation with the parameter a:
3 2
P 3(x)= x + 3x – a = 0.
The Sturm system for this equation has the form
3 2
P 3(x)= f 0(x)= x + 3x – a,
2

[P 3(x)] x = f 1(x)= 3x + 6x,
f 2(x)= 2x + a,
3
f 3(x)= a(4 – a).
4
Case 0 < a < 4.Let us find the number of sign alterations in the Sturm system for x =–∞ and x = ∞:

x f 0(x) f 1(x) f 2(x) f 3(x) number of sign alterations
–∞ – + – + 3
∞ + + + + 0
It follows that N(–∞)– N(∞)= 3. Therefore, for 0 < a < 4, the given polynomial has three real roots.
Case a < 0 or a > 4.Let us find the number of sign alterations in the Sturm system:
x f 0(x) f 1(x) f 2(x) f 3(x) number of sign alterations
–∞ – + – – 2
∞ + + + – 1
It follows that N(–∞)– N(∞)= 1, and therefore for a < 0 or a > 4, the given polynomial has one real root.

Remark 2. If equation P n(x)= 0 has multiple roots, then P n(x)and P n (x)haveacommon divisor and
the multiple roots are found by equating to zero this divisor. In this case, f n(x) is nonconstant and N(a)– N(b)
is the number of roots between a and b, each multiple root counted only once.