Section 2

Reversals of Sign of a Function

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Problem 5.22

Let the function f(x)f(x) be different from 00 and of constant sign both in a certain neighbourhood of the point aa and in a certain neighbourhood of the point bb. The interval a<x<ba < x < b contains an even or an odd number of reversals of sign of the function f(x)f(x), according as f(a)f(a) and f(b)f(b) have the same sign or opposite signs.

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Remark.
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We say that the function f(x)f(x) is of constant sign in an interval in the following two cases: (1) f(x)≤0f(x) \leq 0 in the interval, (2) f(x)≥0f(x) \geq 0 in the interval. Assume that the interval a<x<ba < x < b can be divided into R+1R+1 sub-intervals such that (1) f(x)f(x) is not identically zero in any sub-interval, (2) f(x)f(x) is of constant sign in each sub-interval, (3) f(x)f(x) is of opposite sign in adjacent sub-intervals. Under these three conditions we say that f(x)f(x) has RR reversals of sign in the interval a<x<ba < x < b. A zero of an analytic function does or does not induce a reversal of sign (is or is not a crossing point'') according as the multiplicity of the zero is odd or even. However the concept reversal of sign'' can usefully be applied also to certain non-analytic functions.

Problem 5.23

If the function f(x)f(x) has RR reversals of sign in the interval 0<x<∞0 < x < \infty, then the function ∫0xf(t) dt\int_0^x f(t)\, dt has in this interval RR or fewer reversals of sign.

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Problem 5.24

If RR is the number of reversals of sign and ZZ the number of zeros of f(x)f(x) in the same open interval, then Z−RZ-R is a non-negative even number.

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Problem 5.25

If f(a)=f(b)=0f(a) = f(b) = 0 and f(x)≠0f(x) \neq 0 for a<x<ba < x < b, then in the interval a<x<ba < x < b there are an odd number of reversals of sign of the derivative f′(x)f'(x).

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Problem 5.26

Let A1,A2,…,AnA_1, A_2, \ldots , A_n be non-zero real numbers and let a1<a2<a3<⋯<ana_1 < a_2 < a_3 < \cdots < a_n. One can prove in various cases that the rational function

f(x)=A1x−a1+A2x−a2+⋯+Anx−an f(x) = \frac{A_1}{x-a_1} + \frac{A_2}{x-a_2} + \cdots + \frac{A_n}{x-a_n}

has only real zeros, in particular in the following cases:

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(1)

A1>0,A2>0,…,An−1>0A_1 > 0, \quad A_2 > 0, \quad \ldots , \quad A_{n-1} > 0;

(2)

A1>0,A2>0,…,Ak−1>0,Ak+1>0,…,An>0,A_1 > 0, \quad A_2 > 0, \quad \ldots , \quad A_{k-1} > 0, \quad A_{k+1} > 0, \quad \ldots , \quad A_n > 0,

A1+A2+⋯+An<0,1<k<n. A_1 + A_2 + \cdots + A_n < 0, \quad 1 < k < n.
Problem 5.27

The trigonometric polynomial

f(x)=a0+a1cos⁡x+a2cos⁡2x+⋯+ancos⁡nx f(x) = a_0 + a_1 \cos x + a_2 \cos 2x + \cdots + a_n \cos nx

with real coefficients a0,a1,a2,…,ana_0, a_1, a_2, \ldots , a_n has only real zeros if

∣a0∣+∣a1∣+∣a2∣+⋯+∣an−1∣<an. \left|a_0\right| + \left|a_1\right| + \left|a_2\right| + \cdots + \left|a_{n-1}\right| < a_n.
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