Section 2

Precise Determination of the Number of Zeros by Descartes' Rule of Signs

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Problem 5.183

Along with the polynomial

f(z)=a0+a1z+a2z2+⋯+amzm f(z) = a_0+a_1z+a_2z^2+\cdots +a_mz^m

of degree mm with real coefficients a0,a1,…,ama_0, a_1, \ldots , a_m, am≥0a_m \geq 0, we consider the following three associated polynomials:

P(z,ω)=a0+a1z+a2z(z−ω)+⋯+amz(z−ω)…(z−m−1‾ ω), P(z,\omega ) = a_0+a_1z+a_2z(z-\omega )+\cdots +a_mz(z-\omega )\ldots (z-\overline{m-1}\, \omega ), Q(z,ω)=  a0(1+z−m−1‾ ω)(1+z−m−2‾ ω)…(1+z−ω)(1+z)+  a1(1+z−m−1‾ ω)(1+z−m−2‾ ω)…(1+z−ω)z+  a2(1+z−m−1‾ ω)(1+z−m−2‾ ω)…(z−ω)z\dotfill+  am−1(1+z−m−1‾ ω)(z−m−2‾ ω)…(z−ω)z+  am(z−m−1‾ ω)(z−m−2‾ ω)…(z−ω)z, \begin{aligned} Q(z,\omega ) = \; & a_0 \quad (1+z-\overline{m-1}\, \omega )(1+z-\overline{m-2}\, \omega )\ldots (1+z-\omega )(1+z) \\ + \; & a_1 \quad (1+z-\overline{m-1}\, \omega )(1+z-\overline{m-2}\, \omega )\ldots (1+z-\omega )z \\ + \; & a_2 \quad (1+z-\overline{m-1}\, \omega )(1+z-\overline{m-2}\, \omega )\ldots (z-\omega )z \\ & \dotfill \\ + \; & a_{m-1}(1+z-\overline{m-1}\, \omega )(z-\overline{m-2}\, \omega ) \quad \ldots (z-\omega )z \\ + \; & a_m \quad (z-\overline{m-1}\, \omega ) \quad (z-\overline{m-2}\, \omega ) \quad \ldots (z-\omega )z, \end{aligned} R(z,ω)=  a0(1−z+m−1‾ ω)(1−z+m−2‾ ω)…(1−z+ω)(1−z)+  a1(1−z+m−1‾ ω)(1−z+m−2‾ ω)…(1−z+ω)z+  a2(1−z+m−1‾ ω)(1−z+m−2‾ ω)…(z−ω)z\dotfill+  am−1(1−z+m−1‾ ω)(z−m−2‾ ω)…(z−ω)z+  am(z−m−1‾ ω)(z−m−2‾ ω)…(z−ω)z. \begin{aligned} R(z,\omega ) = \; & a_0 \quad (1-z+\overline{m-1}\, \omega )(1-z+\overline{m-2}\, \omega )\ldots (1-z+\omega )(1-z) \\ + \; & a_1 \quad (1-z+\overline{m-1}\, \omega )(1-z+\overline{m-2}\, \omega )\ldots (1-z+\omega )z \\ + \; & a_2 \quad (1-z+\overline{m-1}\, \omega )(1-z+\overline{m-2}\, \omega )\ldots (z-\omega )z \\ & \dotfill \\ + \; & a_{m-1}(1-z+\overline{m-1}\, \omega )(z-\overline{m-2}\, \omega ) \quad \ldots (z-\omega )z \\ + \; & a_m \quad (z-\overline{m-1}\, \omega ) \quad (z-\overline{m-2}\, \omega ) \quad \ldots (z-\omega )z. \end{aligned}

The three associated polynomials depend on a parameter ω\omega. If we simultaneously replace

z by −z,ω by −ω,aν by (−1)νaν,ν=0,1,2,…,m,then z \text{ by } -z, \qquad \omega \text{ by } -\omega , \qquad a_{\nu } \text{ by } (-1)^{\nu }a_{\nu }, \quad \nu =0,1,2,\ldots ,m, \qquad \text{then} Q(z,ω) is transformed into R(z,ω). Q(z,\omega ) \text{ is transformed into } R(z,\omega ).

If we set

f(z)ekz=∑n=0∞knn!An(k)zn,k>0, f(z)e^{kz} = \sum _{n=0}^{\infty } \frac{k^n}{n!}A_n^{(k)}z^n, \qquad k>0, f(z)(1−z)−k−1=∑n=0∞Γ(k+n+1−m)Γ(k+1)kmn!Bn(k)zn,k integral,k>m−1, f(z)(1-z)^{-k-1} = \sum _{n=0}^{\infty } \frac{\Gamma (k+n+1-m)}{\Gamma (k+1)}\frac{k^m}{n!}B_n^{(k)}z^n, \qquad k \text{ integral}, \quad k>m-1, f(z)(1+z)k−1=∑n=0k+m−1(k−1)!kmn!(k−n+m−1)!Cn(k)zn,k integral,k≥1, f(z)(1+z)^{k-1} = \sum _{n=0}^{k+m-1} \frac{(k-1)!k^m}{n!(k-n+m-1)!}C_n^{(k)}z^n, \qquad k \text{ integral}, \quad k \geq 1,

then the coefficients An(k)A_n^{(k)}, Bn(k)B_n^{(k)} and Cn(k)C_n^{(k)} may be expressed in terms of the associated polynomials of f(z)f(z), P(z,ω)P(z,\omega ), Q(z,ω)Q(z,\omega ), and R(z,ω)R(z,\omega ).

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Problem 5.184

Prove the formula, valid for ω>0\omega >0, −1+(m−1)ω<Re(z)<0-1+(m-1)\omega <\mathrm{Re}\left(z\right)<0,

Q(z,ω)ωm=Γ(1ω+1)Γ(1+zω−m+1)Γ(−zω)∫01(1−t)1+zωt−z+ωωf(tt−1) dt \frac{Q(z,\omega )}{\omega ^m} = \frac{\Gamma \left(\dfrac {1}{\omega }+1\right)}{\Gamma \left(\dfrac {1+z}{\omega }-m+1\right)\Gamma \left(-\dfrac {z}{\omega }\right)} \int _0^1 (1-t)^{\frac{1+z}{\omega }}t^{-\frac{z+\omega }{\omega }}f\left(\frac{t}{t-1}\right)\, dt

and find analogous formulae for P(z,ω)P(z,\omega ) and R(z,ω)R(z,\omega ).

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Problem 5.185

Denote by

Fab,Pab,Qab,Rab \mathfrak {F}_a^b, \quad \mathfrak {P}_a^b, \quad \mathfrak {Q}_a^b, \quad \mathfrak {R}_a^b

the number of zeros of the polynomials

f(z),P(z,ω),Q(z,ω),R(z,ω), f(z), \quad P(z,\omega ), \quad Q(z,\omega ), \quad R(z,\omega ),

respectively, in the open interval a<z<ba<z<b. We have the following inequalities:

F0∞≤P0∞,F−∞0≥P−∞0, \mathfrak {F}_0^{\infty } \leq \mathfrak {P}_0^{\infty }, \qquad \mathfrak {F}_{-\infty }^0 \geq \mathfrak {P}_{-\infty }^0, F01≤Q0∞,F−∞0≥Q−1+(m−1)ω0, \mathfrak {F}_0^1 \leq \mathfrak {Q}_0^{\infty }, \qquad \mathfrak {F}_{-\infty }^0 \geq \mathfrak {Q}_{-1+(m-1)\omega }^0, F0∞≤R01+(m−1)ω,F−10≥R−∞0,F−∞−1≥R1+(m−1)ω∞. \mathfrak {F}_0^{\infty } \leq \mathfrak {R}_0^{1+(m-1)\omega }, \qquad \mathfrak {F}_{-1}^0 \geq \mathfrak {R}_{-\infty }^0, \qquad \mathfrak {F}_{-\infty }^{-1} \geq \mathfrak {R}_{1+(m-1)\omega }^{\infty }.

Here we have assumed throughout that ω>0\omega >0; in the inequalities in the second line we require moreover that (m−1)ω<1(m-1)\omega <1, and in the last inequality of the first column we require that ω−1\omega^{-1} be an integer. [38, 80.]

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Problem 5.186

Assuming that f(z)f(z) does not vanish at the end-points of the corresponding interval (that is at z=0,z=1,z=−1z=0, z=1, z=-1 respectively) we can add to the above inequalities the statement that the difference between the two sides is an even number (possibly 00).---Why must no additional assumptions be made concerning the point z=+∞z=+\infty?

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Problem 5.187

The three power series considered in 183 have the following properties:

(1) They have not fewer changes of sign than the polynomial f(z)f(z) has zeros in the interior of the intervals

(0,+∞),(0,1),(0,+∞), (0,+\infty ), \qquad (0,1), \qquad (0,+\infty ),

respectively.

(2) The number of changes of sign decreases or remains unaltered as kk increases.

(3) The number of changes of sign attains the number of zeros of f(z)f(z) in the corresponding interval, if kk is sufficiently large.

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Problem 5.188

Along with the real polynomial f(z)f(z) of degree mm, consider the fourth associated polynomial

J(z,ω)=f(mω)−(m1)f(m−1‾ ω)z+(m2)f(m−2‾ ω)z2−⋯+(−1)mf(0)zm, J(z,\omega ) = f(m\omega )-\binom {m}{1}f(\overline{m-1}\, \omega )z+\binom {m}{2}f(\overline{m-2}\, \omega )z^2-\cdots +(-1)^mf(0)z^m,

ω>0\omega >0. Analogously to the notation of 185, let Jab\mathfrak {J}_a^b denote the number of zeros of J(z,ω)J(z,\omega ) in the open interval a<z<ba<z<b. Then we have the inequalities

Fmω∞≤J01,F−∞0≤J1∞,F0mω≥J−∞0. \mathfrak {F}_{m\omega }^{\infty } \leq \mathfrak {J}_0^1, \qquad \mathfrak {F}_{-\infty }^0 \leq \mathfrak {J}_1^{\infty }, \qquad \mathfrak {F}_0^{m\omega } \geq \mathfrak {J}_{-\infty }^0.

If we assume that f(0)≥0f(0) \geq 0, f(mω)≥0f(m\omega ) \geq 0, then in addition we can assert that the difference between the two sides is an even number.

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Problem 5.189

The number of zeros of the real polynomial f(z)f(z) of degree mm in the interval mω<z<∞m\omega <z<\infty is not greater than the number of zeros of the polynomial

Δmf(0)+(m1)Δm−1f(0)z+(m2)Δm−2f(0)z2+⋯+f(0)zm \Delta ^mf(0)+\binom {m}{1}\Delta ^{m-1}f(0)z+\binom {m}{2}\Delta ^{m-2}f(0)z^2+\cdots +f(0)z^m

in the interval 0<z<10<z<1; here we use the notation

Δνf(0)=f(νω)−(ν1)f(ν−1‾ ω)+(ν2)f(ν−2‾ ω)−⋯+(−1)νf(0),ν=0,1,2,…,m. \Delta ^{\nu }f(0) = f(\nu \omega )-\binom {\nu }{1}f(\overline{\nu -1}\, \omega )+\binom {\nu }{2}f(\overline{\nu -2}\, \omega )-\cdots +(-1)^{\nu }f(0), \qquad \nu =0,1,2,\ldots ,m.
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Problem 5.189.1

The polynomial

f(x)=a0+a1x+a2x2+a3x3+⋯+amxm f(x) = a_0+a_1x+a_2x^2+a_3x^3+\cdots +a_mx^m

with real coefficients a0,a1,…,ama_0, a_1, \ldots , a_m has not more zeros in the interval 0<x<10<x<1 than the polynomial

S(x)=a0+a1x+a2x4+a3x9+⋯+amxm2 S(x) = a_0+a_1x+a_2x^4+a_3x^9+\cdots +a_mx^{m^2}

has in the interval 0<x<10<x<1, even if in the case of the latter only distinct zeros with odd multiplicity are taken into account.

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Problem 5.189.2

The entire function

D(x)=a11−x+a22−x+a33−x+⋯+amm−x D(x) = a_11^{-x}+a_22^{-x}+a_33^{-x}+\cdots +a_mm^{-x}

has not more zeros in the interval 0<x<∞0<x<\infty than the polynomial f(x)−a0f(x)-a_0 has in the interval 0<x<10<x<1, even if in the case of the latter only distinct zeros with odd multiplicity are taken into account.

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Problem 5.189.3

Let F(x1,x2,…,xm)F(x_1, x_2, \ldots , x_m) denote a homogeneous polynomial with real coefficients in the variables x1,x2,…,xmx_1, x_2, \ldots , x_m and assume that

F(x1,x2,…,xm)>0 F(x_1, x_2, \ldots , x_m)>0

where

x1≥0,x2≥0,…,xm≥0,x1+x2+⋯+xm>0. x_1 \geq 0, \quad x_2 \geq 0, \quad \ldots , \quad x_m \geq 0, \quad x_1+x_2+\cdots +x_m>0.

Then the coefficients of the polynomial

(x1+x2+⋯+xm)kF(x1,x2,…,xm) (x_1+x_2+\cdots +x_m)^kF(x_1, x_2, \ldots , x_m)

are non-negative for all sufficiently large positive integers kk. (The particular case m=2m=2 coincides with a particular case of 187 (3) concerning the third expression considered in 183 when F0∞=0\mathfrak {F}_0^{\infty }=0.)

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