Section 7.2

Power Series Expansion of Rational Functions

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Problem 7.17

Let the power series a0+a1z+a2z2+⋯a_{0} + a_{1} z + a_{2} z^{2} + \cdots represent a rational function whose denominator is of precise degree qq and whose numerator is of precise degree p−1p-1. Assume that the numerator and the denominator have no common divisor. Setting d=max⁡(0,p−q)d = \max (0, p-q) we have

Ad(q+1)=Ad+1(q+1)=Ad+2(q+1)=⋯=0. A_{d}^{(q+1)} = A_{d+1}^{(q+1)} = A_{d+2}^{(q+1)} = \cdots = 0.
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Remark.
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In Problems 17--34 we consider the Hankel (or recurrent) determinants

∣anan+1an+2…an+r−1an+1an+2an+3…an+ran+2an+3an+4…an+r+1\multicolumn5c\dotfillan+r−1an+ran+r+1…an+2r−2∣=An(r), \begin{vmatrix} a_{n} & a_{n+1} & a_{n+2} & \ldots & a_{n+r-1} \\ a_{n+1} & a_{n+2} & a_{n+3} & \ldots & a_{n+r} \\ a_{n+2} & a_{n+3} & a_{n+4} & \ldots & a_{n+r+1} \\ \multicolumn {5}{c}{\dotfill } \\ a_{n+r-1} & a_{n+r} & a_{n+r+1} & \ldots & a_{n+2r-2} \end{vmatrix} = A_{n}^{(r)},

formed from the coefficients of the power series a0+a1z+a2z2+⋯+anzn+⋯a_{0} + a_{1} z + a_{2} z^{2} + \cdots + a_{n} z^{n} + \cdots; ana_{n} is the leading entry, rr is the number of rows.

Problem 7.18

Let d,qd, q be non-negative integers. If we have

Ad(q)≠0,Ad+1(q)≠0,Ad+2(q)≠0,Ad+3(q)≠0,… A_{d}^{(q)} \neq 0, \quad A_{d+1}^{(q)} \neq 0, \quad A_{d+2}^{(q)} \neq 0, \quad A_{d+3}^{(q)} \neq 0, \quad \ldots Ad(q+1)=0,Ad+1(q+1)=0,Ad+2(q+1)=0,… A_{d}^{(q+1)} = 0, \quad A_{d+1}^{(q+1)} = 0, \quad A_{d+2}^{(q+1)} = 0, \quad \ldots

then the power series may be represented as the quotient of two polynomials, where the denominator is of degree qq and the numerator is of degree ≤q+d−1\leq q+d-1.

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Remark.
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Investigate the dependence of the linear forms

Ln(x)=anx0+an+1x1+an+2x2+⋯+an+qxq. L_{n}(x) = a_{n} x_{0} + a_{n+1} x_{1} + a_{n+2} x_{2} + \cdots + a_{n+q} x_{q}.
Problem 7.19
An(r)An+2(r)−An(r+1)An+2(r−1)=(An+1(r))2. A_{n}^{(r)} A_{n+2}^{(r)} - A_{n}^{(r+1)} A_{n+2}^{(r-1)} = (A_{n+1}^{(r)})^{2}.
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Problem 7.20

If we have

Am(q+1)=Am+1(q+1)=Am+2(q+1)=Am+3(q+1)=⋯=Am+t−1(q+1)=0, A_{m}^{(q+1)} = A_{m+1}^{(q+1)} = A_{m+2}^{(q+1)} = A_{m+3}^{(q+1)} = \cdots = A_{m+t-1}^{(q+1)} = 0,

then the tt determinants Am+1(q),Am+2(q),Am+3(q),…,Am+t(q)A_{m+1}^{(q)}, A_{m+2}^{(q)}, A_{m+3}^{(q)}, \ldots , A_{m+t}^{(q)} are either all =0=0, or all ≠0\neq 0.

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Problem 7.21

The triangular array

A0(1)A1(1)A2(1)A3(1)A4(1)∗∗∗∗]0.5ex]A0(2)A1(2)A2(2)∗∗∗∗∗A0(3)∗∗∗An+2(r−1)∗∗]0.5ex]∗∗An(r)An+1(r)An+2(r)∗∗∗An(r+1)∗∗ \begin{array}{ccccccccc} A_{0}^{(1)} & A_{1}^{(1)} & A_{2}^{(1)} & A_{3}^{(1)} & A_{4}^{(1)} & * & * & * & * \\ ]0.5ex] & A_{0}^{(2)} & A_{1}^{(2)} & A_{2}^{(2)} & * & * & * & * & * \\[0.5ex]& & A_{0}^{(3)} & * & * & * & A_{n+2}^{(r-1)} & * & * \\ ]0.5ex] & & & * & * & A_{n}^{(r)} & A_{n+1}^{(r)} & A_{n+2}^{(r)} & * \\[0.5ex]& & & & * & * & A_{n}^{(r+1)} & * & * \end{array}

is arranged so that a right angle opening upwards and bisected by the vertical contains only minors of the determinant standing at the vertex of the angle.

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Problem 7.22

In the array of Problem 21, An(r)A_{n}^{(r)} and An+1(r)A_{n+1}^{(r)} constitute a horizontal pair, An+1(r)A_{n+1}^{(r)} and An(r+1)A_{n}^{(r+1)} constitute a vertical pair, An(r)A_{n}^{(r)} and An+1(r+1)A_{n+1}^{(r+1)} constitute a diagonal pair, An+1(r)A_{n+1}^{(r)} and An−1(r+1)A_{n-1}^{(r+1)} also form a diagonal pair, cross-wise to An(r)A_{n}^{(r)} and An+1(r+1)A_{n+1}^{(r+1)}.

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(1)

If a diagonal pair vanishes, then the diagonal pair cross-wise to it also vanishes.

(2)

If a horizontal pair vanishes, then either the horizontal pair lying above it or the one lying below it also vanishes.

(3)

If a vertical pair vanishes, then either the neighbouring vertical pair on the left or the one on the right also vanishes.

Theorems (1), (2), (3) also hold on the slanting boundary line of the array of Problem 21.

Problem 7.23

If of the infinitely many determinants

A0(k+1),A1(k+1),A2(k+1),…,An(k+1),… A_{0}^{(k+1)}, \quad A_{1}^{(k+1)}, \quad A_{2}^{(k+1)}, \quad \ldots , \quad A_{n}^{(k+1)}, \quad \ldots

only finitely many are different from 00, then the power series a0+a1z+a2z2+⋯+anzn+⋯a_{0} + a_{1} z + a_{2} z^{2} + \cdots + a_{n} z^{n} + \cdots represents a rational function whose denominator is of degree not greater than kk. [Problems 20, 18.]

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Problem 7.24

If of the infinitely many determinants

A0(1),A0(2),A0(3),…,A0(n),… A_{0}^{(1)}, \quad A_{0}^{(2)}, \quad A_{0}^{(3)}, \quad \ldots , \quad A_{0}^{(n)}, \quad \ldots

only finitely many are different from 00, then the power series a0+a1z+a2z2+⋯+anzn+⋯a_{0} + a_{1} z + a_{2} z^{2} + \cdots + a_{n} z^{n} + \cdots represents a rational function.

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Problem 7.25

If of the infinitely many determinants

A0(1),A1(2),A2(3),A3(4),…,An(n+1),… A_{0}^{(1)}, \quad A_{1}^{(2)}, \quad A_{2}^{(3)}, \quad A_{3}^{(4)}, \quad \ldots , \quad A_{n}^{(n+1)}, \quad \ldots A1(1),A2(2),A3(3),…,An(n),… A_{1}^{(1)}, \quad A_{2}^{(2)}, \quad A_{3}^{(3)}, \quad \ldots , \quad A_{n}^{(n)}, \quad \ldots

only finitely many are different from 00, then the power series a0+a1z+a2z2+⋯+anzn+⋯a_{0} + a_{1} z + a_{2} z^{2} + \cdots + a_{n} z^{n} + \cdots represents a rational function.

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Problem 7.26

Show that the sufficient criteria given in Problems 23, 24, 25 for the power series a0+a1z+a2z2+⋯+anzn+⋯a_{0} + a_{1} z + a_{2} z^{2} + \cdots + a_{n} z^{n} + \cdots to represent a rational function are also necessary.

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Problem 7.27

The rank of the matrix H\mathfrak {H} is finite if and only if the power series a0+a1z+a2z2+⋯+anzn+⋯a_{0} + a_{1} z + a_{2} z^{2} + \cdots + a_{n} z^{n} + \cdots represents a rational function.

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Remark.
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An infinite matrix

(a00a01a02…a10a11a12…\multicolumn4c\dotfill) \begin{pmatrix} a_{00} & a_{01} & a_{02} & \ldots \\ a_{10} & a_{11} & a_{12} & \ldots \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix}

is said to be of finite rank rr if all determinants of order r+1r+1 contained in it vanish but not all those of order rr.

In the infinite Hankel matrix H\mathfrak {H}

H=(a0a1a2…a1a2a3…a2a3a4…\multicolumn4c\dotfill) \mathfrak {H} = \begin{pmatrix} a_{0} & a_{1} & a_{2} & \ldots \\ a_{1} & a_{2} & a_{3} & \ldots \\ a_{2} & a_{3} & a_{4} & \ldots \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix}

(aλμ=aλ+μa_{\lambda \mu } = a_{\lambda +\mu }) we term the rank just defined also the gross rank of H\mathfrak {H}. By striking out the first kk columns (or rows) we obtain from H\mathfrak {H} the matrix Hk\mathfrak {H}_{k}

Hk=(akak+1ak+2…ak+1ak+2ak+3…ak+2ak+3ak+4…\multicolumn4c\dotfill) \mathfrak {H}_{k} = \begin{pmatrix} a_{k} & a_{k+1} & a_{k+2} & \ldots \\ a_{k+1} & a_{k+2} & a_{k+3} & \ldots \\ a_{k+2} & a_{k+3} & a_{k+4} & \ldots \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix}

whose rank we shall denote by rkr_{k}. We have H0=H\mathfrak {H}_{0} = \mathfrak {H}, r0=rr_{0} = r, r0≥r1≥r2≥⋯r_{0} \geq r_{1} \geq r_{2} \geq \cdots. The minimum of the sequence of numbers r0,r1,r2,…r_{0}, r_{1}, r_{2}, \ldots, which is necessarily attained after finitely many steps, we term the net rank of H\mathfrak {H}.

Problem 7.28

The gross rank of H\mathfrak {H} is equal to the order of the last non-vanishing determinant in the infinite sequence

A0(1),A0(2),A0(3),A0(4),…. A_{0}^{(1)}, \quad A_{0}^{(2)}, \quad A_{0}^{(3)}, \quad A_{0}^{(4)}, \quad \ldots .

If this is A0(p)A_{0}^{(p)} then the net rank is equal to the order of the first non-vanishing determinant in the finite sequence

A1(p),A2(p−1),A3(p−2),…,Ap(1). A_{1}^{(p)}, \quad A_{2}^{(p-1)}, \quad A_{3}^{(p-2)}, \quad \ldots , \quad A_{p}^{(1)}.

(The net rank =0=0 if these pp determinants all vanish.)

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Problem 7.29

The net rank of H\mathfrak {H} (assumed finite) is equal to the degree of the denominator of the rational function represented by a0+a1z+a2z2+⋯+anzn+⋯a_{0} + a_{1} z + a_{2} z^{2} + \cdots + a_{n} z^{n} + \cdots. The rational function is a proper fraction if and only if the gross rank is equal to the net rank. If the gross rank is greater than the net rank, then the gross rank exceeds the degree of the numerator by one. (The numerator and the denominator of the rational function are assumed to have no common divisors; the degree here is the precise degree.)

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Problem 7.30

The power series

a0+a1z1!+a2z22!+⋯+anznn!+⋯ a_{0} + \frac{a_{1} z}{1!} + \frac{a_{2} z^{2}}{2!} + \cdots + \frac{a_{n} z^{n}}{n!} + \cdots

satisfies a linear homogeneous differential equation with constant coefficients if and only if the determinants

A0(1),A0(2),A0(3),…,A0(n),…, A_{0}^{(1)}, \quad A_{0}^{(2)}, \quad A_{0}^{(3)}, \quad \ldots , \quad A_{0}^{(n)}, \quad \ldots ,

apart from a finite number, all vanish.

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Problem 7.31

Let Qn(z)Q_{n}(z) be a polynomial of degree nn, Qn(0)=1Q_{n}(0) = 1, n=0,1,2,…n = 0, 1, 2, \ldots. For brevity set

a0+a1z+a2z2+⋯=f(z),Qk(z)f(z)=□ka0+□ka1z+□ka2z2+⋯ , a_{0} + a_{1} z + a_{2} z^{2} + \cdots = f(z), \qquad Q_{k}(z) f(z) = \square _{k} a_{0} + \square _{k} a_{1} z + \square _{k} a_{2} z^{2} + \cdots , Qk(z)Ql(z)f(z)=□k□la0+□k□la1z+□k□la2z2+⋯ Q_{k}(z) Q_{l}(z) f(z) = \square _{k} \square _{l} a_{0} + \square _{k} \square _{l} a_{1} z + \square _{k} \square _{l} a_{2} z^{2} + \cdots

(□k□lan\square_{k} \square_{l} a_{n} is a homogeneous linear expression in an,an−1,an−2,…,an−k−la_{n}, a_{n-1}, a_{n-2}, \ldots , a_{n-k-l}).

We have

A0(n+1)=∣a0□1a1…□nan□1a1□1□1a2…□1□nan+1\multicolumn4c\dotfill□nan□n□1an+1…□n□na2n∣. A_{0}^{(n+1)} = \begin{vmatrix} a_{0} & \square _{1} a_{1} & \ldots & \square _{n} a_{n} \\ \square _{1} a_{1} & \square _{1} \square _{1} a_{2} & \ldots & \square _{1} \square _{n} a_{n+1} \\ \multicolumn {4}{c}{\dotfill } \\ \square _{n} a_{n} & \square _{n} \square _{1} a_{n+1} & \ldots & \square _{n} \square _{n} a_{2n} \end{vmatrix}.
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Problem 7.32

Let a0/z+a1/z2+a2/z3+⋯a_{0}/z + a_{1}/z^{2} + a_{2}/z^{3} + \cdots be the power series expansion of a rational function whose denominator (with no common divisors with the numerator) is zq−c1zq−1−c2zq−2−⋯−cqz^{q} - c_{1} z^{q-1} - c_{2} z^{q-2} - \cdots - c_{q}. Between the matrices

Am=(amam+1…am+q−1am+1am+2…am+q\multicolumn4c\dotfillam+q−1am+q…am+2q−2),C=(0000…0cq1000…0cq−10100…0cq−20010…0cq−3\multicolumn7c\dotfill0000…1c1) \mathfrak {A}_{m} = \begin{pmatrix} a_{m} & a_{m+1} & \ldots & a_{m+q-1} \\ a_{m+1} & a_{m+2} & \ldots & a_{m+q} \\ \multicolumn {4}{c}{\dotfill } \\ a_{m+q-1} & a_{m+q} & \ldots & a_{m+2q-2} \end{pmatrix}, \qquad \mathfrak {C} = \begin{pmatrix} 0 & 0 & 0 & 0 & \ldots & 0 & c_{q} \\ 1 & 0 & 0 & 0 & \ldots & 0 & c_{q-1} \\ 0 & 1 & 0 & 0 & \ldots & 0 & c_{q-2} \\ 0 & 0 & 1 & 0 & \ldots & 0 & c_{q-3} \\ \multicolumn {7}{c}{\dotfill } \\ 0 & 0 & 0 & 0 & \ldots & 1 & c_{1} \end{pmatrix}

there exists the relation

Am=A0Cm,m=0,1,2,…. \mathfrak {A}_{m} = \mathfrak {A}_{0} \mathfrak {C}^{m}, \qquad m = 0, 1, 2, \ldots .
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Problem 7.33

In a system of finitely many power series there are rr linearly independent series if the rank of the corresponding matrix is rr, and any power series of the system is linearly dependent on those rr power series.

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Remark.
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Let the rank of the infinite matrix M\mathfrak {M} with finitely many (=m=m) rows

M=(a10a11a12…a1n…a20a21a22…a2n…\multicolumn6c\dotfillam0am1am2…amn…) \mathfrak {M} = \begin{pmatrix} a_{10} & a_{11} & a_{12} & \ldots & a_{1n} & \ldots \\ a_{20} & a_{21} & a_{22} & \ldots & a_{2n} & \ldots \\ \multicolumn {6}{c}{\dotfill } \\ a_{m0} & a_{m1} & a_{m2} & \ldots & a_{mn} & \ldots \end{pmatrix}

be denoted by r0r_{0} and let the rank of the matrix that is formed from M\mathfrak {M} by deleting the first nn columns (in whose left upper corner we have the element a1na_{1n}) be rnr_{n}. We have r0≥r1≥r2≥⋯r_{0} \geq r_{1} \geq r_{2} \geq \cdots, and we term lim⁡n→∞rn\lim_{n \rightarrow \infty } r_{n} the net rank of M\mathfrak {M}.

The matrix M\mathfrak {M} corresponds to the system of mm power series

f1(z)=a10+a11z+a12z2+⋯+a1nzn+⋯ f_{1}(z) = a_{10} + a_{11} z + a_{12} z^{2} + \cdots + a_{1n} z^{n} + \cdots f2(z)=a20+a21z+a22z2+⋯+a2nzn+⋯ f_{2}(z) = a_{20} + a_{21} z + a_{22} z^{2} + \cdots + a_{2n} z^{n} + \cdots fm(z)=am0+am1z+am2z2+⋯+amnzn+⋯ . f_{m}(z) = a_{m0} + a_{m1} z + a_{m2} z^{2} + \cdots + a_{mn} z^{n} + \cdots .

We say that these are linearly dependent if there exist constants c1,c2,…,cmc_{1}, c_{2}, \ldots , c_{m} satisfying ∣c1∣+∣c2∣+⋯+∣cm∣>0\left|c_{1}\right| + \left|c_{2}\right| + \cdots + \left|c_{m}\right| > 0, such that we have

c1f1(z)+c2f2(z)+⋯+cmfm(z)=0 c_{1} f_{1}(z) + c_{2} f_{2}(z) + \cdots + c_{m} f_{m}(z) = 0

identically in zz. They are said to be quasilinearly dependent if there exist constants c1,c2,…,cmc_{1}, c_{2}, \ldots , c_{m} satisfying ∣c1∣+∣c2∣+⋯+∣cm∣>0\left|c_{1}\right| + \left|c_{2}\right| + \cdots + \left|c_{m}\right| > 0, such that we have

c1f1(z)+c2f2(z)+⋯+cmfm(z)=P(z) c_{1} f_{1}(z) + c_{2} f_{2}(z) + \cdots + c_{m} f_{m}(z) = P(z)

identically in zz, where P(z)P(z) is a polynomial.

The power series $f(z), zf(z), z^{2} f(z), \ldots , z^{m} f(z)$ are quasilinearly dependent for sufficiently large $m$'' and f(z)f(z) represents a rational function'': these two statements are equivalent. We say that f(z)f(z) represents an algebraic function if the (m+1)2(m+1)^{2} power series zμ[f(z)]νz^{\mu } [f(z)]^{\nu } (μ,ν=0,1,2,…,m\mu , \nu = 0, 1, 2, \ldots , m) are linearly dependent for sufficiently large mm. We say that f(z)f(z) satisfies an algebraic differential equation of the rrth order, if the (m+1)r+2(m+1)^{r+2} power series

zμ[f(z)]ν[f′(z)]ν1[f′′(z)]ν2…[f(r)(z)]νr(μ,ν,ν1,ν2,…,νr=0,1,2,…,m) z^{\mu } [f(z)]^{\nu } [f'(z)]^{\nu _{1}} [f''(z)]^{\nu _{2}} \ldots [f^{(r)}(z)]^{\nu _{r}} \qquad (\mu , \nu , \nu _{1}, \nu _{2}, \ldots , \nu _{r} = 0, 1, 2, \ldots , m)

are linearly dependent for sufficiently large mm.

Problem 7.34

In a system of finitely many power series there are rr quasilinearly independent series if the net rank of the corresponding matrix is rr, and any power series of the system is quasilinearly dependent on those rr power series.

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