Part 7

Determinants and Quadratic Forms

[84/100%]
Show
LaTeX
Problem 7.1

Let the nn vertices of a polyhedron be numbered in a definite order. Define a determinant of nnth order as follows:

If the λ\lambdath and the μ\muth vertices are the two end-points of an edge of the polyhedron, let aλμ=aμλ=1a_{\lambda \mu } = a_{\mu \lambda } = 1.

If the line joining the λ\lambdath and the μ\muth vertices is not an edge of the polyhedron, let aλμ=0a_{\lambda \mu } = 0. In particular aλλ=0a_{\lambda \lambda } = 0, λ=1,2,…,n\lambda = 1, 2, \ldots , n.

Show that the value of this determinant is independent of the manner of numbering the vertices. Form and calculate the determinants for the regular tetrahedron, hexahedron and octahedron.

?
Problem 7.1.1

(continued). If exactly pp edges end in each vertex the determinant is divisible by pp.

?
Problem 7.1.1

(continued). If exactly pp edges end in each vertex the determinant is divisible by pp.

?
Problem 7.2

Calculate

∣111…1b1a1a1…a1b1b2a2…a2\multicolumn5c\dotfillb1b2b3…an∣ \begin{vmatrix} 1 & 1 & 1 & \ldots & 1 \\ b_{1} & a_{1} & a_{1} & \ldots & a_{1} \\ b_{1} & b_{2} & a_{2} & \ldots & a_{2} \\ \multicolumn {5}{c}{\dotfill } \\ b_{1} & b_{2} & b_{3} & \ldots & a_{n} \end{vmatrix}
?
Problem 7.3

Prove the identity:

∣1aλ+bμ∣1n=∏j>k1,2,…,n(aj−ak)(bj−bk)∏λ,μ1,2,…,n(aλ+bμ). \left|\frac{1}{a_{\lambda } + b_{\mu }}\right|_{1}^{n} = \frac{\prod _{j > k}^{1, 2, \ldots , n} (a_{j} - a_{k})(b_{j} - b_{k})}{\prod _{\lambda , \mu }^{1, 2, \ldots , n} (a_{\lambda } + b_{\mu })}.
?
Problem 7.4

Denoting the determinant of the quadratic form

∑λ=1n∑μ=1nxλxμλ+μ \sum _{\lambda =1}^{n} \sum _{\mu =1}^{n} \frac{x_{\lambda } x_{\mu }}{\lambda + \mu }

by DnD_{n}, we have

Dn=[1!2!…(n−1)!]3n!(n+1)!(n+2)!…(2n)!. D_{n} = \frac{[1! 2! \ldots (n-1)!]^{3} n!}{(n+1)! (n+2)! \ldots (2n)!}.

Also calculate the determinant Dn(α)D_{n}(\alpha ) of the quadratic form

∑λ=1n∑μ=1nxλxμλ+μ+α,α>−2. \sum _{\lambda =1}^{n} \sum _{\mu =1}^{n} \frac{x_{\lambda } x_{\mu }}{\lambda + \mu + \alpha }, \qquad \alpha > -2.
?
Problem 7.5
∣(aλ−bμ)n−1∣1n=∏ν=1n−1(n−ν)n−2ν∏j>k1,2,…,n(aj−ak)(bj−bk). \left|(a_{\lambda } - b_{\mu })^{n-1}\right|_{1}^{n} = \prod _{\nu =1}^{n-1} (n-\nu )^{n-2\nu } \prod _{j > k}^{1, 2, \ldots , n} (a_{j} - a_{k})(b_{j} - b_{k}).
?
Problem 7.6

Derive Theorem 86 of Part V from Theorem 48 of Part V by means of a transformation of the determinant ∣F(αλβμ)∣\left|F(\alpha_{\lambda } \beta_{\mu })\right|.

?
Problem 7.7

Setting f(x)=(r1−x)(r2−x)…(rn−x)f(x) = (r_{1} - x)(r_{2} - x) \ldots (r_{n} - x), we have

∣r1aa…abr2a…abbr3…a\multicolumn5c\dotfillbbb…rn∣=af(b)−bf(a)a−b. \begin{vmatrix} r_{1} & a & a & \ldots & a \\ b & r_{2} & a & \ldots & a \\ b & b & r_{3} & \ldots & a \\ \multicolumn {5}{c}{\dotfill } \\ b & b & b & \ldots & r_{n} \end{vmatrix} = \frac{a f(b) - b f(a)}{a - b}.
?
Remark.
?

Add to all the n2n^{2} elements the variable xx. The determinant formed in this way is a linear function of xx and as such can be determined from two particular values.

Problem 7.8

Set Δ=ad−bc\Delta = ad - bc. Then the functional determinant

∂(aΔ,bΔ,cΔ,dΔ)∂(a,b,c,d)=3Δ4. \frac{\partial (a\Delta , b\Delta , c\Delta , d\Delta )}{\partial (a, b, c, d)} = 3\Delta ^{4}.
?
Problem 7.9

Prove that ρ−2\rho - 2 is a divisor of the expression

∣ρlm+mlnl+ln]1ex]lm+mlρmn+nmnl+lnmn+nmρ∣,l≠0,  m≠0,  n≠0. \begin{vmatrix} \rho & \dfrac {l}{m} + \dfrac {m}{l} & \dfrac {n}{l} + \dfrac {l}{n} \\ ]1ex] \dfrac {l}{m} + \dfrac {m}{l} & \rho & \dfrac {m}{n} + \dfrac {n}{m} \\[1ex] \dfrac {n}{l} + \dfrac {l}{n} & \dfrac {m}{n} + \dfrac {n}{m} & \rho \end{vmatrix}, \qquad l \neq 0, \; m \neq 0, \; n \neq 0.

Determine the remaining factors.

?
Problem 7.10

The determinant

∣1,  xν,  xν2,  …,  xνn−q−1,  xνn−q+1,  …,  xνn−1,  xνn∣,ν=1,2,…,n \left|1, \; x_{\nu }, \; x_{\nu }^{2}, \; \ldots , \; x_{\nu }^{n-q-1}, \; x_{\nu }^{n-q+1}, \; \ldots , \; x_{\nu }^{n-1}, \; x_{\nu }^{n}\right|, \qquad \nu = 1, 2, \ldots , n

is an alternating rational entire function of the nn numbers x1,x2,…,xnx_{1}, x_{2}, \ldots , x_{n} and thus divisible by the product of differences

∏j>k1,2,…,n(xj−xk). \prod _{j > k}^{1, 2, \ldots , n} (x_{j} - x_{k}).

Show that the quotient SqS_{q} is equal to the qqth elementary symmetric function of the nn numbers x1,x2,…,xnx_{1}, x_{2}, \ldots , x_{n}.

?
Problem 7.11

Let the numbers a0,a1,a2,…,ana_{0}, a_{1}, a_{2}, \ldots , a_{n} be different from 00. Then we have identically in zz

a0∣z+a1a0a2a1a3a2…an−1an−2anan−1]1ex]−a1a0z0…000−a2a1z…00\multicolumn6c\dotfill000…−an−1an−2z∣=a0zn+a1zn−1+⋯+an. a_{0} \begin{vmatrix} z + \dfrac {a_{1}}{a_{0}} & \dfrac {a_{2}}{a_{1}} & \dfrac {a_{3}}{a_{2}} & \ldots & \dfrac {a_{n-1}}{a_{n-2}} & \dfrac {a_{n}}{a_{n-1}} \\ ]1ex] -\dfrac {a_{1}}{a_{0}} & z & 0 & \ldots & 0 & 0 \\[1ex] 0 & -\dfrac {a_{2}}{a_{1}} & z & \ldots & 0 & 0 \\ \multicolumn {6}{c}{\dotfill } \\ 0 & 0 & 0 & \ldots & -\dfrac {a_{n-1}}{a_{n-2}} & z \end{vmatrix} = a_{0} z^{n} + a_{1} z^{n-1} + \cdots + a_{n}.
?
Problem 7.11.1

Let a0,a1,a2,…,b0,b1,b2,…a_{0}, a_{1}, a_{2}, \ldots , b_{0}, b_{1}, b_{2}, \ldots denote constants, zz a variable. Suppose that a0≠0a_{0} \neq 0 and that we have identically

(a0+a1z+a2z2+⋯ )(b0+b1z+b2z2+⋯ )=1. (a_{0} + a_{1} z + a_{2} z^{2} + \cdots )(b_{0} + b_{1} z + b_{2} z^{2} + \cdots ) = 1.

Then

a0n+1bn=(−1)n∣a1a000…0a2a1a00…0a3a2a1a0…0\multicolumn6c\dotfillanan−1an−2an−3…a1∣. a_{0}^{n+1} b_{n} = (-1)^{n} \begin{vmatrix} a_{1} & a_{0} & 0 & 0 & \ldots & 0 \\ a_{2} & a_{1} & a_{0} & 0 & \ldots & 0 \\ a_{3} & a_{2} & a_{1} & a_{0} & \ldots & 0 \\ \multicolumn {6}{c}{\dotfill } \\ a_{n} & a_{n-1} & a_{n-2} & a_{n-3} & \ldots & a_{1} \end{vmatrix}.

(This serves as a method of computing the reciprocal of a given power series. The supposed identity may be merely formal; no considerations of convergence are involved.)

?
Problem 7.11.1

Let a0,a1,a2,…,b0,b1,b2,…a_{0}, a_{1}, a_{2}, \ldots , b_{0}, b_{1}, b_{2}, \ldots denote constants, zz a variable. Suppose that a0≠0a_{0} \neq 0 and that we have identically

(a0+a1z+a2z2+⋯ )(b0+b1z+b2z2+⋯ )=1. (a_{0} + a_{1} z + a_{2} z^{2} + \cdots )(b_{0} + b_{1} z + b_{2} z^{2} + \cdots ) = 1.

Then

a0n+1bn=(−1)n∣a1a000…0a2a1a00…0a3a2a1a0…0\multicolumn6c\dotfillanan−1an−2an−3…a1∣. a_{0}^{n+1} b_{n} = (-1)^{n} \begin{vmatrix} a_{1} & a_{0} & 0 & 0 & \ldots & 0 \\ a_{2} & a_{1} & a_{0} & 0 & \ldots & 0 \\ a_{3} & a_{2} & a_{1} & a_{0} & \ldots & 0 \\ \multicolumn {6}{c}{\dotfill } \\ a_{n} & a_{n-1} & a_{n-2} & a_{n-3} & \ldots & a_{1} \end{vmatrix}.

(This serves as a method of computing the reciprocal of a given power series. The supposed identity may be merely formal; no considerations of convergence are involved.)

?
Problem 7.11.2

Compute the coefficient of znz^{n} in the expansion of [(1+z)a−xz]−1[(1+z)^{a} - xz]^{-1} first from Problem 11.1 then directly.

?
Problem 7.11.2

Compute the coefficient of znz^{n} in the expansion of [(1+z)a−xz]−1[(1+z)^{a} - xz]^{-1} first from Problem 11.1 then directly.

?
Problem 7.11.3

Analogously to Problem 11.2, compute the coefficient of znz^{n} in the expansion of [ez−xz]−1[e^{z} - xz]^{-1}.

?
Problem 7.11.3

Analogously to Problem 11.2, compute the coefficient of znz^{n} in the expansion of [ez−xz]−1[e^{z} - xz]^{-1}.

?
Problem 7.12

Let a1,a2,a3,b1,b2,b3,ca_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}, c be real numbers. The system

−a3x2+a2x3=b1,a3x1+a2x2−a1x3=b2,−a2x1+a1x2−a1x3=b3,a1x1+a2x2+a3x3=c \begin{aligned} -a_{3} x_{2} + a_{2} x_{3} & = b_{1}, \\ a_{3} x_{1} \phantom{{} + a_{2} x_{2}} - a_{1} x_{3} & = b_{2}, \\ -a_{2} x_{1} + a_{1} x_{2} \phantom{{} - a_{1} x_{3}} & = b_{3}, \\ a_{1} x_{1} + a_{2} x_{2} + a_{3} x_{3} & = c \end{aligned}

can be consistent in two different ways. In one case the unknowns x1,x2,x3x_{1}, x_{2}, x_{3} are completely undetermined, in the other case they are completely determined.

?
Problem 7.13

Let the entire function

1+c1z+c2z2+c3z3+⋯ 1 + c_{1} z + c_{2} z^{2} + c_{3} z^{3} + \cdots

have all distinct zeros a1,a2,…,an,…,a_{1}, a_{2}, \ldots , a_{n}, \ldots ,

0<∣a1∣<∣a2∣<⋯<∣an∣<⋯ . 0 < \left|a_{1}\right| < \left|a_{2}\right| < \cdots < \left|a_{n}\right| < \cdots .

Consider the system of nn equations

1+a1u1(n)+a12u2(n)+⋯+a1nun(n)=0,1+a2u1(n)+a22u2(n)+⋯+a2nun(n)=0,\dotfill1+anu1(n)+an2u2(n)+⋯+annun(n)=0 \begin{aligned} 1 + a_{1} u_{1}^{(n)} + a_{1}^{2} u_{2}^{(n)} + \cdots + a_{1}^{n} u_{n}^{(n)} & = 0, \\ 1 + a_{2} u_{1}^{(n)} + a_{2}^{2} u_{2}^{(n)} + \cdots + a_{2}^{n} u_{n}^{(n)} & = 0, \\ & \dotfill \\ 1 + a_{n} u_{1}^{(n)} + a_{n}^{2} u_{2}^{(n)} + \cdots + a_{n}^{n} u_{n}^{(n)} & = 0 \end{aligned}

that completely determine the uk(n)u_{k}^{(n)}. If

1∣a1∣+1∣a2∣+⋯+1∣an∣+⋯ \frac{1}{\left|a_{1}\right|} + \frac{1}{\left|a_{2}\right|} + \cdots + \frac{1}{\left|a_{n}\right|} + \cdots

is convergent, then lim⁡n→∞uk(n)\lim_{n \rightarrow \infty } u_{k}^{(n)} exists, but it is not necessarily true that

lim⁡n→∞uk(n)=ck. \lim _{n \rightarrow \infty } u_{k}^{(n)} = c_{k}.
?
Remark.
?

One solution of the infinite system

1+aνu1+aν2u2+⋯=0,ν=1,2,3,… 1 + a_{\nu } u_{1} + a_{\nu }^{2} u_{2} + \cdots = 0, \qquad \nu = 1, 2, 3, \ldots

is: u1=c1,u2=c2,u3=c3,…u_{1} = c_{1}, u_{2} = c_{2}, u_{3} = c_{3}, \ldots.

Problem 7.14

In the equations

c11z1+c12z2+⋯+c1nzn=0,c21z1+c22z2+⋯+c2nzn=0,\dotfillcn1z1+cn2z2+⋯+cnnzn=0 \begin{aligned} c_{11} z_{1} + c_{12} z_{2} + \cdots + c_{1n} z_{n} & = 0, \\ c_{21} z_{1} + c_{22} z_{2} + \cdots + c_{2n} z_{n} & = 0, \\ & \dotfill \\ c_{n1} z_{1} + c_{n2} z_{2} + \cdots + c_{nn} z_{n} & = 0 \end{aligned}

let the coefficients and the unknowns be complex:

cλμ=aλμ+ibλμ,zμ=xμ+iyμ, c_{\lambda \mu } = a_{\lambda \mu } + i b_{\lambda \mu }, \qquad z_{\mu } = x_{\mu } + i y_{\mu },

aλμ,bλμ,xμ,yμa_{\lambda \mu }, b_{\lambda \mu }, x_{\mu }, y_{\mu } real. For the equations to admit not only the identically zero solution

z1=z2=⋯=zn=0,i.e.x1=x2=⋯=xn=y1=y2=⋯=yn=0, z_{1} = z_{2} = \cdots = z_{n} = 0, \qquad \text{i.e.} \quad x_{1} = x_{2} = \cdots = x_{n} = y_{1} = y_{2} = \cdots = y_{n} = 0,

it is necessary and sufficient that the determinant ∣cλμ∣1n\left|c_{\lambda \mu }\right|_{1}^{n} vanishes. This yields two equations in the 2n22n^{2} real numbers aλμ,bλμa_{\lambda \mu }, b_{\lambda \mu }. On the other hand the given equations may be written as 2n2n linear homogeneous equations for 2n2n real unknowns. The necessary and sufficient condition for the existence of a not identically vanishing solution now consists in the vanishing of a real determinant, i.e. in one equation in the aλμ,bλμa_{\lambda \mu }, b_{\lambda \mu }. How can this be correct?

?
Problem 7.15

The six terms in the expansion of a third-order determinant cannot all be positive.

?
Problem 7.16

The rule for the expansion of a determinant consists of two parts. The first part specifies what products are to be formed from the elements, while the second part determines the sign of these products.

The second part may be simplified in the case of second-order determinants in the following manner:

Associate with the elements

(a11a12a21a22) the signs (++−+) respectively. \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \text{ the signs } \begin{pmatrix} + & + \\ - & + \end{pmatrix} \text{ respectively.}

Then form the products prescribed in the first part from the elements, each with its fixed associated sign.

Now prove that a corresponding simplification is impossible in the case of a determinant of order higher than two, i.e. it is impossible to associate with the n2n^{2} elements n2n^{2} fixed signs in such a way that, if we form the products prescribed in the first part of the expansion rule from the elements each with its associated sign, we automatically obtain the correct sign in all the products.

?
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.
?
Remark.
?

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.

?
Remark.
?

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}.
?
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.

?
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.

?
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)}.

?
(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.]

?
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.

?
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.

?
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.

?
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.

?
Remark.
?

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.)

?
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.)

?
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.

?
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}.
?
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 .
?
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.

?
Remark.
?

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.

?
Problem 7.35

Consider the two quadratic forms

∑λ=1n∑μ=1naλμxλxμand∑λ=1n∑μ=1nbλμxλxμ. \sum _{\lambda =1}^{n} \sum _{\mu =1}^{n} a_{\lambda \mu } x_{\lambda } x_{\mu } \qquad \text{and} \qquad \sum _{\lambda =1}^{n} \sum _{\mu =1}^{n} b_{\lambda \mu } x_{\lambda } x_{\mu }.

If they are positive, then the quadratic form

∑λ=1n∑μ=1naλμbλμxλxμ \sum _{\lambda =1}^{n} \sum _{\mu =1}^{n} a_{\lambda \mu } b_{\lambda \mu } x_{\lambda } x_{\mu }

is also positive. If furthermore one of the given forms is definite and if in the matrix of the other the entries on the principal diagonal are different from 00, then the third form is also definite.

?
Problem 7.36

Consider the two symmetric matrices

(a11a12…a1na21a22…a2n\multicolumn4c\dotfillan1an2…ann),(ea11ea12…ea1nea21ea22…ea2n\multicolumn4c\dotfillean1ean2…eann). \begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \multicolumn {4}{c}{\dotfill } \\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{pmatrix}, \qquad \begin{pmatrix} e^{a_{11}} & e^{a_{12}} & \ldots & e^{a_{1n}} \\ e^{a_{21}} & e^{a_{22}} & \ldots & e^{a_{2n}} \\ \multicolumn {4}{c}{\dotfill } \\ e^{a_{n1}} & e^{a_{n2}} & \ldots & e^{a_{nn}} \end{pmatrix}.

If the quadratic form corresponding to the matrix (aλμ)(a_{\lambda \mu }) is positive, then the quadratic form corresponding to (eaλμ)(e^{a_{\lambda \mu }}) is also positive. If moreover among the rows of (aλμ)(a_{\lambda \mu }) there are no two identical, then the form corresponding to (eaλμ)(e^{a_{\lambda \mu }}) is in fact definite.

?
Remark.
?

See Problem 76 of Part V.

Problem 7.37

Let the power series

p0+p1x+p2x2+⋯=F(x) p_{0} + p_{1} x + p_{2} x^{2} + \cdots = F(x)

have no negative coefficients and converge for x=a11,a12,…,annx = a_{11}, a_{12}, \ldots , a_{nn}. If the quadratic form corresponding to the symmetric matrix (aλμ)(a_{\lambda \mu }) of order nn is positive, then the quadratic form corresponding to (F(aλμ))(F(a_{\lambda \mu })) is also positive. If moreover there are among the coefficients p0,p2,p4,…p_{0}, p_{2}, p_{4}, \ldots at least nn different from 00 and if of the rows of (aλμ2)(a_{\lambda \mu }^{2}) no two are identical, then the form corresponding to (F(aλμ))(F(a_{\lambda \mu })) is in fact definite.

?
Problem 7.38

Let the real numbers a0,a1,a2,…,a2na_{0}, a_{1}, a_{2}, \ldots , a_{2n} have the following property: If f(x)f(x) denotes an arbitrary polynomial, with real coefficients of degree at most 2n2n, that does not vanish identically and does not take a negative value for any real value of xx, then also

a0f(x)+a11!f′(x)+a22!f′′(x)+⋯+a2n(2n)!f(2n)(x)≥0(or>0) a_{0} f(x) + \frac{a_{1}}{1!} f'(x) + \frac{a_{2}}{2!} f''(x) + \cdots + \frac{a_{2n}}{(2n)!} f^{(2n)}(x) \geq 0 \qquad (\text{or} > 0)

for all real values of xx.

Show that for this to be true it is necessary and sufficient that the quadratic form

∑λ=0n∑μ=0naλ+μxλxμ \sum _{\lambda =0}^{n} \sum _{\mu =0}^{n} a_{\lambda +\mu } x_{\lambda } x_{\mu }

is positive (or positive definite, respectively).

?
Problem 7.39

Let the numbers a0,a1,…,ana_{0}, a_{1}, \ldots , a_{n}, n≥1n \geq 1 have the following property: If f(x)f(x) is an arbitrary polynomial with real coefficients of degree at most nn, that does not vanish identically and is non-negative for non-negative values of xx, then also

a0f(x)+a11!f′(x)+a22!f′′(x)+⋯+ann!f(n)(x)≥0(or>0) a_{0} f(x) + \frac{a_{1}}{1!} f'(x) + \frac{a_{2}}{2!} f''(x) + \cdots + \frac{a_{n}}{n!} f^{(n)}(x) \geq 0 \qquad (\text{or} > 0)

for all non-negative values of xx (xx real).

Show that for this to be true it is necessary and sufficient that the two quadratic forms

∑λ=0[n/2]∑μ=0[n/2]aλ+μxλxμ,∑λ=0[(n−1)/2]∑μ=0[(n−1)/2]aλ+μ+1xλxμ \sum _{\lambda =0}^{[n/2]} \sum _{\mu =0}^{[n/2]} a_{\lambda +\mu } x_{\lambda } x_{\mu }, \qquad \sum _{\lambda =0}^{[(n-1)/2]} \sum _{\mu =0}^{[(n-1)/2]} a_{\lambda +\mu +1} x_{\lambda } x_{\mu }

are positive (or positive definite, respectively).

?
Problem 7.40

Let the numbers a0,a1,a2,…,ana_{0}, a_{1}, a_{2}, \ldots , a_{n}, n≥1n \geq 1 have the following property: If f(x)f(x) denotes a polynomial with real coefficients of degree ≤n\leq n that does not vanish identically and is non-negative in the interval −1≤x≤1-1 \leq x \leq 1, then also

a0f(x)+a11!f′(x)+a22!f′′(x)+⋯+ann!f(n)(x)≥0 a_{0} f(x) + \frac{a_{1}}{1!} f'(x) + \frac{a_{2}}{2!} f''(x) + \cdots + \frac{a_{n}}{n!} f^{(n)}(x) \geq 0

for −1≤x≤1-1 \leq x \leq 1. Show that the only set of numbers a0,a1,a2,…,ana_{0}, a_{1}, a_{2}, \ldots , a_{n} of this kind is determined by the conditions

a0≥0,a1=a2=⋯=an=0. a_{0} \geq 0, \qquad a_{1} = a_{2} = \cdots = a_{n} = 0.
?
Problem 7.41

Let the two quadratic forms

∑λ=0n∑μ=0naλ+μxλxμ,∑λ=0n∑μ=0nbλ+μxλxμ \sum _{\lambda =0}^{n} \sum _{\mu =0}^{n} a_{\lambda +\mu } x_{\lambda } x_{\mu }, \qquad \sum _{\lambda =0}^{n} \sum _{\mu =0}^{n} b_{\lambda +\mu } x_{\lambda } x_{\mu }

be positive. If we set

cν=a0bν+(ν1)a1bν−1+(ν2)a2bν−2+⋯+aνb0, c_{\nu } = a_{0} b_{\nu } + \binom {\nu }{1} a_{1} b_{\nu -1} + \binom {\nu }{2} a_{2} b_{\nu -2} + \cdots + a_{\nu } b_{0},

then the quadratic form

∑λ=0n∑μ=0ncλ+μxλxμ \sum _{\lambda =0}^{n} \sum _{\mu =0}^{n} c_{\lambda +\mu } x_{\lambda } x_{\mu }

is also positive, and it is in fact definite if at least one of the two preceding forms is definite and the other not identically vanishing.

?
Problem 7.42

Let

cν=a0bν+(ν1)a1bν−1+(ν2)a2bν−2+⋯+aνb0. c_{\nu } = a_{0} b_{\nu } + \binom {\nu }{1} a_{1} b_{\nu -1} + \binom {\nu }{2} a_{2} b_{\nu -2} + \cdots + a_{\nu } b_{0}.

From the positiveness of the four quadratic forms

∑λ=0m∑μ=0maλ+μxλxμ,∑λ=0m−1∑μ=0m−1aλ+μ+1xλxμ, \sum _{\lambda =0}^{m} \sum _{\mu =0}^{m} a_{\lambda +\mu } x_{\lambda } x_{\mu }, \qquad \sum _{\lambda =0}^{m-1} \sum _{\mu =0}^{m-1} a_{\lambda +\mu +1} x_{\lambda } x_{\mu }, ∑λ=0m∑μ=0mbλ+μxλxμ,∑λ=0m−1∑μ=0m−1bλ+μ+1xλxμ \sum _{\lambda =0}^{m} \sum _{\mu =0}^{m} b_{\lambda +\mu } x_{\lambda } x_{\mu }, \qquad \sum _{\lambda =0}^{m-1} \sum _{\mu =0}^{m-1} b_{\lambda +\mu +1} x_{\lambda } x_{\mu }

follows the positiveness of the following two forms:

∑λ=0m∑μ=0mcλ+μxλxμ,∑λ=0m−1∑μ=0m−1cλ+μ+1xλxμ. \sum _{\lambda =0}^{m} \sum _{\mu =0}^{m} c_{\lambda +\mu } x_{\lambda } x_{\mu }, \qquad \sum _{\lambda =0}^{m-1} \sum _{\mu =0}^{m-1} c_{\lambda +\mu +1} x_{\lambda } x_{\mu }.

Furthermore, from the positiveness of the four quadratic forms

∑λ=0m∑μ=0maλ+μxλxμ,∑λ=0m∑μ=0maλ+μ+1xλxμ, \sum _{\lambda =0}^{m} \sum _{\mu =0}^{m} a_{\lambda +\mu } x_{\lambda } x_{\mu }, \qquad \sum _{\lambda =0}^{m} \sum _{\mu =0}^{m} a_{\lambda +\mu +1} x_{\lambda } x_{\mu }, ∑λ=0m∑μ=0mbλ+μxλxμ,∑λ=0m∑μ=0mbλ+μ+1xλxμ \sum _{\lambda =0}^{m} \sum _{\mu =0}^{m} b_{\lambda +\mu } x_{\lambda } x_{\mu }, \qquad \sum _{\lambda =0}^{m} \sum _{\mu =0}^{m} b_{\lambda +\mu +1} x_{\lambda } x_{\mu }

follows the positiveness of the following two forms:

∑λ=0m∑μ=0mcλ+μxλxμ,∑λ=0m∑μ=0mcλ+μ+1xλxμ. \sum _{\lambda =0}^{m} \sum _{\mu =0}^{m} c_{\lambda +\mu } x_{\lambda } x_{\mu }, \qquad \sum _{\lambda =0}^{m} \sum _{\mu =0}^{m} c_{\lambda +\mu +1} x_{\lambda } x_{\mu }.
?
Problem 7.43

Let ϑ\vartheta be real and let the complex numbers

c−n,c−n+1,…,c−1,c0,c1,…,cn−1,cn,c−ν=cˉν,ν=0,1,2,…,n c_{-n}, c_{-n+1}, \ldots , c_{-1}, c_{0}, c_{1}, \ldots , c_{n-1}, c_{n}, \qquad c_{-\nu } = \bar{c}_{\nu }, \quad \nu = 0, 1, 2, \ldots , n

have the following property: If

g(ϑ)=α0+2(α1cos⁡ϑ+β1sin⁡ϑ+α2cos⁡2ϑ+β2sin⁡2ϑ+⋯+αncos⁡nϑ+βnsin⁡nϑ) g(\vartheta ) = \alpha _{0} + 2(\alpha _{1} \cos \vartheta + \beta _{1} \sin \vartheta + \alpha _{2} \cos 2\vartheta + \beta _{2} \sin 2\vartheta + \cdots + \alpha _{n} \cos n\vartheta + \beta _{n} \sin n\vartheta ) =∑ν=−nnγνe−iνϑ,γν=γˉ−ν=αν+iβν,ν=0,1,2,…,n;β0=0, = \sum _{\nu =-n}^{n} \gamma _{\nu } e^{-i\nu \vartheta }, \qquad \gamma _{\nu } = \bar{\gamma }_{-\nu } = \alpha _{\nu } + i\beta _{\nu }, \quad \nu = 0, 1, 2, \ldots , n; \quad \beta _{0} = 0,

denotes an arbitrary trigonometric polynomial of degree at most nn, that does not vanish identically and does not have a negative value for any value of ϑ\vartheta, let

∑ν=−nncνγνe−iνϑ≥0(or>0). \sum _{\nu =-n}^{n} c_{\nu } \gamma _{\nu } e^{-i\nu \vartheta } \geq 0 \qquad (\text{or} > 0).

Show that for this to be true, it is necessary and sufficient that the Hermitian form

∑λ=0n∑μ=0ncμ−λxλxˉμ \sum _{\lambda =0}^{n} \sum _{\mu =0}^{n} c_{\mu -\lambda } x_{\lambda } \bar{x}_{\mu }

is positive (or positive definite, respectively).

?
Problem 7.43.1

Let G(z)G(z) be an entire function that takes real values for real zz, with G(0)=1G(0) = 1, and

−G′(z)G(z)=s1+s2z+s3z2+⋯+smzm−1+⋯ . -\frac{G'(z)}{G(z)} = s_{1} + s_{2} z + s_{3} z^{2} + \cdots + s_{m} z^{m-1} + \cdots .

If G(z)G(z) has infinitely many zeros, all real, and is of the form given in Problem 165 of Part V, then, for n=1,2,3,…n = 1, 2, 3, \ldots,

∣s2s3s4…sn+1s3s4s5…sn+2s4s5s6…sn+3\multicolumn5c\dotfillsn+1sn+2sn+3…s2n∣>0. \begin{vmatrix} s_{2} & s_{3} & s_{4} & \ldots & s_{n+1} \\ s_{3} & s_{4} & s_{5} & \ldots & s_{n+2} \\ s_{4} & s_{5} & s_{6} & \ldots & s_{n+3} \\ \multicolumn {5}{c}{\dotfill } \\ s_{n+1} & s_{n+2} & s_{n+3} & \ldots & s_{2n} \end{vmatrix} > 0.

(If G(z)G(z) of the form in Problem 165 of Part V has only a finite number of zeros, then only a finite number of these recurrent determinants can be ≠0\neq 0 [Problem 27].)

?
Problem 7.43.1

Let G(z)G(z) be an entire function that takes real values for real zz, with G(0)=1G(0) = 1, and

−G′(z)G(z)=s1+s2z+s3z2+⋯+smzm−1+⋯ . -\frac{G'(z)}{G(z)} = s_{1} + s_{2} z + s_{3} z^{2} + \cdots + s_{m} z^{m-1} + \cdots .

If G(z)G(z) has infinitely many zeros, all real, and is of the form given in Problem 165 of Part V, then, for n=1,2,3,…n = 1, 2, 3, \ldots,

∣s2s3s4…sn+1s3s4s5…sn+2s4s5s6…sn+3\multicolumn5c\dotfillsn+1sn+2sn+3…s2n∣>0. \begin{vmatrix} s_{2} & s_{3} & s_{4} & \ldots & s_{n+1} \\ s_{3} & s_{4} & s_{5} & \ldots & s_{n+2} \\ s_{4} & s_{5} & s_{6} & \ldots & s_{n+3} \\ \multicolumn {5}{c}{\dotfill } \\ s_{n+1} & s_{n+2} & s_{n+3} & \ldots & s_{2n} \end{vmatrix} > 0.

(If G(z)G(z) of the form in Problem 165 of Part V has only a finite number of zeros, then only a finite number of these recurrent determinants can be ≠0\neq 0 [Problem 27].)

?
Problem 7.43.2

Let G(z)G(z) be an entire function of the form given in Problem 165 of Part V, but not of the form eβze^{\beta z}, and

1G(z)=c0+c11!z+c22!z2+c33!z3+⋯ . \frac{1}{G(z)} = c_{0} + \frac{c_{1}}{1!} z + \frac{c_{2}}{2!} z^{2} + \frac{c_{3}}{3!} z^{3} + \cdots .

Then, for n=0,1,2,…n = 0, 1, 2, \ldots,

∣c0c1c2…cnc1c2c3…cn+1c2c3c4…cn+2\multicolumn5c\dotfillcncn+1cn+2…c2n∣>0. \begin{vmatrix} c_{0} & c_{1} & c_{2} & \ldots & c_{n} \\ c_{1} & c_{2} & c_{3} & \ldots & c_{n+1} \\ c_{2} & c_{3} & c_{4} & \ldots & c_{n+2} \\ \multicolumn {5}{c}{\dotfill } \\ c_{n} & c_{n+1} & c_{n+2} & \ldots & c_{2n} \end{vmatrix} > 0.
?
Problem 7.43.2

Let G(z)G(z) be an entire function of the form given in Problem 165 of Part V, but not of the form eβze^{\beta z}, and

1G(z)=c0+c11!z+c22!z2+c33!z3+⋯ . \frac{1}{G(z)} = c_{0} + \frac{c_{1}}{1!} z + \frac{c_{2}}{2!} z^{2} + \frac{c_{3}}{3!} z^{3} + \cdots .

Then, for n=0,1,2,…n = 0, 1, 2, \ldots,

∣c0c1c2…cnc1c2c3…cn+1c2c3c4…cn+2\multicolumn5c\dotfillcncn+1cn+2…c2n∣>0. \begin{vmatrix} c_{0} & c_{1} & c_{2} & \ldots & c_{n} \\ c_{1} & c_{2} & c_{3} & \ldots & c_{n+1} \\ c_{2} & c_{3} & c_{4} & \ldots & c_{n+2} \\ \multicolumn {5}{c}{\dotfill } \\ c_{n} & c_{n+1} & c_{n+2} & \ldots & c_{2n} \end{vmatrix} > 0.
?
Problem 7.44

Let the n2n^{2} elements aλμa_{\lambda \mu } of a determinant of nnth order be independent variables. Show that, of the n!n! terms ±a1k1a2k2…ankn\pm a_{1k_{1}} a_{2k_{2}} \ldots a_{nk_{n}} in the expansion of the determinant, only N=n2−2n+2N = n^{2} - 2n + 2 are independent, and determine NN terms by means of which all the remaining terms may be expressed rationally.

?
Problem 7.45

In the expansion of a symmetric nnth order determinant (with arbitrary elements) let sn′s_{n}' be the number of distinct positive terms and sn′′s_{n}'' the number of distinct negative terms. For sn=sn′+sn′′s_{n} = s_{n}' + s_{n}'' we have already in the works of Cayley the recurrence relation

sn+1=(n+1)sn−(n2)sn−2. s_{n+1} = (n+1) s_{n} - \binom {n}{2} s_{n-2}.

Show that for dn=sn′−sn′′d_{n} = s_{n}' - s_{n}'' we have the recurrence relation

dn+1=−(n−1)dn−(n2)dn−2, d_{n+1} = -(n-1) d_{n} - \binom {n}{2} d_{n-2},

and that

lim⁡n→∞n1/2snn!=e3/4π,lim⁡n→∞(−1)n−1n3/2dnn!=e−3/42π. \lim _{n \rightarrow \infty } \frac{n^{1/2} s_{n}}{n!} = \frac{e^{3/4}}{\sqrt{\pi }}, \qquad \lim _{n \rightarrow \infty } \frac{(-1)^{n-1} n^{3/2} d_{n}}{n!} = \frac{e^{-3/4}}{2\sqrt{\pi }}.
?
Problem 7.46

In the expansion of a symmetric nnth order determinant ∣aλμ∣\left|a_{\lambda \mu }\right|, in which the nn elements aλλa_{\lambda \lambda } in the leading diagonal are equal to zero, let σn′\sigma_{n}' be the number of distinct positive terms and σn′′\sigma_{n}'' the number of distinct negative terms. Setting

σn=σn′+σn′′,δn=σn′−σn′′, \sigma _{n} = \sigma _{n}' + \sigma _{n}'', \qquad \delta _{n} = \sigma _{n}' - \sigma _{n}'',

we have

σn+1=nσn+nσn−1−(n2)σn−2,δn+1=−nδn−nδn−1−(n2)δn−2 \sigma _{n+1} = n\sigma _{n} + n\sigma _{n-1} - \binom {n}{2} \sigma _{n-2}, \qquad \delta _{n+1} = -n\delta _{n} - n\delta _{n-1} - \binom {n}{2} \delta _{n-2}

and

lim⁡n→∞n1/2σnn!=e−1/4π,lim⁡n→∞(−1)n−1n3/2δnn!=e1/42π. \lim _{n \rightarrow \infty } \frac{n^{1/2} \sigma _{n}}{n!} = \frac{e^{-1/4}}{\sqrt{\pi }}, \qquad \lim _{n \rightarrow \infty } \frac{(-1)^{n-1} n^{3/2} \delta _{n}}{n!} = \frac{e^{1/4}}{2\sqrt{\pi }}.
?
Problem 7.46.1

Let the capital letters A,B,…,I,…,ZA, B, \ldots , I, \ldots , Z denote n×nn \times n matrices. In particular, let II denote the identity matrix with entries 11 along the main diagonal and 00 elsewhere, and ZZ denote the matrix in which all entries are 11. Show that

Z⋅Z=n⋅Z. Z \cdot Z = n \cdot Z.
?
Problem 7.46.1

Let the capital letters A,B,…,I,…,ZA, B, \ldots , I, \ldots , Z denote n×nn \times n matrices. In particular, let II denote the identity matrix with entries 11 along the main diagonal and 00 elsewhere, and ZZ denote the matrix in which all entries are 11. Show that

Z⋅Z=n⋅Z. Z \cdot Z = n \cdot Z.
?
Problem 7.46.2

(continued). If all the entries of AA are positive numbers,

A⋅B=I, A \cdot B = I,

and n>1n > 1, then there are among the entries of BB at least nn positive numbers and at least nn negative numbers.

Prove this, and show by examples that the two bounds for the number of positive entries, nn and n2−nn^{2} - n, can be attained.

?
Problem 7.46.2

(continued). If all the entries of AA are positive numbers,

A⋅B=I, A \cdot B = I,

and n>1n > 1, then there are among the entries of BB at least nn positive numbers and at least nn negative numbers.

Prove this, and show by examples that the two bounds for the number of positive entries, nn and n2−nn^{2} - n, can be attained.

?
Problem 7.47

Denote the difference product ∏j>k1,2,…,n(xj−xk)\prod_{j>k}^{1,2,\ldots ,n} (x_{j} - x_{k}) by Δ\Delta. It is well known that an expression of the form

Φ=1Δ∑±φ(xλ1,xλ2,…,xλn) \Phi = \frac{1}{\Delta } \sum \pm \varphi (x_{\lambda _{1}}, x_{\lambda _{2}}, \ldots , x_{\lambda _{n}})

is a symmetric function of x1,x2,…,xnx_{1}, x_{2}, \ldots , x_{n}. Here the sum is to be taken over all the n!n! permutations of 1,2,…,n1, 2, \ldots , n and for the even permutations the positive sign is to be chosen and for the odd permutations the negative sign. Prove that, if

φ=∏ν=1n11−x1x2…xν, \varphi = \prod _{\nu =1}^{n} \frac{1}{1 - x_{1} x_{2} \ldots x_{\nu }},

then

Φ=1∏ν=1n(1−xν)∏j>k1,2,…,n(1−xjxk). \Phi = \frac{1}{\prod _{\nu =1}^{n} (1-x_{\nu }) \prod _{j>k}^{1,2,\ldots ,n} (1 - x_{j} x_{k})}.
?
Problem 7.48

Let the characteristic equation of the matrix ∥aλμ∥\left\| a_{\lambda \mu }\right\|, λ,μ=1,2,…,n\lambda , \mu = 1, 2, \ldots , n,

χ(z)=∣a11−za12…a1na21a22−z…a2n\multicolumn4c\dotfillan1an2…ann−z∣=0 \chi (z) = \begin{vmatrix} a_{11} - z & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} - z & \ldots & a_{2n} \\ \multicolumn {4}{c}{\dotfill } \\ a_{n1} & a_{n2} & \ldots & a_{nn} - z \end{vmatrix} = 0

have the roots α1,α2,…,αn\alpha_{1}, \alpha_{2}, \ldots , \alpha_{n}, that need not necessarily be all distinct. Denote the minors of order n−1n-1 of χ(z)\chi (z) by χλμ(z)\chi_{\lambda \mu }(z). Prove that the characteristic equation of the matrix χλμ(z)\chi_{\lambda \mu }(z) has the roots

ζρ=χ(z)αρ−z,ρ=1,2,…,n. \zeta _{\rho } = \frac{\chi (z)}{\alpha _{\rho } - z}, \qquad \rho = 1, 2, \ldots , n.
?
Problem 7.49

The linear transformations

Sα:yp=sin⁡απn+1∑q=0nxqsin⁡πn+1(p−q+α),p=0,1,…,n, S_{\alpha }: \qquad y_{p} = \frac{\sin \alpha \pi }{n+1} \sum _{q=0}^{n} \frac{x_{q}}{\sin \dfrac {\pi }{n+1} (p-q+\alpha )}, \qquad p = 0, 1, \ldots , n,

α\alpha arbitrary, form a group in the following sense: We have

SαSβ=Sα+β. S_{\alpha } S_{\beta } = S_{\alpha +\beta }.
?
Problem 7.50

Let g(ϑ)g(\vartheta ) be a trigonometric polynomial of order at most nn, with purely real coefficients. Determine the conditions that g(ϑ)g(\vartheta ) must satisfy, in order that the linear transformations

Sα:yp=∑q=0nxqg(πn+1(p−q+α)),p=0,1,…,n, S_{\alpha }: \qquad y_{p} = \sum _{q=0}^{n} x_{q} g\left( \frac{\pi }{n+1} (p-q+\alpha ) \right), \qquad p = 0, 1, \ldots , n,

α\alpha arbitrary, form a group in the following sense: We have

SαSβ=Sα+β. S_{\alpha } S_{\beta } = S_{\alpha +\beta }.
?
Problem 7.51

(continued). The determinant of the linear transformation SαS_{\alpha } vanishes for all values of α\alpha. The single exceptional case, when it is identically =1=1, is

g(ϑ)=1n+1sin⁡(n+1)ϑsin⁡ϑ. g(\vartheta ) = \frac{1}{n+1} \frac{\sin (n+1)\vartheta }{\sin \vartheta }.
?
Problem 7.52

(Continuation of Problem 49.) The transformation SαS_{\alpha } is orthogonal, i.e. we have identically in x0,x1,x2,…,xnx_{0}, x_{1}, x_{2}, \ldots , x_{n} that

y02+y12+y22+⋯+yn2=x02+x12+x22+⋯+xn2. y_{0}^{2} + y_{1}^{2} + y_{2}^{2} + \cdots + y_{n}^{2} = x_{0}^{2} + x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2}.

We say that the linear transformation

y1=a11x1+a12x2+⋯+a1nxn+⋯y2=a21x1+a22x2+⋯+a2nxn+⋯=\dotfillym=am1x1+am2x2+⋯+amnxn+⋯=\dotfill \begin{aligned} y_{1} & = a_{11} x_{1} + a_{12} x_{2} + \cdots + a_{1n} x_{n} + \cdots \\ y_{2} & = a_{21} x_{1} + a_{22} x_{2} + \cdots + a_{2n} x_{n} + \cdots \\ & \phantom{{}={}} \dotfill \\ y_{m} & = a_{m1} x_{1} + a_{m2} x_{2} + \cdots + a_{mn} x_{n} + \cdots \\ & \phantom{{}={}} \dotfill \end{aligned}

or that the corresponding matrix ∥amn∥\left\| a_{mn}\right\| is orthogonal, if the relations

am12+am22+am32+⋯+amn2+⋯=1,m=1,2,3,…, a_{m1}^{2} + a_{m2}^{2} + a_{m3}^{2} + \cdots + a_{mn}^{2} + \cdots = 1, \qquad m = 1, 2, 3, \ldots , aλ1aμ1+aλ2aμ2+aλ3aμ3+⋯+aλnaμn+⋯=0,λ≤μ,λ,μ=1,2,3,… a_{\lambda 1} a_{\mu 1} + a_{\lambda 2} a_{\mu 2} + a_{\lambda 3} a_{\mu 3} + \cdots + a_{\lambda n} a_{\mu n} + \cdots = 0, \qquad \lambda \leq \mu , \quad \lambda , \mu = 1, 2, 3, \ldots

and also the further relations

a1n2+a2n2+a3n2+⋯+amn2+⋯=1,n=1,2,3,…, a_{1n}^{2} + a_{2n}^{2} + a_{3n}^{2} + \cdots + a_{mn}^{2} + \cdots = 1, \qquad n = 1, 2, 3, \ldots , a1λa1μ+a2λa2μ+a3λa3μ+⋯+amλamμ+⋯=0,λ≤μ,λ,μ=1,2,3,… a_{1\lambda } a_{1\mu } + a_{2\lambda } a_{2\mu } + a_{3\lambda } a_{3\mu } + \cdots + a_{m\lambda } a_{m\mu } + \cdots = 0, \qquad \lambda \leq \mu , \quad \lambda , \mu = 1, 2, 3, \ldots

are satisfied. We define similarly the orthogonality of a matrix that goes to infinity in four directions:

(\multicolumn7c\dotfill…a−m,−n…a−m,−1a−m,0a−m,1…\multicolumn7c\dotfill…am,−n…am,−1am,0am,1…\multicolumn7c\dotfill). \begin{pmatrix} \multicolumn {7}{c}{\dotfill } \\ \ldots & a_{-m,-n} & \ldots & a_{-m,-1} & a_{-m,0} & a_{-m,1} & \ldots \\ \multicolumn {7}{c}{\dotfill } \\ \ldots & a_{m,-n} & \ldots & a_{m,-1} & a_{m,0} & a_{m,1} & \ldots \\ \multicolumn {7}{c}{\dotfill } \end{pmatrix}.
?
Problem 7.53

The Fibonacci numbers 0,1,1,2,3,5,8,13,21,34,…0, 1, 1, 2, 3, 5, 8, 13, 21, 34, \ldots are defined as follows: Set u0=0u_{0} = 0, u1=1u_{1} = 1, un+un+1=un+2u_{n} + u_{n+1} = u_{n+2} for n=0,1,2,…n = 0, 1, 2, \ldots. The linear transformation

yn=u1x1+u2x2+⋯+unxn−unxn+1unun+2,n=1,2,3,… y_{n} = \frac{u_{1} x_{1} + u_{2} x_{2} + \cdots + u_{n} x_{n} - u_{n} x_{n+1}}{\sqrt{u_{n} u_{n+2}}}, \qquad n = 1, 2, 3, \ldots

is orthogonal.

?
Problem 7.54

Let α\alpha be real and not integral. The linear transformation

ym=sin⁡αππ∑n=−∞∞xnm+n−α,m=…,−2,−1,0,1,2,… y_{m} = \frac{\sin \alpha \pi }{\pi } \sum _{n=-\infty }^{\infty } \frac{x_{n}}{m+n-\alpha }, \qquad m = \ldots , -2, -1, 0, 1, 2, \ldots

is orthogonal.

?
Problem 7.54.1

In problems 54.1--54.4 we consider infinite triangular matrices

(a11a21a22a31a32a33\multicolumn4c\dotfill), \begin{pmatrix} a_{11} & & & \\ a_{21} & a_{22} & & \\ a_{31} & a_{32} & a_{33} & \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix},

where it is understood that all elements above the principal diagonal are 00. An equation between such matrices is regarded as valid if, and only if, it is valid between the corresponding submatrices with n2n^{2} elements obtained by deleting all rows and columns except the first nn, for n=1,2,3,…n = 1, 2, 3, \ldots.

Prove the relation, involving binomial coefficients,

(1111211331\multicolumn4c\dotfill)(1−111−21−13−31\multicolumn4c\dotfill)=(1010010001\multicolumn4c\dotfill). \begin{pmatrix} 1 & & & \\ 1 & 1 & & \\ 1 & 2 & 1 & \\ 1 & 3 & 3 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix} \begin{pmatrix} 1 & & & \\ -1 & 1 & & \\ 1 & -2 & 1 & \\ -1 & 3 & -3 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix} = \begin{pmatrix} 1 & & & \\ 0 & 1 & & \\ 0 & 0 & 1 & \\ 0 & 0 & 0 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix}.
?
Problem 7.54.1

In problems 54.1--54.4 we consider infinite triangular matrices

(a11a21a22a31a32a33\multicolumn4c\dotfill), \begin{pmatrix} a_{11} & & & \\ a_{21} & a_{22} & & \\ a_{31} & a_{32} & a_{33} & \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix},

where it is understood that all elements above the principal diagonal are 00. An equation between such matrices is regarded as valid if, and only if, it is valid between the corresponding submatrices with n2n^{2} elements obtained by deleting all rows and columns except the first nn, for n=1,2,3,…n = 1, 2, 3, \ldots.

Prove the relation, involving binomial coefficients,

(1111211331\multicolumn4c\dotfill)(1−111−21−13−31\multicolumn4c\dotfill)=(1010010001\multicolumn4c\dotfill). \begin{pmatrix} 1 & & & \\ 1 & 1 & & \\ 1 & 2 & 1 & \\ 1 & 3 & 3 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix} \begin{pmatrix} 1 & & & \\ -1 & 1 & & \\ 1 & -2 & 1 & \\ -1 & 3 & -3 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix} = \begin{pmatrix} 1 & & & \\ 0 & 1 & & \\ 0 & 0 & 1 & \\ 0 & 0 & 0 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix}.
?
Problem 7.54.2

Prove the relation, involving Stirling numbers skns_{k}^{n} and SknS_{k}^{n} of the first and second kind respectively (see the introductions to Problems 197 and 186 of Part I),

(1111311761\multicolumn4c\dotfill)(1−112−31−611−61\multicolumn4c\dotfill)=(1010010001\multicolumn4c\dotfill). \begin{pmatrix} 1 & & & \\ 1 & 1 & & \\ 1 & 3 & 1 & \\ 1 & 7 & 6 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix} \begin{pmatrix} 1 & & & \\ -1 & 1 & & \\ 2 & -3 & 1 & \\ -6 & 11 & -6 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix} = \begin{pmatrix} 1 & & & \\ 0 & 1 & & \\ 0 & 0 & 1 & \\ 0 & 0 & 0 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix}.

In other words, the matrices (Skn)(S_{k}^{n}) and ((−1)n−kskn)((-1)^{n-k} s_{k}^{n}) are inverse to each other.

?
Problem 7.54.2

Prove the relation, involving Stirling numbers skns_{k}^{n} and SknS_{k}^{n} of the first and second kind respectively (see the introductions to Problems 197 and 186 of Part I),

(1111311761\multicolumn4c\dotfill)(1−112−31−611−61\multicolumn4c\dotfill)=(1010010001\multicolumn4c\dotfill). \begin{pmatrix} 1 & & & \\ 1 & 1 & & \\ 1 & 3 & 1 & \\ 1 & 7 & 6 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix} \begin{pmatrix} 1 & & & \\ -1 & 1 & & \\ 2 & -3 & 1 & \\ -6 & 11 & -6 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix} = \begin{pmatrix} 1 & & & \\ 0 & 1 & & \\ 0 & 0 & 1 & \\ 0 & 0 & 0 & 1 \\ \multicolumn {4}{c}{\dotfill } \end{pmatrix}.

In other words, the matrices (Skn)(S_{k}^{n}) and ((−1)n−kskn)((-1)^{n-k} s_{k}^{n}) are inverse to each other.

?
Problem 7.54.3

Prove

(11−11−211−33−11−46−41\multicolumn5c\dotfill)2=(101001000100001\multicolumn5c\dotfill). \begin{pmatrix} 1 & & & & \\ 1 & -1 & & & \\ 1 & -2 & 1 & & \\ 1 & -3 & 3 & -1 & \\ 1 & -4 & 6 & -4 & 1 \\ \multicolumn {5}{c}{\dotfill } \end{pmatrix}^{2} = \begin{pmatrix} 1 & & & & \\ 0 & 1 & & & \\ 0 & 0 & 1 & & \\ 0 & 0 & 0 & 1 & \\ 0 & 0 & 0 & 0 & 1 \\ \multicolumn {5}{c}{\dotfill } \end{pmatrix}.
?
Remark.
?

Cf. Problem 34.1 of Part I and Problem 54.1.

Problem 7.54.3

Prove

(11−11−211−33−11−46−41\multicolumn5c\dotfill)2=(101001000100001\multicolumn5c\dotfill). \begin{pmatrix} 1 & & & & \\ 1 & -1 & & & \\ 1 & -2 & 1 & & \\ 1 & -3 & 3 & -1 & \\ 1 & -4 & 6 & -4 & 1 \\ \multicolumn {5}{c}{\dotfill } \end{pmatrix}^{2} = \begin{pmatrix} 1 & & & & \\ 0 & 1 & & & \\ 0 & 0 & 1 & & \\ 0 & 0 & 0 & 1 & \\ 0 & 0 & 0 & 0 & 1 \\ \multicolumn {5}{c}{\dotfill } \end{pmatrix}.
?
Remark.
?

Cf. Problem 34.1 of Part I and Problem 54.1.

Problem 7.54.4

Generalizing Problem 54.3, find all infinite sequences of numbers a0,a1,a2,…a_{0}, a_{1}, a_{2}, \ldots such that the system of relations

yn=anx0−(n1)an−1x1+(n2)an−2x2−⋯+(−1)na0xn,n=0,1,2,3,…, y_{n} = a_{n} x_{0} - \binom {n}{1} a_{n-1} x_{1} + \binom {n}{2} a_{n-2} x_{2} - \cdots + (-1)^{n} a_{0} x_{n}, \qquad n = 0, 1, 2, 3, \ldots ,

should imply the same system with the roles of xx and yy reversed, that is,

xn=any0−(n1)an−1y1+(n2)an−2y2−⋯+(−1)na0yn x_{n} = a_{n} y_{0} - \binom {n}{1} a_{n-1} y_{1} + \binom {n}{2} a_{n-2} y_{2} - \cdots + (-1)^{n} a_{0} y_{n}

for n=0,1,2,3,…n = 0, 1, 2, 3, \ldots.

?
Remark.
?

See Problem 34.1 of Part I.

Problem 7.54.4

Generalizing Problem 54.3, find all infinite sequences of numbers a0,a1,a2,…a_{0}, a_{1}, a_{2}, \ldots such that the system of relations

yn=anx0−(n1)an−1x1+(n2)an−2x2−⋯+(−1)na0xn,n=0,1,2,3,…, y_{n} = a_{n} x_{0} - \binom {n}{1} a_{n-1} x_{1} + \binom {n}{2} a_{n-2} x_{2} - \cdots + (-1)^{n} a_{0} x_{n}, \qquad n = 0, 1, 2, 3, \ldots ,

should imply the same system with the roles of xx and yy reversed, that is,

xn=any0−(n1)an−1y1+(n2)an−2y2−⋯+(−1)na0yn x_{n} = a_{n} y_{0} - \binom {n}{1} a_{n-1} y_{1} + \binom {n}{2} a_{n-2} y_{2} - \cdots + (-1)^{n} a_{0} y_{n}

for n=0,1,2,3,…n = 0, 1, 2, 3, \ldots.

?
Remark.
?

See Problem 34.1 of Part I.

Problem 7.55

We term the determinant

∣f1(x)f1′(x)f1′′(x)…f1(n−1)(x)f2(x)f2′(x)f2′′(x)…f2(n−1)(x)f3(x)f3′(x)f3′′(x)…f3(n−1)(x)\multicolumn5c\dotfillfn(x)fn′(x)fn′′(x)…fn(n−1)(x)∣=W[f1(x),f2(x),…,fn(x)] \begin{vmatrix} f_{1}(x) & f_{1}'(x) & f_{1}''(x) & \ldots & f_{1}^{(n-1)}(x) \\ f_{2}(x) & f_{2}'(x) & f_{2}''(x) & \ldots & f_{2}^{(n-1)}(x) \\ f_{3}(x) & f_{3}'(x) & f_{3}''(x) & \ldots & f_{3}^{(n-1)}(x) \\ \multicolumn {5}{c}{\dotfill } \\ f_{n}(x) & f_{n}'(x) & f_{n}''(x) & \ldots & f_{n}^{(n-1)}(x) \end{vmatrix} = W[f_{1}(x), f_{2}(x), \ldots , f_{n}(x)]

the Wronskian determinant of the system of functions f1(x),f2(x),…,fn(x)f_{1}(x), f_{2}(x), \ldots , f_{n}(x).

If cλμc_{\lambda \mu } are constants, then we have

W(c11f1+c12f2+⋯+c1nfn,  c21f1+c22f2+⋯+c2nfn,  …,  cn1f1+cn2f2+⋯+cnnfn)=∣cλμ∣1n⋅W(f1,f2,…,fn). W(c_{11} f_{1} + c_{12} f_{2} + \cdots + c_{1n} f_{n}, \; c_{21} f_{1} + c_{22} f_{2} + \cdots + c_{2n} f_{n}, \; \ldots , \; c_{n1} f_{1} + c_{n2} f_{2} + \cdots + c_{nn} f_{n}) = \left|c_{\lambda \mu }\right|_{1}^{n} \cdot W(f_{1}, f_{2}, \ldots , f_{n}).
?
Problem 7.56
W[f1(φ(x)),f2(φ(x)),…,fn(φ(x))]=φ′(x)n(n−1)2W[f1(y),f2(y),…,fn(y)], W[f_{1}(\varphi (x)), f_{2}(\varphi (x)), \ldots , f_{n}(\varphi (x))] = \varphi '(x)^{\frac{n(n-1)}{2}} W[f_{1}(y), f_{2}(y), \ldots , f_{n}(y)],

setting y=φ(x)y = \varphi (x) on the right-hand side.

?
Problem 7.57
W(φf1,φf2,…,φfn)=φnW(f1,f2,…,fn). W(\varphi f_{1}, \varphi f_{2}, \ldots , \varphi f_{n}) = \varphi ^{n} W(f_{1}, f_{2}, \ldots , f_{n}).
?
Problem 7.58
1f1nW(f1,f2,…,fn)=W[(f2f1)′,(f3f1)′,…,(fnf1)′]. \frac{1}{f_{1}^{n}} W(f_{1}, f_{2}, \ldots , f_{n}) = W\left[ \left( \frac{f_{2}}{f_{1}} \right)', \left( \frac{f_{3}}{f_{1}} \right)', \ldots , \left( \frac{f_{n}}{f_{1}} \right)' \right].
?
Problem 7.59
ddxW(f1,…,fn−2,fn)W(f1,…,fn−2,fn−1)=W(f1,…,fn−2)W(f1,…,fn−2,fn−1,fn)[W(f1,…,fn−2,fn−1)]2. \frac{d}{dx} \frac{W(f_{1}, \ldots , f_{n-2}, f_{n})}{W(f_{1}, \ldots , f_{n-2}, f_{n-1})} = \frac{W(f_{1}, \ldots , f_{n-2}) W(f_{1}, \ldots , f_{n-2}, f_{n-1}, f_{n})}{[W(f_{1}, \ldots , f_{n-2}, f_{n-1})]^{2}}.
?
Problem 7.60

If W(f1,f2,…,fn−1,fn)W(f_{1}, f_{2}, \ldots , f_{n-1}, f_{n}) vanishes at every point and W(f1,f2,…,fn−1)W(f_{1}, f_{2}, \ldots , f_{n-1}) at no point of the interval a,ba, b, then there exist n−1n-1 constants c1,c2,…,cn−1c_{1}, c_{2}, \ldots , c_{n-1} such that in the whole interval a,ba, b

fn(x)=c1f1(x)+c2f2(x)+⋯+cn−1fn−1(x). f_{n}(x) = c_{1} f_{1}(x) + c_{2} f_{2}(x) + \cdots + c_{n-1} f_{n-1}(x).
?
Problem 7.61

If f1(x),f2(x),…,fn(x)f_{1}(x), f_{2}(x), \ldots , f_{n}(x) are nn linearly independent solutions of the homogeneous linear nnth order differential equation

y(n)+φ1(x)y(n−1)+φ2(x)y(n−2)+⋯+φn(x)y=0, y^{(n)} + \varphi _{1}(x) y^{(n-1)} + \varphi _{2}(x) y^{(n-2)} + \cdots + \varphi _{n}(x) y = 0,

then we have for an arbitrary function yy

y(n)+φ1(x)y(n−1)+φ2(x)y(n−2)+⋯+φn(x)y=W(f1,f2,…,fn,y)W(f1,f2,…,fn). y^{(n)} + \varphi _{1}(x) y^{(n-1)} + \varphi _{2}(x) y^{(n-2)} + \cdots + \varphi _{n}(x) y = \frac{W(f_{1}, f_{2}, \ldots , f_{n}, y)}{W(f_{1}, f_{2}, \ldots , f_{n})}.
?
Problem 7.62

(continued). Set

1=W0,f1=W1,W(f1,f2)=W2,…,W(f1,f2,…,fn)=Wn. 1 = W_{0}, \qquad f_{1} = W_{1}, \qquad W(f_{1}, f_{2}) = W_{2}, \qquad \ldots , \qquad W(f_{1}, f_{2}, \ldots , f_{n}) = W_{n}.

Then we have for an arbitrary function yy

y(n)+φ1(x)y(n−1)+φ2(x)y(n−2)+⋯+φn(x)y y^{(n)} + \varphi _{1}(x) y^{(n-1)} + \varphi _{2}(x) y^{(n-2)} + \cdots + \varphi _{n}(x) y ≡WnWn−1⋅ddxWn−12Wn−2Wn⋯ddxW22W1W3⋅ddxW12W0W2⋅ddxyW1. \equiv \frac{W_{n}}{W_{n-1}} \cdot \frac{d}{dx} \frac{W_{n-1}^{2}}{W_{n-2} W_{n}} \cdots \frac{d}{dx} \frac{W_{2}^{2}}{W_{1} W_{3}} \cdot \frac{d}{dx} \frac{W_{1}^{2}}{W_{0} W_{2}} \cdot \frac{d}{dx} \frac{y}{W_{1}}.

This representation of the linear differential expression, in terms of nn independent solutions of the homogeneous differential equation, is analogous to the factorization of a polynomial of degree nn in terms of its nn zeros.

?
Problem 7.63

Let f1(x),f2(x),…,fn(x)f_{1}(x), f_{2}(x), \ldots , f_{n}(x) be continuous real functions, defined in the interval a,ba, b. The determinant

∣∫abfλ(x)fμ(x) dx∣λ,μ=1,2,…,n \left|\int _{a}^{b} f_{\lambda }(x) f_{\mu }(x) \, dx\right|_{\lambda , \mu = 1, 2, \ldots , n}

is never negative. It vanishes if and only if between the functions f1(x),f2(x),…,fn(x)f_{1}(x), f_{2}(x), \ldots , f_{n}(x) there exists a homogeneous linear relation with constant coefficients, not all 00.

?
Problem 7.64

(continued). The determinant

\begin{vmatrix} \displaystyle \int _{a}^{b} (f_{2}^{2}+f_{3}^{2}+\cdots +f_{n}^{2}) \, dx & -\displaystyle \int _{a}^{b} f_{1} f_{2} \, dx & \ldots & -\displaystyle \int _{a}^{b} f_{1} f_{n} \, dx \\ ]2ex] -\displaystyle \int _{a}^{b} f_{2} f_{1} \, dx & \displaystyle \int _{a}^{b} (f_{1}^{2}+f_{3}^{2}+\cdots +f_{n}^{2}) \, dx & \ldots & -\displaystyle \int _{a}^{b} f_{2} f_{n} \, dx \\[2ex] \multicolumn {4}{c}{\dotfill } \\ ]1ex] -\displaystyle \int _{a}^{b} f_{n} f_{1} \, dx & -\displaystyle \int _{a}^{b} f_{n} f_{2} \, dx & \ldots & \displaystyle \int _{a}^{b} (f_{1}^{2}+f_{2}^{2}+\cdots +f_{n-1}^{2}) \, dx \end{vmatrix} \[ is never negative. It vanishes if and only if the functions $f_{1}(x), f_{2}(x), \ldots , f_{n}(x)$ differ from one another only by constant factors, or, in other words, if they are all multiples of one of them. \]
?
Problem 7.65

(continued). The determinant

∣e∫abfλ(x)fμ(x) dx∣λ,μ=1,2,…,n \left|e^{\int _{a}^{b} f_{\lambda }(x) f_{\mu }(x) \, dx}\right|_{\lambda , \mu = 1, 2, \ldots , n}

is never negative. It vanishes if and only if among the functions f1(x),f2(x),…,fn(x)f_{1}(x), f_{2}(x), \ldots , f_{n}(x) there are two identically equal.

?
Problem 7.66

A theorem mentioned in passing by Gauss in his Theoria combinationis observationum [Werke, Vol. 4, p. 12, cf. the last two lines of Art. 11] may be generalized as follows:

Let f(x)f(x) be defined for x>0x > 0, non-increasing there, f(x)≥0f(x) \geq 0, but not identically =0=0. Moreover let the real numbers a1,a2,…,ana_{1}, a_{2}, \ldots , a_{n} be distinct. Assuming the existence of all the integrals that occur, the determinant

∣(aλ+aμ+1)∫0∞xaλ+aμf(x) dx∣λ,μ=1,2,…,n \left|(a_{\lambda }+a_{\mu }+1) \int _{0}^{\infty } x^{a_{\lambda }+a_{\mu }} f(x) \, dx\right|_{\lambda , \mu = 1, 2, \ldots , n}

is never negative. It is =0=0 if and only if f(x)f(x) is a piecewise constant function with not more than n−1n-1 jump discontinuities.

?
Problem 7.67

Let the function f(ϑ)f(\vartheta ) be properly integrable in the interval 0≤ϑ<2π0 \leq \vartheta < 2\pi. Let its Fourier series (Section 4 of Part VI) be

f(ϑ)∼a0+2∑n=1∞(ancos⁡nϑ+bnsin⁡nϑ). f(\vartheta ) \sim a_{0} + 2 \sum _{n=1}^{\infty } (a_{n} \cos n\vartheta + b_{n} \sin n\vartheta ).

The Hermite forms

Tn(f)=∑λ=0n∑μ=0ncμ−λxλxˉμ T_{n}(f) = \sum _{\lambda =0}^{n} \sum _{\mu =0}^{n} c_{\mu -\lambda } x_{\lambda } \bar{x}_{\mu }

formed from the constants c0=a0c_{0} = a_{0}, cn=an+ibnc_{n} = a_{n} + ib_{n}, c−n=cˉnc_{-n} = \bar{c}_{n}, n=1,2,3,…n = 1, 2, 3, \ldots, are called the Toeplitz forms associated with f(ϑ)f(\vartheta ). (Cf. Problem 43).

Form the Toeplitz forms that belong to the following functions:

f(ϑ)=c=const.,f(ϑ)=a0+2(a1cos⁡ϑ+b1sin⁡ϑ), f(\vartheta ) = c = \text{const.}, \qquad f(\vartheta ) = a_{0} + 2(a_{1} \cos \vartheta + b_{1} \sin \vartheta ), f(ϑ)=1−r21−2rcos⁡ϑ+r2,0<r<1. f(\vartheta ) = \frac{1-r^{2}}{1-2r\cos \vartheta +r^{2}}, \qquad 0 < r < 1.
?
Problem 7.68

If the function f(ϑ)f(\vartheta ) is positive in the interval 0≤ϑ<2π0 \leq \vartheta < 2\pi, then all its associated Toeplitz forms are positive definite. More precisely: If f(ϑ)f(\vartheta ) lies between the bounds m≤f(ϑ)≤Mm \leq f(\vartheta ) \leq M in the interval 0≤ϑ<2π0 \leq \vartheta < 2\pi, then we have

m≤Tn(f)≤M, m \leq T_{n}(f) \leq M,

provided that the variables x0,x1,…,xnx_{0}, x_{1}, \ldots , x_{n} satisfy the condition Tn(1)=1T_{n}(1) = 1. The equality sign can hold here only if f(ϑ)=const.f(\vartheta ) = \text{const.} at every point of continuity of f(ϑ)f(\vartheta ).

?
Problem 7.69

Calculate the determinant Dn(f)D_{n}(f) of the form Tn(f)T_{n}(f) for the functions in Problem 67 and show that, if f(ϑ)f(\vartheta ) is everywhere positive,

lim⁡n→∞Dn(f)n+1=e12π∫02πlog⁡f(ϑ) dϑ=G(f). \lim _{n \rightarrow \infty } \sqrt[n+1]{D_{n}(f)} = e^{\frac{1}{2\pi } \int _{0}^{2\pi } \log f(\vartheta ) \, d\vartheta } = \mathfrak {G}(f).
?
Problem 7.70

Let f(ϑ)f(\vartheta ) be a trigonometric polynomial of first order and let hh be a parameter. The determinant Dn(f−h)D_{n}(f-h) of the Toeplitz form Tn(f−h)T_{n}(f-h) associated with f(ϑ)−hf(\vartheta )-h is a polynomial of degree n+1n+1 in hh with only real zeros. Calculate these zeros.

?
Problem 7.71

(continued). The zeros h0n,h1n,…,hnnh_{0n}, h_{1n}, \ldots , h_{nn} of Dn(f−h)D_{n}(f-h) all lie between the minimum and the maximum of f(ϑ)f(\vartheta ), that is m≤hνn≤Mm \leq h_{\nu n} \leq M, if m≤f(ϑ)≤Mm \leq f(\vartheta ) \leq M in the interval 0,2π0, 2\pi. Moreover if F(h)F(h) denotes an arbitrary properly integrable function in the interval m≤h≤Mm \leq h \leq M, then we have

lim⁡n→∞F(h0n)+F(h1n)+⋯+F(hnn)n+1=12π∫02πF[f(ϑ)] dϑ. \lim _{n \rightarrow \infty } \frac{F(h_{0n})+F(h_{1n})+\cdots +F(h_{nn})}{n+1} = \frac{1}{2\pi } \int _{0}^{2\pi } F[f(\vartheta )] \, d\vartheta .
?
Problem 7.72

Let the Hermitian form

∑λ=0n∑μ=0ncλ−μxλxˉμ,c−ν=cˉν, \sum _{\lambda =0}^{n} \sum _{\mu =0}^{n} c_{\lambda -\mu } x_{\lambda } \bar{x}_{\mu }, \qquad c_{-\nu } = \bar{c}_{\nu },

be positive definite. Then all the zeros of the polynomial

∣c0c1c2…cnc−1c0c1…cn−1c−2c−1c0…cn−2\multicolumn5c\dotfillc−n+1c−n+2c−n+3…c11zz2…zn∣ \begin{vmatrix} c_{0} & c_{1} & c_{2} & \ldots & c_{n} \\ c_{-1} & c_{0} & c_{1} & \ldots & c_{n-1} \\ c_{-2} & c_{-1} & c_{0} & \ldots & c_{n-2} \\ \multicolumn {5}{c}{\dotfill } \\ c_{-n+1} & c_{-n+2} & c_{-n+3} & \ldots & c_{1} \\ 1 & z & z^{2} & \ldots & z^{n} \end{vmatrix}

lie in the interior of the unit circle.

?