Miscellaneous Problems

Deduce from

that $\cos z$ and $\sin z$ have no non-real zeros.

Deduce from

that no derivative of $e^{-z^2}$ has non-real zeros.

The Bessel function of order $0$

has no non-real zeros. One can obtain from the four different representations

1. $J_0(z) = \lim \limits_{n \to \infty } (-1)^n\left(1+\dfrac {z^2}{4n^2}\right)^n P_n\left(\dfrac {z^2-4n^2}{z^2+4n^2}\right)$ (problem 85 of Part Six),

2. $J_0(z) = \lim \limits_{n \to \infty } L_n\left(\dfrac {z^2}{4n}\right)$ (problem 99(b) of Part Six’s solution),

3. $J_0(z) = \dfrac {1}{2\pi }\displaystyle \int_{-\pi }^{\pi } e^{iz\sin \vartheta }\, d\vartheta$ (problems 148, 56 of Part Three),

4. $J_0(z) = \dfrac {2}{\pi }\displaystyle \int_0^1 \dfrac {\cos zt}{\sqrt{1-t^2}}\, dt$ (problem 205 of Part Three) four different proofs of this result.

Let $q$ be an integer, $q \geq 2$. The entire function

has no non-real zeros. One can obtain from 59 two different proofs of this result.

Let $\alpha \geq 0$, $0<\alpha_1 \leq \alpha_2 \leq \alpha_3 \leq \cdots$, and $1/\alpha_1+1/\alpha_2+1/\alpha_3+\cdots$ be convergent. The transcendental entire function

may be represented as the limit of a sequence of polynomials with only real positive zeros.

If

then the polynomials

have only real, positive zeros [63].

In the interval $0<x \leq \alpha_1$

and furthermore we have there more generally

[Thus $g(x)$ is enveloped by its Maclaurin series for $0<x \leq \alpha_1$; 55, 72.]

Assume $\alpha <1$. The entire function of $z$ defined by the integral

has only real zeros [63].

Let $\alpha , \beta , \beta_1, \beta_2, \beta_3, \ldots$ be real, $\alpha \geq 0$, $\beta_{\nu } \neq 0$, and the series $1/\beta_1^2+1/\beta_2^2+1/\beta_3^2+\cdots$ convergent. The transcendental entire function

may be represented as the limit of a sequence of polynomials with only real zeros.

If

then

If $G(z)$ has no positive zeros and $a_0, a_1, a_2, \ldots , a_n$ are real numbers, then the polynomial

can have not more non-real zeros than the polynomial

[68, 69 are special cases.]

Prove from 167 that the Bessel function $J_0(z)$ has no non-real zeros (problem 31 of Part Two).

Prove from 167 that the entire functions considered in 160 have no non-real zeros.

Let $\alpha$ be an even integer, $\alpha \geq 2$. Then the integral

represents an entire function that has no non-real zeros. [167.]

If we have only $\alpha >1$, then the integral still represents an entire function. If $\alpha$ is not an even integer, this function has only finitely many real zeros.

Let $\alpha_1, \alpha_2, \ldots , \alpha_s$ denote the zeros of $G(z)$ in $[0,m]$,

and set

Then argue as follows:

(1)

has precisely $m$ changes of sign (by 47 and the definition of $G^*(z)$) and precisely $m$ positive zeros (by a slight modification of 167---make use of 67 more completely).

(2)

has precisely $m-1$ changes of sign (attention to the various possible cases) and so by 36 at most $m-1$ positive zeros. Yet it has (by a slight modification of solution 66) at least $m-1$, and so finally exactly $m-1$, positive zeros;

(3) Introduce successively $\alpha_2, \alpha_3, \ldots , \alpha_s$ and repeat appropriately the arguments of (2) (attention at each step to the changes wrought by the preceding steps) to arrive at the desired proposition.

Let $\alpha_1, \alpha_2, \ldots , \alpha_s$ denote the zeros of $G(z)$ in $[0,m]$,

and set

Then argue as follows:

(1)

has precisely $m$ changes of sign (by 47 and the definition of $G^*(z)$) and precisely $m$ positive zeros (by a slight modification of 167---make use of 67 more completely).

(2)

has precisely $m-1$ changes of sign (attention to the various possible cases) and so by 36 at most $m-1$ positive zeros. Yet it has (by a slight modification of solution 66) at least $m-1$, and so finally exactly $m-1$, positive zeros;

(3) Introduce successively $\alpha_2, \alpha_3, \ldots , \alpha_s$ and repeat appropriately the arguments of (2) (attention at each step to the changes wrought by the preceding steps) to arrive at the desired proposition.

We make the following assumptions: The entire function

is a limit of polynomials with only positive zeros, but not itself a polynomial [161].

The entire function $G(z)$ is either defined as in 165 or it is equal to the function defined in 165 multiplied by $z$; it has precisely $s$ non-negative zeros which are simple and the distance between any two of them is not less than $1$. (We obtain the present $G(z)$ from the $G(z)$ of 171.1 by making $m$ arbitrarily large.)

Then the entire function

has precisely $s$ non-positive zeros.

We make the following assumptions: The entire function

is a limit of polynomials with only positive zeros, but not itself a polynomial [161].

The entire function $G(z)$ is either defined as in 165 or it is equal to the function defined in 165 multiplied by $z$; it has precisely $s$ non-negative zeros which are simple and the distance between any two of them is not less than $1$. (We obtain the present $G(z)$ from the $G(z)$ of 171.1 by making $m$ arbitrarily large.)

Then the entire function

has precisely $s$ non-positive zeros.

If $\nu \geq 0$ the entire function

has precisely $[\nu ]$ non-positive zeros. (This contains the result of 159.)

If $\nu \geq 0$ the entire function

has precisely $[\nu ]$ non-positive zeros. (This contains the result of 159.)

Let $k, l, m$ and $q$ be non-negative integers,

If all the zeros of the polynomial

are negative, then the zeros of

are also all negative. (This implies, for $n \to \infty$, the result of 160.)

Let $k, l, m$ and $q$ be non-negative integers,

If all the zeros of the polynomial

are negative, then the zeros of

are also all negative. (This implies, for $n \to \infty$, the result of 160.)

If all the zeros of the polynomial

are negative, then the zeros of

are also all negative; $k, l, m$ and $q$ are subject to the same conditions as in 171.4.

If all the zeros of the polynomial

are negative, then the zeros of

are also all negative; $k, l, m$ and $q$ are subject to the same conditions as in 171.4.

The equation

has only real roots. [26.]

Let $f(t)$ be twice continuously differentiable, $f(t)>0$, $f'(t)<0$, $f''(t)<0$ for $0 \leq t \leq 1$. Then the even entire function

has infinitely many, and in fact only real, zeros. (Related to but different from problem 205 of Part Three.) [26, problem 165 of Part Three.]

Let $\varphi (t)$ be properly integrable for $0 \leq t \leq 1$. If we have

then the entire function

has only real zeros. [27.]

Let $f(t)$ be real and continuously differentiable for $0 \leq t \leq 1$. If we have

then the entire function

has only real zeros. (The condition is satisfied in particular if $f(0) \geq 0$, $f'(t) \geq 0$ for $0 \leq t \leq 1$; cf. problem 205 of Part Three.)

Let $a \geq 2$. The entire function

as well as all the partial sums of its power series, has only real negative simple zeros. (Problem 200 of Part Three.)

Let the function $f(t)$ be continuously differentiable and positive for $0<t<1$, and also let $\int_0^1 f(t)\, dt$ exist. The entire function defined by the integral

has no zeros

[Not analogously to problem 205 of Part Three through a limiting process from problem 22 of Part Three, but by appropriate translation of the proof of problem 22 of Part Three.]

Deduce problem 189 of Part Three from 177.

The remainder of the exponential series

vanishes for $z=0$, but otherwise at no point of the half-plane $\mathrm{Re}\left(z\right) \leq n$, if $n=1,2,3,\ldots$; $n=0$ is an exception: Here the zeros all lie on the boundary of the half-plane $\mathrm{Re}\left(z\right) \leq 0$. (This contains 73.) [177.]

If $\sum_{n=0}^{\infty } \left|\dfrac {a_{n+1}}{a_n}\right|^2$ is convergent, then

is an entire function. If we denote the zeros of $F(z)$ by $z_1, z_2, \ldots , z_n, \ldots$, then

is convergent.

If the zeros of a polynomial with real coefficients are all real and simple, then in the interval between two successive zeros lies only one zero of the derivative. Is this theorem also true for transcendental entire functions?

Let the degree of the polynomial $H(x)$ be $\geq 3$. If

then at least one of the two functions $dF/dx$ and $d^2F/dx^2$ has not only real zeros.

Along with the polynomial

of degree $m$ with real coefficients $a_0, a_1, \ldots , a_m$, $a_m \geq 0$, we consider the following three associated polynomials:

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

If we set

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

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

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

Denote by

the number of zeros of the polynomials

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

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

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

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

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

respectively.

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

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

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

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

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

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

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

The polynomial

with real coefficients $a_0, a_1, \ldots , a_m$ has not more zeros in the interval $0<x<1$ than the polynomial

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

The polynomial

with real coefficients $a_0, a_1, \ldots , a_m$ has not more zeros in the interval $0<x<1$ than the polynomial

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

The entire function

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

The entire function

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

Let $F(x_1, x_2, \ldots , x_m)$ denote a homogeneous polynomial with real coefficients in the variables $x_1, x_2, \ldots , x_m$ and assume that

where

Then the coefficients of the polynomial

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

Let $F(x_1, x_2, \ldots , x_m)$ denote a homogeneous polynomial with real coefficients in the variables $x_1, x_2, \ldots , x_m$ and assume that

where

Then the coefficients of the polynomial

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

We can represent any given polynomial as a quotient of two polynomials such that the denominator has no changes of sign and the number of changes of sign of the numerator is equal to the number of positive zeros of the given polynomial.

Let the polynomial $f(x)$ of degree $m$ be such that, if $P(x)$ is an arbitrary polynomial, $P(x)f(x)$ has at least $m$ more changes of sign than $P(x)$. For this it is necessary and sufficient that $f(x)$ has only real positive zeros.

Let the polynomial $f(x)$ be such that, if $P(x)$ is an arbitrary polynomial, $P(x)f(x)+P'(x)$ has not more non-real zeros than $P(x)$.

For this it is necessary and sufficient, that $f(x)=a-bx$, where $a$ is arbitrary, $b \geq 0$.

If the polynomial $f(x)$ has the property that for all positive $p$ the equation $f(x)+p=0$ has only real roots, then $f(x)$ is at most of the second degree. If the roots of the same equation are all real and all of the same sign, then $f(x)$ is of the first degree.

This is evident geometrically; we require a proof based upon other principles.

The expression

formed from the real constants $a_0, a_1, a_2, b_0, b_1, b_2$, is negative if and only if the two polynomials

have real and simple zeros that separate each other, i.e. that lie in such a manner that the two zeros of one polynomial contain between them one and only one zero of the other polynomial.

Let $P(x)$ be a polynomial with only real zeros. Then in general there exists no primitive function of $P(x)$ that has only real zeros. More precisely: If $n$ denotes the precise degree of $P(x)$, determine a real number $a$ such that the polynomial of degree $n+1$

has a maximum number of real zeros. Then we can assert that this maximum number is

and we cannot assert more than this, for the equality sign can actually hold for given $n$ for a suitable $P(x)$.