Operators preserving real zeros (Laguerre-Pólya theory)

Let $p$ be a nonconstant polynomial with real coefficients and only real roots. Prove that for each real number $r$, the polynomial $p - rp'$ has only real roots.

All the roots of the polynomial $a_{0}x^{n}+a_{1}x^{n-1}+\ldots +a_{n}$ are real. Prove that all the roots of the polynomial

are real as well.

All the roots of the polynomial $a_{0}x^{n}+a_{1}x^{n-1}+\ldots +a_{n}$ are real. Prove that the roots of the equation

are real as well.

All the roots of the polynomial $f(x)$ are real. Prove that the roots of the polynomial $f(x)+\lambda f'(x)$ are also real, if $\lambda$ is real.

Under the same condition (see problem 2539), suppose the polynomial $f(x)+m$ has $k$ imaginary¹ roots. Prove that the equation $f'(x)^{2}-f(x)f''(x)-mf'''(x)=0$ has at least $k$ real roots.

Prove the theorem: if all the roots of the polynomial $f(x)$ are real, then for $\lambda > 0$ all the roots of the polynomial $xf'(x)+\lambda f(x)$ are real as well. (Laguerre.)

Prove the theorem: if all the roots of the polynomial $f(x)=a+a_{1}x+a_{2}x^{2}+\ldots +a_{n}x^{n}$ are real, and all the roots of the polynomial $\varphi (\nu )$ are real and nonpositive, then all the roots of the polynomial $F(x)=\sum a_{\nu }\varphi (\nu )x^{\nu }$ are real as well. (Laguerre.)

Prove the theorem: if all the roots of the polynomial $f(x)=a+a_{1}x+a_{2}x^{2}+\ldots +a_{n}x^{n}$ are real, then all the roots of the polynomial $F(x)=\sum a_{\nu }\psi (\nu )x^{\nu }$ are real as well, where $\psi (\nu )=e^{A\nu }\prod \left(1+\dfrac {\nu }{a_{\nu }}\right)e^{-\frac{\nu }{a_{\nu }}}$, the numbers $a_{\nu }$ are positive, and $A$ is real. (Laguerre.)

If the function $\Phi (x)=f(x)+af'(x)+a^{2}f''(x)+\ldots +a^{n}f^{(n)}(x)$, where $f(x)$ is a polynomial of degree $n$, has real roots, then so does the polynomial $f(x)$.

We assume that all the zeros of the polynomial

are in the upper half-plane $\mathrm{Im}\left(z\right) > 0$. Let $a_{\nu } = \alpha_{\nu } + i\beta_{\nu }$, $\alpha_{\nu }, \beta_{\nu }$ real, $\nu = 0, 1, 2, \ldots , n$, and

Then the polynomials $U(z)$ and $V(z)$ have only real zeros.

If a polynomial $f(z)$ whose coefficients are all real has only real zeros then this is true also for its derivative $f'(z)$. If $f(z)$ has complex zeros then they appear in pairs, the two zeros forming a pair are mirror images to each other with respect to the real axis; they are complex conjugates. We draw all those disks the ``vertical'' diameters of which are the line segments connecting the conjugate zeros of such pairs. If $f'(z)$ has any complex zeros they lie in these disks.

The derivative of a polynomial has not more non-real zeros than the polynomial itself.

Let $\mu , \nu$ be integers satisfying $0 \leq \mu < \mu +\nu \leq n$. None of the polynomials

has more nonreal zeros than the polynomial

(Generalization of 53.)

Let $\alpha$ be real and $P(x)$ a polynomial. The polynomial $\alpha P(x)+P'(x)$ has not more non-real zeros than the polynomial $P(x)$ itself. (Generalization of 53.)

Assume that the equation

has only real roots. Then the polynomial

has not more non-real zeros than the polynomial $P(x)$ itself.

has not more non-real zeros than the polynomial $P(x)$ itself.

If the equation

has only real roots, then the following equation

has also only real roots.

Let $P(x)$ be a polynomial of $n$th degree, and let the real number $\alpha$ lie outside the interval $[-n,0]$. Then $\alpha P(x)+xP'(x)$ has not more non-real zeros than $P(x)$. (A generalization of 53 different from 60 and 62.)

Let $Q(x)$ be a polynomial whose zeros are all real and lie outside the interval $[0,n]$. Then the equation

has not more non-real roots than the equation

Let $0<q<1$. The equation

has not more non-real roots than the equation

Let $q > 0$. The equation

has not more non-real roots than the equation

Let $0<a<b$. If $f(x)$ vanishes at $n+1$ points of the interval $[a,b]$ and if all the zeros of the polynomial with real coefficients $a_0+a_1x+a_2x^2+\cdots +a_nx^n$ are real, then at an interior point $\xi$ of $[a,b]$

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.]

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 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.

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, $b \geq 0$.

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.

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.)

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.

Consider the polynomials $P(z), Q(z), R(z)$ and

Assume that

Then

has no more non-real zeros than $P(z)$,

has only real zeros, and

has only zeros whose absolute value equals $1$.

Assume further that $a_0 \neq 0$ and consider the power series

Then the polynomial

has not less non-real zeros than the polynomial $P(x)$ itself.

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

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

(If $G(z)$ 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 $\neq 0$ [Problem 27].)

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

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

Let $a$ be an arbitrary real number. Prove that if the polynomial $P(z) = z^{n}+a_{1} z^{n-1}+\ldots +a_{n}$ has $n$ real roots, then the polynomial $Q(z) = P(z+i a)+P(z-i a)$ also has $n$ real roots.