An Elementary Proof of Takagi’s Theorem on the Differential Composition of Polynomials

Abstract

I give a short and completely elementary proof of Takagi’s 1921 theorem on the zeros of a composite polynomial $f(d/dz)g(z)$.

MSC:

• Primary 30C15
• Primary 30C10

Many theorems in the analytic theory of polynomials [2, 8, 10, 11] are concerned with locating the zeros of composite polynomials. More specifically, let f and g be polynomials (with complex coefficients) and let h be a polynomial formed in some way from f and g; under the assumption that the zeros of f (respectively, g) lie in a subset S (respectively, T) of the complex plane, we wish to deduce that the zeros of h lie in some subset U. The theorems are distinguished by the nature of the operation defining h, and the nature of the subsets S, T, U under consideration.

Here we shall be concerned with differential composition: $h(z)=f(d/dz)g(z)$, or $h=f(D)g$ for short. In detail, if $f(z)={\sum }_{i=1}^m{a}_i{z}^i$ and $g(z)={\sum }_{j=1}^n{b}_j{z}^j$, then $h(z)={\sum }_{i=1}^m{a}_i{g}^{(i)}(z)$; and D denotes the differentiation operator, i.e., $Dg=g\prime$. The following important result was found by Takagi [13] in 1921, subsuming many earlier results:¹

Theorem 1

(Takagi). Let f and g be polynomials with complex coefficients, with $\mathrm{deg}f=m$ and $\mathrm{deg}g=n$. Let f have an r-fold zero at the origin ( $0\le r\le m$), and let the remaining zeros (with multiplicity) be ${\alpha }_1,\dots ,{\alpha }_{m-r}\ne 0$. Let K be the convex hull of the zeros of g. Then either $f(D)g$ is identically zero, or its zeros lie in the set $K+{\sum }_{i=1}^{m-r}[0,n-r]{\alpha }_i^{-1}$.

Here we have used the notations $A+B=\{a+b:a\in A\mathrm{}\text{and}\mathrm{}b\in B\}$ and $AB=\{ab:a\in A\mathrm{}\text{and}\mathrm{}b\in B\}$.

Takagi’s proof was based on Grace’s apolarity theorem [3], a fundamental but somewhat enigmatic result in the analytic theory of polynomials.² This proof is also given in the books of Marden [8, Section 18], Obrechkoff [10, pp. 135–136], and Rahman and Schmeisser [11, Sections 5.3 and 5.4]. Here I give a short and completely elementary proof of Takagi’s theorem.

The key step—as Takagi [13] observed—is to understand the case of a degree-1 polynomial $f(z)=z-\alpha$:

Proposition 2

(Takagi). Let g be a polynomial of degree n, and let K be the convex hull of the zeros of g. Let $\alpha \in ℂ$, and define $h=g\prime -\alpha g$. Then either h is identically zero, or all the zeros of h are contained in K if α = 0, and in $K+[0,n]{\alpha }^{-1}$ if $\alpha \ne 0$.

The case α = 0 is the celebrated theorem of Gauss and Lucas [8, Section 6], [10, Chapter V], and [11, Section 2.1], which is the starting point of the modern analytic theory of polynomials. My proof for general α will be modeled on Cesàro’s [1] 1885 proof of the Gauss–Lucas theorem [11, pp. 72–73], with a slight twist to handle the case $\alpha \ne 0$.

Proof of Proposition 2.

Clearly, h is identically zero if and only if either (a) $g\equiv 0$ or (b) g is a nonzero constant and α = 0. Moreover, if g is a nonzero constant and $\alpha \ne 0$, then the zero set of h is empty. So we can assume that $n\ge 1$.

Let ${\beta }_1,\dots ,{\beta }_n$ be the zeros of g (with multiplicity), so that $g(z)=b_n{\prod }_{i=1}^n(z-{\beta }_i)$ with $b_n\ne 0$. If $z\notin K$, then $g(z)\ne 0$, and we can consider$$\frac{h(z)}{g(z)}=\frac{g\prime (z)-\alpha g(z)}{g(z)}=\sum _{i=1}^n\frac{1}{z-{\beta }_i}-\alpha .$$

If this equals zero, then by taking complex conjugates we obtain$$0=\sum _{i=1}^n\frac{1}{\overline{z}-{\overline{\beta }}_i}-\overline{\alpha }=\sum _{i=1}^n\frac{z-{\beta }_i}{|z-{\beta }_i{|}^2}-\overline{\alpha },$$which can be rewritten as $z={\sum }_{i=1}^n{\lambda }_i{\beta }_i+\kappa \overline{\alpha }$ where$${\lambda }_i=\frac{|z-{\beta }_i{|}^{-2}}{\sum _{j=1}^n|z-{\beta }_j{|}^{-2}},\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\kappa =\frac{1}{\sum _{j=1}^n|z-{\beta }_j{|}^{-2}}.$$

Then ${\lambda }_i>0$ and ${\sum }_{i=1}^n{\lambda }_i=1$, so ${\sum }_{i=1}^n{\lambda }_i{\beta }_i\in K$; and of course $\kappa >0$. Moreover, by the Schwarz inequality we have$$|\alpha {|}^2={\left|\sum _{i=1}^n\frac{1}{z-{\beta }_i}\right|}^2\le n\sum _{i=1}^n|z-{\beta }_i{|}^{-2}=\frac{n}{\kappa },$$

so $\kappa \le n|\alpha {|}^{-2}$. This implies that $\kappa \overline{\alpha }\in [0,n]{\alpha }^{-1}$ and hence that $z\in K+[0,n]{\alpha }^{-1}$. ■

We can now handle polynomials f of arbitrary degree by iterating Proposition 2:

Proof of Theorem 1.

From $f(z)=a_m\left({\prod }_{i=1}^{m-r}(z-{\alpha }_i)\mathrm{}\right)z^r$ it is easy to see that $f(D)=a_m\left({\prod }_{i=1}^{m-r}(D-{\alpha }_i)\mathrm{}\right)D^r$. We first apply D^r to g, yielding a polynomial of degree n – r whose zeros also lie in K (by the Gauss–Lucas theorem); then we repeatedly apply (in any order) the factors $D-{\alpha }_i$, using Proposition 2. ■

Remark.

When α = 0, the zeros of $h=g\prime$ lie in K; so one might expect that when α is small, the zeros of $h=g\prime -\alpha g$ should lie near K. But when α is small and nonzero, the set $K+[0,n]{\alpha }^{-1}$ arising in Proposition 2 is in fact very large. What is going on here?

Here is the answer: Suppose that $\mathrm{deg}g=n$. When α = 0, the polynomial $h=g\prime$ has degree n – 1; but when $\alpha \ne 0$, the polynomial $h=g\prime -\alpha g$ has degree n. So, in order to make a proper comparison of their zeros, we should consider the polynomial $g\prime$ corresponding to the case α = 0 as also having a zero “at infinity.” This zero then moves to a value of order ${\alpha }^{-1}$ when α is small and nonzero.

This behavior is easily seen by considering the example of a quadratic polynomial $g(z)=z^2-{\beta }^2$. Then the zeros of $g\prime -\alpha g$ are$$\begin{array}{c}z=\frac{1\pm \sqrt{1+{\alpha }^2{\beta }^2}}{\alpha }\\ =-\frac{{\beta }^2}{2}\alpha +O({\alpha }^3),\hspace{1em}2{\alpha }^{-1}+O(\alpha ).\end{array}$$

So there really is a zero of order ${\alpha }^{-1}$, as Takagi’s theorem recognizes.

In the context of Proposition 2, one expects that $g\prime -\alpha g$ has one zero of order ${\alpha }^{-1}$ and n – 1 zeros near K (within a distance of order α). More generally, in the context of Theorem 1, one would expect that h has m – r zeros of order ${\alpha }^{-1}$, with the remaining zeros near K. It is a very interesting problem — and one that is open, as far as I know — to find strengthenings of Takagi’s theorem that exhibit these properties. There is an old result that goes in this direction [8, Corollary 18.1], [11, Corollary 5.4.1(ii)], but it is based on a disc D containing the zeros of g, which might in general be much larger than the convex hull K of the zeros.

Postscript. A few days after finding this proof of Proposition 2, I discovered that an essentially identical argument is buried in a 1961 paper of Shisha and Walsh [12, pp. 127–128 and 147–148] on the zeros of infrapolynomials. I was led to the Shisha–Walsh paper by a brief citation in Marden’s book [8, pp. 87–88, Exercise 11]. So the proof given here is not new; but it deserves to be better known.

Acknowledgment

This research was supported in part by U.K. Engineering and Physical Sciences Research Council grant EP/N025636/1.

Additional information

Funding

This research was supported in part by U.K. Engineering and Physical Sciences Research Council grant EP/N025636/1.

Notes

1 See Honda [4], Iyanaga [5, 6], Kaplan [7], and Miyake [9] for biographies of Teiji Takagi (inlinefigure, Takagi Teiji, 1875–1960). Takagi’s papers published in languages other than Japanese (namely, English, German, and French) have been collected in [14].

2 For discussion of Grace’s apolarity theorem and its equivalents—notably Walsh’s coincidence theorem and the Schur–Szegő composition theorem—see Marden [8, Chapter IV], Obrechkoff [10, Chapter VII], and especially Rahman and Schmeisser [11, Chapter 3].