E. I. Krupitskii

ON A CLASS OF POLYNOMIALS LEAST DEVIATING FROM ZERO ON TWO INTERVALS

(Presented by Academician V. I. Smirnov, January 7, 1961)

1. The aim of the present note is to find and investigate polynomials with fixed leading coefficient that deviate least from zero on the two intervals \([-1,-\lambda]\) and \([\lambda,1]\) with weight \(q(x)=\sqrt{1-x^2}\). Polynomials of this kind, but for the constant weight \(q(x)=1\), were constructed and studied by N. I. Akhiezer \((^{1,2})\) and are a generalization to the case of two intervals of the Chebyshev polynomials of the first kind \(T_n(x)\). In contrast to them, the polynomials considered below are the corresponding generalization of the Chebyshev polynomials of the second kind \(U_n(x)\). Following the indicated analogy, we shall call them Akhiezer polynomials of the second kind and denote them by \(B_n(x,\lambda)\). Here the polynomials \(B_n(x,\lambda)\) will be defined as polynomials least deviating from zero on the two intervals \([-1,-\lambda]\) and \([\lambda,1]\) with weight \(\sqrt{1-x^2}\), under the condition that

\[ \max_{x\in[\pm\lambda,\pm1]}\left|\sqrt{1-x^2}B_n(x,\lambda)\right|=1. \]

Passing to the construction of the polynomials \(B_n(x,\lambda)\), we note that the cases of even and odd \(n\) differ essentially.

I. The case of even \(n=2m\). In this case the required polynomial is constructed in an elementary way. Indeed, consider the function

\[ y_{2m}(u)=\sqrt{1-u^2}\,U_{2m}(u),\quad -1\leq u\leq +1, \tag{1} \]

where \(U_{2m}(u)\) is the Chebyshev polynomial of the second kind of degree \(2m\), normalized in such a way that

\[ \max_{u\in[-1,1]}\left|\sqrt{1-u^2}\,U_{2m}(u)\right|=1. \]

The function \(y_{2m}(u)\) is extremal on the segment \([-1,1]\) in the class of functions of the form \(\sqrt{1-u^2}\,P_{2m}(u)\), where \(P_{2m}(u)\) is a polynomial of degree \(2m\).

Using the transformation

\[ u=\sqrt{\frac{x^2-\lambda^2}{1-\lambda^2}}, \tag{2} \]

we obtain

\[ B_{2m}(x,\lambda)=\frac{1}{\sqrt{1-\lambda^2}}\,U_{2m}\left(\sqrt{\frac{x^2-\lambda^2}{1-\lambda^2}}\right). \tag{3} \]

For the limiting values \(\lambda=0\) and \(\lambda=1\) we have

\[ B_{2m}(x,0)=U_{2m}(x); \tag{4} \]

\[ B_{2m}(x,1)=(1-x^2)^m. \tag{5} \]

By means of polynomial (3) the following is solved:

Problem 1. Among all polynomials of degree \(2m\) with leading coefficient equal to unity, find the one that deviates least from zero on the two intervals \([-1,-\lambda]\) and \([\lambda,1]\) with weight \(\sqrt{1-x^2}\).

The required polynomial will be

\[ \dot B_{2m}(x,\lambda)=\frac{L_{2m}}{\sqrt{1-\lambda^2}}\, U_{2m}\left(\sqrt{\frac{x^2-\lambda^2}{1-\lambda^2}}\right), \tag{6} \]

where the corresponding minimal deviation \(L_{2m}\) is determined by the expression

\[ L_{2m}=\frac{(1-\lambda^2)^{m+1/2}}{2^{2m}}. \tag{7} \]

II. The case of odd \(n=2m-1\). Let us first set up the differential equation for the function

\[ y_{2m-1}(x)=\sqrt{1-x^2}\,B_{2m-1}(x,\lambda). \tag{8} \]

In accordance with the general theory, the function \(y_{2m-1}(x)\) must, on each of the segments \([-1,-\lambda]\) and \([\lambda,1]\), attain \(m\) times its maximum value, equal to unity, successively changing sign. Consequently, the polynomial

\[ Q_{4m}(x)=[y_{2m-1}(x)]^2-\frac{1}{2} \tag{9} \]

must satisfy the equation

\[ \frac{1/4-[Q_{4m}(x)]^2}{(1-x^2)(x^2-\gamma^2)(x^2-\lambda^2)} = \frac{[Q'_{4m}(x)]^2}{(4m)^2(x^2-\delta^2)^2}, \qquad 0<\gamma\leq \delta\leq \lambda<1, \tag{10} \]

since the function \(1/4-[Q_{4m}(x)]^2\) has simple zeros at the points \(x=\pm\gamma\), \(x=\pm\lambda\), \(x=\pm1\), and also double zeros at \(4m-3\) points of the interval \([-1,1]\), where \(Q'_{4m}(x)\) also has simple zeros. Moreover, \(Q_{4m}(x)\) has two more simple zeros at the points \(x=\pm\delta\), where the function \(Q_{4m}(x)\) attains an absolute maximum on the segment \([-1,1]\).

As a result we obtain the equation

\[ \frac{1-[y_{2m-1}(x)]^2}{(1-x^2)(x^2-\gamma^2)(x^2-\lambda^2)} = \frac{[y'_{2m-1}(x)]^2}{(2m)^2(x^2-\delta^2)^2}, \tag{11} \]

whose solution we obtain in the form

\[ y_{2m-1}(x)=\sin[2m\,\varphi(x)]; \tag{12} \]

\[ \varphi(x)=\int_{-1}^{x} \frac{(\delta^2-t^2)\,dt} {\sqrt{(1-t^2)(\gamma^2-t^2)(\lambda^2-t^2)}} , \tag{13} \]

where the parameters \(\gamma\) and \(\delta\) are determined by means of the conditions \(y_{2m-1}(0)=0\), \(y_{2m-1}(\pm\lambda)=\pm1\), and \(y_{2m-1}(\pm\gamma)=\pm1\). In practice it is more convenient to pass to an expression for \(y_{2m-1}(x)\) in parametric form through elliptic functions. Following the method of N. I. Akhiezer \((^1)\), putting

\[ x^2=\frac{\operatorname{sn}^2(K/2m)\operatorname{cn}^2 u} {\operatorname{sn}^2(K/2m)-\operatorname{sn}^2 u}; \qquad \operatorname{sn}\frac{K}{2m}=\lambda; \qquad \lambda\sqrt{\frac{1-k^2}{1-k^2\lambda^2}}=\gamma \tag{14} \]

and taking into account that the function \(y_{2m-1}(x)\) has only two poles of multiplicity \(2m-1\), corresponding to \(x=\pm\infty\), we find

\[ \varphi(u)=\frac{i}{2}\ln\frac{H(K/2m+u)}{H(K/2m-u)}; \tag{15} \]

\[ \delta=\left[\lambda \sqrt{\frac{1-\lambda^{2}}{1-k^{2}\lambda^{2}}}\,\frac{H_{1}'(K/2m)}{H_{1}(K/2m)}\right]^{1/2}, \tag{16} \]

where \(H(u)\) and \(H_{1}(u)\) are theta-functions.

Finally we have

\[ y_{2m-1}(x)=\frac{i}{2}\left\{\left[\frac{H(K/2m+u)}{H(K/2m-u)}\right]^{m} -\left[\frac{H(K/2m-u)}{H(K/2m+u)}\right]^{m}\right\}, \tag{17} \]

\[ B_{2m-1}(x,\lambda)=\frac{i}{2}\sqrt{\frac{\operatorname{sn}^{2}u-\lambda^{2}} {\operatorname{sn}^{2}u-\lambda^{2}\operatorname{sn}^{2}u}} \left\{\left[\frac{H(K/2m+u)}{H(K/2m-u)}\right]^{m} -\left[\frac{H(K/2m-1)}{H(K/2m+u)}\right]^{m}\right\}; \tag{18} \]

\[ \operatorname{sn}u=\lambda\sqrt{\frac{1-x^{2}}{\lambda^{2}-x^{2}}}; \qquad \operatorname{sn}\frac{K}{2m}=\lambda. \tag{19} \]

It should be noted that the function (17) is a solution of equation (11), which is also satisfied by the odd Akhiezer polynomials of the first kind \(A_{2m-1}(x,\lambda)\) \((^{3})\).

For \(0<k<1\), formulas (17) and (18) determine the extremal functions for \(1>\lambda>\sin(\pi/4m)\). For \(0\leq \lambda \leq \sin(\pi/4m)\) we obtain

\[ y_{2m-1}(x)=\sin(2m\arccos x)=\sqrt{1-x^{2}}\,U_{2m-1}(x); \qquad B_{2m-1}(x,\lambda)=U_{2m-1}(x), \tag{20} \]

and for \(\lambda=1\)

\[ B_{2m-1}(x,1)=x(1-x^{2})^{m-1}. \tag{21} \]

With the aid of the polynomials \(B_{2m-1}(x,\lambda)\) the following is solved:

Problem 2. Among all polynomials of degree \(2m-1\) with leading coefficient equal to one, find the one that deviates least from zero on the two intervals \([-1,-\lambda]\) and \([\lambda,1]\) with weight \(\sqrt{1-x^{2}}\).

Taking (18) into account, the required polynomial will be

\[ \dot{B}_{2m-1}(x,\lambda)=\frac{iL_{2m-1}}{2} \sqrt{\frac{\operatorname{sn}^{2}u-\lambda^{2}} {\operatorname{sn}^{2}u-\lambda^{2}\operatorname{sn}^{2}u}} \left\{\left[\frac{H(K/2m+u)}{H(K/2m-u)}\right]^{m} -\right. \]

\[ \left. -\left[\frac{H(K/2m-u)}{H(K/2m+u)}\right]^{m}\right\}, \tag{22} \]

where the minimal deviation \(L_{2m-1}\) is determined by the relation

\[ L_{2m-1}=\frac{1}{2^{2m-1}} \left[\frac{\theta(0)\,\theta_{1}(0)} {\theta(K/2m)\,\theta_{1}(K/2m)}\right]^{2m}; \tag{23} \]

here \(\theta(u)\) and \(\theta_{1}(u)\) are the corresponding theta-functions.

1. By using the polynomials \(B_{2m-1}(x,\lambda)\) constructed above, the following is easily solved:

Problem 3. Among all trigonometric polynomials of the form

\[ P_{m}(\varphi)=a_{m}\sin m\varphi+a_{m-1}\sin(m-1)\varphi+\ldots+a_{1}\sin\varphi \tag{24} \]

with given coefficient \(a_{m}\), find the one that deviates least from zero on the two intervals \([-\pi,-\varphi_{0}]\) and \([\varphi_{0},\pi]\).

Indeed, putting \(x=\sin(\varphi/2)\), we obtain

\[ y_{2m-1}(x)=P_{m}(2\arcsin x)=\sqrt{1-x^{2}}\sum_{k=1}^{m}(-1)^{k-1}a_{k}U_{2k-1}(x); \tag{25} \]

\[ -1\leq x\leq -\lambda;\qquad \lambda\leq x\leq 1;\qquad \lambda=\sin\frac{\varphi_{0}}{2}. \]

Consequently, the extremal function (25) is determined from the condition

\[ \sum_{k=1}^{m}(-1)^{k-1}a_k U_{2k-1}(x)=(-1)^{m-1}l_m \dot{B}_{2m-1}(x,\lambda). \tag{26} \]

Equating the coefficients of \(x^{2m-1}\), we find

\[ l_m=2^{2m-1}a_m L_{2m-1} =a_m\left[\frac{\theta(0)\theta_1(0)}{\theta(K/2m)\theta_1(K/2m)}\right]^{2m}. \tag{27} \]

Thus, the required trigonometric polynomial will be

\[ P_m(\varphi,\varphi_0)=(-1)^{m-1}l_m\cos\frac{\varphi}{2}\, B_{2m-1}\left(\sin\frac{\varphi}{2},\sin\frac{\varphi_0}{2}\right), \tag{28} \]

where \(l_m\) is the minimum deviation, determined by formula (27). For the coefficients \(a_k\) we have

\[ \begin{aligned} a_k&=(-1)^{k-1}\frac{2}{\pi} \int_{-1}^{1}\sqrt{1-x^2}\,B_{2m-1}(x,\lambda)\,U_{2k-1}(x)\,dx \\ &=(-1)^{k-1}k\sum_{p=k}^{m}\frac{b_p C_{2(p-k)}^{k}}{2^{2p-1}p}, \end{aligned} \tag{29} \]

where \(b_p\) are the coefficients of the polynomial \(B_{2m-1}(x,\lambda)\).

Expression (28) determines the extremal polynomial for \(\pi/2m<\varphi_0<\pi\). For \(0\leq \varphi_0\leq \pi/2m\), in accordance with (20), we have

\[ P_m(\varphi,\varphi_0)=(-1)^{m-1}l_m\cos\frac{\varphi}{2} U_{2m-1}\left(\sin\frac{\varphi}{2}\right)=a_m\sin m\varphi, \tag{30} \]

and for \(\varphi_0=\pi\)

\[ P_m(\varphi,\pi)=2^{2m-1}a_m\sin\varphi(1-\cos\varphi)^{m-1}. \tag{31} \]

In conclusion we note that the corresponding problem for the polynomial

\[ Q_m(\varphi)=a_m\cos m\varphi+a_{m-1}\cos(m-1)+\cdots+a_0 \]

is solved in an analogous way, but with the use of the polynomial

\[ A_{2m}(x,\lambda)=T_{2m}\left(\sqrt{\frac{x^2-\lambda^2}{1-\lambda^2}}\right), \]

found by N. I. Akhiezer in work (1).

Received
28 XII 1960

References Cited

1. N. I. Akhiezer, Izv. Kazan. Phys.-Math. Soc., 3, issue 2, 3 (1928).
2. N. I. Akhiezer, Lectures on the Theory of Approximation, Moscow–Leningrad, 1947, p. 303.
3. S. N. Bernstein, Izv. Acad. Sci. USSR, Ser. Math., 13, 111 (1949).