Abstract
Full Text
E. I. Krupitskii
ON A CLASS OF POLYNOMIALS LEAST DEVIATING FROM ZERO ON TWO INTERVALS
(Presented by Academician V. I. Smirnov, January 7, 1961)
- 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(x)) attains an absolute maximum on the segment ([-1,1]).}(x)) has two more simple zeros at the points (x=\pm\delta), where the function (Q_{4m
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}}{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.
- 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
- N. I. Akhiezer, Izv. Kazan. Phys.-Math. Soc., 3, issue 2, 3 (1928).
- N. I. Akhiezer, Lectures on the Theory of Approximation, Moscow–Leningrad, 1947, p. 303.
- S. N. Bernstein, Izv. Acad. Sci. USSR, Ser. Math., 13, 111 (1949).