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T. Popoviciu, Remarque sur les polynômes de meilleure approximation, Buletinul Soc. de Ştiinţe din Cluj, 5 (1930), pp.279-286 (in French).

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1930 c -Popoviciu- Bull. Soc. Sc. Cluj - Remark on best approximation polynomials
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BULETINUL SOCIETĂTII DE ŞTINTE DIN CLUJ (ROMÂNIA) BULLETIN OF THE SOCIETY OF SCIENCES OF CLUJ (ROMANIA) Volume V, 1 D 1 1^("re ")1^{\text {re }}1Part, pp. 279-286. June 5, 1930.

NOTE ON BEST APPROXIMATION POLYNOMIALS

1 per

Tiberiu Popoviciu

Student at the École Normale Supérieure, Paris.
Admitted on February 21, 1930.
  1. We consider the class (b) of functions defined and continuous in the finite and closed interval ( has , b ) ; ( has < b ) ( has , b ) ; ( has < b ) (a,b);(a < b)(a, b); (a<b)(has,b);(has<b). Regarding the properties of these functions, let us remember that they can vanish at a finite or infinite number of points, but the set of these points always forms a finite number of intervals which can also be reduced to a single point.
Let us now consider a system of n + 1 n + 1 n+1n+1n+1functions of class (b)
(1)
f 1 , f 2 , , f n f 1 , f 2 , , f n f_(1),f_(2),dots,f_(n)f_{1}, f_{2}, \ldots, f_{n}f1,f2,,fn
linearly independent, that is to say an equality such as
(2)
P = c 0 f 0 + c 1 f 1 + + c n f n = 0 P = c 0 f 0 + c 1 f 1 + + c n f n = 0 P=c_(0)f_(0)+c_(1)f_(1)+cdots+c_(n)f_(n)=0\mathrm{P}=c_{0} f_{0}+c_{1} f_{1}+\cdots+c_{n} f_{n}=0P=c0f0+c1f1++cnfn=0
c 0 , c 1 , , c n c 0 , c 1 , , c n c_(0),c_(1),dots,c_(n)c_{0}, c_{1}, \ldots, c_{n}c0,c1,,cnbeing constants, cannot take place everywhere in ( has , b ) ( has , b ) (a,b)(a, b)(has,b)that if
c 0 = c 1 = = c n = 0 c 0 = c 1 = = c n = 0 c_(0)=c_(1)=cdots=c_(n)=0c_{0}=c_{1}=\cdots=c_{n}=0c0=c1==cn=0
Note that system (1) may not be linearly independent in an interval ( α , β ) ( α , β ) (alpha, beta)(\alpha, \beta)(α,β)interior to ( has , b ) ( has , b ) (a,b)(a, b)(has,b)[such as | has α | + | β b | 0 | has α | + | β b | 0 |a-alpha|+|beta-b|!=0|a-\alpha|+|\beta-b| \neq 0|hasα|+|βb|0].
Any homogeneous linear expression such as (2) will be called a polynomial of the system (1). To any polynomial P P PPPwe can make a point correspond M M MMMcoordinates c 0 , c 1 , , c n c 0 , c 1 , , c n c_(0),c_(1),dots,c_(n)c_{0}, c_{1}, \ldots, c_{n}c0,c1,,cnin ordinary space n + 1 n + 1 n+1n+1n+1dimensions. We call the point M the image of the corresponding polynomial P.
The fact that the functions (1) are linearly independent can be expressed as follows:
He y y yyya one-to-one correspondence between polynomials P P PPPand their images M M MMM.
For the points where P P PPPIf this is cancelled, we make the following agreements: ( 1 ) 1 (^(1))\left({ }^{1}\right)(1)
10. We call a simple zero a point inside the interval ( has , b has , b a,ba, bhas,b) where P becomes zero by changing sign, or an endpoint ( has has hashashasor b) where P is zero.
2 0 2 0 2^(0)2^{0}20We call a double zero, and we count it as two single zeros, an interior point where P P PPPcancels out without changing sign.
A system (1) is a Chebyshev system or system (T) if any polynomial P P PPPcannot have more n n nnnzeros. It follows that for a system not to be a system (T), it is necessary and sufficient that there be at least one polynomial having at least n + 1 n + 1 n+1n+1n+1zeros.
For example, the system
1 , x , x 2 , , x n 1 , x , x 2 , , x n 1,x,x^(2),dots,x^(n)1, x, x^{2}, \ldots, x^{n}1,x,x2,,xn
is a system (T) in any finite interval. Conversely, the system
1 , si x , cos x , , si n x , cos n x 1 , si x , cos x , , si n x , cos n x 1,sin x,cos x,dots,sin nx,cos nx1, \sin x, \cos x, \ldots, \sin nx, \cos nx1,six,cosx,,sinx,cosnx
is not a system (T) in every interval ( 0 , 2 π 0 , 2 π 0.2ft0.2 \pi0,2π), but there is one in an interval of shorter length than 2 π 2 π 2ft2 π2π2.
We will demonstrate the following auxiliary property:
If (1) is not a system ( T T TTT) we can find n + 1 n + 1 n+1n+1n+1distinct points x 1 , x 2 , , x n 1 x 1 , x 2 , , x n 1 x_(1),x_(2),dots,x_(n_(-1))x_{1}, x_{2}, \ldots, x_{n_{-1}}x1,x2,,xn1completely inside ( has , b has , b a,ba, bhas,b) such as the determinant:
(3) | f 0 ( x 1 ) f 1 ( x 1 ) f n ( x 1 ) f 0 ( x 2 ) f 1 ( x 2 ) f n ( x 2 ) f 0 ( x n + 1 ) f 1 ( x n + 1 ) f n ( x n + 1 ) | (3) f 0 x 1 f 1 x 1 f n x 1 f 0 x 2 f 1 x 2 f n x 2 f 0 x n + 1 f 1 x n + 1 f n x n + 1 {:(3)|[f_(0)(x_(1)),f_(1)(x_(1)),dots,f_(n)(x_(1))],[f_(0)(x_(2)),f_(1)(x_(2)),do ts,f_(n)(x_(2))],[*,*,*,*],[f_(0)(x_(n+1)),f_(1)(x_(n+1)),dots,f_(n)(x_(n+1))]|:}\left|\begin{array}{cccc} f_{0}\left(x_{1}\right) & f_{1}\left(x_{1}\right) & \ldots & f_{n}\left(x_{1}\right) \tag{3}\\ f_{0}\left(x_{2}\right) & f_{1}\left(x_{2}\right) & \ldots & f_{n}\left(x_{2}\right) \\ \cdot & \cdot & \cdot & \cdot \\ f_{0}\left(x_{n+1}\right) & f_{1}\left(x_{n+1}\right) & \ldots & f_{n}\left(x_{n+1}\right) \end{array}\right|(3)|f0(x1)f1(x1)fn(x1)f0(x2)f1(x2)fn(x2)f0(xn+1)f1(xn+1)fn(xn+1)|
is zero.
Indeed, there exists a polynomial having at least n + 1 n + 1 n+1n+1n+1zeros. If such a polynomial vanishes at at least n + 1 n + 1 n+1n+1n+1distinct points, none of which coincides with an endpoint, the property is obvious.
The points where P vanishes can be arranged into 3 categories.
10. Or P vanishes while changing sign; let
(4) x 1 , x 2 , , x m (4) x 1 , x 2 , , x m {:(4)x_(1)","x_(2)","dots","x_(m):}\begin{equation*} x_{1}, x_{2}, \ldots, x_{m} \tag{4} \end{equation*}(4)x1,x2,,xm
these points.
2 0 2 0 2^(0)2^{0}20Or P vanishes without changing sign; let
(5)
x 1 , x 2 , , x k x 1 , x 2 , , x k x_(1)^('),x_(2)^('),dots,x_(k)^(')x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{k}^{\prime}x1,x2,,xk
(1) S. Bernstein "Lectures on Extremal Properties ... etc." p. 1; J.W. Young "General Theory of Approximation by Functions Involving a Given Number of Arbitrary Parameters" Transactions of the Amer. Math. Soc. 8 (1907), p. 331.
These points. Finally:
3 0 3 0 3^(0)3^{0}30The extremes where P vanishes
x 1 , x 2 , , x i ( i = 0 , 1 , 2 ) x 1 , x 2 , , x i ( i = 0 , 1 , 2 ) x^('')_(1),x^('')_(2),dots,x^('')_(i)quad(i=0,1,2)x^{\prime \prime}{ }_{1}, x^{\prime \prime}{ }_{2}, \ldots, x^{\prime \prime}{ }_{i} \quad(i=0,1,2)x1,x2,,xi(i=0,1,2)
We have, by hypothesis:
m + 2 k + i n + 1 m + 2 k + i n + 1 m+2k+i >= n+1m+2 k+i \geqq n+1m+2k+in+1
Suppose that:
m + k < n + 1 m + k < n + 1 m+k < n+1m+k<n+1m+k<n+1
therefore, all the more so
k < n + 1 k < n + 1 k < n+1k<n+1k<n+1
Let us first assume that
and consider the table:
k < n k < n k < nk<nk<n
(7) f 0 ( x 1 ) f 1 ( x 1 ) f n ( x 1 ) f 0 ( x 2 ) f 1 ( x 2 ) f n ( x 2 ) f 0 ( x k ) f 1 ( x k ) f n ( x k ) f 0 ( x 1 ) f 1 ( x 1 ) f n ( x 1 ) f 0 ( x i ) f 1 ( x i ) f n ( x i ) (7) f 0 x 1 f 1 x 1 f n x 1 f 0 x 2 f 1 x 2 f n x 2 f 0 x k f 1 x k f n x k f 0 x 1 f 1 x 1 f n x 1 f 0 x i f 1 x i f n x i {:(7)||[f_(0)(x^(')_(1)),f_(1)(x^(')_(1)),dots,f_(n)(x^(')_(1))],[f_(0)(x^(')_(2)),f_(1)(x^(')_(2)),dots,f_(n)(x^(')_(2))],[vdots,,,],[f_(0)(x^(')_(k)),f_(1)(x^(')_(k)),dots,f_(n)(x^(')_(k))],[f_(0)(x^('')_(1)),f_(1)(x^('')_(1)),dots,f_(n)(x^('')_(1))],[f_(0)(x^('')_(i)),f_(1)(x^('')_(i)),dots,f_(n)(x^('')_(i))]||:}\left\|\begin{array}{cccc} f_{0}\left(x^{\prime}{ }_{1}\right) & f_{1}\left(x^{\prime}{ }_{1}\right) & \ldots & f_{n}\left(x^{\prime}{ }_{1}\right) \\ f_{0}\left(x^{\prime}{ }_{2}\right) & f_{1}\left(x^{\prime}{ }_{2}\right) & \ldots & f_{n}\left(x^{\prime}{ }_{2}\right) \\ \vdots & & & \\ f_{0}\left(x^{\prime}{ }_{k}\right) & f_{1}\left(x^{\prime}{ }_{k}\right) & \ldots & f_{n}\left(x^{\prime}{ }_{k}\right) \tag{7}\\ f_{0}\left(x^{\prime \prime}{ }_{1}\right) & f_{1}\left(x^{\prime \prime}{ }_{1}\right) & \ldots & f_{n}\left(x^{\prime \prime}{ }_{1}\right) \\ f_{0}\left(x^{\prime \prime}{ }_{i}\right) & f_{1}\left(x^{\prime \prime}{ }_{i}\right) & \ldots & f_{n}\left(x^{\prime \prime}{ }_{i}\right) \end{array}\right\|(7)f0(x1)f1(x1)fn(x1)f0(x2)f1(x2)fn(x2)f0(xk)f1(xk)fn(xk)f0(x1)f1(x1)fn(x1)f0(xi)f1(xi)fn(xi)
which contains at least as many columns as rows.
If in table (7) all the determinants formed by k + i k + i k+ik+ik+iIf any columns are zero, the desired property follows. But suppose there is a non-zero determinant, for example:
(8) f 0 ( x 1 ) f 1 ( x 1 ) f k + i ( x 1 ) f 0 ( x 2 ) f 1 ( x 2 ) f k + i ( x 2 ) f 0 ( x i ) f 1 ( x i ) f k + i ( x i ) | 0 (8) f 0 x 1 f 1 x 1 f k + i x 1 f 0 x 2 f 1 x 2 f k + i x 2 f 0 x i f 1 x i f k + i x i 0 {:(8){:[f_(0)(x_(1)^(')),f_(1)(x_(1)^(')),dots,f_(k+i)(x_(1)^('))],[f_(0)(x_(2)^(')),f_(1)(x_(2)^(')),dots,f_(k+i)(x_(2)^('))],[*,*,*,*],[f_(0)(x_(i)^('')),f_(1)(x_(i)^('')),dots,f_(k+i)(x_(i)^(''))]|!=0:}\left.\begin{array}{cccc} f_{0}\left(x_{1}^{\prime}\right) & f_{1}\left(x_{1}^{\prime}\right) & \ldots & f_{k+i}\left(x_{1}^{\prime}\right) \tag{8}\\ f_{0}\left(x_{2}^{\prime}\right) & f_{1}\left(x_{2}^{\prime}\right) & \ldots & f_{k+i}\left(x_{2}^{\prime}\right) \\ \cdot & \cdot & \cdot & \cdot \\ f_{0}\left(x_{i}^{\prime \prime}\right) & f_{1}\left(x_{i}^{\prime \prime}\right) & \ldots & f_{k+i}\left(x_{i}^{\prime \prime}\right) \end{array} \right\rvert\, \neq 0(8)f0(x1)f1(x1)fk+i(x1)f0(x2)f1(x2)fk+i(x2)f0(xi)f1(xi)fk+i(xi)|0
At each point x ; x ; x^(');x^{\prime} ;x;And x ; x ; x^('');x^{\prime \prime} ;x;let's assign a non-zero number a ; a ; a^(');a^{\prime} ;has;resp. a j a j a^('')_(j)a^{\prime \prime}{ }_{j}hasjsuch that its sign is that of the polynomial P in the neighborhood of x j x j x^(')_(j)x^{\prime}{ }_{j}xjresp. x x x^('')x^{\prime \prime}x; [the neighborhood of an endpoint is counted only towards the interior of the interval]. We can then determine a system of non-zero numbers
such that
λ 0 , λ 1 , , λ k + 1 λ 0 , λ 1 , , λ k + 1 lambda_(0),lambda_(1),dots,lambda_(k+1)\lambda_{0}, \lambda_{1}, \ldots, \lambda_{k+1}λ0,λ1,,λk+1
λ 0 f 0 ( x 1 ) + λ 1 f 1 ( x 1 ) + + λ k + i f k + i ( x 1 ) = a 1 λ 0 f 0 ( x 2 ) + λ 1 f 1 ( x 2 ) + + λ k + i f k + i ( x 2 ) = a 2 λ 0 f 0 x 1 + λ 1 f 1 x 1 + + λ k + i f k + i x 1 = a 1 λ 0 f 0 x 2 + λ 1 f 1 x 2 + + λ k + i f k + i x 2 = a 2 {:[lambda_(0)f_(0)(x_(1)^('))+lambda_(1)f_(1)(x_(1)^('))+cdots+lambda_(k+i)f_(k+i)(x_(1)^('))=a_(1)^(')],[lambda_(0)f_(0)(x_(2)^('))+lambda_(1)f_(1)(x_(2)^('))+cdots+lambda_(k+i)f_(k+i)(x_(2)^('))=a_(2)^(')],[ cdots*********]:}\begin{aligned} & \lambda_{0} f_{0}\left(x_{1}^{\prime}\right)+\lambda_{1} f_{1}\left(x_{1}^{\prime}\right)+\cdots+\lambda_{k+i} f_{k+i}\left(x_{1}^{\prime}\right)=a_{1}^{\prime} \\ & \lambda_{0} f_{0}\left(x_{2}^{\prime}\right)+\lambda_{1} f_{1}\left(x_{2}^{\prime}\right)+\cdots+\lambda_{k+i} f_{k+i}\left(x_{2}^{\prime}\right)=a_{2}^{\prime} \\ & \cdots \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \end{aligned}λ0f0(x1)+λ1f1(x1)++λk+ifk+i(x1)=has1λ0f0(x2)+λ1f1(x2)++λk+ifk+i(x2)=has2
λ 0 f 0 ( x i ) + λ 1 f 1 ( x ) + + λ k + i f k + i ( x i ) = a i λ 0 f 0 x i + λ 1 f 1 x + + λ k + i f k + i x i = a i lambda_(0)f_(0)(x_(i)^(''))+lambda_(1)f_(1)(x^(''))+cdots+lambda_(k+i)f_(k+i)(x^('')_(i))=a_(i)^('')\lambda_{0} f_{0}\left(x_{i}^{\prime \prime}\right)+\lambda_{1} f_{1}\left(x^{\prime \prime}\right)+\cdots+\lambda_{k+i} f_{k+i}\left(x^{\prime \prime}{ }_{i}\right)=a_{i}^{\prime \prime}λ0f0(xi)+λ1f1(x)++λk+ifk+i(xi)=hasi
Either ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0and consider the polynomial
P 1 = P ε ( λ 0 f 0 + λ 1 f 1 + + λ k + i f k + i ) P 1 = P ε λ 0 f 0 + λ 1 f 1 + + λ k + i f k + i P_(1)=P-epsi(lambda_(0)f_(0)+lambda_(1)f_(1)+cdots+lambda_(k+i)f_(k+i))\mathrm{P}_{1}=\mathrm{P}-\varepsilon\left(\lambda_{0} f_{0}+\lambda_{1} f_{1}+\cdots+\lambda_{k+i} f_{k+i}\right)P1=Pε(λ0f0+λ1f1++λk+ifk+i)
If ε = 0 ε = 0 epsi=0\varepsilon=0ε=0We have P 1 P P 1 P P_(1)-=P\mathrm{P}_{1} \equiv \mathrm{P}P1P. If ε ε epsi\varepsilonεis small but not zero. P 1 P 1 P_(1)\mathrm{P}_{1}P1is close to P. We are sure that if we take ε ε epsi\varepsilonεsmall enough, at each zero x j x j x_(j)x_{j}xjP will correspond to a zero x j x j x_(j)x_{j}xjof P 1 P 1 P_(1)\mathrm{P}_{1}P1neighbor of x j x j x_(j)x_{j}xjWe can also see this because of the choice of numbers. a j , a j a j , a j a^(')_(j),a^('')_(j)a^{\prime}{ }_{j}, a^{\prime \prime}{ }_{j}hasj,hasjthat at each zero x j x j x^(')_(j)x^{\prime}{ }_{j}xjcorrespond to at least two simple zeros for P 1 P 1 P_(1)\mathrm{P}_{1}P1and to zero x j x j x^('')_(j)x^{\prime \prime}{ }_{j}xjcorresponds to at least one zero inside ( a , b a , b a,ba, bhas,b).
So, finally, ε ε epsi\varepsilonεbeing sufficiently small, the polynomial P 1 P 1 P_(1)\mathrm{P}_{1}P1cancels out at least m + 2 k + i m + 2 k + i m+2k+im+2 k+im+2k+itimes; the property results immediately. In the case k = n k = n k=nk=nk=nit suffices to introduce at most one point x l x l x_(l)^('')x_{l}{ }^{\prime \prime}xLsion
:
3. Let f ( x ) f ( x ) f(x)f(x)f(x)a function of class (b) and consider the express-
I ( c 0 , c 1 , , c n ) = Max ( a , b ) | f c 0 f 0 c 1 f 1 c n f n | I c 0 , c 1 , , c n = Max ( a , b ) f c 0 f 0 c 1 f 1 c n f n I(c_(0),c_(1),dots,c_(n))=Max_((a,b))|f-c_(0)f_(0)-c_(1)f_(1)-cdots-c_(n)f_(n)|\mathrm{I}\left(c_{0}, c_{1}, \ldots, c_{n}\right)=\operatorname{Max}_{(a, b)}\left|f-c_{0} f_{0}-c_{1} f_{1}-\cdots-c_{n} f_{n}\right|I(c0,c1,,cn)=Max(has,b)|fc0f0c1f1cnfn|
To simplify the writing we can write
I ( c 0 , c 1 , , c n ) = Max ( a , b ) | f P | = I ( M ) I c 0 , c 1 , , c n = Max ( a , b ) | f P | = I ( M ) I(c_(0),c_(1),dots,c_(n))=Max_((a,b))|f-P|=I(M)\mathrm{I}\left(c_{0}, c_{1}, \ldots, c_{n}\right)=\operatorname{Max}_{(a, b)}|f-\mathrm{P}|=\mathrm{I}(\mathrm{M})I(c0,c1,,cn)=Max(has,b)|fP|=I(M)
Let M be the image of the polynomial P. Then, the function I(M) is continuous with respect to the coefficients. c 0 , c 1 , , c n c 0 , c 1 , , c n c_(0),c_(1),dots,c_(n)c_{0}, c_{1}, \ldots, c_{n}c0,c1,,cnfor everything a x b a x b a <= x <= ba \leq x \leq bhasxb.
The expression I(M) has a lower limit, and we know that this limit is attained by at least one polynomial P, therefore by at least one finite value system. γ 0 , γ 1 , , γ n γ 0 , γ 1 , , γ n gamma_(0),gamma_(1),dots,gamma_(n)\gamma_{0}, \gamma_{1}, \ldots, \gamma_{n}γ0,γ1,,γnquantities c 0 , c 1 , c n c 0 , c 1 , c n c_(0),c_(1),dotsc_(n)c_{0}, c_{1}, \ldots c_{n}c0,c1,cn:
lim I ( M ) = I ( M 1 ) = I ( γ 0 , γ 1 , , γ n ) lim I ( M ) = I M 1 = I γ 0 , γ 1 , , γ n limI(M)=I(M_(1))=I(gamma_(0),gamma_(1),dots,gamma_(n))\lim \mathrm{I}(\mathrm{M})=\mathrm{I}\left(\mathrm{M}_{1}\right)=\mathrm{I}\left(\gamma_{0}, \gamma_{1}, \ldots, \gamma_{n}\right)limitI(M)=I(M1)=I(γ0,γ1,,γn)
M 1 M 1 M_(1)\mathrm{M}_{1}M1is then a minimizing point and the corresponding polynomial is a minimizing polynomial. In general, it is difficult to decide whether the minimizing polynomial is unique or not, but:
If there are two distinct minimizing polynomials, there are infinitely many.
Indeed M 1 , M 2 M 1 , M 2 M_(1),M_(2)M_{1}, M_{2}M1,M2two distinct minimizing points and consider the point
M 3 = M 1 + λ M 2 1 + λ , λ 0 M 3 = M 1 + λ M 2 1 + λ , λ 0 M_(3)=(M_(1)+lambdaM_(2))/(1+lambda),quad lambda >= 0M_{3}=\frac{M_{1}+\lambda M_{2}}{1+\lambda}, \quad \lambda \geq 0M3=M1+λM21+λ,λ0
on the segment M 1 M 2 M 1 M 2 M_(1)M_(2)M_{1} M_{2}M1M2The symbolic representation above being clear.
It is easy to see that:
I ( M 3 ) I ( M 1 ) + λ I ( M 2 ) 1 + λ = I ( M 1 ) I M 3 I M 1 + λ I M 2 1 + λ = I M 1 I(M_(3)) <= (I(M_(1))+lambdaI(M_(2)))/(1+lambda)=I(M_(1))\mathrm{I}\left(\mathrm{M}_{3}\right) \leq \frac{\mathrm{I}\left(\mathrm{M}_{1}\right)+\lambda \mathrm{I}\left(\mathrm{M}_{2}\right)}{1+\lambda}=\mathrm{I}\left(\mathrm{M}_{1}\right)I(M3)I(M1)+λI(M2)1+λ=I(M1)
but by hypothesis,
therefore
I ( M 3 ) I ( M 1 ) I ( M 3 ) = I ( M 1 ) I M 3 I M 1 I M 3 = I M 1 {:[I(M_(3)) >= I(M_(1))],[I(M_(3))=I(M_(1))]:}\begin{aligned} & I\left(M_{3}\right) \geq I\left(M_{1}\right) \\ & I\left(M_{3}\right)=I\left(M_{1}\right) \end{aligned}I(M3)I(M1)I(M3)=I(M1)
It follows that any point of the segment M 1 M 2 M 1 M 2 M_(1)M_(2)M_{1} M_{2}M1M2is minimizing. On the other hand, every minimizing point is at a finite distance, because for a point with unbounded coordinates I(M) becomes infinite and at the same time:
lim I ( M ) Max ( a , b ) | f | = quantité bornée. lim I ( M ) Max ( a , b ) | f | =  quantité bornée.  limI(M) <= Max_((a,b))|f|=" quantité bornée. "\lim \mathrm{I}(\mathrm{M}) \leq \operatorname{Max}_{(a, b)}|f|=\text { quantité bornée. }limitI(M)Max(has,b)|f|= bounded quantity. 
We can therefore state the property:
Minimizing points form a closed and bounded convex domain.
4. We propose to prove the following property:
The necessary and sufficient condition for there to be only one minimizing polynomial, regardless of the function f f fffof class (b), is that the system (1) is a system (T).
We know that this condition is sufficient ( 1 1 ^(1){ }^{1}1We will show that if the system is not a Chebyshev system, we can find a function f f fffadmitting an infinite number of minimizing polynomials.
We know that there is a polynomial
P = c 0 f 0 + c 1 f 1 + + c n f n P = c 0 f 0 + c 1 f 1 + + c n f n P=c_(0)f_(0)+c_(1)f_(1)+cdots+c_(n)f_(n)\mathrm{P}=c_{0} f_{0}+c_{1} f_{1}+\cdots+c_{n} f_{n}P=c0f0+c1f1++cnfn
canceling out in at least n + 1 n + 1 n+1n+1n+1distinct points within the interval ( a , b a , b a,ba, bhas,b). If x 1 , x 2 , , x n + 1 x 1 , x 2 , , x n + 1 x_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}x1,x2,,xn+1are n + 1 n + 1 n+1n+1n+1points satisfying this condition, the determinant (3) is zero because we assume, of course, that P ≡≡ 0 P ≡≡ 0 P≡≡0\mathrm{P} \equiv \equiv 0P≡≡0.
We can then find n + 1 n + 1 n+1n+1n+1Numbers a 1 , a 2 , , a n + 1 a 1 , a 2 , , a n + 1 a_(1),a_(2),dots,a_(n+1)a_{1}, a_{2}, \ldots, a_{n+1}has1,has2,,hasn+1such as the system
E 1 = c 0 f 0 ( x 1 ) + c 1 f 1 ( x 1 ) + + c n f n ( x 1 ) a 1 = 0 E 2 = c 0 f 0 ( x 2 ) + c 1 f 1 ( x 2 ) + + c n f n ( x 2 ) a 2 = 0 E n + 1 = c 0 f 0 ( x n + 1 ) + c 1 f 1 ( n + 1 ) + + c n f n ( x n + 1 ) a n + 1 = 0 E 1 = c 0 ¯ f 0 x 1 + c 1 ¯ f 1 x 1 +      + c n ¯ f n x 1 a 1 = 0 E 2 = c 0 ¯ f 0 x 2 + c 1 ¯ f 1 x 2 +      + c n ¯ f n x 2 a 2 = 0 E n + 1 = c 0 ¯ f 0 x n + 1 + c 1 ¯ f 1 ( n + 1 ) + + c n ¯ f n x n + 1 a n + 1 = 0 {:[E_(1)= bar(c_(0))f_(0)(x_(1))+ bar(c_(1))f_(1)(x_(1))+cdots,+ bar(c_(n))f_(n)(x_(1))-a_(1)=0],[E_(2)= bar(c_(0))f_(0)(x_(2))+ bar(c_(1))f_(1)(x_(2))+cdots,+ bar(c_(n))f_(n)(x_(2))-a_(2)=0],[E_(n+1)= bar(c_(0))f_(0)(x_(n+1))+ bar(c_(1))f_(1)(n+1)+cdots+ bar(c_(n))f_(n)(x_(n+1))-a_(n+1)=0]:}\begin{array}{ll} \mathrm{E}_{1}=\overline{c_{0}} f_{0}\left(x_{1}\right)+\overline{c_{1}} f_{1}\left(x_{1}\right)+\cdots & +\overline{c_{n}} f_{n}\left(x_{1}\right)-a_{1}=0 \\ \mathrm{E}_{2}=\overline{c_{0}} f_{0}\left(x_{2}\right)+\overline{c_{1}} f_{1}\left(x_{2}\right)+\cdots & +\overline{c_{n}} f_{n}\left(x_{2}\right)-a_{2}=0 \\ \mathrm{E}_{n+1}=\overline{c_{0}} f_{0}\left(x_{n+1}\right)+\overline{c_{1}} f_{1}(n+1)+\cdots+\overline{c_{n}} f_{n}\left(x_{n+1}\right)-a_{n+1}=0 \end{array}E1=c0f0(x1)+c1f1(x1)++cnfn(x1)has1=0E2=c0f0(x2)+c1f1(x2)++cnfn(x2)has2=0En+1=c0f0(xn+1)+c1f1(n+1)++cnfn(xn+1)hasn+1=0
is incompatible. In this case, the expression
Max ( | E 1 | , | E 2 | , , | E n + 1 | ) ( 2 ) Max E 1 , E 2 , , E n + 1 2 Max(|E_(1)|,|E_(2)|,dots,|E_(n+1)|)(^(2))\operatorname{Max}\left(\left|E_{1}\right|,\left|E_{2}\right|, \ldots,\left|E_{n+1}\right|\right)\left({ }^{2}\right)Max(|E1|,|E2|,,|En+1|)(2)
which is a continuous function of c ¯ 0 , c ¯ 1 , , c ¯ n c ¯ 0 , c ¯ 1 , , c ¯ n bar(c)_(0), bar(c)_(1),dots, bar(c)_(n)\bar{c}_{0}, \bar{c}_{1}, \ldots, \bar{c}_{n}c¯0,c¯1,,c¯n, admits a non-zero positive minimum. Let m m mmmthe value of this minimum; it is attained for an infinite number of value systems c ¯ 0 , c ¯ 1 , , c ¯ n c ¯ 0 , c ¯ 1 , , c ¯ n bar(c)_(0), bar(c)_(1),dots, bar(c)_(n)\bar{c}_{0}, \bar{c}_{1}, \ldots, \bar{c}_{n}c¯0,c¯1,,c¯nbecause if
c i = c i i i = 0 , 1 , , n c i ¯ = c i i i = 0 , 1 , , n bar(c_(i))=c_(i)^(**)i quad i=0,1,dots,n\overline{c_{i}}=c_{i}^{*} i \quad i=0,1, \ldots, nci=ci*ii=0,1,,n
(') S. Bernstein loc. cit. p. 3; JW Joung loc. cit.
( 2 2 ^(2){ }^{2}2) Max ( λ 1 , λ 2 , , λ n Max λ 1 , λ 2 , , λ n Max(lambda_(1),lambda_(2),dots,lambda_(n):}\operatorname{Max}\left(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\right.Max(λ1,λ2,,λn) means the value of the largest quantity λ i λ i lambda i\lambda iλi.
is a minimizing system, the systems
c i = c i + λ c i i = 0 , 1 , 2 , . . , n c i = c i + λ c i i = 0 , 1 , 2 , . . , n c_(i)=c_(i)^(**)+lambdac_(i)quad i=0,1,2,..,nc_{i}=c_{i}^{*}+\lambda c_{i} \quad i=0,1,2, . ., nci=ci*+λcii=0,1,2,..,n
will also minimize λ λ lambda\lambdaλbeing arbitrary.
Now if f f fffis a continuous function taking the values a i a i a_(i)a_{i}hasito the points x i ( i = 1 , 2 , , n + 1 ) x i ( i = 1 , 2 , , n + 1 ) x_(i)(i=1,2,dots,n+1)x_{i}(i=1,2, \ldots, n+1)xi(i=1,2,,n+1), we will have for every polynomial Q Q QQQ
Max | f Q | m [ dans ( a , b ) ] . Max | f Q | m [ dans ( a , b ) ] . Max|f-Q| >= m quad[dans(a,b)].\operatorname{Max}|f-Q| \geq m \quad[\operatorname{dans}(a, b)] .Max|fQ|m[In(has,b)].
Let's ask
P = c 0 f 0 + c 1 f 1 + + c n f n P = c 0 f 0 + c 1 f 1 + + c n f n P^(**)=c_(0)^(**)f_(0)+c_(1)^(**)f_(1)+cdots+c_(n)^(**)f_(n)\mathrm{P}^{*}=c_{0}^{*} f_{0}+c_{1}^{*} f_{1}+\cdots+c_{n}^{*} f_{n}P*=c0*f0+c1*f1++cn*fn
[one can also have P 0 P 0 P^(**)-=0\mathrm{P}^{*} \equiv 0P*0
To clarify, let's assume that the polynomial P, taken initially, vanishes at a finite number of points; these are first the points x 1 , x 2 , , x n + 1 x 1 , x 2 , , x n + 1 x_(1),x_(2),dots,x_(n+1)x_{1}, x_{2}, \ldots, x_{n+1}x1,x2,,xn+1and then some other points x 1 , x 2 , , x r x 1 , x 2 , , x r x^(')_(1),x^(')_(2),dots,x^(')_(r)x^{\prime}{ }_{1}, x^{\prime}{ }_{2}, \ldots, x^{\prime}{ }_{r}x1,x2,,xr(which may not exist at all).
Let's take the function f f fffsuch that:
(9)
(10) f ( x i ) = P ( x i ) a i i = 1 , 2 , , n + 1 f ( x i ) = 0 i = 1 , 2 , , r (10) f x i = P x i a i i = 1 , 2 , , n + 1 f x i = 0 i = 1 , 2 , , r {:(10){:[f(x_(i))=P^(**)(x_(i))-a_(i),i=1","2","dots","n+1],[f(x_(i)^('))=0,i=1","2","dots","r]:}:}\begin{array}{cc} f\left(x_{i}\right)=\mathrm{P}^{*}\left(x_{i}\right)-a_{i} & i=1,2, \ldots, n+1 \\ f\left(x_{i}^{\prime}\right)=0 & i=1,2, \ldots, r \tag{10} \end{array}(10)f(xi)=P*(xi)hasii=1,2,,n+1f(xi)=0i=1,2,,r
Les points x i , x i partagent l'intervalle ( a , b ) en un certain nombre us-in-in  Les points  x i , x i  partagent l'intervalle  ( a , b )  en un certain nombre   us-in-in  {:[" Les points "x_(i)","x_(i)^(')" partagent l'intervalle "(a","b)" en un certain nombre "],[" us-in-in "]:}\begin{aligned} & \text { Les points } x_{i}, x_{i}^{\prime} \text { partagent l'intervalle }(a, b) \text { en un certain nombre } \\ & \text { us-in-in } \end{aligned} The points xi,xi share the interval (has,b) in a number  us-in-in 
of subintervals ( n + r , n + r + 1 n + r , n + r + 1 n+r,n+r+1n+r, n+r+1n+r,n+r+1Or n + r + 2 n + r + 2 n+r+2n+r+2n+r+2) such that in each of them the polynomial P retains a constant sign.
Either
Max ( a , b ) | P | = A Max ( a , b ) | P | = A Max_((a,b))|P|=A\operatorname{Max}_{(a, b)}|\mathrm{P}|=\mathrm{A}Max(has,b)|P|=HAS
Let's take
0 > λ > m A 0 > λ > m A 0 > lambda > -(m)/((A))0>\lambda>-\frac{m}{\mathrm{~A}}0>λ>m HAS
SO
| λ P | < m | λ P | < m |lambdaP| < m|\lambda \mathrm{P}|<m|λP|<m
We determine f f fffin each subinterval such that it is continuous and such that (9) and (10) are satisfied. To do this, we will examine the various kinds of subintervals that can arise.
10. Interval ( a , x 1 ) a , x 1 (a,x^(')_(1))\left(a, x^{\prime}{ }_{1}\right)(has,x1)Within such an interval, we will take
f = m 2 A P f = m 2 A P f=(m)/(2(A))Pf=\frac{m}{2 \mathrm{~A}} \mathrm{P}f=m2 HASP
then in ( a , x 1 a , x 1 a,x^(')_(1)a, x^{\prime}{ }_{1}has,x1)
And
λ < 0 , | f λ P | = | f | + | λ P | = { m 2 A + | λ | } | P | λ < 0 , | f λ P | = | f | + | λ P | = m 2 A + | λ | | P | lambda < 0,quad|f-lambdaP|=|f|+|lambdaP|={(m)/(2(A))+|lambda|}|P|\lambda<0, \quad|f-\lambda \mathrm{P}|=|f|+|\lambda \mathrm{P}|=\left\{\frac{m}{2 \mathrm{~A}}+|\lambda|\right\}|\mathrm{P}|λ<0,|fλP|=|f|+|λP|={m2 HAS+|λ|}|P|
| f λ P | < m | f λ P | < m |f-lambdaP| < m|f-\lambda \mathrm{P}|<m|fλP|<m
provided that
m 2 A < λ < 0 m 2 A < λ < 0 -(m)/(2(A)) < lambda < 0-\frac{m}{2 \mathrm{~A}}<\lambda<0m2 HAS<λ<0
The same construction is used for an interval ( x r , b x r , b x^(')_(r),bx^{\prime}{ }_{r}, bxr,b).
2 0 2 0 2^(0)2^{0}20. Interval ( a , x 1 ) a , x 1 (a,x_(1))\left(a, x_{1}\right)(has,x1)[Or ( x n + 1 , b ) ] x n + 1 , b {:(x_(n+1),b)]\left.\left(x_{n+1}, b\right)\right](xn+1,b)]. We have
| P ( x 1 ) a 1 | m P x 1 a 1 m |P^(**)(x_(1))-a_(1)| <= m\left|P^{*}\left(x_{1}\right)-a_{1}\right| \leq m|P*(x1)has1|m
We will take from ( a , x 1 a , x 1 a,x_(1)a, x_{1}has,x1)
f = P ( x 1 ) a 1 m A P si P [ P ( x 1 ) a 1 ] > 0 f = P ( x 1 ) a 1 si P [ P ( x 1 ) a 1 ] < 0 f = P x 1 a 1 m A P       si       P P x 1 a 1 > 0 f = P x 1 a 1       si       P P x 1 a 1 < 0 {:[f=P^(**)(x_(1))-a_(1)-(m)/((A))P," si ",P[P^(**)(x_(1))-a_(1)] > 0],[f=P^(**)(x_(1))-a_(1)," si ",P[P^(**)(x_(1))-a_(1)] < 0]:}\begin{array}{lll} f=\mathrm{P}^{*}\left(x_{1}\right)-a_{1}-\frac{m}{\mathrm{~A}} \mathrm{P} & \text { si } & \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]>0 \\ f=\mathrm{P}^{*}\left(x_{1}\right)-a_{1} & \text { si } & \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]<0 \end{array}f=P*(x1)has1m HASP if P[P*(x1)has1]>0f=P*(x1)has1 if P[P*(x1)has1]<0
we then see that
(11)
Max | f + λ P | m Max | f + λ P | m Max|f+lambdaP| <= m\operatorname{Max}|f+\lambda \mathrm{P}| \leq mMax|f+λP|m
provided that
(12)
0 > λ > m A 0 > λ > m A 0 > lambda > -(m)/((A))0>\lambda>-\frac{m}{\mathrm{~A}}0>λ>m HAS
  1. Interval ( x 1 , x 1 x 1 , x 1 x_(1),x_(1)^(')x_{1}, x_{1}^{\prime}x1,x1).
We can take
f = ( x x 1 ) P ( x 1 ) a 1 x 1 x 1 m A P si P [ P ( x 1 ) a 1 ] > 0 f = ( x x 1 ) P ( x 1 ) a 1 x 1 x 1 , si P [ P ( x 1 ) a 1 ] < 0 f = x x 1 P x 1 a 1 x 1 x 1 m A P       si  P P x 1 a 1 > 0 f = x x 1 P x 1 a 1 x 1 x 1 ,       si  P P x 1 a 1 < 0 {:[f=(x-x_(1)^('))(P^(**)(x_(1))-a_(1))/(x_(1)-x_(1)^('))-(m)/((A))P," si "quadP[P^(**)(x_(1))-a_(1)] > 0],[f=(x-x_(1)^('))(P^(**)(x_(1))-a_(1))/(x_(1)-x_(1)^('))","," si "P[P^(**)(x_(1))-a_(1)] < 0]:}\begin{array}{ll} f=\left(x-x_{1}^{\prime}\right) \frac{\mathrm{P}^{*}\left(x_{1}\right)-a_{1}}{x_{1}-x_{1}^{\prime}}-\frac{m}{\mathrm{~A}} \mathrm{P} & \text { si } \quad \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]>0 \\ f=\left(x-x_{1}^{\prime}\right) \frac{\mathrm{P}^{*}\left(x_{1}\right)-a_{1}}{x_{1}-x_{1}^{\prime}}, & \text { si } \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]<0 \end{array}f=(xx1)P*(x1)has1x1x1m HASP if P[P*(x1)has1]>0f=(xx1)P*(x1)has1x1x1, if P[P*(x1)has1]<0
and we still have (11) under conditions (12).
4). Interval ( x 1 , x 2 ) x 1 , x 2 (x^(')_(1),x^(')_(2))\left(x^{\prime}{ }_{1}, x^{\prime}{ }_{2}\right)(x1,x2)All you have to do is take
f = m A P f = m A P f=-(m)/((A))Pf=-\frac{m}{\mathrm{~A}} \mathrm{P}f=m HASP
and (11) will be satisfied with (12).
50. Interval ( x 1 , x 2 x 1 , x 2 x_(1),x_(2)x_{1}, x_{2}x1,x2Let's assume
g = x [ P ( x 1 ) a 1 P ( x 2 ) + a 2 ] + x 2 P ( x 1 ) a 1 x 1 [ P ( x 2 ) a 2 x 1 x 1 g = x P x 1 a 1 P x 2 + a 2 + x 2 P x 1 a 1 x 1 P x 2 a 2 x 1 x 1 g=(x[P^(**)(x_(1))-a_(1)-P^(**)(x_(2))+a_(2)]+x_(2)|__P^(**)(x_(1))-a_(1)~|-x_(1)[P^(**)(x_(2))-a_(2)~|)/(x_(1)-x_(1))g=\frac{x\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}-\mathrm{P}^{*}\left(x_{2}\right)+a_{2}\right]+x_{2}\left\lfloor\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right\rceil-x_{1}\left[\mathrm{P}^{*}\left(x_{2}\right)-a_{2}\right\rceil}{x_{1}-x_{1}}g=x[P*(x1)has1P*(x2)+has2]+x2P*(x1)has1x1[P*(x2)has2x1x1
We then ask
f = g m A P si P [ P ( x 1 ) a 1 ] > 0 , P [ P ( x 2 ) a 2 ] > 0 ou bien si [ P ( x 1 ) a 1 ] [ P ( x 2 ) a 2 ] < 0 f = g si P [ P ( x 1 ) a 1 ] > 0 , P [ P ( x 2 ) a 2 ] < 0 . . f = g m A P       si  P P x 1 a 1 > 0 , P P x 2 a 2 > 0       ou bien si  P x 1 a 1 P x 2 a 2 < 0 f = g       si  P P x 1 a 1 > 0 , P P x 2 a 2 < 0 . . {:[f=g-(m)/((A))Pquad," si "P[P^(**)(x_(1))-a_(1)] > 0","P[P^(**)(x_(2))-a_(2)] > 0],[," ou bien si "[P^(**)(x_(1))-a_(1)][P^(**)(x_(2))-a_(2)] < 0],[f=g," si "P[P^(**)(x_(1))-a_(1)] > 0","P[P^(**)(x_(2))-a_(2)] < 0..]:}\begin{array}{ll} f=g-\frac{m}{\mathrm{~A}} \mathrm{P} \quad & \text { si } \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]>0, \mathrm{P}\left[\mathrm{P}^{*}\left(x_{2}\right)-a_{2}\right]>0 \\ & \text { ou bien si }\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]\left[\mathrm{P}^{*}\left(x_{2}\right)-a_{2}\right]<0 \\ f=g & \text { si } \mathrm{P}\left[\mathrm{P}^{*}\left(x_{1}\right)-a_{1}\right]>0, \mathrm{P}\left[\mathrm{P}^{*}\left(x_{2}\right)-a_{2}\right]<0 . . \end{array}f=gm HASP if P[P*(x1)has1]>0,P[P*(x2)has2]>0 or if [P*(x1)has1][P*(x2)has2]<0f=g if P[P*(x1)has1]>0,P[P*(x2)has2]<0..
We can still determine one λ 1 > 0 λ 1 > 0 lambda_(1) > 0\lambda_{1}>0λ1>0such that if
0 > λ > λ 1 0 > λ > λ 1 0 > lambda > -lambda_(1)0>\lambda>-\lambda_{1}0>λ>λ1
we still have
in the interval ( x 1 , x 2 x 1 , x 2 x_(1),x_(2)x_{1}, x_{2}x1,x2().
Max | f λ P | m Max | f λ P | m Max|f-lambdaP| <= m\operatorname{Max}|f-\lambda \mathrm{P}| \leq mMax|fλP|m
This property results from the following observation:
Let g ( x ) g ( x ) g(x)g(x)g(x)a continuous function in ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1)(we take the inter-nalle ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1)to simplify the explanation) positive (negative) in this interval and
g ( 0 ) = g ( 1 ) = 1 g ( 0 ) = g ( 1 ) = 1 g(0)=g(1)=1g(0)=g(1)=1g(0)=g(1)=1
So, to clarify things, a > 0 , b < 0 a > 0 , b < 0 a^(') > 0,b^(') < 0a^{\prime}>0, b^{\prime}<0has>0,b<0We can then determine a λ 1 > 0 λ 1 > 0 lambda_(1) > 0\lambda_{1}>0λ1>0such as for 0 < λ < λ 1 0 < λ < λ 1 0 < lambda < lambda_(1)0<\lambda<\lambda_{1}0<λ<λ1we have
a + x ( b a ) + λ g < a a + x b a + λ g < a a^(')+x(b^(')-a^('))+lambda g < aa^{\prime}+x\left(b^{\prime}-a^{\prime}\right)+\lambda g<ahas+x(bhas)+λg<has
In ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1)The demonstration is immediate.
The function f f fffConstructed in this way, it answers the question, because λ λ lambda\lambdaλbeing between
0 , λ 1 0 , λ 1 0,-lambda_(1)0,-\lambda_{1}0,λ1
Or λ 1 λ 1 lambda_(1)\lambda_{1}λ1is a positive number, we have
| f λ P | m | f λ P | m |f-lambdaP| <= m|f-\lambda \mathrm{P}| \leq m|fλP|m
In ( a , b ) ( a , b ) (a,b)(a, b)(has,b)and at least in one respect x i x i x_(i)x_{i}xion al'égalité
| f λ P | = m . | f λ P | = m . |f-lambdaP|=m.|f-\lambda \mathrm{P}|=m .|fλP|=m.
On the other hand, any other polynomial gives
| f Q | m . | f Q | m . |f-Q| >= m.|f-Q| \geq m .|fQ|m.
5. If P were to vanish throughout an entire interval ( α , β α , β alpha,beta\alpha, \betaα,β(For example, we could still do the construction very easily.) The points x 1 , x 2 x 1 , x 2 x_(1),x_(2)x_{1}, x_{2}x1,x2 . x n + 1 . x n + 1 dots.x_(n+1)\ldots . x_{n+1}.xn+1can be taken inside ( α , β ) ( α , β ) (alpha,beta)(\alpha, \beta)(α,β)So apart from ( α , β ) ( α , β ) (alpha,beta)(\alpha, \beta)(α,β)We are doing the same construction as before. In ( α , β α , β alpha,beta\alpha, \betaα,β(we still keep the conditions (9); and in each interval x i , x i + 1 x i , x i + 1 x_(i),x_(i+1)x_{i}, x_{i+1}xi,xi+1we take the function t t tttlinear; and constant in ( α , x 1 ) α , x 1 (alpha,x_(1))\left(\alpha, x_{1}\right)(α,x1)And( x n + 1 , β x n + 1 , β x_(n+1),betax_{n+1}, \betaxn+1,βThe function thus obtained still answers the question.
1930

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