On the proof of the Weierstrass’s theorem using interpolation polynomials

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T. Popoviciu, Asupra demonstraţiei teoremei lui Weierstrass cu ajutorul polinoamelor de interpolare, Lucrările ses. generale a Acad. R.P.R., pp. 1664-1667 (in Romanian).

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republished in English, in 1998: T. Popoviciu, On the proof of Weierstrass’ theorem using interpolation polynomials, East J. Approximations, 4 (1998) no. 1, pp. 107-110 (translated by D. Kacsó)

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1950 a -Popoviciu- Lucr. His. Gen. St. Acad. RPR - On the demonstration of Weierstrass's theorem with
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ON THE PROOF OF WEIERSTRASS'S THEOREM WITH THE HELP OF INTERPOLATION POLYNOMIALS

OFPROFESSOR TIBERIU POPOVICIU,MLLMBRE CORISSPONDISNT AI, ACATBIGHEII RLRCommunication presented at the meeting of June 7, 1950.

I. Either [ has , b ] [ has , b ] [a,b][a, b][has,b]an interval ends and closed, which, as the case may be, can be considered reduced to [ 0 , I ] [ 0 , I ] [0,I][0, \mathrm{I}][0,I]or [ I , I ] [ I , I ] [-I,I][-\mathrm{I}, \mathrm{I}][I,I], which does not restrict the generality of the issues addressed.
Let us consider a triangular array of nodes in [ has , b ] [ has , b ] [a,b][a, b][has,b]
(I) x n , the , x n , 1 , , x n , n , n = 1 , 2 , 3 , (I) x n , the , x n , 1 , , x n , n , n = 1 , 2 , 3 , {:(I)x_(n)","o","x_(n,1)","dots","x_(n,n)","quad n=1","2","3","dots:}\begin{equation*} x_{n}, o, x_{n, 1}, \ldots, x_{n, n}, \quad n=1,2,3, \ldots \tag{I} \end{equation*}(I)xn,the,xn,1,,xn,n,n=1,2,3,
and a triangular array of polynomials in x x xxx,
(2) P n , 0 , P n , 1 , , P n , n , n = 1 , 2 , 3 , ( P n , m = P n , m ( x ) ) (2) P n , 0 , P n , 1 , , P n , n , n = 1 , 2 , 3 , P n , m = P n , m ( x ) {:(2)P_(n,0)","quadP_(n,1)","dots","P_(n,n)","quad n=1","2","3","dots(P_(n,m)=P_(n,m)(x)):}\begin{equation*} P_{n, 0}, \quad P_{n, 1}, \ldots, P_{n, n}, \quad n=1,2,3, \ldots\left(P_{n, m}=P_{n, m}(x)\right) \tag{2} \end{equation*}(2)Pn,0,Pn,1,,Pn,n,n=1,2,3,(Pn,m=Pn,m(x))
A function f ( x ) f ( x ) f(x)f(x)f(x)uniform [ has , b ] [ has , b ] [a,b][a, b][has,b]attach the side of polynomial of interpolation (in a general sense)
Q n [ f ; x ] = i = 0 n P n , t ( x ) f ( x n , t ) , n = 1 , 2 , 3 , Q n [ f ; x ] = i = 0 n P n , t ( x ) f x n , t , n = 1 , 2 , 3 , Q_(n)[f;x]=sum_(i=0)^(n)P_(n,t)(x)f(x_(n,t)),quad n=1,2,3,dotsQ_{n}[f ; x]=\sum_{i=0}^{n} P_{n, t}(x) f\left(x_{n, t}\right), \quad n=1,2,3, \ldotsQn[f;x]=i=0nPn,t(x)f(xn,t),n=1,2,3,
The problem of determining the arrays (1) and (2) such that
lim n Q n [ f ; x ] = f ( x ) lim n Q n [ f ; x ] = f ( x ) lim_(n rarr oo)Q_(n)[f;x]=f(x)\lim _{n \rightarrow \infty} Q_{n}[f ; x]=f(x)limnQn[f;x]=f(x)
uniform in [ has , b ] [ has , b ] [a,b][a, b][has,b]for any function f ( x ) f ( x ) f(x)f(x)f(x)continue in [ has , b ] [ has , b ] [a,b][a, b][has,b], put by AND AND AND\mathbb{E}AND. Bore1 [I], a most elegant solution of S. Bernstein [2] by polynomials (written for the interval [ 0 , I ] [ 0 , I ] [0,I][0, I][0,I]).
B n [ f ; x ] = i = 0 n f ( i n ) ( n i ) x i ( 1 x ) n L B n [ f ; x ] = i = 0 n f i n n i x i ( 1 x ) n L B_(n)[f;x]=sum_(i=0)^(n)f((i)/(n))((n)/(i))x^(i)(1-x)^(nl)B_{n}[f ; x]=\sum_{i=0}^{n} f\left(\frac{i}{n}\right)\left(\frac{n}{i}\right) x^{i}(1-x)^{nl}Bn[f;x]=i=0nf(in)(ni)xi(1x)nL
  1. If in particular we have
    a) P n , t ( x ) 0 , i = 0 , 1 , , n ; n = 1 , 2 , 3 , ; x [ has , b ] P n , t ( x ) 0 , i = 0 , 1 , , n ; n = 1 , 2 , 3 , ; x [ has , b ] P_(n,t)(x) >= 0,i=0,1,dots,n;n=1,2,3,dots;x in[a,b]P_{n, t}(x) \geqq 0, i=0.1, \ldots, n ; n=1,2,3, \ldots ; x \in[a, b]Pn,t(x)0,i=0,1,,n;n=1,2,3,;x[has,b]
  1. i = 0 n i P n , i ( x ) = 1 , n = 1 , 2 , 3 , i = 0 n i P n , i ( x ) = 1 , n = 1 , 2 , 3 , sum_(i=0)^(n_(i))P_(n,i)(x)=1,quad n=1,2,3,dots\sum_{i=0}^{n_{i}} P_{n, i}(x)=1, \quad n=1,2,3, \ldotsi=0niPn,i(x)=1,n=1,2,3,
    difference Q n [ f ; x ] f ( x ) Q n [ f ; x ] f ( x ) Q_(n)[f;x]-f(x)Q_{n}[f ; x]-f(x)Qn[f;x]f(x)is immediately delimited by the formula
| Q n [ f ; x ] f ( x ) | 2 oh ( HAS n ) Q n [ f ; x ] f ( x ) 2 oh HAS n |Q_(n)[f;x]-f(x)| <= 2omega(A_(n))\left|Q_{n}[f ; x]-f(x)\right| \leqq 2 \omega\left(A_{n}\right)|Qn[f;x]f(x)|2oh(HASn)
under
HAS n = sup [ has , b ] i = 0 r | x n , i x | P n , i ( x ) HAS n = sup [ has , b ] i = 0 r x n , i x P n , i ( x ) A_(n)=su p_([a,b])sum_(i=0)^(r)|x_(n,i)-x|P_(n,i)(x)A_{n}=\sup _{[a, b]} \sum_{i=0}^{r}\left|x_{n, i}-x\right| P_{n, i}(x)HASn=sup[has,b]i=0r|xn,ix|Pn,i(x)
if oh ( d ) oh ( d ) omega(delta)\omega(\delta)oh(d)This oscillating module and function f ( x ) f ( x ) f(x)f(x)f(x)If
HAS n 0 HAS n 0 A_(n)rarr0A_{n} \rightarrow 0HASn0, for n n n rarr on \rightarrow \inftyn, the problem of him E. Bo re 1 is solved. This occurs in particular in the case of S. Bernstein polynomials, as I have shown elsewhere [3].
To prove the Weierstrass theorem, in addition to the conditions a a alpha\alphaa), b b beta\betab), it is enough to have
( HAS n ) B n = sup [ has , b ] i = 0 n ( x n , i x ) 2 P n , i ( x ) 0 , for n HAS n B n = sup [ has , b ] i = 0 n x n , i x 2 P n , i ( x ) 0 ,  for  n (A_(n) <= )B_(n)=sqrt(su p_([a,b])sum_(i=0)^(n)(x_(n,i)-x)^(2)P_(n,i)(x))rarr0," pentra "n rarr oo\left(A_{n} \leqq\right) B_{n}=\sqrt{\sup _{[a, b]} \sum_{i=0}^{n}\left(x_{n, i}-x\right)^{2} P_{n, i}(x)} \rightarrow 0, \text { pentru } n \rightarrow \infty(HASn)Bn=sup[has,b]i=0n(xn,ix)2Pn,i(x)0, for n
This is what happened and it was made by S. Bernstein, this is what happened B n = I 2 n B n = I 2 n B_(n)=(I)/(2sqrtn)B_{n}=\frac{I}{2 \sqrt{n}}Bn=I2n
3. Weierstrass's theorem can also be proven using the polynomials introduced by I 1 I 1 I_(1)I_{1}I1. Head r r rrr[4] and which derives from the interpolation polynomials of him Hermit.
If
L n ( x ) = c ( x x n , 0 ) ( x x n , t ) ( x x n , n ) , L n ( x ) = c x x n , 0 x x n , t x x n , n , l_(n)(x)=c(x-x_(n,0))(x-x_(n,t))dots(x-x_(n,n)),l_{n}(x)=c\left(x-x_{n, 0}\right)\left(x-x_{n, t}\right) \ldots\left(x-x_{n, n}\right),Ln(x)=c(xxn,0)(xxn,t)(xxn,n),
under c c cccThis is constant and suitable.
Let us consider the Hermit polynomial
(3) i = 0 n h n , t ( x ) f n , i + i = 0 n k n , t ( x ) f n , t (3) i = 0 n h n , t ( x ) f n , i + i = 0 n k n , t ( x ) f n , t {:(3)sum_(i=0)^(n)h_(n,t)(x)f_(n,i)+sum_(i=0)^(n)k_(n,t)(x)f_(n,t)^('):}\begin{equation*} \sum_{i=0}^{n} h_{n, t}(x) f_{n, i}+\sum_{i=0}^{n} k_{n, t}(x) f_{n, t}^{\prime} \tag{3} \end{equation*}(3)i=0nhn,t(x)fn,i+i=0nkn,t(x)fn,t
of degree 2 n + 1 2 n + 1 2n+12 n+12n+1, which reduces to f n , t f n , t f_(n,t)f_{n, t}fn,tfor x = x n , t x = x n , t x=x_(n,t)x=x_{n, t}x=xn,tand whose derivative: it reduces to f n , i f n , i f_(n,i)^(')f_{n, i}^{\prime}fn,ifor x = x n , i , i = 0 , 1 , , n x = x n , i , i = 0 , 1 , , n x=x_(n,i),i=0,1,dots,nx=x_{n, i}, i=0,1, \ldots, nx=xn,i,i=0,1,,n.
Then the condition β β beta\betab) is achieved through i = 0 n h n , i ( x ) = 1 i = 0 n h n , i ( x ) = 1 sum_(i=0)^(n)h_(n,i)(x)=1\sum_{i=0}^{n} h_{n, i}(x)=1i=0nhn,i(x)=1 We then have
i = 0 n h n , t ( x ) ( x x n , t ) 2 = 2 i = 0 n k n , t ( x ) ( x x n , t ) = 2 l n 2 ( x ) i = 0 n 1 [ l n ( x n , t ) ] 2 i = 0 n h n , t ( x ) x x n , t 2 = 2 i = 0 n k n , t ( x ) x x n , t = 2 l n 2 ( x ) i = 0 n 1 l n x n , t 2 sum_(i=0)^(n)h_(n,t)(x)(x-x_(n,t))^(2)=2sum_(i=0)^(n)k_(n,t)(x)(x-x_(n,t))=2l_(n)^(2)(x)sum_(i=0)^(n)(1)/([l_(n)^(')(x_(n),t)]^(2))\sum_{i=0}^{n} h_{n, t}(x)\left(x-x_{n, t}\right)^{2}=2 \sum_{i=0}^{n} k_{n, t}(x)\left(x-x_{n, t}\right)=2 l_{n}^{2}(x) \sum_{i=0}^{n} \frac{1}{\left[l_{n}^{\prime}\left(x_{n}, t\right)\right]^{2}}i=0nhn,t(x)(xxn,t)2=2i=0nkn,t(x)(xxn,t)=2Ln2(x)i=0n1[Ln(xn,t)]2
provided α α alpha\alphaa) is achieved if we take h n , i ( x ) = P n , i ( x ) h n , i ( x ) = P n , i ( x ) h_(n,i)(x)=P_(n,i)(x)h_{n, i}(x)=P_{n, i}(x)hn,i(x)=Pn,i(x)as shown. L. Fejér, in the particular case for the Ceb âşev polynomial (written for the interval [ I , I ] [ I , I ] [-I,I][-\mathbf{I}, \mathbf{I}][I,I]),
l n ( x ) = T n + 1 ( x ) = cos ( n + 1 ) arc cos x l n ( x ) = T n + 1 ( x ) = cos ( n + 1 ) arc cos x l_(n)(x)=T_(n+1)(x)=cos(n+1)arc cos xl_{n}(x)=T_{n+1}(x)=\cos (n+1) \operatorname{arc} \cos xLn(x)=Tn+1(x)=cos(n+1)bowcosx
In this case
i = 0 n I [ l n ( x n , i ) ] 2 = i = 0 n I [ T n + 1 ( x n , i ) ] 2 = I ( n + I ) 2 i = 0 n ( I x n , i 2 ) = I 2 ( n + I ) i = 0 n I l n x n , i 2 = i = 0 n I T n + 1 x n , i 2 = I ( n + I ) 2 i = 0 n I x n , i 2 = I 2 ( n + I ) sum_(i=0)^(n)(I)/([l_(n)^(')(x_(n),i)]^(2))=sum_(i=0)^(n)(I)/([T_(n+1)^(')(x_(n,i))]^(2))=(I)/((n+I)^(2))sum_(i=0)^(n)(I-x_(n,i)^(2))=(I)/(2(n+I))\sum_{i=0}^{n} \frac{\mathrm{I}}{\left[l_{n}^{\prime}\left(x_{n}, i\right)\right]^{2}}=\sum_{i=0}^{n} \frac{\mathrm{I}}{\left[T_{n+1}^{\prime}\left(x_{n, i}\right)\right]^{2}}=\frac{\mathrm{I}}{(n+\mathrm{I})^{2}} \sum_{i=0}^{n}\left(\mathrm{I}-x_{n, i}^{2}\right)=\frac{\mathrm{I}}{2(n+\mathrm{I})}i=0nI[Ln(xn,i)]2=i=0nI[Tn+1(xn,i)]2=I(n+I)2i=0n(Ixn,i2)=I2(n+I)
It follows that in this case
B n = 1 n + 1 B n = 1 n + 1 B_(n)=(1)/(sqrt(n+1))B_{n}=\frac{1}{\sqrt{n+1}}Bn=1n+1
So, for L. Fejér's polynomials,
F n [ t ; x ] = i = 0 n h n , i ( x ) f ( x n , i ) F n [ t ; x ] = i = 0 n h n , i ( x ) f x n , i F_(n)[t;x]=sum_(i=0)^(n)h_(n,i)(x)f(x_(n,i))F_{n}[t ; x]=\sum_{i=0}^{n} h_{n, i}(x) f\left(x_{n, i}\right)Fn[t;x]=i=0nhn,i(x)f(xn,i)
under h n , t ( x ) h n , t ( x ) h_(n,t)(x)h_{n, t}(x)hn,t(x)are the fundamental polynomials of the first kind corresponding to Hermite polynomials (3) and Chebâşev knots x n , i x n , i x_(n,i)x_{n, i}xn,i, check the inequality
| F n [ f ; x ] f ( x ) | 2 ω ( I n + I ) , x [ I , I ] F n [ f ; x ] f ( x ) 2 ω I n + I , x [ I , I ] |F_(n)[f;x]-f(x)| <= 2omega((I)/(sqrt(n+I))),quad x in[-I,I]\left|F_{n}[f ; x]-f(x)\right| \leqq 2 \omega\left(\frac{\mathrm{I}}{\sqrt{n+\mathrm{I}}}\right), \quad x \in[-\mathrm{I}, \mathrm{I}]|Fn[f;x]f(x)|2oh(In+I),x[I,I]
It follows from here that the approximation given by L. F ej é r's polynomials is at least of the same order as the approximation given by S. Bexnstein's polynomials.
The above observations raise various new problems on interpolation polynomials, which we will treat in a more detailed paper.

SUMMARY

Delay non-slip locks for approximation polynomials i = 0 n P n , i ( x ) f ( x n , i ) i = 0 n P n , i ( x ) f x n , i sum_(i=0)^(n)P_(n,i)(x)f(x_(n),i)\sum_{i=0}^{n} P_{n, i}(x) f\left(x_{n}, i\right)i=0nPn,i(x)f(xn,i), generalizing the polynomials of S. N. Bernstein (2). Along with the conditions α α alpha\alphaa), β β beta\betab) very light organization used by the cable module functions f ( x ) f ( x ) f(x)f(x)f(x). It is shown that L. Fejer's polynomials (4) lead to Weierstrass's theorem with an approximation reaching the approximation of S. N. Bernstein's polynomials.

SUMMARY

Some remarks on approximation polynomials of the form n P n , i ( x ) f ( x n , i ) n P n , i ( x ) f x n , i sum_(sum)^(n)P_(n,i)(x)f(x_(n),i)\sum_{\sum}^{n} P_{n, i}(x) f\left(x_{n}, i\right)nPn,i(x)f(xn,i), generalizing the polynomials of S. Bernstein [2]. Under the conditions α α alpha\alphaa), β β beta\betab) we can easily delimit the error using the oscillation modulus of the function f ( x ) f ( x ) f(x)f(x)f(x). As an application it is shown that the polynomials of L. Fejér [4] lead to the Weierstrass theorem, with an approximation which reaches that of the polynomials of S. Bernstein.

BIBLIOGRAPHY

  1. E. Borel, Lessons on functions of real variables. 1905.
  2. S. Bernstein, Proof of Weierstrass's theorem based on the calculus of probabilities. Communication Soc. Matem. of Kharkov, 2, 13, 1-2, 1912.
  3. T. Popovieiu, On the approximation of convex functions of higher order. Mathematica, 10, 49-54, 1934.
  4. L. Fejér, On Weierstrass approximation, especially by Hermitian interpolation. Math. Ann., 102, 707-725, 1930.
1950

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