Question

Part I: Show that (y − y ∗ 0 )(y − y ∗ 1 ). . .(y − y ∗ n ) = 5 n+1 2 n Tn+1(x), where x = y/5

Part II: It can be shown that there exists R > 0 such that |f (n) (y)| ≤ Rn for all y ∈ [−5, 5]. Assuming this, show that limn→∞ max{|f(y) − Pn(y)|, y ∈ [−5, 5]} = 0

Ij = COS Problem 1: Recall that the Chebyshev nodes x*, x1,...,x* are determined on the interval (-1, 1) as the zeros of Tn+1(x) = cos((n + 1) arccos(x)) and are given by 2j +17 j = 0,1,...n. n +1 2 Consider now interpolating the function f(x) = 1/(1 + x2) on the interval (-5,5). We have seen in lecture that if equispaced nodes are used, the error grows unbound- edly as more points are used. The purpose of this problem is to show that if the Chebyshev nodes are used, this problem disapears. First denote by y* = 5 x x and let Pn(y) denote the polynomial of degree n that interpolates f at these points. Then the error estimate for polynomial interpolation gives for y € (-5,5] \f(y) – Pn(y) \fm+1C)||(y – yö)(y – yi)... (y – yn) (n + 1)! for some ce € (-5,5). Part 1: Show that 5n+1 (y - y1)(y – yi)... (y – y*) -Tn+1(x), where x = y/5 2n Part II: It can be shown that there exists R > 0 such that | f(n)(y) = RM for all y E1-5,5). Assuming this, show that lim max{\f(y) – Pn(y)], y € (-5,5)} = 0 no

Answer 1

Solution:-

Given that

The interval [-1, 1] as the zeros of

Tn+1(C) = cos((n +1) arccos(2)

T; COS 2j +17 n+1 2 j = 0,1,..., n

on the interval t = 5, 5

FC) = (1+2

degree m that interpolates f at these polats. Then the error estimate for polynomial interpolation gives for

ye -5,5

f(y) – Pn(y) If+C) (n+1)! (y - 3)(y – yi)...(y- yn)

for some

e -5,5

(y - y0) y - yi)...(y - yn)

C yo = 5x0, y1 = 5.21....Yn = 5.3.7 5

y = 5.2 (5.r – 507) (5.x – 5.21)....(5.2 – 21

........(A)

(y - y0)(y- yi)...(y - y) = 5"+1(1 – 27)(x - 27...(2-2))

To (2) = 1, T2) =

Tn+1(2) 2xTn (2) - Tn-1(2)

In+1() = (x-)(x - 21)... (1-21

from (1):-

For m = 0

T1(2) = (- 2040

T40 = 2

40

m = 0, j = 0

7T -20 = cos( 1.0

.ro = 0

m = 1:-

T2() = cos(2 cos (2)

cos (2) (32) = 0

LE SOD

cos(20) = 2 cos- 1

  

= 2.c -1

T2(2) 2.0 1

From (1)

T2(x) = A1(x - 2)(x -x1)

j = 1, m = 1

co E| COS 22

C 37 COS

21 = cos(T- 4.

21 = - cos

21 2

j = 0, m = 1

COS El

10 COS 4

2

T. (r) - : A1 (x (x + v2 2

T2(2) A1 (x²

For m = 2:-

From (1)

Tz(x) = (2-x)(2-21) (2 - 0) (A2)

j = 0, m = 2

E COS 3

03 = COS

3 27 2

j = 1, m = 2

7 COS سنت انت E |

= 0

j = 2, m = 2

El COS 3

5 -x; = cos(- 6

2 = cos(TT 6

- cos

23 V3 2

T3 (2) = cos(3 cos (2)

cos (2) (32) = 0

LE SOD

cos(30) = 4 cos 0 - 3 cos 0

  $=4x^3-3x$

$4x^3-3x=(x-\frac{\sqrt{3}}{2})(x-0)(x+\frac{\sqrt{3}}{2})A_2$

A2 = 4

So, proceeding thus we have $A_n=2^n$

$T_{n+1}(x)=[(x-x^*_0)(x-x^*_1)...(x-x^*_n)]2^n$ .......(2)

Part I:-

So, from (A) we have

$(y-y^*_0)(y-y^*_1)...(y-y^*_n)=\frac{5^{n+1}T_{n+1}(x)}{2^n}$

where $n=\frac{4}{5}$

Part II:-

It can be shown that there exist R > 0

such that

$|f^n(y)|\leq R^n\ \forall\ y\in [-5,5]$

and we know,

error estimate

$|f(y)-pn(y)|=\frac{f^{n+1}(c)}{(n+1)!}|(y-y^*_0)(y-y^*_1)...(y-y^*_n)|$

$\leq \frac{R^{n+1}}{(n+1)!}\frac{5^{n+1}}{2^n}T_{n+1}(x)$

[by part (1) and ]

f"(y)<R"

$|f(y)-pn(y)|=\leq \frac{R^{n+1}}{(n+1)!}.\frac{5^{n+1}}{2^n}\cos ((n+1) \cos^{-1}(x))$

  $\leq \frac{R^{n+1}}{(n+1)!}.\frac{5^{n+1}}{2^n}[|\cos (n+1) \cos^{-1}(x)|\leq 1]$

$\lim_{m\rightarrow \infty}max|f(y)-pn(y)|=\lim_{m\rightarrow \infty}\frac{R^{n+1}5^{n+1}}{(m+1)!2^n}$

$\lim_{m\rightarrow \infty}\frac{R^{n+1}5^{n+1}}{(m+1)!2^n}$

$\lim_{n\rightarrow \infty}max|f(y)-pn(y)|=0\ for\ y\in[-5,5]$

$5R\lim_{m\rightarrow \infty}\frac{(\frac{5R}{2})^n}{(m+1)!}$

[$\therefore \frac{5R}{2}=p$]

$=5R\lim_{n\rightarrow \infty}\frac{p^n}{(n+1)!}$

= 0

as is increases very fast as n grows large

Hence $\lim_{n\rightarrow \infty}\frac{p^n}{(n+1)!}=0$

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