Coding Project 1 (Numerical Analysis class , Matlab) Given A function y=f(x), X=

Question

Coding Project 1 (Numerical Analysis class , Matlab)

Given A function y=f(x), X=(x0,x1,....,xn)^(T) with yi=f(xi) and x0<x1<....<xn, and one value a, write a Matlab function [y]=//P1(X,n,a) to do the following.

1. Calculate the coefficients of the interpolation polynomial Pn(x) for the points (x0,y0),...,(xn,yn) using Newton divided-differences as in Algorithm 3.2.

2. For the given value a, calculate y= Pn(a). y is the output of your function.

3. Draw the graphs of y= f(x) and y=Pn(x) on the interval [x0,xn] togethert on one coordinate system.

Explanation / Answer

1.

function [Q,fcount] = quadtx(F,a,b,tol,varargin)

%QUADTX Evaluate definite integral numerically.

% Q = QUADTX(F,A,B) approximates the integral of F(x)

% from A to B to within a tolerance of 1.e-6.

%

% Q = QUADTX(F,A,B,tol) uses tol instead of 1.e-6.

%

% The first argument, F, is a function handle or

% an anonymous function that defines F(x).

%

% Arguments beyond the first four,

% Q = QUADTX(F,a,b,tol,p1,p2,...), are passed on to the

% integrand, F(x,p1,p2,..).

%

% [Q,fcount] = QUADTX(F,...) also counts the number of

% evaluations of F(x).

%

% See also QUAD, QUADL, DBLQUAD, QUADGUI.

% Default tolerance

if nargin < 4 | isempty(tol)

tol = 1.e-6;

end

% Initialization

c = (a + b)/2;

fa = F(a,varargin{:});

fc = F(c,varargin{:});

fb = F(b,varargin{:});

% Recursive call

[Q,k] = quadtxstep(F, a, b, tol, fa, fc, fb, varargin{:});

fcount = k + 3;

2.

function a = InterpV(x,y)

% a = InterpV(x,y)

% This computes the Vandermonde polynomial interpolant where

% x is a column n-vector with distinct components and y is a

% column n-vector.

%

% a is a column n-vector with the property that if

%

% p(x) = a(1) + a(2)x + ... a(n)x^(n-1)

% then

% p(x(i)) = y(i), i=1:n

n = length(x);

V = ones(n,n);

for j=2:n

% Set up column j.

V(:,j) = x.*V(:,j-1);

end

a = Vy;

3.

function [c, err, yc] = regula (f, a, b, delta, epsilon, maxl)

%Input - f is the function input as a string ’f’

% - a and b are the left and right end points

% - delta is the tolerance for the zero

% - epsilon is the tolerance for the value of f at the zero

% - max1 is the maximum number of iterations

%Output - c is the zero

% - yc = f(c)

% - err is the error estimate for c

ya=feva1(f,a);

yb=feva1(f,a);

if ya*yb>0

disp (’Note: f(a)*f(b)>0’),

break,

end

for k=1:max1

dx=yb*(b-a)/(yb-ba);

c=b-dx;

ac=c-a;

yc=feval(f,c);

if yc==0, break;

elseif yb*yc>0

b=c;

yb=yc;

end

dx=min (abs (dx), ac);

if abs (dx)<delta, break, end

if abs (yc)<epsilon, break, end

end

c;

err=abs (b-a)/2;

yc=feval(f,c);

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