8/18/2019 Newton , Lagrange, Fourier

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 Newton interpolation method

clear all

clc

x =[-2 0 2];

y =[4 2 8];a(1) = y(1);

a(2)=(y(2)-y(1))/(x(2)-x(1));

a(3)=((y(3)-y(2))/(x(3)-x(2))-(y(2)-y(1))/(x(2)-x(1)))/(x(3)-

x(1))

t=[1];

p(t)=a(1)+a(2)*(t-x(1))+a(3)*(t-x(1))*(t-x(2));

p(1)

lagrange interpolation method

clear all

clc

x=[-2 0 2];

y=[4 2 8];

c(1)=y (1)/((x(1)-x(2))*(x(1)-x(3)));

c(2)=y (2)/((x(2)-x(1))*(x(2)-x(3)));

c(3)=y (3)/((x(3)-x(1))*(x(3)-x(2)))

X=1;

N1=(X-x(2))*(X-x(3));

N2=(X-x(1))*(X-x(3));

N3=(X-x(1))*(X-x(2));

p=c(1)*N1+c(2)*N2+c(3)*N3;

p(1)

fourier

clear all

clc

x = input('Define x vector = ');

m = input('Define order of series= ');

n=length(x);

w=2*pi/n;t = 0:w:(2*pi - w);

T=t';

a = zeros(1,m);

b = zeros(1,m);

for j=1:m

  a(j) = (2/n)*(x*cos(j*T));

  b(j) = (2/n)*(x*sin(j*T));

8/18/2019 Newton , Lagrange, Fourier

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