Is anything wrong with Eval function? I get value of pn =1 for all x values.
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function HW_4_8()
y= @(x)(1/(1+25*x.^2));
%x=linspace(-1,1,41);
for i=0:5
x=cos(i*pi/40);
end
fprintf('Coefficient of the newton interpolation\n\n');
a= Coef(x,y);
for i=1:length(a)
fprintf('%5.0f\t%15.10f\n',i-1,a(i));
end
xp=linspace(-1,1,200);
fprintf('Approximate value of y(x)\n');
fprintf(' n x p(x)');
fprintf( ' y(x) diff\n');
for i=1:length(xp)
p=xp(i);
funval=(1./(1+25*p.^2));
% fplot(funval,[-1,1]);
pn= Eval(x,a,p);
D=abs(funval-pn);
% disp(D);
% fplot(diff,[-1,1]);
fprintf('%5.0f\t%5.10f\t%15.10f\t%15.10f\t%15.10f\n',i,p,pn,funval,D);
end
%fplot(D,[-1,1]);
end
function[a]=Coef(x,y)
mx=length(x);
my=length(y);
% if mx ~= my
% error('must be same lenght');
% end
T=zeros(mx,mx);
T(:,1)=(my)';
for j=2:mx
for i=1:(mx-j+1)
T(i,j)=(T(i+1,j-1)-T(i,j-1))/(x(i+j-1)-x(i));
end
end
a=T(1,:);
end
function pn=Eval(x,a,p)
m=length(a);
sum=0;
for i=1:m
temo=1;
for j=1:i-1
temo=temo*(p-x(j));
end
sum =sum+a(i)*temo;
end
pn=sum;
end
4 Comments
Are Mjaavatten
on 9 Apr 2018
Thanks for the feedbacks from both of you.
Guillaume: This started out as a brief comment on the x vector, but it sort of grew as I wrote. I should have considered changing it to it an answer.
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