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DaliaViola
Posts: **3**Member

in Matlab

Hi, I'm Italian, so I apologize for my bad english...i'm writing here because in this period I'm trying to study matlab, but I think I denied...so I have a program and I ask you what tests can I do to verify its operation, the program is this:

[code]% The function y allows you to calculate the value of the % interpolating polynomial, given some starting points in % arbitrary values of the variable

% independent.

%It accepts INPUT% in three vectors:

% x: whose components constitute the abscissas of the %points interpolating;

% y whose components are the ordinates of the %interpolating points;

% t: tab vector, whose components are the values on which

% we want to calculate the polynomial interpolating the %points (x (i), f (i))

% and a scalar:

% TOLL: tolerance within which must be the absolute value % of

% difference of the abscissa (or ordinate) of the points % of interpolation.

% In OUTPUT instead provides the vector y whose %components %are the calculated values of the polynomial

% in the components of the vector t tab.

function y= Newton(x, f, t, TOLL)

n= length(x); m=length(f); k =length(t); y=1:1:k;

if n~= m % length control vectors x and f

error('error:interpolation points unspecified');

end

A=zeros(n,n); % initialize matrix designed to contain %the divided differences

for i=1:1:n

A(i,i)=f(i);

end

for i=2:1:n

for j=i-1 1

if abs(x(i)-x(j))<= TOLL && abs(f(i)-f(j))<= TOLL

error('coincident points');

elseif abs(x(i)-x(j))<= TOLL

error('abscissas coincident')

end

A(i,j)= (A(i,j+1) - A(i-1,j))/(x(i)-x(j));

% the matrix A at the end of the cycle will be %inferiorly triangular

% and is such that for i> j %A(i,j)=f[x_j,x_(j+1),....,x_i]

%with f [, ..] is the divided difference

end

end

for h=1:1:k % calculation of the interpolating %polynomial in the components of t

p=A(n,1);

for i=n-11

p=(t(h) - x(i))*p + A(i,1);

end

y(h)=p;

end

end[/code]

I hope you can help me, thanks a lot! Byeeee!!!

[code]% The function y allows you to calculate the value of the % interpolating polynomial, given some starting points in % arbitrary values of the variable

% independent.

%It accepts INPUT% in three vectors:

% x: whose components constitute the abscissas of the %points interpolating;

% y whose components are the ordinates of the %interpolating points;

% t: tab vector, whose components are the values on which

% we want to calculate the polynomial interpolating the %points (x (i), f (i))

% and a scalar:

% TOLL: tolerance within which must be the absolute value % of

% difference of the abscissa (or ordinate) of the points % of interpolation.

% In OUTPUT instead provides the vector y whose %components %are the calculated values of the polynomial

% in the components of the vector t tab.

function y= Newton(x, f, t, TOLL)

n= length(x); m=length(f); k =length(t); y=1:1:k;

if n~= m % length control vectors x and f

error('error:interpolation points unspecified');

end

A=zeros(n,n); % initialize matrix designed to contain %the divided differences

for i=1:1:n

A(i,i)=f(i);

end

for i=2:1:n

for j=i-1 1

if abs(x(i)-x(j))<= TOLL && abs(f(i)-f(j))<= TOLL

error('coincident points');

elseif abs(x(i)-x(j))<= TOLL

error('abscissas coincident')

end

A(i,j)= (A(i,j+1) - A(i-1,j))/(x(i)-x(j));

% the matrix A at the end of the cycle will be %inferiorly triangular

% and is such that for i> j %A(i,j)=f[x_j,x_(j+1),....,x_i]

%with f [, ..] is the divided difference

end

end

for h=1:1:k % calculation of the interpolating %polynomial in the components of t

p=A(n,1);

for i=n-11

p=(t(h) - x(i))*p + A(i,1);

end

y(h)=p;

end

end[/code]

I hope you can help me, thanks a lot! Byeeee!!!

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