Gauss forward interpolation formula

n = length(x)
I created the difference table (tbl) using the following code. difference table is in attached file
tbl=y';
for j=2:n
for i=1:n-j+1
tbl(i,j)=tbl(i+1,j-1)-tbl(i,j-1);
end
end
p = (xp – x0) / h
yp = y0 + p.dy0 + p(p-1) (d2 y-1 + d3 y-1 ) / 2! + p(p-1) (p-2) (d3 y-1 + d4 y-1 ) / 3! +
p(p-1) (p-2)(p-3) (d4 y-1 + d5 y-1 ) / 4! + p(p-1) (p-2)(p-3)(p-4) (d5 y-1 + d6 y-1 ) / 5!
I created a vector with the terms 1, p, p(p-1), p(p-1)(p-2)…p(p-1)(p-2)..(p-(n-2)) as t=cumprod([1,p-(0:n-2)]);
Now, suggest me the code to create a vector with the following terms
y0 , dy0 , (d2 y-1 + d3 y-1 ) / 2! , (d3 y-1 + d4 y-1 ) / 3! , (d4 y-1 + d5 y-1 ) / 4! ,
(d5 y-1 + d6 y-1 ) / 5! , …….

댓글 수: 4

John D'Errico
John D'Errico 2018년 10월 17일
Since this is your homework, why not make an effort? Try it, THEN ask for help in fixing what you tried. You learn nothing by just giving up and asking for someone to provide the answer.
John D'Errico
John D'Errico 2018년 10월 18일
편집: John D'Errico 2018년 10월 18일
Please don't answer your question with an answer that only adds information. Learn how to use comments instead. Answer moved to a comment:
My code is given below:
function [ yval ] = gauss_p( xd,yd,xp )
n=length(xd);
if(length(yd)==n)
tbl=yd';
for j=2:n
for i=1:n-j+1
tbl(i,j)=tbl(i+1,j-1)-tbl(i,j-1);
end
end
tbl
h=xd(2)-xd(1);
if rem(n,2)==0
k=n/2+1;
else
k=n/2+0.5;
end
p=(xp-xd(k))/h
pt=cumprod([1,p-(0:n-3)])
dt=[tbl(k,1) tbl(k,2) tbl(k-1,3)+tbl(k-1,4) tbl(k-1,4)+tbl(k-1,5) tbl(k-1,5)+tbl(k-1,6) tbl(k-1,6)+tbl(k-1,7)]./factorial(0:n-2)
yval=sum(pt.*dt)
end
Suggest me to simplify the calculation of 'dt' vector with any predefined function
Anton Gluchshenko
Anton Gluchshenko 2020년 11월 18일
편집: Anton Gluchshenko 2020년 11월 18일
Sorry. But It is not your code... or?
Good one sir, can u show me the code of Newton backwards interpolation... Thank u sir..

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2018년 10월 17일

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2021년 3월 18일

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