フィルターのクリア

What is the code for lagrange interpolating polynomial for a set of given data?

263 ビュー (過去 30 日間)
I have tried this code. My teacher recommended to use poly and conv function. But I dont get the point of using unknown 'x' in poly. But still it's giving a result which is incorrect.
x = [0 1 2 3 4 5 6 ];
y = [0 .8415 0.9093 0.1411 -0.7568 -0.9589 -0.2794];
sum = 0;
for i = 1:length(x)
p=1;
for j=1:length(x)
if j~=i
c = poly((x-x(j)))/(x(i)-x(j)) ;
p = conv(c, p);
end
end
sum = sum + y(i)*p;
end
  1 件のコメント
Aurangzaib laghari
Aurangzaib laghari 2022 年 9 月 20 日
function [P,R,S] = lagrangepoly(X,Y,XX)
%LAGRANGEPOLY Lagrange interpolation polynomial fitting a set of points
% [P,R,S] = LAGRANGEPOLY(X,Y) where X and Y are row vectors
% defining a set of N points uses Lagrange's method to find
% the N-1th order polynomial in X that passes through these
% points. P returns the N coefficients defining the polynomial,
% in the same order as used by POLY and POLYVAL (highest order first).
% Then, polyval(P,X) = Y. R returns the x-coordinates of the N-1
% extrema of the resulting polynomial (roots of its derivative),
% and S returns the y-values at those extrema.
%
% YY = LAGRANGEPOLY(X,Y,XX) returns the values of the polynomial
% sampled at the points specified in XX -- the same as
% YY = POLYVAL(LAGRANGEPOLY(X,Y)).
%
% Example:
% To find the 4th-degree polynomial that oscillates between
% 1 and 0 across 5 points around zero, then plot the interpolation
% on a denser grid inbetween:
% X = -2:2; Y = [1 0 1 0 1];
% P = lagrangepoly(X,Y);
% xx = -2.5:.01:2.5;
% plot(xx,polyval(P,xx),X,Y,'or');
% grid;
% Or simply:
% plot(xx,lagrangepoly(X,Y,xx));
%
% Note: if you are just looking for a smooth curve passing through
% a set of points, you can get a better fit with SPLINE, which
% fits piecewise polynomials rather than a single polynomial.
%
% See also: POLY, POLYVAL, SPLINE
% 2006-11-20 Dan Ellis dpwe@ee.columbia.edu
% $Header: $
% For more info on Lagrange interpolation, see Mathworld:
% http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html
% Make sure that X and Y are row vectors
if size(X,1) > 1; X = X'; end
if size(Y,1) > 1; Y = Y'; end
if size(X,1) > 1 || size(Y,1) > 1 || size(X,2) ~= size(Y,2)
error('both inputs must be equal-length vectors')
end
N = length(X);
pvals = zeros(N,N);
% Calculate the polynomial weights for each order
for i = 1:N
% the polynomial whose roots are all the values of X except this one
pp = poly(X( (1:N) ~= i));
% scale so its value is exactly 1 at this X point (and zero
% at others, of course)
pvals(i,:) = pp ./ polyval(pp, X(i));
end
% Each row gives the polynomial that is 1 at the corresponding X
% point and zero everywhere else, so weighting each row by the
% desired row and summing (in this case the polycoeffs) gives
% the final polynomial
P = Y*pvals;
if nargin==3
% output is YY corresponding to input XX
YY = polyval(P,XX);
% assign to output
P = YY;
end
if nargout > 1
% Extra return arguments are values where dy/dx is zero
% Solve for x s.t. dy/dx is zero i.e. roots of derivative polynomial
% derivative of polynomial P scales each power by its power, downshifts
R = roots( ((N-1):-1:1) .* P(1:(N-1)) );
if nargout > 2
% calculate the actual values at the points of zero derivative
S = polyval(P,R);
end
end

サインインしてコメントする。

採用された回答

Mohammad Ehsanul Hoque
Mohammad Ehsanul Hoque 2016 年 10 月 2 日
After 3 days i have found the answer myself. The poly function takes arguments as roots of a polynomial. Like if x-2=0 is the equation, poly(2) is enough to find the polynomial matrix.
So, we dont need to put extra 'x' in poly. Except this problem, the code is alright. Now using polyval function we can plot smooth curve from limited data.
Thanks
x=[0 1 2 3 4 5 6];
y=[0 .8415 .9093 .1411 -.7568 -.9589 -.2794];
sum=0;
for i=1:length(x)
p=1;
for j=1:length(x)
if j~=i
c = poly(x(j))/(x(i)-x(j));
p = conv(p,c);
end
end
term = p*y(i);
sum= sum + term;
end
disp(sum);
  4 件のコメント
Sebastian Quintanar
Sebastian Quintanar 2021 年 9 月 15 日
when i copy this code the sum ends up turning into a single value, why is that?
Ahmed J. Abougarair
Ahmed J. Abougarair 2024 年 3 月 20 日
Unfortunately, the written code is incorrect and needs to be reworked

サインインしてコメントする。

その他の回答 (7 件)

MD. ABU SAYED
MD. ABU SAYED 2018 年 7 月 4 日
You can solve lagrange interpolating polynomial for a set of given data this way (most simplest implementation).
x = [12 13 14 16];
y = [5 6 9 11];
sum = 0;
a = 12.5;
for i = 1:length(x)
u = 1;
l = 1;
for j = 1:length(x)
if j ~= i
u = u * (a - x(j));
l = l * (x(i) - x(j));
end
end
sum= sum + u / l * y(i);
end
disp(sum);
I am hopeful this will be helpful for anyone.
  3 件のコメント
George Vigilaios
George Vigilaios 2020 年 9 月 3 日
this code works perfectly!!! thnx!!!
is there a way to see coefficients of the polynomial ?
thnx in advance!
Samson Onyambu
Samson Onyambu 2021 年 6 月 24 日
to get the coefficients you can use newtons dividied difference
%Newton Divided Difference Interpolation Method
%Computes coefficients of interpolating polynomial
%Input: x and y are vectors containing the x and y coordinates
% of the n data points
%Output: coefficients c of interpolating polynomial in nested form
%Use with nest.m to evaluate interpolating polynomial
function c=newtdd(x,y,n)
for j=1:n
v(j,1)=y(j); % Fill in y column of Newton triangle
end
for i=2:n % For column i,
for j=1:n+1-i % fill in column from top to bottom
v(j,i)=(v(j+1,i-1)-v(j,i-1))/(x(j+i-1)-x(j));
end
end
for i=1:n
c(i)=v(1,i); % Read along top of triangle
end % for output coefficients

サインインしてコメントする。


Vincent Naudot
Vincent Naudot 2020 年 9 月 25 日
It is always better to avoid loops!
Here is what you can do
function qlloc=ql(rs,ry,x) %rs stand for the x-node, ry for the y-nodes
mlocx=rs'*ones(1,length(rs));
msave=mlocx;
mloci=mlocx;
mlocx=-mlocx+x;
mlocx=mlocx-diag(diag(mlocx))+diag(ones(1,length(rs)));
mloci=-mloci+msave';
mloci=mloci-diag(diag(mloci))+diag(ones(1,length(rs)));
px=prod(mlocx);
pi=prod(mloci);
polyvect=px./pi;
qlloc=dot(ry,polyvect);
end
  3 件のコメント
Kutlu Yigitturk
Kutlu Yigitturk 2020 年 12 月 27 日
Thank you very much, sir for your efforts!
image-pro
image-pro 2022 年 4 月 18 日
how to apply this method for image

サインインしてコメントする。


Mohamed Ashraf
Mohamed Ashraf 2021 年 5 月 6 日
arr = input('Enter the x values: ');
fx = input('Enter the y values: ');
x = input('Enter a value: ')
lnth = length(arr);
p = 0;
for i = 1 : lnth
prdct = 1;
for j = 1 : lnth
if j ~= i
prdct= prdct*((x-arr(i))/(arr(i)-arr(j)));
end
end
p = p + fx(i)*prdct;
end
display(p);

Hiren Rana
Hiren Rana 2021 年 11 月 11 日
x = [0 1 2 3 4 5 6 ];
y = [0 .8415 0.9093 0.1411 -0.7568 -0.9589 -0.2794];
sum = 0; for i = 1:length(x) p=1; for j=1:length(x) if j~=i c = poly((x-x(j)))/(x(i)-x(j)) ; p = conv(c, p); end end sum = sum + y(i)*p; end

Trevor Sakwa
Trevor Sakwa 2022 年 1 月 20 日
this is what i got to find coefficients of polynomial functions using lagrange formulae
x=[ -3 0 2 5];
y=[ 528 1017 1433 2312];
sum=0;
for i=1:length(x)
p=1;
for j=1:length(x)
if j~=i
c = poly(x(j))/(x(i)-x(j));
p = conv(p,c);
end
end
term = p*y(i);
sum= sum + term;
end
disp(sum);

Aurangzaib laghari
Aurangzaib laghari 2022 年 9 月 20 日
function [P,R,S] = lagrangepoly(X,Y,XX)
%LAGRANGEPOLY Lagrange interpolation polynomial fitting a set of points
% [P,R,S] = LAGRANGEPOLY(X,Y) where X and Y are row vectors
% defining a set of N points uses Lagrange's method to find
% the N-1th order polynomial in X that passes through these
% points. P returns the N coefficients defining the polynomial,
% in the same order as used by POLY and POLYVAL (highest order first).
% Then, polyval(P,X) = Y. R returns the x-coordinates of the N-1
% extrema of the resulting polynomial (roots of its derivative),
% and S returns the y-values at those extrema.
%
% YY = LAGRANGEPOLY(X,Y,XX) returns the values of the polynomial
% sampled at the points specified in XX -- the same as
% YY = POLYVAL(LAGRANGEPOLY(X,Y)).
%
% Example:
% To find the 4th-degree polynomial that oscillates between
% 1 and 0 across 5 points around zero, then plot the interpolation
% on a denser grid inbetween:
% X = -2:2; Y = [1 0 1 0 1];
% P = lagrangepoly(X,Y);
% xx = -2.5:.01:2.5;
% plot(xx,polyval(P,xx),X,Y,'or');
% grid;
% Or simply:
% plot(xx,lagrangepoly(X,Y,xx));
%
% Note: if you are just looking for a smooth curve passing through
% a set of points, you can get a better fit with SPLINE, which
% fits piecewise polynomials rather than a single polynomial.
%
% See also: POLY, POLYVAL, SPLINE
% 2006-11-20 Dan Ellis dpwe@ee.columbia.edu
% $Header: $
% For more info on Lagrange interpolation, see Mathworld:
% http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html
% Make sure that X and Y are row vectors
if size(X,1) > 1; X = X'; end
if size(Y,1) > 1; Y = Y'; end
if size(X,1) > 1 || size(Y,1) > 1 || size(X,2) ~= size(Y,2)
error('both inputs must be equal-length vectors')
end
N = length(X);
pvals = zeros(N,N);
% Calculate the polynomial weights for each order
for i = 1:N
% the polynomial whose roots are all the values of X except this one
pp = poly(X( (1:N) ~= i));
% scale so its value is exactly 1 at this X point (and zero
% at others, of course)
pvals(i,:) = pp ./ polyval(pp, X(i));
end
% Each row gives the polynomial that is 1 at the corresponding X
% point and zero everywhere else, so weighting each row by the
% desired row and summing (in this case the polycoeffs) gives
% the final polynomial
P = Y*pvals;
if nargin==3
% output is YY corresponding to input XX
YY = polyval(P,XX);
% assign to output
P = YY;
end
if nargout > 1
% Extra return arguments are values where dy/dx is zero
% Solve for x s.t. dy/dx is zero i.e. roots of derivative polynomial
% derivative of polynomial P scales each power by its power, downshifts
R = roots( ((N-1):-1:1) .* P(1:(N-1)) );
if nargout > 2
% calculate the actual values at the points of zero derivative
S = polyval(P,R);
end
end

Marco Bertola
Marco Bertola 2023 年 8 月 16 日
One less loop
x = [0 1 2 3 4 5 6 ]; %The nodes
N= size(x,2); %The number of nodes
J=1:N;
P=zeros(1,N);
LL=poly(x);
LLder=polyder(LL);
L=zeros(N);
for j=1:N
L(j,:)= poly(x(J~=j))/polyval(LLder,x(j));
end
%The rows of L are the coeffs of the Lagrange interpolation polynomials; it
%is computed only in terms of the nodes, x. E.g. polyval(L(2,:),x(2))=1 and
%polyval (L(2,:), x(3))=0...
y = [0 .8415 0.9093 0.1411 -0.7568 -0.9589 -0.2794];
P=y*L
P = 1×7
-0.0002 -0.0031 0.0732 -0.3577 0.2255 0.9038 0
%P is the interpolating polynomial: polyval(P,x)=y.
polyval(P,x)
ans = 1×7
0 0.8415 0.9093 0.1411 -0.7568 -0.9589 -0.2794
y
y = 1×7
0 0.8415 0.9093 0.1411 -0.7568 -0.9589 -0.2794

カテゴリ

Help Center および File ExchangeInterpolation についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by