Plotting Newton Interpolating Polynomial

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Zayd Islam
Zayd Islam 2021-3-19
评论： Zayd Islam 2021-3-19
Currently I have these values for x & y and the calculated coefficients (z). Of a newton polynomial
I have attached an image of the function i want to calculate, can someone help me put this function in a nested loop format. And then that function can be plotted aswell. also this loop and function cretaed should be feasible and therefore useable for any degree polynomial
In the image y1 y21 y321 y4321 and y54321 is the same as z(1) z(2) z(3) z(4) z(5) and in the image x is just x-axis on graph
also in the image x1 x2 x3 x4 corresponds to x(1) x(2) x(3) & x(4).
Thank You for your help
x = [-1.9021 -1.1756 0 1.1756 1.9021]
y = [0.0268 0.2362 0.5 0.2362 0.0268]
z = [0.0268 0.2882 -0.0335 -0.0511 0.0269]

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回答（2 个）

Alan Stevens
Alan Stevens 2021-3-19
Do you mean something like this?
xconsts = [-1.9021 -1.1756 0 1.1756 1.9021];
y = [0.0268 0.2362 0.5 0.2362 0.0268];
yconsts = [0.0268 0.2882 -0.0335 -0.0511 0.0269];
x = linspace(-1.9021,1.9021,5);
ypoly = zeros(1,numel(x));
for j = 1:numel(x)
ypoly(j) = npoly(x(j),xconsts,yconsts);
end
disp(ypoly)
subplot(2,1,1)
plot(x,ypoly,'o-'),grid
xlabel('x'),ylabel('y from polynomial')
subplot(2,1,2)
plot(x,ypoly-y,'*-'),grid
xlabel('x'),ylabel('y-ypoly')
function y = npoly(x,xconsts,yconsts)
xi = 1;
y = yconsts(1);
for i = 1:(numel(xconsts)-1)
xi = xi*(x-xconsts(i));
y = y + yconsts(i+1)*xi;
end
end
  2 个评论
Zayd Islam
Zayd Islam 2021-3-19
not entirely Alan anychance you could join a meeting on zoom or teams with me?
this is for my coursework project by the way
Alan Stevens
Alan Stevens 2021-3-19
Can you explain why this isn't what you were looking for?

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Zayd Islam
Zayd Islam 2021-3-19
Hi Alan basically what i was looking for was a function to take in my X and Y values and the newton coefficients (Z) and of course the degree of the polynomial which is 1 less the number of data points so in this case it was degree 4. I would like the function to return a polynomial with a function handle so like this. Hope you understand what i mean :)
%o is just any letter
yourpolynomial_is = @(o) z(1) + z(2)*(o-x(1)) etc etc etc etc
  2 个评论
Alan Stevens
Alan Stevens 2021-3-19
I've not written the function as an anonymous function, but it might still work the way you want I think.
If I understand you correctly, you want a symbolic expression returned. If you define your o as syms o, the function I've written might produce what you want. Unfortunately, I don't have the symbolic toolbox, so I can't check that.

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