plotting a natural cubic spline

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Justin Howard
Justin Howard 2020 年 3 月 24 日
コメント済み: Justin Howard 2020 年 3 月 26 日
Hello, im a bit confused on how to approach this problem im supposed to find and plot the cubic spline S satisfying S(0) = 1,S(1) = 3,S(2) = 3,S(3) = 4,S(4) = 2 and with S''(0) = S''(4) = 0.
  2 件のコメント
darova
darova 2020 年 3 月 24 日
What does S(0) mean? Can you make a simple drawing? DO you have a picture or something?
Justin Howard
Justin Howard 2020 年 3 月 24 日
it'
it's refering to the function/equation we're trying to find. like the function S(x)=1 when x is zero.

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回答 (2 件)

John D'Errico
John D'Errico 2020 年 3 月 24 日
編集済み: John D'Errico 2020 年 3 月 24 日
Are you allowed to use existing code? So if you have the curve fitting toolbox, you can just use csape.
x = 0:4;
y = [1 3 3 4 2];
spl = csape(x,y,'variational')
spl =
struct with fields:
form: 'pp'
breaks: [0 1 2 3 4]
coefs: [4×4 double]
pieces: 4
order: 4
dim: 1
fnplt(fnder(spl,2))
yline(0);
As you can see, the second derivative as plotted is zero at the ends, and the second derivative curve is piecewise linear. So csape did as was needed, producing a natural cubic spline interpolant.
Plotting the plsine is pretty easy too.
fnplt(spl)
hold on
plot(x,y,'ro')
However, I would not be remotely surprised if your question is to actually formulate the equations and then solve for the spline yourself. After all, this is surely homework. So is that your need? (I'm not at all sure why you would be doing this for any other purpose than homework.)
If your need is to formulate the plsine completely from scratch, then my hope is you are not asking someone to do that part for you. It is easy enough to do anyway. Start by reading here:
  4 件のコメント
Justin Howard
Justin Howard 2020 年 3 月 26 日
ah okay got it thank you.
Justin Howard
Justin Howard 2020 年 3 月 26 日
I know how to do it by hand without the use of matlab , but our professor wanted us to redo it on matlab, and i just wasn't sure where to start to properly construct on there.

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Robert U
Robert U 2020 年 3 月 24 日
Hi Justin Howard,
I guess there is a mistake concerning the second derivative at the start and end points of the requested spline interpolation. Usually just the slope is given which would be denoted S'.
You can use the function spline() to provide a cubic spline interpolation of the given points including the slopes at start and end points.
x = [0,1,2,3,4];
y = [0,1,3,3,4,2,0];
x_spline = 0:0.1:4;
y_spline = spline(x,y,x_spline);
fh = figure;
ah = axes(fh);
hold(ah,'on')
plot(ah,x,y(2:end-1),'o');
plot(ah,x_spline,y_spline)
Kind regards,
Robert
  2 件のコメント
John D'Errico
John D'Errico 2020 年 3 月 24 日
Robert - actually, this is a request for a NATURAL cubic spline. This is the case where the SECOND derivative is forced to zero at each end point, not the first derivative. Here the name natural probably arises from the calculus of variations, where the name natural end conditions are exactly as indicated.
Robert U
Robert U 2020 年 3 月 24 日
Ah. Thanks for the clarification.

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