How to solve polynomial for a number of values x?

Question

Jelle van Zuijlen 2015-2-16

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编辑： Jelle van Zuijlen 2015-2-16

I have a data set with x and y values. (x=discharge and y = water level). Now I made a 3rd order polynomial fit and I need the value x given y for y = 0.00, 0.05, ..., 14.00 How do I solve the polynomial for x and get the x values in a nice matrix?

x = [5965 7867 9459 11763 13881 14794 16000 16619 20000] y = [3.7 6.34 7.04 7.89 8.61 8.91 9.25 9.42 10.19]

plot(x,y,'o')

[p,~,mu] = polyfit(x,y,3); y2 = polyval(p,x,[],mu); hold on plot(x,y2) hold off

h = [0.00:0.05:14.00]; %h are the values of y for which I want to know x

so the result should be a matrix like: [y1 x1 y2 x2 y3 x3 etc]

I'm only interested in real solutions

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Answer 1

Torsten 2015-2-16

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x0=1;
for i=1:length(h)
X(i)=fzero(@(x)polyval(p,x)-h(i),x0);
x0=X(i)
end

Best wishes

Torsten.

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Torsten 2015-2-16

Maybe you have to work with the scaled values:

X(i)=fzero(@(x)polyval(p,x,[],mu)-h(i),x0);

?

Best wishes

Torsten.

Jelle van Zuijlen 2015-2-16

You, Sir, are my hero for today. Many thanks.

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Answer 2

John D'Errico 2015-2-16

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在 MATLAB Online 中打开

Nothing personal, but that cubic polynomial is a relatively poor fit.

See that the cubic has some lack of fit, but that far more importantly, in extrapolation, it starts to do some nasty stuff, that does not seem indicated by your data.

Whereas, with essentially no options chosen at all, my SLM toolbox provides this result:

slm = slmengine(x,y,'plot','on','knots',[3000:3000:24000]);

Which seems to represent the data a bit more cleanly.

The inverse function is also easily done in one call to slmeval.