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How to solve polynomial for a number of values x?

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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

Risposta accettata

Torsten
Torsten il 16 Feb 2015
x0=1;
for i=1:length(h)
X(i)=fzero(@(x)polyval(p,x)-h(i),x0);
x0=X(i)
end
Best wishes
Torsten.
  3 Commenti
Torsten
Torsten il 16 Feb 2015
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
Jelle van Zuijlen il 16 Feb 2015
You, Sir, are my hero for today. Many thanks.

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Più risposte (1)

John D'Errico
John D'Errico il 16 Feb 2015
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.
  1 Commento
Jelle van Zuijlen
Jelle van Zuijlen il 16 Feb 2015
Modificato: Jelle van Zuijlen il 16 Feb 2015
Can you explain how you achieved that? What is this slmengine ? I'm new to Matlab so I don't know about this
Edit: I've downloaded your slmengine package, will check if I can come to this fit.
Thanks anyways

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