Finding values from Cubicinterp fit

If I have performed a cubic interp to some data,

x = linspace(1,maxFwhm+1,maxFwhm+1)'
f=fit(x,y,'cubicinterp')
hold on
plot(f,'k--')
coeffvalues(f)

How can I now calculate the x value that corresponds with the y value of 0.5?

Thanks Jason

Answer 1

[Steven Lord]

Use fzero to solve f(x) = 0.5. Keep in mind you can evaluate the fit by passing it a value.

load census
populationFit = fit(cdate, pop, 'poly2');
populationIn1925 = populationFit(1925)

You can check that it evaluated the fit correctly.

plot(cdate, pop)
hold on
plot(populationFit)
plot(repmat(1925, 1, 2), ylim)
plot(xlim, repmat(populationIn1925, 1, 2))
plot(1925, populationIn1925, 'o')

[Jason]

Hi, Im not sure I follow how your suggesting I use fzero.

I have tried:

fun=f-0.5;
x0 = [3 5];
myx = fzero(fun,x0)

But didn't work.

[Steven Lord]

You need to make sure you evaluate the fit for a specific value that fzero will pass into your function.

fun = @(x) f(x)-0.5;

[Jason]

Thankyou

Answer 2

[John D'Errico]

Download my SLM Toolbox . Since I don't have your data, I'll make some up. Not a very creative function, I know, but it will suffice.

Here, for which values of x does the function equal 0.5?

x = (0:5)';
y = rand(size(x));
y = rand(size(x))
y =
      0.79221
      0.95949
      0.65574
     0.035712
      0.84913
      0.93399

f=fit(x,y,'cubicinterp');

slmsolve(f.p,0.5)
ans =
       2.1967        3.671

The results are as good as can be had (using the function roots), not an approximation.

[Jason]

Thankyou John for your suggestion