So, you have this equation:
y = 2*t*(1-cos(x)*(n-1)/(n*p-p*(1-cos(x)))
I've rewritten it above, putting in multiplies where you used . or you used nothing at all. Thus cos(x)(n-1) is not a valid expression. You should get used to writing things so they are readable.
The problem is, this is not valid either, as I wrote it, since the parens are unbalanced. See that there are 6 ( characters in that line, and only 5 ) in there. I never added or deleted any parens. That means your expression makes no mathematical sense.
IF you show me how the parens should be written, then I can show you how to compute n. For example, should it have been this:
y = 2*t*(1-cos(x)*(n-1))/(n*p-p*(1-cos(x)))
or this, where the extra parens was added at the end?
y = 2*t*(1-cos(x)*(n-1)/(n*p-p*(1-cos(x))))
The two expressions are significantly different.
In the first case, solving for n, I would get this relation:
n = (2*t*(1 + cos(x)) + p*y*(1 - cos(x)))/(p*y + 2*t*cos(x))
In the second case (assuming my algebra was correct at this time of day) I might have gotten this:
n = (p*y*(1 - cos(x)) + 2*t*cos(x) - 2*p*t*(1 - cos(x)))/(p*y - 2*p*t + 2*t*cos(x))
But you have multiple values for x and y, thus pairs of points such that these equations should hold true.
So load in your data.
load F:\michel.dat
x=michel(:,1);
y=michel(:,2);
Then compute the mean, but in those expressions, use ./ instead of /, and .* instead of * in the correct expression. The dotted operators are element-wise multiplies and divides.
You don't need .* between two scalar values, but whenever you have vectors in the expression, you need to use the element-wise operators.
So in the first expression, I would get:
n = mean((2*t*(1 + cos(x)) + p*y.*(1 - cos(x)))./(p*y + 2*t*cos(x)));
In the second case, it should be this:
n = mean((p*y.*(1 - cos(x)) + 2*t*cos(x) - 2*p*t*(1 - cos(x)))./(p*y - 2*p*t + 2*t*cos(x)));
The idea is that the simplest estimator of n is simply the arithmetic mean of all of those possible results for n, given all the sets of x and y.
In fact though, if your original equation had noise in it, I pulled a fast one, because the best estimator of n should really be estimated using a nonlinear least squares estimate, based on the original equation. We would get a subtly different result from computing the mean as I did above.
Had you attached your data as a .mat file to a comment, or to your original question, I could now show how to do all of this more properly as a nonlinear least squares estimation. But first, tell me where the extra paren was supposed to be, as I don't want to do it all twice.
