There are several issues here.
1. Using this form is a bad idea:
Parameters = inv(G'*G) * G'* d;
Sigh. It gets used over, and over again. And people still teach others to do it. Instead, use this:
Parameters = G\d;
MATLAB has the ability to solve a least squares problem with a non-square matrix, and do so efficiently. You need not turn it into a square matrix so you can use inv on it. That is a BAD idea numerically.
When you use the form that you did, it squares the condition number of the problem. So even if your problem is only moderately poorly conditioned to start with, squaring that now makes it impossible to solve. You get numerical garbage.
2. There are at least two circumstances where you can get a negative R^2. One is if you have no constant term in the model. As I recall, one way to describe what R^2 tells you, is how well your model performs, when compared to a model that has only a constant term. That is, if the default model was just to take the mean of your data and use that for prediction, then does your chosen model work better?
https://en.wikipedia.org/wiki/Coefficient_of_determination
So, as long as your model includes a constant term, then all such models must in theory be better predictors than a model that uses only the mean of your data, because that is in effect a constant term.
If you read the link I show above, you will find a form for R^2 as:
R^2 = 1 - ss_res/ss_tot
That is, ss_tot is the sum of squares that results from effectively treating your model as the mean, a constant term. ss_res is the sum of squares that results from your model. So if ss_tot is LESS than ss_res, that result will be NEGATIVE.
I'll give an example where it will be clear that you can easily get a negative R^2.
n = 100;
x = rand(n,1);
y = rand(n,1);
mu_y = mean(y); % a model that assumes a constant result.
% a model with a constant term has a column of 1's in it.
my_model = [x,x.^2]\y;
ss_tot = sum((y - mu_y).^2);
ss_res = sum(([x,x.^2]*my_model - y).^2);
R2 = 1 - ss_res/ss_tot
R2 =
-0.44063
So, I VERY carefully chose data for which the only good predictor is the constant y=0.5. No polynomial signal should exist at all.
However, my model was:
y = a1*x + a2*x^2
Clearly that will never fit better than the model that just presumed a constant term. As well, it has no constant term in it. Therefore? we get a negative R^2.
Having said all that, that it is easy to get a negative R^2, your model has a constant term in it.
That tells me, along with your comment that the model yields a singular matrix, that the second reason why it came out negative, was because your model is numerical GARBAGE. The matrix is singular. So you cannot possibly expect reasonable results from it. The computation will be pure garbage, therefore? Negative R^2 can result.
3. Next, what are you doing? Why is your model producing garbage in the first place? Lets look at what you did:
for period = 0:0.5:1
% Assigning the data provided, sealevel to "d"
d = sealevel{1};
t = timesea{1};
% Finding the size of the data d
[Rdata , Cdata] = size(d);
...
% Creating third variable for equation
thirdco = cos((2*pi/period) * t);
What happens when period == 0??????
Garbage. What happens when you divide by zero? inf results. Your matrix will be singular. Your model will be worthless. Your R^2 will probably be negative. Lots of bad things will happen.
I won't go much further on this, because we now know that bad things are happening.
