# Cosine model design matrix non-lineal model

I need to applied the T- student testing to the parameters $$\beta_0, \beta_1, \beta_2,\beta_3,\beta_4$$ which model is:

$$y = \beta_0 + \beta_1 t + \beta_2 Cos(\beta_3 t+\beta_4)$$

to do that, I need the design matrix X, and then $$(X'X)^{-1}$$ to get $$c_{jj}$$ diagonal elements to find $$Se(B_i)$$

I have tried to take the matrix like:

$$$$\beta_0 \quad \beta_1 \quad \; \beta_2 \quad \quad \beta_3 \quad \beta_4$$$$

$$$$\begin{bmatrix} 1 & t_1 & Cos(t_1)& t_1 & 1\\ 1 & t_2 & Cos(t_2)& t_2 & 1\\ \vdots & \vdots & \vdots& \vdots& \vdots\\ 1 & t_n & Cos(t_3)& t_n & 1 \end{bmatrix}$$$$

or even;

$$$$\beta_0 \quad \beta_1 \qquad \quad \beta_2 \qquad \quad \beta_3 \quad \beta_4$$$$

$$$$\begin{bmatrix} 1 & t_1 & Cos(\beta_3 t_1 + \beta_4)& t_1 & 1\\ 1 & t_2 & Cos(\beta_3 t_2 + \beta_4)& t_2 & 1\\ \vdots & \vdots & \vdots& \vdots& \vdots\\ 1 & t_n & Cos(\beta_3 t_3 + \beta_4)& t_n & 1 \end{bmatrix}$$$$

but with that form of the matrix is not possible to calculate $$(X'X)^{-1}$$

• I doubt that the way you take in your question would solve the problem. I could propose another method simpler and more robust. But I need to check if it is compatible with your model equation according to your kind of data. To check I need at least one example of data (numerical data possibly reduced to be not too big). Mar 12, 2023 at 13:55