# Tag Info

1

Here is a very simple-minded way. First, set $V = \max(|y_i|)$. Then, let $\phi_i =\arcsin(y_i/V)-x_i$. Finally, $\phi =\frac1{n}\sum \phi_i$. Note: If the data is over multiple sinusodial cycles, adjust $\phi_{i+1}$ so it differs from $\phi_i$ by less than $\pi$.

1

For illustration purposes of Marty Cohen's answer, I generated $50$ data points ($i$ from $1$ to $50$), $x_i=i$, $y_i=12.34 \sin (x_i+2.345)+2(-1)^i$ (the error is quite large). The largest absolute value of the $y_i$'s is $14.3129$. From there, the sum of the $\phi_i$'s equals $-1275.08$ so $\phi =-25.5017$; adding $9\pi$, this gives $\phi=2.7726$. Using ...

1

For the first question, the model is $$y=\frac{1}{a+b~x^c}$$ which is nonlinear with respect to all parameters. I suppose that you have $n$ data points$(x_i,y_i)$ based on which you want to adjust parameters $a,b,c$ and, as usual, the problem with nonlinear regression is to start with reasonable values. For the time, generate values $z_i=\frac 1 {y_i}$ ...

1

To see why first lets take an ordinary regression: $y_i=\alpha + X_i'\beta +\epsilon_i \space \space \space \space$ (1) Therefore we have $\hat{y_i} = \hat{\alpha} + X_i'\hat{\beta}$ But by definition, $\hat{\alpha} = \bar{y}-\bar{X}'\hat{\beta}$ and thus $\hat{y_i} = \bar{y} + (X_i-\bar{X})'\hat{\beta} = \bar{y} + X_d'\hat{\beta}$ ...

1

It is certainly possible (in mostly silly cases). For example, suppose you sample the same $x$ value twice and get two experimental $y$ values that differ. One such case might be getting the points $(0, -1)$ and $(0, 1)$ if we sample $x=0$ twice. Then, any line passing through the origin has SSE=2 which is the minimum possible. But of course there are ...

Only top voted, non community-wiki answers of a minimum length are eligible