I want to find the parameters in logistic function. I read the guide at here. It very clear to explain. But it did not has final solution that I need. Now, we will consider a basis logistic function $$h_\theta(x^{i})=\frac{1}{1+e^{-\theta_0-\theta_1x}}$$ We want to find $\theta_0$ and $\theta_1$ subject to minimum cost function $$J(\theta)=-\frac{1}{m}\sum_{i=1}^{m}y^{i}\log(h_\theta(x^{i}))+(1-y^{i})\log(1-h_\theta(x^{i}))$$
We use the notation
$$\theta x^i:=\theta_0+\theta_1 x^i_1 $$
Then
$$\log h_\theta(x^i)=\log\frac{1}{1+e^{-\theta x^i} }=-\log ( 1+e^{-\theta x^i} ),$$ $$\log(1- h_\theta(x^i))=\log(1-\frac{1}{1+e^{-\theta x^i} })=\log (e^{-\theta x^i} )-\log ( 1+e^{-\theta x^i} )=-\theta x^i-\log ( 1+e^{-\theta x^i} ),$$
and
$$J(\theta)=-\frac{1}{m}\sum_{i=1}^m \left[y_i\theta x^i-\theta x^i-\log(1+e^{-\theta x^i})\right]=-\frac{1}{m}\sum_{i=1}^m \left[y_i\theta x^i-\log(1+e^{\theta x^i})\right],~~(*)$$
where the second equality follows from
$$-\theta x^i-\log(1+e^{-\theta x^i})= -\left[ \log e^{\theta x^i}+ \log(1+e^{-\theta x^i} ) \right]=-\log(1+e^{\theta x^i}). $$
All you need now is to compute the partial derivatives of $(*)$ w.r.t. $\theta_j$. As
$$\frac{\partial}{\partial \theta_J}y_i\theta x^i=y_ix^i_j$$ $$\frac{\partial}{\partial \theta_j}\log(1+e^{\theta x^i})=\frac{x^i_je^{\theta x^i}}{1+e^{\theta x^i}}=x^i_jh_\theta(x^i), $$
OK, The above steps are correct. But it did not have the solution of $\theta_0$ and $\theta_1$. We can take the derivative and put it equal $0$. Namely,
$$\frac{\partial J(\theta)}{\partial \theta_0}=y-\frac{e^{\theta_0+\theta_1 x}}{1+e^{\theta_0+\theta_1 x}}=0~~(*)$$ $$\frac{\partial J(\theta)}{\partial \theta_1}=yx-\frac{xe^{\theta_0+\theta_1 x}}{1+e^{\theta_0+\theta_1 x}}=0 ~~(**) $$ $(*)$ and (**) are they correct? Assume that $y$ and $x$ are known. How to represent $\theta_0$ and $\theta_1$ by $y$ and $x$ . Thank you so much
