@nrailgun 2016-03-01T09:33:57.000000Z 字数 833 阅读 1823

UFLDL Summary

机器学习

Linear regression

J (θ) = 1 2 \sum i (θ T x (i) - y (i)) 2

$J(\theta) = \frac 1 2 \sum_i \left(\theta^T x^{(i)} - y^{(i)} \right)^2$

\partial J ( θ ) \partial θ j = \sum i x (i) j (θ T x (i) - y (i))

$\frac {\partial J(\theta)} {\partial \theta_j} = \sum_i x_j^{(i)} \left(\theta^T x^{(i)} - y^{(i)} \right)$

P (y = 1 ∣ x) = h θ (x) = 1 1 + exp ( - θ T x )

$P(y = 1 \mid x) = h_\theta(x) = \frac 1 {1 + \exp(-\theta^T x)}$

P (y = 0 ∣ x) = 1 - h θ (x)

$P(y = 0 \mid x) = 1 - h_\theta(x)$

J (θ) = - \sum i (y (i) log (h θ (x (i))) + (1 - y (i)) log (1 - h θ (x (i))))

$J(\theta) = - \sum_i \left( y^{(i)} \log(h_\theta(x^{(i)})) + (1 - y^{(i)}) \log(1 - h_\theta(x^{(i)})) \right)$

\nabla θ J (θ) = \sum i x (i) (h θ (x (i)) - y (i))

$\nabla_\theta J(\theta) = \sum_i x^{(i)} (h_\theta(x^{(i)}) - y^{(i)})$

Σ = 1 m \sum i m x (i) (x (i)) T

$\Sigma = \frac 1 m \sum_i^m x^{(i)} (x^{(i)})^T$

$U = \left[ u_1, u_2, \dots \right]$ is the eigenvector matrix, $u_1^T x$ is the length of projection of $x$ onto $u_1$ .

$x = U x_{\rm rot}$ , thus $x_{\rm rot} = U^T x$ .