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minimum point, but rather hover around a minimum point.Below, the objective function is solved. The first is biased for W and b respectively:$\TRIANGLEDOWN_WL (w,b) =-\sum \limits_{x_i \in m}y_ix_i$$\TRIANGLEDOWN_BL (w,b) =-\sum \limits_{x_i \in m}y_i$Randomly select a point of error to update W and B:$w \gets W+\eta y_ix_i$$b \gets B+\eta y_i$The algorithm ends until there are no missed points in the training set.Logistic regression :Target function:$min \ L (w) $$L (W) =-\sum_{i=1}^{n}[y_i* (

^{(L)}-y) ^2\) . (PS: Here l=4, some textbooks calculate mean square error when multiplied by \ (1/2\))Then we want to solve the gradient of \ (\omega\) and \ (b\) .Here is an example of \ (\frac{c_0}{\partial \omega^{(L)}}\) :To find the gradient, which is the sensitivity of the cost function to the change of parameters.It can be found that changing \ (\omega^{(l)}\)first affects the \ (z^{(l)}\), then affects the \ (a^{(l)}\), and finally affects \

is 0.5, the positive and negative classes can be separated according to the vertical bar of the magenta, no problem;However, when adding a sample, in the Green Fork, the regression line becomes a green linear, when the selection of 0.5 is a threshold, the above 4 Red forks (positive Class) into the negative class inside, the problem is very large;In addition, in the two classification problem, y=0 or y=1, and in linear regression, $h _\theta (x) $ can be greater than 1 or less than 0, which is

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the program by observing the time to complete a batch of tasks under different conditions.This is the key to the task itself is the processing logic, since we are talking about the CPU load, the task must be CPU-intensive tasks. Then, the processing time of a single task should not be too short, otherwise the scheduling process will become the bottleneck of the program, reflecting the CPU load problem; On the other hand, the processing time of a single task should not be too long, otherwise the

1.(1) Twice corner re-division integral(2)\[\int (\arcsin x) ^2 dx = (\arcsin x) ^2 x-2\int \frac{x}{\sqrt{1-x^2}} \arcsin x dx\]and\[\int \frac{x}{\sqrt{1-x^2}} \arcsin x dx=-\int \arctan x D (\sqrt{1-x^2}) =-\arcsin x \sqrt{1-x^2}-+\int dx=-\arcsin x \sqrt{1-x^2} +x +c.\]So\[\int (\arcsin x) ^2 dx= (\arcsin x) ^2 x+2\arcsin x \sqrt{1-x^2} -2x +c.\](3)\[\int x \tan^2 x dx = \int x (\sec^2 x-1) dx= \int x D

1. Point out the order of the following equation and determine whether it is linear, or nonlinear, if it is linear, indicating whether it is homogeneous or non-homogeneous.(1). $u _t-(U_{xx}+u_{yy}) +1=0.$Solution: This is the second order linear non-homogeneous equation.(2). $u _t-u_{xx}+xu=0$.Answer: This is the second order linear homogeneous equation.(3). $u _t-u_{xxt}+uu_x=0$.Answer: This is a third-order semilinear equation.(4). $u _x^2+uu_y=0$.Solution: This is a first order complete nonl

Inverse propagation algorithmWe already know how forward propagation is calculated in the previous section. That is, given how the x calculates y through each node.Then there is the question of how we can determine the weights of each neuron, or how to train a neural network.In the traditional machine learning algorithm we use the gradient descent algorithm to do the weight update:\[\theta_j:=\theta_j-\alpha\frac\delta{\delta\theta_j}j (θ) \]Even if t

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Tags: blog Io OS AR for SP Div on Art $ \ Sum _ {n = 1} ^ {\ infty} \ frac {1} {n ^ s} = \ prod _ {P \ In \ mathcal {p }}\ frac {1} {1-P ^ {-s }}. $ the Cauchy-Schwarz inequality \ [\ left (\ sum _ {k = 1} ^ n A_k B _k \ right) ^ 2 \ Leq \ left (\ sum _ {k = 1} ^ n A_k ^ 2 \ right) \ left (\ sum _ {k = 1} ^ n B _k ^ 2 \ right) \] A cross product formula \ [\ Mathbf {v} _ 1 \ times \ mathbf {v} _ 2 = \

) d; -(Void) setDenominator: (int) d; -(Int) numerator; -(Int) denominator; -(Void) setNumerator: (int) n ddd: (int) d; -(Void) setNumerator: (int) n: (int) d :( int); // Here, three setNumerator functions are reloaded. @ End (2) Fraction. m #import"Fraction.h" #import @implementationFraction -(void)print{ printf("%i/%i",numerator,denominator); } -(void)setNumerator:(int)n{ nnumerator=n; } -(void)setDenominator:(int)d{ ddenominator=d;

{eqalign}& \frac{a}{b}=a_0+\frac{r_1}{b} \\& \frac{b}{r_1}=a_1+\frac{r_2}{r_1} \\& \frac{r_1}{r_2}=a_2+\frac{r_3}{r_2} \\& \cdots \\& \frac{r_{N-2}}{r_

) \) is relatively simple and worth discussing. \ (q (x) \) has an obvious combination of meaning, each coefficient of absolute value is partial division of the number of divisions, the meaning of the symbol is the number of parts of odd and even time division of the difference. It is not difficult to obtain the (q (m) \) satisfying formula (4) by using this combinatorial meaning to assist the discussion (the process is shown in the textbook). The recursive relation (5) of the \ (p (m) \) is the

many problems, such as we want to sample the distribution defined in \ ([0,\infty) \).To enable the use of asymmetric recommended distributions, the Metropolis-hastings algorithm introduces an additional modifier (c\), defined by the recommended distribution:\ (c = \frac{q (x^{(t-1)}|x^*)}{q (x^*|x^{(t-1)})}\)The correction factor adjusts the transfer operator so that the transfer probability of the \ (x^{(t-1)}\rightarrow x^{(t)}\) is equal to the t

Set $k, i,j$ are natural numbers, and $k =i+j$, try to find the series $\dps{\vsm{n}\frac{1}{(Kn-i) (kn+j)}}$.Solution: Former $N $ for the original series and for $$\beex \bea \sum_{n=1}^n \frac{1}{(kn-i) (kn+j)} =\frac{1}{k}\sum_{n=1}^n \sex{\frac{1}{kn-i}-\ FRAC{1}{KN+J}}

Note:$\alpha$ and $\beta$ are known to be commonly used (unlike the LDA EM algorithm)1. Why is it availableLDA The goal of model solving is to get $\phi$ and $\theta$ Assuming that the subject of each word is now known $z $, you can obtain the $\theta$ distribution, and expect $E (\theta) $ as the subject of each document$E (\THETA_{MK}) =\frac{n_m^k+\alpha_k}{n_m+\alpha_k}$Similarly, the posterior distribution of $\phi$ can be obtained, expecting $

the iterative updating of the e-step and M-step of several rounds, we can get the suitable approximate hidden variable distribution $\theta,\beta, z$ and the $\alpha,\eta$ of the model posteriori parameters, and then we get the LDA document topic distribution and the keyword distribution we need.It is shown that to fully understand LDA's variational inference em algorithm, it is necessary to clarify the process of it in the course of the e-step variational inference and the EM algorithm after t

In the vectorization section of the second week, the vectorization process of the gradient descent method is not very clear at first, and it was later deduced and recorded here.The following is the parametric recursive formula for gradient descent (assuming n=2):Equation 1:$\theta_0: = \theta_0-\alpha \frac{1}{m}\sum_{i=1}^{m} (H_\theta (x^{(i)})-y^{(i)}) x^{(i)}_0$$\theta_1: = \theta_1-\alpha \frac{1}{m}\s

Ducomet, Bernard; Ne?asová,šárka; Vasseur, Alexis. On global motions of a compressible barotropic and selfgravitating gas with density-dependent viscosities. Z. Angew. Math. Phys. (+), No. 3, 479--491.by Eq. (+), we see readily that the authors concerns about the finite mass case, and thus $$\bex \int, \rho\rd x\leq C. \e ex$$ Moreover, $$\bex \int \rho \ln \rho =\int \rho \frac{1}{\ve}\ln \rho^\ve \leq \int \fra