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Reference:alex Graves [supervised Sequence labelling with Recurrentneural Networks]Alex is the most famous variant of Rnn, lstm inventor Jürgen Schmidhuber Gaotu, is now joined University of Toronto, apprentice Hinton.Statistical language model and Sequence learning 1.1 language model based on frequency statisticsThe most famous language model in the field of NLP is N-gram.It is based on the Markov hypothesis, of course, which is a 2-gram (Bi-gram) Model:The probability that any word $w_{i}$ app

= \ phi) $ The independence of random events is the basic premise assumption of various mathematical models. 2. Regularity of random events-Probability 　　Frequency definition: n tests were conducted under the same conditions. In the n tests, the number of times event A occurred $ n_A $ is called the frequency of event, the ratio $ \ frac {n_A} {n} $ is the frequency of event A, and is recorded as $ f_n (A) $ Frequency is not Probability Probabilit

$\mathcal{f}ш$ solution process is as follows$\begin{align*}==\sum_{k=-\infty}^{\infty}\mathcal{f}\varphi (k) \ \=\sum_{k=-\infty}^{\infty}\varphi (k) \qquad (the\ poisson\ sum\ Formula) \ \=\end{align*}$So$\mathcal{f}ш=ш$Fourier transform of $ш_p$First, convert the $ш_p$ into a $ш$ form.$\begin{align*}Ш_p=\sum_{k=-\infty}^{\infty}\delta (X-KP) \ \=\sum_{k=-\infty}^{\infty}\delta (P (\frac{x}{p}-k) \ \=\sum_{k=-\infty}^{\infty}\

1. Set the $A, b,c$ is a subset of the collection $M $, please prove: $$\bex (C\subset A) \wedge (C\subset b) \lra (C\subset a\cap b). \eex$$Proof: clearly established.2. Set the set $X $ meet $\bar{\bar \bbn}\leq \bar{\bar x}$. Please prove: Collection $Y =x\cup\bbn$ meet $\bar{\bar X}=\bar{\bar y}$.Proof: Apparently $\bar{\bar X}\leq \bar{\bar y}$. On the other hand, by $\bar{\bar N}\leq \bar{\bar x}$ know $X $ by a subset $A $, $$\bex y=x\cup\bbn = (X\bs a) \cup (A\CUP\BBN) \sim (X\bs a) \cup

In memory allocation, it is generally necessary to use the memory size based on memory alignment, such as the general 32-bit platform for 4-byte alignment, and 64-bit platform using 8-byte alignment and so on.The general algorithm used is to use the formula first$int (\frac{a + b-1} {b}) $ (where a is the actual memory used and B is the alignment value)Then the corresponding alignment value can be obtained by multiplying the value by B.Formula derivat

Title Link: [CQOI2007] Remainder summationTest instructions: Ask $\sum_{i=1}^{n}k\ mod \ i$The deformation of the formula is more conventional$$\sum_{i=1}^{n}k\ mod \ i=\sum_{i=1}^{n}{(K-\lfloor{\frac{k}{i}}\rfloor *i)}=k*n-\sum_{i=1}^n{\lfloor{\frac{k}{i} \rfloor}*i}$$Note that the $\lfloor{\frac{k}{i}\rfloor}$ of the value of the ladder-like increment, there is

Mathematical Analysis 1. known functions $ f (x) = \ ln x-Ax $, where $ A $ is a constant. if $ f (x) $ has two zeros $ x_1, x_2 $. test Certificate: $ x_1x_2> E ^ 2 $. 2. set the equation $ \ SiN x-x \ Cos x = 0 $ in $ (0, + \ infty) $ to resolve the $ N $ to $ x_n $. proof: $ \ Bex n \ PI + \ cfrac {\ PI} {2}-\ cfrac {1} {n \ PI} 3. discuss the consistency of $ f (x) = x \ SiN x $, $ g (x) = x \ ln x $ on $ [1, \ infty) $. 4. set $ f (x) $ to have a second-order continuous derivative on

algorithm is the same as that of the DNN, that is, by means of the gradient descent method, a suitable RNN model parameter $u,w,v,b,c$ is obtained. Since we are based on time-reverse propagation, the reverse propagation of RNN is sometimes called BPTT (back-propagation through times). Of course, the BPTT and DNN here are also very different, that is, all of the $u,w,v,b,c$ in the sequence are shared, and we update the same parameters when we reverse the propagation.To simplify the description,

;\gamma$ for some $\gamma>0$ and then the training error drops exponentially fast. Nevertheless, because of ITS tendency to focus on training examples that is misclassified, Adaboost algorithm can be quite susceptible to Over-fitt Ing. We'll give a new simple proof of \ref{ada1} and \REF{ADA2}; Additionally, we try to explain what the parameter $\alpha_t=\frac{1}{2}\cdot\log\frac{1-\epsilon_t}{\epsilon_t}$

Study the convergence of the following integrals:(1). $\dps{\int_{-\infty}^{+\infty} x^ne^{-\sex{x^2+\frac{1}{x^2}}}\rd x}$ ($n $ for natural number).(2). $\dps{\int_0^{+\infty} \sin^2\sez{\pi\sex{x+\frac{1}{x}}}\rd x}$.Answer:(1). $$\bex \int_{-\infty}^{+\infty}x^ne^{-\sex{x^2+\frac{1}{x^2}}}\rd x =\int_{-\infty}^{-1}+\int_{-1}^0 +\int_0^1 +\int_1^{+\infty} x^ne

Don't want to work today ... But I'm going to finish today's mission.2. (1) Solution: function $y =e^{\arctan x}$ on the definition field $ (-\infty,+\infty) $ on a continuous, and\[Y ' = E^{\arctan x} \frac{1}{1+x^2}, \qquadY ' = E^{\arctan x} \frac{1}{(1+x^2) ^2}-e^{\arctan x}\FRAC{2X}{1+X^2}.\]On $ (-\infty,+\infty) $ $y "$ exists, but does not keep the same s

This lesson proves that the ratio of the first fit algorithm to bin packing problem is 1.7.Averaging volumeIf there are $n $ items, the volume of each item is 0.51, we can analyze the lower bound of the optimal objective function value at least about $n/2$. But the Nether is too loose (in fact the optimal target function value is $n $) and can only be used to prove that the approximate ratio is 2. How can we prove that the ratio is 1.7?The clever mathematicians somehow thought of the method of "

After PAE is turned on, the structure of the 32-bit linear address has changed, with the following structure30-31-bit: Page directory pointer table index21-29-bit: Page Catalog index12-20-bit: Page table index0-11-bit: in-page offsetAfter PAE is turned on, the address in the table is a physical address, and the size of all table entries becomes 8Byte, in the following format:Analyzing PAE's address conversion mechanism in conjunction with address analysis of numbers in Calc.exe in Windows Server

Why is the area of the circle \ (S = \pi r^2 \)? How to testify?This formula is important because it is the basis for all the volume of the rotating body (cylinder, cone, round table, ball, etc.), it is better to be more rigorous proof, rather than approximate to calculate the end.There may be many, but the well-known "evidence law" is somewhat problematic.The evidence law of primary SchoolCut watermelon slices? Too tight to be convincing:The arc of others is clearly curved, why do you say that

Set $ f \ in C ^ {n + 1} (\ BBR) $, test certificate: For $ \ forall \ A \ In \ BBR $, $ \ Bex \ frac {\ RD ^ n} {\ RD x ^ n} \ SEZ {\ frac {f (x)-f ()} {X-A }}_{ x = A }=\ frac {f ^ {(n + 1)} (a)} {n + 1 }. \ EEx $ Proof: mathematical induction is used for $ N $. when $ n = 1 $, $ \ beex \ Bea \ frac {\ RD} {\ RD x}