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-1}\cdot Dx=\int_0^1x^n\cdot Dx=\frac 1{n+1}\]Multiply the number of combinations to get the answer to this simplification question for \ (\frac n{n+1}\)ExtendedWe have now obtained the value of the maximum (number of n\ ) is expected to be \ (\frac n{n+1}\), the same can be calculated as the minimum number (number of ( 1\) of the numerical value is expected to b

Link to original topicThe LCM (b) Represents the least common multiple of a and B, and A (n) represents the average of the LCM (n,i) (1 For example: A (4) = (LCM (1,4) + LCM (2,4) + LCM (3,4) + LCM)/4 = (4 + 4 + 12 + 4)/4 = 6. F (A, b) = A (a) + a (A + 1) + ... A (b). (F (A, B) =∑a (k), a For example: F (2, 4) = A (2) + A (3) + A (4) = 2 + 4 + 6 = 12. given a, B, calculate F (A, b), as the result may be large, output f (A, B)% 1000000007 results can be. Should we say that this is a better way to

No Shadow Random ThoughtsDate: January 2016.Source: http://www.zhaokv.com/2016/01/softmax.htmlDisclaimer: Copyright, reprint please contact the author and indicate the sourceSoftmax is one of the most common output functions in machine learning, and there is a lot of information on the web about what it is and how it is used, but there is no data to describe the rationale behind it. This paper first briefly introduces the Softmax, and then focuses on the mathematical analysis of the principle be

Section I.1-4: Slightly5. Note: It is generally proposed that a $1/n$, and then observe\[\lim_{n\to \infty} (\frac{n}{n^2+1} +\frac{n}{n^2+2^2}+\cdots + \frac{n}{n^2+n^2})=\lim_{n\to \infty} (\frac{1}{1+1/n^2} +\frac{1}{1+ (2/n) ^2} +\cdots +\

): \ [{J _ \ theta} (X) =\ frac {1 }{{ 2 m }}\ sum \ limits _ {I = 1} ^ m {{{ ({h _ \ theta} ({x ^ {( I )}}) -{y ^ {(I )}})} ^ 2 }}=\ frac {1} {2 m }}\ sum \ limits _ {I = 1} ^ m {{{ ({\ theta ^ T} \ bullet {x ^ {(I )}} -{y ^ {(I )}})} ^ 2 }}=\ frac {1} {2 m }}\ sum \ limits _ {I = 1} ^ m {{{ ({x ^ {(I) T }}\ bullet \ theta-{y ^ {(I)} ^ 2 }}\] We can also use the

. Because the infinite connected fraction is infinite, we cannot directly calculate its results. To calculate the approximate result of an infinite continuous fraction, the simple method is to calculate the first K items, as shown below: In this way, the computation of an infinite connected fraction can be completed through K-times. Of course, the computation result is an approximate number rather than an accurate number. The question requires us to complete a process named cont-

, and a system became moreis orderly, the information entropy is lower, conversely, the more chaotic a system, the higher the information entropy. So information entropy can be thought of as an orderly system.A measure of the degree of\[h (x) =-\sum_{i=1}^{n} p_{i} log_{2} p_{i} \]Third, information gain information GainThe information gain is for one characteristic, that is, to see a characteristic, the system has it and the amount of information when it is not, bothThe difference is the amount

Topic Link: The Beauty of circulationThis question feels very beautiful ... It's not easy to have such a beautiful face and a higher level of thinking ...For the sake of convenience, let me first speak about two symbols. $[a]$ indicates that if $a$ is true, return $1$, otherwise return $0$; $a \perp b$ represents $a$ and $b$ coprime.First, we need to consider what conditions a score needs to be met to become pure repeating decimal.Let's think about how we use division to determine whether a frac

Logistic regressionTypically a two-tuple classifier (also available for multivariate classification), such as the following classification problems Email:spam/not spam Tumor:malignant/benign Suppose (hypothesis): $ $h _\theta (x) = g (\THETA^TX) $$ $ $g (z) = \frac{1}{1+e^{-z}}$$ where g (z) is called the sigmoid function, and its function image is shown, you can see the predicted value $y$ The value range is (0, 1), so for $h _\thet

function Grammar Show Not good. (\frac{1}{2}) A little better. \left (\frac{1}{2} \right) You can use \left and \right to display different parentheses: function Grammar Show parentheses, parentheses \left( \frac{a}{b} \right)

One.1.22. $1+\frac{1}{e^2}$3. $y =\frac13 e x$4. $y =\frac \pi 4$Two. 1. A 2. D 3. C 4. B 5. BThree.1. Left limit\[\lim_{x\to 0^-} \frac{-\sin x}{x}=-1,\]Right limit\[\lim_{x\to 0^+} \frac{\sin x}{x}=1,\]So the original limit does not exist.2.\[\frac{dy}{dx}=-e^{2t}-te^{2t},

. This polynomial base is designed like this:$ $l _j (t) =\frac{(t-t_1) (t-t_2) ... (T-t_{j-1}) (T-t_{j+1}) ... (T-t_n)} {(t_j-t_1) (t_j-t_2) ... (T_j-t_{j-1}) (T_j-t_{j+1}) ... (T_j-t_n)} =\frac{\prod\limits_{k=1,k\neq j}^n (T-t_k)}{\prod\limits_{k=1, K\neq j}^n (t_j-t_k)}$$ so there are:$ $l _j (t_i) =\delta_{ij}, i,j=1,2,... n $$ where $\delta$ is Kronecker (Kronecker) notation, when two subscripts are

proof $\dps{\int_0^\infty f\sez{\sex{ax-\frac{b}{x}}^2}\rd x=\frac{1}{a}\int_0^\infty f (y^2) \rd y}$ (where left and right integrals exist, and $A, b>0$). Proof: $$\beex \bea i\equiv \int_0^\infty f\sez{\sex{ax-\frac{b}{x}}^2}\rd x\\ =\int_0^\infty f\sez{\sex{\frac{ B}{t}-at}^2}\f