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}} '}^{i}}={{y}^{ i}} ({{{{w} '}^{t}}{{x}^{i}}+{b} ') =k{{\hat{\gamma}}^{i}}$.In addition, the function interval for defining a collection of $m$ samples is $\hat{\gamma}=\underset{x}{\mathop{\min}}\,{{\hat{\gamma}}^{i}}$, which is the smallest function interval in all samples.The meaning of the set interval is much clearer, which is the geometrical distance of the sample to the ${{w}^{t}}x+b=0$ plane.Point A coordinate is recorded as Vector ${{x}_{a}}$, it to the plane ${{w}^{t}}x+b=0$ projecti

1. Solution: Function $y = (x-1) \cdot \sqrt[3]{x^2}$ within the defined domain $ (-\infty,+\infty) $, and\[F ' (x) = \frac53 x^{2/3}-\frac23 X^{-1/3}.\]Make $f ' (x) =0$ $x =\frac25$, and notice that when $x =0$, the function is not directed. The list is examined as follows:\begin{center}\begin{tabular}{|c|c |c | c | c|c |}\hline$x $ $ (-\infty,0) $ 0 $ (0,\frac25) $ \mbox{$\frac25$} $ (\frac25,+\infty) $ \\[2ex]$f ' (x) $ + \mbox{not present} $-$ 0 + \\[2ex]$f (x) $ $\nearrow$ \mbox

1. (1) 0 (2) 0(Parity and periodicity)2.(1) because $x >x^2$, $x \in (0,1) $, so $e ^x > e^{x^2}$, so\[\int_0^1 e^x dx > \int_0^1 e^{x^2} dx.\](2)\[\INT_0^{2\PI} x\sin x dx = \int_0^{\pi} x\sin x dx + \int_\pi^{2\pi} x\sin x dx\]Notice that the second integral on the right changes from $\pi$ to $2\pi$, the absolute value of $\sin x$ and the change from $\pi $ to $0$, but the sign is the minus sign, and at this point the $x $ is obviously larger than the $x $ of the original range, so the integra

4. (G.M. krause) to $ \ bex \ lm_1 = 1, \ quad \ lm_2 = \ frac {4 + 5 \ sqrt {3} I} {13 }, \ quad \ lm_3 = \ frac {-1 + 2 \ sqrt {3} I} {13 }, \ quad v = \ sex {\ sqrt {\ frac {5} {8 }}, \ frac {1} {2 }, \ sqrt {\ frac {1} {8 }}^ T. \ eex $ then make $ \ bex A = \ diag (\ lm

indefinite equation such as \ (ax + by = c\)First of all, if you have a little thought, you have to find that the condition must be satisfied \ (c \mid \ gcd (A, b) \) The original equation has an integer solutionTherefore, we can turn the primitive into a more easily discussed form\ (AX+BY=GCD (a, b) \ \ (a>=b) \)Because of the above thinking, it is not difficult to find\ (\because gcd (b) =gcd (b,a\ mod\ b) \)Ditto, we still discuss its special case, when \ ( b=0\) , that is \ (ax + by=a\) ,

derivatives, Limits, sums and integrals The expressions are obtained in LaTeX by typing \frac{du}{dt} and \frac{d^2 u}{dx^2. The mathematical symbol is produced using \partial. Thus the Heat equation is obtained in LaTeX by typing \[\frac{\partial u}{\partial T} = h^2 \left (\frac{\partial^2 u}{\partial x^2}

statistical description of the dataIn order to grasp the full picture of the data, we focus on the data center trend measurement, data dissemination and graphic display.Center Trend MeasurementCenter trend measures the central or central location of the data distribution, or, given an attribute, where does the majority of its values fall?Mean value (mean)The most common and most effective are the arithmetic mean values:\[\overline{x} = \frac{\sum_{i=

1. Calculate the following limits (7 points per sub-topic, total 28 points)(1). $\displaystyle \lim\limits_{x\to 0} \frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^{2}x}$.(2) $\displaystyle \lim\limits_{n\to \infty} \left (\frac{1^{p}+2^{p}+\cdots +n^{p}}{n^{p}}-\frac{n}{p+1}\right) $,$ ( P\in N,p\ge 1) $.(3) $\displaystyle \lim\limits_{x\to +\infty}\

Candidates Note:1. This paper is divided into paragraphs Ⅰ and Ⅱ, Volume I is a choice, and Volume II is a non-choice question.2. This paper is full of 100 points and the exam time is 90 minutes.3. Please fill in the answers to the answer volume corresponding position, the exam is over, candidates only hand in the answer book.Volume Ⅰ (total 40 points for the selected question)First, a total of 10 small questions, each small topic 4 points. In each of the four options given in the question, ther

${\large 1.} $ COMPUTE Limit $$\lim\limits_{n\rightarrow\infty}\frac{[1^p+2^p+\cdots+ (2n-1) ^p]^{q+1}}{[1^q+2^q+\cdots+ (2n-1) ^q]^{p+1}}$$${\BF Solution:}$\begin{align*}Make \lim\limits_{n\rightarrow\infty}\frac{[1^p+2^p+\cdots+ (2n-1) ^p]^{q+1}}{[1^q+2^q+\cdots+ (2n-1) ^q]^{p+1}}=e^a\\Then a=\lim_{n\rightarrow\infty}[(q+1) \ln (1^p+2^p+\cdots+ (2n-1) ^p)-(q+1) \ln (1^q+2^q+\cdots+ (2n-1) ^q)]\\=\lim_{n\r

Stolz theorem: If $ (1) Y _ {n + 1}> Y_n \ qquad (n = 1, 2, \ cdots); \ (2) \ lim \ limits _ {n \ To \ infty} Y_n = + \ infty; $ $ (3) \ lim \ limits _ {n \ To \ infty} \ frac {X _ {n + 1}-X_n} {Y _ {n + 1}-Y_n} $, Then $ \ lim \ limits _ {n \ To \ infty} \ frac {x_n} {Y_n} = \ lim \ limits _ {n \ To \ infty} \ frac {X _ {n + 1}-X_n} {Y _ {n + 1}-Y_n }. $ Cer

R} r^pc^q\]\ (m_{0,0}\) is the area of the regionWhen \ (p+q\ge1\) . Moment divided by area to get normalized momentThe normalized moment is determined by the position of the image. It is often useful to make features that do not change depending on the location of the area in the image. These center moments are implemented by the following formula at \ ((p+q\ge2) \) :\[\mu_{p,q}=\frac{1}{a}\sum (r-n_{1,0}) ^p (c-n_{0,1}) ^q\]The second-order center

}\) is the Beatty series .Below is a description of the following Beatty sequence and beatty theorem :Take positive irrational number \ (\alpha,\beta\), make \ (\frac{1}{\alpha}+\frac{1}{\beta}=1\)Constructs two series \ (a_{n},b_{n}\), their entries are \ (a_{n}=\lfloor{\alpha n}\rfloor,b_{n}=\lfloor{\beta n}\rfloor\)Then the sequence is obviously a positive integer, and theBeatty theorem states that two s

Environment Settings Complete environment settings by referring to the Titanium command line tool. Create a module To create a module, we need to pass some parameters to the titanium command line tool. Module name ($ MODULE_NAME) and ID ($ MODULE_ID) Target running platform of the module (android) Android SDK installation directory ($ ANDROID_SDK) such as/opt/android-sdk For an Android system, we create a module with the following parameters:Reference Titanium create -- pl

In this paper, there are several common gradient-based methods in unconstrained optimization, mainly gradient descent and Newton method, BFGS and L-BFGS algorithm, the problem form of unconstrained optimization is as follows, for $x \in \mathbb{r}^n$, the objective function is: \[\MIN_XF (x) \] Taylor series The gradient-based approach involves the Taylor series, a brief introduction to the Taylor series, which means that the function $f (x) $ in the neighborhood of the point $x _0$ has $n +1$ d

Let $A =\{a_1$$\overline{\varepsilon} (A) =\limsup\limits_{n\to \infty}\frac{\log| a\cap\{1,2,\cdots,n\}|} {\log n}.$$We can similarly define the lower exponential density.We have the following result claimed in gaps and the exponent of convergence for an integer sequence.Main Result: $$\overline{\varepsilon} (a) =\tau (a) =\limsup\limits_{n\to \infty}\frac{\log n}{\log a_n}.$$Proof. Let $\overline{\varepsi

1. $$\bex \al\in\bbr\ra \int_0^\infty \frac{\rd x}{(1+x^2) (1+x^\al)}=? \eex$$Answer: $$\beex \bea \int_0^\infty \frac{\rd x}{(1+x^2) (1+x^\al)}=\int_0^1\cdots+\int_1^\infty\cdots\\ =\int_1 ^\infty \frac{x^\al \rd x}{\cdots}+\int_1^\infty \frac{\rd x}{\cdots}\\ =\int_1^\infty \frac