subnetting exercises and solutions

Known points $\dps{\int_0^\infty \frac{\sin \beta x}{x}\rd x=\frac{\pi}{2}\sgn \beta}$ (see Example 7.1.38), quadrature $\dps{\int_0^\infty \frac {\sin X\cos XT} {X}\rd x}$. (North China Electric Power Institute)Solution: $$\beex \bea \int_0^\infty \frac{\sin x\cos xt}{x}\rd x =\frac{1}{2}\int_0^\infty \frac{\sin x (t-1) +\sin x (1-T)} {x}\rd x\\ =\frac{1}{2}\cdot \frac{\pi}{2}\sez{\sgn (t+1) +\sgn (1-t)}\\ =\sedd{\ba{ll} 0,|t|>1,\\ \ Cfrac{\pi}{4},|t|=1,\\ \cfrac{\pi}{2},|t|[Typical problems an

Proof: $$\bex \int_0^\infty \frac{\rd x}{1+x^4}=\int_0^\infty \frac{x^2}{1+x^4}\rd x=\frac{\pi}{2\sqrt{2}}. \eex$$ (Beihang University)Proof: $$\beex\bea i\equiv \int_0^\infty \frac{\rd x}{1+x^4} =\int_0^\infty \frac{1}{1+\sex{\frac{1}{t}}^4}\cdot \ Frac{1}{t^2}\rd t\quad\sex{t=\frac{1}{x}}\\ =\int_0^\infty \frac{t^2}{1+t^4}\rd T =\frac{1}{2}\int_0^\infty \ Frac{1+t^2}{1+t^4}\rd T =\frac{1}{2}\int_0^\infty \frac{\frac{1}{t^2}+1}{\frac{1}{t^2}+t^2}\rd t\\ =\frac{1}{2 }\int_0^\infty \frac{1}{\sex

Set $f (x) $ for any finite interval $[0,a]\ (a>0) $ on normal integrable, on $[0,\infty) $ on absolute integrable, then $$\bex \vlm{n}\int_0^\infty f (x) |\sin nx|\rd x =\frac{2} {\pi}\int_0^\infty f (x) \rd x. \eex$$ (Nanjing University)Solution: Take $g in Example 4.5.32 (x) =|\sin x|$, $x \in [0,\pi]$, $$\bex \frac{1}{\pi}\int_0^\pi |\sin x|\rd X=\frac{2}{\pi} \eex$$ the conclusion . In addition, it can be proved by example 5.4.18 (the method of Fourier series).[Typical problems and methods

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}\frac{b}{at^2}\rd t\quad\sex{t=\frac{b}{ax}}\\ =\frac{1}{2}\int_0^\infty f\sez{\sex{Ax-\frac{B}{ x}}^2} \sex{1+\frac{b}{ax^2}}\rd x\\ =\frac{1}{2a}\int_0^\infty f\sez{\sex{ax-\frac{b}{x}}^2} \rd \sex{Ax-\frac B {x}} \ =\

If the $\forall\ i,j$ has a $ (a_i-a_j) (B_i-b_j) \geq 0$, then $a _i,b_i$ is called quasi-sequential. If the constant has the opposite inequality, it is called the inverse order. Test: $a _i,b_i$ $$\bex \sum_{i=1}^n a_i\cdot \sum_{i=1}^n b_i\leq n \sum_{i=1}^n a_ib_i, \eex$$ $a _i,b_i$ reverse order inequality inverse number. equals sign if and only if $a _1=\cdots=a_n$ or $b _1=\cdots =b_n$. (Chebyshev)Solution: $a _i,b_i$, $$\bex 0\leq \sum_{i,j=1}^n (A_i-a_j) (b_i-b_j) =2\sex{n \sum_{i=1}^n

If $u _1,u_2,\cdots,u_n\geq 0$, $u _1\cdot u_2\cdots u_n=1$, there is $u _1+u_2+\cdots+u_n\geq n$. Try to prove this conclusion and derive theorem 3 (mean theorem) from it.Proof: h\ "Older inequality is easily promoted to $$\bex \sum_{i=1}^n \frac{1}{p_i}=1,\quad 1[Typical problems and methods in mathematical analysis of Periven exercises reference Solutions]4.4.7

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 $$\beex \bea \sum_{n=1}^n \frac{1}{(kn-i) (kn+j)} =\frac{1}{k}\sum_{n=1}^n \sex{\frac{1}{kn-i}-\ FRAC{1}{KN+J}} =\frac{1}{k}\sez{\sum_{n=1}^n \frac{1}{kn-i}-\sum_{n=1}^n \frac{1}{k (n+1)-i}}\\ =\frac{1}{k}\ Sez{\frac{1}{k-i}-\frac{1}{k (n+1)-i}}. \eea \eeex$$ so the series and for $\dps{\frac{1}{k (k-i)}}$.[Typical problems and method

{2k+\frac{1}{4}},\quad b_k=f^{-1}\sex{2k+\frac{3}{4}, \eex$$ have $$ \bex A_k\leq X\leq B_k\ra 2k+\frac{1}{4}\leq f (x) =x+\frac{1}{x}\leq 2k+\frac{3}{4} \ra \sin^2\sez{\pi\sex{x+\frac{1}{x }}}\geq \frac{1}{2}, \eex$$ $$\beex \bea \int_{a_k}^{b_k}\sin^2\sez{\pi\sex{x+\frac{1}{x}}}\rd x \geq \frac{1}{2} (B_k-a_k) \ =\frac{1}{2}\sez{f^{-1} (Z_k)-f^{-1} (y_k)}\quad\sex{z_k=2k+\frac{3}{4},\ y_k=2k+\frac{1}{4}}\\ =\frac{1}{4}\sez{z_k-y_k+\sqrt{z^2-4}-\sqrt{y_k^2-4}}\\ \geq \frac{1}{4} (z_k-y_k) \ = \

algorithms using Python,openai Gymand. I separated them into chapters (with brief summaries) and exercises, and solutions so, can use them to supplement T He theoretical material above.all of the ' is ' in the Github repository. Some of the more time-intensive algorithms are still work and progress. I ' ll update this post as I implement them. Table of Contents Introduction to RL problems, OpenAI gym MDPs