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Set $f (x) $ is a nonnegative continuous function on the $0\leq x(1). There is a bounded derivative $f ' (x) $ on the $0\leq x(2). $\dps{\int_0^\infty f (x) \rd xProof: Set $|f ' |\leq m$, then by the Lagrange mean value theorem, $$\bex |f (x)-F (Y) |=|f ' (\xi) |\cdot |x-y|\leq m|x-y|. \eex$$ and $f $ Lipschitz continuous, consistent continuous. The conclusion is established by an example of 4.5.24.[Typical problems and methods in mathematical analysis of Periven

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

Example 4.5.37 the inverse proposition is not tenable, that is $f (x) $ in $ (0,1) $ within the monotony, $\dps{\vlm{n}\frac{1}{n}\sum_{i=1}^{n-1} f\sex{\frac{i}{n}}}$ exist, $\dps{\int_0^ 1 f (x) \rd x}$ can not converge.Solution: Take $$\bex F (x) =\frac{1}{x}-\frac{1}{1-x}\ra F ' (x) =-\frac{1}{x^2}-\frac{1}{(1-x) ^2}[Typical problems and methods in mathematical analysis of Periven exercises reference 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

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

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

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

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