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November 24, 2018 12:57
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طراحی الگوریتم
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| \documentclass[12pt,twoside,a4paper]{article} | |
| \usepackage{tikz} | |
| \usepackage{geometry} | |
| \geometry{ | |
| a4paper, | |
| total={210mm,297mm}, | |
| left=10mm, | |
| right=10mm, | |
| top=15mm, | |
| bottom=15mm, | |
| } | |
| \usepackage{tabularx} | |
| \usepackage{float} | |
| \usepackage{hyperref} | |
| \usepackage{gensymb} | |
| \usepackage{xepersian} | |
| \newcommand*\circled[1]{\tikz[baseline=(char.base)]{ | |
| \node[shape=circle,draw,inner sep=2pt] (char) {#1};}} | |
| \settextfont{XB Niloofar} | |
| %\pagenumbering{arabic} | |
| %\pagestyle{headings} | |
| \title{طراحی الگوریتم} | |
| \author{احسان قاسملو} | |
| \date{} | |
| \begin{document} | |
| $$\sum_{i=1}^{n}i=1+2+...+n=\frac{n(n+1)}{2}\approx\frac{1}{2}n^2=\theta(n^2)$$ | |
| $$lim \frac{1+2+3+..+n}{3n^2+2n+6}=lim \frac{\frac{1}{2}n^2}{3n2}=\frac{1}{6}$$ | |
| $$\sum_{i=1}^{n}i^2=1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6} \approx \frac{1}{3}n^3=\theta(n^3)$$ | |
| $$\sum_{i=1}^{n}i^3=1^3+2^3+...+n^3=(\frac{n(n+1)}{2})^2 \approx \frac{1}{4}n^4=\theta(n^4)$$ | |
| $$\sum_{i=1}^{n}i^p=1^p+2^p+...+n^p \approx \frac{1}{p+1}n^{p+1}=\theta(n^{p+1})$$ | |
| $$\sum_{i=1}^{n}f(x)\approx \int_{1}^{n} f(x)dx$$ | |
| $$\int_{1}^{n} i^pdi = \frac{1}{p+1}i^{p+1}]^n$$ | |
| $$\sum_{i=1}^{n}i^p\approx \theta(n^{p+1} \log n)$$ | |
| $$\sum_{i=1}^{n}\frac{1}{i}= 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\approx \int_{1}^{n}\frac{1}{i}=\ln i ]^n=\ln n - \ln 1=\ln n =\theta(\ln n)=\theta(\log_a n)=\theta(\lg n)$$ | |
| $$\sum_{i=1}^{10}(6i^2+4i+5)=6\sum_{1}^{10}i^2+4\sum_{1}^{10}i+\sum_{1}^{10}5=6\frac{10*11*21}{6}+4\frac{10*11}{2}+50$$ | |
| $$\sum_{i=m}^{n}a=a+a+...+a=(n-m+1)a$$ | |
| $$a+ax+ax^2+...+ax^{n-1}=\sum_{1}^{n-1}ax^i=a\frac{x^n-1}{x-1}$$ | |
| $$a+ax+ax^2+...=\frac{a}{1-x}$$ | |
| $$n \rightarrow \infty , |x|<1$$ | |
| $$a+(a+d)+(a+2d)+(a+3d)+...++(a+(n-1)d)=\frac{(a+a+(n-1)d)n}{2}$$ | |
| $$\sum_{i=1}^{n}\sqrt{i}=\sum_{i=1}^{n}i^{\frac{1}{2}}=\theta(n^{\frac{1}{2}+1})=\theta(n\sqrt{n})$$ | |
| $$\sum_{i=1}^{n}\frac{1}{\sqrt{i}}\approx\int_{1}^{n}\frac{1}{\sqrt{i}}=\int_{1}^{n}i^{-\frac{1}{2}}=\frac{1}{-\frac{1}{2}+1}i^{-\frac{1}{2}+1}|^n_1=\theta(\sqrt{n})$$ | |
| $$\log_b a=c \rightarrow b^c=a $$ | |
| $$a>0 , b>0 , b\neq 1$$ | |
| $$\log_2 8=3 \rightarrow 2^3=8 $$ | |
| $$\log_0.1 1000=x \rightarrow (0.1)^x=1000$$ | |
| \end{document} |
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