Skip to content

Instantly share code, notes, and snippets.

@id4ehsan
Created November 24, 2018 12:57
Show Gist options
  • Select an option

  • Save id4ehsan/372a432a91f4b8325fc54526dae5369b to your computer and use it in GitHub Desktop.

Select an option

Save id4ehsan/372a432a91f4b8325fc54526dae5369b to your computer and use it in GitHub Desktop.
طراحی الگوریتم
\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}
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment