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Latex

Example	LaTeX code	Result
Polynomial	{tex}a^2+b^2=c^2{/tex}	{tex}a^2+b^2=c^2{/tex}
Quadratic formula	{tex}x={-b\pm\sqrt{b^2-4ac} \over 2a}}{/tex}	{tex}x={-b\pm\sqrt{b^2-4ac} \over 2a}}{/tex}
Tall parentheses and fractions	{tex}2 = \left( \frac{\left(3-x\right) \times 2}{3-x} \right){/tex}	{tex}2 = \left( \frac{\left(3-x\right) \times 2}{3-x} \right){/tex}
	{tex}S_{\text{new}} = S_{\text{old}} - \frac{ \left( 5-T \right) ^2} {2}{/tex}	{tex}S_{\text{new}} = S_{\text{old}} - \frac{ \left( 5-T \right) ^2} {2}{/tex}
Integrals	\int_a^x \!\!\!\int_a^s f(y)\,dy\,ds = \int_a^x f(y)(x-y)\,dy{/tex}	\int_a^x \!\!\!\int_a^s f(y)\,dy\,ds = \int_a^x f(y)(x-y)\,dy{/tex}
Summation	{tex}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2\,n} {3^m\left(m \,3^n+n \,3^m\right)}{/tex}	{tex}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2\,n} {3^m\left(m \,3^n+n \,3^m\right)}{/tex}
Differential equation	{tex}u'' + p(x)u' + q(x)u = f(x),\quad x>a{/tex}	{tex}u'' + p(x)u' + q(x)u = f(x),\quad x>a{/tex}
Complex numbers	{tex}|\bar{z}| = |z|, |(\bar{z})^n| = |z|^n, \arg(z^n) = n \arg(z){/tex}	{tex}|\bar{z}| = |z|, |(\bar{z})^n| = |z|^n, \arg(z^n) = n \arg(z){/tex}
Limits	{tex}\lim_{z\rightarrow z_0} f(z)=f(z_0){/tex}	{tex}\lim_{z\rightarrow z_0} f(z)=f(z_0){/tex}
Integration with limits and fractions	{tex}\int\limits_{ -\infty }^{a} \frac{e^{-(x-\mu)/s}} {s(1+e^{-(x-\mu)/s})^2} \,dx = {1 \over 1+e^{-(a-\mu)/s}}{/tex}	{tex}\int\limits_{ -\infty }^{a} \frac{e^{-(x-\mu)/s}} {s(1+e^{-(x-\mu)/s})^2} \,dx = {1 \over 1+e^{-(a-\mu)/s}}{/tex}
Continuation and cases	{tex}f(x) = \begin{cases} 1 & -1 \le x < 0 \\ \frac{1}{2} & x = 0 \\ 1 - x^2 & \mbox{otherwise} \end{cases}{/tex}	{tex}f(x) = \begin{cases} 1 & -1 \le x < 0 \\ \frac{1}{2} & x = 0 \\ 1 - x^2 & \mbox{otherwise} \end{cases}{/tex}
Prefixed subscript	{tex}{}_pF_q(a_1, \dots, a_p;c_1, \dots, c_q;z) \\ ~\qquad = \sum_{n=0}^\infty \frac {(a_1)_n \cdots (a_p)_n}{(c_1)_n \cdots (c_q)_n} \frac {z^n}{n!}{/tex}	{tex}{}_pF_q(a_1, \dots, a_p;c_1, \dots, c_q;z) \\ ~\qquad = \sum_{n=0}^\infty \frac {(a_1)_n \cdots (a_p)_n}{(c_1)_n \cdots (c_q)_n} \frac {z^n}{n!}{/tex}

Asymptotic growth of function (Comparision)

Rules for comparing asymptotic growth of functions
1)f(n) < g(n) if

2) if f(n) < g(n)  then 1/g(n) < 1/f(n)

Ordering functions according to their asymptotic growth

1) n^α < n^β 
whenever β>α
Examples
n^1.2 < n²
n^0.02 < n^0.2

2) 1 < log(log n) < log n < n^ε < n^C < n^{log n} < Cⁿ < nⁿ < C^Cn

where 0 < ε < 1 and C > 1
because  f(n)/g(n) --> infinity  as n --> infinity  for each f(n) < g(n)
Examples
n^{log n} < 2ⁿ
log n < n^0.0001
nⁿ < 2²ⁿ

3)Using the reciprocal rule in above hierarchy we can deduce that
1/n^ε < 1 / log n 
1/n^ε < 1
and so on

4) log n!  <  n log n

5)2ⁿ  <  eⁿ  <  n!  < 2²ⁿ

Reference
Concrete Mathematics, Graham Knuth Pattasnik
Algorithms, Cormen   Leiserson   Rivest