Difference between revisions of "Math Diagnostic"

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hello
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hello  
  
<math>\pi=\frac{3}{4} \sqrt{3}+24 \int_0^{1/4}{\sqrt{x-x^2}dx}</math>
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<math>\pi=\frac{3}{4} \sqrt{3}+24 \int_0^{1/4}{\sqrt{x-x^2}dx}</math>  
  
 
<math>\text{cost}= \text{base}  
 
<math>\text{cost}= \text{base}  
\times 2^{\text{level}-1}</math>
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\times 2^{\text{level}-1}</math>
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<br> <math>
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  \operatorname{erfc}(x) =
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  \frac{2}{\sqrt{\pi}} \int_x^{\infty} e^{-t^2}\,dt =
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  \frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n)!}{n!(2x)^{2n}}</math>
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<math>\begin{bmatrix}
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0      & \cdots & 0      \\
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\vdots & \ddots & \vdots \\
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0      & \cdots & 0
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\end{bmatrix}</math>
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<math>
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\left . \frac{A}{B} \right \} \to X</math>
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<math>{\color{Blue}x^2}+{\color{YellowOrange}2x}-{\color{OliveGreen}1}</math>
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<math>x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>
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<math>\int_a^x \!\!\!\int_a^s f(y)\,dy\,ds
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= \int_a^x f(y)(x-y)\,dy</math>
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<math>|\bar{z}| = |z|,
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|(\bar{z})^n| = |z|^n,
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\arg(z^n) = n \arg(z)</math>
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<math>\lim_{z\rightarrow z_0} f(z)=f(z_0)</math>
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<math>\phi_n(\kappa) =
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\frac{1}{4\pi^2\kappa^2} \int_0^\infty
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\frac{\sin(\kappa R)}{\kappa R}
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\frac{\partial}{\partial R}
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\left[R^2\frac{\partial D_n(R)}{\partial R}\right]\,dR</math>
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<math>{}_pF_q(a_1,\dots,a_p;c_1,\dots,c_q;z)
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= \sum_{n=0}^\infty
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\frac{(a_1)_n\cdots(a_p)_n}{(c_1)_n\cdots(c_q)_n}
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\frac{z^n}{n!}</math>
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<math>\varpi \varrho \varsigma \varphi</math>
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<math>\left \uparrow \frac{a}{b} \right \downarrow \quad \left \Uparrow \frac{a}{b} \right \Downarrow \quad \left \updownarrow \frac{a}{b} \right \Updownarrow</math>
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<math>1/\tau(\omega)</math>
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<math>\tau</math>
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<math>\kappa</math>
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<math>\phi</math>
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<math>1/\kappa(\omega)</math>
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<math>1/\tau_0 + \kappa\omega^2/2 \approx 1/\tau_0 \lambda_0\omega^2/\omega_\lambda \Rightarrow \omega_\lambda = 2\lambda_0/\kappa</math>
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<math>\lambda(\omega,T) = \frac{\lambda_0}{1 + (\kappa\omega/2\lambda_0)^2 + (T/T_\lambda)^2}</math>
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<math>1/\tau(\omega)</math>
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*Making this assumption gives <math>1/\kappa(\omega) = 1/\kappa + \kappa\omega^2/2 \approx 1/\kappa \lambda_0\omega^2/\omega_\lambda \Rightarrow \omega_\lambda = 2\lambda_0/\kappa,</math> so that:<math>\lambda(\omega,T) = \frac{\lambda_0}{1 + (\kappa\omega/2\lambda_0)^2 + (T/T_\lambda)^2}.</math>

Latest revision as of 21:29, 14 April 2016

hello




  • Making this assumption gives so that: