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- === <br> 3. Practice considerations === ==== 3.1 Log-likelihood ====13 KB (1,966 words) - 10:50, 22 January 2015
- <center>[[Image:runyan3.jpg|frame|none|alt=Alt text|<font size= 4> '''Figure 3''' </font size>]] </center> <br /> # Repeat steps 2 and 3 until the value of <math>J</math> no longer changes8 KB (1,350 words) - 10:57, 22 January 2015
- \begin{cases} \end{cases}5 KB (790 words) - 10:01, 14 March 2015
- <font size = 3>The purpose of Upsampling is to manipulate a signal in order to artificiall <font size = 3>3 KB (565 words) - 10:01, 14 March 2015
- ...solely on the 'r' value that is contained in 'z'. The ROC is one of three cases; :3. The ROC is the space in between two circles centered at the origin.6 KB (1,019 words) - 18:11, 23 February 2015
- =QE2013_AC-3_ECE580-3= ...3_AC-3_ECE580-1|Part 1]],[[QE2013_AC-3_ECE580-2|2]],[[QE2013_AC-3_ECE580-3|3]],[[QE2013_AC-3_ECE580-4|4]],[[QE2013_AC-3_ECE580-5|5]]8 KB (1,016 words) - 12:19, 25 March 2015
- ...3_AC-3_ECE580-1|Part 1]],[[QE2013_AC-3_ECE580-2|2]],[[QE2013_AC-3_ECE580-3|3]],[[QE2013_AC-3_ECE580-4|4]],[[QE2013_AC-3_ECE580-5|5]] ...iplier approach, which is more complicated but would apply to more general cases. Solution 1 is not as general but is simpler for the given problem. They2 KB (330 words) - 12:22, 25 March 2015
- | <math> \int x^{2} ch ax dx=(\dfrac{x^{2}}{a^{2}}+\dfrac{2}{a^{3}}) sh ax-\dfrac{2x}{a^{2}} ch ax +C</math> ...int\dfrac{dx}{(ch ax+1)^{2}}=\dfrac{1}{2a}th\dfrac{ax}{2}-\dfrac{1}{6a}th^{3}\dfrac{ax}{2} +C</math>8 KB (1,479 words) - 17:44, 26 February 2015
- | <math> \int\arg ch\dfrac{a}{x}dx=\begin{cases} ...{x}+\arcsin\dfrac{x}{a}}{x\arg ch\dfrac{a}{x}-\arcsin\dfrac{x}{a}} & .\end{cases} +C</math>3 KB (624 words) - 17:52, 26 February 2015
- |<math> \int x^2 \arcsin \frac {x }{ a}dx = \frac{x^3}{3}\arcsin \frac {x}{a} + \frac {\left( x^2+2a^2 \right) \sqrt { a^2-x^2 }}{9} ...ot 3}+ \frac{1 \cdot 3(x/a)^5}{2 \cdot 4 \cdot 5 \cdot 5} + \frac {1 \cdot 3 \cdot 5 (x/a)^7}{2 \cdot 4 \cdot 6 \cdot 7 \cdot 7} + \cdot \cdot \cdot +C<8 KB (1,433 words) - 17:05, 26 February 2015
- ...rac {ax}{1 \cdot 1!} + \frac {(ax)^2}{2 \cdot 2!} + \frac {(ax)^3}{3 \cdot 3!} + \cdot \cdot \cdot +C</math> \begin{cases}3 KB (513 words) - 17:09, 26 February 2015
- | <math> \int x^{2} sh ax dx=(\dfrac{x^{2}}{a^{2}}+\dfrac{2}{a^{3}}) ch ax-\dfrac{2x}{a^{2}} sh ax +C</math> | <math> \int\dfrac{sh ax}{x} dx=ax+\dfrac{(ax)^{3}}{3\cdot3!}+\dfrac{(ax)^{5}}{5\cdot5!}+\cdots +C</math>7 KB (1,378 words) - 17:42, 26 February 2015
- \begin{cases} \end{cases} </math>4 KB (826 words) - 18:06, 26 February 2015
- ...{ ax + b} = \frac {(ax+b)^2}{2a^3} - \frac {2b(ax+b) }{a^3} + \frac{b^2}{a^3} \ln (ax +b)+C</math> ...{3a^4} - \frac {3b(ax+b)^2 }{2a^4} + \frac{3b^2(ax+b)}{a^4} - \frac{b^3}{a^3}\ln (ax +b)+C</math>7 KB (1,373 words) - 18:07, 26 February 2015
- ...h> \int x^2 \sin a x d x = \frac {2 x}{a^2} \sin a x + \left ( \frac {2}{a^3} - \frac {x^2}{a} \right)\cos a x +C</math> ...}{a^2} - \frac{6}{a^4}\right)\sin a x + \left ( \frac {6x}{a^3} - \frac {x^3}{a} \right)\cos a x +C</math>14 KB (2,809 words) - 16:12, 26 February 2015
- ...that the distribution of stars within a galaxy is accurately modeled by a 3-dimensional homogeneous Poisson process for which the following two facts a ...ius }r\right\} \right)</math><math class="inline">=1-e^{-\frac{4}{3}\pi r^{3}\lambda}.</math>2 KB (384 words) - 00:22, 10 March 2015
- ...ing i.i.d , <math class="inline">\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots</math> each have finite mean <math class="inline">\mu</math> , and <math class="inline">E\left[\mathbf{X}_{i}\mathbf{X}_{j}\right]=\begin{cases}4 KB (699 words) - 11:08, 10 March 2015
- \begin{cases} \end{cases} </math>29 KB (4,417 words) - 15:53, 12 March 2015
- ...n-1|Part 1]],[[ECE-QE_CS1-2011_solusion-2|2]],[[ECE-QE_CS1-2011_solusion-3|3]] \begin{cases}8 KB (1,336 words) - 01:53, 31 March 2015
- =[[HW3ECE38F15|Homework 3]] Solution, [[ECE438]], [[2015_Fall_ECE_438_Boutin|Fall 2015]], [[user:mbou ==Question 3==7 KB (1,181 words) - 19:17, 19 October 2015
- ...X_5[0]e^{j2\pi n0/5} + X_5[1]e^{j2\pi n1/5} + X_5[2]e^{j2\pi n2/5} + X_5[3]e^{j2\pi n3/5} + X_5[4]e^{j2\pi n4/5} \right ) \\ X_5[1]=5 \mbox{, and } X_5[0]=X_5[2]=X_5[3]=X_5[4]=09 KB (1,594 words) - 15:36, 20 October 2015
- \mathcal{Y}(\omega) =\frac{1}{4} \sum_{k=0}^{3} \mathcal{X} \left (\frac{\omega-k2\pi}{4} \right ) ==Question 3==6 KB (945 words) - 11:40, 19 October 2015
- ...modeled as an impulse train passing through the vocal tract. While in both cases vocal tract selects the noise heard, in voiced sound the impulse train dete ...ally eliminated almost completely. Since the noise is not periodic in most cases, any substantial noise is effectively multiplied by a zero. The probability9 KB (1,777 words) - 23:23, 21 November 2015
- \begin{cases} \end{cases}5 KB (766 words) - 22:18, 7 March 2016
- <math> f(d) = 3 </math> <br /> * Analysis of social structures (special cases may include schools, military, parties, events, etc.).13 KB (2,051 words) - 22:06, 24 April 2016
- *In some cases, we need pre-processing to transform the data in some way to get the statis ==3. algorithm assumption==8 KB (1,405 words) - 22:56, 27 November 2016
- ...essed signals. In this project, I want to explore what will happen in both cases. In order to feel it more directly, I decided to process some music signals ==3. Result==3 KB (553 words) - 22:41, 26 November 2016
- \bar{E}=\begin{cases} \end{cases}} \left(\frac{V}{m}\right)4 KB (646 words) - 23:24, 25 April 2017
- <math>\quad rank=2\ne \mbox 3</math> \qquad rank<3 \qquad must\;contain\;\lambda=-14 KB (588 words) - 00:10, 21 May 2017
- <math>\begin{cases} \end{cases}</math>4 KB (557 words) - 01:01, 21 May 2017
- <math>\begin{cases} \end{cases}</math>7 KB (1,126 words) - 05:45, 22 May 2017
- <math>\begin{cases} \end{cases}</math>3 KB (393 words) - 11:33, 18 June 2017
- & = \begin{cases}Ar\cos\theta, r<a \hspace{3cm} \leftarrow\text{ no singularity at }r=0\\ \end{cases}4 KB (702 words) - 17:02, 17 June 2017
- ...{2cm}r\le a\\-E_0r\cos\theta+\frac{B\cos\theta}{r^2}\hspace{2cm}r\ge a\end{cases} <math>Aa = -E_0a+\frac{B}{a^2}\to A = -E_0 + \frac{B}{a^3}</math>4 KB (642 words) - 10:44, 18 June 2017
- ...ega}\bigg)\\\sigma\to\infty\to\epsilon_c\to\infty\\\text{ then }M_c=0 \end{cases}</math> 3) <math>\nabla\times\bar{H} = \bar{J}+\frac{\partial D}{\partial t}\hspace{13 KB (591 words) - 11:21, 18 June 2017
- ...ega}\bigg)\\\sigma\to\infty\to\epsilon_c\to\infty\\\text{ then }M_c=0 \end{cases}</math> 3) <math>\nabla\times\bar{H} = \bar{J}+\frac{\partial D}{\partial t}\hspace{13 KB (591 words) - 11:24, 18 June 2017
- \rho=\begin{cases} qN_{D2} & x_{n1}\le x\le x_{n2}\text{ region 3}\\2 KB (385 words) - 10:52, 6 August 2017
- In both cases ‘R’ refers to the radius if the graph were to be revolved around the gi ==3. References==3 KB (576 words) - 20:37, 3 October 2017
- In both cases ‘'''<math>R</math>'''’ refers to the "'''radius'''" if the graph were t ==3. References==4 KB (664 words) - 09:42, 19 January 2018
- <math>=\begin{cases} undefined, & if & s≤2 \end{cases} </math>.6 KB (1,071 words) - 18:26, 22 November 2017
- ...ac{dx}{dt}}{t} - \frac{x}{t^2}</math>, <math>\frac{d^2y}{dt^2}=\frac{2x}{t^3} - \frac{2\frac{dx}{dt}}{t^2} + \frac{\frac{d^2y}{dt^2}}{t}</math>. Substitute them into ODE, hence <math>t^2 (lnt+1) (\frac{2x}{t^3} - \frac{2\frac{dx}{dt}}{t^2} + \frac{\frac{d^2x}{dt^2}}{t}) + t (2lnt+1) (7 KB (1,254 words) - 19:49, 22 November 2017
- <math>y(t)=\begin{cases} \end{cases}</math><br />5 KB (985 words) - 12:38, 30 November 2018
- <small>''Figure 3 Merma, 2004 P80 The logistic equation Adapted from Northwestern University The following graph demonstrates four different cases with θ>1 or θ<1.10 KB (1,532 words) - 22:51, 2 December 2018
- Question 3: Optimization6 KB (899 words) - 01:04, 24 February 2019
- ...ve all permutations pertaining to who draws which name and determine which cases will have every person draw someone else's name. ...ginal" spot. These two permutations are called derangements of <math>\{1,2,3\}</math>.6 KB (996 words) - 00:53, 3 December 2018
- ...ta^2(n+5)^2]}+\dfrac{sin[(1-\beta)\pi(n+0.5)/2]}{\pi[(n+.5)-4\beta^2(n+.5)^3]},-\infty<n<\infty</math> with <math>\beta=0.5</math> <br> \begin{cases}4 KB (738 words) - 15:34, 19 February 2019
- Question 3: Optimization <math>\begin{cases}1 KB (178 words) - 11:45, 25 February 2019
- Question 3: Optimization Subject to <math>x_1+x_2-2<=0</math> and <math>x_1+2x_2-3<=0</math><br>2 KB (247 words) - 11:48, 25 February 2019
- \begin{cases} \end{cases}1 KB (190 words) - 16:32, 19 February 2019
- [[File:windowSmall.png|400px|thumb|right|Filtered Image using LPF (Window = 3 pixels)]] B = imData(:,:,3);7 KB (1,006 words) - 19:22, 7 April 2019
