I have a set of code in which I have   \begin{IEEEeqnarray*}{rCl} f(y,x) &amp; = &amp; \frac{1}{(2\pi)\left | \Sigma_{xx} \right |^{\frac{1}{2} \cdot        \left | (\Sigma_{yy} - \Sigma_{yx}\Sigma_{xx}^{-1}\Sigma_{yx}^{\prime})^{-1} \right |^{\frac{1}{2} \\ &amp;&amp; \cdot exp \left \{ -\frac{1}{2} \left \lbrack (y - \mu_y)^{\prime} (\Sigma_{yy} - \Sigma{yx}\Sigma_{ya}\Sigma_{xx}^{-1}\Sigma_{yx}^{\prime})^{-1}(y - \mu_y) \\ &amp;&amp; -(x - mu_x)^{\prime}\Sigma_{xx}^{-1}\Sigma_{yx}^{\prime} (\Sigma_{yy} - \Sigma_{yx}\Sigma_{xx}^{-1}\Sigma_{yx}^{\prime})^{-1}(y-\mu_y) \right \rbrack \right \}   It works if I put it under the second line but not the third.  Any suggestions much appreciated.
2773
I have a multi-line equation that I want to have matching sized parentheses. The advice I read online was to use \left( equation \right. \\ left. rest-of-equation \right) . However, this syntax leaves first half of the equation unaware of what is in the second half. For example, if you include a square root term after the break, the second parenthesis will be larger than the opening parenthesis. When you are using the aligned environment to split a long equation this can look somewhat offputting.    Compare the top equation to the bottom one. The parentheses match in size for the bottom equation, but the left hand parentheses for the top one is  noticeably smaller.  Top Code:   \begin{aligned}[t]   \gamma_{\text{u}} + \frac{1}{108}   \left(57\gamma_{\text{er}} \right. &amp;+  38\gamma_{\text{se}}\\ &amp; {}\pm  \left.\sqrt{2601 \gamma_{\text{er}}^2 + 3648  \gamma_{\text{er}}\gamma_{\text{se}} + 1444 \gamma_{\text{se}}^2}  \right)   \end{aligned}   Bottom Code:  \begin{aligned}[t] \gamma_{\text{u}} + \frac{1}{16} \left(9\gamma_{\text{er}}  +  6\gamma_{\text{se}} \pm  \sqrt{49 \gamma_{\text{er}}^2 + 76 \gamma_{\text{er}}\gamma_{\text{se}} + 36 \gamma_{\text{se}}^2} \right) \end{aligned}
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Equation exceeds one column of IEEE pattern in LaTeX how should I correct it? Please help, the code is attached herewith.  \begin{equation}\label{eq11}         p\left( {v,h|u,y,...} \right) = {\rm{RelU}}\left( {\left[ {\hat b + {w^h}\left\{ {\left( {{v^T}{w^v}} \right) \circ \left( {{y^T}{w^y}} \right)} \right\}} \\ \right]*\left[ {\hat a + {w^v}\left\{ {\left( {{h^T}{w^h}} \right) \circ \left( {{y^T}{w^y}} \right)} \right\}} \right]} \right) \end{equation}
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I am looking for suggestions on how to typeset complex numbers in the modulo-argument form, sometimes called phasor notation. Have already checked  but nobody mentions this in particular. I am explicitly excluding exponential and sine-cosine notations. My MWE is as follows  \documentclass{article} \begin{document}     $z = r \angle \phi$ \end{document}   I think it looks ugly, specially if the angle has many digits, ie  $z = 1.19 \angle -78.2039^{\circ}$   I was thinking of an \angle replacement which maybe would extend the lower segment up to the last digit. What do you think?  EDIT: It looks like the steinmetz package doesn't do a good job with the vertical space if the \phasor{} argument has a \fraction{}{} with parenthesis, see:  \documentclass{article} \usepackage{steinmetz}  \begin{document}     Something.     \[I = I \phase{\left(\frac{A_0}{B_0}\right)}\]     This line is too close to the upper equation, don't you think?      \[I = I \phase{\left(\frac{A_0}{B_0}\right)}\]     Whereas this line is NOT so close to the upper equation. \end{document}   Strangely, commenting the first line (% Something.) fixes the problem. Is there any way to fix this?
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I am trying to put these two matrices next to each other in the same empheq environment, separated by a comma and the space allotted by \qquad \quad.  \documentclass[10pt]{amsart}  \usepackage{nccmath} \usepackage{empheq} \usepackage{amssymb}  \begin{document}  \begin{empheq}[left = \empheqlbrack, right =\empheqrbrack]{equation*} \renewcommand{\arraystretch}{1.75} \begin{array}{@{\,}ccc|l@{\,}} 1 &amp; 0 &amp; -1 &amp; a \\ 0 &amp; 1 &amp; -1 &amp; a + b \\ 0 &amp; 0 &amp; \phantom{-}1 &amp; -\mfrac{1}{2} \, a - \mfrac{1}{2} \, b + \mfrac{1}{2} \, c \end{array} \end{empheq}  \begin{empheq}[left = \empheqlbrack, right =\empheqrbrack]{equation*} \renewcommand{\arraystretch}{1.75} \begin{array}{@{\,}ccc|l@{\,}} 1   &amp;   0   &amp;   0   &amp;   \mfrac{1}{2} \, a - \mfrac{1}{2} \, b + \mfrac{1}{2} \, c \\ 0   &amp;   1   &amp;   0   &amp;   \mfrac{1}{2} \, a + \mfrac{1}{2} \, b + \mfrac{1}{2} \, c \\ 0   &amp;   0   &amp;   1   &amp;   -\phantom{-}\mfrac{1}{2} \, a - \mfrac{1}{2} \, b + \mfrac{1}{2} \, c \end{array} \end{empheq}  \end{document}
75391
In the 70's I had a stack of old Amateur Radio magazines (50s-60s), and for a long time I saved an article about using the  to combine a number of resistors to achieve a specific value. Does anyone recall and have a copy of this article, or know how the Euclidean algorithm is applied to solve this problem?
32812
Change the order of integration in:  $$\int_0^{2a}\int_0^\sqrt{2ax-x^2}\frac{\phi'(y)(x^2+y^2)x }{\sqrt{4a^2x^2-(x^2+y^2)^2}}dxdy$$ and hence evaluate it.    I changed the order of integration and got the limits $y=0$ to $a$ and $x=a-\sqrt{a^2-y^2}$ to $a+\sqrt{a^2-y^2}$. I then tried changing to polar coordinates, but it is only making the integrand more complex.
69652
I use at the picture above:  \DeclareMathOperator{\vo}{vol} And this is the complete line:    \[ \lim_{a}=\vo_{a} \]   I can make that the a will be below the vol? (like the lim).  Thank you!!
34889
I have performed Gaussian elimination on this matrix to reduce it to   $$ \left[ \begin{array}{@{}ccc|c@{}} -3&amp;-1&amp;2 &amp; 1 \\ 0&amp; \frac{-5}{3}&amp; \frac{10}{3} &amp; \frac{8}{3} \\ 0&amp;0&amp; a+2 &amp; b + \frac{6}{5} \\ \end{array} \right] $$  I thought that setting $a$ equal to $-2$ and having $b$ not equal to $-\frac{6}{5}$ would be the answer to this problem, but it apparently isn't. Could someone please explain why?
67013
I am using TexStudio 2.9.4 on Windows 7. There is a problem that has been raised here at least once. It is that TexStudio flags unrecognised command  to  \begin{IEEEeqnarray}{rCl} and then within the environment flags expressions like \mathrm as being used outside the math environment. The fix (on Windows) is to manually edit the file C:\Users\USER\AppData\Roaming\texstudio\IEEEtrantools.cwl adding:  \begin{IEEEeqnarray}{cols}#\math,array \end{IEEEeqnarray} \begin{IEEEeqnarray*}{cols}#\math,array \end{IEEEeqnarray*}   in place of the \IEEEeqnarray entry there.   This works but is there a separate file that I could add (eg myfixes.cwl) which TexStudio would read? Burying the fix in a big autogenerated file is not ideal.  BTW is there any documentation on the format of .cwl files?
62186
I am working with the ideals $\mathfrak{p}=\left&lt;2,1+\sqrt{-5}\right&gt;, \mathfrak{q}=\left&lt;3,1+\sqrt{-5}\right&gt;, \mathfrak{t}=\left&lt;3,1-\sqrt{-5}\right&gt;$ and I am trying to prove that they are prime in $\mathbb{Z}[\sqrt{-5}]$.  I understand a good method to do this involves taking norms of each element that generates the ideals. This gives, for instance, the norms of the generators of $\mathfrak{p}=4,6$. These are both divisible by $2$. I have also calculated $\mathfrak{p}^2=\left&lt;2\right&gt;$. I get the feeling that since both norms are divisible by the generator of the ideal squared this proves it is prime, but I am not quite sure why.   In a similar vein, the norms of the generators of $\mathfrak{q,t}=9,6$ which are both divisible by $3$. I know $\mathfrak{qt}=\left&lt;3\right&gt;$, but I am not sure if this is relevant, since it is not either ideal squared. I think the norm of the ideal itself being prime implies the ideal is prime, but I am not sure how to find the norm of the ideal from the norm of its generating elements.  This question is similar to: , but I don't quite understand the reasoning behind the chosen answer. I am not sure how the answerer deduces that the example there is prime either. Following their reasoning as far as I can, the  fact that $2$ divides both norms means that $\left&lt;2\right&gt; \subset \mathfrak{p} \subset \mathbb{Z}[\sqrt{-5}]$ but I'm not sure how, and then I don't know how this proves it is prime.
65440
When I type $\overset{\circ}{Z}^\lambda$ the lambda is placed slightly above and to the right of the circle. Is there a way to make it appear as a superscript on the Z? I tried $\overset{\circ}{Z^\lambda}$, but then the circle is too high and to the right.  As a bizarre aside: a friend uses the exact same code with no problem. However, when I type it into MathOverflow it looks the same as in my compiler (i.e. bad).
88043
We are given two independent standard normal random variables $X$ and $Y$. We need to find out the M.G.F of $XY$.  I tried as follows : \begin{align} M_{XY}(t)&amp;=E\left(e^{(XY)t}\right)\\&amp;=\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(XY)t}f_X(x)f_y(y)dxdy \\ &amp;=\dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(xy)t}e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &amp;= \dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-\dfrac{(x-ty)^2}{2}}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &amp;= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dy \\ &amp;= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{-y^{2}(\frac{1}{2}+t^2)}dy \\ &amp;= \dfrac{1}{\sqrt{1+2t^2}} \end{align} Is this correct ?
78122
Please take a look on the following pseudo-code:  boolean blocked[2]; int turn; void P(int id) {       while(true) {              blocked[id] = true;              while(turn != id) {                     while(blocked[1-id])                     /* do nothing */;                     turn = id;              }              /* critical section */              blocked[id] = false;              /* remainder */       } } void main() {       blocked[0] = false;       blocked[1] = false;       turn = 0;       parbegin(P(0), P(1)); //RUN P0 and P1 parallel }   I thought that a could implement a simple Mutual - Exclution solution using the code above. But it's not working. Has anyone got an idea why?  Any help would really be appreciated!
18287
How to write this Piecewise function using LaTeX?  I tried $ \begin{array}{cc}   \{ &amp;      \begin{array}{cc}       0 &amp; x\leq 0 \\       \frac{100-x}{100} &amp; 0\leq x\leq 100 \\       0 &amp; 100\leq x     \end{array} \end{array} $  but it didn't work perfectly, the bracket &quot;{&quot; looks small, see:
99779
The following code works fine  \begin{theorem}[Euler] Blah \end{theorem}   But whenever I try to use square brackets in the description I get an error.  \begin{theorem}[The set $[0, 1]$ is compact] Blah \end{theorem}   Can anybody tell me how to avoid the error, please?
34219
In the vein of , I'm now at a roadblock on question 3 of this sheet:    (note: it's not my intention to ask every question I get stuck on here, merely the ones which have interesting general results; it just so happens that the two I've struggled with so far fit this criteria)  It goes as follows:  Let $N$ be a non-negative integer-valued random variable with mean $\mu_1$ and variance $\sigma_1^2$, and let $X_1, X_2, ...$ be random variables, each with mean $\mu_2$ and variance $\sigma_2^2$; furthermore, assume that $N, X_1, X_2,  . . .$ are independent. Without using generating functions, calculate the mean and variance of the random variable $S_N = X_1 + ... + X_N$ (when $N=0$ interpret $S_N$ as $0$).  I have the answer to calculating the mean, which I'll write up as an answer for future reference. Here's where I've got to with the variance bit:  $\mathbb{E}(S_N^2) = 0.P(N=0) + \sum_{r=1}^\infty\mathbb{E}((\sum_{i=1}^rX_i)^2).P(N=r)$  For $r=1$, we have the inner expectation as: $\mathbb{E}(X_1^2)$ which is equal to $\sigma_2^2+\mu_2^2$.  For $r &gt; 1$, we have the inner expectation equal to: $r(\sigma_2^2 + 2\mu_2^2)$  So $\mathbb{E}(S_N^2) = (\sigma_2^2+\mu_2^2)P(N=1) + (\sigma_2^2 + 2\mu_2^2)\sum_{r=2}^\infty rP(N=r)$  which is almost in the form where I can use the same trick as in calculating the mean, but not quite.  Any help is much appreciated.
74314
I want to lower bound \begin{align}  \mathbb{P} \left( -\epsilon \leq  \sum_{i=1}^n X_i \leq 0 \right) &amp;= \mathbb{P} \left( - \frac{\epsilon}{\sqrt{\sum_{i=1}^n \sigma_i^2}} \leq  \frac{\sum_{i=1}^n X_i}{\sum_{i=1}^n \sigma_i^2} \leq 0 \right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \end{align} where $\{X_i\}$ are mean zero, independent random variables with $\mathbb{E}[X_i^2] = \sigma_i^2 \leq H^2$ and $\mathbb{E}[|X_i|^3] = \lambda_i^3 \leq H^3$. From Equation $3$ in , I have that \begin{align} \sup_{x \in \mathbb{R}} |F_n(x) - \Phi(x)| \leq C_0 \left( \sum_{i=1}^n \sigma_i^2 \right)^{-3/2} \sum_{i=1}^n \lambda_i^3 \end{align} Loosening the bound above gives \begin{align} \sup_{x \in \mathbb{R}} |F_n(x) - \Phi(x)| \leq \frac{C_0}{\sqrt{n}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2) \end{align} Now I can write $(1)$ as \begin{align} F_n(0) - F_n \left(-  \frac{\epsilon}{\sqrt{\sum_{i=1}^n \sigma_i^2}} \right) \end{align} which by $(2)$ can be lower bounded as \begin{align} \Phi(0) - \frac{2C_0}{\sqrt{n}} - \Phi\left ( -  \frac{\epsilon}{\sqrt{\sum_{i=1}^n \sigma_i^2}} \right ) &amp;\geq \frac{1}{2} - \frac{2C_0}{\sqrt{n}} - \Phi\left ( -  \frac{\epsilon}{H \sqrt{n}} \right ) \end{align} Can the last expression be improved to something like $K/\sqrt{n}$ for some constant $K$ for large $n$? I feel like I need some approximations to normal CDF $\Phi(x)$ for small $x$.