\begin{align}   \mathcal{L}_{\lambda} =  u_{1}(h) - u_{1}(h^{*}) + T \geq 0 \text{(foc$\lambda$)}, \label{eqn:foca1}\\   \mathcal{L}_{T} = -1 + \lambda = 0 \Leftrightarrow \lambda = 1 \text{(focT)} ,\label{eqn:focb1}\\   \mathcal{L}_{h} =- u'_{2}(h) + \lambda u'_{1}(h) = 0 \Leftrightarrow \lambda = \frac{u'_{2}(h)}{u'_{1}(h)} \text{(foch)}.\label{eqn:focc1}\\   \lambda \geq 0 \text{(nn)}, \label{eqn:focd1}\\   \lambda \mathcal{L}_{\lambda} =  \lambda [u_{1}(h) - u_{1}(h^{*}) + T] = 0 \text{(cs$\lambda$)} .\label{eqn:foce1} \end{align}   I want to have foc$\lambda$, \text{(focT)}, etc, as an equation number on the other sides if that makes sense?
19795
When I have equations, as in  \begin{equation} ... \end{equation}   Can I arbitrarily assign the numbers to the equations? Instead of the usual order (1), (2), (3), ..., I'd like something like (1), (2.1), (2.2), (2.3), (3.1), (3.2), (4), ...
7104
If $\textsf{V}$ is a finite-dimensional complex vector space, $\textsf{T}\in\mathcal{L}(\textsf{V})^1$, $\lambda$ is arbitrary in $\mathbb{C}$ and if    $$\textsf{V} = \text{null}(\textsf{T}-\lambda\textsf{I})  \oplus \text{range}(\textsf{T}-\lambda\textsf{I})$$   then prove that $\textsf{T}$ is diagonalizable.   Attempt : I am solving Axler's 3 edition book in Exercise $5c.$ The book hasn't introduced the Jordan normal form or the generalized eigenvectors. Could someone please give a direction to move ahead.  Thanks a lot for your help.    $^1$ $\mathcal{L}(\textsf{V})$ is the set of all linear maps from $\textsf{V}$ to itself.
59966
Here is my code:   \begin{align*} (a-bu)e^{-\frac{1}{4}(v^2+u^2)(1+c)+\frac{1}{2}(1-c)u|v|}  &amp;=(a-bx-by)e^{-\frac{1}{4}((x-y)^2+(x+y)^2)(1+c)+\frac{1}{2}(1-c)(x+y)(y-x)}\\ &amp;=(a-bx-by)e^{-(x^2+cy^2)} \\ &amp;=f(x,y). \end{align*}   I have a long equation. My question is as follows: I want to isolate the expression   (a-bu)e^{-\frac{1}{4}(v^2+u^2)(1+c)+\frac{1}{2}(1-c)u|v|}   on the first line so that I don't have an equation on the first line and I want this expression left justified with the other lines (that have the equal sign). How can I do this?
16381
I would want to do something like:  &gt;&gt;&gt; lst = [1, 2, 3, 4, 5] &gt;&gt;&gt; lst.find(lambda x: x % 2 == 0) 2 &gt;&gt;&gt; lst.findall(lambda x: x % 2 == 0) [2, 4]   Is there anything nearing such behavior in Python's standard libraries?  I know it's very easy to roll-your-own here, but I'm looking for a more standard way.
121835
I believe I am very close to finishing this proof, but I cannot figure out the last part. If anybody could check my work and maybe give me a little hint, it would be greatly appreciated!  Let $V$ be a finite-dimensional vector space and $T \in \mathcal{L}(V)$, and let $\mathbf{u,v} \in V$ be eigenvectors of $T$.  Claim:  If $\mathbf{u} + \mathbf{v}$ is an eigenvector of $T$, then $\mathbf{u}, \mathbf{v}$, and $\mathbf{u+v}$ all correspond to the same eigenvalue.  Proof (So far!): Suppose $T(\mathbf{u}) = \lambda_1 \mathbf{u}$ and $T(\mathbf{v}) = \lambda_2 \mathbf{v}$ with $\mathbf{u}, \mathbf{v} \neq \mathbf{0}$.   Now suppose $T(\mathbf{u+v}) = \lambda_3(\mathbf{u+v})$ with $\mathbf{u+v}\neq \mathbf{0}$.  Then, $$ T(\mathbf{u}) + T(\mathbf{v}) = \lambda_3 \mathbf{u} + \lambda_3 \mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} =  \lambda_3 \mathbf{u} + \lambda_3\mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} - \lambda_3 \mathbf{u} -\lambda_3\mathbf{v} = \mathbf{0}\\ (\lambda_1 - \lambda_3) \mathbf{u} + (\lambda_2 - \lambda_3)\mathbf{v} = \mathbf{0} $$  Now I know in order to show that $\lambda_1 = \lambda_2 = \lambda_3$, I must show that the only solution to the last line is the trivial one. This would imply that $\mathbf{u}$ and $\mathbf{v}$ are linearly independent which I am unconvinced of!  The only information I have to my advantage I haven't used yet is the fact that $\mathbf{u}, \mathbf{v}, \mathbf{u+v} \neq \mathbf{0}$. I really cannot see how this information can help me though.  Perhaps I am going about this wrong, but that's why I want to ask! Thanks for your help.
60027
I was thinking this integral : $$I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$$ What I do is use a Reciprocal subsitution, easy to show that: $$I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{\lambda ^2x^2+\lambda x+1}\text{d}x =\frac{1}{\lambda}\int_0^{\infty}\frac{(\ln x-\ln \lambda)^2}{x^2+x+1}\text{d}x \\ =\frac{\ln ^2\lambda}{\lambda}\int_0^{\infty}\frac{1}{x^2+x+1}\text{d}x+\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2x}{x^2+x+1}\text{d}x \\ =\frac{\ln ^2\lambda}{\lambda}\frac{2\pi}{3\sqrt{3}}+\frac{2}{\lambda}\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x$$ But I hav no idea on process the remaining integral:( Anyone knows how to solve it?  THX guys! I came with another method might work for this: Recall the handy GF $$\frac{1}{x^2+x+1}=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k$$ Then we have :  $$\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\frac{2}{\left(k+1\right)^3}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k$$ With $$\sum_{k=1}^{\infty}\frac{\sin (kx)}{k^3}=\frac{\pi ^2}{6}x-\frac{\pi}{4}x^2+\frac{x^3}{12}$$ What can we arrived?
22150
I have the following equations: \begin{equation} {1\over f(x,y)} {\partial f(x,y) \over \partial x} \alpha(x,y) + {1\over f(x,y)} {\partial f(x,y) \over \partial y} \beta(x,y) = \gamma(x,y) \end{equation} Where $\alpha(x,y)$, $\beta(x,y)$ and $\gamma(x,y)$ are exactly known functions. What I need to find is $f(x,y)$. Can anyone help me find a solution? Or are there any assumptions (other than $f(x,y) = g(x)h(y)$ that will allow me to solve this analytically? Is it a standard equation that I haven't come across? Many thanks guys!
95019
I would like to construct an explicit bijection from $\alpha^*$ to $\alpha$, where $\alpha$ is an infinite ordinal. I have been told a possible bijection would be:  \begin{equation} f(\beta) = \begin{cases} 0 &amp; \text{if } \beta = \alpha \\ \beta^* &amp; \text{if } \beta \text{ is finite} \\ \beta &amp; \text{otherwise} \end{cases} \end{equation}  I don't understand how this is constructed, or why it is a bijection.
22956
Let $C = \{0, 1, 2, \ldots, \aleph_0, \aleph_1, \aleph_2, \ldots\}$. What is $\left|C\right|$? Or is it even well-defined?
31934
Let $\lambda , \mu$ $\in$ $\mathbb{C}$. Define $T:$ $\mathbb{C}$ $\to$ $\mathbb{C}$ by $T(z)=\lambda z + \mu \bar{z}$ for every $z \in \mathbb{C}$. Show that if  $\vert T(z) \vert = \vert z \vert$ then $\lambda\mu =0$  I'm doing the first thing that comes to mind:  $\vert \lambda z+\mu\bar{z}\vert ^2=\vert z \vert ^2$ So we have $(\lambda\mu+\mu \bar{z})(\overline \lambda \overline z + \bar{\mu}z)=z\bar{z}$  How can I conclude the proof? Thanks.
112405
I want to write a numbered system of equations. Right now, this is what I do and I am satisfied with the output, but I have a feeling it's not the best way to do it.  \begin{equation}   \begin{array}{ccl}     a + b + c &amp; = &amp; d \\     e + f     &amp; = &amp; g \\     h         &amp; = &amp; i   \end{array} \end{equation}   How should I handle system of equations with a single number associated to it?  UPDATE :  I have read the  and I feel more comfortable with what it offers. However, are there still some situations in which using the array environment is still relevant. For example, I would not know how to produce this result otherwise :  \begin{equation}   \left\{   \begin{array}{lcl}     x_1 + x_2 + x_3 &amp; = &amp; 1 \\     x_1 + x_2       &amp; = &amp; 2 \\     x_1             &amp; = &amp; 3   \end{array}   \right. \end{equation}   In the case the spacing seems correct, and it would not bee desirable to have the first column aligned to the = sign.
124299
Let $\lambda_i \in \mathbb C$  $(i=1,2,\cdots,n)$ be $n$ different complex numbers and $p_i \in \mathbb C[t]$ $(i=1,2,\cdots,n)$ polynomials. If we have the relation $${e^{{\lambda _1}t}}{p_1}(t) + {e^{{\lambda _2}t}}{p_2}(t) +  \cdots  + {e^{{\lambda _n}t}}{p_n}(t) = 0$$ for all $t \geq 0$, can we conclude that all $p_i(t)=0$?
88581
So as I have learned from a previous post that the  \begin{aligned}... \end{aligned} with the usage of "&amp;" before "=" will help arrange neatly the equations where the equal sign of each equation will be just below the equal sign of the following equation and so on.  My question is if I have a total of 8 equations, how to number each equation and at the same time preserve the alignment of the equal signs.  Note if I use \begin{equation}... \end{equation}, I get only one number for the collection of 8 equations at the very end which is not what I want.
30931
Does anyone know how to get from (1) to (2) in the system  $$    \begin{align}     \mathrm{g}^{\mu\nu}_{,\rho}+     \mathrm{g}^{\sigma\nu}{{\Gamma}}^{\mu}_{\sigma\rho}+     \mathrm{g}^{\mu\sigma}{{\Gamma}}^{\nu}_{\rho\sigma}     -\frac{1}{2}(     {{\Gamma}}^{\sigma}_{\rho\sigma}+{{\Gamma}}^{\sigma}_{\sigma\rho}     )     \mathrm{g}^{\mu\nu}     &amp;=0, \tag1   \\     \mathrm{g}^{[\mu\nu]}_{,\nu}     +\frac{1}{2}(     {{\Gamma}}^{\rho}_{\rho\nu}-{{\Gamma}}^{\rho}_{\nu\rho}     )     \mathrm{g}^{(\mu\nu)}     &amp;=0, \tag2     \end{align} $$  by contracting equation (1) once with respect to ($\mu,\rho$), then with respect to ($\nu,\rho$)?  Where $\Gamma$ is not symmetric with respect to the lower Indices.  My attempt so far to solve this problem is: Well, when contracting with respect to μ and ρ I get: $$ -\frac{1}{2} g^{\rho \nu } {\Gamma} _{a\rho}^{a}-\frac{1}{2} g^{\rho    \nu} \Gamma _{\rho a}^{a}+g^{a\nu} \Gamma _{a\rho}^{\rho}+g^{\rho a} \Gamma _{\rho a}^{\nu}+g_{,\rho}^{\rho\nu}=0 $$ and when contracting with respect to nu and ρ I get: $$ -\frac{1}{2} g^{\mu \rho } \Gamma _{a\rho}^a-\frac{1}{2} g^{\mu    \rho} \Gamma _{\rho a}^a+g_{}^{a\rho } \Gamma _{a\rho}^{\mu }+g^{\mu a} \Gamma _{\rho a}^{\rho    }+g_{,\rho }^{\mu \rho }=0 $$ when subtracting these two equations I get: $$ \frac{1}{2} g_{}^{\mu \rho } \Gamma _{a\rho }^a-\frac{1}{2} g_{}^{\rho    \nu } \Gamma _{a\rho }^a+\frac{1}{2} g_{}^{\mu \rho } \Gamma    _{\rho a}^a-\frac{1}{2} g_{}^{\rho \nu } \Gamma _{\rho    a}^a+g_{}^{a\nu } \Gamma _{a\rho }^{\rho    }-g_{}^{a\rho } \Gamma _{a\rho }^{\mu }-g_{}^{\mu a}    \Gamma _{\rho a}^{\rho }+g_{}^{\rho a} \Gamma _{\rho    a}^{\nu }-g_{,\rho }^{\mu \rho }+g_{,\rho }^{\rho \nu }=0 $$ I cant see how this is equal to equation (2)
57706
I am asked to solve $$u_x+u_y=1$$ If is was homogeneous i.e., $u_x+u_y$ the answer would be $u(x,y)=f(y-x)$ where $f$ is an arbitrary function. I have found the following set of solutions: $$u(x,y)=\lambda x +(1-\lambda)y$$ where $\lambda$ is an arbitrary constant(real or imaginary). I just have no idea what method other then trial and error would have lead me here. Any ideas? Thanks!
122424
This is an equation given in my book.     The question is why have they used a negative sign on the LHS?   Now, if you try to derive the mirror equation with simple geometry, you get  1/v +1/u =1/f . I got that equation without applying any signs there!  So what is the need to apply the signs ‘here'?   If i try to derive this equation ( the one in the photo ) i get the exact answer, but with the positive sign in the LHS. WHY we use the negative sign in the derivation, and then  again we use conventions while solving the questions?
58066
I try to reproduce the output that is generated by:  \begin{equation}     \begin{aligned}     (x-x_1)^2 + (y-y_1)^2 &amp; = r_1^2 \\     (x-x_2)^2 + (y-y_2)^2 &amp; = r_2^2 \\     (x-x_3)^2 + (y-y_3)^2 &amp; = r_3^2 \\     \end{aligned}     \label{label a} \end{equation} Additional lines: \begin{equation}     \begin{aligned}     (x-x_k)^2 + (y-y_k)^2 &amp; = r_k^2          \end{aligned}     \label{label b} \end{equation}   but make the two aligned blocks use the same alignment (as if it was just one align block. I tried to do this with just one align block + intertext for the line of text in the middle. But this gave me 4 lines of equation with 4 individual numbers which is not desired.   Thanks for your help and sorry for my bad english.