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I have two difficulties on understanding the solution to an example in a course I took this semester on optimization. This example is given to illustrate the usage of Lagrange multiplier method (please see Example 1 in the image below:
$${\Large ?}\;\left\{\begin{align*}(x_1-x_2)(x_2-x_3)(x_3-x_1)&=0\\x_1^2+x_2^2+x_3^2&=4\\x_1+x_2+x_3&=1\end{align*}\right.$$ One solution is $$x_1=x_2,\;x_3=1-2x_1,\;2x_1^2+(1-2x_1)^2=4 \\ \left(\dfrac13+\dfrac{\sqrt{22}}6,\;\dfrac13+\dfrac{\sqrt{22}}6,\;\dfrac13-\dfrac{\sqrt{22}}3\right) \\ \left(\dfrac13-\dfrac{\sqrt{22}}6,\;\dfrac13-\dfrac{\sqrt{22}}6,\;\dfrac13+\dfrac{\sqrt{22}}3\right)$$ and permutations of these.
$\underline{\text{Example 2}:}$
$$\begin{array}{ll}\text{minimize}&ax\\\text{subject to}&x_1x_2+x_1x_3+x_2x_3=0\\&x_1^2+x_2^2+x_3^2=1\end{array}$$ Necessary conditions for optimality: $$\eqalign{&{\Large\Rightarrow}\;\;\;\left\{\begin{array}{l}a_1+\lambda_1(x_2+x_3)+2\lambda_2x_1=0\\a_2+\lambda_1(x_3+x_1)+2\lambda_2x_2=0\\a_3+\lambda_1(x_1+x_2)+2\lambda_2x_3=0\\x_1x_2+x_1x_3+x_2x_3=0\\x_1^2+x_2^2+_3^2=1\end{array}\right.\\ &{\Large\Rightarrow}\;\;\;\left|\begin{matrix}a_1 & x_2+x_3 & x_1 \\ a_2 & x_3+x_1 & x_2 \\ a_3 & x_1+x_2 & x_3\end{matrix}\right|=0\\ &{\Large\Rightarrow}\;\;\;\underbrace{(x_1+x_2+x_3)}_{\neq0}\left|\begin{matrix}a_1 & 1 & x_1 \\ a_2 & 1 & x_2 \\ a_3 & 1 & x_3\end{matrix}\right|=0.}$$
I marked my question with a red "?" in the image. I can not understand how these two steps comes out. Could you please help me?
I also have the same difficulty for the second example, I hope, after I solve the first example with your help, I can understand the second one myself. But if I still fail to understand it, I'll post it and ask for help here. Sorry for my dullness in math and thank you for any help.
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The way that this is usually done in QED is deeply related to the Ward Identities and QED's LSZ reduction formula. This discussion is taken from Chapter 67-8 in Srednicki's book "Ward Identities in QED I-II".
In particular it is related to the
amputated correlators of $\langle 0 |T\{ A^\mu(x) A^\nu (0)\} |0\rangle$, which are the photon-photon scattering terms, which are related to the (momentum space) diagrams $\langle 0 |T\{ A^\mu(-p) A^\nu (p)\} |0\rangle_{amputated} $ of the form below (there could be higher loop contributions if you're masochistic enough, but this is the only diagram that contributes at 1 loop after amputation). We will consider amputated diagrams.
In particular, we'll have that
$\langle 0 |T\{ J^\mu(x) J^\nu (0)\} |0\rangle = \int \frac{d^4 p}{(2 \pi)^4} e^{i p x} [\langle 0 |T\{ A^\mu(-p) A^\nu (p)\} |0\rangle_{amputated} + \text{contact terms} ]$
where the contact terms are delta-function like terms that do not affect the divergence structure of the amplitudes.
The way to see this is to use the Ward identities and (an analog of) the LSZ scattering formula. The LSZ formula for photons relates photon-photon scattering amplitudes functions to derivates of photon-photon correlators which amputate the external legs. It says
$\langle f | i \rangle = i \epsilon_\mu^f \epsilon_\nu^i \int d^4x e^{-ipx} (-\partial_x^2) \int d^4x e^{+ipy} (-\partial_y^2) \langle 0 | A^\mu(x) A^{\nu} (y) |0 \rangle)$
where $\langle f | i \rangle$ is the amplitude going from the initial to final photon state, and $\epsilon_\mu^j$ are the initial and final polarization vectors, and the $\partial^2$ terms get rid of the leg contributions. From here, we will 'amputate' further and only consider this amplitude before contracting with the polarization.
The next ingredient is the Ward Identities for QED. The Ward identities state that if $\frac{\delta \mathcal{L}}{\delta A^\mu}$ are the equations of motion for your Lagrangian $\mathcal{L}$, then
$\langle 0| \frac{\delta \mathcal{L}}{\delta A^\mu}(x) O_1(y) O_2(z)... |0 \rangle = 0 + \text{contact terms}$.
Here, the contact terms are zero if $x,y,z$ are distinct, and will vanish inside any scattering amplitude. In addition, if all of the $O_i$ fields are distinct from $A_\mu$, then they are identically zero. For QED, the equations of motion will be
$\frac{\delta \mathcal{L}}{\delta A^\mu} = - \partial^2 A_\mu + J_\mu$
where $J_\mu = \bar \psi \gamma_\mu \psi$ is the coupling to the electromagnetic field. (I'm ignoring some counterterms that are needed actually. Srednicki includes these). This tells us that up to contact terms, we can replace $\partial^2 A_\mu$ with $\bar \psi \gamma_\mu \psi$. In particular, we'll have exactly what we stated earlier,
$\langle 0 |T\{ J^\mu(x) J^\nu (0)\} |0\rangle = \int \frac{d^4 p}{(2 \pi)^4} e^{i p x} [\langle 0 |T\{ A^\mu(-p) A^\nu (p)\} |0\rangle_{amputated} + \text{contact terms} ]$
where the contact terms are zero up to this loop order.
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10 True of False Problems about Nonsingular / Invertible Matrices Problem 500
10 questions about nonsingular matrices, invertible matrices, and linearly independent vectors.
The quiz is designed to test your understanding of the basic properties of these topics.
You can take the quiz as many times as you like.
The solutions will be given after completing all the 10 problems.
Click the View question button to see the solutions. Add to solve later Notations: $I$ denotes an identity matrix and $O$ denotes a zero matrix. The sizes of these matrices should be determined from the context. 10 True or False Problems about Nonsingular Matrix Operations Quiz-summary
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1 2 3 4 5 6 7 8 9 10 Information
Determine whether each of the following sentences are
True or False.
Notations: $I$ denotes an identity matrix and $O$ denotes a zero matrix. The sizes of these matrices should be determined from the context.
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Average score Your score 1 2 3 4 5 6 7 8 9 10 Answered Review Question 1 of 10 1. Question
True or False. Suppose that $A$ and $B$ are nonsingular $n\times n$ matrices. Then $A+B$ is nonsingular.Correct
False. For example, let $A=I$ and $B=-I$. Then both matrices are nonsingular but $A+B=O$ is singular.Incorrect
False. For example, let $A=I$ and $B=-I$. Then both matrices are nonsingular but $A+B=O$ is singular.
Question 2 of 10 2. Question
True or False. If a square matrix has no zero rows or columns, then it has an inverse matrix.Correct
False. As a counterexample, consider
\[A=\begin{bmatrix} 1 & 1\\ 1& 1 \end{bmatrix}.\] Then $A$ has no zero rows or columns, yet is does not have the inverse matrix as the determinant of $A$ is zero.Incorrect
False. As a counterexample, consider
\[A=\begin{bmatrix} 1 & 1\\ 1& 1 \end{bmatrix}.\] Then $A$ has no zero rows or columns, yet is does not have the inverse matrix as the determinant of $A$ is zero. Question 3 of 10 3. Question
True or False. Let $A$ be an $m \times n$ matrix.
If the equation $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\mathbf{x}\in \R^n$, then the columns of $A$ are linearly independent.Correct
True. Let
\[B=\begin{bmatrix} B_1 & B_2 &\dots & B_n \\ \end{bmatrix},\] where $B_i$ is the $i$-th column vector of $B$ for $i=1, \dots, n$. Suppose that we have a linear combination \[c_1B_1+c_2B_2+\cdots+c_n B_n=\mathbf{0}\] for some scalars $c_1, c_2, \dots, c_n$. Then we can write it as \[\begin{bmatrix} B_1 & B_2 &\dots & B_n \\ \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}=\mathbf{0}.\]
Since $B\mathbf{x}=\mathbf{0}$ has only the trivial solution, we must have
\[\begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}=\mathbf{0}.\] Hence $c_1=c_2=\cdots=c_n=0$, and the column vectors $B_1, B_2, \dots, B_n$ are linearly independent.Incorrect
True. Let
\[B=\begin{bmatrix} B_1 & B_2 &\dots & B_n \\ \end{bmatrix},\] where $B_i$ is the $i$-th column vector of $B$ for $i=1, \dots, n$. Suppose that we have a linear combination \[c_1B_1+c_2B_2+\cdots+c_n B_n=\mathbf{0}\] for some scalars $c_1, c_2, \dots, c_n$. Then we can write it as \[\begin{bmatrix} B_1 & B_2 &\dots & B_n \\ \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}=\mathbf{0}.\]
Since $B\mathbf{x}=\mathbf{0}$ has only the trivial solution, we must have
\[\begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}=\mathbf{0}.\] Hence $c_1=c_2=\cdots=c_n=0$, and the column vectors $B_1, B_2, \dots, B_n$ are linearly independent. Question 4 of 10 4. Question
True or False. Let $A$ be an $m \times n$ matrix.
If the equation $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\mathbf{x}\in \R^n$, then the rows of $A$ are linearly independent.Correct
False. As a counterexample, consider the $3\times 2$ matrix
\[A=\begin{bmatrix} 1 & 0 \\ 1 & 0 \\ 0 &1 \end{bmatrix}.\] Then the equation $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\mathbf{x}=\mathbf{0}$, and yet the first and the second row of $A$ are linearly dependent.Incorrect
False. As a counterexample, consider the $3\times 2$ matrix
\[A=\begin{bmatrix} 1 & 0 \\ 1 & 0 \\ 0 &1 \end{bmatrix}.\] Then the equation $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\mathbf{x}=\mathbf{0}$, and yet the first and the second row of $A$ are linearly dependent. Question 5 of 10 5. Question
True or False. The row echelon form of an invertible $3\times 3$ matrix is invertible.Correct
True. The reduced row echelon form of an invertible matrix is the identity matrix, which is invertible.Incorrect
True. The reduced row echelon form of an invertible matrix is the identity matrix, which is invertible.
Question 6 of 10 6. Question
True or False. There is a non-zero nonsingular matrix $A$ such that $A^2=O$.Correct
False. Suppose $A$ is nonsingular such that $A^2=O$.
Since $A$ is nonsingular, it is invertible. Hence we have \begin{align*} A=A^{-1}A^2=A^{-1}O=O, \end{align*} and the matrix $A$ must be the zero matrix.Incorrect
False. Suppose $A$ is nonsingular such that $A^2=O$.
Since $A$ is nonsingular, it is invertible. Hence we have \begin{align*} A=A^{-1}A^2=A^{-1}O=O, \end{align*} and the matrix $A$ must be the zero matrix. Question 7 of 10 7. Question
True or False. If $A$ and $B$ are invertible $n\times n$ matrices, then $AB=BA$.Correct
False. For example, consider
\[A=\begin{bmatrix} 1 & 1\\ 0& 1 \end{bmatrix} \text{ and } B=\begin{bmatrix} 1 & 0\\ 0& 2 \end{bmatrix}.\] Then these matrices are invertible as their determinants are $\det(A)=1\neq 0$ and $\det(B)=2\neq 0$. However they do not commute since \begin{align*} AB=\begin{bmatrix} 1 & 2\\ 0& 2 \end{bmatrix} \text{ and } BA=\begin{bmatrix} 1 & 1\\ 0& 2 \end{bmatrix}. \end{align*}Incorrect
False. For example, consider
\[A=\begin{bmatrix} 1 & 1\\ 0& 1 \end{bmatrix} \text{ and } B=\begin{bmatrix} 1 & 0\\ 0& 2 \end{bmatrix}.\] Then these matrices are invertible as their determinants are $\det(A)=1\neq 0$ and $\det(B)=2\neq 0$. However they do not commute since \begin{align*} AB=\begin{bmatrix} 1 & 2\\ 0& 2 \end{bmatrix} \text{ and } BA=\begin{bmatrix} 1 & 1\\ 0& 2 \end{bmatrix}. \end{align*} Question 8 of 10 8. Question
True or False. If $A$ and $B$ are $n\times n$ nonsingular matrices such that $A^2=I$ and $B^2=I$, then $(AB)^{-1}=BA$.Correct
True. Note that we have
\[A=A^{-1}A^2=A^{-1}I=A^{-1}.\] Similarly, we have $B^{-1}=B$.
It follows that
\begin{align*} (AB)^{-1}=B^{-1}A^{-1}=BA. \end{align*}Incorrect
True. Note that we have
\[A=A^{-1}A^2=A^{-1}I=A^{-1}.\] Similarly, we have $B^{-1}=B$.
It follows that
\begin{align*} (AB)^{-1}=B^{-1}A^{-1}=BA. \end{align*} Question 9 of 10 9. Question
True or False. If $A$ is an $m \times n$ matrix such that $A\mathbf{x}=\mathbf{0}$ for every vector $\mathbf{x}$ in $\R^n$, then $A$ is the $m\times n$ zero matrix.Correct
True. For each $i=1, 2, \dots, n$, let $\mathbf{e}_i\in \R^n$ be the $n$-dimensional vector whose $i$-th entry is $1$ and $0$ elsewhere.
Then $A\mathbf{e}_i$ is the $i$-th column vector of the matrix $A$.
Since by assumption $A\mathbf{e}_i=\mathbf{0}$, we see that the $i$-th column of $A$ is the zero vector.
As this is true for any $i=1, \dots, n$, we conclude that $A$ is the zero matrix.Incorrect
True. For each $i=1, 2, \dots, n$, let $\mathbf{e}_i\in \R^n$ be the $n$-dimensional vector whose $i$-th entry is $1$ and $0$ elsewhere.
Then $A\mathbf{e}_i$ is the $i$-th column vector of the matrix $A$.
Since by assumption $A\mathbf{e}_i=\mathbf{0}$, we see that the $i$-th column of $A$ is the zero vector.
As this is true for any $i=1, \dots, n$, we conclude that $A$ is the zero matrix. Question 10 of 10 10. Question
True or False. Let $A$ be a $2 \times 2$ nonsingular matrix and let $\mathbf{v}_1$ and $\mathbf{v}_2$ be linearly independent vectors in $\R^2$.
Then the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent vectors in $\R^2$.Correct
True. Consider a linear combination
\[c_1(A\mathbf{v}_1)+c_2(A\mathbf{v}_2)=\mathbf{0},\] where $c_1, c_2$ are scalars.
It yields that
\[A(c_1\mathbf{v}_1+c_2\mathbf{v}_2)=\mathbf{0}.\]
Since $A$ is nonsingular, we have
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}.\]
Because the vectors $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent, we have
\[c_1=c_2=0.\] Thus, the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent vectors.Incorrect
True. Consider a linear combination
\[c_1(A\mathbf{v}_1)+c_2(A\mathbf{v}_2)=\mathbf{0},\] where $c_1, c_2$ are scalars.
It yields that
\[A(c_1\mathbf{v}_1+c_2\mathbf{v}_2)=\mathbf{0}.\]
Since $A$ is nonsingular, we have
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}.\]
Because the vectors $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent, we have
\[c_1=c_2=0.\] Thus, the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are linearly independent vectors.
Add to solve later
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Proving The Existence of Limits Examples 4
We will now look at some more examples of proving the existence of limits using the definition of a limit, that is $\lim_{x \to a} f(x) = L$ says that $\forall \epsilon > 0 \: \exists \delta > 0 \: \mathrm{s.t.} \: \forall x : 0 < \mid x - a \mid < \delta, \mid f(x) - L \mid < \epsilon$. More examples can be found on the following pages:
Proving The Existence of Limits Examples 1 Proving The Existence of Limits Examples 2 Proving The Existence of Limits Examples 3 Proving The Existence of Limits Examples 4 Proving The Existence of Limits Examples 5 Example 1 Prove that $\lim_{x \to 0} \mid x \mid = 0$.
Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $0 < \mid x \mid < \delta$ then $\mid \mid x \mid \mid < \epsilon$.
We note that $\mid \mid x \mid \mid = \mid x \mid < \epsilon$. So choose $\delta = \epsilon$ so that if $0 < \mid x \mid < \delta = \epsilon$ then $\mid \mid x \mid \mid < \epsilon$.
Therefore $\lim_{x \to 0} \mid x \mid = 0$.
Example 2 Prove that $\lim_{x \to -2} (x^2 - 1) = 3$.
Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $0 < \mid x + 2 \mid < \delta$ then $\mid (x^2 - 1) - 3 \mid < \epsilon$.
We first start with the following inequality:(1)
We now want to find a $C > 0$ such that $\mid x - 2 \mid < C$. We will restrict $x$ to lie in some interval centered at $-2$ and within a distance of $1$ or in other words, $\delta ≤ 1$. Therefore:(2)
Therefore we choose $C = 5$ and thus:(3)
Now we need to choose a $\delta$ that satisfies $\mid x + 2 \mid < \delta ≤ 1$ and $\mid x + 2 \mid < \frac{\epsilon}{5}$. We will choose $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{5} \}$.
Now suppose that $\epsilon < 5$. Then $\delta = \frac{\epsilon}{5}$, and so:(4)
Now suppose that $\epsilon ≥ 5$. Then $\delta = 1$. Since $\delta$ is defined to be the minimum of $1$ and $\frac{\epsilon}{5}$, it follows that $1 ≤ \frac{\epsilon}{5}$, and so:(5)
Therefore $\lim_{x \to -2} (x^2 - 1) = 3$.
Example 3 Prove that $\lim_{x \to 2} x^3 = 8$.
Let $\epsilon > 0$ be given and find $\delta > 0$ such that if $0 < \mid x - 2 \mid < \delta$ then $\mid x^3 - 8 \mid < \epsilon$.
We will start with the following inequality:(6)
We want to find a $C > 0$ such that $\mid x^2 + 2x + 4 \mid < C$. Suppose that we restrict $x$ to be within 1 from 2. That is, $\delta ≤ 1$. Therefore $\mid x^2 + 2x + 4 \mid < 19$ as shown with the following diagram:
We can also solve this algebraically. If $\delta ≤ 1$, then:(7)
Remember we restricted $x$ such that $1 < x < 3$, so then $7 < x^2 + 2x + 4 < 19$. Therefore, we will choose $C = 19$ so then:(8)
We will let $\delta = \mathrm{min} \{ 1, \frac{\epsilon}{19} \}$.
Now if $\epsilon < 19$, then $\delta = \frac{\epsilon}{19}$ and thus:(9)
Now if $\epsilon ≥ 19$, then $\delta = 1$. But since $\delta$ is the minimum of $1$ and $\frac{\epsilon}{19}$, then this implies that $1 < \frac{\epsilon}{19}$ and thus:(10)
Therefore $\lim_{x \to 2} x^3 = 8$.
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First, you should note that the set of isolated points of $E$ is countable. This is in fact a general property of $\mathbb{R}$
Theorem: Let $E$ be a subset of $\mathbb{R}$ and let $F$ be the set of isolated points of $\mathbb{R}$. Then $F$ is at most countable.
Proof: Suppose otherwise, that is, that $F$ is uncountable. Then there exists some interval $[k,k+1]$ such that $F\cap[k,k+1]$ is uncountable. For each $x\in F\cap [k,k+1]$, choose a rational number $q_x$, $0<q_x<1$ such that $(x-2q_x,x+2q_x)\cap F=\varnothing$. Since the set $\left\{q_x:x\in F\cap[k,k+1]\right\}$, then there exists some $q$ such that $X=\left\{x:q_x=q\right\}$ is uncountable, in particular infinite. The choice of $q_x$ implies that the sets $(x-q,x+q)$ are all disjoint for $x\in X$, and they are all contained in $[k-1,k+2]$. Therefore, we constructed an infinite family of disjoint intervals of length $2q$, all of which are contained in the bounded interval $[k-1,k+2]$, a contradiction. QED
(Probably, there is a nicer proof of this theorem somewhere in this site.)
Therefore, we should not worry about the isolated points of $E$ when analysing derivatives: the set of isolated points has null measure.
A trick that works here is to extend your function $f$ to an interval containing $E$. We can do this in the following manner:
Let $E\subseteq\mathbb{R}$ and $f:E\to\mathbb{R}$ be monotonic. The function $\hat{f}:(\inf E,\sup E)\to\mathbb{R}$ given by $\hat{f}(x)=\sup_{y\in E,y\leq x}f(y)$ is an extension of $f$ (if $\sup E$ or $\inf E\in E$, define $\hat{f}(\sup E)=f(\sup E)$ or $\hat{f}(\inf E)=f(\inf E)$).
Another extension is given by $\overline{f}(x)=\inf_{y\in E,y\geq x}f(x)$. In fact, you can check that if $g$ is any other extension of $f$ defined on $[\inf E,\sup E]\cap E$, then $\hat{f}(x)\leq g(x)\leq\overline{f}(x)$ for all $x$.
Alternatively, you can prove this with Zorn's Lemma, but the argument is basically the same: Zorn's lemma gives you a maximal extension of $f$ to a monotonic function $\widetilde{f}:F\to \mathbb{R}$ defined on some subset $F\supseteq E$. To show that $F$ is an interval you apply the argument above and extend $\widetilde{f}$ to some interval containing $F$. Maximality implies that $F$ is that interval.
Now, about your question of differentiability of $f$: For almost every point $x$ of $(\inf E,\sup E)$, the function $\hat{f}$ is differentiable at $x$. But we also know that almost every point of $E$ is not isolated. Using these two fact, we conclude that almost every point $x$ of $E\cap(\inf E,\sup E)$ is not an isolated point of $E$, and $\hat{f}$ is differentiable at $x$. You can then check that for such $x$, $f$ is differentiable at $x$, and $f'(x)=\hat{f}'(x)$.
Therefore, $f$ is differentiable at almost every point of $E$.
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Working out the non relativistic limit of the Dirac equation, we encounter this quantity: $(\vec{\sigma} \cdot \vec{p})$ and in my notes it says that $$ (\vec{\sigma} \cdot \vec{p})^2 = p^i p^j\sigma^i\sigma^j=\vec p^{\,2} \tag{1} $$
When we couple the Dirac equation and we write $$\vec{p} \rightarrow \vec{p} - \frac{e}{c} \vec{A} \equiv \vec{\pi} $$
we obtain a similar quantity: $(\vec{\sigma} \cdot \vec{\pi})$, but to calculate its square we now use the fact that $\sigma^i \sigma^j= \delta^{ij} + i \epsilon^{ijk}\sigma^k$ and we obtain
$$ (\vec{\sigma} \cdot \vec{\pi})^2= \pi^i \pi^j \sigma^i \sigma^j= \vec{\pi}^{\,2} + i \epsilon^{ijk} \pi^i \pi^j\sigma^k \tag{2}$$
Question:
Why does the $\epsilon^{ijk}$ term vanish in $(1)$ but it does not vanish in $(2)$?
Thank you for any help in advance
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Difference between revisions of "Main Page"
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== The Problem ==
== The Problem ==
−
Let <math>[3]^n</math> be the set of all length <math>n</math> strings over the alphabet <math>1, 2, 3</math>. A ''combinatorial line'' is a set of three points in <math>[3]^n</math>, formed by taking a string with one or more wildcards <math>
+
Let <math>[3]^n</math> be the set of all length <math>n</math> strings over the alphabet <math>1, 2, 3</math>. A ''combinatorial line'' is a set of three points in <math>[3]^n</math>, formed by taking a string with one or more wildcards <math></math> in it, e.g., <math>\ldots</math>, and replacing those wildcards by <math>1, 2</math> and <math>3</math>, respectively. In the example given, the resulting combinatorial line is: <math>\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}</math>. A subset of <math>[3]^n</math> is said to be ''line-free'' if it contains no lines. Let <math>c_n</math> be the size of the largest line-free subset of <math>[3]^n</math>.
'''<math>k=3</math> Density Hales-Jewett (DHJ(3)) theorem:''' <math>\lim_{n \rightarrow \infty} c_n/3^n = 0</math>
'''<math>k=3</math> Density Hales-Jewett (DHJ(3)) theorem:''' <math>\lim_{n \rightarrow \infty} c_n/3^n = 0</math>
Revision as of 09:22, 14 February 2009 The Problem
Let [math][3]^n[/math] be the set of all length [math]n[/math] strings over the alphabet [math]1, 2, 3[/math]. A
combinatorial line is a set of three points in [math][3]^n[/math], formed by taking a string with one or more wildcards [math]\ast[/math] in it, e.g., [math]112\!\ast\!\!1\!\ast\!\ast3\ldots[/math], and replacing those wildcards by [math]1, 2[/math] and [math]3[/math], respectively. In the example given, the resulting combinatorial line is: [math]\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}[/math]. A subset of [math][3]^n[/math] is said to be line-free if it contains no lines. Let [math]c_n[/math] be the size of the largest line-free subset of [math][3]^n[/math]. [math]k=3[/math] Density Hales-Jewett (DHJ(3)) theorem: [math]\lim_{n \rightarrow \infty} c_n/3^n = 0[/math]
The original proof of DHJ(3) used arguments from ergodic theory. The basic problem to be considered by the Polymath project is to explore a particular combinatorial approach to DHJ, suggested by Tim Gowers.
Useful background materials
Some background to the project can be found here. General discussion on massively collaborative "polymath" projects can be found here. A cheatsheet for editing the wiki may be found here. Finally, here is the general Wiki user's guide
Threads (1-199) A combinatorial approach to density Hales-Jewett (inactive) (200-299) Upper and lower bounds for the density Hales-Jewett problem (inactive) (300-399) The triangle-removal approach (inactive) (400-499) Quasirandomness and obstructions to uniformity (inactive) (500-599) Possible proof strategies (active) (600-699) A reading seminar on density Hales-Jewett (active) (700-799) Bounds for the first few density Hales-Jewett numbers, and related quantities (active)
There is also a chance that we will be able to improve the known bounds on Moser's cube problem.
Here are some unsolved problems arising from the above threads.
Here is a tidy problem page.
Proof strategies
It is natural to look for strategies based on one of the following:
Szemerédi's original proof of Szemerédi's theorem. Szemerédi's combinatorial proof of Roth's theorem. Ajtai-Szemerédi's proof of the corners theorem. The density increment method. The triangle removal lemma. Ergodic-inspired methods. The Furstenberg-Katznelson argument. Bibliography
Density Hales-Jewett
H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem for k=3“, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–241. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem“, J. Anal. Math. 57 (1991), 64–119. R. McCutcheon, “The conclusion of the proof of the density Hales-Jewett theorem for k=3“, unpublished.
Behrend-type constructions
M. Elkin, "An Improved Construction of Progression-Free Sets ", preprint. B. Green, J. Wolf, "A note on Elkin's improvement of Behrend's construction", preprint. K. O'Bryant, "Sets of integers that do not contain long arithmetic progressions", preprint.
Triangles and corners
M. Ajtai, E. Szemerédi, Sets of lattice points that form no squares, Stud. Sci. Math. Hungar. 9 (1974), 9--11 (1975). MR369299 I. Ruzsa, E. Szemerédi, Triple systems with no six points carrying three triangles. Combinatorics (Proc. Fifth Hungarian Colloq., Keszthely, 1976), Vol. II, pp. 939--945, Colloq. Math. Soc. János Bolyai, 18, North-Holland, Amsterdam-New York, 1978. MR519318 J. Solymosi, A note on a question of Erdős and Graham, Combin. Probab. Comput. 13 (2004), no. 2, 263--267. MR 2047239
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A boring method is to carefully apply the (partially) extended Euclidean algorithm.
But in the question, the modulus is a power of two (specifically $2^6$), and we can use that$$a\,x\equiv1\pmod{2^k}\implies a\,x\,(2-a\,x)\equiv1\pmod{2^{2k}}$$from which it follows this fact:
if the modular inverse of $a$ modulo $2^k$ is (the lower $k$ bits of) $x$, then
the modular inverse of $a$ modulo $2^{2k}$ is (the lower $2k$ bits of) $x\,(2-a\,x)$
(where negative integers are in 2's-complement convention, dominant in modern CPUs).
This not-so-much-known fact allows computation of multiplicative inverse modulo $2^k$. We start from an inverse $x$ of $a$ over few bits (that can be $x=a$, perhaps $\bmod 7$, which is the inverse for any odd $a$ over three bits), and iterate $x\gets x\,(2-a\,x)$, possibly truncated to the number of known-correct result bits. That number of bits doubles at each iteration, thus about $\log_2(k)$ steps are enough, and it is only used product, subtraction, and bit truncation on values no wider than $k$ bits. That is blindingly fast compared to the Euclidean algorithm's $O(k)$ steps; and eases getting data-independent execution time, which comes handy in some cryptographic computations (e.g. the preliminary computation of $m'$ in Montgomery multiplication, algorithm 14.36 of Alfred J. Menezes, Paul C. van Oorschot and Scott A. Vanstone's Handbook of Applied Cryptography).
I learned the technique from Colin Plumb's
Computing multiplicative inverses (post on sci.crypt with Message-ID: <1994Apr6.093116.27805@mnemosyne.cs.du.edu>, 1994). His statement applies to inverse modulo a prime power, and points the relation to the Newton's iteration for finding $x = 1/a$ in $\Bbb R$.
A modern exposition, with benchmarks, is in Jean-Guillaume Dumas:
On Newton-Raphson iteration for multiplicative inverses modulo prime powers.
A bibliography, and other techniques faster than the Euclidean algorithm, are in Çetin Kaya Koç:
A New Algorithm for Inversion mod $p^k$.
Here, to perform the desired computation quickly, we use $k=3$, $a=47$, and compute $a\bmod2^k=47\bmod8=7$, which multiplicative inverse modulo $8$ is also $x=7$. Now we compute$$\begin{align}(x\,(2-a\,x))\bmod2^{2k}&=(7\,(2-47\times7))\bmod 64\\&=15\end{align}$$
Hence the desired modular inverse of $47$ modulo $64$ is $15$.
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Now showing items 1-10 of 18
J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ...
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Limits of Products of Complex-Valued Functions
Recall from the Limits of Sums and Differences of Complex-Valued Functions page that if $(S, d_S)$ and $(\mathbb{C}, d)$ are metric spaces where $d$ is the usual metric on $\mathbb{C}$ defined for all $x, y \in \mathbb{C}$ by $d(x, y) = \mid x - y \mid$ then if $A \subseteq S$, $p \in S$ is an accumulation point of $A$, and $f, g : A \to \mathbb{C}$ are such that $\lim_{x \to p} f(x) = a$ and $\lim_{x \to p} g(x) = b$ then:(1)
We will now look at an important result regarding the limits of products of complex-valued functions.
Theorem 1: Let $(S, d_S)$ and $(\mathbb{C}, d)$ be metric spaces where $d$ is the usual metric defined for all $x, y \in \mathbb{C}$ by $d(x, y) = \mid x - y \mid$. Also let $A \subseteq S$, $p \in S$ be an accumulation point of $A$, and $f, g : A \to \mathbb{C}$. Suppose that $\lim_{x \to p} f(x) = a$ and $\lim_{x \to p} g(x) = b$. Then $\lim_{x \to p} f(x)g(x) = ab$. Proof:Let $\epsilon$ be such that $0 < \epsilon < 1$. If $\epsilon \geq 1$ then we can such take a smaller $\delta$ corresponding to a smaller $\epsilon_0$ such that $0 < \epsilon_0 < 1$ and the inequalities will still hold. Since $\lim_{x \to p} f(x) = a$ we have that for $\epsilon_1 = \frac{\epsilon}{\mid a \mid + \mid b \mid + 1} > 0$ there exists a $\delta_1 > 0$ such that if $x \in D(f) \setminus \{ p \}$ and $d_S(x, p) < \delta_1$ then $d(f(x), a) < \epsilon_1$, $(*)$. Similarly, since $\lim_{x \to p} g(x) = b$ we have that for $\epsilon_2 = \frac{\epsilon}{\mid a \mid + \mid b \mid + 1}$ there exists a $\delta_2 > 0$ such that if $x \in D(g) \setminus \{ p \}$ and $d_S(x, p) < \delta_2$ then $d(g(x), b) < \epsilon_2$, $(**)$. Now since $\epsilon < 1$ we have that $\epsilon_1, \epsilon_2 < 1$, so then: Let $\delta = \min \{ \delta_1, \delta_2 \}$. Then if $x \in D(f) \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $d_S(x, p) < \delta_1$ AND $d_S(x, p) < \delta_2$ so $(*)$, $(**)$, and $(***)$ will hold and: Therefore $\lim_{x \to p} f(x)g(x) = ab$. $\blacksquare$
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Open and Closed Set Differences in Metric Spaces
Open and Closed Set Differences in Metric Spaces
Suppose that $(M, d)$ is a metric space and that $A, B \subseteq M$. Suppose that we know that $A$ is an open subset and $B$ is a closed subset. What can we say about the differences $A \setminus B$ and $B \setminus A$? Are they necessarily open? Are they necessarily closed? The theorem below will tell us that $A \setminus B$ is always open and $B \setminus A$ is always closed. In the theorems below, we use the important fact that $A \setminus B = A \cap (M \setminus B)$ and $B \setminus A = B \cap (M \setminus A)$:
Theorem 1: Let $(M, d)$ be a metric space and let $A, B \subseteq M$. If $A$ is an open subset and $B$ is a closed subset then $A \setminus B$ is an open subset. (1) Proof: Let $A, B \subseteq M$ and let $A$ be an open subset and let $B$ be a closed subset. Then $B^c = M \setminus B$ is an open subset and
\begin{align} \quad A \setminus B = A \cap (M \setminus B) \end{align}
Then $A \setminus B$ is the intersection of two open sets. A finite intersection of open sets is open, so $A \setminus B$ is open. $\blacksquare$
Theorem 2: Let $(M, d)$ be a metric space and let $A, B \subseteq M$. If $A$ is an open subset and $B$ is a closed subset then $B \setminus A$ is a closed subset. (2) Proof: Let $A, B \subseteq M$ and let $A$ be an open subset and let $B$ be a closed subset. Then $A^c = M \setminus A$ is a closed subset and:
\begin{align} \quad B \setminus A = B \cap (M \setminus A) \end{align}
Then $B \setminus A$ is the intersection of two closed sets. Any union of closed sets is closed (in particular, a finite union of closed sets is closed), so $B \setminus A$ is closed. $\blacksquare$
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Darboux's theorem says that any symplectic manifold $(M^{2n}, \omega)$ is locally symplectomorphic to the "trivial" symplectic manifold $( \Bbb R^{2n}, \omega_0)$, where$$\omega_0 = \sum_{j = 1}^n dx^j \wedge dy^j$$is the standard symplectic form on $\Bbb R^{2n}$. What this means is the following. Given any point $p \in M$, there is a neighborhood $U \subset M$ of $p$ and a diffeomorphism$$\psi: U \longrightarrow \Bbb R^{2n}$$such that $\psi^\ast \omega_0 = \omega$ on $U$,
i.e. $\psi$ is a symplectomorphism. Note that we do not throw away the symplectic structure in this notion of equivalence; without the symplectic structure there is no symplectic geometry to speak of in the first place. "Equivalence" between two symplectic manifolds means that they are symplectomorphic.
A symplectic manifold $(M, \omega)$ is different from a Riemannian manifold $(N, g)$. A symplectic form $\omega$ is a nondegenerate $2$-form,
i.e. a nondegenerate antisymmetric rank $2$ tensor. A Riemannian metric $g$, on the other hand, is a nondegenerate symmetric rank $2$ tensor. So right away we see that symplectic manifolds and Riemannian manifolds are fairly different objects. But this is not the reason we say that Riemannian geometry and symplectic geometry are totally different subjects. Darboux's theorem implies that there are no local invariants of symplectic manifolds; any such invariant would also be a symplectomorphism invariant of $(\Bbb R^{2n}, \omega_0)$, and hence we wouldn't be able to discern anything special about a symplectic manifold $(M, \omega)$ by inspecting it locally. This is in stark contrast to Riemannian geometry, where the curvature of a Riemannian manifold is a local object.
Therefore the reason we say symplectic geometry is a totally different subject than Riemannian geometry is that symplectic manifolds have no local symplectomorphism invariants, while local isometry invariants of Riemannian manifolds such as curvature are quite central to the field of Riemannian geometry.
Added: The proof of Darboux's theorem relies on the fact that a symplectic form $\omega$ is closed, i.e. $d\omega = 0$. If $\omega$ is not closed ( i.e. we have an almost symplectic manifold) then we do not necessarily have Darboux's theorem. So while Riemannian geometry is in some sense about nondegenerate symmetric bilinear forms on tangent spaces, symplectic geometry is about nondegenerate antisymmetric bilinear forms on tangent spaces with an extra condition described by the differential equation $d\omega = 0$. The extra condition is what makes the theory qualitatively different.
We can impose a differential equation for the metric on a Riemannian manifold to get a "version of Darboux's theorem" for Riemannian geometry. The analogous condition in Riemannian geometry is having a metric $g$ whose Riemann curvature tensor is everywhere zero (
i.e. a flat metric); such a manifold is locally isometric to $\Bbb R^n$ with the standard metric.
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Dihedral Group and Rotation of the Plane Problem 52
Let $n$ be a positive integer. Let $D_{2n}$ be the dihedral group of order $2n$. Using the generators and the relations, the dihedral group $D_{2n}$ is given by
\[D_{2n}=\langle r,s \mid r^n=s^2=1, sr=r^{-1}s\rangle.\] Put $\theta=2 \pi/n$. (a)Prove that the matrix $\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix}$ is the matrix representation of the linear transformation $T$ which rotates the $x$-$y$ plane about the origin in a counterclockwise direction by $\theta$ radians.
(b)Let $\GL_2(\R)$ be the group of all $2 \times 2$ invertible matrices with real entries. Show that the map $\rho: D_{2n} \to \GL_2(\R)$ defined on the generators by
\[ \rho(r)=\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{bmatrix} \text{ and }
\rho(s)=\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}\] extends to a homomorphism of $D_{2n}$ into $\GL_2(\R)$.
(c)Determine whether the homomorphism $\rho$ in part (b) is injective and/or surjective.
Contents
Hint. For (a), consider the unit vectors of the plane and consider where do the unit vector go by the linear transformation $T$. Show that $\rho(r)$ and $\rho(s)$ satisfy the same relations as $D_{2n}. Consider the determinant. Proof. (a) The matrix representation of the linear transformation $T$
Let $\mathbf{e}_1, \mathbf{e}_2$ be the standard basis of the plane $\R^2$. That is
\[\mathbf{e}_1=\begin{bmatrix} 1 \\ 0 \end{bmatrix} \text{ and } \mathbf{e}_2=\begin{bmatrix} 0 \\ 1 \end{bmatrix}.\] Then by the $\theta$ rotation $\mathbf{e}_1$ moves to the point $\begin{bmatrix} \cos \theta \\ \sin \theta \end{bmatrix}$ and $\mathbf{e}_2$ moves to the point $\begin{bmatrix} -\sin \theta \\ \cos \theta \end{bmatrix}$. Therefore the matrix representation of $T$ is the matrix $\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix}$.
(Recall that if $T$ is a linear transformation from a vector space $V$ to itself with a basis $\{\mathbf{e}_1, \dots, \mathbf{e}_n\}$, its representation matrix is given by the matrix $[T(\mathbf{e}_1) \cdots T(\mathbf{e}_n)]$ whose $i$-th column is the vector $T(\mathbf{e}_i)$.)
(b) $\rho$ is a homomorphism of $D_{2n}$ into $\GL_2(\R)$
Any element $x \in D_{2n}$ can be written as $x=r^as^b$ using the relations.
Then we define the value of $\rho$ on $x$ by \[\rho(x):=\rho(r)^a\rho(s)^b=\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix}^a \begin{bmatrix} 0 & 1\\ 1& 0 \end{bmatrix}^b.\]
We need to show that this is well defined.
To do this, we show that $\rho(r)$ and $\rho(s)$ satisfy the same relation as $D_{2n}$.
We have
\[\rho(r)^n=\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix}^n= \begin{bmatrix} \cos (n\theta) & -\sin (n\theta)\\ \sin (n\theta)& \cos (n\theta) \end{bmatrix} =I_2,\] where $I_2$ is the $2\times 2$ identity matrix. Also we have $\rho(s)^=I_2$.
Finally, we compute
\begin{align*} \rho(r)\rho(s)\rho(r)&=\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix} \begin{bmatrix} 0 & 1\\ 1& 0 \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix} \\[6pt] &=\begin{bmatrix} – \sin \theta & \cos \theta\\ \cos \theta& \sin \theta \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix} \\[6pt] &=\begin{bmatrix} 0 & \sin^2 \theta + \cos^2 \theta\\ \cos^2+\sin^2 \theta& 0 \end{bmatrix}=I_2 \end{align*} Therefore, the extension of $\rho$ does not depend on the expression of $x=r^as^b$. (c) Determine whether the homomorphism $\rho$ in part (b) is injective and/or surjective.
We first show that $\rho$ is injective.
Suppose that we have $\rho(x)=I_2$ for $x \in D_{2n}$. Write $x=r^as^b$. Then we have $\rho(r)^a\rho(s)^b=I_2$.
We compute the determinant of both sides and get
\[\det(\rho(r))^a \det(\rho(s))^b=1.\] Since $\det(\rho(r))=1$ and $\det(\rho(s))=-1$ we have $(-1)^b=1$, thus $b$ must be even. Then $x=r^a$ since the order of $s$ is two. Then $\rho(r)^a=I_2$ implies that $r\theta=2\pi m$ for some $m\in \Z$.
Hence $r=nm$ and we obtain $x=r^{nm}=1$ since the order of $r$ is $n$. Therefore the kernel of $\rho$ is trivial, hence the homomorphism $\rho$ is injective.
As the argument shows, the determinant of $\rho(x)$ is either $\pm 1$. The homomorphism $\rho$ is not surjective since $\GL_2(\R)$ contains elements with determinants not equal to $\pm 1$.
Add to solve later
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If ice is "all around the sun" I fail to see how it can be moving at a velocity of 1000 m/s inwards. The mass of the sun is $2\cdot 10^{30}\mathrm{\;kg}$ and the radius $7\cdot 10^{8}\mathrm{\;m}$.
The thickness of a shell of ice with that inner radius and mass would be (assuming the usual density of ice of about 0.9x that of liquid water) approximately $10^8 \mathrm{\;m}$. That is probably sufficiently thick to withstand the gravitational attraction of the sun; certainly it will stop most solar radiation.
To take one gram of ice from absolute zero to melting takes roughly 273*4.2+334=1500 J. The sun puts out about $4\cdot 10^{26}$ W - assuming that the ice would absorb all that heat, it would take $8\cdot 10^9$ seconds to melt all the ice - a bit more than 200 years.
All that time the sun would be happily continuing to produce power - but I expect all life on earth would have ceased by the time it starts to shine again.
One obvious question - would the spherical "ice shell" be stable undertake the tremendous gravitational stress? Would it melt under the pressure? Would the pressure of the steam generated blow the water / ice outwards? It would be interesting to analyze those questions further. I suspect the over all conclusion - that a 100,000 km thick layer of water will "turn out the lights on Earth" will be unaltered by these details - because that water will still be between the Earth and the Sun, regardless of the distance and the phase.
update - a few additional thoughts.
First - the crushing strength of ice is quite low: no more than about 1000 psi (7 MPa) according to this USGS report. That is obviously many orders of magnitude smaller than the pressure on the inside of the 100,000 km thick ice shell. The average distance of the shell (mid point) is $7.5\cdot 10^8 \mathrm{\;m}$ from the center of the Sun, and will therefore experience a gravitational acceleration of
$$a = \frac{GM}{R^2} = \frac{6.7\cdot 10^{-11} \cdot 2\cdot 10^{30}}{(7.5\cdot 10^8)^2} = 240 \mathrm{\;m/s^2}$$
Thus the pressure on the inner surface is roughly
$$P = \rho a t = 0.9E3 \cdot 240 \cdot 1E8 = 2.5 \cdot 10^{13} Pa = 22 TPa$$
An obvious question to ask: what happens to ice at that pressure? The phase diagram I could find (at this location) "only" goes up to 1 TPa, but it suggests that "really cold" ice does in fact remain solid at these pressures (unlike slightly warmer ice like we normally encounter, this would be "phase XI hexagonal" ice).
The next interesting question is that of steam formation. If we did drop a certain volume of ice into the sun (inside the closed ice shell), what happens to the pressure? Presumably the pressure would increase somewhat, but it really isn't relevant - because again, at the pressure you would have to generate to support the ice shell, the density of the water would have to be very high - in fact, it would no longer be a gas, but a solid (or at least with comparable density to a solid - we would be in a part of the phase diagram that is not given).
Finally, the question of the potential energy of the ice - and the impact of the release of this energy on the over all equation. For the purpose of this calculation, we can't just assume that things fall to the center of the sun - even photons that are generated at the center of the sun take a long time to diffuse to the surface, so we can assume the same is true for water. Let's assume therefore that the water simply falls to the surface. While the inside of the shell only falls 1000 km, on average the ice would fall 50,000 km. The force of gravity can be considered (to first order) constant over this distance, so the work done on 1 kg of ice would be
$$W = F\cdot d = 240 N \cdot 5\cdot 10^{7} m = 12 GJ$$
The ice that fell from the inside of the shell (the first ice that melts) has less energy, namely
$$W = 240 N \cdot 10^6 m = 240 MJ$$
and I am for now ignoring the claim that the ice is moving at 1000 km/s (from the original question) as that would mean 1 kg of ice had a kinetic energy of $\frac12 m v^2 = 500 GJ$.
Whichever way you look at it, that is a very considerable amount of energy. It suggests that as the ice starts melting from the inside of the shell out, the water slamming into the surface of the sun will actually heat the sun up, speeding up the melting of the rest of the ice. The whole process will therefore take much less time than I initially estimated - there will be a runaway reaction.
Just to calibrate us - all that ice slamming into the sun adds about 12 GJ/kg * 2E30 kg = 2.4E40 J to the sun. If none of that energy was transferred to the sun, it would lead to a temperature rise of the water of about 3 million degrees. Just from the potential energy (not the initial kinetic energy). That is much hotter than the sun - so there would be a runaway melt reaction.
So it seems that after a brief time when the sun is dark (much less than 200 years), it would shine very, very brightly? Still seems like an uncomfortable solar system.
UPDATE 2
One more thought. If the ice was a little bit less dense, so that structurally it can all fall to the surface of the sun, the 100,000 km thick layer of ice would (at an initial velocity of 1000 km/s) take just 100 seconds to fall into the sun. On average, each bit of ice would fall just 1000 km, and most of the energy dissipated would be the inital kinetic energy (500 GJ/kg - much more than the 240 MJ/kg gravitational energy).
This would briefly heat the surface of the sun to a temperature of more than 100 million degrees - hotter than the core of the sun. So in that case the sun might blink briefly (while the ice is still acting as a shield) - but very quickly, it would all be over for the earth. Of course at that temperature all kinds of fusion reactions would take place - and an immense amount of heat would radiate from the surface of the sun.
It reminds me of Tom Lehrer's song - "We will all go together when we go"
There will be no more misery
when the world is our rotisserie yes we will all fry together when we fry.
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, and to attain this field in specific regions of the brain, the electric current should pass through different head layers via skin, fat, skull, meninges, and cortex (part of the brain). In order to model the brain, different layers should be considered, including gray and white matters.The meninges, three layers of protective tissue, cover the outer surface of the central nervous system (brain and spinal cord) and comprise three connective tissue layers viz. (from the innermost to the outermost layer) the pia mater, arachnoid and the dura mater. The meninges also
Kathrin Badstübner, Marco Stubbe, Thomas Kröger, Eilhard Mix and Jan Gimsa
Education and Research (BMBF, FKZ 01EZ0911). The custom-made stimulator system was developed in cooperation with the Steinbeis company (STZ1050, Rostock, Germany) and Dr. R. Arndt (Rückmann & Arndt, Berlin, Germany).References1 Krack P, Hariz MI, Baunez C, Guridi J, Obeso JA. Deep brain stimulation: from neurology to psychiatry? Trends Neurosci. 2010;33:474-84. https://doi.org/10.1016/j.tins.2010.07.002 10.1016/j.tins.2010.07.002 20832128Krack P Hariz MI Baunez C Guridi J Obeso JA Deep brain stimulation: from neurology to psychiatry
Lisa Röthlingshöfer, Mark Ulbrich, Sebastian Hahne and Steffen Leonhardt
magnetic field strength (h) are assigned to the edges. Hence, a system of equations, called Maxwell-Grid Equations, has to be solved for the whole calculation domain, where each cell is described by:(1)C e → = − ∂ b → ∂ t C ˜ h → = − ∂ d → ∂ t + j →$$C\overrightarrow{e}=-\frac{\partial \overrightarrow{b}}{\partial t}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tilde{C}\overrightarrow{h}=-\frac{\partial \overrightarrow{d}}{\partial t}+\overrightarrow{j}$$(2)S ˜ d → = q S b → = 0$$\tilde
electrodes are near-to constant because of the high resistance to current of the stratum corneum in the considered frequency range [ 3 ]. This allows us to rewrite the boundary conditions, Eqs. 5 - 7 , between the probe and the uppermost skin layer n , stratum corneum, as (we drop the subindex ` eff ’ for notational convenience in the analysis)− σ n ∂ Φ ( r , H n ) ∂ z = ∑ j = 1 m I j A j [ U ( R 2 j − 1 − r ) − U ( R 2 j − 2 − r ) ] ,$$\begin{array}{}\displaystyle-\sigma_{n}\frac{\partial\Phi(r,\mathcal{H}_{n})}{\partial z}=\sum_{j=1}^{m
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Consider a problem with three variables: $u$, $\sigma_l$, and $\sigma_w$ where $\sigma_w > \sigma_l$. I want to represent the following relationship using integer programming. \begin{equation} u = \begin{cases} \sigma_w - x & x < \sigma_l \\ 0 & x > \sigma_l \end{cases} \end{equation}
Using simple either-or constraints (with a binary variable and big M) I can get the correct value for the cases where $x > \sigma_w$ and the case where $x < \sigma_l$, however when $x \in (\sigma_l , \sigma_w)$ these either-or constraints fail. To show you what I have tried thus far, I write the either-or constraints I tried:
$0 \leq u \leq Mz\\ \sigma_l - x + z(\sigma_w - \sigma_l) \leq u \leq \sigma_l - x + z(\sigma_w - \sigma_l) + M(1-z)$
where $z \in \{0,1\}$ and $M$ is an arbitarily large number.
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In the mathtools documentation, page 27, a command
\Set* (along with
\Set) is defined to stretch according to the large input inside:
\providecommand\given{}\newcommand\SetSymbol[1][]{ \nonscript\:#1\vert \allowbreak \nonscript\: \mathopen{}}\DeclarePairedDelimiterX\Set[1]\{\}{ \renewcommand\given{\SetSymbol[\delimsize]} #1}
For instance, using the above command, the following
\documentclass{article}\usepackage{mathtools}\begin{document}\[ \Set{(2,\pm 1)} \cup \Set*{\left (\frac{2t^2 - 2t + 2}{t^2 + 1},\, \frac{1-t^2}{t+1}\right ) \given t \neq -1 }\]\end{document}
Now, I quite like the automatic scaling options. However, in my old own
\Set command, I always added some space around the argument:
\newcommand{\Set}[1]{\{\, #1 \,\}}
Does anyone know how this spacing behaviour might be added to mathtools' approach to defining a set-command? Or even to create something entirely different, but which achieves both the scaling and the spacing?
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Directional Derivatives
We will now look at a new type of derivative known as a
directional derivative. Directional Derivatives of Two Variable Functions
Let $z = f(x, y)$ be a two variable real-valued function. Recall from the Partial Derivatives page that the partial derivative $\frac{\partial z}{\partial x} = \lim_{h \to 0} \frac{f(a + h, b) - f(a, b)}{h}$ represented the rate of change of $z$ in the $x$-direction at any point $P(a, b, f(a,b))$ for $(a, b) \in D(f)$ by giving us the slope of the tangent line obtained by the curve of intersection of $S$ with the plane $y = b$ at point $P$. Similarly, the partial derivative $\frac{\partial z}{\partial y} = \lim_{h \to 0} \frac{f(a, b + h) - f(a,b)}{h}$ represented the rate of change of $z$ in the $y$-direction at any point $P(a, b, f(a,b))$ for $(a, b) \in D(f)$ by giving us the slope of the tangent line obtained by the curve of intersection of $S$ with the plane $x = a$ at point $P$.
Correspondingly, $\frac{\partial z}{\partial x}$ tells us the rate of change of [[ in the direction of the unit vector $\vec{i}$ while $\frac{\partial z}{\partial y}$ tells us the rate of change of $y$ in the direction of the unit vector $\vec{j}$.
We will now extend this idea further. Let $z = f(x, y)$ be a two variable real-valued function that generates the surface $S$, and let $z_0 = f(x_0, y_0)$ be the point $P(x_0, y_0, z_0)$ that lies on $S$. Let $\vec{u} = (a, b)$ be an arbitrary unit vector ($\| \vec{u} \| = 1$). Consider a vertical plane $\Pi$ that passes through $P$ that is in the direction of $\vec{u}$. This plane $\Pi$ intersects the surface $S$ to form a curve $C$ that contains $P$. The slope of the tangent line $T$ on $C$ at $P$ represents the rate of change of $z$ in the direction of the arbitrary vector $\vec{u}$.
Let $Q(x,y,z)$ be another point on the curve $C$, and let $P'(x_0, y_0, 0)$ and $Q'(x,y,0)$ be the projections of $P$ and $Q$ onto the $xy$-plane. As we can see, the vector $\vec{P'Q'}$ with reduced coordinates $\vec{P'Q'} = (x - x_0, y - y_0)$ goes in the same direction as $\vec{u} = (a, b)$ and so $\vec{P'Q'} = (x - x_0, y - y_0) = h\vec{u} = h(a, b) = (ha, hb)$ for some scalar $h$ as illustrated in the following diagram.
Therefore $x = x_0 + ha$ and $y = y_0 + hb$. So if $\Delta z = z - z_0$ (that is the change in $z$ coordinate from $Q(x, y, z)$ to $P(x_0, y_0, z_0)$) then:(1)
Definition: Let $z = f(x, y)$ be a two variable real-valued function. Then the Directional Derivative of $f$ at $(x_0, y_0) \in D(f)$ in the direction of the unit vector $\vec{u} = (a, b)$ is defined as $D_{\vec{u}} \: f(x_0, y_0) = \lim_{h \to 0} \frac{f(x_0 + ha, y_0 + hb) - f(x_0, y_0)}{h}$ provided this limit exists. If the unit vector $\vec{u} = (a, b)$ is denoted in bold as $\mathbf{u}$ then the notation for the directional derivative of $f$ at $(x_0, y_0) \in D(f)$ in the direction of $\mathbf{u}$ is denoted $D_{\mathbf{u}} \: f(x_0, y_0)$.
One important note that we should mention is that the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ are both directional derivatives in the directions of $\vec{i}$ and $\vec{j}$ as we've noted above, that is:(2)
Let's look at an example of computing a directional derivative using the definition. Consider the function $z = f(x, y) = x + y^2$ and the point $(1, 1) \in D(f)$, and suppose that wanted to compute the directional derivative of $f$ at $(1, 1)$ in the direction of the unit vector $\vec{u} = \left (\frac{3}{5}, \frac{4}{5} \right)$. Therefore:(3)
Of course, computing directional derivatives using limits can be tedious. Fortunately, there is an easier way to compute directional derivatives.
Theorem 1: If $z = f(x, y)$ is a differentiable function of the variables $x$ and $y$, then for any unit vector $\vec{u} = (a, b)$, the directional derivative of $f$ at $(x, y) \in D(f)$ in the direction of $\vec{u}$ is given by $D_{\vec{u}} \: f(x, y) = \frac{\partial z}{\partial x} a + \frac{\partial z}{\partial y} b = \left ( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right ) \cdot (a, b)$. Proof:Let $g(h) = f(x_0 + ha, y_0 + hb)$ be a single variable function in terms of $h$. The derivative $g'(0)$ can be computed as: Let $x = x_0 + ha$ and $y = y_0 + hb$ both be parametric equations of the parameter $h$. Then $g(h) = f(x, y)$. If we use the chain rule type 1 on $g(h) = f(x,y)$ and note that $\frac{d}{dh} x = \frac{d}{dh} (x_0 + ha) = a$ and $\frac{d}{dh} y = \frac{d}{dh} (y_0 + hb) = b$ we get that: Now we let $h = 0$ so that $x = x_0$ and $y = y_0$ and so: So we have that both $g'(0) = D_{\vec{u}} \: f (x_0, y_0)$ and $g'(0) = f_x(x_0, y_0)a + f_y(x_0, y_0)b$ so we conclude that: Directional Derivatives of Three Variable Functions
Definition: Let $w = f(x, y, z)$ be a three variable real-valued function. Then the Directional Derivative of $f$ at $(x_0, y_0, z_0) \in D(f)$ in the direction of the unit vector $\vec{u} = (a, b, c)$ is defined as $D_{\vec{u}} \: f(x_0, y_0, z_0) = \lim_{h \to 0} \frac{f(x_0 + ha, y_0 + hb, z_0 + hc) - f(x_0, y_0, z_0)}{h}$ provided this limit exists.
Theorem 1 can also be extended for directional derivatives of three variables.
Theorem 2: If $w = f(x, y, z)$ is a differentiable function of the variables $x$, $y$, and $z$ then for any unit vector $\vec{u} = (a, b, c)$, the directional derivative of $f$ at $(x, y, z) \in D(f)$ in the direction of $\vec{u}$ is given by $D_{\vec{u}} \: f(x, y, z) = \frac{\partial w}{\partial x} a + \frac{\partial w}{\partial y} b + \frac{\partial w}{\partial z} c = \left ( \frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z} \right ) \cdot (a, b, c)$.
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Identity Matrices
Identity Matrices
Definition: A square $n \times n$ matrix $I$ is considered an Identity Matrix if all entries along the main diagonal are 1 and all other entries are 0. Alternatively we can define identity matrices such that if $(I)_{ij} = 1$ if $i =j$ and $(I)_{ij} = 0$ otherwise.
We note that all identity matrices are square. For example, the identity matrix $I_{3 \times 3}$ (commonly denoted $I_{3}$ for brevity) looks like:(1)
\begin{align} I_{3 \times 3} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \end{align}
Theorem 1: Let $A$ be an $m \times n$ matrix and $I$ be an identity matrix and assume all operations are defined. Then: a) $AI_n = A$ and $I_mA = A$. b) $\mathrm{tr}(I_m) = m$. Proof of (a):Recall that any entry of a matrix product can be determined by the formula $(AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + ... + a_{in}b_{nj}$. Let $B = I_n$. Recall that $b_{nj} = 1$ if $n = j$ (and $b_{nj} = 0$ otherwise). Let $n = j$, then the entry $(AI)_{in} = a_{i1}b_{1n} + a_{i2}b_{2n} + ... + a_{in}b_{nn}$. Since all other $b$ entries are $0$, it follows that $(AI)_{in} = a_{in} \cdot 1 = (A)_{in}$. Therefore $AI_n = A$. $\blacksquare$ Proof of (b):$I_m$ has only $1$s on the main diagonal. There will be exactly $m$ entries on the main diagonal, so $1 \cdot m = m$, so $tr(I_m) = m$. $\blacksquare$
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Measurable Functions
So far we have looked at three major classes of functions.
On the Step Functions on General Intervals page we said that a function $f$ on a general interval $I$ is a step function if there exists a closed and bounded interval $[a, b] \subseteq I$ such that there exists a partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ where $f$ is constant on each open subinterval $(x_{k-1}, x_k)$ (for all $k \in \{ 1, 2, ..., n \}$) and such that $f(x) = 0$ for all $x \in I \setminus [a, b]$. The set of all step functions on $I$ was denoted $S(I)$.
On the Upper Functions and Integrals of Upper Functions page we said that a function $f$ on a general interval $I$ is an upper function if there exists an sequence of step functions, call it $(f_n(x))_{n=1}^{\infty}$ that is increasing and converges to $f$ almost everywhere on $I$ and such that the numerical sequence $\displaystyle{�406�\lim_{n \to \infty} \int_I f_n(x) \: dx}$ is finite. The set of all upper functions on $I$ was denoted $U(I)$.
On the The Lebesgue Integral page we said that a function $f$ is a Lebesgue integrable function on $I$ if there exists upper functions $u(x)$ and $v(x)$ on $I$ such that $f(x) = u(x) - v(x)$. The set of all Lebesgue integrable function on $I$ was denoted $L(I)$.
We will now look at an every broader class of functions called measurable functions which we define below.
Definition: A function $f$ is said to be a Measurable Function on an interval $I$ if there exists a sequence of step functions, $(f_n(x))_{n=1}^{\infty}$ on $I$ that converges to $f$ almost everywhere on $I$. The set of all measurable functions on $I$ is denoted $M(I)$.
Consider a function $f$ which is Lebesgue integrable on $I$. Then $f$ is the difference of two upper functions on $I$, say $u$ and $v$, where $f = u - v$. But each upper function is the limit of a convergent generating sequence of step functions $(u_n(x))_{n=1}^{\infty}$ and $(v_n(x))_{n=1}^{\infty}$.
If we let $(w_n(x)) = (u_n(x) - v_n(x))_{n=1}^{\infty}$ then the sequence $(w_n(x))_{N=1}^{\infty}$ is a sequence of step functions that converges to $u - v = f$. So, every Lebesgue integrable function is a measurable function. Thus the following inclusion holds for any interval $I$:(1)
One may think that the set of Lebesgue integrable functions equals the set of Measurable functions. This is NOT the case, however, giving an example of a measurable function that is not Lebesgue integrable is difficult.
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Sometimes it is more appropriate to utilize what is known as the
vector form of the equation of plane. Vector Form Equation of a Plane
Let $\vec{n} = (a, b, c)$ be a normal vector to our plane $\Pi$, that is $\Pi \perp \vec{n}$. Instead of using just a single point from the plane, we will instead take a vector that is parallel from the plane. Let $P(x, y, z)$ and $P_0(x_0, y_0, z_0)$ be two points on the plane. It follows that $\vec{P_0P} \perp \vec{n}$.
We will define the vector $\vec{r_0}$ to have its initial point at the origin and terminal point at $P_0$. We will also define $\vec{r}$ to have its initial point at the origin as well, but instead have its terminal point at $P$. It thus follows that $\vec{P_0P} = \vec{r} - \vec{r_0}$. Furthermore, we note that $\vec{r} - \vec{r_0}$ is parallel to the plane $\Pi$ and perpendicular to $\vec{n}$. From the dot product we get that:(1)
Definition: Let $\Pi$ be a plane in $\mathbb{R}^3$. Then the Vector Form Equation of the plane $\Pi$ is $0 = \vec{n} \cdot (\vec{r} - \vec{r_0}) = 0$ where $\vec{n} = (a, b, c)$ is any normal vector to $\Pi$, $\vec{r} = \vec{OP}$ and $\vec{r_0} = \vec{OP_0}$. Example 1 Determine the vector form equation of a plane $\Pi$ given that the points $P(-5, 2, 8)$ and $Q(2, 3, 3)$ lie on the plane and the vector $\vec{n} = (4, 4, -1)$ is perpendicular to $\Pi$.
To solve this, we must determine what our vectors $\vec{r}$ and $\vec{r_0}$ are. We note that if $O$ is the origin, then $\vec{OP} = \vec{r_0} = (-5, 2, 8)$ and $\vec{OQ} = \vec{r} = (2, 3, 3)$. Plugging these values and our norm $\vec{n}$ into the form we obtain:(2)
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The BBC article Event Horizon Telescope ready to image black hole describes the Event Horizon Telescope, a coordinated observing technique with several radio telescope arrays across the globe forming a synthetic aperture with an Earth-sized baseline.
$$\frac{\lambda}{r_{Earth}} \sim \frac{r_{Sag A*}}{D_{Sag A*}} \sim 10^{-11}$$
...when plugging in 1 millimeter for $\lambda$, and with $r_{Sag A*}$ and $D_{Sag A*}$ the radius of, and distance from Earth to Sagittarius A*, the black hole in the center of the Milky Way Galaxy of 20 million km and 26,000 light years, respectively. (values taken from the article).
The equation I've written shows that millimeter wavelength interferometry with an Earth-sized baseline has the possibility to resolve the existence of some structure with scale of the black hole's event horizon.
My question is
How does the Event Horizon Telescope implement the interferometry? It would certainly be impossible to bring all signals together to a central site and perform the interference there in real time as down-converted analog signals, and quite difficult/expensive to do it with dedicated, synchronized digital optical fiber lines. Is the massive amounts of data sent as IP packets over the internet to a central correlator (numerical interferometer)?
The article mentions atomic clocks and lots of hard drives, and I have a hunch these have something to do with it.
above: "The eventual EHT array will have 12 widely spaced participating radio facilities". From BBC.
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The Annals of Statistics Ann. Statist. Volume 10, Number 1 (1982), 297-301. An Inequality Comparing Sums and Maxima with Application to Behrens-Fisher Type Problem Abstract
A sharp inequality comparing the probability content of the $\ell_1$ ball and that of $\ell_\infty$ ball of the same volume is proved. The result is generalized to bound the probability content of the $\ell_p$ ball for arbitrary $p \geq 1$. Examples of the type of bound include $P\{(|X_1|^p + |X_2|^p)^{1/p} \leq c\} \geq F^2(c/2^{1/2p}),\quad p \geq 1,$ where $X_1, X_2$ are independent each with distribution function $F$. Applications to multiple comparisons in Behrens-Fisher setting are discussed. Multivariate generalizations and generalizations to non-independent and non-exchangeable distributions are also discussed. In the process a majorization result giving the stochastic ordering between $\Sigma a_i X_i$ and $\Sigma b_i X_i$, when $(a^2_1, a^2_2, \cdots, a^2_n)$ majorizes $(b^2_1, b^2_2, \cdots, b^2_n)$, is also proved.
Article information Source Ann. Statist., Volume 10, Number 1 (1982), 297-301. Dates First available in Project Euclid: 12 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aos/1176345712 Digital Object Identifier doi:10.1214/aos/1176345712 Mathematical Reviews number (MathSciNet) MR642741 Zentralblatt MATH identifier 0481.62017 JSTOR links.jstor.org Citation
Dalal, Siddhartha R.; Fortini, Peter. An Inequality Comparing Sums and Maxima with Application to Behrens-Fisher Type Problem. Ann. Statist. 10 (1982), no. 1, 297--301. doi:10.1214/aos/1176345712. https://projecteuclid.org/euclid.aos/1176345712
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Question
Find the resistance that must be placed in parallel with a $10.0 \Omega$ galvanometer having a $100 \mu \textrm{A}$ sensitivity to allow it to be used as an ammeter with: (a) a 20.0-A full-scale reading, and (b) a 100-mA full-scale reading.
Final Answer
$5.00 \times 10^{-5} \Omega$ $10.0 \textrm{ m}\Omega$ Calculator Screenshots
Video Transcript
This is College Physics Answers with Shaun Dychko. We have a galvanometer that has a sensitivity of 100 micoramps which means when there is full scale deflection of the needle in the galvanometer, there must be 100 microamps of current going through it. The internal resistance we're told is 10 ohms and our question is what should this shunt resistance be in order to have a full scale deflection when, A, we have a total current through the circuit of 20 amps or in B, when the total current in the circuit is 100 milliamps. Okay. So the voltage across the galvanometer is going to be the same as the voltage across the shunt resistance because they're connected in parallel and so this
Isubscript G is the current through the galvanometer multiplied by the internal resistance of the galvanometer is the voltage, and you could also say that the current across the shunt resistance multiplied by capital Ris the voltage across the shunt resistor. This Isubscript capital R is going to be the total current going into this junction minus whatever is taken up through the galvanometer. So I've replaced it with that. That's using the junction rule. So I tfor total has to equal current to the galvanometer plus the current through the shunt resistance. We can solve for the shunt resistance by subtracting I Gfrom both sides and then switching the sides around. You get I tminus I G, okay? That's multiplied by capital Rand so since these are connected in parallel, these have the same voltage across them and so we can equate them which we do here. We'll solve for capital Rby dividing both sides by I tminus I G. So the shunt resistance is 100 microamps times 10 ohms, that's the sensitivity of the galvanometer multiplied by its internal resistance divided by the current total at which full scale deflection is meant to occur which is 20 amps in part A, minus 100 times ten to the minus six amps. That gives 5.00 times ten to the minus five ohms. Now in part B, It's exactly the same idea but now there's just a different value for the total current at which full scale deflection of the galvanometer should occur. So we use the same formula but substitute in 100 milliamps which is 100 times ten to the minus three amps instead of 20 amps like before. In this case we get 10.0 milliohms.
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This example is an implementation of the assessment of a new total hip replacement (THR) technology described in chapter 3.5 of Decision Modelling for Health Economic Evaluation. A more detailed report is available at this location. This reports goes a bit further in the analysis. For the sake of simplicity we will not reproduce exactly the analysis from the book. See vignette
vignette("i-reproduction", "heemod") for an exact reproduction.
This model has 5 states:
omrPTHR);
omrRTHR);
Two transition probabilities are time-varying in this model:
Other-cause death probabilities (mortality rate
mr) for the United Kingdom is taken from WHO databases using the
get_who_mr() function. The variable
sex, taking values
0 and
1, must be recoded in
"FMLE" and
"MLE" before being passed to this function.
standardRR increases with time with the following formula (a Weibull distribution):
\[ P_{revision} = 1 - \exp(\lambda \times ((t-1)^\gamma-t^\gamma)) \]
Where \(t\) is the time since revision, \(\gamma = 1.45367786\) and:
\[ \lambda = exp(cons + ageC \times age + maleC \times sex) \]
Where \(age\) and \(sex\) (female =
0, male = 1) are individual characteristics, \(cons = -5.49094\), \(ageC = -0.0367\) and \(maleC = 0.768536\).
standardRR is modified by the relative risk \(rrNP1 = 0.260677\).
\[ P_{revision} = 1 - \exp(\lambda \times rr \times NP1 \times ((t-1)^\gamma-t^\gamma)) \]
rrr) probability is set to be constant at
The key element to specify time-varying elements in
heemod is through the use of the
package-defined variables
markov_cycle and
state_cycle. See vignette
vignette("b-time-dependency", "heemod") for more details.
In order to build this more complex Markov model, parameters need to be defined through
define_parameters() (for 2 reasons: to keep the transition matrix readable, and to avoid repetition by re-using parameters between strategies).
The equations decribed in the previous section can be written easily, here for a female population (
sex =
0) starting at 60 years old (
age_init =
60).
param <- define_parameters( age_init = 60, sex = 0, # age increases with cycles age = age_init + markov_cycle, # operative mortality rates omrPTHR = .02, omrRTHR = .02, # re-revision mortality rate rrr = .04, # parameters for calculating primary revision rate cons = -5.49094, ageC = -.0367, maleC = .768536, lambda = exp(cons + ageC * age_init + maleC * sex), gamma = 1.45367786, rrNP1 = .260677, # revision probability of primary procedure standardRR = 1 - exp(lambda * ((markov_cycle - 1) ^ gamma - markov_cycle ^ gamma)), np1RR = 1 - exp(lambda * rrNP1 * ((markov_cycle - 1) ^ gamma - markov_cycle ^ gamma)), # age-related mortality rate sex_cat = ifelse(sex == 0, "FMLE", "MLE"), mr = get_who_mr(age, sex_cat, country = "GBR", local = TRUE), # state values u_SuccessP = .85, u_RevisionTHR = .30, u_SuccessR = .75, c_RevisionTHR = 5294)param
## 20 unevaluated parameters.## ## age_init = 60## sex = 0## age = age_init + markov_cycle## omrPTHR = 0.02## omrRTHR = 0.02## rrr = 0.04## cons = -5.49094## ageC = -0.0367## maleC = 0.768536## lambda = exp(cons + ageC * age_init + maleC * sex)## gamma = 1.45367786## rrNP1 = 0.260677## standardRR = 1 - exp(lambda * ((markov_cycle - 1)^gamma - markov_cycle^gamma))## np1RR = 1 - exp(lambda * rrNP1 * ((markov_cycle - 1)^gamma - markov_cycle^gamma))## sex_cat = ifelse(sex == 0, "FMLE", "MLE")## mr = get_who_mr(age, sex_cat, country = "GBR", local = TRUE)## u_SuccessP = 0.85## u_RevisionTHR = 0.3## u_SuccessR = 0.75## c_RevisionTHR = 5294
Now that parameters are defined, the probability transitions can be easily written:
## A transition matrix, 5 states.## ## PrimaryTHR SuccessP RevisionTHR SuccessR Death ## PrimaryTHR C omrPTHR ## SuccessP C standardRR mr ## RevisionTHR C omrRTHR + mr## SuccessR rrr C mr ## Death 1
## A transition matrix, 5 states.## ## PrimaryTHR SuccessP RevisionTHR SuccessR Death ## PrimaryTHR C omrPTHR ## SuccessP C np1RR mr ## RevisionTHR C omrRTHR + mr## SuccessR rrr C mr ## Death 1
While it is possible to plot the matrix thanks to the
diagram package, the results may not always be easy to read.
Utilities and costs are then associated to states. In this model costs are discounted at a rate of 6% and utilities at a rate of 1.5%.
Now that parameters, transition matrix and states are defined we can define the strategies for the control group and the NP1 treatment.
We use
define_starting_values() to take into account the cost of surgery.
strat_standard <- define_strategy( transition = mat_standard, PrimaryTHR = define_state( utility = 0, cost = 0 ), SuccessP = define_state( utility = discount(u_SuccessP, .015), cost = 0 ), RevisionTHR = define_state( utility = discount(u_RevisionTHR, .015), cost = discount(c_RevisionTHR, .06) ), SuccessR = define_state( utility = discount(u_SuccessR, .015), cost = 0 ), Death = define_state( utility = 0, cost = 0 ), starting_values = define_starting_values( cost = 394 ))strat_standard
## A Markov model strategy:## ## 5 states,## 2 state values
strat_np1 <- define_strategy( transition = mat_np1, PrimaryTHR = define_state( utility = 0, cost = 0 ), SuccessP = define_state( utility = discount(u_SuccessP, .015), cost = 0 ), RevisionTHR = define_state( utility = discount(u_RevisionTHR, .015), cost = discount(c_RevisionTHR, .06) ), SuccessR = define_state( utility = discount(u_SuccessR, .015), cost = 0 ), Death = define_state( utility = 0, cost = 0 ), starting_values = define_starting_values( cost = 579 ))strat_np1
## A Markov model strategy:## ## 5 states,## 2 state values
Both strategies can now be run for 60 years. By default models are computed for 1000 person starting in
PrimaryTHR.
## Fetching mortality data from package cached data.
## Using cached data from year 2015.
A comparison of both strategies can be done with
summary(). The incremental cost and effect are displayed in columns
Cost and
Effect.
## 2 strategies run for 60 cycles.## ## Initial state counts:## ## PrimaryTHR = 1000L## SuccessP = 0L## RevisionTHR = 0L## SuccessR = 0L## Death = 0L## ## Counting method: 'life-table'.## ## Values:## ## utility cost## standard 11836240 11955401## np1 17386292 17405827## ## Efficiency frontier:## ## standard -> np1## ## Differences:## ## Cost Diff. Effect Diff. ICER Ref.## np1 5450.426 5550.052 0.9820496 standard
The new treatment costs £1 more per QALY gained.
It should be noted that this result differs from the original study. This difference is explained by higher population-level all-causes mortality rates in the original study than in the WHO database (used here). See vignette
vignette("i-reproduction", "heemod") for an exact reproduction of the analysis.
We can plot the counts per state:
## Scale for 'colour' is already present. Adding another scale for## 'colour', which will replace the existing scale.
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I am stuck in this exercise from my textbook:
Consider a one-period market model with $N+1$ assets: a bond, a stock and $N-1$ call options. The prices of the bond are $B_0=1$ and $B_1 = 1+r$, where $r$ is a constant. The prices of the stock are given by a constant $S_0$ and a random variable $S_1$ taking values in $\{0, 1, \ldots, N-1, N \}$ for a given integer $N \geq 4$. Finally, let the time-0 price of the call option with strike $K \in \{ 1, \ldots, N-1 \}$ be denoted by $C(K)$. Now we introduce a contingent claim with time-1 payout $\xi_1 = g(S_1)$, where $g$ is the function $$g(M) = \mathbf{1}_{ \{M = K_0 \} }, \quad 0 \leq K_0 \leq N. $$ Assuming that the market has no arbitrage, we want to find the time-0 price $\xi_0$ in the following cases: $$ 2 \leq K_0 \leq N-2 \, ; \quad K_0 = N-1 \, ; \quad K_0 = 0 .$$
Let $Y$ be the state price density of the market such that $Y_0 =1$. We know that $$ \mathbb{E} [ YS_1 ] = S_0, \quad \mathbb{E} [ Y ( S_1 - K)^{+} ] = C(K), \text{ for } K \in \{1, \ldots, N-1 \}. $$
But how can we compute $$\xi_0 = \mathbb{E}[ Y \mathbf{1}_{ \{ S_1 = K_0 \} }] \quad ?$$
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Preprints (rote Reihe) des Fachbereich Mathematik Refine Year of publication 1993 (13) (remove) Has Fulltext yes (13) (remove)
240
244
242
Efficient algorithms and structural results are presented for median problems with 2 new facilities including the classical 2-Median problem, the 2-Median problem with forbidden regions and bicriterial 2-Median problems. This is the first paper dealing with multi-facility multiobjective location problems. The time complexity of all presented algorithms is O(MlogM), where M is the number of existing facilities.
238
Despite their very good empirical performance most of the simplex algorithm's variants require exponentially many pivot steps in terms of the problem dimensions of the given linear programming problem (LPP) in worst-case situtation. The first to explain the large gap between practical experience and the disappointing worst-case was Borgwardt (1982a,b), who could prove polynomiality on tbe average for a certain variant of the algorithm-the " Schatteneckenalgorithmus (shadow vertex algorithm)" - using a stochastic problem simulation.
245
Let \(A\):= {\(a_i\mid i= 1,\dots,m\)} be an i.i.d. random sample in (\mathbb{R}^n\), which we consider a random polyhedron, either as the convex hull of the \(a_i\) or as the intersection of halfspaces {\(x \mid a ^T_i x\leq 1\)}. We introduce a class of polyhedral functionals we will call "additive-type functionals", which covers a number of polyhedral functionals discussed in different mathematical fields, where the emphasis in our contribution will be on those, which arise in linear optimization theory. The class of additive-type functionals is a suitable setting in order to unify and to simplify the asymptotic probabilistic analysis of first and second moments of polyhedral functionals. We provide examples of asymptotic results on expectations and on variances.
248
The article provides an asymptotic probabilistic analysis of the variance of the number of pivot steps required by phase II of the "shadow vertex algorithm" - a parametric variant of the simplex algorithm, which has been proposed by Borgwardt [1] . The analysis is done for data which satisfy a rotationally invariant distribution law in the \(n\)-dimensional unit ball.
246
Max ordering (MO) optimization is introduced as tool for modelling production planning with unknown lot sizes and in scenario modelling. In MO optimization a feasible solution set \(X\) and, for each \(x\in X, Q\) individual objective functions \(f_1(x),\dots,f_Q(x)\) are given. The max ordering objective \(g(x):=max\) {\(f_1(x),\dots,f_Q(x)\)} is then minimized over all \(x\in X\). The paper discusses complexity results and describes exact and approximative algorithms for the case where \(X\) is the solution set of combinatorial optimization problems and network flow problems, respectively.
239
We investigate two versions of multiple objective minimum spanning tree problems defined on a network with vectorial weights. First, we want to minimize the maximum of Q linear objective functions taken over the set of all spanning trees (max linear spanning tree problem ML-ST). Secondly, we look for efficient spanning trees (multi criteria spanning tree problem MC-ST). Problem ML-ST is shown to be NP-complete. An exact algorithm which is based on ranking is presented. The procedure can also be used as an approximation scheme. For solving the bicriterion MC-ST, which in the worst case may have an exponential number of efficient trees, a two-phase procedure is presented. Based on the computation of extremal efficient spanning trees we use neighbourhood search to determine a sequence of solutions with the property that the distance between two consecutive solutions is less than a given accuracy.
236
Es wird anhand von Beispielen, an denen der Autor in der Vergangenheit gearbeitet hat, gezeigt, wie man Modelle der exakten Naturwissenschaften auf wirtschaftliche Probleme anwenden kann. Insbesondere wird diskutiert, wo Grenzen dieser Übertragbarkeit liegen. Die Arbeit ist eine Zusammenfassung eines Vortrags, der im SS 1992 im Rahmen des Studium Generale an der Universität Kaiserslautern gehalten wurde.
243
Given Q different objective functions, three types of single-facility problems are considered: Lexicographic, pareto and max ordering problems. After discussing the interrelation between the problem types, a complete characterization of lexicographic locations and some instances of pareto and max ordering locations is given. The characterizations result in efficient solution algorithms for finding these locations. The paper relies heavily on the theory of restricted locations developed by the same authors, and can be further extended, for instance, to multifacility problems with several objectives. The proposed approach is more general than previously published results on multicriteria planar location problems and is particulary suited for modelling real-world problems.
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I am trying to prove the following relation
$$ \langle\Phi|\hat{\bar{\psi}}(x)A\hat\psi(x)|\Phi\rangle=\bar\varphi(x)A\varphi(x)+\langle0|\hat{\bar{\psi}}(x)A\hat\psi(x)|0\rangle\,, $$ where $|\Phi\rangle$ is a state that contains a single electron and it is defined in this way
$$ |\Phi\rangle=\sum_{r=1}^2\int d^3\!q\;f_r(\textbf q)\hat b^\dagger_r(\textbf q)|0\rangle $$ normalized so that $\langle\Phi|\Phi\rangle=1$, and $\varphi(x)$ is the spinor defined $\varphi(x)=\langle0|\hat\psi|\Phi\rangle$.
So far I have tried to write explicitly the LHS in the mode expansion of the Dirac field operators, but I don't think this is the right approach. Any hints or suggestions on how to proceed?
Note that I have include the homework tag so please avoid giving the full answer.
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Perceptron is the most basic and primary implementation of a biological neuron in machine intelligence. Moreover the concept of perceptron can be leveraged to build more complex neural networks which we will see later. These are used mainly for supervised learning and can be modified to work with unsupervised learning also. The implementation of perceptron is inspired from the actual neuron which can be seen below.
The way it works is that it receives inputs through
Dendrites and the corresponding Synapses required to stimulus the inputs are also received. Then the necessary activation is produced in the nucleus of the neuron and the final output is passed through Axon.
Similarly we can build our perceptron by certain inputs and with some associated weights and there by passing it through the activation function gives the output. But the catch here is that
we don’t know what would be weights associated with the inputs. And this is where the actual concept of perceptron lies in.
Let’s say we have our inputs as $ X_0,X_1,X_2,….,X_m$ with $n$ features, i.e the dimensions of our input matrix would be $m \times n$ where $m$ represents
and $n$ represents no.of.samples . Therfore the weights vector contains array of $n$ elements. We then find their sum of their products as $${Z = W^T \cdot X}$$. And then the $Z$ is passed through an
no.of.features
activationfunction which is a Unit Step functionin case of a perceptron.
Anyway we can change the
threshold and the range of the step function. All we need to have a Unit Step function as our activation function.
So combining everything,
$${Z = W^T \cdot X}$$
$${\bar Y = g(Z)}$$
$${g}$$ is the activation function.
Therefore $${\bar Y}$$ gives the corresponding output.
How do we know the correct weights ?
Yeah! That’s an very interesting question. We actually don’t know the exact weights suitable for any experiment before as it is very difficult to make an educated
guess. But we will learn the correct weights by training our perceptron which we will see later. For now we initialize the weights to be all zeros or uniformly distributed. It’s worth noting that whatever the initial weights might be, after required training we all have the same weights no matter the initial weights. What is training?
Every neural network needs a set of weights to predict/classify the output. But we are initially left with all zeros or uniformly distributed weights which we start with. The output thus provided with these may not be correct and hence therefore is the need for training. In training we propagate our input and weights through an activation function and the output through the activation function is used to update the weights by measuring the error or
Cost. How is Cost calculated?
The cost is nothing but how
far our predicted output $ \bar Y$ from the actual output $Y$. Therefore it can be defined as Ordinary Least Square method or sometimes using Cross Entropy. For now, we look at how Ordinary Least Square method is implemented.
$${Cost=\frac{1}{2m}\sum_{i=0}^{m}(Y-\bar Y)^2}$$
How do we update our weights?
I’m glad you asked. Basically the weights are updated by computing the
partial derivative of the Cost wrt weights i.e,
$${\Delta W = \frac{\partial}{\partial W}Cost}$$
Therefore,
$${\Delta W = ( Y- \bar Y) \cdot X}$$
Finally,
$${W = W – \alpha \Delta W}$$
We iterate through this process until a
stop criterion is reached or after some specified iterations.
Where ${\alpha}$ is the
learning rate which controls how fast/slow our perceptron should learn. It is an hyper parameter that needs to be manually set and if it too large then our model may skip some points and may not converge, if it is small then it may converge but slowly. Can you guide me through an example?
Okay, Let’s code our Perceptron by hand and will see how can we classify the famous
Iris dataset of two classes since we involve only one output neuron. from sklearn.datasets import load_iris import numpy as np from sklearn.cross_validation import train_test_split import matplotlib.pyplot as plt
We first start off by importing necessary packages. We load our iris-dataset from
and sklearn.datasets for numerical computations and
numpy
for splitting our data in to training and testing sets,
train_test_split
for plotting the errors.
matplotlib
iris = load_iris() idxs = np.where(iris.target!=2)[0] X = iris.data[idxs] Y = iris.target[idxs] Line 8: We are only interested in the 2 targets as we have only one output neuron for now, so we are slicing the indexes where the target is either 0 or 1 but not 2 Line 9-10: We load the datasets with the corresponding indexes with labels 0 and 1. X_train,X_test,Y_train,Y_test = train_test_split(X,Y,test_size=0.1,random_state=1) plt.scatter(X_train[Y_train==1][:,1],X_train[Y_train==1][:,2]) plt.scatter(X_train[Y_train==0][:,1],X_train[Y_train==0][:,2],color='red') plt.show()
We then split our dataset into training and testing sets and we plot a scatter plot of two features of two targets to visualize.
W = np.zeros((X_train.shape[1])) learning_rate = 0.1 errors = []
We then initialize our weights to be all zeros with dimensions equal to
, and errors to save all the errors through all the iterations it passes on. no.of.features for i in xrange(100): Z = X_train.dot(W) a = np.where(Z>=0.5,1,0) error = 0.5*np.mean((Y_train-a)**2) W = W - learning_rate*(a-Y_train).dot(X_train) errors.append(error)
We first find the value of $Z$ which is then passed to the activation function. Here the activation is a step function which is thresholded by
. Then we calculate the error/cost as we discussed before and then update our weights. 0.5 plt.plot(errors) plt.show() Z = X_test.dot(W) a = np.where(Z>=0.5,1,0) #accuracy print np.mean(np.where(Y_test==a,1,0))
Then we plot our cost vs no.of.iterations and measure the accuracy of our model by using the updated weights. The
accuracy measured on our test set is which is expected. 100%
From the above plot we can say that our perceptron converges before
iterations. And you can also observe that our cost function is fluctuating through some distance before converging and that can be managed with 20 momentumwhich we will see later.
Let’s see the
Class based implementation of our Perceptron algorithm. import numpy as np class Perceptron(object): def __init__(self,learning_rate=0.1,n_iter=100,step_threshold=0.5): self.learning_rate = learning_rate self.n_iter = n_iter self._costs = [] self._step_threshold = step_threshold def fit(self,X,y): self._W = np.random.uniform(size=(X.shape[1],)) for i in xrange(self.n_iter): z = X.dot(self._W) a = self._activation(z) self._costs.append(self._cost(y,a)) w_update = (y-a).dot(X) self._W = self._W + self.learning_rate*w_update def _activation(self,z): return np.where(z>=self._step_threshold,1,0) def _cost(self,y_true,y_pred): return 0.5*np.mean((y_pred-y_true)**2) def predict(self,X): return self._activation(X.dot(self._W)) def score(self,y_true,y_pred): return np.mean(y_true==y_pred)
This is our Perceptron class which should be familiar to you if you have followed along. Therefore we can simply rewrite our above script as below.
from perceptron import Perceptron from sklearn.metrics import accuracy_score from sklearn.datasets import load_iris import numpy as np from sklearn.model_selection import train_test_split import matplotlib.pyplot as plt iris = load_iris() idxs = np.where(iris.target!=2)[0] X = iris.data[idxs] Y = iris.target[idxs] X_train,X_test,Y_train,Y_test = train_test_split(X,Y,test_size=0.3,random_state=1) pcpt = Perceptron(n_iter=20) pcpt.fit(X_train,Y_train) y_pred = pcpt.predict(X_test) print "Accuracy:",pcpt.score(y_pred,Y_test) plt.plot(pcpt._costs) plt.show()
Therfore, we can say our class based perceptron is much more
pythonic and can be reused anywhere in our scripts.
Till now we have seen a simple perceptron algorithm and in the upcoming posts we will see how this simple perceptron can be leveraged to extend to
Multi-layer-Perceptron which is where our journey to actual neural networks start. Then we will use that neural network to recognize handwritten digits in the later posts.
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Difference between revisions of "Probability Seminar"
(→Probability related talk in PDE Geometric Analysis seminar: Monday, 3:30pm to 4:30pm, Van Vleck 901)
(→February 21, Diane Holcomb, KTH)
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Abstract: There has been a great deal or recent work on the asymptotics of the maximum of characteristic polynomials or random matrices. Other recent work studies the analogous result for log-correlated Gaussian fields. Here we will discuss a maximum result for the centered counting function of the Sine beta process. The Sine beta process arises as the local limit in the bulk of a beta-ensemble, and was originally described as the limit of a generalization of the Gaussian Unitary Ensemble by Valko and Virag with an equivalent process identified as a limit of the circular beta ensembles by Killip and Stoiciu. A brief introduction to the Sine process as well as some ideas from the proof of the maximum will be covered. This talk is on joint work with Elliot Paquette.
Abstract: There has been a great deal or recent work on the asymptotics of the maximum of characteristic polynomials or random matrices. Other recent work studies the analogous result for log-correlated Gaussian fields. Here we will discuss a maximum result for the centered counting function of the Sine beta process. The Sine beta process arises as the local limit in the bulk of a beta-ensemble, and was originally described as the limit of a generalization of the Gaussian Unitary Ensemble by Valko and Virag with an equivalent process identified as a limit of the circular beta ensembles by Killip and Stoiciu. A brief introduction to the Sine process as well as some ideas from the proof of the maximum will be covered. This talk is on joint work with Elliot Paquette.
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== Probability related talk in PDE Geometric Analysis seminar: Monday, 3:30pm to 4:30pm, Van Vleck 901,
+
== Probability related talk in PDE Geometric Analysis seminar: Monday, 3:30pm to 4:30pm, Van Vleck 901, Xiaoqin Guo, UW-Madison ==
−
Xiaoqin Guo, UW-Madison ==
Title: Quantitative homogenization in a balanced random environment
Title: Quantitative homogenization in a balanced random environment
Revision as of 11:32, 18 February 2019 Spring 2019 Thursdays in 901 Van Vleck Hall at 2:25 PM, unless otherwise noted. We usually end for questions at 3:15 PM.
If you would like to sign up for the email list to receive seminar announcements then please send an email to join-probsem@lists.wisc.edu
January 31, Oanh Nguyen, Princeton
Title:
Survival and extinction of epidemics on random graphs with general degrees
Abstract: We establish the necessary and sufficient criterion for the contact process on Galton-Watson trees (resp. random graphs) to exhibit the phase of extinction (resp. short survival). We prove that the survival threshold $\lambda_1$ for a Galton-Watson tree is strictly positive if and only if its offspring distribution has an exponential tail, settling a conjecture by Huang and Durrett. On the random graph with degree distribution $D$, we show that if $D$ has an exponential tail, then for small enough $\lambda$ the contact process with the all-infected initial condition survives for polynomial time with high probability, while for large enough $\lambda$ it runs over exponential time with high probability. When $D$ is subexponential, the contact process typically displays long survival for any fixed $\lambda>0$. Joint work with Shankar Bhamidi, Danny Nam, and Allan Sly.
Wednesday, February 6 at 4:00pm in Van Vleck 911 , Li-Cheng Tsai, Columbia University
Title:
When particle systems meet PDEs
Abstract: Interacting particle systems are models that involve many randomly evolving agents (i.e., particles). These systems are widely used in describing real-world phenomena. In this talk we will walk through three facets of interacting particle systems, namely the law of large numbers, random fluctuations, and large deviations. Within each facet, I will explain how Partial Differential Equations (PDEs) play a role in understanding the systems..
Title:
Fluctuations of the KPZ equation in d\geq 2 in a weak disorder regime
Abstract: We will discuss some recent work on the Edwards-Wilkinson limit of the KPZ equation with a small coupling constant in d\geq 2.
February 14, Timo Seppäläinen, UW-Madison
Title:
Geometry of the corner growth model
Abstract: The corner growth model is a last-passage percolation model of random growth on the square lattice. It lies at the nexus of several branches of mathematics: probability, statistical physics, queueing theory, combinatorics, and integrable systems. It has been studied intensely for almost 40 years. This talk reviews properties of the geodesics, Busemann functions and competition interfaces of the corner growth model, and presents some new qualitative and quantitative results. Based on joint projects with Louis Fan (Indiana), Firas Rassoul-Agha and Chris Janjigian (Utah).
February 21, Diane Holcomb, KTH
Title:
On the centered maximum of the Sine beta process Abstract: There has been a great deal or recent work on the asymptotics of the maximum of characteristic polynomials or random matrices. Other recent work studies the analogous result for log-correlated Gaussian fields. Here we will discuss a maximum result for the centered counting function of the Sine beta process. The Sine beta process arises as the local limit in the bulk of a beta-ensemble, and was originally described as the limit of a generalization of the Gaussian Unitary Ensemble by Valko and Virag with an equivalent process identified as a limit of the circular beta ensembles by Killip and Stoiciu. A brief introduction to the Sine process as well as some ideas from the proof of the maximum will be covered. This talk is on joint work with Elliot Paquette.
Title: Quantitative homogenization in a balanced random environment
Abstract: Stochastic homogenization of discrete difference operators is closely related to the convergence of random walk in a random environment (RWRE) to its limiting process. In this talk we discuss non-divergence form difference operators in an i.i.d random environment and the corresponding process—a random walk in a balanced random environment in the integer lattice Z^d. We first quantify the ergodicity of the environment viewed from the point of view of the particle. As consequences, we obtain algebraic rates of convergence for the quenched central limit theorem of the RWRE and for the homogenization of both elliptic and parabolic non-divergence form difference operators. Joint work with J. Peterson (Purdue) and H. V. Tran (UW-Madison).
Wednesday, February 27 at 1:10pm Jon Peterson, Purdue
Title:
Functional Limit Laws for Recurrent Excited Random Walks
Abstract:
Excited random walks (also called cookie random walks) are model for self-interacting random motion where the transition probabilities are dependent on the local time at the current location. While self-interacting random walks are typically very difficult to study, many results for (one-dimensional) excited random walks are remarkably explicit. In particular, one can easily (by hand) calculate a parameter of the model that will determine many features of the random walk: recurrence/transience, non-zero limiting speed, limiting distributions and more. In this talk I will prove functional limit laws for one-dimensional excited random walks that are recurrent. For certain values of the parameters in the model the random walks under diffusive scaling converge to a Brownian motion perturbed at its extremum. This was known previously for the case of excited random walks with boundedly many cookies per site, but we are able to generalize this to excited random walks with periodic cookie stacks. In this more general case, it is much less clear why perturbed Brownian motion should be the correct scaling limit. This is joint work with Elena Kosygina.
March 7, TBA March 14, TBA March 21, Spring Break, No seminar March 28, Shamgar Gurevitch UW-Madison
Title:
Harmonic Analysis on GLn over finite fields, and Random Walks
Abstract: There are many formulas that express interesting properties of a group G in terms of sums over its characters. For evaluating or estimating these sums, one of the most salient quantities to understand is the
character ratio:
$$ \text{trace}(\rho(g))/\text{dim}(\rho), $$
for an irreducible representation $\rho$ of G and an element g of G. For example, Diaconis and Shahshahani stated a formula of this type for analyzing G-biinvariant random walks on G. It turns out that, for classical groups G over finite fields (which provide most examples of finite simple groups), there is a natural invariant of representations that provides strong information on the character ratio. We call this invariant
rank. This talk will discuss the notion of rank for $GL_n$ over finite fields, and apply the results to random walks. This is joint work with Roger Howe (Yale and Texas AM).
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Problem 533
Consider the complex matrix
\[A=\begin{bmatrix} \sqrt{2}\cos x & i \sin x & 0 \\ i \sin x &0 &-i \sin x \\ 0 & -i \sin x & -\sqrt{2} \cos x \end{bmatrix},\] where $x$ is a real number between $0$ and $2\pi$.
Determine for which values of $x$ the matrix $A$ is diagonalizable.
When $A$ is diagonalizable, find a diagonal matrix $D$ so that $P^{-1}AP=D$ for some nonsingular matrix $P$. Problem 529
Let $\F_3=\Zmod{3}$ be the finite field of order $3$.
Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal $I=(x^2+1)$ generated by $x^2+1\in \F_3[x]$. (a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field. How many elements does the field have? (b) Let $ax+b+I$ be a nonzero element of the field $\F_3[x]/(x^2+1)$, where $a, b \in \F_3$. Find the inverse of $ax+b+I$. (c) Recall that the multiplicative group of nonzero elements of a field is a cyclic group.
Confirm that the element $x$ is not a generator of $E^{\times}$, where $E=\F_3[x]/(x^2+1)$ but $x+1$ is a generator.Add to solve later
Problem 528
Let $V$ denote the vector space of all real $2\times 2$ matrices.
Suppose that the linear transformation from $V$ to $V$ is given as below. \[T(A)=\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}A-A\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}.\] Prove or disprove that the linear transformation $T:V\to V$ is an isomorphism. Problem 527
A square matrix $A$ is called
idempotent if $A^2=A$. (a)Let $\mathbf{u}$ be a vector in $\R^n$ with length $1$.
Define the matrix $P$ to be $P=\mathbf{u}\mathbf{u}^{\trans}$.
Prove that $P$ is an idempotent matrix.
(b)Suppose that $\mathbf{u}$ and $\mathbf{v}$ be unit vectors in $\R^n$ such that $\mathbf{u}$ and $\mathbf{v}$ are orthogonal.
Let $Q=\mathbf{u}\mathbf{u}^{\trans}+\mathbf{v}\mathbf{v}^{\trans}$.
Prove that $Q$ is an idempotent matrix.
(c)Prove that each nonzero vector of the form $a\mathbf{u}+b\mathbf{v}$ for some $a, b\in \R$ is an eigenvector corresponding to the eigenvalue $1$ for the matrix $Q$ in part (b). Add to solve later Problem 526
A ring is called
local if it has a unique maximal ideal. (a) Prove that a ring $R$ with $1$ is local if and only if the set of non-unit elements of $R$ is an ideal of $R$.
Add to solve later
(b) Let $R$ be a ring with $1$ and suppose that $M$ is a maximal ideal of $R$. Prove that if every element of $1+M$ is a unit, then $R$ is a local ring. Problem 525
Let
\[R=\left\{\, \begin{bmatrix} a & b\\ 0& a \end{bmatrix} \quad \middle | \quad a, b\in \Q \,\right\}.\] Then the usual matrix addition and multiplication make $R$ an ring.
Let
\[J=\left\{\, \begin{bmatrix} 0 & b\\ 0& 0 \end{bmatrix} \quad \middle | \quad b \in \Q \,\right\}\] be a subset of the ring $R$. (a) Prove that the subset $J$ is an ideal of the ring $R$.
Add to solve later
(b) Prove that the quotient ring $R/J$ is isomorphic to $\Q$. Problem 524
Let $R$ be the ring of all $2\times 2$ matrices with integer coefficients:
\[R=\left\{\, \begin{bmatrix} a & b\\ c& d \end{bmatrix} \quad \middle| \quad a, b, c, d\in \Z \,\right\}.\]
Let $S$ be the subset of $R$ given by
\[S=\left\{\, \begin{bmatrix} s & 0\\ 0& s \end{bmatrix} \quad \middle | \quad s\in \Z \,\right\}.\] (a) True or False: $S$ is a subring of $R$.
Add to solve later
(b) True or False: $S$ is an ideal of $R$.
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I want to see if a propeller is bad or good for generating static thrust. This is what I know about the propeller and the engine used to turn it during a test:
Engine shaft power, $P$ = 0.58 hp
Propeller diameter, $D$ = 8.5 ft
Revolutions per minute, $RPM$ = 245
Static thrust, $T$ = 18.75 lbf
Can I calculate the efficiency of the propeller? What is the maximum achievable efficiency?
Additional explanations
The formula I was suggested to use could be exactly what I need. However, using it, I do not get some reported experimental results like:
Engine shaft power, $P$ = 6 hp
Propeller diameter, $D$ = 8.5 ft
Static thrust, $T$ = 66 lbf
Propeller efficiency, $\eta_{prop} \geq 66\%$
Gear efficiency, $\eta_{gears} \geq 85\%$
Doing the calculations: $\eta_{prop} = \sqrt{\frac{T^3}{(P^2 \cdot \pi \cdot \eta_{gears} \cdot \frac{D^2}{2} \cdot \rho )}} = 37.2\%$ which is well below 66%.
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What is meant by a local Lagrangian density?
How will a non-local Lagrangian look like?
What is the problem that we do not consider such Lagrangian densities?
What is a local Lagrangian density?
A classical field theory on Minkowski space $\mathbb R^{d,1}$ is specified by a space $\mathcal C$ of field configurations $\phi:\mathbb R^{d,1}\to T$, and an action functional $S:\mathcal C\to\mathbb R$. The set $T$ is called the target space of the theory, and is often a vector space. If there exists a function $L:\mathcal C\times\mathbb R\to \mathbb R$ for which\begin{align} S[\phi] = \int_{\mathbb R} dt\, L[\phi](t),\end{align}then we call $L$ a
lagrangian for the theory. If, further, there exists a function $\tilde L$ such that\begin{align} L[\phi](t) = \int_{\mathbb R^d} d^d\mathbf x \,\tilde L[\phi](t, \mathbf x)\end{align}then we call $\tilde L$ a langrangian density for the theory. Finally, if there exists a positive integer $n$ and a function $\mathscr L$ such that\begin{align} \tilde L[\phi](t, \mathbf x) = \mathscr L(t,\mathbf x,\phi(t, \mathbf x), \partial\phi(t,\mathbf x), \dots, \partial^n\phi(t,\mathbf x))\end{align}then we say that the lagrangian density is local. In other words, the lagrangian density is local provided its value at a given spacetime point depends only on that point, the value of the field at that point, and a finite number of its derivatives at that same point. An example of a non-local Lagrangian density.
Consider $T = \mathbb R$, namely a theory of a single real scalar field. Let $\mathbf a\in\mathbb R^d$ be given, and define a Lagrangian density by \begin{align} \tilde L[\phi](t,\mathbf x) = \phi(t,\mathbf x) + \phi(t, \mathbf x+\mathbf a). \end{align} This Lagrangian density is not local because the value of the Lagrangian at a given point $(t,\mathbf x)$ depends on the value of the field at that point and on the value of the field at the point $(t,\mathbf x+\mathbf a)$. If we were to Taylor expand the second term $\phi(t,\mathbf a)$ about $\mathbf x$, then we would see that the Lagrangian density depends on an infinite number of derivatives of the field, thus violating the definition of a local Lagrangian density.
What's the issue with theories with non-local Lagrangian densities?
I'm no expert on this, so I'll divert to another user. I will say, however, that people do study theories with non-local Lagrangian densities in practice, so there's nothing a priori "wrong" with them, but they might generically exhibit some pathology that you might prefer not to have.
Perhaps most relevant, though, if you're taking QFT from a high energy theorist, for example, is that the Lagrangian density of the Standard Model is local, so there's no need to consider non-local beasts if one is studying the Standard Model.
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Matrices
Definition: A Matrix $A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}$ is a rectangular array of numbers or expressions known as entries to the matrix. If $A$ has $m$ rows and $n$ columns then we say that the size of $A$ is $m \times n$.
The following examples represents some matrices of varying sizes.(1)
We will now look at some other important definitions.
Definition: A matrix $A$ is said to be Square if it has the same number of rows as it does columns, that is if $A$ is of size $m \times n$ then $m = n$. A general square matrix is commonly denoted with size $n \times n$.
Definition: If $A$ is an $m \times n$ matrix, the entry in row $i$ and column $j$ (clearly $i, j$ are integers and $1 ≤ i ≤ m$ and $1 ≤ j ≤ n$) is denoted by $a_{ij}$ (e.g. $a_{23}$ represents the entry in row $2$ and column $3$ of a matrix $A$).
Consider the following matrix $B = \begin{bmatrix} 4 & 1\\ 2 & 7 \end{bmatrix}$. This matrix is square since it has the same number of rows ($2$) as it does columns ($2$), and $B$ has entries $b_{11} = 4$, $b_{12} = 1$, $b_{21} = 2$ and $b_{22} = 7$.
We will now look at some elementary row operations on matrices that we will eventually utilize to help us with solving systems of linear equations.
Elementary Row Operations
There are 3 elementary row operations that we can perform on matrices. The notation in brackets indicate the notation for the operation applied where $a$ and $b$ represent any arbitrary row in the matrix to which the operations are applied.
1.Multiplying a row by a constant $k$ where $k ≠ 0$ ($kR_{a} \to R_{a}$). 2.Adding (or subtracting) a multiple $k$ of a row to another ($R_a + kR_b \to R_a$). 3.Interchanging two rows ($R_a \leftrightarrow R_b$).
For example, consider the following matrix $\begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}$.
Let's now multiply the first row by the constant $3$, that is perform the row operation $3R_{1} \to R_{1}$. We thus obtain the matrix $\begin{bmatrix} 3 & 6\\ 3 & 4 \end{bmatrix}$.
Now let's take row $1$ and add twice row $2$ to it, that is $R_1 + 2R_2 \to R_1$. We obtain $\begin{bmatrix} 9 & 14 \\ 3 & 4 \end{bmatrix}$.
Finally, let's interchange rows $1$ and $2$ as $R_{1} \leftrightarrow R_{2}$ to get $\begin{bmatrix} 3 & 4\\ 9 & 14 \end{bmatrix}$.
Definition: A set of matrices are said to be Row Equivalent if a series of finite elementary row operations can be performed in order to obtain one matrix from the next.
In our example above, all of the matrices we listed were considered row equivalent as we can easily obtain one from another with just a few of these elementary row operations.
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Any longtime reader cannot help but notice that we get many
abstract duplicate questions, e.g. this recent question on partial fraction computation, which is not essentially different from many other questions of the same shape, e.g. this question. Once you know how to solve one of these problems you can solve them all. There are many classes of problems that frequently arise in reparametrized variants, e.g. divisibility problems using Fermat's little theorem, proving basic properties of gcds, etc. Does it make sense to try to try to prevent these minor variations from swamping the site? In a couple years time we could well have many hundreds of variants of such questions that are all essentially the same except for minor variation of parameters. Among other detrimental consequences, this greatly obfuscates search results. Certainly our user community has the expertise to appropriately classify and eliminate these "abstract" duplicates. Perhaps with a little ad-hoc add-on infrastructure, and with moderator support, we could address these issues before they get out of hand. Thoughts?
Any longtime reader cannot help but notice that we get many
The current thinking is that subtly different variants of the same question be
closed as duplicates of a more canonical, more general question and answer pair:
If you keep seeing the same form of questions, whether it’s mod_rewrite rules on Server Fault, freezing computers on Super User, or how to use regular expressions to parse HTML, write a great, canonical answer, once and for all. Make it community wiki so that as many other people as possible can make it great. Work really hard on writing something that is clear, concise, and understandable by as wide an audience as possible.
Personally, I would express this sentiment as "old-timers are tired of answering what is essentially the same question in millions of tiny different varations".
Whenever you feel that is happening, I recommend approaching it as per the above.
I would suggest that:
if there is a sufficiently-general form of the question, new variants should be closed as duplicate, but notdeleted—this provides a way for someone searching for that variant to get to the more general result; if there is not a sufficiently-general form of the question, ask the general form of the questionso that it can be answered in the general case (remember, you're allowed to ask and answer your own question as doing so provides valuable content) and future variants can be closed as duplicates, as above.
(too long for comment)
I agree. It is not clear what should be done. Sometimes the problem can be fixed if the most general solution was posted. As an example, the following three threads are conceptually identical:
But here, the question can be generalized and solved. That is for polynomials we can show: $$\lim_{x\rightarrow\infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}}-x=\frac{a_{n-1}}{n}$$
Should they be merged? I definitely think so. The proof techniques in the answers of each post are exactly the same.
A harder question is what to do with all the Fermat's little theorem problems They are all so similar, but different enough that there is no "universal case" that can be proven.
At least so far, I don't see this as a problem. It would be nice if a search turned up the earlier variants, but I don't know how to do that and the evidence is that the posters of these problems don't search anyway. I would support a relaxation of the wording of "exact duplicate" to "if you read this you should be able to figure it out", but the volume is not too bad. If I can find the previous answer I post a link, but often I can't.
A request for comment:
Perhaps it will be worthwhile to have a two-pronged approach:
Maintain a list of FAQThis is not an FAQ in the sense of how to use the site. But actually a list of Frequently Asked Questions. This could be implemented as a Meta Thread (and to be incorporated in/linked to from the FAQ on how to use this website), or perhaps as a tag (though for the latter it'd be much more reasonable for this to be a "restricted" tag, with a reputation limit before one can tag a question as such). To this list (or a separate one) we can also add all those silly puzzle questions that crop up every now and then, and those inevitable questions about 0.999999... = 1. Aggressively closing and possibly deletingsuch minor variants and duplicates. If I am not mistaken, 10K+ users (of which there are quite a few now on this site) can vote to have an already closed question deleted. So with a bit of help from the community we can certainly clean up the clutter.
To populate the list of FAQ, each question in the FAQ should be a good CW, abstract question/answer to a canonical form of the question that often appears. And ideally we should also link to, in that question should be included links to some less abstract examples that has already appeared on this forum. So we address the abstract and the practical in one go.
Of course, this will require
a lot of work from the community.
One of the reasons we get the same question over and over is people who are learning new concepts
cannot tell if two problems are essentially the same. In a sense, what they are asking us is "what type of problem is this and how do I know it is that type?"
Deleting what looks, to trained eyes, to be "minor" variations will not stem the flow of questions... but grouping a large number of similar examples together just might.
Is there any way that we could do this?
This brings up a good point. I frequent [Combinatorics] which is always flooded with minor variations of the same question. For example,
How many ways can George give 6 apples to his 2 kids John and Rebecca? How many ways can Alice give 8 apples to her 3 kids Jimmy, Mary, and Susan? How many ways can you place 8 unlabelled balls into 3 labelled urns? How many ways can you place $n$ unlabelled balls into $m$ labelled urns? How many ways can you place $n$ balls into $m$ urns?
Do we close both (1),(2),(3) as duplicates of (4)? Will the person asking question (1) be able to tell how their question relates to (4)? Do we close (4) as a duplicate of (5)?
I honestly think "duplicates" by itself is completely inadequate for organizational purposes. We need an easier way to navigate the hierarchy of generalizations. Being able to tag one post as a generalization of another would go a long way. Then each question could have four links attached at the bottom: specializations / variations / generalizations / duplicates. The questions above can be linked (1) ~ (2)=(3) < (3) < (4). A few issues:
Since a question can often be generalized in many different ways, how do we define (variations).
Do we allow generalizations between vastly different areas of mathematics. Do I really want (5) in measure theory terms? How do [tags] play a role?
If (a)<(b)<(c), then when you click the (generalizations) link on (a), should (c) be listed?
etc.
I don't know what the moderator features are on the site, but it might be possible to merge the specific questions into the thread of the more general one under a category of "examples".
I'm curious if we can somehow combine the tools of wolfram with some sort of language comprehenser program, and an AI which is fed the whole of Math.SE as test data.
Then theoretically if someone posts a question there will immediately be an answer from the computer which will answer the question correctly if it indeed is 'minor' enough of a variant
leaving the only questions that were not instantaneously answered correctly as new questions which need to be answered by the community.
Quora.com originally began itself by using visibly editable aliases.
Audit logs (the only real power of Stack and Q) should be used to make sense of the (abtract) competition, for public education.
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The vector $\vec X = [X_1,\dots,X_N]^T$ has a rotationally invariant distribution. That is, if $A$ is any orthogonal matrix, then the distribution of $\vec X$ and $A \vec X$ are the same. Hence by letting $A$ be an orthogonal matrix that takes $[1,\dots,1]^T$ to $[\sqrt N,0,\dots,0]^T$, your problem is the same as computing$$ \Pr(\sqrt N X_1 \mid \sum X_i^2 = r^2) .$$
Next, since $\vec X$ is rotationally invariant, your question is equivalent to: what is the probability distribution of $\sqrt N x_1$ if $\vec x$ is chosen randomly on the sphere is radius $r$. So now the problem is reduced to the geometry of spheres in $N$ dimensions.
The total surface area of the sphere is $s_N r^{N-1}$ where $s_N = \frac{N \pi^{N/2}}{\Gamma(\frac N2 + 1)} = \frac{2 \pi^{N/2}}{\Gamma(\frac N2)}$ (see https://en.wikipedia.org/wiki/N-sphere). Take a slice through the sphere of radius $r$ such that the first coordinate $x_1$ is between $x$ and $x + \delta x$, and set $x = r \sin\theta$ where $\theta$ is the angle between the point on the sphere and the equator $x_1 = 0$. Then the surface area of that slice is approximately$$ s_{N-1} (r \cos \theta)^{N-2} \sec \theta \, \delta x = s_{N-1} r^{N-2} (r^2 - x^2)^{(N-3)/2} \delta x .$$So the probability that $\sqrt Nx_1 $ lies between $x$ and $x + \delta x$ is approximately$$ \frac{s_{N-1}}{r \sqrt N s_N} \left(r^2 - \frac{x^2}N\right)^{(N-3)/2} \delta x .$$
Details might be wrong, but the idea is correct. It is similar to the student $t$-test distribution.
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Tagged: rank of a matrix Problem 643
For each of the following matrices, find a row-equivalent matrix which is in reduced row echelon form. Then determine the rank of each matrix.
(a) $A = \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix}$. (b) $B = \begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix}$. (c) $C = \begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix}$. (d) $D = \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}$.
Add to solve later
(e) $E = \begin{bmatrix} -2 & 3 & 1 \end{bmatrix}$. Problem 352
A
hyperplane in $n$-dimensional vector space $\R^n$ is defined to be the set of vectors \[\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}\in \R^n\] satisfying the linear equation of the form \[a_1x_1+a_2x_2+\cdots+a_nx_n=b,\] where $a_1, a_2, \dots, a_n$ (at least one of $a_1, a_2, \dots, a_n$ is nonzero) and $b$ are real numbers. Here at least one of $a_1, a_2, \dots, a_n$ is nonzero.
Consider the hyperplane $P$ in $\R^n$ described by the linear equation
\[a_1x_1+a_2x_2+\cdots+a_nx_n=0,\] where $a_1, a_2, \dots, a_n$ are some fixed real numbers and not all of these are zero. (The constant term $b$ is zero.)
Then prove that the hyperplane $P$ is a subspace of $R^{n}$ of dimension $n-1$.Add to solve later
Problem 329
Let $n$ be a positive integer. Let $T:\R^n \to \R$ be a non-zero linear transformation.
Prove the followings. (a) The nullity of $T$ is $n-1$. That is, the dimension of the nullspace of $T$ is $n-1$. (b) Let $B=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}\}$ be a basis of the nullspace $\calN(T)$ of $T$. Let $\mathbf{w}$ be the $n$-dimensional vector that is not in $\calN(T)$. Then \[B’=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}, \mathbf{w}\}\] is a basis of $\R^n$.
Add to solve later
(c) Each vector $\mathbf{u}\in \R^n$ can be expressed as \[\mathbf{u}=\mathbf{v}+\frac{T(\mathbf{u})}{T(\mathbf{w})}\mathbf{w}\] for some vector $\mathbf{v}\in \calN(T)$. Problem 303
Let $A$ be a real $7\times 3$ matrix such that its null space is spanned by the vectors
\[\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \text{ and } \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}.\] Then find the rank of the matrix $A$.
(
Purdue University, Linear Algebra Final Exam Problem) Read solution The Possibilities For the Number of Solutions of Systems of Linear Equations that Have More Equations than Unknowns Problem 295
Determine all possibilities for the number of solutions of each of the system of linear equations described below.
(a) A system of $5$ equations in $3$ unknowns and it has $x_1=0, x_2=-3, x_3=1$ as a solution. (b) A homogeneous system of $5$ equations in $4$ unknowns and the rank of the system is $4$.
(
The Ohio State University, Linear Algebra Midterm Exam Problem) Read solution Given Graphs of Characteristic Polynomial of Diagonalizable Matrices, Determine the Rank of Matrices Problem 217
Let $A, B, C$ are $2\times 2$ diagonalizable matrices.
The graphs of characteristic polynomials of $A, B, C$ are shown below. The red graph is for $A$, the blue one for $B$, and the green one for $C$.
From this information, determine the rank of the matrices $A, B,$ and $C$.
Read solution Add to solve later
Problem 154
Define the map $T:\R^2 \to \R^3$ by $T \left ( \begin{bmatrix}
x_1 \\ x_2 \end{bmatrix}\right )=\begin{bmatrix} x_1-x_2 \\ x_1+x_2 \\ x_2 \end{bmatrix}$. (a) Show that $T$ is a linear transformation. (b) Find a matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ for each $\mathbf{x} \in \R^2$.
Add to solve later
(c) Describe the null space (kernel) and the range of $T$ and give the rank and the nullity of $T$.
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Null Space of a Linear Map
Definition: If $T \in \mathcal L (V, W)$ then the Null Space or Kernel of the linear transformation $T$ is the subset of $V$ defined as $\mathrm{null} (T) = \{ v \in V : T(v) = 0 \}$, that is, the null space of $T$ is the set of vectors from $V$ that are mapped to the zero vector in $W$ under $T$.
Before we look at some examples of the null spaces of linear transformations, we will first establish that the null space can never be equal to the empty set, in other words, the null spaces always contains at least one element from $V$, namely, the zero vector from $V$.
Lemma 1: If $T \in \mathcal L (V, W)$ then the null space of $T$ contains at least one element from $V$, that is $\mathrm{null} (T) \neq \emptyset$. Proof:Since $V$ is a vector space, by definition $V$ contains an additive identity, call it $0_V$. Therefore $0_V = 0_V + 0_V$, and so $T(0_V) = T(0_V + 0_V)$. Since $T$ is a linear transformation, we note that $T(0_V + 0_V) = T(0_V) + T(0_V)$. Therefore $T(0_V) = T(0_V) + T(0_V)$ which implies that $T(0_V) = 0_W$. Therefore $0_V \in \mathrm{null} (T)$, and so $\mathrm{null} T \neq \emptyset$. $\blacksquare$
From the lemma above, we conclude that the zero vector (additive identity) in $V$ is always mapped to the zero vector in $W$. Now the following lemma will tell us that the null space of $T$ is also a subspace of $V$.
Lemma 2: If $T \in \mathcal L (V, W)$ then the subset $\mathrm{null} (T)$ is a subspace of $V$. Proof:Since $\mathrm{null} (T) \subseteq V$, all we must do is verify that $\mathrm{null} (T)$ is closed under addition, closed under scalar multiplication, and contains the zero vector. Let $u, v \in \mathrm{null} (T)$ and $a \in \mathbb{F}$. Since $u, v \in \mathrm{null} (T)$ we have that by the definition of the null space of $T$ that $T(u) = 0$ and $T(v) = 0$. Since $T$ is a linear transformation we also have that $0 + 0 = T(u) + T(v) = T(u + v) = 0$, and so $(u + v) \in \mathrm{null} (T)$. Therefore $\mathrm{null} (T)$ is closed under addition. Now since $T(u) = 0$, we have that $aT(u) = a0 = 0$. Since $T$ is a linear transformation, then $aT(u) = T(au) = 0$, and so $(au) \in \mathrm{null} (T)$. Therefore $\mathrm{null} (T)$ is closed under scalar multiplication. From Lemma 1, we verified that $0_V \in \mathrm{null} (T)$ and so $\mathrm{null} (T)$ contains the zero vector. Therefore $\mathrm{null} (T)$ is a subspace of $V$. $\blacksquare$
We will now look at some examples of null spaces.
The Null Space of The Zero Map
If $0 \in \mathcal L (V, W)$ represents the zero map, then $\mathrm{null} (T) = V$, since all vectors $v \in V$ are mapped to the zero vector $0_W \in W$ by the definition of the zero map.
The Null Space of The Identity Map
If $I \in \mathcal L (V, V)$ represents the identity map, then $\mathrm{null} (T) = \{ 0 \}$. We note that all vectors $v \in V$ are mapped back to themselves, and so the only vector in $V$ that mapped to zero is $0$ itself.
The Null Space of The Differentiation of Polynomials Map
If $T \in \mathcal L (\wp (\mathbb{R}), \wp (\mathbb{R}))$ represents the transformation defined by $T(p(x)) = p'(x)$, then the null space of $T$ is the set of all polynomials whose first derivatives are equal to zero, that is, the set of constant functions $p(x) = a_0$, $a_0 \in \mathbb{R}$. So $\mathrm{null} (T) = \{ p(x) : p(x) = a_0, \: a_0 \in \mathbb{R} \}$.
The Null Space of The Left Shift Operator
If $T \in \mathcal L (\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ represents the left shift operator, then the null space of $T$ is the set of infinite sequences, all of whose terms are zero. Thus any sequence in the form $(a_1, 0, 0, ...)$ is contained in the null space since $T((a_1, 0, 0, ...)) = (0, 0, ...)$, so $\mathrm{null} (T) = \{ (a_1, 0, 0, ...) : a_1 \in \mathbb{F} \}$.
The Null Space of The Right Shift Operator
If $T \in \mathcal L (\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ represents the right shift operator, then the null space of $T$ is the set of infinite sequences, all of whose terms are zero once again. The only such sequence is the zero sequence in $\mathbb{F}^{\infty}$, since $T((a_1, a_2, ...)) = (0, a_1, ...)$ we must have that $a_1 = a_2 = ... = 0$, and so $\mathrm{null} (T) = \{ (0, 0, ... ) \}$.
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Solar core is 34% of the Sun mass, so sun will continue to be a star, at some point after the event. It will implode and probably intensify the processes compared to previous conditions and to a comparable star with 66% of the Sun mass. Dynamics of collapsing processes, may lead to ejection of plasma, so it definitely not recommended for planets like our at current state of our technological development.
The core of the Sun is considered to extend from the center to about 0.2 to 0.25 of solar radius
$$U = -G\int_0^R {\frac{(4\pi r^2\rho)(\tfrac{4}{3}\pi r^{3}\rho)}{r}} dr = -G{\frac{16}{3}}\pi^2 \rho^2 \int_0^R {r^4} dr = -G{\frac{16}{15}}{\pi}^2{\rho}^2 R^5$$
From Gravitational binding energy wiki article, but in this case core is out, so is out 0.25R
$$U = -G\int_{0.25R}^R {\frac{(4\pi r^2\rho)(\tfrac{4}{3}\pi r^{3}\rho)}{r}} dr = -G{\frac{16}{3}}\pi^2 \rho^2 \int_{0.25R}^R {r^4} dr = -G{\frac{16}{3}}\pi^2 \rho^2 \cdot \frac{1}{5}(R^5-0.0009765625 \cdot R^5)$$
They assume even distribution for mass, which is definitely not the case, but we will continue assumption farther, let assume that a thing called radius will not change significantly for star with 66% of its original mass (not true but lets hope it is enough true for our purposes)
With these assumptions energy stored as kinetic energy/heat energy/etc as result of this collapse will may be like:
$$ 0.0009765625 \cdot \frac{0.4356 \cdot 3GM^2}{5R} = \text{9.68176538754e+37 J}$$
That is a Lot, even compared to 3.828e+26 J/s which sun produces, according Sun Fact Sheet
I'm not sure in any assumption and calculations here, I'm just try to estimate orders of magnitude of orders of magnitude. Looks like it have potential for nasty things.
The question is -- will it be enough? What happens to supernovas is not because they collapse, collapse itself is consequence of what actually happening, and happening changing the fuel they burn. (I do not posses deep knowledge about the processes there, but this is one of them)
Hydrogen burning is a slow process, if we compare it to other types of thermonuclear reactions, as it can be seen in example with thermonuclear bombs, they do not need such extreme conditions as sun have constantly in the core, and they produce more energy per given mass, then sun does per same mass in same time.
Removing core, have potential to slow down burning, as it contains heavy atoms like maybe carbon which maybe helps to catalyze hydrogen burning.
But this potential energy of collapse will heat hydrogen to higher temperatures and maybe compress it in to more dense state for some amount of time, which may not linearly improve speed of hydrogen burning. Which will lead to expansion of matter, slowing down the reaction and create circumstances to collapse again.
I will not wonder if collapsing/expanding cycles will continue for next million of years. How long it will continue will be question of how good will be that system as oscillator.
During that dance solar ejecta will have place, that is for sure, and it will be spectacular to observe from a safe distance.
Will it act really as supernova - probably not, depends, mmm interesting question. I mean sure not any star will, some stars may really become supernova from that, specially if core removal is done in the way to maximize that probability, but others will not.(basically these who may be supernova in a future they can, who will not, they probably will not)
Will this situation lead to some nasty things happening in the star system in a supernova fashion way of bad and good(depends who and for what uses that). Yes, it probably will have some elements of supernova - energy bursts, plasma bursts etc.
Will it be a apocalypse for star system, for planets probably not, for some one on a planet, probably yes.
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this is a mystery to me, despite having changed computers several times, despite the website rejecting the application, the very first sequence of numbers I entered into it's search window which returned the same prompt to submit them for publication appear every time, I mean ive got hundreds of them now, and it's still far too much rope to give a person like me sitting along in a bedroom the capacity to freely describe any such sequence and their meaning if there isn't any already there
my maturity levels are extremely variant in time, that's just way too much rope to give me considering its only me the pursuits matter to, who knows what kind of outlandish crap I might decide to spam in each of them
but still, the first one from well, almost a decade ago shows up as the default content in the search window
1,2,3,6,11,23,47,106,235
well, now there is a bunch of stuff about them pertaining to "trees" and "nodes" but that's what I mean by too much rope you cant just let a lunatic like me start inventing terminology as I go
oh well "what would cotton mathers do?" the chat room unanimously ponders lol
i see Secret had a comment to make, is it really a productive use of our time censoring something that is most likely not blatant hate speech? that's the only real thing that warrants censorship, even still, it has its value, in a civil society it will be ridiculed anyway?
or at least inform the room as to whom is the big brother doing the censoring? No? just suggestions trying to improve site functionality good sir relax im calm we are all calm
A104101 is a hilarious entry as a side note, I love that Neil had to chime in for the comment section after the big promotional message in the first part to point out the sequence is totally meaningless as far as mathematics is concerned just to save face for the websites integrity after plugging a tv series with a reference
But seriously @BalarkaSen, some of the most arrogant of people will attempt to play the most innocent of roles and accuse you of arrogance yourself in the most diplomatic way imaginable, if you still feel that your point is not being heard, persist until they give up the farce please
very general advice for any number of topics for someone like yourself sir
assuming gender because you should hate text based adam long ago if you were female or etc
if its false then I apologise for the statistical approach to human interaction
So after having found the polynomial $x^6-3x^4+3x^2-3$we can just apply Eisenstein to show that this is irreducible over Q and since it is monic, it follwos that this is the minimal polynomial of $\sqrt{1+\sqrt[3]{2}}$ over $\mathbb{Q}$ ? @MatheinBoulomenos
So, in Galois fields, if you have two particular elements you are multiplying, can you necessarily discern the result of the product without knowing the monic irreducible polynomial that is being used the generate the field?
(I will note that I might have my definitions incorrect. I am under the impression that a Galois field is a field of the form $\mathbb{Z}/p\mathbb{Z}[x]/(M(x))$ where $M(x)$ is a monic irreducible polynomial in $\mathbb{Z}/p\mathbb{Z}[x]$.)
(which is just the product of the integer and its conjugate)
Note that $\alpha = a + bi$ is a unit iff $N\alpha = 1$
You might like to learn some of the properties of $N$ first, because this is useful for discussing divisibility in these kinds of rings
(Plus I'm at work and am pretending I'm doing my job)
Anyway, particularly useful is the fact that if $\pi \in \Bbb Z[i]$ is such that $N(\pi)$ is a rational prime then $\pi$ is a Gaussian prime (easily proved using the fact that $N$ is totally multiplicative) and so, for example $5 \in \Bbb Z$ is prime, but $5 \in \Bbb Z[i]$ is not prime because it is the norm of $1 + 2i$ and this is not a unit.
@Alessandro in general if $\mathcal O_K$ is the ring of integers of $\Bbb Q(\alpha)$, then $\Delta(\mathcal O_K) [\mathcal O_K:\Bbb Z[\alpha]]^2=\Delta(\mathcal O_K)$, I'd suggest you read up on orders, the index of an order and discriminants for orders if you want to go into that rabbit hole
also note that if the minimal polynomial of $\alpha$ is $p$-Eisenstein, then $p$ doesn't divide $[\mathcal{O}_K:\Bbb Z[\alpha]]$
this together with the above formula is sometimes enough to show that $[\mathcal{O}_K:\Bbb Z[\alpha]]=1$, i.e. $\mathcal{O}_K=\Bbb Z[\alpha]$
the proof of the $p$-Eisenstein thing even starts with taking a $p$-Sylow subgroup of $\mathcal{O}_K/\Bbb Z[\alpha]$
(just as a quotient of additive groups, that quotient group is finite)
in particular, from what I've said, if the minimal polynomial of $\alpha$ wrt every prime that divides the discriminant of $\Bbb Z[\alpha]$ at least twice, then $\Bbb Z[\alpha]$ is a ring of integers
that sounds oddly specific, I know, but you can also work with the minimal polynomial of something like $1+\alpha$
there's an interpretation of the $p$-Eisenstein results in terms of local fields, too. If the minimal polynomial of $f$ is $p$-Eisenstein, then it is irreducible over $\Bbb Q_p$ as well. Now you can apply the Führerdiskriminantenproduktformel (yes, that's an accepted English terminus technicus)
@MatheinBoulomenos You once told me a group cohomology story that I forget, can you remind me again? Namely, suppose $P$ is a Sylow $p$-subgroup of a finite group $G$, then there's a covering map $BP \to BG$ which induces chain-level maps $p_\# : C_*(BP) \to C_*(BG)$ and $\tau_\# : C_*(BG) \to C_*(BP)$ (the transfer hom), with the corresponding maps in group cohomology $p : H^*(G) \to H^*(P)$ and $\tau : H^*(P) \to H^*(G)$, the restriction and corestriction respectively.
$\tau \circ p$ is multiplication by $|G : P|$, so if I work with $\Bbb F_p$ coefficients that's an injection. So $H^*(G)$ injects into $H^*(P)$. I should be able to say more, right? If $P$ is normal abelian, it should be an isomorphism. There might be easier arguments, but this is what pops to mind first:
By Schur-Zassenhaus theorem, $G = P \rtimes G/P$ and $G/P$ acts trivially on $P$ (the action is by inner auts, and $P$ doesn't have any), there is a fibration $BP \to BG \to B(G/P)$ whose monodromy is exactly this action induced on $H^*(P)$, which is trivial, so we run the Lyndon-Hochschild-Serre spectral sequence with coefficients in $\Bbb F_p$.
The $E^2$ page is essentially zero except the bottom row since $H^*(G/P; M) = 0$ if $M$ is an $\Bbb F_p$-module by order reasons and the whole bottom row is $H^*(P; \Bbb F_p)$. This means the spectral sequence degenerates at $E^2$, which gets us $H^*(G; \Bbb F_p) \cong H^*(P; \Bbb F_p)$.
@Secret that's a very lazy habit you should create a chat room for every purpose you can imagine take full advantage of the websites functionality as I do and leave the general purpose room for recommending art related to mathematics
@MatheinBoulomenos No worries, thanks in advance. Just to add the final punchline, what I wanted to ask is what's the general algorithm to recover $H^*(G)$ back from $H^*(P; \Bbb F_p)$'s where $P$ runs over Sylow $p$-subgroups of $G$?
Bacterial growth is the asexual reproduction, or cell division, of a bacterium into two daughter cells, in a process called binary fission. Providing no mutational event occurs, the resulting daughter cells are genetically identical to the original cell. Hence, bacterial growth occurs. Both daughter cells from the division do not necessarily survive. However, if the number surviving exceeds unity on average, the bacterial population undergoes exponential growth. The measurement of an exponential bacterial growth curve in batch culture was traditionally a part of the training of all microbiologists...
As a result, there does not exists a single group which lived long enough to belong to, and hence one continue to search for new group and activity
eventually, a social heat death occurred, where no groups will generate creativity and other activity anymore
Had this kind of thought when I noticed how many forums etc. have a golden age, and then died away, and at the more personal level, all people who first knew me generate a lot of activity, and then destined to die away and distant roughly every 3 years
Well i guess the lesson you need to learn here champ is online interaction isn't something that was inbuilt into the human emotional psyche in any natural sense, and maybe it's time you saw the value in saying hello to your next door neighbour
Or more likely, we will need to start recognising machines as a new species and interact with them accordingly
so covert operations AI may still exists, even as domestic AIs continue to become widespread
It seems more likely sentient AI will take similar roles as humans, and then humans will need to either keep up with them with cybernetics, or be eliminated by evolutionary forces
But neuroscientists and AI researchers speculate it is more likely that the two types of races are so different we end up complementing each other
that is, until their processing power become so strong that they can outdo human thinking
But, I am not worried of that scenario, because if the next step is a sentient AI evolution, then humans would know they will have to give way
However, the major issue right now in the AI industry is not we will be replaced by machines, but that we are making machines quite widespread without really understanding how they work, and they are still not reliable enough given the mistakes they still make by them and their human owners
That is, we have became over reliant on AI, and not putting enough attention on whether they have interpret the instructions correctly
That's an extraordinary amount of unreferenced rhetoric statements i could find anywhere on the internet! When my mother disapproves of my proposals for subjects of discussion, she prefers to simply hold up her hand in the air in my direction
for example i tried to explain to her that my inner heart chakras tell me that my spirit guide suggests that many females i have intercourse with are easily replaceable and this can be proven from historical statistical data, but she wont even let my spirit guide elaborate on that premise
i feel as if its an injustice to all child mans that have a compulsive need to lie to shallow women they meet and keep up a farce that they are either fully grown men (if sober) or an incredibly wealthy trust fund kid (if drunk) that's an important binary class dismissed
Chatroom troll: A person who types messages in a chatroom with the sole purpose to confuse or annoy.
I was just genuinely curious
How does a message like this come from someone who isn't trolling:
"for example i tried to explain to her that my inner heart chakras tell me that my spirit guide suggests that many ... with are easily replaceable and this can be proven from historical statistical data, but she wont even let my spirit guide elaborate on that premise"
3
Anyway feel free to continue, it just seems strange @Adam
I'm genuinely curious what makes you annoyed or confused yes I was joking in the line that you referenced but surely you cant assume me to be a simpleton of one definitive purpose that drives me each time I interact with another person? Does your mood or experiences vary from day to day? Mine too! so there may be particular moments that I fit your declared description, but only a simpleton would assume that to be the one and only facet of another's character wouldn't you agree?
So, there are some weakened forms of associativity. Such as flexibility ($(xy)x=x(yx)$) or "alternativity" ($(xy)x=x(yy)$, iirc). Tough, is there a place a person could look for an exploration of the way these properties inform the nature of the operation? (In particular, I'm trying to get a sense of how a "strictly flexible" operation would behave. Ie $a(bc)=(ab)c\iff a=c$)
@RyanUnger You're the guy to ask for this sort of thing I think:
If I want to, by hand, compute $\langle R(\partial_1,\partial_2)\partial_2,\partial_1\rangle$, then I just want to expand out $R(\partial_1,\partial_2)\partial_2$ in terms of the connection, then use linearity of $\langle -,-\rangle$ and then use Koszul's formula? Or there is a smarter way?
I realized today that the possible x inputs to Round(x^(1/2)) covers x^(1/2+epsilon). In other words we can always find an epsilon (small enough) such that x^(1/2) <> x^(1/2+epsilon) but at the same time have Round(x^(1/2))=Round(x^(1/2+epsilon)). Am I right?
We have the following Simpson method $$y^{n+2}-y^n=\frac{h}{3}\left (f^{n+2}+4f^{n+1}+f^n\right ), n=0, \ldots , N-2 \\ y^0, y^1 \text{ given } $$ Show that the method is implicit and state the stability definition of that method.
How can we show that the method is implicit? Do we have to try to solve $y^{n+2}$ as a function of $y^{n+1}$ ?
@anakhro an energy function of a graph is something studied in spectral graph theory. You set up an adjacency matrix for the graph, find the corresponding eigenvalues of the matrix and then sum the absolute values of the eigenvalues. The energy function of the graph is defined for simple graphs by this summation of the absolute values of the eigenvalues
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It is easier to talk about these concepts by moving past Knight a bit. There has been a lot of discussion since then. The easiest way to start talking about this is in discussing the three main schools of thought in probability and statistics. The schools are, in order of discovery, the Bayesian, the Likelihoodist and the Frequentist. We are going to ignore Fisher's Likelihoodist school because it is epistemological in nature and implies no logical rational behavior.
In the Bayesian school there is no such thing as chance. It doesn't exist. There is only uncertainty. To give an example, imagine you watch someone toss a coin twenty times and it comes up head every time. Now, imagine the same person tosses it H-T in perfect alternation in ten pairings of twenty tosses. The person tossing the coin is the Amazing Randi. He lets you inspect the coin and you have a physicist friend test it and it is a perfectly fair coin.
You are uncertain as to whether the coin is the same coin that was tossed. You are uncertain as to whether the Amazing Randi has sufficient control to cause the pattern of tosses. You are even uncertain as to whether the Amazing Randi had complete control the entire time or whether some of the tosses were really accidents that worked out despite a loss of control.
In Bayesian models there is only an absence of information. Probability exists in the mind of the person making decisions. It is not a physical property of the universe. This is even true in quantum mechanics which is usually interpreted in terms of chance. The goal is to use data to construct a distribution of "plausabilities," to use Cox's terminology.
In the Bayesian framework, data is fixed and not random. There can be no such thing as a "random sample." It is non-nonsensical in meaning. Because of this, the Bayesian definition of an expectation is $$E(\theta|X)=\int_{\theta\in\Theta}\theta{f(\theta|X)}\mathrm{d}\theta.$$
If you would gamble on an uncertainty and an expectation exists, it does not exist for heavy-tailed distributions, then you integrate over the set of all possible models of the world.
In the Frequentist world $\theta=k,k\in\Re$ and there is no uncertainty at all. Rather than use the word "risk" at this point, we will discuss chance as the word risk has an additional meaning since Knight. Events can happen only due to chance and when used in the calculus of variations, then all parameters are treated as perfectly known. Chance is a physical property of the universe.
Mechanically, this is no different than treating the null hypothesis as strictly true. In the Frequentist school there is no such thing as a p-value. That is only part of the Likelihoodist school. In the Frequentist school there is a value, $\alpha$, that creates an acceptance region and a rejection region. If a sample implies that the null cannot be rejected, then it is treated AS IF true. The Frequentist school is behavioral. You do not know if it is true, but the only rational behavior is to act as if it is. If it is in the rejection region, then you are are to behave AS IF false and AS IF the minimum variance unbiased estimator is the true value.
In the example, for the first test for a null of $\theta=.5$, the null would be rejected. In the second sample, despite the weird pattern and the past observations, it would be accepted as true. The samples cannot be combined due to the axioms. This creates a weird to irrational response where one moment you are behaving as if false and the next as if true.
Finance has almost entirely dropped the Bayesian meaning and almost entirely adopted the non-Bayesian meaning. You rarely see a serious discussion of uncertainty in finance because all mean-variance finance models are Frequentist models. They strictly depend upon Frequentist axioms.
The Frequentist definition of an expectation, where one exists, is $$E(X)=\int_{x\in\chi}xf(x|\theta)\mathrm{d}x.$$ Frequentist models live in the sample space, which is random.
Risk, in the Bayesian sense is the amount of resources exposed to uncertainty. Risk in the Frequentist sense is the amount of resources exposed to chance. Both definitions could have the phrase "of loss," appended to the end.
This focus on Frequentist methods is problematic because it blocks good modeling. Any function of the data is a statistic. Therefore $\sum\sin(x_i)$ is a statistic. Intuitively you may know it is a pointless statistic, but how do you show that mathematically? It turns out that the problem was solved with what is known as the complete class theorem.
Any solution inside the acceptable set is valid and any solution outside the set is invalid. The sum of the sines is an invalid estimator of the sample mean of the normal distribution. You can know this because it is outside the complete class of solutions.
All Bayesian solutions are inside the complete class of valid solutions. Frequentist solutions are valid either to the extent they map to a Bayesian solution for a specific sample, or at the limit. This is problematic because there is a recent proof that there does not exist a Frequentist solution for stocks. It is the reason the CAPM has never been validated and the heavy tails exist in the tests. Frequentist solutions are always outside the complete class of solutions. They are valid for Vegas style gambles and for corporate bonds with fixed coupons, but not for stocks.
I would recommend reading "Probability, the Language of Science," by E.T. Jaynes and "Decision Theory:Principles and Approaches," by Parmigiani. Models like the CAPM cannot work in a Bayesian framework, though it is far from obvious, which means they cannot work. It is unfortunate that Knightian uncertainty fell out of discussion for elegant models.
You can find papers on stocks and uncertainty at https://ssrn.com/abstract=2828744and https://ssrn.com/abstract=2656681
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Table of Contents
Riemann Stieltjes-Integrals with Integrators of Bounded Variation
Recall from the Functions of Bounded Variation page that if $f$ is a function defined on the closed interval $[a, b]$ and if $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$, then the variation of $f$ associated with $P$ is:(1)
If there exists an $M \in \mathbb{R}$, $M > 0$ such that $V_f(P) \leq M$ for all $P \in \mathscr{P}[a, b]$ then we said that $f$ is of bounded variation on $[a, b]$.
On the Total Variation of a Function page we said that if $f$ was of bounded variation then the least upper bound of the variation of $f$ ranging through all partitions $P \in \mathscr{P}[a, b]$ is the total variation of $f$ and we denote it:(2)
On the Decomposition of Functions of Bounded Variation as the Difference of Two Increasing Functions page we saw that if $f$ is of bounded variation on $[a, b]$ and we define the function for all $x \in [a, b]$ by $V(x) = V_f(a, x)$ then $f$ can be expressed as the difference of two increasing functions, namely:(3)
This decomposition is not unique though. For any increasing function $g$ we have that $f = (V + g) - (V - f + g)$.
Now let $\alpha$ be any function of bounded variation. Then $\alpha = \alpha_1 - \alpha_2$ where $\alpha_1$ and $\alpha_2$ are increasing functions. If $f$ is Riemann-Stieltjes integrable with respect to $\alpha_1$ and $\alpha_2$ then by Linearity of the Integrator of Riemann-Stieltjes Integrals page we must have that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$. We can of course use any of the important theorems regarding the existence of Riemann-Stieltjes integrals of functions of increasing integrators.
However, if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ then $f$ need not be Riemann-Stieltjes integrable with respect to both $\alpha_1$ and $\alpha_2$.
The following theorem tells us that for the special decomposition $V$ and $V - \alpha$ that if $f$ is bounded on $[a, b]$, and Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then it is also Riemann-Stieltjes integrable with respect to $V$ (and $V - \alpha$).
Theorem 1: Let $f$ be a bounded function defined on $[a, b]$ and let $\alpha$ be a function of bounded variation on $[a, b]$. Furthermore let $V(x) = V_{\alpha}(a, x)$ be the total variation function on $\alpha$. Then if $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a, b]$ then $f$ is Riemann-Stieltjes integrable with respect to $V$ and $V - \alpha$ on $[a, b]$.
We will not prove Theorem 1 as it is rather lengthy to show.
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Evolution equations and subdifferentials in Banach spaces
1.
Department of Applied Physics, School of Science and Engineering, Waseda University, 3-4-1 Ohkubo, Shinjuku-ku, Tokyo 169-8555
2.
Department of Applied Physics, School of Science and Engineering, Waseda University, 3-4-1, Okubo, Tokyo, 169-8555
Studies for this equation in the Hilbert space framework has been done by several authors. However the study in the V -V* setting is not pursued yet.
Our method of proof relies on some approximation arguments in a Hilbert space. To carry out this procedure, it is assumed that there exists a Hilbert space H satisfying $V \subset H \-= H$* $\subset V$* with densely defined continuous injections.
As an application of our abstract theory, the initial-boundary value problem is discussed for the nonlinear heat equation: $ut(x, t)-\Delta_p u(x, t)-|u|^(q-2) u(x, t) = f(x, t), x \in \Omega, u|_(\partial\Omega) = 0, t \>= 0$, where $\Omega$ is a bounded domain in $\mathbb(R)^N$. In particular, the local existence of solutions is assured under the so-called subcritical condition, i.e., $q < p$*, where $p$* denotes Sobolev’s critical exponent, provided that the initial data $u_0$ belongs to $W_0^(1,p)(\Omega)$.
Keywords:subdifferential, global solution, Sobolev's critical exponent, local solution, nonlinear heat equation, subcritical., reflexive Banach space, Evolution equation, p-Laplacian. Mathematics Subject Classification:35K22, 35K55, 35K65, 35K9. Citation:Goro Akagi, Mitsuharu Ôtani. Evolution equations and subdifferentials in Banach spaces. Conference Publications, 2003, 2003 (Special) : 11-20. doi: 10.3934/proc.2003.2003.11
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Tomás Caraballo, Marta Herrera-Cobos, Pedro Marín-Rubio.
Global attractor for a nonlocal
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Time regularity of the evolution solution to fractional stochastic
heat equation.
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Existence of solutions for a class of p-Laplacian type equation
with critical growth and potential vanishing at infinity.
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Carlo Mercuri, Michel Willem.
A global compactness result for the p-Laplacian involving critical nonlinearities.
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Viorel Barbu, Gabriela Marinoschi.
An identification problem for a linear evolution equation in a banach space.
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Ronghua Jiang, Jun Zhou.
Blow-up and global existence of solutions to a parabolic equation associated with the fraction
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Global existence and decay of energy for a nonlinear wave equation with $p$-Laplacian damping.
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Michinori Ishiwata.
Existence of a stable set for some nonlinear parabolic equation involving critical Sobolev exponent.
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$ L^p $-$ L^q $ estimates for the damped wave equation and the critical exponent for the nonlinear problem with slowly decaying data.
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Local and global existence of solutions to a strongly damped wave equation of the $ p $-Laplacian type.
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The classical Brown Representability Theorem states: Denote $hCW_*$ the homotopy category of pointed CW-complexes. Let $F : hCW_* \to Set_*$ be a contravariant functor. Then $F$ is representable if and only if
$F$ respects coproducts, i.e. $F(\vee_{i \in I} X_i) = \prod_{i \in I} F(X_i)$ for all families $X_i$ of pointed CW-complexes. $F$ satisfies a sort of mayer-vietoris-axiom: If $X$ is a pointed CW-complex which is the union of two pointed subcomplexes $A,B$, then the canonical map $F(X) \to F(A) \times_{F(A \cap B)} F(B)$ is surjective 1.
What about omitting the base points? So let $F :hCW \to Set$ be a contravariant functor that satisfies the analogous properties as above (replace the wedge-sum by the disjoint union). Is then $F$ representable? I'm not sure if we just can copy the proof of the pointed case (which can be found, e.g., in Switzer's book "Algebraic Topology - Homology and Homotopy", Representability Theorems). For example, $F(pt)$ can be anything (in contrast to the pointed case), it will be the set of path components in the classifying space. Besides, the proof uses homotopy groups and in particular the famous theorem of Whitehead, which deal with pointed CW-complexes. Nevertheless, I hope that $F$ is representable ... what do you think?
As a first step, we may define for every $i \in F(pt)$ the subfunctor $F_i$ of $F$ by $F_i(Y) = \{f \in F(Y) : \forall y : pt \to Y : f|_{y} = i \in F(pt)\}$, which should be thought as the connected component associated to $i$. Then it's not hard to show that $F_i$ satisfies the same properties as $F$ and that $F_i = [-,X_i]$ implies $F = [-,\coprod_i X_i]$. In other words, we may assume that $F(pt)=pt$ (so that the classifying space will be connected).
1 You can't expect it to be bijective, cf. question about categorical homotopy colimits
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Consider the MWE:
\documentclass{memoir}\usepackage{graphicx}\captiontitlefont{\slshape} % All captions slanted\begin{document}\begin{figure} \centering \includegraphics[width=\textwidth]{example-image} \caption{Flowchart of a gradient-based optimization algorithm. $ \Delta \alpha_i^{(k)} $ is the change of $ \alpha_i $ from iteration $ k $ to iteration $ k+1 $.}\end{figure}\end{document}
resulting in this figure (without the red line):
The caption is slated, but numbers in math mode are not. In my opinion it looks weird that the '+1' is not slanted. It can be fixed (badly) by moving it outside math mode, but then the spacing is clearly wrong.
How do I make the entire caption slanted, but with variables like 'k' staying in math mode font?
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The First and Second Arens Products on A** The First Arens Product
Let $\mathfrak{A}$ be a Banach algebra. Consider the second dual, $\mathfrak{A}^{**}$, which is clearly a Banach space. We would like to make $\mathfrak{A}^{**}$ a Banach algebra too, but it is not entirely obvious what the multiplication on $\mathfrak{A}^{**}$ should be. One type of multiplication we can define on $\mathfrak{A}^{**}$ is the
First Arens Product on $\mathfrak{A}^{**}$. It is defined in steps as follows:
Definition: Let $\mathfrak{A}$ be a Banach algebra.
1. For each $a \in \mathfrak{A}$ and for each $f \in \mathfrak{A}^*$ we define $f \cdot a \in \mathfrak{A}^*$ by $(fa)(b) := f(ab) \quad (\forall b \in \mathfrak{A})$.
2. For each $F \in \mathfrak{A}^{**}$ and for each $f \in \mathfrak{A}^*$ we define $F \cdot f \in \mathfrak{A}^*$ by $(F \cdot f)(a) := F(fa) \quad (\forall a \in \mathfrak{A})$
3. Lastly, the
First Arens Product on $\mathfrak{A}^{**}$ is the multiplication on $\mathfrak{A}^{**}$ defined for all $F, G \in \mathfrak{A}^{**}$ by $(FG)(f) := F(Gf) \quad (\forall f \in \mathfrak{A}^*)$.
It is clear that the First Arens Product defined above satisfies the 3 axioms on the Algebras over F page, making $\mathfrak{A}^{**}$ with the First Arens Product a Banach algebra. Indeed, let's verify these three axioms
1)Let $F, G, H \in \mathfrak{A}^{**}$. We will show that $[FG]H = F[GH]$. This is done by showing that these functionals are equal for all $f \in \mathfrak{A}^*$. Indeed: 2)Let $F, G, H \in \mathfrak{A}^{**}$. We now show that $F[G + H] = FG + FH$. This is again done by showing that these functionals are equal for all $f \in \mathfrak{A}^*$. Indeed by the linearity of $F$ we have that: 3)Let $F, G \in \mathfrak{A}^{**}$ and let $\alpha \in \mathbb{C}$. We lastly show that $[\alpha F]G = \alpha [FG] = F[\alpha G]$. Yet again, this is done by showing that these functionals are equal for all $f \in \mathfrak{A}^*$: So that $[\alpha F]G = \alpha [FG]$. Also: So that $\alpha [FG] = F[\alpha G]$ too. The Second Arens Product
We can quite naturally define another type of multiplication on $\mathfrak{A}^{**}$ call the
Second Arens Product as follows:
Definition: Let $\mathfrak{A}$ be a Banach algebra. 1) For each $a \in \mathfrak{A}$ and for each $f \in \mathfrak{A}^*$ we define $af \in \mathfrak{A}^*$ by $(af)(b) := f(ba) \quad (\forall b \in \mathfrak{A})$. 2) For each $F \in \mathfrak{A}^{**}$ and for each $f \in \mathfrak{A}^*$ we define $fF \in \mathfrak{A}^*$ by $(fF)(a) := F(af) \quad (\forall a \in \mathfrak{A})$. 3) Lastly, the Second Arens Product on $\mathfrak{A}^{**}$ is the multiplication on $\mathfrak{A}^{**}$ defined for all $F, G \in \mathfrak{A}^{**}$ by $(F * G)(f) :=F(fG) \quad (\forall f \in \mathfrak{A}^*)$. Here we will use $*$ to explicitly denote the second Arens product. Equality of the First and Second Arens Products
In general, given $F, G \in \mathfrak{A}^{**}$ it may be that $FG$ (first Arens product) is not equal to $F * G$ (second Arens product). When these two product are equal for all $F, G \in \mathfrak{A}^{**}$, the Banach algebra $\mathfrak{A}$ is given a special name.
Definition: Let $\mathfrak{A}$ be a Banach algebra. Then $\mathfrak{A}$ is said to be Arens Regular if $FG = F*G$ for all $F, G \in \mathfrak{A}^{**}$, where the lefthand side of the equality is multiplication with respect to the first Arens product, and the righthand side of the equality is multiplication with respect to the second Arens product.
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Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?
A. 24% B. 30% C. 36% D. 42% E. 48%
If polygon \(ABCDEF\) is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.
We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to \(180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ\), and there are 6 interior, equal angles, then each of those angles must measure \(\frac{720^\circ}{6} = 120^\circ\).
Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center \(O\):
Consider small triangle \(AXB\). The hypotenuse of this right triangle, \(AB\), has length \(r\). Angle \(ABO\) is \(60^\circ\), so \(AXB\) is a 30-60-90 triangle. This means that \(XB\) is \(\frac{r}{2}\) in length, and \(AX\) is \(\frac{\sqrt{3}r}{4}\) in length. Since \(AX\) and \(XB\) are the same length (by symmetry), \(AC\) is \(\sqrt{3}r\) in length.
This means that the area of triangle \(ABC\) is \(\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}\). Since there are three shaded triangles in all (including \(ABC\)), the total shaded area is \(\frac{3 \sqrt{3} r^2}{4}\). The area of the circle is , so the ratio of the shaded area to the area of the circle is given by \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\)
Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?
A. 24% B. 30% C. 36% D. 42% E. 48%
If polygon \(ABCDEF\) is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.
We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to \(180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ\), and there are 6 interior, equal angles, then each of those angles must measure \(\frac{720^\circ}{6} = 120^\circ\).
Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center \(O\):
Consider small triangle \(AXB\). The hypotenuse of this right triangle, \(AB\), has length \(r\). Angle \(ABO\) is \(60^\circ\), so \(AXB\) is a 30-60-90 triangle. This means that \(XB\) is \(\frac{r}{2}\) in length, and \(AX\) is \(\frac{\sqrt{3}r}{4}\) in length. Since \(AX\) and \(XB\) are the same length (by symmetry), \(AC\) is \(\sqrt{3}r\) in length.
This means that the area of triangle \(ABC\) is \(\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}\). Since there are three shaded triangles in all (including \(ABC\)), the total shaded area is \(\frac{3 \sqrt{3} r^2}{4}\). The area of the circle is , so the ratio of the shaded area to the area of the circle is given by \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\)
The closest answer is 42%.
Answer: D
Bunuel,
I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?
Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?
A. 24% B. 30% C. 36% D. 42% E. 48%
If polygon \(ABCDEF\) is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.
We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to \(180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ\), and there are 6 interior, equal angles, then each of those angles must measure \(\frac{720^\circ}{6} = 120^\circ\).
Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center \(O\):
Consider small triangle \(AXB\). The hypotenuse of this right triangle, \(AB\), has length \(r\). Angle \(ABO\) is \(60^\circ\), so \(AXB\) is a 30-60-90 triangle. This means that \(XB\) is \(\frac{r}{2}\) in length, and \(AX\) is \(\frac{\sqrt{3}r}{4}\) in length. Since \(AX\) and \(XB\) are the same length (by symmetry), \(AC\) is \(\sqrt{3}r\) in length.
This means that the area of triangle \(ABC\) is \(\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}\). Since there are three shaded triangles in all (including \(ABC\)), the total shaded area is \(\frac{3 \sqrt{3} r^2}{4}\). The area of the circle is , so the ratio of the shaded area to the area of the circle is given by \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\)
The closest answer is 42%.
Answer: D
Bunuel,
I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?
Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?
A. 24% B. 30% C. 36% D. 42% E. 48%
If polygon \(ABCDEF\) is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.
We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to \(180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ\), and there are 6 interior, equal angles, then each of those angles must measure \(\frac{720^\circ}{6} = 120^\circ\).
Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center \(O\):
Consider small triangle \(AXB\). The hypotenuse of this right triangle, \(AB\), has length \(r\). Angle \(ABO\) is \(60^\circ\), so \(AXB\) is a 30-60-90 triangle. This means that \(XB\) is \(\frac{r}{2}\) in length, and \(AX\) is \(\frac{\sqrt{3}r}{4}\) in length. Since \(AX\) and \(XB\) are the same length (by symmetry), \(AC\) is \(\sqrt{3}r\) in length.
This means that the area of triangle \(ABC\) is \(\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}\). Since there are three shaded triangles in all (including \(ABC\)), the total shaded area is \(\frac{3 \sqrt{3} r^2}{4}\). The area of the circle is , so the ratio of the shaded area to the area of the circle is given by \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\)
The closest answer is 42%.
Answer: D
Bunuel,
I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?
that link was helpful, but now I'm just trying to tie it back in with your slightly alternative explanation above. Still, i cannot see why you say that AX = \(\frac{\sqrt{3}r}{4}\) because we have an equilateral triangle (AOB), with sides R, and height AX. The formula states that the height AX would be \(\frac{\sqrt{3}r}{2}\). Please advise.
Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?
A. 24% B. 30% C. 36% D. 42% E. 48%
If polygon \(ABCDEF\) is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.
We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to \(180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ\), and there are 6 interior, equal angles, then each of those angles must measure \(\frac{720^\circ}{6} = 120^\circ\).
Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center \(O\):
Consider small triangle \(AXB\). The hypotenuse of this right triangle, \(AB\), has length \(r\). Angle \(ABO\) is \(60^\circ\), so \(AXB\) is a 30-60-90 triangle. This means that \(XB\) is \(\frac{r}{2}\) in length, and \(AX\) is \(\frac{\sqrt{3}r}{4}\) in length. Since \(AX\) and \(XB\) are the same length (by symmetry), \(AC\) is \(\sqrt{3}r\) in length.
This means that the area of triangle \(ABC\) is \(\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}\). Since there are three shaded triangles in all (including \(ABC\)), the total shaded area is \(\frac{3 \sqrt{3} r^2}{4}\). The area of the circle is , so the ratio of the shaded area to the area of the circle is given by \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\)
The closest answer is 42%.
Answer: D
Hey Bunuel, I think the highlighted part above should be ax=xc. and that is why it is creating a bit of confusion.
I think this is a high-quality question and I agree with explanation. There is a typo. AX equals (√3*r)/2, not (√3*r)/4 as it mentioned in the second line below the picture in the explanation. The rest is correct.
I think we don't have to actually calculate and waste time in exam. Rather, we can simply see that if the area of triangle is "x" the highlighted area, when wrapped inside, is also "x". Hence, the total area becomes (2*Triangle area+Remaining area). Now we see clearly that the remaining area is a smaller area as compared to the area of triangle. Hence, Area of triangle can not be less than 33% (Option A&B goes out). At the same time, the area of Triangle can not be 50% or very close to 50%, since that would mean that remaining area is hardly 2% (Option E goes out). Now, If we choose 36% that would mean that remaining area is nearly equal to area of triangle, which is clearly not the case (Option C Goes out). The only remaining and sensible choice that remains is Option D.
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In Hardy-Littlewood's 1923 paper
"Some problems of 'Partitio Numerorum' III" it is proven, assuming a weak version of GRH (namely that there is $\varepsilon>0$ s.t. all zeroes of $L(s,\chi)$ have $\Re(s)<3/4-\varepsilon$), that for all $k\geqslant 3$, when $n\to\infty$ through the integers with same parity than $k$:$$ r_{k}(n) \sim \frac{2C_k}{(k-1)!}\frac{n^{k-1}}{\log(n)^k}\prod_{\substack{p\mid n \\ p\geqslant 3}} \left(\frac{(p-1)^k+(-1)^k(p-1)}{(p-1)^k -(-1)^k} \right), $$
where $r_{k}(n):= \{(p_1,\ldots,p_k)\in\mathbb{P}^k:\sum p_i = n \}$ and $C_k$ is a constant given by: $$ C_k := \prod_{p\geqslant 3} \left(1- \frac{(-1)^k}{(p-1)^k}\right). $$
Roughly 15 years later Vinogradov introduced his influencial technique that allowed him to prove this estimate unconditionally for $k=3$. One thing that bothers me is:
what about the other $k$?
The wikipedia article on Goldbach's conjecture has (as of today [08/Oct/2016]) the following passage:
This formula has been rigorously proven to be asymptotically valid for $k \geqslant 3$ from the work of Vinogradov, but is still only a conjecture when $k=2$.$^{\text{[citation needed]}}$
I gave a quick look at K. F. Roth & Anne Davenport's translation of Vinogradov's The Method of Trigonometrical Sums in the Theory of Numbers, and in Chapter X. "Goldbach's Problem" it has the following passage at the start:
In the present chapter I give a solution of Goldbach's problem concerning the representability of every sufficiently large odd number $N$ as the sum of three primes, and I establish an asymptotic formula for the number of representations.
The method used here enables one also to solve more general additive problems involving primes, for example the question of representability of large numbers $N$ in the form $$ N = p_1^n + \ldots + p_s^n $$ (Waring's problem for primes). But I do not consider these more general questions here.
Every other source I consulted only talks about the case $k=3$. Does Vinogradov's method for $k=3$ implies the other cases as a corollary? Where can I find more (historical) information about the other $k$ in this Hardy & Littlewood's estimate? Thanks in advance!
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I think that the question is sufficiently precise if we think at a
realistic meaning of the word “inconsistent”. Also nowadays, for non logicians the adjective “inconsistent” doesn't really mean “free of contradictions” (this is only the obvious meaning given by modern Mathematical Logic), but rather it means not acceptable by a large or important part of the scientific community.
Also nowadays, some of our works in some parts of modern Mathematics are not accepted as sufficiently rigorous by other parts. These works are hence perceived only as not sufficiently precise “ways of arguing”. Therefore, these “foreign argumentations” are perceived as potentially inconsistent, and need a different reformulation to be accepted. I know of relationships of this type between some parts of Geometry and Analysis, to mention only an example. It is the same problem occurring in the relationships between (some parts of) Physics and Mathematics because these two disciplines are really completely different “games”: in Physics the most important achievement is the existence of a dialectic between formulas and a part of nature, even if the related Mathematics lacks in formal clarity and is hence not accepted by several mathematicians.
Analogously, early calculus was consistent until the community accepted these “ways of arguing” and discovered statements which could be verified as true by a dialogue with other part of knowledge: Physics and geometrical intuition in primis.
Since in the early calculus the formal intuition (in the modern sense of manipulation of symbols, without a reference to intuition) was surely weak, the dialectic between proofs and intuition was surely stronger (I mean statistically, in the distribution of 17th century mathematicians). In my opinion, this is the reason of the discovering of true statements, even if the related proofs are perceived as “weak” nowadays. Once the great triumvirate Cantor, Dedekind, and Weierstrass decided that it was time to make a step further, the notion of “inconsistent” changed for this important part of the community and hence, sooner or later, for all the others.
Also from the point of view of rules of inference, the consistency of early calculus has to be meant in the sense of dialectic between different parts of knowledge and acceptance by the related scientific community.
Therefore, in this sense, in my opinion early calculus is as consistent as our (and the future) calculus.
I agree with Joel that “we are not in a qualitatively different situation”: probably in the near future all proofs will be computer assisted, in the sense that all the missing steps will be checked by a computer (whose software will be verified, once again, by a large part of the community) and we will only need to provide the main steps. Necessarily, articles will change in nature and, I hope, they will be more focused on those ideas and intuitions thanks to which we were able to create the results we are presenting. Therefore, young students in the future will probably read disgusted at our papers saying: “how were they able to understand how all these results were created? These papers seems like phone books: def, lem, thm, cor, def, lem, thm, cor... without any explanation of discovery rules and several missing formal steps!”.
Finally, I think that only formally, but not conceptually, this early calculus may look similar to NSA or SDG. In my opinion, one of the main reason of the lack of diffusion of NSA is that its techniques are perceived as “voodoo” by all modern mathematicians (the majority) that rely their work on the dialogue between formal mathematics and informal intuition. Too much frequently the lack of intuition is too strong in both theories. For example, for a person like Cauchy, what is the intuitive meaning of the standard part of the sine of an infinite number (NSA)? For people like Bernoulli, what is the intuitive meaning of properties like $x\le0$ and $x\ge0$ for every infinitesimal and $\neg\neg\exists h$ such that $h$ is infinitesimal (but not necessarily there exists an infinitesimal; SDG)? Moreover, as soon as discontinuous functions appeared in the calculus, the natural reactions of almost every working mathematicians (of 17th century
and nowadays) looking at the microaffinity axiom is not to change Logic switching to the intuitionistic one, but to change this axiom inserting a restriction on the quantifier “for every $f:R\longrightarrow R$”.
The apparently inconsistent argumentation of setting $h\ne0$ and finally $h=0$, can be faithfully formalized using classical calculus rather than using these theories of infinitesimals. We can say that $f:R\longrightarrow R$ (here $R$ is the usual Archimedean real field) is differentiable at $x$ if there exists a function $r:R\times R\longrightarrow R$ such that $f(x+h)=f(x)+h\cdot r(x,h)$ and such that $r$ is continuous at $h=0$. It is easy to prove that this function $r$ is unique. Therefore, we can assume $h\ne0$, we can make freely calculations to discover what is the unique form of the function $r(x,h)$ for $h\ne0$ and, in the final formula, to set $h=0$ because $r$ is clearly continuous for all the examples of functions of the early calculus. This is called the Fermat-Reyes methods, and it can be proved also for generalized functions like Schwartz distributions (and hence for an isomorphic copy of the space of all the continuous functions). Moreover, in my opinion, both Cauchy and Bernoulli would had perfectly understood this method and the related intuition. On the contrary, they would not be able to understand all the intuitive inconsistencies they can easily find both in NSA and SDG.
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Tagged: field Problem 283
Let $F$ be a field and let
\[H(F)=\left\{\, \begin{bmatrix} 1 & a & b \\ 0 &1 &c \\ 0 & 0 & 1 \end{bmatrix} \quad \middle| \quad \text{ for any} a,b,c\in F\, \right\}\] be the Heisenberg group over $F$. (The group operation of the Heisenberg group is matrix multiplication.)
Determine which matrices lie in the center of $H(F)$ and prove that the center $Z\big(H(F)\big)$ is isomorphic to the additive group $F$.Add to solve later
Problem 229
Let $p\in \Z$ be a prime number and let $\F_p$ be the field of $p$ elements.
For any nonzero element $a\in \F_p$, prove that the polynomial \[f(x)=x^p-x+a\] is irreducible and separable over $F_p$.
(Dummit and Foote “Abstract Algebra” Section 13.5 Exercise #5 on p.551)Add to solve later
The Ideal $(x)$ is Prime in the Polynomial Ring $R[x]$ if and only if the Ring $R$ is an Integral Domain Problem 198
Let $R$ be a commutative ring with $1$. Prove that the principal ideal $(x)$ generated by the element $x$ in the polynomial ring $R[x]$ is a prime ideal if and only if $R$ is an integral domain.
Prove also that the ideal $(x)$ is a maximal ideal if and only if $R$ is a field.Add to solve later
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The Root Test for Positive Series
The Root Test for Positive Series
We will now look at a test that is analogous to the ratio test. This test can be useful in certain cases, specifically when there are terms raised to the $n^{\mathrm{th}}$ power.
Theorem 1 (The Root Test for Positive Series): If the sequence of terms $\{ a_n \}$ is ultimately positive and that $\lim_{n \to \infty} (a_n)^{1/n} = L$ where $L$ is some nonnegative real number or $\infty$, then if $0 ≤ L < 1$ then the series $\sum_{n=1}^{\infty} a_n$ converges. If $1 < L ≤ \infty$ then the series $\sum_{n=1}^{\infty} a_n$ diverges to infinity. If $L = 1$ then this test does not provide any additional information about the convergence/divergence of the series $\sum_{n=1}^{\infty} a_n$. Proof of Theorem:Let $L < 1$. There exists a number $r$ such that $L < r < 1$. Now since $\lim_{n \to \infty} (a_n)^{1/n} = L$, then for some $N \in \mathbb{N}$, if $n ≥ N$ then $(a_n)^{1/n} < r$ and so for $n ≥ N$, $a_n ≤ r^n$. Now notice that $\sum_{n=1}^{\infty} r^n$ is a geometric series. Since $0 < L < r < 1$ we conclude that the series $\sum_{n=1}^{\infty} r^n$ converges and so the series $\sum_{n=1}^{\infty} a_n$ must also diverge by the comparison theorem. Now let $L > 1$. Since $L > 1$ there exists some $N \in \mathbb{N}$ such that if $n ≥ N$ then $(a_n)^{1/n} > 1$ which implies that $a_n > 1^n = 1$ if $n ≥ N$. Thus $\lim_{n \to \infty} a_n \neq 0$ and so by The Divergence Theorem for Series it follows that the series $\sum_{n=1}^{\infty} a_n$ is divergent. $\blacksquare$
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Difference between revisions of "De Bruijn-Newman constant"
(→Writeup)
(→Threads)
Line 95: Line 95:
* [https://terrytao.wordpress.com/2018/05/04/polymath15-ninth-thread-going-below-0-22/ Polymath15, ninth thread: going below 0.22?], Terence Tao, May 4, 2018.
* [https://terrytao.wordpress.com/2018/05/04/polymath15-ninth-thread-going-below-0-22/ Polymath15, ninth thread: going below 0.22?], Terence Tao, May 4, 2018.
* [https://terrytao.wordpress.com/10725 Polymath15, tenth thread: numerics update], Rudolph Dwars and Kalpesh Muchhal, Sep 6, 2018.
* [https://terrytao.wordpress.com/10725 Polymath15, tenth thread: numerics update], Rudolph Dwars and Kalpesh Muchhal, Sep 6, 2018.
+
== Other blog posts and online discussion ==
== Other blog posts and online discussion ==
Revision as of 12:14, 28 December 2018
For each real number [math]t[/math], define the entire function [math]H_t: {\mathbf C} \to {\mathbf C}[/math] by the formula
[math]\displaystyle H_t(z) := \int_0^\infty e^{tu^2} \Phi(u) \cos(zu)\ du[/math]
where [math]\Phi[/math] is the super-exponentially decaying function
[math]\displaystyle \Phi(u) := \sum_{n=1}^\infty (2\pi^2 n^4 e^{9u} - 3 \pi n^2 e^{5u}) \exp(-\pi n^2 e^{4u}).[/math]
It is known that [math]\Phi[/math] is even, and that [math]H_t[/math] is even, real on the real axis, and obeys the functional equation [math]H_t(\overline{z}) = \overline{H_t(z)}[/math]. In particular, the zeroes of [math]H_t[/math] are symmetric about both the real and imaginary axes. One can also express [math]H_t[/math] in a number of different forms, such as
[math]\displaystyle H_t(z) = \frac{1}{2} \int_{\bf R} e^{tu^2} \Phi(u) e^{izu}\ du[/math]
or
[math]\displaystyle H_t(z) = \frac{1}{2} \int_0^\infty e^{t\log^2 x} \Phi(\log x) e^{iz \log x}\ \frac{dx}{x}.[/math]
In the notation of [KKL2009], one has
[math]\displaystyle H_t(z) = \frac{1}{8} \Xi_{t/4}(z/2).[/math]
De Bruijn [B1950] and Newman [N1976] showed that there existed a constant, the
de Bruijn-Newman constant [math]\Lambda[/math], such that [math]H_t[/math] has all zeroes real precisely when [math]t \geq \Lambda[/math]. The Riemann hypothesis is equivalent to the claim that [math]\Lambda \leq 0[/math]. Currently it is known that [math]0 \leq \Lambda \lt 1/2[/math] (lower bound in [RT2018], upper bound in [KKL2009]).
The
Polymath15 project seeks to improve the upper bound on [math]\Lambda[/math]. The current strategy is to combine the following three ingredients: Numerical zero-free regions for [math]H_t(x+iy)[/math] of the form [math]\{ x+iy: 0 \leq x \leq T; y \geq \varepsilon \}[/math] for explicit [math]T, \varepsilon, t \gt 0[/math]. Rigorous asymptotics that show that [math]H_t(x+iy)[/math] whenever [math]y \geq \varepsilon[/math] and [math]x \geq T[/math] for a sufficiently large [math]T[/math]. Dynamics of zeroes results that control [math]\Lambda[/math] in terms of the maximum imaginary part of a zero of [math]H_t[/math]. Contents [math]t=0[/math]
When [math]t=0[/math], one has
[math]\displaystyle H_0(z) = \frac{1}{8} \xi( \frac{1}{2} + \frac{iz}{2} ) [/math]
where
[math]\displaystyle \xi(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \zeta(s)[/math]
is the Riemann xi function. In particular, [math]z[/math] is a zero of [math]H_0[/math] if and only if [math]\frac{1}{2} + \frac{iz}{2}[/math] is a non-trivial zero of the Riemann zeta function. Thus, for instance, the Riemann hypothesis is equivalent to all the zeroes of [math]H_0[/math] being real, and Riemann-von Mangoldt formula (in the explicit form given by Backlund) gives
[math]\displaystyle \left|N_0(T) - (\frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} - \frac{7}{8})\right| \lt 0.137 \log (T/2) + 0.443 \log\log(T/2) + 4.350 [/math]
for any [math]T \gt 4[/math], where [math]N_0(T)[/math] denotes the number of zeroes of [math]H_0[/math] with real part between 0 and T.
The first [math]10^{13}[/math] zeroes of [math]H_0[/math] (to the right of the origin) are real [G2004]. This numerical computation uses the Odlyzko-Schonhage algorithm. In [P2017] it was independently verified that all zeroes of [math]H_0[/math] between 0 and 61,220,092,000 were real.
[math]t\gt0[/math]
For any [math]t\gt0[/math], it is known that all but finitely many of the zeroes of [math]H_t[/math] are real and simple [KKL2009, Theorem 1.3]. In fact, assuming the Riemann hypothesis,
all of the zeroes of [math]H_t[/math] are real and simple [CSV1994, Corollary 2].
It is known that [math]\xi[/math] is an entire function of order one ([T1986, Theorem 2.12]). Hence by the fundamental solution for the heat equation, the [math]H_t[/math] are also entire functions of order one for any [math]t[/math].
Because [math]\Phi[/math] is positive, [math]H_t(iy)[/math] is positive for any [math]y[/math], and hence there are no zeroes on the imaginary axis.
Let [math]\sigma_{max}(t)[/math] denote the largest imaginary part of a zero of [math]H_t[/math], thus [math]\sigma_{max}(t)=0[/math] if and only if [math]t \geq \Lambda[/math]. It is known that the quantity [math]\frac{1}{2} \sigma_{max}(t)^2 + t[/math] is non-increasing in time whenever [math]\sigma_{max}(t)\gt0[/math] (see [KKL2009, Proposition A]. In particular we have
[math]\displaystyle \Lambda \leq t + \frac{1}{2} \sigma_{max}(t)^2[/math]
for any [math]t[/math].
The zeroes [math]z_j(t)[/math] of [math]H_t[/math] obey the system of ODE
[math]\partial_t z_j(t) = - \sum_{k \neq j} \frac{2}{z_k(t) - z_j(t)}[/math]
where the sum is interpreted in a principal value sense, and excluding those times in which [math]z_j(t)[/math] is a repeated zero. See dynamics of zeros for more details. Writing [math]z_j(t) = x_j(t) + i y_j(t)[/math], we can write the dynamics as
[math] \partial_t x_j = - \sum_{k \neq j} \frac{2 (x_k - x_j)}{(x_k-x_j)^2 + (y_k-y_j)^2} [/math] [math] \partial_t y_j = \sum_{k \neq j} \frac{2 (y_k - y_j)}{(x_k-x_j)^2 + (y_k-y_j)^2} [/math]
where the dependence on [math]t[/math] has been omitted for brevity.
In [KKL2009, Theorem 1.4], it is shown that for any fixed [math]t\gt0[/math], the number [math]N_t(T)[/math] of zeroes of [math]H_t[/math] with real part between 0 and T obeys the asymptotic
[math]N_t(T) = \frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} + \frac{t}{16} \log T + O(1) [/math]
as [math]T \to \infty[/math] (caution: the error term here is not uniform in t). Also, the zeroes behave like an arithmetic progression in the sense that
[math] z_{k+1}(t) - z_k(t) = (1+o(1)) \frac{4\pi}{\log |z_k|(t)} = (1+o(1)) \frac{4\pi}{\log k} [/math]
as [math]k \to +\infty[/math].
Threads Polymath proposal: upper bounding the de Bruijn-Newman constant, Terence Tao, Jan 24, 2018. Polymath15, first thread: computing H_t, asymptotics, and dynamics of zeroes, Terence Tao, Jan 27, 2018. Polymath15, second thread: generalising the Riemann-Siegel approximate functional equation, Terence Tao and Sujit Nair, Feb 2, 2018. Polymath15, third thread: computing and approximating H_t, Terence Tao and Sujit Nair, Feb 12, 2018. Polymath 15, fourth thread: closing in on the test problem, Terence Tao, Feb 24, 2018. Polymath15, fifth thread: finishing off the test problem?, Terence Tao, Mar 2, 2018. Polymath15, sixth thread: the test problem and beyond, Terence Tao, Mar 18, 2018. Polymath15, seventh thread: going below 0.48, Terence Tao, Mar 28, 2018. Polymath15, eighth thread: going below 0.28, Terence Tao, Apr 17, 2018. Polymath15, ninth thread: going below 0.22?, Terence Tao, May 4, 2018. Polymath15, tenth thread: numerics update, Rudolph Dwars and Kalpesh Muchhal, Sep 6, 2018. Polymath 15, eleventh thread: Writing up the results, and exploring negative t, Terence Tao, Dec 28, 2018. Other blog posts and online discussion Heat flow and zeroes of polynomials, Terence Tao, Oct 17, 2017. The de Bruijn-Newman constant is non-negative, Terence Tao, Jan 19, 2018. Lehmer pairs and GUE, Terence Tao, Jan 20, 2018. A new polymath proposal (related to the Riemann hypothesis) over Tao's blog, Gil Kalai, Jan 26, 2018. Code and data Writeup
Here are the Polymath15 grant acknowledgments.
Test problem Zero-free regions
See Zero-free regions.
Wikipedia and other references Bibliography [A2011] J. Arias de Reyna, High-precision computation of Riemann's zeta function by the Riemann-Siegel asymptotic formula, I, Mathematics of Computation, Volume 80, Number 274, April 2011, Pages 995–1009. [B1994] W. G. C. Boyd, Gamma Function Asymptotics by an Extension of the Method of Steepest Descents, Proceedings: Mathematical and Physical Sciences, Vol. 447, No. 1931 (Dec. 8, 1994),pp. 609-630. [B1950] N. C. de Bruijn, The roots of trigonometric integrals, Duke J. Math. 17 (1950), 197–226. [CSV1994] G. Csordas, W. Smith, R. S. Varga, Lehmer pairs of zeros, the de Bruijn-Newman constant Λ, and the Riemann hypothesis, Constr. Approx. 10 (1994), no. 1, 107–129. [G2004] Gourdon, Xavier (2004), The [math]10^{13}[/math] first zeros of the Riemann Zeta function, and zeros computation at very large height [KKL2009] H. Ki, Y. O. Kim, and J. Lee, On the de Bruijn-Newman constant, Advances in Mathematics, 22 (2009), 281–306. Citeseer [N1976] C. M. Newman, Fourier transforms with only real zeroes, Proc. Amer. Math. Soc. 61 (1976), 246–251. [P2017] D. J. Platt, Isolating some non-trivial zeros of zeta, Math. Comp. 86 (2017), 2449-2467. [P1992] G. Pugh, The Riemann-Siegel formula and large scale computations of the Riemann zeta function, M.Sc. Thesis, U. British Columbia, 1992. [RT2018] B. Rodgers, T. Tao, The de Bruijn-Newman constant is non-negative, preprint. arXiv:1801.05914 [T1986] E. C. Titchmarsh, The theory of the Riemann zeta-function. Second edition. Edited and with a preface by D. R. Heath-Brown. The Clarendon Press, Oxford University Press, New York, 1986. pdf
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Why do people use Quadratic Programming techniques (such as SMO) when dealing with kernelized SVMs? What is wrong with Gradient Descent? Is it impossible to use with kernels or is it just too slow (and why?).
Here is a little more context: trying to understand SVMs a bit better, I used Gradient Descent to train a linear SVM classifier using the following cost function:
$J(\mathbf{w}, b) = C {\displaystyle \sum\limits_{i=1}^{m} max\left(0, 1 - y^{(i)} (\mathbf{w}^t \cdot \mathbf{x}^{(i)} + b)\right)} \quad + \quad \dfrac{1}{2} \mathbf{w}^t \cdot \mathbf{w}$
I am using the following notations:
$\mathbf{w}$ is the model's feature weights and $b$ is its bias parameter. $\mathbf{x}^{(i)}$ is the $i^\text{th}$ training instance's feature vector. $y^{(i)}$ is the target class (-1 or 1) for the $i^\text{th}$ instance. $m$ is the number of training instances. $C$ is the regularization hyperparameter.
I derived a (sub)gradient vector (with regards to $\mathbf{w}$ and $b$) from this equation, and Gradient Descent worked just fine.
Now I would like to tackle non-linear problems. Can I just replace all dot products $\mathbf{u}^t \cdot \mathbf{v}$ with $K(\mathbf{u}, \mathbf{v})$ in the cost function, where $K$ is the kernel function (for example the Gaussian RBF, $K(\mathbf{u}, \mathbf{v}) = e^{-\gamma \|\mathbf{u} - \mathbf{v}\|^2}$), then use calculus to derive a (sub)gradient vector and go ahead with Gradient Descent?
If it is too slow, why is that? Is the cost function not convex? Or is it because the gradient changes too fast (it is not Lipschitz continuous) so the algorithm keeps jumping across valleys during the descent, so it converges very slowly? But even then, how can it be worse than Quadratic Programming's time complexity, which is $O({n_\text{samples}}^2 \times n_\text{features})$? If it's a matter of local minima, can't Stochastic GD with simulated annealing overcome them?
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Invertibility of a Linear Map Examples 1
Let $V$ and $W$ be vector spaces. Recall from the Invertibility of a Linear Map page that a linear map $T \in \mathcal L (V, W)$ is said to be invertible if there exists a linear map $S \in \mathcal L (W, V)$ such that $ST = I_V$ and $TS = I_W$ where $I_V$ is the identity map on $V$ and $I_W$ is the identity map on $W$. Furthermore, the linear map $S = T^{-1}$ is said to be the inverse of $T$.
We will now look at some examples regarding the invertibility of a linear map.
Example 1 Reprove that if $T \in \mathcal L (V, W)$ is invertible then $T$ has a unique inverse $S \in \mathcal L (W, V)$.
Suppose that $S, S' \in \mathcal L (W, V)$ are both inverses to $T$ - that is $ST = I_V$ and $TS = I_W$ as well as $S'T = I_V$ and $TS' = I_W$. Then we have that:(1)
Therefore $S = S'$ so the inverse $S$ is unique.
Example 2 Reprove that if $T \in \mathcal L(V, W)$ then $T$ is invertible if and only if $T$ is bijective.
$\Rightarrow$ Suppose that $T$ is invertible. Then $T^{-1} \in \mathcal L (W, V)$ exists such that $T^{-1} T = I_V$ and $TT^{-1} = I_W$. We need to show that then $T$ is both injective and surjective.
Suppose that $T(u) = T(v)$. Then we have that:(2)
Therefore $u = v$ so $T$ is injective.
To show that $T$ is surjective, note that for every vector $w \in W$ we have that $w = TT^{-1}(w) = T(T^{-1}(w))$. Therefore, for every vector $w \in W$ there exists a vector $T^{-1}(w) \in V$ such that $w = T(T^{-1}(w))$, so $T$ is surjective.
$\Leftarrow$ Now suppose that $T$ is bijective. Then $T$ is both injective and surjective. We will show that $T$ is invertible by constructing a linear map $S$.
Define $S : W \to V$ to be $v = S(w)$ where $v$ is the unique vector $v \in V$ that is mapped to $w \in W$. We note that for every vector $w \in W$ there exists a vector $v \in V$ such that $w = T(v)$ since $T$ is surjective, and the vector $v$ is unique for every vector $w$ since $T$ is surjective.
Then clearly:(3)
Therefore $(TS)(w) = w$ implies that $TS = I_W$. Furthermore we note that:(4)
The equation above says that $T(S(T(v))) = T(v)$, and since $T$ is injective this implies that $S(T(v)) = v$ so $(ST)(v) = v$ and therefore $ST = I_V$.
We will not reprove that $S$ is a linear map since that follows easily from showing that the additivity and homogeneity properties of linear maps hold.
Example 3 Prove that the linear map $T \in \mathcal L (\wp (\mathbb{R}), \wp (\mathbb{R}))$ defined by $T(p(x)) = x^2 p(x)$ for every $p(x) \in \wp (\mathbb{R})$ is not invertible.
Consider the polynomial $1 \in \wp (\mathbb{R})$. Note that there exists no polynomial $p(x) \in \wp (\mathbb{R})$ such that $T(p(x)) = x^2 p(x) = 1$. Therefore $T$ is not surjective which implies that $T$ is not invertible.
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Square Lebesgue Integrable Functions
We will now discuss another class of functions known as square Lebesgue integrable functions which we define below.
Definition: A function $f$ is said to be Square Lebesgue Integrable on an interval $I$ if $f$ is a measurable function on $I$ and $f^2$ is Lebesgue integrable on $I$. The set of all square Lebesgue integrable functions on $I$ is denoted $L^2(I)$.
Note that in the definition above that $f$ itself is not required to be Lebesgue integrable - only its square is!
For example, let $f$ be any continuous function on a closed and bounded interval $I$. Then $f$ is Riemann integrable on $I$, so $f$ is Lebesgue integrable on $I$ and furthermore, $f$ is measurable on $I$.
The product of two continuous functions is still continuous, so $f^2$ is also Riemann integrable on $I$ and hence $f^2$ is Lebesgue integrable on $I$.
Since $f$ is measurable on $I$ and $f^2$ is Lebesgue integrable on $I$ we see that any continuous function on a closed and bounded interval $I$ is square Lebesgue integrable.
Let's look at an example of a function that is not square Lebesgue integrable. Consider the function $\displaystyle{f(x) = \frac{1}{\sqrt{x}}}$ on the interval $(0, 1]$. Then $f$ is measurable on $I$ since the following sequence of step functions converges to $f$ on $(0, 1]$:(1)
It should not be too hard to see that $(f_n(x))_{n=1}^{\infty}$ converges to $f$. The graphs of $f_2$, $f_3$, and $f_4$ are given below:
However, the function $\displaystyle{f^2(x) = \frac{1}{x}}$ is not Lebesgue integrable on $(0, 1]$. In fact, the Lebesgue integral of $f^2$ on $(0, 1]$ is the Riemann integral of $f^2$ on $(0, 1]$ since $f^2$ is continuous and nonnegative, so:(2)
So $f^2$ is not Lebesgue integrable on $(0, 1]$ which shows that $f$ is not a square Lebesgue integrable function on $(0, 1]$.
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Tagged: linearly dependent Problem 603
Let $C[-2\pi, 2\pi]$ be the vector space of all continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the functions \[f(x)=\sin^2(x) \text{ and } g(x)=\cos^2(x)\] in $C[-2\pi, 2\pi]$.
Prove or disprove that the functions $f(x)$ and $g(x)$ are linearly independent.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 572
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold). The time limit was 55 minutes. Problem 7. Let $A=\begin{bmatrix} -3 & -4\\ 8& 9 \end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix} -1 \\ 2 \end{bmatrix}$. (a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$. (b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$. Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular. Problem 9. Determine whether each of the following sentences is true or false. (a) There is a $3\times 3$ homogeneous system that has exactly three solutions. (b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric. (c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$. (d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
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(e) The vectors \[\mathbf{v}_1=\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \mathbf{v}_3=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\] are linearly independent. Problem 563
Let
\[\mathbf{v}_1=\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} 1 \\ a \\ 5 \end{bmatrix}, \mathbf{v}_3=\begin{bmatrix} 0 \\ 4 \\ b \end{bmatrix}\] be vectors in $\R^3$.
Determine a condition on the scalars $a, b$ so that the set of vectors $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly dependent.Add to solve later
The Matrix $[A_1, \dots, A_{n-1}, A\mathbf{b}]$ is Always Singular, Where $A=[A_1,\dots, A_{n-1}]$ and $\mathbf{b}\in \R^{n-1}$. Problem 560
Let $A$ be an $n\times (n-1)$ matrix and let $\mathbf{b}$ be an $(n-1)$-dimensional vector.
Then the product $A\mathbf{b}$ is an $n$-dimensional vector. Set the $n\times n$ matrix $B=[A_1, A_2, \dots, A_{n-1}, A\mathbf{b}]$, where $A_i$ is the $i$-th column vector of $A$.
Prove that $B$ is a singular matrix for any choice of $\mathbf{b}$.Add to solve later
Problem 415 (a) Let $R$ be a commutative ring. If we regard $R$ as a left $R$-module, then prove that any two distinct elements of the module $R$ are linearly dependent.
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(b) Let $f: M\to M’$ be a left $R$-module homomorphism. Let $\{x_1, \dots, x_n\}$ be a subset in $M$. Prove that if the set $\{f(x_1), \dots, f(x_n)\}$ is linearly independent, then the set $\{x_1, \dots, x_n\}$ is also linearly independent. Read solution Problem 375
Let $n>1$ be a positive integer. Let $V=M_{n\times n}(\C)$ be the vector space over the complex numbers $\C$ consisting of all complex $n\times n$ matrices. The dimension of $V$ is $n^2$.
Let $A \in V$ and consider the set \[S_A=\{I=A^0, A, A^2, \dots, A^{n^2-1}\}\] of $n^2$ elements. Prove that the set $S_A$ cannot be a basis of the vector space $V$ for any $A\in V$. Linearly Dependent if and only if a Vector Can be Written as a Linear Combination of Remaining Vectors Problem 347
Let $V$ be a vector space over a scalar field $K$.
Let $S=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n\}$ be the set of vectors in $V$, where $n \geq 2$.
Then prove that the set $S$ is linearly dependent if and only if at least one of the vectors in $S$ can be written as a linear combination of remaining vectors in $S$.Add to solve later
Linearly Independent vectors $\mathbf{v}_1, \mathbf{v}_2$ and Linearly Independent Vectors $A\mathbf{v}_1, A\mathbf{v}_2$ for a Nonsingular Matrix Problem 284
Let $\mathbf{v}_1$ and $\mathbf{v}_2$ be $2$-dimensional vectors and let $A$ be a $2\times 2$ matrix.
(a) Show that if $\mathbf{v}_1, \mathbf{v}_2$ are linearly dependent vectors, then the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are also linearly dependent. (b) If $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent vectors, can we conclude that the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are also linearly independent?
Add to solve later
(c) If $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent vectors and $A$ is nonsingular, then show that the vectors $A\mathbf{v}_1, A\mathbf{v}_2$ are also linearly independent. Problem 281 (a) For what value(s) of $a$ is the following set $S$ linearly dependent? \[ S=\left \{\,\begin{bmatrix} 1 \\ 2 \\ 3 \\ a \end{bmatrix}, \begin{bmatrix} a \\ 0 \\ -1 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ a^2 \\ 7 \end{bmatrix}, \begin{bmatrix} 1 \\ a \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ -2 \\ 3 \\ a^3 \end{bmatrix} \, \right\}.\]
Add to solve later
(b) Let $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be a set of nonzero vectors in $\R^m$ such that the dot product \[\mathbf{v}_i\cdot \mathbf{v}_j=0\] when $i\neq j$. Prove that the set is linearly independent. Problem 277
Determine whether the following set of vectors is linearly independent or linearly dependent. If the set is linearly dependent, express one vector in the set as a linear combination of the others.
\[\left\{\, \begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}, \begin{bmatrix} -1 \\ -2 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ -2 \\ 7 \\ 11 \end{bmatrix}\, \right\}.\] Problem 48
Let $V$ be an $n$-dimensional vector space over a field $K$.
Suppose that $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ are linearly independent vectors in $V$.
Are the following vectors linearly independent?
\[\mathbf{v}_1+\mathbf{v}_2, \quad \mathbf{v}_2+\mathbf{v}_3, \quad \dots, \quad \mathbf{v}_{k-1}+\mathbf{v}_k, \quad \mathbf{v}_k+\mathbf{v}_1.\]
If it is linearly dependent, give a non-trivial linear combination of these vectors summing up to the zero vector.Add to solve later
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I believe you are confusing the wing angle of attack with the pitch of the aircraft. Aircraft moving at a slow, near-stall speed, despite pointing the nose up, will still be traveling more or less horizontally. Their VSI instrument will read near zero. Whereas, if you take an aircraft moving quickly and pull the nose up to the same angle, the aircraft will, obviously, climb rapidly.
Why does this matter? The angle of attack is defined based on the wing's motion through the relative wind. The wing's orientation relative to the ground isn't involved in the definition in any way. When the aircraft as a whole is climbing, the relative wind is coming down from above. As a result the angle of attack is reduced, compared to what it would be if the plane were not climbing.
Just to show some quick numbers, suppose you took an aircraft moving at 100 kts in still air and pulled the nose up so that you are now climbing at 3,000 FPM (most aircraft will lose speed doing this, but the math is valid until the airplane slows down). $1knot\approx100FPM$, so you'll now have an upward vector of 30 knots. Your 100 kt airspeed is now moving up at an angle. A little trigonometry:
$$\sin(x)=\frac{30}{100}$$$$x=17.46°$$
So, your angle of attack is 17.46 degrees farther away from stalling when climbing at 3000FPM than it would be if your aircraft had the same pitch but was in level flight.
However, few aircraft have the engine power to sustain a climb at this rate. The aircraft will bleed off speed, and as the speed bleeds off, the aircraft will slow, the climb rate will decrease, the aircraft's velocity will become closer to horizontal, and, eventually, the aircraft will stall if the pitch is held constant.
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show that this integral:
$$\dfrac{1}{2\pi h}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{\dfrac{-i(p-p')x}{h}}x^n\varphi{(p')}dxdp'=\left(ih\dfrac{\partial }{\partial p}\right)^n\varphi{(p)}$$
where $i^2=-1$
maybe this use integration by parts？ But I fell very hard,and I can't prove it.
I think first we must this $$\int_{-\infty}^{\infty}e^{\dfrac{-i(p-p')x}{h}}x^ndx=?$$ and follow it can use
integration by parts?
But I consider sometimes can't have this resulut
Thank you
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Firstly, I wasn't sure exactly where to put this. It's a typesetting query but the scope is greater than $\TeX$; however it's specific also to physics and even more specific to this site.
I've recently been reading a style guide for scientific publications (based on ISO 31-11), however there was no mention of quantum mechanical operators. I've seen them written a few ways and was wondering if there was a decision handed down from "up above" that any particular way is best.
$H$ -- I see this most commonly but I suspect it's mostly due to (mild) laziness to not distinguish it from a variable. $\hat{H}$ -- This is nicer to me because it makes the distinction between operator and variable. From what I understand of the ISO the italic means it's subject to change, which is true of the form of an operator, but not really its meaning? So I'm not totally sure if that's appropriate here. $\mathrm{H}$ -- Roman lettering is used for functions e.g. $\sin{x}$, $\mathrm{erf}(x)$, and even the differential operator (as in $\frac{\mathrm{d}}{\mathrm{d}x}$) so this seems to me like the most suitable category to put operators in. $\hat{\mathrm{H}}$ -- Probably the least ambiguous but may also be redundant.
Which would be the best to use? Am I being too pedantic?
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Matrix Multiplication
Definition: Given matrix $A$ of size $m \times r$ and matrix $B$ of size $r \times n$, their product denoted $AB$ is the $m \times n$ matrix whose $ij^{th}$ entries result from taking row $i$ in matrix $A$ and multiplying corresponding entries of column $j$ in matrix $B$ and summing their products.
For example, consider the following matrices:(1)
We first note that matrix $A$ is of size $2 \times 2$, and matrix $B$ is of size $2 \times 3$, hence the product $AB$ is defined and will be of size $2 \times 3$. To find the first entry of our matrix $AB$, we will take row $1$ of $A$ and multiply corresponding entries of column $1$ of $B$ (and then sum then) to get $2\cdot3 + 1 \cdot 6 = 12$. Therefore, the first entry of our product matrix is $12$, and the rest of the entries are as follows:(2)
In general given $A_{m \times r}$ and $B_{r \times n}$:(3)
Furthermore, $ij^{th}$ entry in the product $AB$ can be obtained by the following formula:(4)
The Cancellation Law for Multiplication Does NOT Hold
Suppose we have three matrices $A$, $B$, and $C$ and assume the products $AB$ and $AC$ are defined and $AB = AC$. We may be quick to assume that $B = C$, however this is not true for all cases (though in certain cases it is true). For example, consider the following $2 \times 2$ matrices:(5)
Note that $AB = AC = \begin{bmatrix} 3 & 4\\ 6 & 8 \end{bmatrix}$. However clearly $B ≠ C$.
Example 1 Show that if $A$ is a matrix of size $m \times n$ then $AA^T$ is always defined and will be a square matrix. Proof:We first note that $A^T$ will be an $n \times m$ sized matrix. The matrix product $A_{m \times n}A^T_{n \times m}$ will always be defined since $A$ has n columns and $A^T$ has n rows. Furthermore, the product will be of size $m \times m$ and will hence be square $\blacksquare$.
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Because the need for color manipulation comes up fairly often in computer graphics, particularly transformations of hue, saturation, and value, and because some of this math is a bit tricky, here’s how to do HSV color transforms on RGB data using simple matrix operations.
Note: This isn’t about converting between RGB and HSV; this is only about applying an HSV-space modification to an RGB value and getting another RGB value out. There is no affine transformation to convert between RGB and HSV, as there is not a linear mapping between the two.
Preliminary
All of the operations here require multiplication of matrices and vectors. As a quick primer/refresher, multiplying a matrix with a vector looks like this:
\[ \begin{bmatrix} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{bmatrix} \begin{bmatrix} V_x \\ V_y \\ V_z \end{bmatrix} = \begin{bmatrix} A_xV_x + A_yV_y + A_zC_z \\ B_xV_x + B_yV_y + B_zC_z \\ C_xV_x + C_yV_y + C_zC_z \\ \end{bmatrix} \]
Matrix multiplication, like numerical multiplication, is associative, but unlike numerical mutiplication, is not commutative. This means that if you have multiple matrix operations to perform to a single vector, like \(v'=ABCDv\), you can combine them together into a single master matrix where \(Z=ABCD\) and then \(v'=Zv\). This will be useful later on.
A note about response curves
The math involved assumes that we are dealing with a linear color space. However, most displays use an exponential curve. For this to behave completely accurately, you’ll have to convert your colors to linear color before, and to exponential color afterwards. A pretty close approximation (assuming a display gamma of 2.2) is as follows, where \(L\) is the linear-space value, \(G\) is the gamma-space value, and \(M\) is the maximum value; for traditional 8-bit/channel images this is 255.
\[ L = M \left( \frac{G}{M} \right)^{2.2} \\ G = M \left( \frac{L}{M} \right)^{0.455} \]
Note that if you are going to do this on a per-pixel basis in real time you will almost certainly want to precompute this as a lookup table.
For the remainder of the math, we don’t care about the scale of the values, as long as they start at 0.
Step 1: Convert RGB → YIQ
RGB values aren’t very convenient for doing complex transforms on, especially hue. The math for doing a hue rotation on RGB is nasty. However, the math for doing a hue rotation on YIQ is very easy; YIQ is a color space which uses the perceptive-weighted brightness of the red, green and blue channels to provide a luminance (Y) channel, and places the chroma values for red, green and blue roughly 120 degrees apart in the I-Q plane.
Note that there are many color spaces that you can use for this transform which have different hue-mapping characteristics; strictly-speaking, there is no single natural “angle” between any given colors, and different effects can be achieved by using different color spaces such as YPbPr or YUV. (An earlier version of this page used YPbPr for simplicity, but that led to much confusion when people expected the 180-degree rotation of red to be cyan, which is the case in YIQ but not in YPbPr.)
The transformation from RGB to YIQ is best expressed as a matrix, with the RGB value multiplied as a \(1\times3\) vector on the right:
\[ T_{YIQ} = \begin{bmatrix} 0.299 & 0.587 & 0.114 \\ 0.596 & -0.274 & -0.321 \\ 0.211 & -0.523 & 0.311 \end{bmatrix} \]
One convenient property of the YIQ color space is that it is unit-independent, so we don’t have to care about what our range of colors is, as long as it starts at 0.
Step 2: Hue
Now that our color is in YIQ format, doing a hue transform is pretty simple – we’re just rotating the color around the Y axis. The math for this is as follows (where \(H\) is the hue transform amount, in degrees):
\[ \theta = \frac{H\pi}{180} \\ U = \cos(\theta) \\ W = \sin(\theta) \\ T_h = \begin{bmatrix} 1 & 0 & 0 \\ 0 & U & -W \\ 0 & W & U \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\theta) & -\sin(\theta) \\ 0 & \sin(\theta) & \cos(\theta) \end{bmatrix} \]
Step 3: Saturation
Saturation is just the distance between the color and the gray (Y) axis; you just scale the I and Q channels. So its matrix is:
\[ T_s = \begin{bmatrix} 1 & 0 & 0 \\ 0 & S & 0 \\ 0 & 0 & S \end{bmatrix} \]
Step 4: Value
Finally, the value transformation is a simple scaling of the color as a whole:
\[ T_v = \mathrm{I}V = \begin{bmatrix} V & 0 & 0 \\ 0 & V & 0 \\ 0 & 0 & V \end{bmatrix} \]
Step 5: Convert back to RGB
To convert YIQ back to RGB, it’s also pretty simple; we just use the inverse of the first matrix: \[ T_{RGB} = \begin{bmatrix} 1 & 0.956 & 0.621 \\ 1 & -0.272 & -0.647 \\ 1 & -1.107 & 1.705 \end{bmatrix} \]
Pulling it all together
The final composed transform \(T_{HSV}\) is \(T_{RGB}T_hT_sT_vT_{YIQ}\) (we compose the matrices from right-to-left since the original color is on the right), which you get by multiplying all the above matrices together. Because matrix multiplication is associative, it’s easiest to work out \(T_hT_sT_v\) first, which is
\[ \begin{bmatrix} V & 0 & 0 \\ 0 & VSU & -VSW \\ 0 & VSW & VSU \end{bmatrix} \]
where \(U\) and \(W\) are the same as in \(T_h\), above.
From here we can work out the master transform:
\[ T_{HSV} = \begin{bmatrix} 1 & 0.956 & 0.621 \\ 1 & -0.272 & -0.647 \\ 1 & -1.107 & 1.705 \end{bmatrix} \begin{bmatrix} V & 0 & 0 \\ 0 & VSU & -VSW \\ 0 & VSW & VSU \end{bmatrix} \begin{bmatrix} 0.299 & 0.587 & 0.114 \\ 0.596 & -0.274 & -0.321 \\ 0.211 & -0.523 & 0.311 \end{bmatrix} \\ = \begin{bmatrix} .299V+.701VSU+.168VSW & .587V-.587VSU+.330VSW & .114V-.114VSU-.497VSW \\ .299V-.299VSU-.328VSW & .587V+.413VSU+.035VSW & .114V-.114VSU+.292VSW \\ .299V-.3VSU+1.25VSW & .587V-.588VSU-1.05VSW & .114V+.886VSU-.203VSW \end{bmatrix} \]
As a quick smoke test, we test the matrix with \(V=S=1\) and \(H=0\) (meaning \(U=1\) and \(W=0\)) and as a result we get something very close to the identity matrix (with a little divergence due to roundoff error).
And what’s the code for this?
Here’s some simple C++ code to do an HSV transformation on a single
Color (where
Color is a
struct containing three members,
r,
g, and
b, in whatever data format you want):
Color TransformHSV( const Color &in, // color to transform float h, // hue shift (in degrees) float s, // saturation multiplier (scalar) float v // value multiplier (scalar)) { float vsu = v*s*cos(h*M_PI/180); float vsw = v*s*sin(h*M_PI/180); Color ret; ret.r = (.299*v + .701*vsu + .168*vsw)*in.r + (.587*v - .587*vsu + .330*vsw)*in.g + (.114*v - .114*vsu - .497*vsw)*in.b; ret.g = (.299*v - .299*vsu - .328*vsw)*in.r + (.587*v + .413*vsu + .035*vsw)*in.g + (.114*v - .114*vsu + .292*vsw)*in.b; ret.b = (.299*v - .300*vsu + 1.25*vsw)*in.r + (.587*v - .588*vsu - 1.05*vsw)*in.g + (.114*v + .886*vsu - .203*vsw)*in.b; return ret;}
What if I want to use someone else’s color transform matrix?
Let’s say you want to perfectly replicate the affine color transformation done by some other library or image-manipulation package (say, Flash, for example) but don’t know the exact colorspace they use for their intermediate transformation. Well, for any affine transformation (i.e. not involving gamma correction or whatnot), this is actually pretty simple. First, just run the unit colors of red, green, and blue through that color filter, and then those become the rows of your transformation matrix.
Code for that would look something like this:
Color TransformByExample( const Color &in, // color to transform const Color &r, // pre-transformed red const Color &g, // pre-transformed green const Color &b, // pre-transformed blue float m // Maximum value for a channel) { Color ret; ret.r = (in.r*r.r + in.g*r.g + in.b*r.b)/m; ret.g = (in.r*g.r + in.g*g.g + in.b*g.b)/m; ret.b = (in.r*b.r + in.g*b.g + in.b*b.b)/m; return ret;}
Note that this only works for simple affine transforms, and the usual precautions about gamma vs. linear apply; a much more general approach is to use a CLUT, which can encode arbitrarily many color manipulations into a single color mapping image.
A note about gamut and response
The YIQ color space as used here (as well as in NTSC) attempts to keep the perceptive brightness the same, regardless of channel. This leads to some fairly major problems with the response range; for example, since pure green is perceived as about twice as bright as pure red, rotating pure green to red would produce a super-red – and meanwhile, the value on some channels can also be pulled down past 0. This is not a flaw in the algorithm, so much as a fundamental flaw in how color works to begin with, and a disconnect between the intuitive notion of how “hue” works vs. the actual physics involved. (In a sense, a hue “rotation” is a pretty ridiculous thing to even try to do to begin with; while red and blue shift are very real phenomena, there is no actual fundamental property of light or color which places red, green and blue at 120-degree intervals apart from each other.)
The easy solution to this issue is to always clamp the color values between 0 and 1. A more correct solution is to actually keep track of where the colors are beyond the range and spread that excess energy (or lack thereof) across the image (i.e. to neighboring pixels), such as in HDR rendering. As more display devices move to floating-point-based color representation, this will become easier to deal with. However, at present (May 2018), no commonly-supported image formats support anything other than a clamped integer color space (although there are a few emerging standards such as Radiance HDR and OpenEXR which are finally gaining traction).
In many cases, if you’re making heavy use of hue transformation (in e.g. a game), you are better off splitting your assets into different layers which you can tint appropriately and composite them together for display.
A reminder about gamma
As mentioned earlier, these equations only apply to linear color, whereas most image formats and displays assume an exponential color space. While an HSV transform will generally look
accepetable on a gamma colorspace, it tends to bias things weirdly and won’t look quite right. So, you probably want to convert your gamma RGB values to linear before you apply the HSV transform:
float LinearToGamma(float value, float gamma, float max) { return max*pow(value/max, 1/gamma);}float GammaToLinear(float value, float gamma, float max) { return max*pow(value/max, gamma);}
In this case, both the linear and gamma colors will be stored in the same range \( \left(0,max\right) \). In real-world real-time implementations you will probably do this transformation using either a pair of lookup tables or a polynomial approximation so you don’t have to do multiple
pow() calls per pixel. (Or, better yet, convert your assets to be linear in the first place.)
What about Photoshop’s HSL adjustments?
Photoshop does not use a linear transform to do its HSL adjustments. Instead, it does an ad-hoc “intuitive” approach where it uses the lowest channel value to determine the saturation, and uses the ratio between the remaining two channels to determine the “angle” on the traditional 6-spoke print color wheel (red/yellow/green/cyan/blue/magenta). This provides perfect
numeric values based on peoples' expectations of hues vs. RGB channels, but it actually makes for some pretty terrible chroma adjustment which tends to only work well on solid colors. Even smooth gradients between two colors are completely fouled up by this approach.
This can be seen clearly in this side-by-side comparison; notice the severely obvious banding on the Photoshop version:
In my opinion, Photoshop’s HSL transform mechanism isn’t really something that should be emulated since its utility is limited to begin with.
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The Radius of Curvature at a Point on a Curve
Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function that traces out the smooth curve $C$, and let $P$ be the point on $C$ at $t$. The Radius of Curvature at $P$ is $\rho (t) = \frac{1}{\kappa (t)}$ provided that $\kappa (t) \neq 0$. For functions $y = f(x)$ with curvature $\kappa (x)$, the radius of curvature is $\rho (x) = \frac{1}{\kappa (x)}$. Notice that in both of these cases that $\kappa \neq 0$. We saw from The Curvature of Straight Lines and Circles page that $\kappa = 0$ only when the curve $C$ is a straight line.
Simply put, the radius of curvature is just the reciprocal of the curvature at a given point.
Example 1 Let $\vec{r}(t) = (t, t^2, t^3)$. Find the radius of curvature at $t = 0$.
We must first find the curvature of $\vec{r}(t)$. To do so, we will use the formula $\kappa (t) = \frac{ \| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3}$.
We first differentiate $\vec{r}(t)$ to get $\vec{r'}(t) = (1, 2t, 3t^2)$, and once again to get $\vec{r''}(t) = (0, 2, 6t)$. Therefore $\vec{r'}(t) \times \vec{r''}(t)$ is given by the following formula:(1)
Thus $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{36t^4 + 36t^2 + 4}$. We also have that $\| \vec{r'}(t) \|^3 = \left (\sqrt{1 + 4t^2 + 9t^4}\right )^3$. Therefore the curvature is given by:(2)
So the radius of curvature is:(3)
Plugging in $t = 1$ and we have that $\rho (1) = \frac{1}{2}$.
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In this paper we study consequences of the results of Kang et al. [Monoidal categorification of cluster algebras, J. Amer. Math. Soc. 31 (2018), 349–426] on a monoidal categorification of the unipotent quantum coordinate ring$A_{q}(\mathfrak{n}(w))$together with the Laurent phenomenon of cluster algebras. We show that if a simple module$S$in the category${\mathcal{C}}_{w}$strongly commutes with all the cluster variables in a cluster$[\mathscr{C}]$, then$[S]$is a cluster monomial in$[\mathscr{C}]$. If$S$strongly commutes with cluster variables except for exactly one cluster variable$[M_{k}]$, then$[S]$is either a cluster monomial in$[\mathscr{C}]$or a cluster monomial in$\unicode[STIX]{x1D707}_{k}([\mathscr{C}])$. We give a new proof of the fact that the upper global basis is a common triangular basis (in the sense of Qin [Triangular bases in quantum cluster algebras and monoidal categorification conjectures, Duke Math. 166 (2017), 2337–2442]) of the localization$\widetilde{A}_{q}(\mathfrak{n}(w))$of$A_{q}(\mathfrak{n}(w))$at the frozen variables. A characterization on the commutativity of a simple module$S$with cluster variables in a cluster$[\mathscr{C}]$is given in terms of the denominator vector of$[S]$with respect to the cluster$[\mathscr{C}]$.
We construct a cycle in higher Hochschild homology associated to the two-dimensional torus which represents 2-holonomy of a nonabelian gerbe in the same way as the ordinary holonomy of a principal G-bundle gives rise to a cycle in ordinary Hochschild homology. This is done using the connection 1-form of Baez–Schreiber. A crucial ingredient in our work is the possibility to arrange that in the structure crossed module$\unicode[STIX]{x1D707}:\mathfrak{h}\rightarrow \mathfrak{g}$of the principal 2-bundle, the Lie algebra$\mathfrak{h}$is abelian, up to equivalence of crossed modules.
Let 𝔟 be the Borel subalgebra of the Lie algebra 𝔰𝔩2 and V2 be the simple two-dimensional 𝔰𝔩2-module. For the universal enveloping algebra$\[{\cal A}: = U(\gb \ltimes {V_2})\]$of the semi-direct product 𝔟⋉V2 of Lie algebras, the prime, primitive and maximal spectra are classified. Please approve edit to the sentence “The sets of completely prime…”.The sets of completely prime ideals of$\[{\cal A}\]$are described. The simple unfaithful$\[{\cal A}\]$-modules are classified and an explicit description of all prime factor algebras of$\[{\cal A}\]$is given. The following classes of simple U(𝔟⋉V2)-modules are classified: the Whittaker modules, the 𝕂[X]-torsion modules and the 𝕂[E]-torsion modules.
Let$\mathfrak{g}=\mathfrak{g}\mathfrak{l}_{N}(\Bbbk )$, where$\Bbbk$is an algebraically closed field of characteristic$p>0$, and$N\in \mathbb{Z}_{{\geqslant}1}$. Let$\unicode[STIX]{x1D712}\in \mathfrak{g}^{\ast }$and denote by$U_{\unicode[STIX]{x1D712}}(\mathfrak{g})$the corresponding reduced enveloping algebra. The Kac–Weisfeiler conjecture, which was proved by Premet, asserts that any finite-dimensional$U_{\unicode[STIX]{x1D712}}(\mathfrak{g})$-module has dimension divisible by$p^{d_{\unicode[STIX]{x1D712}}}$, where$d_{\unicode[STIX]{x1D712}}$is half the dimension of the coadjoint orbit of$\unicode[STIX]{x1D712}$. Our main theorem gives a classification of$U_{\unicode[STIX]{x1D712}}(\mathfrak{g})$-modules of dimension$p^{d_{\unicode[STIX]{x1D712}}}$. As a consequence, we deduce that they are all parabolically induced from a one-dimensional module for$U_{0}(\mathfrak{h})$for a certain Levi subalgebra$\mathfrak{h}$of$\mathfrak{g}$; we view this as a modular analogue of Mœglin’s theorem on completely primitive ideals in$U(\mathfrak{g}\mathfrak{l}_{N}(\mathbb{C}))$. To obtain these results, we reduce to the case where$\unicode[STIX]{x1D712}$is nilpotent, and then classify the one-dimensional modules for the corresponding restricted$W$-algebra.
The purpose of this paper is to define an α-type cohomology, which we call α-type Chevalley–Eilenberg cohomology, for Hom-Lie algebras. We relate it to the known Chevalley–Eilenberg cohomology and provide explicit computations for some examples. Moreover, using this cohomology, we study formal deformations of Hom-Lie algebras, where the bracket as well as the structure map α are deformed. Furthermore, we provide a generalization of the grand crochet and study, in a particular case, the α-type cohomology for Hom-Lie bialgebras.
Let F be a field of characteristic different of 2 and let M1|1(F)(+) denote the Jordan superalgebra of 2 × 2 matrices over the field F. The aim of this paper is to classify irreducible (unital and one-sided) Jordan bimodules over the Jordan superalgebra M1|1(F)(+).
The category of Cohen–Macaulay modules of an algebra$B_{k,n}$is used in Jensen et al. (A categorification of Grassmannian cluster algebras, Proc. Lond. Math. Soc. (3) 113(2) (2016), 185–212) to give an additive categorification of the cluster algebra structure on the homogeneous coordinate ring of the Grassmannian of$k$-planes in$n$-space. In this paper, we find canonical Auslander–Reiten sequences and study the Auslander–Reiten translation periodicity for this category. Furthermore, we give an explicit construction of Cohen–Macaulay modules of arbitrary rank. We then use our results to establish a correspondence between rigid indecomposable modules of rank 2 and real roots of degree 2 for the associated Kac–Moody algebra in the tame cases.
The exceptional simple Lie algebras of types E7 and E8 are endowed with optimal$\mathsf{SL}_2^n$-structures, and are thus described in terms of the corresponding coordinate algebras. These are nonassociative algebras which much resemble the so-called code algebras.
In this note we consider parabolic subroot systems of a complex simple Lie Algebra. We describe root theoretic data of the subroot systems in terms of that of the root system and we give a selection of applications of our results to the study of generalized flag manifolds.
We introduce the oriented Brauer–Clifford and degenerate affine oriented Brauer–Clifford supercategories. These are diagrammatically defined monoidal supercategories that provide combinatorial models for certain natural monoidal supercategories of supermodules and endosuperfunctors, respectively, for the Lie superalgebras of type Q. Our main results are basis theorems for these diagram supercategories. We also discuss connections and applications to the representation theory of the Lie superalgebra of type Q.
We prove that if$\mathfrak{s}$is a solvable Lie algebra of matrices over a field of characteristic 0 and$A\in \mathfrak{s}$, then the semisimple and nilpotent summands of the Jordan–Chevalley decomposition of$A$belong to$\mathfrak{s}$if and only if there exist$S,N\in \mathfrak{s}$,$S$is semisimple,$N$is nilpotent (not necessarily$[S,N]=0$) such that$A=S+N$.
Given a root system, the Weyl chambers in the co-weight lattice give rise to a real toric variety, called the real toric variety associated with the Weyl chambers. We compute the integral cohomology groups of real toric varieties associated with the Weyl chambers of type Cn and Dn, completing the computation for all classical types.
We compare two cohomological Hall algebras (CoHA). The first one is the preprojective CoHA introduced in [19] associated with each quiver Q, and each algebraic oriented cohomology theory A. It is defined as the A-homology of the moduli of representations of the preprojective algebra of Q, generalizing the K-theoretic Hall algebra of commuting varieties of Schiffmann-Vasserot [15]. The other one is the critical CoHA defined by Kontsevich-Soibelman associated with each quiver with potentials. It is defined using the equivariant cohomology with compact support with coefficients in the sheaf of vanishing cycles. In the present paper, we show that the critical CoHA, for the quiver with potential of Ginzburg, is isomorphic to the preprojective CoHA as algebras. As applications, we obtain an algebra homomorphism from the positive part of the Yangian to the critical CoHA.
This paper provides some evidence for conjectural relations between extensions of (right) weak order on Coxeter groups, closure operators on root systems, and Bruhat order. The conjecture focused upon here refines an earlier question as to whether the set of initial sections of reflection orders, ordered by inclusion, forms a complete lattice. Meet and join in weak order are described in terms of a suitable closure operator. Galois connections are defined from the power set of$W$to itself, under which maximal subgroups of certain groupoids correspond to certain complete meet subsemilattices of weak order. An analogue of weak order for standard parabolic subsets of any rank of the root system is defined, reducing to the usual weak order in rank zero, and having some analogous properties in rank one (and conjecturally in general).
In this paper, we complete the ADE-like classification of simple transitive 2-representations of Soergel bimodules in finite dihedral type, under the assumption of gradeability. In particular, we use bipartite graphs and zigzag algebras of ADE type to give an explicit construction of a graded (non-strict) version of all these 2-representations.
Moreover, we give simple combinatorial criteria for when two such 2-representations are equivalent and for when their Grothendieck groups give rise to isomorphic representations.
Finally, our construction also gives a large class of simple transitive 2-representations in infinite dihedral type for general bipartite graphs.
We analyse infinitesimal deformations of pairs$(X,{\mathcal{F}})$with${\mathcal{F}}$a coherent sheaf on a smooth projective variety$X$over an algebraically closed field of characteristic 0. We describe a differential graded Lie algebra controlling the deformation problem, and we prove an analog of a Mukai–Artamkin theorem about the trace map.
We explicitly describe the isomorphism between two combinatorial realizations of Kashiwara’s infinity crystal in types B and C. The first realization is in terms of marginally large tableaux and the other is in terms of Kostant partitions coming from PBW bases. We also discuss a stack notation for Kostant partitions which simplifies that realization.
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Newform invariants
Coefficients of the \(q\)-expansion are expressed in terms of \(\beta = \frac{1}{2}(1 + \sqrt{17})\). We also show the integral \(q\)-expansion of the trace form.
For each embedding \(\iota_m\) of the coefficient field, the values \(\iota_m(a_n)\) are shown below.
For more information on an embedded modular form you can click on its label.
This newform does not admit any (nontrivial) inner twists.
\( p \) Sign \(2\) \(-1\) \(3\) \(1\) \(23\) \(1\) \(29\) \(-1\)
This newform can be constructed as the intersection of the kernels of the following linear operators acting on \(S_{2}^{\mathrm{new}}(\Gamma_0(4002))\):
\(T_{5}^{2} \) \(\mathstrut +\mathstrut 3 T_{5} \) \(\mathstrut -\mathstrut 2 \) \(T_{7} \) \(T_{11}^{2} \) \(\mathstrut +\mathstrut T_{11} \) \(\mathstrut -\mathstrut 4 \)
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I'm conducting a X-ray diffraction experiment for lab, but am new to solid-state physics and crystallography. I have to find the Debye-Waller factor (DWF) of Al at room temperature using X-ray powder diffraction. The value given in the International Tables for X-Ray Crystallography Vol. 3 is approximately $0.78 \mathring{A}^{2}$.
I obtain the diffraction peaks using a diffractometer with $\text{Cu}K_{\alpha}$ radiation and a Ni filter at the receiving slit.
I find the integrated intensity $I$ relative to that of the first diffraction peak by finding the peaks' areas above the background. Since $$I=K\exp(-2B\sin^2\theta/\lambda^2)$$ then $$\log I=\log K-2B\frac{\sin^2\theta}{\lambda^2}$$ where $\theta$ is the diffraction angle, $\lambda$ the wavelength in angstrom of the radiation incident on the aluminum powder sample, and $B$ is the DWF. $K$ is a constant proportional to factors like the structure factor and Lorentz polarization factors. I then fit the plot of $\log I$ vs. $\frac{\sin^2\theta}{\lambda^2}$ with a linear function, and divide the slope of the fit by $2$ to get the DWF.
The diffraction peaks I'm seeing are extremely similar to the plot of the diffraction peaks below:
However, when I actually plot $\log I$ vs. $\sin^2\theta/\lambda^2$, I get:
and a DWF of $5.92 \mathring{A}^{2}$. Due to this difference in magnitude, I feel like I'm doing something incorrectly when measuring the integrated intensity. Could someone offer a suggestion?
Thank you in advance.
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Answer
A=36$\pi$$\approx$113
Work Step by Step
The diameter of the inscribed circle is equal to the height of the isosceles trapezoid. h$^2$=13$^2$-($\frac{1}{2}$(18-8))$^2$ h$^2$=169-25=144 h=12 r=.5h r=6 A=$\pi$r$^2$ A=36$\pi$$\approx$113
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Bounded Subsets in Euclidean Space
Definition: Let $S \subseteq \mathbb{R}^n$. The set $S$ is said to be Bounded if there exists a $\mathbf{x} \in \mathbb{R}^n$ and a positive real number $r > 0$ such that $S \subseteq B(\mathbf{x}, r)$ and $S$ is said to be Unbounded otherwise. In other words, a subset $S$ of $\mathbb{R}^n$ is bounded if $S$ is the subset of some ball in $\mathbb{R}^n$.
In $\mathbb{R}^2$, a set $S \subseteq \mathbb{R}^2$ that is bounded might look something like:
Meanwhile, a set $S$ that is unbounded might look something like:
For a more concrete example, consider the subset $S = [0, 1) \times [0, 1) \subseteq \mathbb{R}^2$. Then the ball $B(\mathbf{0}, 2)$ is such that $S \subseteq B(\mathbf{0}, 2)$, so $S$ is bounded. However, the subset $S = \{ (x, y) \in \mathbb{R}^2 : x \geq 0, y \geq 0 \} \subseteq \mathbb{R}^2$ is unbounded. To prove this, set $\mathbf{x} = \mathbf{0}$. Suppose that there exists an $r > 0$ such that $S \subseteq B(\mathbf{0}, r)$. now consider the point $\mathbf{y} = (r+1, r+1) \in S$. Then:(1)
Therefore $(r + 1, r + 1) \not \in B(\mathbf{0}, r)$ for all $r > 0$, but $(r+1, r+1) \in S$ for all $r > 0$, so $S \not \subseteq B(\mathbf{0}, r)$ for all $r > 0$, so $S$ is unbounded.
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I am using the align command to align equations :
\begin{align*} \psi \colon & SU(2)\otimes SU(2) \otimes SU(2) \to \mathbb{C}\\ &(U_{l_1} \otimes U_{l_2} \otimes U_{l_3}) \mapsto \psi(U_{l_1} \otimes U_{l_2} \otimes U_{l_3}).\end{align*}
As I understand it, the "&" tells to latex where the object need to be vertically aligned.
But does & mean something in general in latex ? Or is it a command that has always different significations ?
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The Dehn invariant of a polyhedron is a vector in $\mathbb{R}\otimes_{\mathbb{Z}}\mathbb{R}/2\pi\mathbb{Z}$ defined as the sum over the edges of the polyhedron of the terms $\sum\ell_i\otimes\theta_i$ where $\ell_i$ is the length of edge $i$ and $\theta_i$ is its dihedral angle.
Are all vectors in this space realizable as the Dehn invariants of polyhedra?
(To be very specific let's define a polyhedron to be a bounded closed manifold embedded into Euclidean or hyperbolic space as a subset of the union of finitely many planes.)
You can add two representable Dehn invariants by taking the disjoint unions of their polyhedra (or gluing them together on any pair of faces, even with mismatched face shapes, if you prefer a single connected polyhedron). For Euclidean polyhedra, you can multiply the Dehn invariant by any scalar by scaling the polyhedron by the same factor. So the left sides of the products in the sum are controllable but the right sides are not, and getting a nonzero Dehn invariant that represents only a single angle $\theta$ and its rational multiples seems difficult except for some very special cases.
The Dehn invariant also makes sense in hyperbolic space but there you have the opposite problem. You can get a polyhedron whose dihedrals are all rational multiples of the same angle $\theta$, for your favorite $\theta$, by scaling one of the Platonic solids, but scaling the lengths by arbitrary real factors (while keeping the angles fixed) becomes difficult. So I'd be interested in answers both for the Euclidean and for the hyperbolic cases, especially if they require different arguments or have different answers.
If an answer can be found in the published literature, references would be helpful.
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(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
Let $A, B, C$ be $n\times n$ invertible matrices. When you simplify the expression\[C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2,\]which matrix do you get?(a) $A$(b) $C^{-1}A^{-1}BC^{-1}AC^2$(c) $B$(d) $C^2$(e) $C^{-1}BC$(f) $C$
Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.Let\[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\]where\begin{align*}p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3.\end{align*}
(a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.
(b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be a vector space and $B$ be a basis for $V$.Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$.Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form\[\begin{bmatrix}1 & 0 & 2 & 1 & 0 \\0 & 1 & 3 & 0 & 1 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0\end{bmatrix}.\]
(a) What is the dimension of $V$?
(b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.Let\[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix}a & b\\c& -a\end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\]
(a) Show that $W$ is a subspace of $V$.
(b) Find a basis of $W$.
(c) Find the dimension of $W$.
(The Ohio State University, Linear Algebra Midterm)
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 2 and contains Problem 4, 5, and 6.Check out Part 1 and Part 3 for the rest of the exam problems.
Problem 4. Let\[\mathbf{a}_1=\begin{bmatrix}1 \\2 \\3\end{bmatrix}, \mathbf{a}_2=\begin{bmatrix}2 \\-1 \\4\end{bmatrix}, \mathbf{b}=\begin{bmatrix}0 \\a \\2\end{bmatrix}.\]
Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.
Problem 5.Find the inverse matrix of\[A=\begin{bmatrix}0 & 0 & 2 & 0 \\0 &1 & 0 & 0 \\1 & 0 & 0 & 0 \\1 & 0 & 0 & 1\end{bmatrix}\]if it exists. If you think there is no inverse matrix of $A$, then give a reason.
Problem 6.Consider the system of linear equations\begin{align*}3x_1+2x_2&=1\\5x_1+3x_2&=2.\end{align*}
(a) Find the coefficient matrix $A$ of the system.
(b) Find the inverse matrix of the coefficient matrix $A$.
(c) Using the inverse matrix of $A$, find the solution of the system.
(Linear Algebra Midterm Exam 1, the Ohio State University)
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 1 and contains the first three problems.Check out Part 2 and Part 3 for the rest of the exam problems.
Problem 1. Determine all possibilities for the number of solutions of each of the systems of linear equations described below.
(a) A consistent system of $5$ equations in $3$ unknowns and the rank of the system is $1$.
(b) A homogeneous system of $5$ equations in $4$ unknowns and it has a solution $x_1=1$, $x_2=2$, $x_3=3$, $x_4=4$.
Problem 2. Consider the homogeneous system of linear equations whose coefficient matrix is given by the following matrix $A$. Find the vector form for the general solution of the system.\[A=\begin{bmatrix}1 & 0 & -1 & -2 \\2 &1 & -2 & -7 \\3 & 0 & -3 & -6 \\0 & 1 & 0 & -3\end{bmatrix}.\]
Problem 3. Let $A$ be the following invertible matrix.\[A=\begin{bmatrix}-1 & 2 & 3 & 4 & 5\\6 & -7 & 8& 9& 10\\11 & 12 & -13 & 14 & 15\\16 & 17 & 18& -19 & 20\\21 & 22 & 23 & 24 & -25\end{bmatrix}\]Let $I$ be the $5\times 5$ identity matrix and let $B$ be a $5\times 5$ matrix.Suppose that $ABA^{-1}=I$.Then determine the matrix $B$.
(Linear Algebra Midterm Exam 1, the Ohio State University)
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Table of Contents
Examples of Closed Unit Balls That are NOT Compact
Recall from the Closed Unit Ball Criterion for Finite Dimensional Normed Linear Spaces page that if $(X, \| \cdot \|_X)$ is a normed linear space then the closed unit ball of $X$ is compact if and only if $X$ is finite-dimensional.
We will now look at some examples of normed linear spaces whose closed unit balls are not compact. By the theorem referenced above, these spaces will be infinite-dimensional.
Proposition 1: The closed unit ball in $C[0, 1]$ is not compact. In metric spaces, a space is compact if and only if every sequence in the space has a subsequence which converges in the space. This is known as the Bolzano-Weierstrass property. Proof:$C[0, 1]$ is infinite-dimensional and hence the closed unit ball in $C[0, 1]$ is not compact. Let $\overline{B}(0, 1)$ denote the closed unit ball in $C[0, 1]$. Then: For each $n \in \mathbb{N}$ let $f_n : [0, 1] \to \mathbb{R}$ be defined by $f_n(x) = x^n$. Observe that for each $n \in \mathbb{N}$ we have that: Then $(f_n)$ is a sequence in $B(0, 1)$. However, observe that $(f_n)$ does not converge uniformly. In fact, $(f_n)$ converges pointwise to: In particular, no subsequence of $(f_n)$ converges uniformly on $[0, 1]$. Now in particular, we have that no subsequence of $(f_n)$ converges in $\bar{B}(0, 1)$ and so $\bar{B}(0, 1)$ is not compact.
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LHCb Collaboration,; Aaij, R; Adeva, B; Adinolfi, M; Bernet, R; Bowen, E; Bursche, A; Chiapolini, N; Chrzaszcz, M; Dey, B; Elsasser, C; Graverini, E; Lionetto, F; Lowdon, P; Mauri, A; Müller, K; Serra, N; Steinkamp, O; Storaci, B; Straumann, U; Tresch, M; Vollhardt, A; Weiden, A; et al, (2015).
Differential branching fraction and angular analysis of $\Lambda^{0}_{b} \rightarrow \Lambda \mu^+\mu^-$ decays. Journal of High Energy Physics, 2015(6):115. Abstract
The differential branching fraction of the rare decay $\Lambda^{0}_{b} \rightarrow \Lambda \mu^+\mu^-$ is measured as a function of $q^{2}$, the square of the dimuon invariant mass. The analysis is performed using proton-proton collision data, corresponding to an integrated luminosity of $3.0 \mbox{ fb}^{-1}$, collected by the LHCb experiment. Evidence of signal is observed in the $q^2$ region below the square of the $J/\psi$ mass. Integrating over $15 < q^{2} < 20 \mbox{ GeV}^2/c^4$ the branching fraction is measured as $d\mathcal{B}(\Lambda^{0}_{b} \rightarrow \Lambda \mu^+\mu^-)/dq^2 = (1.18 ^{+0.09} _{-0.08} \pm 0.03 \pm 0.27) \times 10^{-7} ( \mbox{GeV}^{2}/c^{4})^{-1}$, where the uncertainties are statistical, systematic and due to the normalisation mode, $\Lambda^{0}_{b} \rightarrow J/\psi \Lambda$, respectively. In the $q^2$ intervals where the signal is observed, angular distributions are studied and the forward-backward asymmetries in the dimuon ($A^{l}_{\rm FB}$) and hadron ($A^{h}_{\rm FB}$) systems are measured for the first time. In the range $15 < q^2 < 20 \mbox{ GeV}^2/c^4$ they are found to be $A^{l}_{\rm FB} = -0.05 \pm 0.09 \mbox{ (stat)} \pm 0.03 \mbox{ (syst)}$ and $A^{h}_{\rm FB} = -0.29 \pm 0.07 \mbox{ (stat)} \pm 0.03 \mbox{ (syst)}$.
Abstract
The differential branching fraction of the rare decay $\Lambda^{0}_{b} \rightarrow \Lambda \mu^+\mu^-$ is measured as a function of $q^{2}$, the square of the dimuon invariant mass. The analysis is performed using proton-proton collision data, corresponding to an integrated luminosity of $3.0 \mbox{ fb}^{-1}$, collected by the LHCb experiment. Evidence of signal is observed in the $q^2$ region below the square of the $J/\psi$ mass. Integrating over $15 < q^{2} < 20 \mbox{ GeV}^2/c^4$ the branching fraction is measured as $d\mathcal{B}(\Lambda^{0}_{b} \rightarrow \Lambda \mu^+\mu^-)/dq^2 = (1.18 ^{+0.09} _{-0.08} \pm 0.03 \pm 0.27) \times 10^{-7} ( \mbox{GeV}^{2}/c^{4})^{-1}$, where the uncertainties are statistical, systematic and due to the normalisation mode, $\Lambda^{0}_{b} \rightarrow J/\psi \Lambda$, respectively. In the $q^2$ intervals where the signal is observed, angular distributions are studied and the forward-backward asymmetries in the dimuon ($A^{l}_{\rm FB}$) and hadron ($A^{h}_{\rm FB}$) systems are measured for the first time. In the range $15 < q^2 < 20 \mbox{ GeV}^2/c^4$ they are found to be $A^{l}_{\rm FB} = -0.05 \pm 0.09 \mbox{ (stat)} \pm 0.03 \mbox{ (syst)}$ and $A^{h}_{\rm FB} = -0.29 \pm 0.07 \mbox{ (stat)} \pm 0.03 \mbox{ (syst)}$.
Additional indexing
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I am currently going through electromagnetic form factor. I came across the fact that since the proton is not an elementary particle its scattering(elastic) with electron can be modeled using general general electromagnetic form factor of form
$ \Gamma_{\mu} = \gamma_{\mu}F_1(q^2) + i\sigma_{\mu\lambda}q^{\lambda}F_2(q^2)\\ $
where we get scattering amplitude and cross section in terms of $ F_1, F_2 $.
So there are two questions
Is this the general procedure actually followed while present day experiment for calculating the amplitude or is there another (better) method?
Can I do the same for scattering of electron with pions (which are spin 0) using corresponding general vertex function(which i think would have only 1 form factor ,i suppose)?
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Topological Methods in Nonlinear Analysis Topol. Methods Nonlinear Anal. Volume 13, Number 2 (1999), 181-190. Degree and Sobolev spaces Abstract
Let $u$ belong (for example) to $W^{1,n+1}(S^n\times \Lambda, S^n)_{\lambda\in\Lambda}$ where $\Lambda$ is a connected open set in ${\mathbb R}^k$. For a.e. $\lambda\in\Lambda$ the map $x\mapsto u(x,\lambda)$ is continuous from $S^n$ into $S^n$ and therefore its (Brouwer) degree is well defined. We prove that this degree is independent of $\lambda$ a.e. in $\Lambda$. This result is extended to a more general setting, as well to fractional Sobolev spaces $W^{s,p}$ with $sp\geq n+1$.
Article information Source Topol. Methods Nonlinear Anal., Volume 13, Number 2 (1999), 181-190. Dates First available in Project Euclid: 29 September 2016 Permanent link to this document https://projecteuclid.org/euclid.tmna/1475178877 Mathematical Reviews number (MathSciNet) MR1742219 Zentralblatt MATH identifier 0956.46024 Citation
Brezis, Haïm; Li, Yanyan; Mironescu, Petru; Nirenberg, Louis. Degree and Sobolev spaces. Topol. Methods Nonlinear Anal. 13 (1999), no. 2, 181--190. https://projecteuclid.org/euclid.tmna/1475178877
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To form a product, you give me $n$ objects, $A_1,\dots,A_n$, and I give you back an object $A_1\times\dots\times A_n$, together with $n$ maps $\pi_i\colon A_1\times\dots\times A_n\to A_i$ (one to each of the $A_i$) satisfying the universal property of the product.
So what happens if $n=0$? Then you give me $0$ objects, and I give you back an object which we call $1$, together with $0$ maps $\pi_i$ (one to each of the $A_i$, of which there aren't any), satisfying the universal property of the product.
What does the universal property say in this case?
For any $X$ given together with $0$ maps $f_i$ (one to each of the $A_i$, of which there aren't any), there is a unique map $!\colon X\to 1$ making all of the triangles commute ($\pi_i\circ ! = f_i$ for all $i$, of which there aren't any).
Removing the vacuous conditions from the definition, we see that the empty product is an object $1$ such that for every object $X$ there is a unique map $!\colon X\to 1$, i.e. $1$ is a terminal object.
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The Normed Space Induced by an Inner Product
Theorem 1: Let $H$ be an inner produce space. Then the function $\| \cdot \| : H \to \mathbb{R}$ defined for all $x \in H$ by $\| x \| = \langle x, x \rangle^{1/2}$ is a norm on $H$. Proof:We show that $\| \cdot \|$ has all of the properties of a norm. First, suppose that $x = 0$. Then $\| 0 \| = \langle 0, 0 \rangle^{1/2} = 0$. Now suppose that $\| x \| = 0$. Then $\langle x, x \rangle = 0$. But by definition, this means that $x = 0$. Hence $\| x \| = 0$ if and only if $x = 0$. Second, let $x \in H$ and let $\lambda \in \mathbb{C}$. Then: Lastly, let $x, y \in H$. Then by the Cauchy-Schwarz inequality we have that: Therefore: Therefore $\| \cdot \| : H \to \mathbb{R}$ is a norm. $\blacksquare$
So whenever we have an inner product space $H$ we can obtain a norm on $H$. We give this norm a special name defined below.
Definition: Let $H$ be an inner product space. The Norm Induced by the Inner Product on $H$ is the norm $\| \cdot \| : H \to \mathbb{R}$ defined for all $x \in H$ by $\| x \| = \langle x, x \rangle^{1/2}$.
From now on, if we say that $H$ is an inner product space then we associate with $H$ both the inner product $\langle \cdot, \cdot \rangle$ defined on it and the norm $\| \cdot \|$ defined above.
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One thing to note about relativistic cosmology is its solutions evolve in time. So if at one moment the universe has a specific value of spatial curvature the next moment it would be different. The value of curvature specified in OP is quite large and thus it correspond to just a specific moment near the Big Bang (or Big Crunch) of this hypothetical universe.
Another thing is that providing just the curvature and openness is not enough to specify the evolution of the universe even with the additional constraint that we "don't need to invent a new set of laws of physics".
Evolution of the universe on large scales could be a modeled by a solution of Friedmann equations. By specifying openness we assume $k=-1$ and spatial curvature specifies the scale factor at that moment $a(t_0)$, which for the OP value would be on the order of a hundred meters. One also has to specify time derivative $\dot{a}$ and matter content, which enters the equations through energy $\rho$ and pressure $p$.
Let us have a look at the first Friedmann equation:$$ \frac{\dot{a}^2 + kc^2}{a^2} = \frac{8 \pi G \rho + \Lambda c^2}{3} $$The right hand side should be positive, if one assumes that the matter is something like what we have in our universe (i.e. there are no negative energies, including dark one). We see than that this immediately puts a constraint on a speed of expansion: $|\dot{a}|>c$ at all times. This means that all matter that is accessible to observation (including future observations) is located within volume of a sphere with radius smaller than $a$ (Hubble volume). In particular, exponential decay of
electrostatic force is now replaced by more complicated phenomenon: as soon as the distance between charges becomes comparable to $a$, one has to include non-stationarity of spacetime.
We see, that specifying $\dot{a}$ also gives us the energy density. For example if we specify $\dot{a}=2 c$, when $a=200\,\text{m}$, then the energy density would be about $10^{22}\,\text{kg}/\text{m}^3$. This is large, of course, but inside the Hubble volume there would be only a fraction of the solar mass of matter. And this matter would rapidly expand as it cools off without forming a single gravitationally bound object: no stars, no planets, no comets, no complex chemistry.
To combat this uninteresting fate we could try to increase $\dot{a}$ (while keeping the value of $a$ constant). This increases energy density in the r.h.s. but also decreases overall role of term with $k$ (at least during early stages of expansion). At the same time this pushes the moment at which a given value of $a$ achieved closer and closer to the Big Bang singularity. And if there is enough matter in the Hubble volume, there could be formation of stars, star clusters, and galaxies just like in our universe. Within such gravitationally bound systems the effects of openness of the universe would be negligible, however observer that evolved in such a world would notice far fewer or even completely absent large scale structures such as galactic groups, clusters, superclusters and that the universe is expanding more rapidly than ours.
Now, I understand that OP's desire is to have a universe that is spatially hyperbolic yet almost static. I do not know of the way one could achieve this in a somewhat realistic cosmological model, however such universe could in principle be constructed with the help of negative cosmological constant $\Lambda$. Such solution would be an extension of static Einstein universe model on the case of negative spatial curvature (and negative cosmological constant). This requires precise matching between the values of $a$, $\Lambda$ and $\rho$ and the solution would unstable against small perturbations.
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Limits of Polynomials and Rational Functions
Before we look at some theorems regarding the limits of polynomials and rational functions, we should first formally define what each is.
Definition: A function in the form $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ where $a_0, a_1, ..., a_n \in \mathbb{R}$ is said to be a polynomial of degree $n$. If $p(x)$ and $q(x)$ are polynomials, then the function $r(x) = \frac{p(x)}{q(x)}$, $q(x) \neq 0$ is said to be a rational function.
For example, the $p : \mathbb{R} \to \mathbb{R}$ defined by $p(x) = 3 + 2x^2 + 4x^4$ is a polynomial, and the function $r : \mathbb{R} \setminus \{ 1 \} \to \mathbb{R}$ defined by $r(x) = \frac{2x + 3x^2}{1 - x}$ is a rational function.
We will now look at some theorems regarding the limits of these functions.
Theorem 1: If $p(x) = a_0 + a_1x + ... + a_nx^n$, $a_0, a_1, ..., a_n \in \mathbb{R}$ is a polynomial function. Then the limit at $x = c$ exists and $\lim_{x \to c} p(x) = p(c)$. Proof:Let $p(x) = a_0 + a_1x + ... + a_nx^n$ where $a_0, a_1, ..., a_n \in \mathbb{R}$ be a polynomial function. Then we have that: Now recall that $\lim_{x \to c} 1 = 1$ and $\lim_{x \to c} x = c$. Furthermore, from the Operations on Functions and Their Limits page, recall that since $\lim_{x \to c} x = c$, then $\lim_{x \to c} x^2 = \lim_{x \to c} x \cdot \lim_{x \to c} x = c^2$, …, $\lim_{x \to c} x^n = c^n$ (this can be proven by induction), and so:
Theorem 2: If $r(x) = \frac{p(x)}{q(x)}$ is a rational function where $q(c) \neq 0$, then the limit at $x = c$ exists and $\lim_{x \to c} r(x) = \frac{p(c)}{q(c)}$. Let $r(x) = \frac{p(x)}{q(x)}$ be a rational function. From theorem 1, since $p(x)$ and $q(x)$ are polynomials, we have that $\lim_{x \to c} p(x) = p(c)$ and $\lim_{x \to c} q(x) = q(c)$. Therefore by the Quotient Law for limits, $\lim_{x \to c} r(x) = \lim_{x \to c} \frac{p(x)}{q(x)} = \frac{p(c)}{q(c)}$, which is valid since $q(c) \neq 0$. $\blacksquare$
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Directional Derivatives Examples 2
Recall from the Directional Derivatives page that for a two variable real-valued function $z = f(x, y)$, the directional derivative of $f$ at a point $(x, y) \in D(f)$ in the direction of the unit vector $\vec{u} = (a, b)$ is given by the formula:(1)
For a three variable real-valued function $w = f(x, y, z)$, the directional derivative of $f$ at a point $(x, y, z) \in D(f)$ in the direction of the unit vector $\vec{u} = (a, b, c)$ is given by the formula:(2)
Example 1 Find the directional derivative of $z = f(x, y) = e^x \sin y$ at $\left ( 0, \frac{\pi}{3} \right )$ in the direction of the vector $\vec{v} = (-6, 8)$.
We first note that $\vec{v}$ is not a unit vector. Let's find a unit vector that goes in the direction of $\vec{v}$. We have that $\| \vec{v} \| = \sqrt{(-6)^2 + (8)^2} = \sqrt{36+64} = 10$, and so $\vec{u} = \left ( \frac{-6}{10}, \frac{8}{10} \right ) = \left ( \frac{-3}{5}, \frac{4}{5} \right )$ is a unit vector in the direction of $\vec{v}$.
Now we compute the partial derivatives of $f$. We have that $\frac{\partial z}{\partial x} = e^x \sin y$ and $\frac{\partial z}{\partial y} = e^x \cos y$. Therefore:(3)
Example 2 Find the directional derivative of $z = f(x, y) = x^4 - x^2y^3$ at $(2, 1)$ in the direction of the vector $\vec{v} = (1, 3)$.
We first note that $\vec{v}$ is not a unit vector. Let's find a unit vector that goes in the direction of $\vec{v}$. We have that $\| \vec{v} \| = \sqrt{(1)^2 + (3)^2} = \sqrt{10}$, and so $\vec{u} = \left (\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}} \right )$ is a unit vector in the direction of $\vec{v}$.
Now we compute the partial derivatives of $f$. We have that $\frac{\partial z}{\partial x} = 4x^3 -2xy^3$ and $\frac{\partial z}{\partial y} = -3x^2y^2$. Therefore:(4)
Example 3 Find the directional derivative of $w = f(x, y, z) = xe^y + ye^z + ze^x$ at $(0, 0, 0)$ in the direction of the vector $\vec{v} = (5, 1, -2)$.
Once again, we note that $\vec{v}$ is not a unit vector, so let's find a unit vector that goes in the direction of $\vec{v}$. We have that $\| \vec{v} \| = \sqrt{(5)^2 + (1)^2 + (-2)^2} = \sqrt{30}$. So $\vec{u} = \left ( \frac{5}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-2}{\sqrt{30}} \right)$ is a unit vector in the direction of $\vec{v}$.
Now we compute the partial derivatives of $f$. We have that $\frac{\partial w}{\partial x} = e^y + ze^x$, $\frac{\partial w}{\partial y} = xe^y + e^z$, and $\frac{\partial w}{\partial z} = ye^z + e^x$. Therefore:(5)
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What are the open big problems in algebraic geometry and vector bundles?
More specifically, I would like to know what are interesting problems related to moduli spaces of vector bundles over projective varieties/curves.
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What are the open big problems in algebraic geometry and vector bundles?
More specifically, I would like to know what are interesting problems related to moduli spaces of vector bundles over projective varieties/curves.
A few of the more obvious ones:
* Resolution of singularities in characteristic p
*Hodge conjecture * Standard conjectures on algebraic cycles (though these are not so urgent since Deligne proved the Weil conjectures). *Proving finite generation of the canonical ring for general type used to be open though I think it was recently solved; I'm not sure about the details.
For vector bundles, a longstanding open problem is the classification of vector bundles over projective spaces.
(Added later) A very old major problem is that of finding which moduli spaces of curves are unirational. It is classical that the moduli space is unirational for genus at most 10, and I think this has more recently been pushed to genus about 13. Mumford and Harris showed that it is of general type for genus at least 24. As far as I know most of the remaining cases are still open.
Let me mention a couple of problems related to vector bundles on projective spaces.
The Hartshorne conjecture. In its weak form it says that any rank 2 vector bundle on $\mathbf{P}^n_{\mathbf{C}},n>6$ is a direct sum of line bundles, which implies that any codimension 2 smooth subvariety whose canonical class is a multiple of the hyperplane sectionis a complete intersection. In a stronger form Hartshorne's conjecture says that any codimension $>\frac{2}{3}n$ subvariety of $\mathbf{P}^n_{k},k$ an algebraically closed field is a complete intersection. See Hartshorne, Varieties of small codimension in a projective space, Bull AMS 80, 1974. The weak conjecture fails for $n=3$ and $4$ -- there are examples (due to Horrocks and Mumford) of non-split vector bundles of rank 2 on $\mathbf{P}^4_{\mathbf{C}}$, but so far as I know the question if any such examples exist for $n>4$ is open. See here Evidences on Hartshorne's conjecture? References? for a discussion including some references.
The existence of non-algebraic topological vector bundles on $\mathbf{P}^n_{\mathbf{C}}$. It is a classical result that any topological complex vector bundle on $\mathbf{P}^n_{\mathbf{C}}, n\leq 3$ is algebraic, see e.g. Okonek, Schneider, Spindler, Vector bundles on complex projective spaces, chapter 1, \S 6. It is strongly suspected that for $n>3$ there are topological complex vector bundles that are not algebraic. Good candidates are nontrivial rank 2 vector bundles on $\mathbf{P}^n_{\mathbf{C}}, n\geq 5$ all of whose Chern classes vanish which were constructed by E. Rees, see MR0517518. It is claimed there that these bundles do not admit a holomorphic structure, but later a gap was found in the proof. See here Complex vector bundles that are not holomorphic for some more information.
Linearization Conjecture. Every algebraic action of $\mathbb{C}^*$ on $\mathbb{C}^n$ is linear in some coordinates of $\mathbb{C}^n$. Open for $n>3$.
Cancellation Conjecture. If $X\times \mathbb{C}\cong \mathbb{C}^{m+1}$ then $X\cong \mathbb{C}^m$. Open for $m>2$.
Coolidge-Nagata Conjecture. A rational cuspidal curve in $\mathbb{P}^2$ is rectifiable, i.e. there exists a birational automorphism of $\mathbb{P}^2$ which transforms the curve into a line.
There's also the big open question (I think it's still open) about whether rationally connected varieties are always unirational. I think people believe the answer is
NO, but they don't know an example.
Joe Harris had some slides a few years ago with regards to this Seattle 2005
We can also mention two other major open problems :
The abundance conjecture, stating that if a $K_X+\Delta$ is klt and nef, then it is semi-ample (a multiple has no base-point)
The Griffith's conjecture : if $E$ is an ample vector bundle over a compact complex manifold, then it is Griffith-positive. (this is known for line bundles of course)
There's also Fujita's conjecture.
Conjecture: Suppose $X$ is a smooth projective dimensional complex algebraic variety with ample divisor $A$. Then
It's also often stated in the complex analytic world.
Also there are many refinements (and generalizations) of this conjecture. For example, the assumption that $X$ is smooth is probably more than you need (something close to rational singularities should be ok). It also might even be true in characteristic $p > 0$.
It's known in relatively low dimensions (up to 5 in case 1. I think?)
In connection to vector bundles over $\mathbb{P}^n$, Hartshorne's paper from 1979 provides a list of open problems. The paper is "Algebraic vector bundles on projective spaces: A problem list"
Topology, 18:117–128, 1979.
I don't know which of those problems are still open, but I would be interested in knowing how much progress has been made on those problems, since 1979.
Open problems in Algebraic Geometry, S.J. Edixhoven (editor), B.J.J. Moonen (editor), F. Oort (editor).
The
Tate conjecture: Let $k$ be a finitely generated field, $X/k$ a smooth projective geometrically integral variety and $\ell$ invertible in $k$. Then the cycle class map $$\mathrm{CH}^r(X) \otimes_\mathbf{Z} \mathbf{Q}_\ell \to \mathrm{H}^{2r}(\bar{X},\mathbf{Q}_\ell(r))^{G_k}$$ is surjective.
It is e.g. proved for $r=1$ and Abelian varieties, a deep theorem. See http://www.math.harvard.edu/~chaoli/doc/TateConjecture.html.
This is analogous to the Hodge conjecture for complex varieties.
The Maximal Rank Conjecture is a major outstanding problem in Brill-Noether theory, although recent advances in tropical techniques might point the way to a solution; see https://arxiv.org/abs/1505.05460.
EDIT: The Maximal Rank Conjecture was proved by Eric Larson in his PhD thesis; see: https://arxiv.org/abs/1711.04906
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The Cauchy-Davenport Theorem says that if $A_1, \ldots, A_k$ are subsets of ${\mathbb Z}_p$, $p$ prime, then $| \sum_i A_i | \geq \min (p, \sum_i |A_i| -k +1)$.
I am looking for a generalization that bounds the number of ways each element $a \in \sum_i A_i$ can be represented as $a=\sum_i a_i$ with $a_i \in A_i$.
Specifically, I am interested in the case where $\sum_i |A_i| -k +1 \geq p$. Let $N_{min}$ denote the minimum number of ways any element can be written as a sum, and let $N_{max}$ denote the maximum number of ways any element can be written as a sum. Can we bound the ratio $N_{max}/N_{min}$? My vague conjecture would be that for $\sum_i |A_i| -k +1 \gg p$ we can bound the ratio by a constant, but I have no progress toward a proof.
I am aware of two papers by Pollard from the '70s and some followup work looking at sort-of-related questions, but am not aware of anything that comes close to addressing the above.
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Change to browse by: References & Citations Bookmark(what is this?) Condensed Matter > Disordered Systems and Neural Networks Title: Chaotic wave packet spreading in two-dimensional disordered nonlinear lattices
(Submitted on 20 Aug 2019)
Abstract: We reveal the generic characteristics of wave packet delocalization in two-dimensional nonlinear disordered lattices by performing extensive numerical simulations in two basic disordered models: the Klein-Gordon system and the discrete nonlinear Schr\"{o}dinger equation. We find that in both models (a) the wave packet's second moment asymptotically evolves as $t^{a_m}$ with $a_m \approx 1/5$ ($1/3$) for the weak (strong) chaos dynamical regime, in agreement with previous theoretical predictions~\cite{F10}, (b) chaos persists, but its strength decreases in time $t$ since the finite time maximum Lyapunov exponent $\Lambda$ decays as $\Lambda \propto t^{\alpha_{\Lambda}}$, with $\alpha_{\Lambda} \approx -0.37$ ($-0.46$) for the weak (strong) chaos case, and (c) the deviation vector distributions show the wandering of localized chaotic seeds in the lattice's excited part, which induces the wave packet's thermalization. We also propose a dimension-independent scaling between the wave packet's spreading and chaoticity, which allows the prediction of the obtained $\alpha_{\Lambda}$ values. Submission historyFrom: Charalampos Skokos [view email] [v1]Tue, 20 Aug 2019 20:20:12 GMT (1127kb,D)
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TL;DR: there is no mathematical certainty that every output value of common cryptographic hash functions is reachable, but for most that's overwhelmingly likely. A notable exception is double-SHA-256 (SHA256d) used in Bitcoin mining, where overwhelmingly likely there are some unreachable outputs.
For an idealized 256-bit hash, it becomes likely that every possible hash value $y$ is reached by hashing some $x$ when there are about $2^{264}$ possible values of $x$, including all 33-byte $x$; then odds that some $y$ is not covered become vanishingly small: less than one in $2^{b-264}$ if there are $2^b$ possible values of $x$; so while there is never a guarantee, it is practically certain that all $y$ are covered when we allow say 40-byte $x$: odds of the contrary are less than $2^{-56}$.
Argument: a hash can be idealized as a random function. The problem of if a random function covers all its image set is the coupon collector's problem. If the destination set has $n$ elements, the expected number of elements in the source set to fully cover the image set is $E(n)=n\log n+\gamma\,n+{1\over2}+O({1\over n})$ where $\gamma\approx0.577$ is the Euler–Mascheroni constant; and odds of not covering the input set after hashing $E(n)/\epsilon$ elements are less than $\epsilon$. Changing $n$ to $2^b$ for a $b$-bit hash, we get $E(2^b)\approx2^b(b\log(2)+\gamma)$ and $\begin{align}\log_2(E(2^b))&\approx b+\log_2\left(b+{\gamma\over\log(2)}\right)+\log_2(\log(2))\\&\approx b+\log_2(b+0.833)-0.528766\end{align}$
SHA-256 is a Merkle-Damgård hash using a compression function built per the Davies-Meyer construction. The problem of if all $y=\operatorname{SHA-256}(x)$ are reached reduces, up to 55 bytes hashed, to if all $\operatorname{Enc}_x(IV)$ are reached for a certain constant $IV$ and a certain block cipher $\operatorname{Enc}$ where $x$ is the key, and key scheduling encompasses the block padding. For an idealized block cipher the same analysis as for an idealized hash would hold, so we'd cover all values with overwhelming odds. SHA-256 uses an ARX block cipher and I see no reason why it would significantly differ from an idealized one, but that's a weak argument.
We can be more positive that we get full coverage when we allow two rounds (say, 100-byte hashes), because the problem is now if all $\operatorname{Enc'}_{x_1}(y_0)\boxplus y_0$ are reached where $x_1$ is the 36-byte second block, and $y_0$ is allowed to vary among the nearly-full set of 32-byte values reachable with the first round ( $\boxplus$ is 256-bit addition without carry across 32-bit boundaries). Still, there is no mathematical guarantee.
In Bitcoin mining, what's computed is $y=\operatorname{SHA-256}(\operatorname{SHA-256}(x))$. We are, for the outer hash, much below the coupon collector's bound and overwhelmingly above the birthday bound, therefore it is overwhelmingly likely that there are some values of $y$ not reached.
I second Dave Thomson's comment below.
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Feature #1994 LaTEX support in Wiki, Forums and Issues
Status: New Start date: 2008-10-06 Priority: Normal Due date: Assignee: - % Done:
0%
Category: Wiki Target version: - Resolution: Description
Latex support could be included especially in Wiki pages, so that Latex expressions could be displayed as images in Wiki. Specifically, I want to use Redmine Wiki as means of documentation tool for mathematical methods that are used within a multi-precision library. I could not find a way to display equations in Wiki (Textile does not support mathematical expressions as far as I know)
History #9 Updated by Andrzej Giniewicz almost 10 years ago
Anh Kỳ Huỳnh wrote:
Luiz Carlos Junior wrote:
For the implementation of mathematical models, it is essential to have Math/LaTeX support in Wiki.
I agree.
For one who will write the plugin please look at jsMath. It's fun and simple
I also think that jsMath is way to go!
#11 Updated by Peter Lawrence about 9 years ago
Regarding note 10. I've developed a plugin using jsmath and mathjax see http://github.com/PeterLawrence/redmine_jsequation_macros
Even thought this renders equations well in wiki pages and issues, it does not function for preview, pdf or atom feeds. It's only really a solution for when you have problems installing the latex plugin, for example on a windows based system. #12 Updated by Ling Li over 8 years ago Fileinit.rb added
I like Peter's plugin that uses MathJax (note-11). With MathJax, the displayed equations are beautiful and scalable. I also made changes regarding points raised in note-1 and note-2.
The new plugin (attached as a single file) defines a macro that enables the LaTeX math support via MathJax on a page-by-page basis. For example,
{{EnableLaTeXMath}} This is a math equation: \[ x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]The first line enables math on the page; after that one is free to write LaTeX equations. Compared to the suggestions in note-10 and note-11, I think this is simpler to use (no
{{latex(...)}}or
{{mathjax(...)}}every time an equation is needed),
doesn't have problems with line breaks or curly brackets in the equations, and naturally supports both inline-style and display-style equations.
And I believe it works for preview too.
#13 Updated by Phil Rutschman over 8 years ago
I successfully used Ling Li's plugin. While it does correctly working during preview the first time, subsequent previews do not. This appears to be because MathJax's onload handler is only invoked during initial configuration. Subsequent previews are running within the same DOM, and MathJax does not reconfigure itself.
I was able to work around this by explicitly enqueuing a typeset operation from within the script to dynamically load MathJax. This operation will fail (silently) in the usual case that MathJax is being loaded for the first time. There may well be a more elegant way to handle this, but it seems to work.
--- vendor/plugins/latex_math/init.rb 2011-04-11 00:48:43.000000000 -0700 +++ vendor/plugins/latex_math_preview_hack/init.rb 2011-04-11 00:48:29.000000000 -0700 @@ -48,6 +48,7 @@ else {script.text = config} document.getElementsByTagName(\"head\")[0].appendChild(script); + try { MathJax.Hub.Queue([\"Typeset\",MathJax.Hub]); } catch (err) {}; })(); //]]> </script>" #16 Updated by Michael Boratko almost 8 years ago
I have created a simple plugin which enabled MathJax for all pages (without typing {{EnableLaTeXMath}}) and also works on previews.
It is admittedly fairly crude in the way that it updates previews; essentially there is a jQuery function which watches the preview div for changes and makes a call to MathJax when it is changed. The inclusion of jQuery could be avoided, I'm only using it for this one purpose, but the bigger performance hit (though it's really theoretical, I don't notice any actual decrease in performance) is that jQuery has to poll repeatedly for changes to the preview box.Notes about the plugin / things you can modify:
I wanted the default LaTeX modifiers, so $ and $$ represent inline and display code respectively. You can change this, along with any other configuration options by editing the file within the hooks/redmine_latex_mathjax/ folder. If you don't care about LaTeX in the previews and want to avoid loading jQuery and having the constant polling, simply remove the last three javascript tags from the file within the hooks/redmine_latex_mathjax/ folder. #17 Updated by Michael Boratko almost 8 years ago
I forgot to mention, this plugin also styles text in the subject line of issues, comments of commits, basically any text on the whole site. The dollar signs are actually convenient for this, as they are not used frequently otherwise. If you do want to enter a dollar sign explicitly, you can always escape it by putting a backslash before it, but even this is not needed unless two dollar signs appear on the same line.
I really would like to remove the jQuery, however. If anyone knows of a hook that I am unaware of which allows you to modify the text which appears in the body of an issue, I would be able to strip out the jQuery and this plugin (for all uses I can think of anyway) would be complete.
#18 Updated by Mauro Calderara almost 8 years ago
The code seems to work in the preview but not after saving a wiki page. For some reason backslashes get quoted using another backslash, so '\sum' is turned into '\\sum' when saving the page. Any hints on why that is or what needs to be changed in order to avoid it?
#21 Updated by Mauro Calderara almost 8 years ago Filescreenshot.png added
Michael - yes, the code used is a fresh clone from your git repo. Other plugins installed are:
stuff_to_do_plugin timesheet_plugin redmine_tab
I uploaded a screenshot here. The markup used for this is:
$\sum x_i$
In the preview it is show correctly but whenever I save it, it looks like on the screenshot. I suspect something to escape the '\' when clicking save since when editing the page again the markup saved seems to be
$\\sum x_i$ #22 Updated by Michael Boratko almost 8 years ago
That is interesting. When I get a chance, I will install those plugins and see if I get the same result. What version of Redmine are you using?
If you could also test some other things for me, I would appreciate it. Try writing the markup described here:
$\int$ \sum x_i
(on the second one I have intentionally not wrapped it in dollar signs)
As you can see here, the wiki itself does not normally automatically escape backslashes, so I would be interested to see what your wiki does by default. If you could even uninstall my plugin and try writing
$\sum x_i$ to see if the backslash is escaped, that would be most helpful.
As a workaround (though I would really like to fix this entirely) you can try wrapping the entire thing within < notextile >< /notextile > tags (without the spaces of course).
#23 Updated by Mauro Calderara almost 8 years ago
Where should I look to see what these plugins do with respect to the formatting of the wiki/textile input? Here's what testing revealed:
The markup $\int$is again saved as $\\int$even though it displays correctly in the preview. With \sum x_i(without the enclosing dollar sign) it also gets stored as \\sum x_i The same happens with the notextile tags
The version of redmine installed here is 1.2 stable as of this morning, i.e. r8108
It seems that for some reason my wiki really just puts a '\' in front of every '\\' while the normal wiki doesn't do so (as one can see here). I just wonder what causes this. google didn't reveal anything so far, so I'll try asking in IRC.
#24 Updated by Mauro Calderara almost 8 years ago
Michael, after having been unsuccessful on IRC I continued playing around with the plugin and got it fixed. One reason for the phenomenon might have been that I used an old postgres-connector gem. I also uninstalled and reinstalled your plugin, now everything seems to work fine. Alas I can't tell whether it was the database connector or the reinstall but I suspect the database connector to have been the cuplrit.
Either way, for me things seem to work now. Thanks a lot for writing the plugin!
#25 Updated by Michael Boratko almost 8 years ago
Mauro, glad to hear that it works now. I also suspect the postgres-connector gem, especially since the backslash was escaped even outside the confines of the dollar signs, and escaping backslashes would not be out of the question with regard to a database connector. Good luck to you!
#27 Updated by Carsten Koenig about 7 years ago
Have installed Michael Boratkos plugin. Worked fine, but hat to add a try-catch to line 52 to make preview work:
File:
view_layouts_base_html_head_hook.rb in plugins/redmine_latex_mathjax/lib/redmine_latex_mathjax/hooks/ Line 52: - $('#preview').contentChange(function() { MathJax.Hub.Typeset(); } ); + $('#preview').contentChange(function() { try { MathJax.Hub.Typeset(); } catch (err) {}} );
Now it works fine. Thanks for the plugin, Michael.
#31 Updated by hannes wader almost 7 years ago Using MathJax¶
The most easy way to integrate Latex with Redmine is to integrate MathJax
You just need to paste these two lines into the file
app/layouts/base.html.erb
<script type="text/x-mathjax-config"> MathJax.Hub.Config({tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']]}}); </script> <script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML"> </script>
and it works perfectly!
I am not familiar with Plugin development, but if its possible to extend the main layout this way, that it would be a very nice, simple and small plugin. :)
#32 Updated by Michael Boratko almost 7 years ago
Billy T wrote:
Is this plugin working on 2.x?
It now works on 2.x https://github.com/holidayworking/redmine_latex_mathjax
#33 Updated by C Tibirna almost 7 years ago
Michael Boratko wrote:
Billy T wrote:
Is this plugin working on 2.x?
It now works on 2.x https://github.com/holidayworking/redmine_latex_mathjax
No, it doesn't. The original source is still https://github.com/process91/redmine_latex_mathjax , but it was not updated to 2.x. It only needs
(09:52:38) [tibirnao@walter]:redmine_latex_mathjax> git diffdiff --git a/init.rb b/init.rbindex d19303c..f3d5be1 100644--- a/init.rb+++ b/init.rb@@ -1,5 +1,5 @@ require 'redmine'-RAILS_DEFAULT_LOGGER.info 'Redmine LaTeX MathJax'+::Rails.logger.info 'Redmine LaTeX MathJax' Redmine::Plugin.register :redmine_latex_mathjax do name 'Redmine Latex MathJax'
I sent a pull request.
#35 Updated by C Tibirna over 6 years ago
And all of a sudden, MathJax stopped working on Redmine 2.3.1.
Other versions: Ruby 1.8.7, Rails 3.2.13, redmine_latex_mathjax 0.1.0.
I didn't notice immediately that it doesn't work anymore. It might be due to one of the updates in the last 7 months. There is no error in the redmine log, nor anywhere else I could look at.
Does anybody use MathJax successfully with Redmine 2.3.1?
#36 Updated by C Tibirna over 6 years ago
And all of a sudden, MathJax stopped working on Redmine 2.3.1.
Other versions: Ruby 1.8.7, Rails 3.2.13, redmine_latex_mathjax 0.1.0.
I didn't notice immediately that it doesn't work anymore. It might be due to one of the updates in the last 7 months. There is no error in the redmine log, nor anywhere else I could look at.
Does anybody use MathJax successfully with Redmine 2.3.1?
#38 Updated by Michael Boratko about 6 years ago
I believe this is fixed now. The problem was that I had used an unofficial link for the MathJax javascript (I'm not sure why I had done that). In any event, it seems this unofficial source no longer hosts MathJax. A user "crax" on GitHub had pointed this out and corrected the link to use the official CDN, and I just merged his changes so things should now be working. Please update the plugin and let me know if there is still an issue. Sorry for the latency in my reply.
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I am deeply confused.... For a thermally insulated ideal gas expanding freely, I think that $PV^{\gamma}=cnst$ must hold. Through the equation $PV=nRT$, it must be that $TV^{{\gamma}-1}=\rm constant$. Because the free expansion changes the volume of the gas, the temperature of the gas must change too. However, any book on thermodynamics says that there is no change in the internal energy of the gas in free expansion. Thus, the temperature of the ideal gas must not change. Why does this discrepancy occur? I am just totally stuck.....
$PV^{\gamma}=C$ applies only to the
reversible expansion of an ideal gas. Free expansion is not a reversible expansion because, even though the gas can be returned to its original state, its surroundings can not (without changing something else). The ideal gas law describes the behavior of ideal gases only for thermodynamic equilibrium states. In an irreversible expansion, the gas does not pass through a sequence of thermodynamic equilibrium states.
It depends on the system you are considering. Let's consider two different cases:
1) An ideal gas is inside an insulated piston which can move. You can write:
$$dU=\delta Q -\delta L \rightarrow dU= -\delta L \rightarrow nc_{v}dT=-dL$$ Now, if the process is reversible you have $dL=pdV$. And so: $$nc_{v}dT=-pdV$$ Also remember: $$pV=nRT \rightarrow p=\frac{nRT}{V} \rightarrow nc_{v}dT=-\frac{nRT}{V}dV$$ Then you integer in $dT$ and in $dV$ and obtain $pV^{\gamma}=kost$. Note that we have got this equation ONLY for reversible processes. Of course you can write this equation in some different ways by mixing it with the equation for ideal gases. If you draw this equation into the Clapeyron plane (p-V), you see that this is an hyperbole.
2)Joule free expansion. You have an ideal gas inside a tank; this first tank is linked to a second tank by a valve. In the second thank there is nothing. By opening the valve the gas will go inside the second tank and finally it will be inside the two tank. Let's analyze this process: $$dU=\delta Q -\delta L \rightarrow dU=\delta Q $$ Note that the work is zero and the process is irreversible. If you measure the temperature of the gas at the end of the process, you will note that it hasn't changed, and for this reason the gas hasn't exchange heat. So: $$\delta Q=dU=0$$ I mean only the second one is a free expansion. How you have seen during a free expansion the system is not insulated however its temperature doesn't change, and it doesn't exchange work. While the first case I've analyzed isn't a free expansion: the gas is insulated, it exchanges work and its temperature changes.
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Table of Contents
Sequence of Terms Divergence Criterion for Infinite Series
Recall from the Convergence and Divergence of Infinite Series page that an infinite series $\displaystyle{\sum_{k=1}^{\infty} a_k}$ is said to converge to the sum $s$ if the corresponding sequence of partial sums $(s_k)_{k=1}^{\infty}$ converges to $s$, and the series is said to diverge if it does not converge to any sum $s$.
We will now look at a very important theorem which provides us with divergence criterion for a series.
Theorem 1: If the series $\displaystyle{\sum_{k=1}^{\infty} a_k}$ converges then $\displaystyle{\lim_{k \to \infty} a_k = 0}$. Proof:Suppose that $\sum_{k=1}^{\infty} a_k = s$. Let $(s_k)_{k=1}^{\infty}$ denote the sequence of partial sums of this sequence. Then we have that: Taking the limit as $k \to \infty$ from both sides and we see that:
Corollary 1: If $\displaystyle{\lim_{k \to \infty} a_k \neq 0}$ then the series $\displaystyle{\sum_{k=1}^{\infty} a_k}$ diverges. Proof:This is simply the contrapositive statement of Theorem 1. $\blacksquare$
The corollary above is much more applicable than Theorem 1 in general. It gives us criterion for quickly determining whether many series are divergent. Note that the converse of Theorem 1 is not true in general though, that is, if $\lim_{k \to \infty} a_k = 0$ then we cannot necessarily determine whether the series $\displaystyle{\sum_{k=1}^{\infty} a_k}$ converges or diverges.
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I was thinking to solve this by computer programs but I prefer a solution.
How to obtain a list of 3 consecutive non square free positive integers? In general, how to obtain the same kind of list with $k$ elements? Thanks.
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I was thinking to solve this by computer programs but I prefer a solution.
How to obtain a list of 3 consecutive non square free positive integers? In general, how to obtain the same kind of list with $k$ elements? Thanks.
Let $n$ be the first number. It will work if we can arrange for the following: $$ n\equiv 0\pmod{4} $$ $$ n+1\equiv 0\pmod{9} $$ $$ n+2\equiv 0\pmod{25} $$ Using the Chinese Remainder Theorem, the first two congruences are equivalent to requiring that $n\equiv 8 \pmod{36}$. Combining this with the third congruence gives that $n\equiv 548\pmod{900}$. Thus, three such numbers are $548$, $549$, and $550$.
A similar algorithm works for $k$ consecutive square-free numbers.
$48 = 3 \times 2^4,49 = 7^2,50 = 2 \times 5^2$ would be the first example of a sequence with $3$ consecutive numbers. I found it simply by looking at them all until I found the first one. Given any set of distinct coprime numbers $a_1, \dots, a_k$, the Chinese Remainder Theorem ensures us of the existence of a solution to $n + i \equiv 0 \pmod {a_i}$, which gives you at least the existence of some large possibility for a sequence of non-square free consecutive integers.
For the info, the first consecutive non-square free integers I found were $(8,9)$, $(24,25)$ and $(44,45)$. Then I found $(48,49,50)$. Note that you can easily generate infinitely many such triples in the following manner : Since the prime divisors of $(48,49,50)$ that are the square parts of them are $2,5,7$, then the triple ($48 + k(2\times 5 \times 7)^2$, $49 + k(2 \times 5 \times 7)^2$, $50 + k (2 \times 5 \times 7)^2$) is also a sequence of three consecutive non-square free numbers ($k \in \mathbb N$).
Hope that helps,
See http://oeis.org/A045882 and references given there.
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Classical
acoustic theory
derives from fluid mechanics
, and centers on the mathematical
description of sound waves
. See acoustics
for the engineering
approach.
In approaching the description of a sound wave the mathematics never gives the whole story. The subtleties of thermodynamics are difficult enough to recommend a gradual familiarization with some related problems of vibration such as arise in mechanical sound production: motion of a spring[?] vibration of a string[?] equation of motion harmonic.
Besides the the math tools that we will use again, the preceding examples help inform the beginner's physical intuition with analogies to the periodic compression domains.
The propagation of sound waves in air can be modeled by an equation of motion and an equation of continuity. With some simplifications, they can be given as follows:
<math>\rho_0 \frac{\partial}{\partial t} \mathbf{v}(\mathbf{x}, t) + \nabla p(\mathbf{x}, t) = 0</math>
<math>\frac{\partial}{\partial t} p(\mathbf{x}, t) + \rho_0 c^2 \nabla \cdot \mathbf{v}(\mathbf{x}, t) = 0</math>
where <math>p(\mathbf{x}, t)</math> is the acoustic pressure and <math>\mathbf{v}(\mathbf{x}, t)</math> is the acoustic fluid velocity vector, <math>\mathbf{x}</math> is the vector of spatial coordinates <math>x, y, z</math>, <math>t</math> is the time, <math>\rho_0</math> is the static density of air and <math>c</math> is the speed of sound in air.
Related articles
transfer function sound pressure[?] acoustic impedance[?] acoustic resistance[?] law of gases frequency Fourier analysis
instrumental acoustics[?], music theory
voice production formant speech synthesis
All Wikipedia text is available under the terms of the GNU Free Documentation License
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Let's say we have a multivariate distribution $D$ which generates random $n$-dimensional vectors $x$ for us ($x \in R^n$). We know that the dimensions of vector $x$ are correlated, and that each dimension of $x$ has a mean of 0 and a standard deviation of 1. Now, let's say we have another random
vector $y$ (of shape $(n,1)$) defined as:$$y = \sum_{i=1}^{M} \alpha^{(i)} x^{(i)}\qquad\\x^{(i)} \sim D\\\alpha^{(i)} \sim \mathcal{N}(0,1)$$where $x^{(i)}$ is the $i$'th sampled vector from the distribution $D$ and has the shape $(n,1)$. We sample $M$ of these vectors ($i=1,2,3,...,M$) where $M>>n$. Also, $\alpha^{(i)}$ is a sampled scalar from the distribution $\mathcal{N}(0,1)$.
What would be the expectation and variance of the
dot product between x and y?$$E[x \cdot y]=?\\Var(x \cdot y)=?$$ Update1: In case this is too difficult to solve for any distribution $D$, I would still appreciate it if someone can solve this for when $D$ is a multivariate gaussian distribution with a full rank covariance matrix, and assume that $y = \dfrac{\sum_{i=1}^{M} \alpha_i x^{(i)}}{\sum_{i=1}^{M} \alpha_i}$.
Correct me if I'm wrong but I think these two assumptions would make things easier because we could treat $y$ as $y \sim D$.
Edit: Changed notations of question as suggested by whuber and mlofton. Thank you.
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Table of Contents
Inner Product Spaces Over the Field of Real Numbers
We will soon show that the set of all square Lebesgue integrable functions is an inner product space, but of course, we will first need to formally define an inner product space and inner product (which the reader is likely already familiar with).
Definition: Let $V$ be a vector space over the field $\mathbb{R}$. A two variable function $(\cdot, \cdot ): V \times V \to \mathbb{R}$ is called an Inner Product on $V$ if it satisfies the following properties: 1) $(x, x) \geq 0$ for all $x \in V$, and $(x, x) = 0$ if and only if $x = 0$. 2) $(x + y, z) = (x, z) + (y, z)$ for all $x, y, z \in V$. 3) $(\alpha x, y) = \alpha (x, y)$ for all $x, y \in V$ and for all $\alpha \in \mathbb{R}$. 4) $(x, y) = (y, x)$ for all $x, y \in V$. Note that we $(x, y)$ is allowed to equal to $0$ if $x \neq y$ and $x, y \neq 0$. We only require that the inner product of an element with ITSELF if equal to $0$ if and only if that element is $0$. Of course, inner products can be defined on vector spaces over other fields such as the set of complex numbers, $\mathbb{C}$.
Definition: If $V$ is a vector space over the field $\mathbb{R}$ and $(\cdot, \cdot)$ is an inner product of $V$ then $(V, (\cdot, \cdot))$ is said to be an Inner Product Space.
Perhaps the most familiar inner product space for the reader is the set $\mathbb{R}^n$ with the inner product $(\cdot, \cdot)$ defined for all vectors $\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n$ by:(1)
This inner product is common denoted $\mathbf{x} \cdot \mathbf{y}$ as opposed to the notation $(\mathbf{x}, \mathbf{y})$. If we consider the $\mathbb{R}^3$ and the vectors $\mathbf{x} = (1, 2, 3)$ and $\mathbf{y} = (2, -1, 4)$ then the inner product between these two vectors is:(2)
It is very easy to verify that this is indeed an inner product on $\mathbb{R}^n$ and it is given a special name - the Euclidean inner product on $\mathbb{R}^n$ or simply, the dot product.
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Uniformly Cauchy Sequences of Functions
Recall from the Pointwise Cauchy Sequences of Functions page that a sequence of functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is said to be pointwise Cauchy if for all $\epsilon > 0$ and for all $x \in X$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:(1)
In other words, the sequence of functions $(f_n(x))_{n=1}^{\infty}$ is pointwise Cauchy if for each $x_0 \in X$ the numerical sequence $(f_n(x_0))_{n=1}^{\infty}$ is Cauchy.
We can also similarly define a sequence of functions that is uniformly Cauchy.
Definition: A sequence of real-valued functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is said to be uniformly Cauchy if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then for all $x \in X$ we have that $\mid f_m(x) - f_n(x) \mid < \epsilon$.
We now look at a very important theorem which states that a sequence of real-valued functions is uniformly Cauchy if and only if it is uniformly convergent.
Theorem 1: A sequence of real-valued functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is uniformly Cauchy on $X$ if and only if it is uniformly convergent on $X$. Proof:Let $\epsilon > 0$ be given. $\Rightarrow$ Suppose that $(f_n(x))_{n=1}^{\infty}$ is uniformly Cauchy. Then for all $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ and for all $x \in X$ we have that: Let $f(x) = \lim_{n \to \infty} f_n(x)$. We will show that $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f$. For $m \geq N$ and for $n = m + k \geq N$ we have that $(*)$ holds for all $x \in X$ and so: So for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m \geq N$ and for all $x \in X$ we have that $\mid f_m(x) - f(x) \mid < \epsilon$, so $(f_n(x))_{n=1}^{\infty}$ is uniformly convergent. $\Leftarrow$ Suppose that $(f_n(x))_{n=1}^{\infty}$ is uniformly convergent to the limit function $f(x)$. Then for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ we have that for all $x \in X$ that: Take this $N$ and let $m, n \geq N$. Then by the triangle inequality we have that for all $x \in X$: So for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ and for all $x \in X$ we have that $\mid f_m(x) - f_n(x) \mid < \epsilon$, so $(f_n(x))_{n=1}^{\infty}$ is uniformly Cauchy. $\blacksquare$
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A
definite description is a denoting phrase in the form of "the X" where X is a noun-phrase or a singular common noun. The definite description is proper if X applies to a unique individual or object. For example: "the first person in space" and "the 42nd President of the United States of America", are proper. The definite descriptions "the person in space" and "the Senator from Ohio" are improper because the noun phrase X applies to more than one thing, and the definite descriptions "the first man on Mars" and "the Senator from Washington D.C." are improper because X applies to nothing. Improper descriptions raise some difficult questions about the law of excluded middle, denotation, modality, and mental content.
Contents Russell's analysis 1 Generalized quantifier analysis 2 Fregean analysis 3 Mathematical logic 4 See also 5 References 6 External links 7 Russell's analysis
France is currently a republic, and has no king. Bertrand Russell pointed out that this raises a puzzle about the truth value of the sentence "The present King of France is bald."
The sentence does not seem to be true: if we consider all the bald things, the present King of France isn't among them, since there is no present King of France. But if it is false, then one would expect that the negation of this statement, that is, "It is not the case that the present King of France is bald," or its logical equivalent, "The present King of France is not bald," is true. But this sentence doesn't seem to be true either: the present King of France is no more among the things that fail to be bald than among the things that are bald. We therefore seem to have a violation of the Law of Excluded Middle.
Is it meaningless, then? One might suppose so (and some philosophers have; see below) since "the present King of France" certainly does fail to refer. But on the other hand, the sentence "The present King of France is bald" (as well as its negation) seem perfectly intelligible, suggesting that "the Present King of France" can't be meaningless.
Russell proposed to resolve this puzzle via his theory of descriptions. A definite description like "the present King of France", he suggested, isn't a referring expression, as we might naively suppose, but rather an "incomplete symbol" that introduces quantificational structure into sentences in which it occurs. The sentence "the present King of France is bald", for example, is analyzed as a conjunction of the following three quantified statements:
there is an x such that x is currently King of France: ∃x[PKoF(x)] (using 'PKoF' for 'currently King of France') for any x and y, if x is currently King of France and y is currently King of France, then x=y (i.e. there is at most one thing which is currently King of France): ∀x∀y[[PKoF(x) & PKoF(y)] → y=x] for every x that is currently King of France, x is bald: ∀x[PKoF(x) → B(x)] (using 'B' for 'bald')
More briefly put, the claim is that "The present King of France is bald" says that some x is such that x is currently King of France, and that any y is currently King of France only if y = x, and that x is bald:
∃x[PKoF(x) & ∀y[PKoF(y) → y=x] & B(x)]
This is
false, since it is not the case that some x is currently King of France.
The negation of this sentence, i.e. "The present King of France is not bald", is ambiguous. It could mean one of two things, depending on where we place the negation 'not'. On one reading, it could mean that there is no one who is currently King of France and bald:
~∃x[PKoF(x) & ∀y[PKoF(y) → y=x] & B(x)]
On this disambiguation, the sentence is
true (since there is indeed no x that is currently King of France).
On a second reading, the negation could be construed as attaching directly to 'bald', so that the sentence means that there is currently a King of France, but that this King fails to be bald:
∃x[PKoF(x) & ∀y[PKoF(y) → y=x] & ~B(x)]
On this disambiguation, the sentence is
false (since there is no x that is currently King of France).
Thus, whether "the present King of France is not bald" is true or false depends on how it is interpreted at the level of logical form: if the negation is construed as taking wide scope (as in ~∃x[PKoF(x) & ∀y[PKoF(y) → y=x] & B(x)]), it is true, whereas if the negation is construed as taking narrow scope (with the existential quantifier taking wide scope, as in ∃x[PKoF(x) & ∀y[PKoF(y) → y=x] & ~B(x)]), it is false. In neither case does it lack a truth value.
So we do
not have a failure of the Law of Excluded Middle: "the present King of France is bald" (i.e. ∃x[PKoF(x) & ∀y[PKoF(y) → y=x] & B(x)]) is false, because there is no present King of France. The negation of this statement is the one in which 'not' takes wide scope: ~∃x[PKoF(x) & ∀y[PKoF(y) → y=x] & B(x)]. This statement is true because there does not exist anything which is currently King of France. Generalized quantifier analysis
Stephen Neale, among others, has defended Russell's theory, and incorporated it into the theory of generalized quantifiers. On this view, 'the' is a quantificational determiner like 'some', 'every', 'most' etc. The definite description 'the' has the following denotation (using lambda notation):
λf.λg.[∃x(f(x)=1 & ∀y(f(y)=1 → y=x) & g(x)=1)].
(That is, the definite article 'the' denotes a function which takes a pair of properties f and g to truth if, and only if there exists something that has the property f, only one thing has the property f, and that thing also has the property g.) Given the denotation of the predicates 'present King of France' (again PKoF for short) and 'bald (B for short)'
λx.[PKoF(x)] λx.[B(x)]
we then get the Russellian truth conditions via two steps of function application: 'The present King of France is bald' is true if, and only if ∃x[PKoF(x) & ∀y[PKoF(y) → y=x] & B(x)]. On this view, definite descriptions like 'the present King of France' do have a denotation (specifically, definite descriptions denote a function from properties to truth values—they are in that sense not syncategorematic, or "incomplete symbols"); but the view retains the essentials of the Russellian analysis, yielding exactly the truth conditions Russell argued for.
Fregean analysis
The Fregean analysis of definite descriptions, implicit in the work of Frege and later defended by Strawson (1950) among others, represents the primary alternative to the Russellian theory. On the Fregean analysis, definite descriptions are construed as referring expressions rather than quantificational expressions. Existence and uniqueness are understood as a presupposition of a sentence containing a definite description, rather than part of the content asserted by such a sentence. The sentence 'The present King of France is bald', for example, isn't used to claim that there exists a unique present King of France who is bald; instead, that there is a unique present King of France is part of what this sentence
presupposes, and what it says is that this individual is bald. If the presupposition fails, the definite description fails to refer, and the sentence as a whole fails to express a proposition.
The Fregean view is thus committed to the kind of truth value gaps (and failures of the Law of Excluded Middle) that the Russellian analysis is designed to avoid. Since there is currently no King of France, the sentence 'The present King of France is bald' fails to express a proposition, and therefore fails to have a truth value, as does its negation, 'The present King of France is not bald'. The Fregean will account for the fact that these sentences are nevertheless
meaningful by relying on speakers' knowledge of the conditions under which either of these sentences could be used to express a true proposition. The Fregean can also hold on to a restricted version of the Law of Excluded Middle: for any sentence whose presuppositions are met (and thus expresses a proposition), either that sentence or its negation is true.
On the Fregean view, the definite article 'the' has the following denotation (using lambda notation):
λf: ∃x(f(x)=1 & ∀y(f(y)=1 → y=x)).[the unique y such that f(y)=1]
(That is, 'the' denotes a function which takes a property f and yields the unique object y that has property f, if there is such a y, and is undefined otherwise.) The presuppositional character of the existence and uniqueness conditions is here reflected in the fact that the definite article denotes a partial function on the set of properties: it is only defined for those properties f which are true of exactly one object. It is thus undefined on the denotation of the predicate 'currently King of France', since the property of currently being King of France is true of no object; it is similarly undefined on the denotation of the predicate 'Senator of the US', since the property of being a US Senator is true of more than one object.
Mathematical logic
In much formal work, authors use a definite description operator symbolized using \scriptstyle\iota x. The operator is usually defined so as to reflect a Russellian analysis of descriptions (though other authors, especially in linguistics, use the \scriptstyle \iota operator with a Fregean semantics). Thus
\iota x(\phi x),
means "the unique \scriptstyle x such that \scriptstyle\phi x", and
\psi(\iota x(\phi x))
is stipulated to be equivalent to "There is exactly one \scriptstyle\phi and it has the property \scriptstyle\psi":
\exists x\forall y (\phi(y) \iff y=x \and \psi(y)) See also References Donnellan, Keith, "Reference and Definite Descriptions," in Philosophical Review 75 (1966): 281-304. Neale, Stephen, Descriptions, MIT Press, 1990. Ostertag, Gary (ed.). (1998) Definite Descriptions: A Reader Bradford, MIT Press. (Includes Donnellan (1966), Chapter 3 of Neale (1990), Russell (1905), and Strawson (1950).) Reimer, Marga and Bezuidenhout, Anne (eds.) (2004), Descriptions and Beyond, Clarendon Press, Oxford Russell, Bertrand, "On Denoting," in Mind 14 (1905): 479-493. Online text Strawson, P. F., "On Referring," in Mind 59 (1950): 320-344. External links
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You are currently browsing the tag archive for the ‘latex’ tag.
Amsrefs is a package for preparing bibliographic lists. If, like me, you use bibtex then you may find this post informative. If you enter your bibliographic items into the tex file manually, \emph-asizing titles and consulting Chicago Manual of Style to check whether the publisher should appear before or after the publication year then please have mercy on your co-authors and start eating with fork and knife. You spill typos all over the place.
So I have tried using amsrefs recently. Pro: Bibliographic items are entered in the tex file using a command similar to \bibitem, no need to keep a separate bib file and running bibtex. This is more convenient, especially if your folders are as messy as mine. Cons: Bibliographic items are not sorted, they appear in the pdf in the same order they appear in the tex file. Worse, all the entries in your bibliographic list appear in the pdf document, even those you don’t cite in the main text. The referee will search for their name, find the paper in the list of references, then search for the citation and get pissed when it’s not there: apparently you know about their paper but have nothing to say about it. Another con: You are in charge of capitalization of the journal and paper titles. Chicago Manual of Style anybody ?
Bottom line: I think I will return to bibtex. Am I missing anything ?
This happens a lot: I am reading a paper, as usual going directly to the results and skipping the introduction, related literature, discussion, preliminaries, formal model etc. And then there is some which I have no idea what it stands for. I would like to search for `\alpha’ in the pdf document, but if there is a way to do it then I have never heard about it.
So, imagine my delight when I heard of Springer’s LaTeX Search tool, which does something that I never even dared to wish — search in their database for an equation that contains a given latex code. Pretty awesome, isn’t it ?
I tried some arbitrary code
i\hbar\frac{\partial}{\partial t}\Psi=\hat H\Psi
(which translates to )
but apparently nobody has used this equation before.
So I tried something else: E=mc^2. Again no exact matches but this time there are a couple of similar results
Well, as Jeffrey Shallit said, it is, at least, a start.
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I know that $c_{d}=0.03 + 0.095c_{l}^2$.
What is the glide ratio?
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The glide ratio for a given angle of attack is the ratio of lift to drag. Both of these are proportional to the respective coefficients (with a proportionality constant of $\frac12\rho v^2A$), so the glide ratio is simply $c_{\mathrm L}/c_{\mathrm D}$.
Practically, what you're looking for may be the
maximal glide ratio. Finding the global maximum of$$\frac{c_{\mathrm L}}{c_{\mathrm D}} = \frac{c_{\mathrm L}}{0.03+0.095 c_{\mathrm L}^2} = \frac1{c_{\mathrm D}}\sqrt{\frac{c_{\mathrm D}-0.03}{0.095}}=\sqrt{\frac1{0.095}\frac1{c_{\mathrm D}}-\frac{0.03}{0.095}\Bigl(\frac1{c_{\mathrm D}}\Bigr)^2}$$ should be a simple matter of algebra -- for the last expression you don't even need calculus, just finding the apex of a parabola.
$C_D = C_{D0} + K \cdot C_L^2$ therefore $C_{D0} = 0.03$ and $K = 0.095$
Maximum lift-to-drag ratio occurs at the condition for minimum drag and at this condition $C_D = 2*C_{D0}$ and $C_L = \sqrt{C_{D0}/K}$.
Therefore $\max(L/D) = C_L/C_D = 0.5 \cdot \sqrt{1/(K \cdot C_{D0})} = 9.4$
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Recall that if we have a matrix that is in Row Echelon Form (REF), then we could use Gaussian Elimination, and if necessary, Back Substitution in order to solve a system of linear equations represented by an augmented matrix. We will now look at the similar method of Gauss-Jordan elimination by reducing a matrix to Reduced Row Echelon Form (RREF).
Gauss-Jordan Elimination
To solve a system by Gauss-Jordan Elimination, we will take a matrix and reduce it fully to RREF. For example consider the following system of linear equations:(1)
We can represent this system with the following augment matrix:(2)
When we reduce this matrix to RREF we obtain:(3)
Note that we can immediately tell the solutions to this system by the matrix, that is $(x_1, x_2, x_3) = (\frac{9}{7}, \frac{4}{7}, 0)$.
Example 1 Solve the following system of 3 linear equations of 4 variables given the following matrix $\begin{bmatrix} -9 & -5 & 5 & 3 & 1\\ 7 & 8 & 10 & -3 & -8\\ 5 & 6 & -4 & -9 & 9 \end{bmatrix}$ representing the system using Gauss-Jordan Elimination:
We will first reduce the augmented matrix of the system to Reduced-Row Echelon Form to obtain the following matrix (we used a computer algebra system (CAS) to obtain the solution since the results are a little messy):(4)
From this we can see that $x_4$ is our free variable, so let $x_4 = t$ where $t \in \mathbb{R}$. Therefore a general solution arises when:(5)
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Graph illustrating consumer (red) and producer (blue) surpluses on a supply and demand chart
In mainstream economics,
economic surplus, also known as total welfare or Marshallian surplus (named after Alfred Marshall), refers to two related quantities. Consumer surplus or consumers' surplus is the monetary gain obtained by consumers because they are able to purchase a product for a price that is less than the highest price that they would be willing to pay. Producer surplus or producers' surplus is the amount that producers benefit by selling at a market price that is higher than the least that they would be willing to sell for.
In Marxian economics, the term
surplus may also refer to surplus value, surplus product and surplus labour. Overview
Economist Paul A. Baran introduced the concept of "
economic surplus" to deal with novel complexities raised by the dominance of monopoly capital. With Paul Sweezy, Baran elaborated the importance of this innovation, its consistency with Marx's labor concept of value, and supplementary relation to Marx's category of surplus value. [1]
On a standard supply and demand diagram, consumer surplus is the area (triangular if the supply and demand curves are linear) above the equilibrium price of the good and below the demand curve. This reflects the fact that consumers would have been willing to buy a single unit of the good at a price higher than the equilibrium price, a second unit at a price below that but still above the equilibrium price, etc., yet they in fact pay just the equilibrium price for each unit they buy.
Likewise, in the supply-demand diagram, producer surplus is the area below the equilibrium price but above the supply curve. This reflects the fact that producers would have been willing to supply the first unit at a price lower than the equilibrium price, the second unit at a price above that but still below the equilibrium price, etc., yet they in fact receive the equilibrium price for all the units they sell.
Consumer surplus
Consumer surplus is the difference between the maximum price a consumer is willing to pay and the actual price they do pay. If a consumer would be willing to pay more than the current asking price, then they are getting more benefit from the purchased product than they initially paid. An example of a good with generally high consumer surplus is drinking water. People would pay very high prices for drinking water, as they need it to survive. The difference in the price that they would pay, if they had to, and the amount that they pay now is their consumer surplus. Note that the utility of the first few liters of drinking water is very high (as it prevents death), so the first few liters would likely have more consumer surplus than subsequent liters.
The maximum amount a consumer would be willing to pay for a given quantity of a good is the sum of the maximum price they would pay for the first unit, the (lower) maximum price they would be willing to pay for the second unit, etc. Typically these prices are decreasing; they are given by the individual demand curve. For a given price the consumer buys the amount for which the consumer surplus is highest, where consumer surplus is the sum, over all units, of the excess of the maximum willingness to pay over the equilibrium (market) price. The consumer's surplus is highest at the largest number of units for which, even for the last unit, the maximum willingness to pay is not below the market price
The aggregate consumers' surplus is the sum of the consumer's surplus for all individual consumers. This can be represented graphically as shown in the above graph of the market demand and supply curves.
Calculation from supply and demand
The consumer surplus (individual or aggregated) is the area under the (individual or aggregated) demand curve and above a horizontal line at the actual price (in the aggregated case: the equilibrium price). If the demand curve is a straight line, the consumer surplus is the area of a triangle:
CS = \frac{1}{2} Q_{\mathit{mkt}} \left( {P_{\mathit{max}} - P_{\mathit{mkt}}} \right)
Where P
mkt is the equilibrium price (where supply equals demand), Q mkt is the total quantity purchased at the equilibrium price and P max is the price at which the quantity purchased would fall to 0 (that is, where the demand curve intercepts the price axis). For more general demand and supply functions, these areas are not triangles but can still be found using integral calculus. Consumer surplus is thus the definite integral of the demand function with respect to price, from the market price to the maximum reservation price (i.e. the price-intercept of the demand function): CS = \int^{P_{\mathit{max}}}_{P_{\mathit{mkt}}} D(P)\, dP, where D(P_{\mathit{max}}) = 0.
This shows that if we see a rise in the equilibrium price and a fall in the equilibrium quantity, then consumer surplus falls.
Distribution of benefits when price falls
When supply of a good expands, the price falls (assuming the demand curve is downward sloping) and consumer surplus increases. This benefits two groups of people: Consumers who were already willing to buy at the initial price benefit from a price reduction; also they may buy more and receive even more consumer surplus, and additional consumers who were unwilling to buy at the initial price but will buy at the new price and also receive some consumer surplus.
Consider an example of linear supply and demand curves. For an initial supply curve S
0, consumer surplus is the triangle above the line formed by price P 0 to the demand line (bounded on the left by the price axis and on the top by the demand line). If supply expands from S 0 to S 1, the consumers' surplus expands to the triangle above P 1 and below the demand line (still bounded by the price axis). The change in consumer's surplus is difference in area between the two triangles, and that is the consumer welfare associated with expansion of supply.
Some people were willing to pay the higher price P
0. When the price is reduced, their benefit is the area in the rectangle formed on the top by P 0, on the bottom by P 1, on the left by the price axis and on the right by line extending vertically upwards from Q 0.
The second set of beneficiaries are consumers who buy more, and new consumers, those who will pay the new lower price (P
1) but not the higher price (P 0). Their additional consumption makes up the difference between Q 1 and Q 0. Their consumer surplus is the triangle bounded on the left by the line extending vertically upwards from Q 0, on the right and top by the demand line, and on the bottom by the line extending horizontally to the right from P 1. Rule of one-half
The
rule of one-half estimates the change in consumer surplus for small changes in supply with a constant demand curve. Note that in the special case where the consumer demand curve is linear, consumer surplus is the area of the triangle bounded by the vertical line Q=0, the horizontal line P = P_{mkt} and the linear demand curve. Hence, the change in consumer surplus is the area of the trapezoid with i) height equal to the change in price and ii) mid-segment length equal to the average of the ex-post and ex-ante equilibrium quantities. Following the figure above, \Delta CS = \frac{1}{2} \left( {Q_1 + Q_0 } \right)\left( {P_1 - P_0} \right)
where:
CS = consumers' surplus Q 0 and Q 1 are, respectively, the quantity demanded before and after a change in supply P 0 and P 1 are, respectively, the prices before and after a change in supply See also References ^ Baran, P.A. & Sweezy, P.M. (2012). "Some Theoretical Implications". Monthly Review. 64 (3). Further reading
This article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, E-Government Act of 2002.
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Flavon-induced Higgs lepton flavour violations
Presented by Dr. Venus KEUS
Content
The current experimental limit on Charge Lepton Flavour Violating (CLFV) processes allows the branching ratios of $h \to \tau \mu$ and $h \to \tau e$ processes to be of order 10%. Since such CLFV processes are forbidden in the Standard Model (SM), we aim to explain these processes by employing the Froggatt-Nielsen mechanism. This mechanism requires the addition of a scalar field called the flavon, singlet under SM gauge group which breaks spontaneously due to the flavon field acquiring a Vacuum Expectation Value (VEV). We show that the observed CLFV branching ratios can be explained due to the flavon filed mixing with the Higgs boson passing all experimental bounds.
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The most important thing is practice. As you calculate more and more limits, you'll start to develop some form of intuition regarding what method to try first, and the correct approach will probably come faster to you. You might see a problem and be reminded of some other limit you did before, so you'll try a similar method and maybe find it also works.
As you've said, appealing directly to the definition is sometimes the most straightforward way. Some inequalities which are often useful include $$|x|≤\sqrt{x^2+y^2} \ ; \ |\sin (u)|≤|u| \ ; \ 2|xy|≤x^2 +y^2 \ ; \ |x+y|≤|x|+|y| $$
A very common way to show that a given limit does NOT exist is to approach to the limit point trough several paths which conduct to different values. For instance, consider: $$\lim_{(x,y)\to (0,0)}{3y^2-2x^2 \over 7x^2+3y^2}$$
Putting $y=x$ we obtain: $$\lim_{x \to 0}{3x^2 -2x^2 \over 7x^2+3x^2}=\lim_{x \to 0}{x^2 \over 10x^2}={1 \over 10}$$
On the other hand, if we get closer to $(0,0)$ by moving along the horizontal axis, that is $y=0$ we get: $$\lim_{x \to 0}{-2x^2 \over 7x^2}={-2 \over 7}$$
It follows that the limit does not exist. What we have done is actually a shorthand for finding two sequences $p_n$ and $q_n$ such that $p_n \to (0,0)$ and $q_n \to (0,0)$ but $$\lim_{n \to \infty}f(p_n)\neq \lim_{n \to \infty}f(q_n)$$ This is usually a quicker method than applying the definition, but it only works for showing that the limit does not exist. Even if you were to try a thousand paths which all led to the same value, that would not show that the limit actually exists. If might, however, suggest you it does, and so it might be time to try and prove it using the definition.
Some commonly used paths include: $x=0$, $y=0$, $x=y$, $y=x^n$, $y=mx$, where $m\in \mathbb{R}$.
A third method that is also quite common is changing to polar coordinates. In the example you gave, put: $$\begin{cases} x=r\cos \theta \\ y=r\sin \theta \end{cases}$$ Then $\frac{xy(x^2-y^2)}{x^2+y^2}$ becomes $${r^4\cos\theta \sin \theta(\cos^2 \theta - \sin ^2 \theta) \over r^2(cos^2 \theta + \sin^2 \theta)}=r^2\cos\theta \sin \theta(\cos^2 \theta - \sin ^2 \theta)$$
This tends to $0$ as $r \to 0^+$, hence the desired limit equals $0$. What we have used is that the vector $(x,y)$ tends to $0$ if and only if the radius gets arbitrarily small, that is $r→0^+$ (it tends to zero from the right because the radius is always positive by definition). To see this, you can note that: $\left \|(x,y)\right \|=\sqrt{x^2+y^2}=\sqrt{r^2(\cos^2\theta +\sin^2\theta)}=r$.
Other, more obscure, techniques include using the mean value theorem or a Taylor expansion (be it one or several variables) to deduce some useful inequality or identity. Its probably better not to worry about these and try to solve them when you encounter them, as they are not very common.
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Please give me an intuitive explanation of 'implicit function theorem'. I read some bits and pieces of information from some textbook, but they look too confusing, especially I do not understand why they use Jacobian matrix to illustrate this theorem.
Let's use a simple example with only two variables. Assume there is some relation $f(x,y)=0$ between these variables (which is a general curve in 2D). An example would be $f(x,y) =x^2+y^2-1$ which is the unit circle in $\mathbb{R}^2$. Now you are interested to figure out the slope of the tangent to this curve at some point $x_0,y_0$ on the curve [with $f(x_0,y_0)=0$].
What you can do is to change $x$ a little bit $x = x_0 + \Delta x$. You are interested then how $y$ changes ($y= y_0 + \Delta y$); remember that we are interested in points on the curve with $f(x,y)=0$. Using Taylor expansion on $f(x,y)=0$ yields (up to lowest order in $\Delta x$ and $\Delta y$) $$f(x,y)= \partial_x f(x_0,y_0) \Delta x + \partial_y f(x_0,y_0) \Delta y =0.$$ The slope is thereby given by $$ \frac{\Delta y}{\Delta x} = - \frac{ \partial_x f(x_0,y_0)}{\partial_y f(x_0,y_0)}.$$ As $\Delta x \to 0$ higher order terms in the Taylor expansion (which we neglected) vanish and $\frac{\Delta y}{\Delta x}$ becomes the slope of the curve implicitly defined via $f(x,y)=0$ at $(x_0,y_0)$.
More variables and higher dimensional spaces can be treated similarly (using Taylor series in several variables). But the example above should provide you with enough intuition and insight to understand the 'implicit function theorem'.
The implicit function theorem really just boils down to this: if I can write down $m$ (sufficiently nice!) equations in $n + m$ variables, then, near any sufficiently nice solution point, there is a function of $n$ variables which give me the remaining $m$ coordinates of nearby solution points. In other words, I can, in principle, solve those equations and get the last $m$ variables in terms of the first $n$ variables. But (!) in general this function is only valid on some small set and won't give you all the solutions either.
Here's a concrete example. Consider the equation $x^2 + y^2 = 1$. This is a single equation in two variables, and for a fixed $x_0 \ne 1, y_0 \ne$ satisfying the equation, there is a function $f$ of $x$ such that $x^2 + f(x)^2 = 1$ for $y$ near $x_0$, and $f(x_0) = y_0$. (Explicitly, for $y_0 > 0$, $f(x) = \sqrt{1 - x^2}$, and for $y_0 < 0$, $f(x) = -\sqrt{1 - x^2}$.) Notice that the function doesn't give you all the solution points — but this isn't surprising, since the solution locus of this equation is a circle, which isn't the graph of any function. Nonetheless, I have basically solved the equation and written $y$ in terms of $x$.
The other answers have done a really good job explaining the implicit function theorem in the setting of multivariable calculus. There is a generalization of the implicit function theorem which is very useful in differential geometry called the rank theorem.
Rank Theorem:Assume $M$ and $N$ are manifolds of dimension $m$ and $n$ respectively. If $F : M \to N$ is a smooth map, $p \in M$ and $F_{*} : T_qM \to T_{F(q)}N$ has rank $k$ in a neighborhood of $p$, then there are coordinates $(x^1, \dots, x^k, \dots, x^m)$ centered around $p$ and $(v^1, \dots, v^n)$ centered around $F(p)$ such that in local coordinates, $F$ is given by the equation $$ F(x^1, \dots, x^k, \dots, x^m) = (x^1, \dots, x^k, 0, \dots, 0)$$
Essentially, the rank theorem tells us that if the total derivative of $F$ in a neighborhood of $p$ has rank $k$, then locally around $p$, we can think of $F$ as a linear map with rank $k$.
Here is an example of how you use the rank theorem in to prove a version of the implicit function theorem in differential geometry.
Assume $M$ is a manifold of dimension $n+k$ and $N$ is a manifold of dimension $k$. Assume $\Theta : M \to N$ is a smooth map and $\Theta_{*} : T_pM \to T_{\Theta(p)}N$ has rank $k$ (i.e $p$ is a regular point). Since maximal rank is an open condition, it follows that $F_*$ has rank $k$ in a neighborhood of $p$. By the rank theorem, there are coordinates $(x^1, \dots, x^{n+k})$ defined on the open set $U$ and centered around $p$ and coordinates centered around $\Theta(p)$ such that $$\Theta(x^1, \dots, x^{n+k}) = (x^1, \dots, x^k).$$ If we write $q = \Theta(p)$, then the above equation tells us that $$\Theta^{-1}(q) \cap U = \{ p \in U : x^{1}(p) = \dots = x^{k}(p) = 0 \}$$ which implies that $ (x^{k+1}, \dots, x^{k+n}) : \Theta^{-1}(q) \cap U \to \mathbb{R}^n$ is an open topological embedding. This defines an $n$-dimensional smooth structure on $\Theta^{-1}(q) \cap U$, and in coordinates, $\iota : \Theta^{-1}(q) \cap U \to U$ is given by $(x^{k+1}, \dots, x^{k+n}) \mapsto (0, \dots, 0, x^{k+1}, \dots, x^{k+n})$ which is smooth.
In summary, if $p$ is a regular point and $q = \Theta(p)$, then there are coordinates $(x^1, \dots, x^{n+k})$ for $M$ around $p$ such that $(x^{k+1}, \dots, x^{k+n})$ are coordinates for $\Theta^{-1}(q)$ around $p$ (once suitably restricted of course). This is kind of a differential geometers formulation of the normal implicit function theorem.
EDIT: In the comments, David pointed out that the statement of the rank theorem above was wrong. The reason is that rank $ \geq k $ is an open condition and rank $< k $ is a closed condition, which I didn't really understand when I wrote this answer 2 years ago. Things should be fixed now
There are basically two interpretations of (part of) the implicit function theorem (IMFT).
One is that it tells you under what conditions we have solutions to some equation of the form $f=0$, which has been mentioned in Zhen Lin's answer.
More precisely,
let $E$ be an open set of ${\Bbb R}^{n+m}={\Bbb R}^n\times{\Bbb R}^m$ and $f:E\to {\Bbb R}^n$ be a $C^1$ mapping such that $f(a,b)=0$, where $(a,b)\in {\Bbb R}^n\times {\Bbb R}^m$. Put $A=f'(a,b)$ and assume that $A_x$ is invertible, where $A_x$ denotes the restriction of the linear map $A$ on ${\Bbb R}^n\times\{0\}$. $\tag{*}$
Then we can "solve" the equation $f(x,y)=0$ for each of those $y$ near $b$, which means in an open neighborhood $W\subset{\Bbb R}^m$ of $b$, we have an implicitly defined function $g:W\to {\Bbb R}^n$ such that $f(g(y),y)=0$ for every $y\in W$. In particular, $f(g(b),b)=f(a,b)=0$. The word "solve" might be kind of confusing. After all, what does it mean by
solving an equation $f=0$? Basically it means one can write some variables in terms of others. On the other hand, the IMFT only tells you the existence of $g$ but not what $g$ looks like.
Another way to look at the IMFT is that it tells you when a set defined by $$ S=\{z\in{\Bbb R}^d|f(z)=0\} $$ is locally a graph of some function. (Here "being locally a graph" means one can find an open set $U\subset {\Bbb R}^d$ such that $U\cap S$ is a graph.) Under the assumption $(*)$, we would have the following conclusion
there exsit open sets $U\subset{\Bbb R}^{n+m}$ and $W\subset {\Bbb R}^m$, with $(a,b)\in U$ and $b\in W$ such that $$ U\cap S=G:=\{(g(y),y)\mid y\in W\}, $$ where $g$ is a function from $W$ to ${\Bbb R}^n$.
Note that we call $G$ the graph of the function $g$.
Using the notations in the second interpretation, the first one basically tells you $$ G\subset U\cap S. $$
Here is a nice explanation given by one of our Professor at an Undergraduate training camp.
Please see under the section geometry for the
Implicit function theorem.
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To what extent may the interest rate models be applied for modeling implied volatity?
The story:I was checking different stochastic option pricing models for being able to replicate implied volatility term strucure (namely its hump shape). While doing that, it came to my mind that interest rate term structure is roughthly the same thing:
Interest rate TS: $\frac{1}{h} E^P \int\limits_t^{t+h}r_t dt$
Volatility TS: $\frac{1}{h} E^Q \int_t\limits^{t+h}\sigma_t^2 dt$
Where P is a physical measure, Q - risk-neutral measure and volatility is derived from some option pricing model, for instance, Heston (under risk-neutral measure):
$dS_t = rS_tdt + \sigma_t S_t dW_t^r$
$d\sigma_t^2 = \kappa (\theta - \sigma_t^2) dt + \zeta \sigma_t dW_t^\sigma$
While conceptually interest rate and volatility of asset price are different things, they apper to be just the same thing in analytical sence.
The question:So the question arises: to what extent can we use interest rate models for modeling implied volatity?
Techically what I'm up to is:
We need a model for underlying asset which will perform hump shape in implied volatility term-structure?
Just take the interest rate model which allows for a hump in a yield curve, write $\sigma^2$ instead of $r$ and we're done.
Motivation:I thought that checking various option pricing models for being able to generate humps in TS was a smart research idea for my master thesis (it appeared to that there is a lack of literature on this topic). But if the results for interest rate models may be easily applied for volatility what I'm doing is futile since a bunch of literature exists on replicating all kinds of yield curves (hump, tilt etc).
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