text stringlengths 256 16.4k |
|---|
Thermodynamics Heat, Internal Energy and Work, First Law of Thermodynamics Heat is the energy that is transferred between a system and its surrounding and heat is a path dependent quantity. Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration. The energy due to molecular motion is internal KE U Kand heat due to molecular configuration U = U k+ U p Change in internal energy (du) is path independent and depends only on the initial and final state du = U f− U i The amount of external work done by a system it expands or contracts \tt \int dw=\int_{V_1}^{V_2}P\ dv work is path dependent. In a cyclic process work done is equal to the area under the cyclic. In this case W is +ve if the cycle is clockwise and W is −ve if the cycle is anticlock wise. According to the first law of thermodynamics heat given to a system (dQ) is equal to the sum of increase in its internal energy (du) and work done (dw) by the system dQ = du + dw dQ is +ve when heat is supplied and −ve when heat rejected. du is +ve when temperature increases and −ve when temperature decreases. dw is +ve when work done by system. dw is −ve when work done on system. First law of thermodynamics is a consequence of law of conservation of energy. View the Topic in this video From 7:00 To 57:40
Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers.
1. The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) is taken into account. It states that
ΔQ = ΔU + ΔW
2. The specific heat capacity of a substance is defined by
s = \frac{1}{m} \frac{\Delta Q}{\Delta T}
3. For gases when heat is absorbed and temperature changes ⇒ ΔQ = μCΔT
At constant pressure (ΔQ) P = μC PΔT At constant volume (ΔQ) V = μC VΔT
4. Internal energy of an ideal gas is totally kinetic and is given by U = U_{k} = \frac{3}{2} \mu RT and change in internal energy \Delta U = \frac{3}{2} \mu R \Delta T
5.
Isothermal Process : In this process, temperature of the system is kept constant during the changing of state. As, Q = nC_{iso}dT \Rightarrow C_{iso} = \frac{Q}{ndT} = \infty
6.
Adiabatic process: In this process, no exchange of heat takes place between the system and surroundings, i.e., dQ = 0 or Q = 0 Work done, W = \frac{nR(T_{1} - T_{2})}{\gamma - 1} = \frac{p_{1}V_{1} - p_{2}V_{2}}{\gamma - 1}
7.
Isobaric process : This is the process in which pressure is kept constant. (i) Molar heat capacity of the process is C p and dQ = nC pdT and dU = nC vdT. (ii) From the first law of thermodynamics, dQ = dU + dW and dW = pdV = nRdT ⇒ dQ = dU + nRdt Process equation is \frac{V}{T} = constant.
8.
Cyclic process: In a cyclic process, the system returns to its initial state. Efficiency of the cycle, \tt \eta = \frac{work \ done}{Heat \ Supplied} |
Edit: Added proof that the 4-simplex generates a free group.
Here's a fairly simple proof that the graph of twin dodecahedra is a tree, which is then used to prove that the 4-simplex generates a free group.
We'll start by working with a graph of twin 4-simplexes, rather than twin dodecahedra. (This is an approach that John Baez suggested.) Given any 4-simplex with five unit quaternions $q_i$ as its vertices, we will define five "twins" of this 4-simplex, relative to each of its five vertices, as the 4-simplexes with vertices:
$$q^{(m)}_i = q_m q_i^{-1} q_m$$
When $q_m = q_i$ we have:
$$q^{(m)}_m = q_m$$
So the simplex and its twin always share a vertex. If we then "twinned" the twin simplex relative to that shared vertex, we would be taken back to the
original simplex:
$$q^{(m)}_m (q^{(m)}_i)^{-1} q^{(m)}_m = q_m q_m^{-1} q_i q_m^{-1} q_m = q_i$$
We will define a map $G: \mathbb{Q}^4 \to \mathbb{Q}(\sqrt{5})^4$ as:
$$G(x) = \left(x_0, x_1 \sqrt{5}, x_2 \sqrt{5}, x_3 \sqrt{5}\right)$$
Now suppose we have unit quaternions:
$$q_a = G(a) \qquad q_b = G(b)$$
where $a, b \in \mathbb{Q}^4$, and $q_a$ and $q_b$ are
distinct vertices of the same 4-simplex, so that $q_a \cdot q_b = -\frac{1}{4}$. Then the quaternion multiplications of the twinning operation simplify to an extraordinary extent, and we have:
$$q_a q_b^{-1} q_a = -\frac{1}{2} G(a+2b)\qquad q_a \ne q_b$$
(If $q_a = q_b$, then of course the twin itself is simply equal to $q_a$.)
Let's further suppose that all $a_i, b_i$ are equal to integers divided by powers of 2, and that we write each unit quaternion $q_a$ and $q_b$ with a common denominator across all four components:
$$q_a = \frac{1}{2^{n_a}} G(r)$$$$q_b = \frac{1}{2^{n_b}} G(s)$$
where $r, s \in \mathbb{Z}^4$, and at least one of the $r_i$ and at least one of the $s_i$ are odd.
We then have:
$$q_a q_b^{-1} q_a = -\frac{1}{2^{n_a+n_b+1}} G(2^{n_b}r+2^{n_a+1}s)$$
If $n_a+1 \gt n_b$, we can cancel a factor of $2^{n_b}$, giving us:
$$q_a q_b^{-1} q_a = -\frac{1}{2^{n_a+1}} G(r+2^{n_a-n_b+1}s)$$
Since at least one of the $r_i$ is odd, no further cancellation is possible.
If $n_a+1 \lt n_b$, we can cancel a factor of $2^{n_a+1}$, giving us:
$$q_a q_b^{-1} q_a = -\frac{1}{2^{n_b}} G(2^{n_b-n_a-1}r+s)$$
Since at least one of the $s_i$ is odd, no further cancellation is possible.
If $n_a+1 = n_b$, we have:
$$q_a q_b^{-1} q_a = -\frac{1}{2^{n_b}} G(r+s)$$
Here it's possible that there will be further cancellation, depending on the details of the $r_i$ and $s_i$. However, it will turn out that we'll never need to make use of this case, because it only shows up if we move backwards through the graph.
So $n_t$, the power of 2 in the denominator of the twin's vertex, will be:
$$n_t=\begin{cases} max(n_a+1,n_b) & n_a+1 \ne n_b \\ n_b - ? & n_a+1 = n_b\end{cases}$$
In the original simplex, one of the vertices has $n=0$ as the power of 2 in its denominator, and the other four vertices have $n=2$. In its twins, we have the cases:
$$\begin{array}{cccc}n_a & n_b & q_a = q_b & n_t \\0 & 0 & T & 0 \\0 & 2 & F & 2 \\2 & 0 & F & 3 \\2 & 2 & F & 3 \\2 & 2 & T & 2 \end{array}$$
So the twin relative to the vertex $q_0=(1,0,0,0)$ has $n_t=0,2,2,2,2$ again, while those relative to the other vertices have $n_t=2,3,3,3,3$. But we only get $n_t=2$ when $q_a=q_b$, so as we extend the graph outwards we will never twin these twins relative to that vertex, since it would just take us back to the original simplex. This means that when we take twins of the twins, we don't encounter the case $n_b = n_a+1$, since in the first level twins we
never set $q_a$ equal to the vertex with $n_t=2$.
As we take twins of twins, any sequence of powers we find starting from the twin with powers $0,2,2,2,2$ will just match one we find starting from the original simplex, so wlog we can assume we extend the graph through a twin with powers $2,3,3,3,3$. The possibilities are then:
$$\begin{array}{cccc}n_a & n_b & q_a = q_b & n_t \\3 & 2 & F & 4 \\3 & 3 & F & 4 \\3 & 3 & T & 3\end{array}$$
Once again, we will never take higher-level twins with the vertex we get from $q_a=q_b$, so again we avoid the case $n_b = n_a+1$. And it's clear now that the pattern will continue like this: every $n$th level twin (that we did not reach via $q_0$) will have powers $n+1,n+2,n+2,n+2,n+2$, while those that we
did reach via $q_0$ will just look like $(n-1)$th level twins, with powers $n, n+1, n+1, n+1, n+1$.
If we choose a sequence $m_1, m_2, ... , m_M$ of vertices, with $m_k \ne m_{k+1}$, and apply the twinning operation repeatedly using the vertex with index $m_k$, for even $M$ we end up with:
$$q^{(m_1 m_2 ... m_M)}_i = q_{m_1} q_{m_2}^{-1} ... q_{m_M}^{-1} \: q_i \: q_{m_M}^{-1} ... q_{m_2}^{-1} q_{m_1}$$
while for odd $M$ we get:
$$q^{(m_1 m_2 ... m_M)}_i = q_{m_1} q_{m_2}^{-1} ... q_{m_M} \: q_i^{-1} \: q_{m_M} ... q_{m_2}^{-1} q_{m_1}$$
Now suppose that our graph was not a tree. Then there would be two different sequences of indexes like this, say $m_1, m_2, ... , m_M$ and $p_1, p_2, ... , p_P$ that resulted in the same simplex. We will define the rotation:
$$R(x) = q_{m_M}^{\pm 1} ... q_{m_2} q_{m_1}^{-1} \: x \: q_{m_1}^{-1} q_{m_2} ... q_{m_M}^{\pm 1}$$
where $q_{m_M}^{\pm 1} = q_{m_M}$ if $M$ is even, and $q_{m_M}^{-1}$ if $M$ is odd.
Applying $R$ to the first simplex will reduce it to either the original simplex, $q_i$, if $M$ is even, or to the twin relative to $q_0$, with vertices $q_i^{-1}$, if $M$ is odd.
If $M$ is even, applying $R$ to the second simplex will map it into the simplex that arises from the sequence $m_M, ..., m_1, p_1, ..., p_P$. If the two original sequences agree at the beginning, then $q_{m_1}^{-1}$ will cancel with $q_{p_1} = q_{m_1}$ in the string of vertex products, and so on for as long as the sequences are the same, but since the sequences are not identical, some non-empty sequence will remain.
If $M$ is odd, applying $R$ to the second simplex will map it into the
inverse of the simplex that arises from the sequence $m_M, ..., m_1, p_1, ..., p_P$, again with some possible cancellations.
We then have the result that either the original simplex (if $M$ is even) or its inverse (if $M$ is odd), both of which have powers $0,2,2,2,2$, is equal to some other simplex in the graph (if $M$ is even), due to some non-empty sequence of twinning operations, or the inverse of that simplex (if $M$ is odd).
The only simplex in the graph, besides the original, with powers $0,2,2,2,2$ is the one due to the index sequence $0$: the twin of the original with respect to $q_0$. But we know this is the
inverse of the original simplex, rather than being equal to it. Whether $M$ is even or odd, the claim ends up being that the original simplex and its inverse are equal, which is false. So our supposition that the graph is not a tree must be false.
Finally, we can relate this to the graph of dodecahedra as follows. Each simplex with vertices $q_i$ can be used to construct a dodecahedron, whose 20 vertices are:
$$d_{i,j} = q_i q_j^{-1} \qquad i \ne j$$
The dodecahedron constructed from a twin simplex has vertices:
$$d^{(m)}_{i,j} = q^{(m)}_i (q^{(m)}_j)^{-1} = q_m q_i^{-1} q_m (q_m q_j^{-1} q_m)^{-1} = q_m q_i^{-1} q_j q_m^{-1}$$
These share the 8 vertices where $i=m, j \ne m$ or $j=m, i \ne m$:
$$d^{(m)}_{m,j} = q_j q_m^{-1} = d_{j,m}$$$$d^{(m)}_{i,m} = q_m q_i^{-1} = d_{m,i}$$
which makes them twin dodecahedra. So the tree of twin simplexes gives rise to a graph of twin dodecahedra.
The reason we can't immediately claim that the
graph of dodecahedra is also a tree is that two non-identical simplexes can be used to construct the same dodecahedron. If we right-multiply all the vertices of a given simplex by the same unit quaternion $q_R$, rotating it into another simplex, then $q_R$ cancels out in the formula for the dodecahedron vertices.
So, could two of the simplexes in our tree be the same up to right multiplication by some quaternion $q_R$? We can show that this is impossible by exploiting the fact that all of these simplexes have their vertices in a particular subset of the golden field, $G(\mathbb{Q}^4) \subset \mathbb{Q}(\sqrt{5})^4$, in which the first coordinate is purely rational and all the other coordinates are rational multiples of $\sqrt{5}$.
Suppose we have three linearly independent, unit quaternions $q_i \in G(\mathbb{Q}^4)$, each of which is described in terms of a 4-vector of rationals, $a_i \in \mathbb{Q}^4$:
$$q_i = G(a_i) \qquad i=1,2,3$$
And suppose we pick some fourth quaternion, $p \in G(\mathbb{Q}^4)$, that we wish to map $q_1$ into by right multiplication, with $p = G(b)$ for some $b \in \mathbb{Q}^4$. The quaternion we need to right-multiply with to achieve this will be:
$$q_R = q_1^{-1} p$$
In general, $q_R$ will not belong to $G(\mathbb{Q}^4)$. The question we want to answer is: what restrictions are imposed on $p$ by requiring the images of the other $q_i$ under right multiplication:
$$q_i q_R = q_i q_1^{-1} p \qquad i=2,3$$
to lie in $G(\mathbb{Q}^4)$. Multiplying this out and setting the appropriate rational and irrational parts of the components of the product to zero, we obtain a set of eight linear equations in the rational parameters $b_j, j=0,1,2,3$:
$$\begin{array}{lcr} b_3 \left(a_{1,2} a_{2,1}-a_{1,1} a_{2,2}\right)+b_2 \left(a_{1,1} a_{2,3}-a_{1,3} a_{2,1}\right)+b_1 \left(a_{1,3} a_{2,2}-a_{1,2} a_{2,3}\right) & = & 0\\ b_3 \left(a_{1,0} a_{2,2}-a_{1,2} a_{2,0}\right)+b_2 \left(a_{1,3} a_{2,0}-a_{1,0} a_{2,3}\right)+b_0 \left(a_{1,2} a_{2,3}-a_{1,3} a_{2,2}\right) & = & 0\\ b_3 \left(a_{1,1} a_{2,0}-a_{1,0} a_{2,1}\right)+b_1 \left(a_{1,0} a_{2,3}-a_{1,3} a_{2,0}\right)+b_0 \left(a_{1,3} a_{2,1}-a_{1,1} a_{2,3}\right) & = & 0\\ b_2 \left(a_{1,0} a_{2,1}-a_{1,1} a_{2,0}\right)+b_1 \left(a_{1,2} a_{2,0}-a_{1,0} a_{2,2}\right)+b_0 \left(a_{1,1} a_{2,2}-a_{1,2} a_{2,1}\right) & = & 0\\ b_3 \left(a_{1,2} a_{3,1}-a_{1,1} a_{3,2}\right)+b_2 \left(a_{1,1} a_{3,3}-a_{1,3} a_{3,1}\right)+b_1 \left(a_{1,3} a_{3,2}-a_{1,2} a_{3,3}\right) & = & 0\\ b_3 \left(a_{1,0} a_{3,2}-a_{1,2} a_{3,0}\right)+b_2 \left(a_{1,3} a_{3,0}-a_{1,0} a_{3,3}\right)+b_0 \left(a_{1,2} a_{3,3}-a_{1,3} a_{3,2}\right) & = & 0\\ b_3 \left(a_{1,1} a_{3,0}-a_{1,0} a_{3,1}\right)+b_1 \left(a_{1,0} a_{3,3}-a_{1,3} a_{3,0}\right)+b_0 \left(a_{1,3} a_{3,1}-a_{1,1} a_{3,3}\right) & = & 0\\ b_2 \left(a_{1,0} a_{3,1}-a_{1,1} a_{3,0}\right)+b_1 \left(a_{1,2} a_{3,0}-a_{1,0} a_{3,2}\right)+b_0 \left(a_{1,1} a_{3,2}-a_{1,2} a_{3,1}\right) & = & 0\end{array}$$
This system has a 1-parameter family of solutions:
$$b_i = \lambda a_{1,i}$$
In other words, the only vectors we can map $q_1$ into by right multiplication are scalar multiples of itself, if we want the images of $q_2$ and $q_3$ to lie in $G(\mathbb{Q}^4)$. So right multiplication can't map one of the simplexes in the tree into another, and each distinct simplex belongs in a distinct equivalence class modulo right multiplication.
This means the dodecahedra created from distinct simplexes in the tree are distinct, and the graph of dodecahedra is a tree.
To prove that the 4-simplex generates a free group, suppose we have any finite product of positive or negative powers of the elements $\{q_1, q_2, q_3, q_4\}$. By inserting the element 1 in the form $q_0$ or $q_0^{-1}$ where necessary, we can write this product as:
$$p=q_{m_1} q_{m_2}^{-1} q_{m_3} ... q_{m_M}^{\pm 1}$$
for some sequence $m_1, m_2, ... , m_M$ with $m_k \ne m_{k+1}$. We then have the vertices of the dodecahedron constructed from the 4-simplex associated with this sequence being, for even $M$:
$$d^{(m_1 m_2 ... m_M)}_{i,j} = p q_i q_j^{-1} p^{-1}$$
and for odd $M$:
$$d^{(m_1 m_2 ... m_M)}_{i,j} = p q_i^{-1} q_j p^{-1}$$
If $p=1$ for some non-empty sequence, then this dodecahedron would also appear elsewhere in the tree, either as the dodecahedron constructed from the original simplex, or the one constructed from its twin relative to $q_0$. But this is impossible, so $p \ne 1$. |
Let\[\mathbf{v}_{1}=\begin{bmatrix}1 \\ 1\end{bmatrix},\;\mathbf{v}_{2}=\begin{bmatrix}1 \\ -1\end{bmatrix}.\]Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?
For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]
For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by\[ (f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S . \]
(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?
(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.
(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.
(d) Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as\[ V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} . \]
(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?
(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.
Let $\mathrm{P}_3$ denote the set of polynomials of degree $3$ or less with real coefficients. Consider the ordered basis\[B = \left\{ 1+x , 1+x^2 , x – x^2 + 2x^3 , 1 – x – x^2 \right\}.\]Write the coordinate vector for the polynomial $f(x) = -3 + 2x^3$ in terms of the basis $B$.
Let $V$ denote the vector space of $2 \times 2$ matrices, and $W$ the vector space of $3 \times 2$ matrices. Define the linear transformation $T : V \rightarrow W$ by\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a+b & 2d \\ 2b – d & -3c \\ 2b – c & -3a \end{bmatrix}.\]
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.
Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,\[T (f) (x) = x f(x).\]
Prove that $T$ is a linear transformation, and find its range and nullspace.
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.
Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$.
(a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.
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)
Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let\[V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}\]be a subset in $C[-1, 1]$.
(a) Prove that $V$ is a subspace of $C[-1, 1]$.
(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.
(c) Prove that\[B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}\]is a basis for $V$. |
The perfect number $6$ is in the middle of the primes $5$ and $7$. It is the only perfect number with this property because odd numbers are not in the middle of two twin primes and even perfect numbers have the form $2^{n-1}(2^n-1)\ ,\ n\ge 2$ with a prime $2^n-1$, so they are greater than $4$, but not divisible by $3$ for $n>2$, hence not the middle of two twin-primes.
The least multi-perfect number $N$, for which $N-1$ is prime (except $6$), is $$N=13794\ 54720=2^8\times3\times5\times7\times19\times37\times73$$
, which is $4$-perfect, that means $\frac{\sigma(n)}{n}=4$ , where $\sigma(n)$ is the sum of the divisors of $n$.
Is there a multi-perfect-number $N$ greater than $6$, such that $N-1$ and $N+1$ are both prime ? If yes, what is the minimal $N$ ? |
The definition of the adjoint operator of an operator $\hat A$ is
\begin{equation} (\vec x|\hat A \vec y) = (\hat A \vec x| \vec y) \quad \forall x, y \in \mathcal{H} \end{equation}
where $(\cdot|\cdot)$ is the inner product of a Hilbert space.
So before this definition came into play, I innocently tried to do the expected values of $\hat a$ and $\hat a^\dagger$ for the coherent state $\left|\alpha\right>$, getting:
\begin{equation} \left<\alpha\right|\hat a \left|\alpha\right> = \alpha \left<\alpha|\alpha\right> = \alpha \end{equation}
and
\begin{equation} \left<\alpha\right|\hat a^\dagger \left|\alpha\right> = \alpha^* \left<\alpha|\alpha\right> = \alpha^* \end{equation}
which I think are correct.
But then, I thought that the expected value of $\hat a$ and $\hat a^\dagger$ should be the same, since one is the transposed complex-conjugated operator of the other, and that
usually means that one is the adjoint of the other, but with the definition of adjoint operator and getting different values for the brakets I did before, it's clear that they're not their adjoint.
So my question is:
Did I suppose or do anything wrong? Is this definition I gave for adjoint operator not applicable to this case?
If it's correct, do $\hat a$ and $\hat a^\dagger$ have any operator corresponding to its adjoint, or if the transposed complex-conjugated operator of $\hat A$ is not the adjoint of $\hat A$ then $$\hat A$ doesn't have an adjoint operator?
If they do have an adjoint, which are they? |
Tagged: group Problem 343
Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.
Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of $G$. Problem 332
Let $G=\GL(n, \R)$ be the
general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices. Consider the subset of $G$ defined by \[\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.\] Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$. The subgroup $\SL(n,\R)$ is called special linear group Problem 322
Let $\R=(\R, +)$ be the additive group of real numbers and let $\R^{\times}=(\R\setminus\{0\}, \cdot)$ be the multiplicative group of real numbers.
(a) Prove that the map $\exp:\R \to \R^{\times}$ defined by \[\exp(x)=e^x\] is an injective group homomorphism.
Add to solve later
(b) Prove that the additive group $\R$ is isomorphic to the multiplicative group \[\R^{+}=\{x \in \R \mid x > 0\}.\] |
I am reading through McCandless "The ABCs of RBCs" this summer to get a preview of what I need to know for the coming Fall semester. It did not take long to find a statement that I can easily accept but cannot prove. On Page 9 after deriving the steady state condition of $(\delta + n)\bar{k} = \sigma A_0 f(\bar{k})$ in a zero-technological growth regime (where $\delta$ is depreciation and $n$ is growth rate of the labor force), the author says that "the stability of the positive stationary state can be seen from the equation $$ k_{t+1} = g(k_t) = \frac{(1-\delta)k_t + \sigma A_0 f(k_t)}{1+n}$$ Notice that between 0 and the positive $\bar{k}$, the function $g(k_t)$ is above the 45 degree line, so that $k_{t+1}$ is greater than k_t." He provides a standard-looking Solow model state diagram where I can verify this graphically, but not analytically.
I'm trying to prove every statement in the book in order to better familiarize myself with the details of macroeconomic theory before diving deeper, but I am absolutely stumped as to how I would prove that $k_{t+1} > k_t$ when $0 < k_t < \bar{k}$ and vice versa. I first tried manipulating the capital motion equation to prove it directly, and I could not reach a proof. The next strategy I have been trying to employ is to differentiate the capital motion equation and prove that its derivative is less than 1 at the steady state point, but it fails:
$$ \frac{\partial k_{t+1}}{\partial k_t} = \frac{(1-\delta) + \sigma A_0 f'(k_t)}{1+n} > \frac{(1-\delta) + \sigma A_0 f'(\bar{k})}{1+n} = \frac{(1-\delta)+(\delta+n)}{1+n} = 1 $$
Does anyone have an alternative strategy I can follow? It's been driving me nuts for two days. |
The central part of my GSoC project is about implementing the Jacobi-Davidson method natively in Julia, available in JacobiDavidson.jl. This method computes a few approximate solutions of the eigenvalue problem $Ax = \lambda Bx$ for large and sparse matrices $A$ and $B$. As it uses iterative solvers internally, much time has gone into improving IterativeSolvers.jl in general. Lastly, as iterative solvers are typically used with preconditioners, I have implemented the incomplete LU factorization for sparse matrices as well in ILU.jl.
The Jacobi-Davidson implementation is ready for use and can be applied to solving the (generalized) eigenvalue problem for non-Hermitian matrices. It’s similar to the
eigs method already available in Julia: it gives you a couple eigenvalues near a specified target in the complex plane.
At this point no official release has been tagged yet, as there is still some work to be done: hopefully the functions for the generalized and ordinary eigenvalue problem can largely be merged as they are very similar. Also, some optimizations for Hermitian problems should yet be implemented; lastly the methods do not yet support generic vectors and numbers.
We have been preparing a new release of IterativeSolvers.jl that improves speed and memory usage of solvers like GMRES, CG, Chebyshev iteration, stationary methods and the Power Method. Also two new methods MINRES and BiCGStab(l) are available, together with efficient implementations of stationary methods for Julia’s
SparseMatrixCSC matrix type.
Additionally the package has been upgraded to Julia 0.6 and the documentation has been restructured and improved.
Iterative methods for linear systems $Ax = b$ such as BiCGStab(l) might not converge quickly on any given matrix $A$. Typically convergence is best if the matrix $A$ is just a perturbation of the identity matrix. If that’s not the case, preconditioners might help: rather than solving $Ax = b$ you could try and solve $(PA)x = Pb$ where $P$ is a preconditioner such that $PA$ is closer to the identity matrix.
A perfect preconditioner would compute the full LU decomposition of $A$, but that’s too much computational work and would require way to much of memory. A well-known trick is to compute the LU factorization only approximately, by dropping small terms during the process. This is called incomplete LU or ILU.
As ILU for the
SparseMatrixCSC type was not yet available in Julia, I’ve implemented it based on the article “Crout versions of ILU for general sparse matrices” by Na Li, Yousef Saad and Edmond Chow.
The package ILU.jl is completely ready for use and is well tested.
Below you can find a few examples on how to use the packages I’ve been working on.
Let’s take a look at a toy example of the generalized eigenvalue problem $Ax = \lambda Bx$ where $A$ and $B$ are diagonal matrices of size $n \times n$ with $A_{kk} = \sqrt{k}$ and $B_{kk} = 1 / \sqrt{k}$. The eigenvalues are just the integers $1, \cdots, n$. Our goal is to find a few eigenvalues right in the interior of the spectrum near $n / 2$.
We implement the action of the matrices $A$ and $B$ matrix-free, using LinearMaps.jl:
using LinearMapsfunction myA!(y, x) for i = 1 : length(x) @inbounds y[i] = sqrt(i) * x[i] endendfunction myB!(y, x) for i = 1 : length(x) @inbounds y[i] = x[i] / sqrt(i) endendn = 100_000A = LinearMap{Complex128}(myA!, n; ismutating = true)B = LinearMap{Complex128}(myB!, n; ismutating = true)
The order of the matrices is
100_000. It turns out that if we target eigenvalues in the interior of the spectrum, iterative solvers used internally in Jacobi-Davidson might have trouble solving very indefinite systems.
In that case we should use a preconditioner for $(A - \tau B)$, where $\tau$ is the target. We will just use the exact inverse, which is a diagonal matrix $P$ with entries $P_{kk} = \sqrt{k} / (k - \tau)$. It can be implemented matrix-free and in-place:
import Base.LinAlg.A_ldiv_B!struct SuperPreconditioner{numT <: Number} target::numTendfunction A_ldiv_B!(p::SuperPreconditioner, x) for i = 1 : length(x) @inbounds x[i] *= sqrt(i) / (i - p.target) endend
Now we call Jacobi-Davidson with the
Near target and pass the preconditioner. We use GMRES as the iterative solver, but we could have used BiCGStabl(l) as well.
using JacobiDavidsonτ = 50_000.1 + 0imtarget = Near(τ)P = SuperPreconditioner(τ)schur, residuals = jdqz(A, B, gmres_solver(n, iterations = 10), preconditioner = P, target = target, pairs = 5, ɛ = 1e-9, min_dimension = 5, max_dimension = 10, max_iter = 200, verbose = true)
It converges to the eigenvalues 49999, 50000, 50001, 50002 and 50004:
50004.00000000014 + 3.5749921718300463e-12im49999.999999986496 - 7.348301591250897e-12im50001.00000000359 - 1.9761169705101647e-11im49998.99999999998 - 1.0866253642291695e-10im50002.00000000171 - 2.3559720511618024e-11im
It does not yet detect 50003, but that might happen when
pairs is increased a bit. As a result of our preconditioner, Jacobi-Davidson converges very quickly:
It’s not easy to construct a preconditioner this good for any given problem, but usually people tend to know what works well in specific classes of problems. If no specific preconditioner is availabe, you can always try a general one such as ILU. The next section illustrates that.
As an example of how ILU can be used we generate a non-symmetric, banded matrix having a structure that typically arises in finite differences schemes of three-dimensional problems:
n = 64N = n^3A = spdiagm((fill(-1.0, n - 1), fill(3.0, n), fill(-2.0, n - 1)), (-1, 0, 1))Id = speye(n)A = kron(A, Id) + kron(Id, A)A = kron(A, Id) + kron(Id, A)x = ones(N)b = A * x
The matrix $A$ has size $64^3 \times 64^3$. We want to solve the problem $Ax = b$ using for instance BiCGStab(2), but it turns out that convergence can get slow when the size of the problem grows. A quick benchmark shows it takes about 2.0 seconds to solve the problem to a reasonable tolerance:
> using BenchmarkTools, IterativeSolvers> my_x = @btime bicgstabl($A, $b, 2, max_mv_products = 2000);2.051 s> norm(b - A * my_x) / norm(b)1.6967043606691152e-9
Now let’s construct the ILU factorization:
> using ILU> LU = crout_ilu(A, τ = 0.1)> nnz(LU) / nnz(A)2.1180353639352374
Using the above drop tolerance $\tau$, our ILU factorization stores only about twice as many entries as the original matrix, which is reasonable. Let’s see what happens when we benchmark the solver again, now with ILU as a preconditioner:
> my_x = @btime bicgstabl($A, $b, 2, Pl = $LU, max_mv_products = 2000);692.187 ms> norm(b - A * my_x) / norm(b)2.133397068536056e-9
It solves the problem 66% faster to the same tolerance. There is of course a caveat, as constructing the preconditioner itself takes time as well:
> LU = @btime crout_ilu($A, τ = 0.1);611.019 ms
So all in all the problem is solved about 36% faster. However, if we have multiple right-hand sides for the same matrix, we can construct the preconditioner only once and use it multiple times. Even when the matrix changes slightly you could reuse the ILU factorization. The latter is exactly what happens in Jacobi-Davidson.
I would really want to thank my GSoC mentor Andreas Noack for the many discussions we had in chat and video calls. Also I would like to thank the Julia community in general for giving me a warm welcome, both online and at JuliaCon 2017. |
Zeno might appreciate this (where “appreciate” is used in the technical sense of “bang head against wall on account of”). Via Isabel comes Eric Schechter’s page of Common Errors in College Math. If you survived calculus, read through it and congratulate yourself on all the mistakes you don’t make anymore!
(See how optimistic I am?)
Schechter provides one of the most inspiring examples of getting the right answer through the wrong method that I’ve ever seen. The problem is to evaluate the following definite integral:
[tex]\int_0^{2\pi} \cos x\, dx. [/tex]
This is how our student started:
[tex]\int_0^{2\pi} \cos x\, dx = \left.\frac{\sin x}{x}\right|_0^{2\pi} = \frac{\sin 2\pi}{2\pi} – \frac{\sin 0}{0}.[/tex]
But wait, there’s more!
[tex]\frac{\sin 2\pi}{2\pi} – \frac{\sin 0}{0} = \sin – \sin = 0.[/tex]
And they say we can’t eliminate sin from the world. |
In number theory,
integer factorization or prime factorization is the decomposition of a composite number into smaller non-trivial divisors, which when multiplied together equal the original integer.
When the numbers are very large, no efficient, non-quantum integer factorization algorithm is known; an effort by several researchers concluded in 2009, factoring a 232-digit number (RSA-768), utilizing hundreds of machines over a span of two years.
[1] However, it has not been proven that no efficient algorithm exists. The presumed difficulty of this problem is at the heart of widely used algorithms in cryptography such as RSA. Many areas of mathematics and computer science have been brought to bear on the problem, including elliptic curves, algebraic number theory, and quantum computing.
Not all numbers of a given length are equally hard to factor. The hardest instances of these problems (for currently known techniques) are semiprimes, the product of two prime numbers. When they are both large, for instance more than two thousand bits long, randomly chosen, and about the same size (but not too close, e.g. to avoid efficient factorization by Fermat's factorization method), even the fastest prime factorization algorithms on the fastest computers can take enough time to make the search impractical; that is, as the number of digits of the primes being factored increases, the number of operations required to perform the factorization on any computer increases drastically.
Many cryptographic protocols are based on the difficulty of factoring large composite integers or a related problem—for example, the RSA problem. An algorithm that efficiently factors an arbitrary integer would render RSA-based public-key cryptography insecure.
Prime decomposition
This image demonstrates the prime decomposition of 864. A shorthand way of writing the resulting prime factors is 2
5 × 3 3
By the fundamental theorem of arithmetic, every positive integer has a unique prime factorization. (A special case for 1 is not needed using an appropriate notion of the empty product.) However, the fundamental theorem of arithmetic gives no insight into how to obtain an integer's prime factorization; it only guarantees its existence.
Given a general algorithm for integer factorization, one can factor any integer down to its constituent prime factors by repeated application of this algorithm. However, this is not the case with a special-purpose factorization algorithm, since it may not apply to the smaller factors that occur during decomposition, or may execute very slowly on these values. For example, if N is the number (2
521 − 1) × (2 607 − 1), then trial division will quickly factor 10 × N as 2 × 5 × N, but will not quickly factor N into its factors. Current state of the art
The most difficult integers to factor in practice using existing algorithms are those that are products of two large primes of similar size, and for this reason these are the integers used in cryptographic applications. The largest such semiprime yet factored was RSA-768, a 768-bit number with 232 decimal digits, on December 12, 2009.
[1] This factorization was a collaboration of several research institutions, spanning two years and taking the equivalent of almost 2000 years of computing on a single-core 2.2 GHz AMD Opteron. Like all recent factorization records, this factorization was completed with a highly optimized implementation of the general number field sieve run on hundreds of machines. Difficulty and complexity
If a large,
b-bit number is the product of two primes that are roughly the same size, then no algorithm has been published that can factor in polynomial time, i.e., that can factor it in time O( b ) for some constant k k. There are published algorithms that are faster than O((1+ε) ) for all positive ε, b i.e., sub-exponential.
The best published asymptotic running time is for the general number field sieve (GNFS) algorithm, which, for a
b-bit number n, is: O\left(\exp\left(\left(\begin{matrix}\frac{64}{9}\end{matrix} b\right)^{1\over3} (\log b)^{2\over3}\right)\right).
For an ordinary computer, GNFS is the best published algorithm for large
n (more than about 100 digits). For a quantum computer, however, Peter Shor discovered an algorithm in 1994 that solves it in polynomial time. This will have significant implications for cryptography if quantum computation is possible. Shor's algorithm takes only O( b 3) time and O( b) space on b-bit number inputs. In 2001, the first seven-qubit quantum computer became the first to run Shor's algorithm. It factored the number 15. [2]
When discussing what complexity classes the integer factorization problem falls into, it's necessary to distinguish two slightly different versions of the problem:
The function problem version: given an integer N, find an integer d with 1 < d < N that divides N (or conclude that N is prime). This problem is trivially in FNP and it's not known whether it lies in FP or not. This is the version solved by most practical implementations. The decision problem version: given an integer N and an integer M with 1 ≤ M ≤ N, does N have a factor d with 1 < d < M? This version is useful because most well-studied complexity classes are defined as classes of decision problems, not function problems. This is a natural decision version of the problem, analogous to those frequently used for optimization problems, because it can be combined with binary search to solve the function problem version in a logarithmic number of queries.
It is not known exactly which complexity classes contain the decision version of the integer factorization problem. It is known to be in both NP and co-NP. This is because both YES and NO answers can be verified in polynomial time given the prime factors (we can verify their primality using the AKS primality test, and that their product is N by multiplication). The fundamental theorem of arithmetic guarantees that there is only one possible string that will be accepted (providing the factors are required to be listed in order), which shows that the problem is in both
UP and co-UP. [3] It is known to be in BQP because of Shor's algorithm. It is suspected to be outside of all three of the complexity classes P, NP-complete, and co-NP-complete. It is therefore a candidate for the NP-intermediate complexity class. If it could be proved that it is in either NP-Complete or co-NP-Complete, that would imply NP = co-NP. That would be a very surprising result, and therefore integer factorization is widely suspected to be outside both of those classes. Many people have tried to find classical polynomial-time algorithms for it and failed, and therefore it is widely suspected to be outside P.
In contrast, the decision problem "is
N a composite number?" (or equivalently: "is N a prime number?") appears to be much easier than the problem of actually finding the factors of N. Specifically, the former can be solved in polynomial time (in the number n of digits of N) with the AKS primality test. In addition, there are a number of probabilistic algorithms that can test primality very quickly in practice if one is willing to accept the vanishingly small possibility of error. The ease of primality testing is a crucial part of the RSA algorithm, as it is necessary to find large prime numbers to start with. Factoring algorithms Special-purpose
A special-purpose factoring algorithm's running time depends on the properties of the number to be factored or on one of its unknown factors: size, special form, etc. Exactly what the running time depends on varies between algorithms.
An important subclass of special-purpose factoring algorithms is the
Category 1 or First Category algorithms, whose running time depends on the size of smallest prime factor. Given an integer of unknown form, these methods are usually applied before general-purpose methods to remove small factors. [4] For example, trial division is a Category 1 algorithm. General-purpose
A general-purpose factoring algorithm, also known as a
Category 2, Second Category, or Kraitchik family algorithm (after Maurice Kraitchik), [4] has a running time which depends solely on the size of the integer to be factored. This is the type of algorithm used to factor RSA numbers. Most general-purpose factoring algorithms are based on the congruence of squares method. Other notable algorithms Heuristic running time
In number theory, there are many integer factoring algorithms that heuristically have expected running time
L_n\left[1/2,1+o(1)\right]=e^{(1+o(1))(\log n)^{\frac{1}{2}}(\log \log n)^{\frac{1}{2}}}
in o and L-notation. Some examples of those algorithms are the elliptic curve method and the quadratic sieve. Another such algorithm is the
class group relations method proposed by Schnorr, [5] Seysen, [6] and Lenstra [7] that is proved under of the Generalized Riemann Hypothesis (GRH). Rigorous running time
The Schnorr-Seysen-Lenstra probabilistic algorithm has been rigorously proven by Lenstra and Pomerance
[8] to have expected running time L_n\left[1/2,1+o(1)\right] by replacing the GRH assumption with the use of multipliers. The algorithm uses the class group of positive binary quadratic forms of discriminant Δ denoted by G Δ. G Δ is the set of triples of integers ( a, b, c) in which those integers are relative prime. Schnorr-Seysen-Lenstra Algorithm
Given is an integer
n that will be factored, where n is an odd positive integer greater than a certain constant. In this factoring algorithm the discriminant Δ is chosen as a multiple of n, Δ= - dn, where d is some positive multiplier. The algorithm expects that for one d there exist enough smooth forms in G Δ. Lenstra and Pomerance show that the choice of d can be restricted to a small set to guarantee the smoothness result.
Denote by
P Δ the set of all primes q with Kronecker symbol \left(\tfrac{\Delta}{q}\right)=1. By constructing a set of generators of G Δ and prime forms f q of G Δ with q in P Δ a sequence of relations between the set of generators and f q are produced. The size of q can be bounded by c_0(\log|\Delta|)^2 for some constant c_0.
The relation that will be used is a relation between the product of powers that is equal to the neutral element of
G Δ. These relations will be used to construct a so-called ambiguous form of G Δ, which is an element of G Δ of order dividing 2. By calculating the corresponding factorization of Δ and by taking a gcd, this ambiguous form provides the complete prime factorization of n. This algorithm has these main steps:
Let
n be the number to be factored. Let Δ be a negative integer with Δ = - dn, where d is a multiplier and Δ is the negative discriminant of some quadratic form. Take the t first primes p_1=2,p_2=3,p_3=5, \dots ,p_t, for some t\in{\mathbb N}. Let f_q be a random prime form of G Δ with \left(\tfrac{\Delta}{q}\right)=1. Find a generating set X of G Δ Collect a sequence of relations between set X and { f ∈ q : q P Δ} satisfying: \left(\prod_{x \in X_{}} x^{r(x)}\right).\left(\prod_{q \in P_\Delta} f^{t(q)}_{q}\right) = 1 Construct an ambiguous form (a, b, c) that is an element f ∈ G Δ of order dividing 2 to obtain a coprime factorization of the largest odd divisor of Δ in which Δ = -4a.c or a(a - 4c) or (b - 2a).(b + 2a) If the ambiguous form provides a factorization of n then stop, otherwise find another ambiguous form until the factorization of n is found. In order to prevent useless ambiguous forms from generating, build up the 2-Sylow group S 2(Δ) of G(Δ).
To obtain an algorithm for factoring any positive integer, it is necessary to add a few steps to this algorithm such as trial division, and the Jacobi sum test.
Expected running time
The algorithm as stated is a probabilistic algorithm as it makes random choices. Its expected running time is at most L_n\left[1/2,1+o(1)\right].
[8] Applications Fast Fourier Transforms
The best-known fast Fourier transform algorithms depend upon integer factorization.
See also Notes ^ a b Kleinjung, et al (2010-02-18). "Factorization of a 768-bit RSA modulus". ^ LIEVEN M. K. VANDERSYPEN, et al (2007-12-27). "NMR quantum computing: Realizing Shor's algorithm". ^ Lance Fortnow (2002-09-13). "Computational Complexity Blog: Complexity Class of the Week: Factoring". ^ a b ^ Schnorr, Claus P. (1982). "Refined analysis and improvements on some factoring algorithms". Journal of Algorithms 3 (2): 101–127. ^ Seysen, Martin (1987). "A probabilistic factorization algorithm with quadratic forms of negative discriminant". Mathematics of Computation 48 (178): 757–780. ^ Lenstra, Arjen K (1988). "Fast and rigorous factorization under the generalized Riemann hypothesis". Indagationes Mathematicae 50: 443–454. ^ a b H.W. Lenstra, and C. Pomerance; Pomerance, Carl (July 1992). "A Rigorous Time Bound for Factoring Integers" (PDF). Journal of the American Mathematical Society 5 (3): 483–516. References Chapter 5: Exponential Factoring Algorithms, pp. 191–226. Chapter 6: Subexponential Factoring Algorithms, pp. 227–284. Section 7.4: Elliptic curve method, pp. 301–313. Donald Knuth. The Art of Computer Programming, Volume 2: Seminumerical Algorithms, Third Edition. Addison-Wesley, 1997. ISBN 0-201-89684-2. Section 4.5.4: Factoring into Primes, pp. 379–417. External links
msieve - SIQS and NFS - has helped complete some of the largest public factorizations known Video on YouTube explaining uniqueness of prime factorization using a lock analogy. A collection of links to factoring programs Richard P. Brent, "Recent Progress and Prospects for Integer Factorisation Algorithms", Computing and Combinatorics", 2000, pp. 3–22. download Manindra Agrawal, Neeraj Kayal, Nitin Saxena, "PRIMES is in P." Annals of Mathematics 160(2): 781-793 (2004). August 2005 version PDF [1] is a public-domain integer factorization program for Windows. It claims to handle 80-digit numbers. See also the web site for this program [2] Eric W. Weisstein, , November 8, 2005 MathWorld Headline News“RSA-640 Factored”
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.
Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles.
By using this site, you agree to the Terms of Use and Privacy Policy. World Heritage Encyclopedia™ is a registered trademark of the World Public Library Association, a non-profit organization. |
If symmetry conditions are met, FIR filters have a linear phase. This is not true for IIR filters.
However, for what applications is it bad to apply filters that do not have this property and what would be the negative effect?
Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It only takes a minute to sign up.Sign up to join this community
A linear phase filter will
preserve the waveshape of the signal or component of the input signal (to the extent that's possible, given that some frequencies will be changed in amplitude by the action of the filter).
This could be important in several domains:
coherent signal processing and demodulation, where the waveshape is important because a thresholding decision must be made on the waveshape (possibly in quadrature space, and with many thresholds, e.g. 128 QAM modulation), in order to decide whether a received signal represented a "1" or "0". Therefore, preserving or recovering the originally transmitted waveshape is of utmost importance, else wrong thresholding decisions will be made, which would represent a bit error in the communications system. radar signal processing, where the waveshape of a returned radar signal might contain important information about the target's properties audio processing, where some believe (although many dispute the importance) that "time aligning" the different components of a complex waveshape is important for reproducing or maintaining subtle qualities of the listening experience (like the "stereo image", and the like)
Let me add the following graphic to the great answers already given.
When a filter has
linear phase, then all the frequencies within that signal will be delayed the same amount in time (as described mathematically in Fat32's answer).
Any signal can be decomposed (via Fourier Series) into separate frequency components. When the signal gets delayed through any channel (such as a filter), as long as all of those frequency components get delayed the same amount, the same signal (signal of interest, within the passband of the channel) will be recreated after the delay.
Consider a square wave, which through the Fourier Series Expansion is shown to be made up of an infinite number of odd harmonic frequencies.
In the graphic above I show the summation of the first three components. If these components are all delayed the same amount, the waveform of interest is intact when these components are summed. However, significant group delay distortion will result if each frequency component gets delayed a different amount in time.
The following may help give additional intuitive insight for those with some RF or analog background.
Consider an ideal lossless broadband delay line (such as approximated by a length of coaxial cable), which can pass wideband signals without distortion.
The transfer function of such a cable is shown in the graphic below, having a magnitude of 1 for all frequencies and a phase negatively increasing in direct linear proportion to frequency. The longer the cable, the steeper the slope of the phase, but in all cases "linear phase".
This makes sense; the phase delay of 1 Hz signal passing through a cable with a 1 second delay will be 360°, while a 2 Hz signal with the same delay will be 720°, etc...
Bringing this back to the digital world, $z^{-1}$ is the z-transform of a 1 sample delay (therefore a delay line), with a similar frequency response to what is shown, just in terms of H(z); a constant magnitude = 1 and a phase that goes linearly from $0$ to $-2\pi$ from f = 0 Hz to f = fs (the sampling rate).
Just to add to what's already been said, you can see this intuitively by looking at the following sinusoid with monotonically increasing frequency.
Shifting this signal to the right or left will change its phase. But note also that the phase change will be larger for higher frequencies, and smaller for lower frequencies. Or in other words, the phase increases linearly with frequency. Thus a constant time shift corresponds to a linear phase change in the frequency domain.
The answer to this question is already been explained clearly in the previous replies. Yet I wish to give it a try to present a mathematical interpretation of the same
Consider a linear time invariant System whose frequency response is governed by $H(w)$.
i.e if the input to this system is $e^{jw_{0}t}$ the output will be $H(w_{0})e^{jw_{0}t}$
Here $H(w_{0})$ is a complex number which has a phase component denoted by $arg(H(w))$ and magnitude component denoted by $|H(w)|$
if the system has linear phase response then $$arg(H(w)) = Kw$$ where $K$ is a constant
If the phase is linear the output of the system for the input $e^{jw_{0}t}$ will be $$y(t) = |H(w)|*e^{jw_{0}t + jKw_{0}}$$ $$ = |H(w)|*e^{jw_{0}(t + K)}$$ which is nothing but a delayed version of input with some scaling applied.
So if the phase is linear then all the frequency components of the signal will undergo the same amount of delay in time-domain which results in shape preservation.
The essence and importance of linear phase property lies in the definition and the effect of
group delay $$\tau(\omega) = - \frac {d\phi(\omega)}{d\omega}$$ on the applied signal $x[n]$, where $\phi(\omega)$ is the phase response of the filter; (phase of its frequency response).
Assume that a filter, with a fixed group delay of $n_0$ samples, is applied a
narrowband input signal $x[n]$. Then the output signal will be (approximately) of the form $y[n] = K x[n-n_0]$ where $K$ is the filter gain evaluated at the center frequency of the narrowband input signal $x[n]$. This means that the input signal will be weighted and shifted intact as a whole by the group delay of the filter. And this can only happen when the group delay is independent of the frequency $\omega$. And this will be the case if the underlying filter has linear phase (or generalized linear phase). Note that if the input signal is of broadband type; i.e., its minimum and maximum frequencies are far from its center frequency, then the approximation is not valid and eventhough the group delay would still be the same for each sinusoidal component in the signal, their relative output amplitudes will differ by the frequency dependent filter gain $K(w)$.
Then what's the effect of a filter with nonlinear phase (or frequency dependent group delay) on the input signal? A simple example would be a complicated input signal considered as a sum of multiple wavepackets at different center frequencies. After the filtering, each packet with a particular center frequency will be
shifted (delayed) differently due to frequency dependent group delay. And this will be resulting in a change in the time-order (or space order) of those wave packets, sometimes drastically, depending on how nonlinear the phase is, which is called as dispersion in communications terminalology. Not only the composite waveshape, but also some event orders may be lost. This kind of dispersive channels have severe effects such as ISI (inter symbol interference) on transmitted data.
This property of linear phase filters, therefore, is also known as
waveform-preserving property, which is applicable to narrowband signals in particular. An example where waveform is important, other than ISI as mentioned above, is in processing of images, where the Fourier transform phase information is of paramount importance compared to magnitude of Fourier transform, for intelligibility of the image. The same, however, cannot be said for perception of sound signals due to a different kind of sensitivity of the ear to the stimulus.
I will just put a summary for these great answers mentioned above: |
Suppose $M$ is a compact manifold so that it admits an embedding $f:M\to N\times\mathbb{R}^l$ with a splitting of the normal bundle as $\nu_f\simeq\nu\oplus\epsilon^l$ where $\nu$ is some $k$-dimensional vector bundle and $\epsilon^l$ is the trivial $l$-dimensional bundle. The mutli-compression theorem of Rourke and Sanderson, implies that $f$ is regularly homotopy to an embedding $g:M\to N\times\mathbb{R}^l$ so that the composition
$$i:M\to N\times\mathbb{R}^l\to N$$
is an immersion with $\nu_i\simeq\nu$. For the theorem, see
Rourke and Sanderson. The compression theorem I. Geom. Topol. 5, 399-429 (2001).
or Theorem 2.2 of
Eccles and Grant. Bordism groups of immersions and classes represented by self-intersections. Algebr. Geom. Topol. 7, 1081-1097 (2007).
My question is the following. It seems to me that having $\nu_f\simeq\nu\oplus\epsilon^l$ does not necessarily imply that the trivial part if all coming from the second component of $f$ and the theorem says that we may continuously, change $f$ so that stays in the same homotopy class and all trivial parts comes from the second component of the deformed map $g$, so that collapsing all of them will not damage $\nu$ ? Does this interpretation make sense? |
Yes. This can be done in $O(d)$ time. As a nice consequence, if we have a balanced binary tree, then $d = O(\lg n)$, so range-deletion can be done in $O(\lg n)$ time, regardless of how large the range is.
This is hard to explain without a picture, but the key insight is this: it suffices to delete a set $S$ of nodes, where $S$ can be chosen so that $|S| \le 2d$. (This will of course delete all of the descendants of every node in $S$.) Since $|S| = O(d)$, the running time to delete the nodes in $S$ will be $O(d)$. In other words, the interval $[i,j]$ can be expressed as a union of $O(d)$ subtrees. Roughly speaking, the set $S$ will be chosen by taking a subset of the siblings of the nodes on the path from $i$ or $j$ to the root.
In particular, we will define
$$S = \{n : L(n) \subseteq [i,j] \text{ and } L(\text{parent}(n)) \not\subseteq [i,j]\},$$
where $L(n)$ is the set of leaves that are descendants of $n$. I will show first that deleting $S$ has the desired effect, and second that that $|S| \le 2d$, from which it will follow that the range-deletion can be done in $O(\lg d)$ time.
Lemma 1. Deleting $S$ will have the effect of deleting leaves $i,i+1,\dots,j-1,j$ but not any of the other leaves.
Proof. Consider any leaf $\ell$ with $i \le \ell \le j$. Then $\ell$ is a descendant of some $n \in S$: if $\ell \notin S$, then some ancestor of $S$ must be in $S$ (imaging walking up from $\ell$ along the path towards the root; consider the last ancestor $n$ of $S$ such that $L(n) \subseteq [i,j]$; since $L(\ell) =\{\ell\} \subseteq [i,j]$, such an ancestor must exist). Conversely, a leaf $\ell$ is deleted only if it is a descendant of some $n \in S$; but by construction, $L(n) \subseteq [i,j]$ and $\ell \in L(n)$, so it follows that if leaf $\ell$ is deleted, $\ell \in [i,j]$.
Now let's show that $|S| \le 2d$. Let $P(\ell)$ denote the set of nodes along the path from $\ell$ to the root. Let $Q(\ell)$ denote the set of siblings of the nodes in $P(\ell)$ ($\ell$'s sibling, $\ell$'s parent's sibling, $\ell$'s parent's parent's sibling, and so on). Then I claim:
Lemma 2. $|Q(\ell)| \le d$ for all $\ell$.
Proof. Since the tree has height $d$, $|P(\ell)| \le d$. Now $|Q(\ell)| \le |P(\ell)| \le d$.
Lemma 3. $S \subseteq Q(i) \cup Q(j)$.
Proof. Suppose $n \in S$, so that $L(n) \subseteq [i,j]$ but $L(\text{parent}(n)) \not\subseteq [i,j]$. Let $m$ be $n$'s sibling. Note that $L(\text{parent}(n)) = L(n) \cup L(m)$. It follows that we must have $L(m) \not\subseteq [i,j]$ (otherwise $L(\text{parent}(n)) \subseteq [i,j]$ and $n$ wouldn't be in $S$). Also $L(n) \cap L(m) = \emptyset$ and $L(n)$ and $L(m)$ are adjacent intervals ($L(n) \cup L(m)$ is itself an interval; there is no gap in between $L(n)$ and $L(m)$). It follows that either $i \in L(m)$ or $j \in L(m)$, i.e., $m$ must be an ancestor of either $i$ or $j$, i.e., either $m \in Q(i)$ or $m \in Q(j)$.
Lemma 4. $|S| \le 2d$.
Proof. Follows from Lemma 2, Lemma 3, and the union bound.
Caveat: I haven't checked whether rebalancing can also be done in $O(d)$ time. For instance, for a balanced binary tree like an AVL tree or a red-black tree, we have $d=O(\lg n)$, and it would be nice if we could delete the entire range in $O(\lg n)$ time. However, it's not clear whether this is possible. While we can indeed delete the nodes in $S$ in $O(\lg n)$ time, I haven't checked whether it's possible to do all necessary subsequent rebalancing operations in $O(\lg n)$ time. Rebalancing is certainly doable in $O((\lg n)^2)$ time, but I don't know whether it can be done in $O(\lg n)$ time. |
This question already has an answer here:
$$\int_0^{\infty} \frac{\sin^2(x)}{x^2(x^2+1)} dx$$
The integral is equals with $\frac{\pi}{4}+\frac{\pi}{4e^2}$, but i can't prove it.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
This question already has an answer here:
$$\int_0^{\infty} \frac{\sin^2(x)}{x^2(x^2+1)} dx$$
The integral is equals with $\frac{\pi}{4}+\frac{\pi}{4e^2}$, but i can't prove it.
Let $$I(a)=\int_{0}^{\infty}\frac{\sin^2 ax}{x^2(x^2+1)}dx=\int_0^\infty \frac{\sin^2 ax}{x^2}dx-\int_0^\infty \frac{\sin^2ax}{(1+x^2)}dx\\=\frac{\pi a}{2}-\int_0^\infty\frac{\sin^2 ax}{1+x^2}dx$$ Here I have used the result $$\int_{0}^\infty \frac{\sin^2 x}{x^2}dx=\pi/2$$ Then, $$dI/da=\int_{0}^\infty \frac{\sin 2ax}{x(x^2+1)}\\\implies d^2I/da^2=2\int_{0}^\infty \frac{\cos 2ax}{x^2+1}dx\\=2\int_{0}^\infty\frac{1-2\sin^2ax}{1+x^2}dx\\=2\pi/2-4\int_{0}^\infty\frac{\sin^2 ax}{1+x^2}dx\\=\pi-4(a\pi/2-I(a))\\\implies d^2I/da^2=4I-2a\pi+\pi$$ The CF is $$C_1 e^{2a}+C_2e^{-2a}$$ and the PI is $$\pi\frac{1}{D^2-4}(1-2a)=\frac{1}{D-2}\pi e^{-2a}\int (1-2a)e^{2a}da\\=\frac{1}{D-2}\pi e^{-2a}(1/2 e^{2a}-ae^{2a}+1/2e^{2a})=\frac{1}{D-2}\pi (1-a)\\=\pi e^{2a}\int e^{-2a}(1-a)da=\pi(-1/2e^{-2a}+a/2e^{-2a}+1/4e^{-2a})e^{2a}\\=\pi(a/2-1/4)$$ So, $$I(a)=C_1e^{2a}+C_2e^{-2a}+\pi(a/2-1/4)$$Now, $$I(0)=0, dI(0)/da=0\\\implies C_1+C_2=\pi/4\\2a(C_1-C_2)=-\pi/2\\\implies C_1=\pi/8-\pi/8a, \ C_2=\pi/8+\pi/8a$$ Hence, the desired integral is $$I(1)=C_1(a=1)e^2+C_2(a=1)e^{-2}+\pi/4=\pi/4+\pi/4e^2$$ |
Yes, it is possible. What follows isn't a completely rigorous proof, but you should be able to make one from that.
The basic idea is to construct a set $A$ such that along one subsequence, $\lim |A \cap I_{a_n}|/a_n = 1$ and along another subsequence, $\lim |A \cap I_{b_n}|/b_n = 0$. We'll do this by adding or skipping consecutive "chunks" of integers.
Start by adding $1$ to $A$. We're at density $1$. Then skip $2$. For $n = 2$, we have $|A \cap I_2|/2 = 1/2$. Now add $3$ and $4$, so that $|A \cap I_4|/4 = 3/4$. Now skip $5$ through $12$, so that $|A \cap I_{12}|/12 = 3/12 = 1/4$.
This should give you an idea of how the following works:
"Add" enough integers so that you're back to density $\geq (2^n - 1)/2^n$ Then "skip" enough integers so that you're below density $\leq 1/2^n$.
Repeat the two steps in order. Along the two subsequences constructed, the limit of the density is either 1 or 0. |
Is energy in every possible inertial frame equal? If not, does is not violate energy conservation?
You are confusing between conserved and invariant quantities.
A quantity is said to be
conserved if it is a constant of motion, i.e. does not depend on time.
A quantiy is said to be
invariant if it remains the same when transformed between different frames of reference.
In Newtonian mechanics energy is not an invariant quantity, but a conserved one. It means that it might change when you move between reference frames. But if you choose a specific frame, then it will be the same in all times.
In special relativity there is, however, an invariant quantity that is related to the energy. In relativity energy and momentum come together in the form of the four-momentum $$p^{\mu}=\begin{pmatrix}\frac{E}{c}\\p_x\\p_y\\p_z \end{pmatrix}$$ The interval of this four-vector (in the case of signature $(1,-1,-1,-1)$ metric) is given by $$p^{\mu}p_{\mu}=m^2c^2$$ and is invariant under Lorentz transformations.
You can think on it in analogy to rotations. Rotations preserve the norm of a three-vector. Lorentz transformations are kind of rotations of space and time, and they preserve the 'norm' of four-vectors.
It is not necessary to think in the context of relativity. In newtonian mechanics it also happens; imagine an object (with mass $m$) moving at a constant velocity $v$, then the kinetic energy is $ \frac{1}{2} m v^2 $, but if you observe the same situation from the point of view of a bird which is flying beside the object at the same velocity, you are going to see a zero kinetic energy.
The value of the energy is frame-dependent, but it is constant for both observers (if the system is isolated and the observers are inertial).
Firstly, take into account that in relativity the energy depends on the inertial frame, because it is not invariant under Lorentz transformations (those that connect inertial frames). The energy is part of a four-vector called four-momentum:
$$ p^{\mu} = (E/c, p_{x}, p_{y}, p_{z}) $$
where $p_{x}, p_{y}, p_{z}$ are the components of the momentum $\vec{p}$. When you switch from one inertial frame to another, the components of this four-vector mix between them. The only thing that is completely independent of the frame is its module:
$$ \eta_{\mu \nu}p^{\mu}p^{\nu} = E^2/c^2 - \vec{p}·\vec{p} $$
which is what we call the
rest mass $m$ of the object:
$$ E^2/c^2 - \vec{p}·\vec{p} = m^2c^2 $$
Secondly, if you choose a particular inertial frame in special relativity, the total four-momentum $p^{\mu}$ is a constant of motion in a (elastic) process with no external actions.
To sum up, in an elastic isolated process:
What is conserved but frame-dependent: the components $p^{\mu}$. What is conserved and frame-independent: the module of $p^{\mu}$ (the rest mass, $m$). |
I got the following question as my homework.
Given $V$ is a vector space with $P \in \operatorname{End} V$. $P \circ P = P$ (
"P is idempotent"). Show that $V = \operatorname{Ker} P \oplus \operatorname{Im} P$.
One $P$ I can imagine is a projection from 3d-space to plane, that just sets some coordinates to zero. For example $\begin{pmatrix} x \\ y \\ z\end{pmatrix} \mapsto \begin{pmatrix}x \\ y \\ 0\end{pmatrix}$. Then $\operatorname{Ker} P$ would give the line $\begin{pmatrix} 0 \\ 0 \\ z\end{pmatrix}$ and $\operatorname{Im} P$ would contain all $\begin{pmatrix}x \\ y \\ 0\end{pmatrix}$. So the result of $\operatorname{Ker} P \oplus \operatorname{Im} P$ is of course $V$.
But how do I prove that in a mathematical way? |
Interpolation is a technique for calculating values between the lines within a table. This is one of the simplest process that is based on Quadratic approximation polynomial. Interpolation is a popular for tabular form function. It is applicable on polynomials even with approximately low degrees. This is an integral part of numerical analysis where values are obtained with the help of an algorithm in mathematics. Also, you can use calculators or built-in function if necessary.
Interpolation is still important for functions that are stored in the tabular format and they are introduced to understand the wider application of finite differences. If you will check the dictionary meaning of word interpolation then this is the estimation of unknown quantities between any two known quantities.
If you know the long-term cycles, seasonality or trends then calculating interpolation values is easy. It is frequently used for regression analysis or series analysis in mathematics. This is a special case of curve fitting and solutions are easily to identify with fairly low degrees.
The formula of quadratic interpolation in mathematics is given as below:
\[\large f(x_{j}+\theta h)\approx f_{j}+ \theta \triangle f_{j}+ \frac{1}{2}\theta (\theta -1)\triangle^{2}f_{j}\]
When you are living in the computer age then working on interpolation technique is easy with pre-built functions and algorithms. But formula is still important when data is arranged in tabular format and applicable for finite differences. During the construction of interpolation polynomials, there is a trade-off between better fit and smooth well-behaved functions. More data points used for the interpolation, higher the degree of resultant polynomials and accuracy of data points will be perfect.
Question: Find the estimate of $cos(80^{\circ}: {35}’)$ by quadratic interpolation ?
Table of $cos x$:
x $80^{\circ}$ 0′ 0.1736 10′ 0.1708 20′ 0.1679 30′ 0.1650 40′ 0.1622 Solution:
Given Table,
x f(x) = $cos x$ $\delta$ $\delta^{2}$ $80^{\circ}$ 0′ 0.1736 $80^{\circ}$ 10′ 0.1708 -28 -1 $80^{\circ}$ 20′ 0.1679 -29 0 $80^{\circ}$ 30′ 0.1650 -29 1 $80^{\circ}$ 40′ 0.1622 -28 -1
$cos 80^{\circ}35’$ = $f(80^{\circ}30′) + 0.5\delta f(80^{\circ}30′) + \frac{1}{2} 0.5(0.5 – 1)\delta^{2}f(80^{\circ}30′)$
$cos 80^{\circ}35’$ = 0.1650 + 0.5(-0.0028) + (0.5)(0.5)(-0.5)(-0.0001)
$cos 80^{\circ}35’$ = 0.1636 |
I have a high pass filter given by the time-domain relationship between input and output:
$x'(t) = x(t) - \alpha \times x(t-1)$.
I would like to get its cut-off frequency. How can I proceed?
Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It only takes a minute to sign up.Sign up to join this community
I have a high pass filter given by the time-domain relationship between input and output:
$x'(t) = x(t) - \alpha \times x(t-1)$.
I would like to get its cut-off frequency. How can I proceed?
First you discretize you equation (
note: I assumed that your $x(t-1)$ means a delay of one samples and not 1 second, but you can adapt easily if that's not the case) :
$$ x'[n] = x[n] - \alpha \times x[n-1] $$
Where $$ x[n] = x(n\times T_s)$$
Then you use the $\mathcal{Z}$ transform : $$X'[z] = X[z] - \alpha \times X[z]\times z^{-1}$$
And then you get the transfer function in the Z domain : $$ \frac{X'[z]}{X[z]} = 1 - \alpha \times z^{-1}$$
This is a causal FIR filter, called a
1-zero filter with its zero at $\alpha$. Your cutoff frequency will depend on your sample frequency. : Illustration : Code for the illustration
figure,hold onfor alpha=0.1:0.1:1 h = freqz([1 0], [1 alpha]); f = 0:1/length(h):1-1/length(h); semilogy(f, 20*log(abs(h))); hold on grid onendlegend('0.1', '0.2', '0.3', '0.4', '0.5', '0.6', '0.7', '0.8', '0.9', '1');title('normalized frequency response') |
I tried to do some math in a blog post of mine and came to one with a floor function. I wasn't sure how to deal with it so I just ignored it, and then added the ceiling function in my final equation as that seemed to give me the result I wanted. I'm wondering what is the correct way of handling these functions in equations?
What I did was this:
$$\begin{align} G(n) &= \left\lfloor n\log{\varphi}-\dfrac{\log{5}}{2}\right\rfloor+1 \\\\ n\log{\varphi} &= G(n)+\dfrac{\log{5}}{2}-1 \\\\ n &= \left\lceil\dfrac{G(n)+\dfrac{\log{5}}{2}-1}{\log\varphi}\right\rceil \end{align}$$
How should I have done this in a correct way? How do I work with the ceiling and floor functions when I shuffle around with equations? |
As a starting point to this, determining seasonality for a given market is as follows:
i) Take several years of historical spot price time series, e.g. TTF spot prices. For year $i$ work out a yearly price $p_{yr,i}$ by taking the arithmetic average of daily spot prices. Do the same in respect of month number $j$ of the same year to get a monthly price $p_{mth,i}^{j}$. The monthly shaping factors $f_{i}$ are then $f_{i}=\frac{p_{mth,i}^{j}}{p_{yr,i}}$. Determine the $f_{i}$ for a number of years (where possible i use at least 3, but that is a judgement call), and use their average. As you say, winter will be more expensive, i.e. you expect $f_{i}>1$ for $i\in\{1,2,3,10,11,12\}$ and $f_{i}<1$ for $i\in\{4,5,6,7,8,9\}$
ii) if you need to use daily shaping, you can determine the ratio of weekday prices (numbered 1 to 7) to the monthly prices. This results in 7 weekday factors for each month, i.e. another 84 factors. This is how it is done in electricity, where intraweek shaping is very pronounced. I guess in gas you might find it sufficient to have only two factors per month, one for the workdays and one for the weekend.
Having determined the seasonality factors, one can turn them into a seasonality-related drift term $\mu(t)$ to describe W/S term structure. |
Let us begin with the Maxwell's equations for electrostatics and magnetostatics.
For electrostatics:$$\begin{equation}\boldsymbol{\nabla\cdot E}=\frac{\rho}{\epsilon_0}\end{equation}$$$$\begin{equation}\boldsymbol{\nabla\times E}=0\end{equation}$$For magnetostatics:$$\begin{equation}\boldsymbol{\nabla\cdot B}=0\end{equation}$$$$\begin{equation}\boldsymbol{\nabla\times B}=\frac{\boldsymbol{j}}{c^2\epsilon_0}\end{equation}$$
The equation $\boldsymbol{\nabla\cdot B}=0$ in magnetostatics says that
. Comparing this with the analogous equation $\boldsymbol{\nabla\cdot E}=\dfrac{\rho}{\epsilon_0}$ in electrostatics, we can conclude that there are no magnetic charges. This implies that the magnetic field lines neither starts nor ends. Then what is the origin of the magnetic field? the divergence of the magnetic field is 'always' zero
Magnetic fields are always associated with electric currents. And the equation $\boldsymbol{\nabla\times B}=\dfrac{\boldsymbol{j}}{c^2\epsilon_0}$ says that the curl is proportional to the electric current density. So the magnetic field lines form closed loops around the current vectors.
Now the question is:
Why don't electric field lines form closed loops?
At first sight, one might think that electric field lines also form closed loops when $\rho=0$. But this is not the case. To understand this, look at the equation $\boldsymbol{\nabla\times E}=0$ which says that
. This simply means that electric field lines can never form closed loops. This fact is closely related to conservative nature of the electrostatic field. the curl of electric field is 'always' zero |
Reynolds-averaged Navier-Stokes equations allow to split the description of a turbulent fluid into an averaged (typically laminar) flow on some length and/or time-scale and separate equations for the turbulent fluctuations. The resulting equations look like this $$\rho\bar{u}_j \frac{\partial \bar{u}_i }{\partial x_j} = \rho \bar{f}_i + \frac{\partial}{\partial x_j} \left[ - \bar{p}\delta_{ij} + \mu \left( \frac{\partial \bar{u}_i}{\partial x_j} + \frac{\partial \bar{u}_j}{\partial x_i} \right) - \rho \overline{u_i^\prime u_j^\prime} \right ]$$ Here the bars denotes the averaged values, and $\rho \overline{u_i^\prime u_j^\prime}$ is called the Reynolds stress tensor characterizing the influence of the turbulent fluctuations on the mean flow.
To actually evaluate the Reynolds stress one usually passes to something called the Boussinesque hypothesis and that is that you can actually
model the stress as a viscous stress tensor with a "turbulent viscosity" $\mu_t$ and an isotropic stress coming from "turbulent kinetic energy" $k$. That is, in Cartesian coordinates$$\rho \overline{u_i^\prime u_j^\prime} = \mu_t \left( \frac{\partial \bar{u}_i}{\partial x_j} + \frac{\partial \bar{u}_j}{\partial x_i} \right) -\frac{2}{3} k \delta_{ij}$$Then there is a number of models for how to compute the quantities $\mu_t$ and $k$. Just for illustration, one of them is the k-epsilon model, where two transport equations for variables $k,\epsilon$ are solved$$\frac{\partial (\rho k)}{\partial t}+ \frac {\partial (\rho k \bar{u}_i)}{\partial x_i}= f(k, \epsilon, \bar{u}_j, \partial \bar{u}_j/\partial x_k,...)$$$$\frac{\partial (\rho \epsilon)}{\partial t}+ \frac {\partial (\rho \epsilon \bar{u}_i)}{\partial x_i}= g(k, \epsilon, \bar{u}_j, \partial \bar{u}_j/\partial x_k,...)$$and turbulent viscosity is then determined as $\mu_t = \mu_t(k,\epsilon)$. Many other models exist.
Of course, in astrophysics we are talking about plasma dynamics, which is modeled by (radiative) compressible magneto-hydrodynamics. However, this set of equations can be Reynolds-averaged in very much the same way as the pure-fluid equations. The equations of models such as the k-epsilon model would have to be generalized by introducing the production of turbulent kinetic energy due to effects such as the magneto-rotational instability but otherwise the models should work in a similar fashion. Possibly, one would also have to include a model for the turbulent magnetic-field fluctuations in the Maxwell stress $\sim \overline{B_i B_j}$.
So now for my question: These Reynolds averaged models seem to have applications only in engineering, but I have never seen them applied in an astrophysical context. Why is this so?
I have instead seen a single, very special model, and that is the Shakura-Sunyaev prescription for turbulent viscosity in steady, thin accretion disks: $\mu_t = \alpha \rho \bar{p}$, where $\alpha$ is a constant. However, I do not see any other context than steady, thin disks where this kind of prescription can be useful. Does one perhaps use more sophisticated prescriptions in other astrophysical contexts such as the theory of stellar structure, intergalactic medium, or the solar wind? |
Difference between revisions of "Conservation of Angular Momentum"
m (made mistake when reformatting maths)
m (Made symbol for angular momentum consistent)
Line 1: Line 1:
The '''conservation of angular momentum''' is a fundamental concept of physics along with other [[conservation law]]s such as those of [[energy]] and [[momentum (physics)|linear momentum]]. It states that the angular momentum of a system remains constant unless changed through an action of external [[force]]s.
The '''conservation of angular momentum''' is a fundamental concept of physics along with other [[conservation law]]s such as those of [[energy]] and [[momentum (physics)|linear momentum]]. It states that the angular momentum of a system remains constant unless changed through an action of external [[force]]s.
−
In [[Newtonian mechanics]], the angular momentum of a point mass about a point is defined as <math>\vec
+
In [[Newtonian mechanics]], the angular momentum of a point mass about a point is defined as <math>\vec= \vec r \times \vec p</math> where <math>\vec{r}</math> is the position [[vector quantity|vector]] of the point mass with respect to the point of reference and <math>\vec{p}</math> is the [[momentum (physics)|linear momentum]] vector of the point mass.
The principle of angular momentum can be applied to a system of particles by summing the angular momentum of each particle about the same point. This can be represented as:
The principle of angular momentum can be applied to a system of particles by summing the angular momentum of each particle about the same point. This can be represented as:
<math>
<math>
−
\vec{
+
\vec{}_{sys} = \sum_i \vec{}_i
</math>
</math>
where
where
−
:<math>\vec{
+
:<math>\vec{}_{sys}</math> is the total angular momentum of the system
−
:<math>\vec{
+
:<math>\vec{}_i</math> is the angular momentum of the i<sup>th</sup> particle
==Proof of conservation==
==Proof of conservation==
Revision as of 15:56, 13 December 2016
The
conservation of angular momentum is a fundamental concept of physics along with other conservation laws such as those of energy and linear momentum. It states that the angular momentum of a system remains constant unless changed through an action of external forces.
In Newtonian mechanics, the angular momentum of a point mass about a point is defined as where is the position vector of the point mass with respect to the point of reference and is the linear momentum vector of the point mass.
The principle of angular momentum can be applied to a system of particles by summing the angular momentum of each particle about the same point. This can be represented as:
where
is the total angular momentum of the system is the angular momentum of the i thparticle Proof of conservation
For a constant radius, the second term is zero. Hence From this, it can be concluded that in the absence of an external moment, angular momentum must be conserved.
Example
Imagine a rod of length and mass suspended from one end vertically, and that a small block mass having velocity and mass collides with the other end and sticks to it. The maximum angle of displacement of the rod from the vertical axis can be calculated using the
conservation of angular momentum:
Thus:
Conservation of energy can be applied after the collision, such that the final potential energy of the rod, (with the sticking block mass) equals the initial kinetic energy, just after the collision:
Now solve the two sides of the above equation separately:
Solving the above four equations yields:
Plugging in for angular velocity from the initial equations above yields:
Calculating the moment of inertia now becomes necessary for a rod of length and mass , with a small block of mass at its end.
An ordinary rod of length has the following moment of inertia relative to an axis of rotation at one end:
The moment of inertia is additive and defined as:
where is the mass at each (perpendicular) distance from the axis of rotation.
Thus the moment of inertia for a rod of length and mass , with a small block of mass at its end is simply:
which is:
Plugging this back into the unsolved equation above yields:
If we complicate the problem further by assuming the small block began with velocity zero from an incline of height h, then applying conservation of energy to the moment in time just prior to its collision with the rod yields the following velocity of impact:
and hence
and thus the solution is:
or |
I don't get your downsample step when you downsampled by factor $M$.Let me go from scratch with the spectrum visualization below, with time domain, continuous frequency domain and discrete frequency domain from left to right.When we reduce the sampling frequency by a factor $k$, the signal spectrum is copied to new replicas at $f_s/k$. The discrete ...
I don't have a real answer but I have the feeling that this result will help you out: Bernstein's inequality says that, if the signal $x(t)$ is bandlimited to $|f|\leq B$, then $$\left| \frac{\textrm{d}x(t)}{\textrm{d}t}\right|\leq 4\pi B \,\textrm{sup}_{\tau\in\mathbb{R}}|x(\tau)| ,\,\,t\in\mathbb{R}$$ where $\textrm{sup}$ stands for "least upper bound".I ...
ObservationsI have used +1 and -1 in the sequence instead of your 1 and 0. With $\alpha=1$, the band-limited continuous function $f_m(T)$ in your first two figures (with the above mentioned modification) is:$$f_m(T) = \sum_{k=1-m}^m \operatorname{sign}\left(\operatorname{sinc}(\pi k - \pi/2)\right)\operatorname{sinc}(\pi T-\pi k),\tag{1}$$where:$$\...
Some recent cell phone models use something like a Cirrus Logic CS42xx series audio IO chip, which seems to use a digital polyphase interpolation filter, a sigma delta modulator, followed by a switched capacitor DAC and low-pass filter.Sinc interpolation (or, given finite hardware, a polyphase FIR kernel similar to a windowed Sinc) is one high quality ...
The key idea is that the random sampling approach enforces more constraints on the resulting signal than the uniform sampling approach does.The POCS (projections onto convex sets) algorithm used for the reconstruction of the randomly sampled signal is the key piece: it enforces:That the signal must be from this spectrum.That the signal is real-valued....
The general mathematical framework for interpolation is approximation theory. I guess the most important result is that for signals with bandwidth limitation, you can have perfect reconstruction via $sinc(\cdot)$ convolution; the famous sampling theorem, that has been mentioned here several times. I guess it is equally well-known that it is not really ...
SLAM(Simultaneous Localization and Mapping) algorithms can be used to for 3D reconstruction. They offer solutions for both monocular as well as stereo cameras. With single camera they estimate depth with few images and reconstruct the scene. You can find some of the open source solutions here.Real time 3D reconstruction can done using ORBSLAM and it is ...
Math and Physics are not the same thing. They can be very close but infinity is hard to realize. Going from analog to digital, a more accurate description is an average of the waveform over a small time window and the average is in turn quantized. Filters don’t have infinite stop bands but 80 dB is usually more than good enough. There is always a small ...
First, a warm welcome to SE!Basically, you have a calibrated 3D reconstruction problem. The typical approach follows a 5-stage pipeline:Identify 2D features in each image along with the associated descriptors. Algorithms such as SURF, SIFT or AKAZE are heavily used and are available in many vision libraries such as OpenCV.Match the extracted keypoints ...
The sampling theorem requires a perfectly bandlimited signal, bandlimited to below twice the sampling frequency. The problem with this is that only an infinite length signal (e.g. exists before the big bang) can be perfectly bandlimited. This is from the Fourier theorem regarding any domain with finite support.Thus all real-world signals are ...
Alright, I'll try to take a shot at giving you an explanation of what's going on with the DWT.So in the CWT, basically what you are doing is you are generating the wavelet coefficients by convolving the signal with each scale and shift of the mother wavelet function. However, the problem with doing this is that when you analyze the low frequency components ...
It sounds like $W_k$ is just a matrix of 1's and 0's. When $W_k$ is a 1, then the corresponding range value in $D_k$ is used. If it's a 0, then that range value is not used.Similarly, $T_k$ should be mostly 1, and 0 when the corresponding $D_k$ values are unreliable.Does that tally with your understanding? If not, can you elaborate on your question ...
It could be useful in the field of OCR as a pre processing step.Think about badly scanned data.You'd like to convert it into binary image.So the first step would be applying some kind of thresholding.Since no thresholding is perfect, There will be some "Holes" / "Gaps" within the text.Closing those "Holes" / "Gaps" can be done using morphological ...
There are some mistakes in the question and your description.When we sample a signal of frequency $f_m$ with a sampling rate $f_s$, the sampled signal contains the frequencies $f=f_m \pm nf_s$, where $n \in \mathbb{Z} $.Here the frequencies available in the sampled signal are calculated as,$f_m=14100 Hz$and $f_s=400 Hz$.we know that,$ f=f_m \pm ...
Because the sampling frequency is above the Nyquist rate, the original signal can be written in terms of its samples $y(n)$:$$x(t)=\sum_{m=-\infty}^{\infty}y(m)\frac{\sin[\pi(t-mT)/T]}{\pi(t-mT)/T}\tag{1}$$By setting $t=nT+T/2$ you get from (1)$$z(n)=\sum_{m=-\infty}^{\infty}y(m)\frac{\sin[\pi(n-m)+\pi/2]}{\pi(n-m)+\pi/2}$$and since$$\sin[\pi(n-m)+\...
One simple approach would be taking the mean square error (MSE) by usingfitness_1 = mean((inputimage(:) - reconstructedimage_1(:)).^2)though, as your image size won't change, you can ommit the mean and use sum instead.fitness_1 = sum((inputimage(:) - reconstructedimage_1(:)).^2)In general you want to have some kind of distance measure. Look at ...
I agree with Jim Clay's answer, but I think it is important to point out two things. First of all, there are no phase distortions due to the hold operation, just a simple delay of half a sampling interval. So nothing needs (and can) be done about the phase. Second, it is important to realize that the gain roll-off due to the sinc shape is relatively mild. ...
What you are describing is the distortion introduced by an ideal digital-to-analog converter (DAC) in the analog domain. Two things are typically done to reduce this distortion:Analog filteringOversamplingAs you note in your question, the distortion is modeled, in the frequency domain, by a sinc rolloff. Increasing the sample rate before converting ...
$x(t)$ must be a band pass signal. Under certain conditions on the sampling frequency and its relation to the lower and upper band edges of the signal, $x(t)$ can be sampled at a frequency that is lower than twice its maximum frequency, without introducing aliasing. Have a look at this article and at this answer to a related question.For the numbers given ...
First, since $t>0$ in your case, you can write your function as\begin{equation}C(t)=1.6925\left(\exp^{-0.136t}- \exp^{-1.192t} \right) u(t)\end{equation}where $u(t)$ is a unit-step function. Then, denote the FT of $C(t)$ as $C(F)$, which is given by\begin{equation}C(f)=1.6925\left(\frac{1}{0.136+2\pi jf}- \frac{1}{1.192+ 2\pi jf} \right)\end{...
The Poisson equation may suffer from ill-posedness if the boundary condition is not sufficient to yield a unique solution to the problem. The most often used boundary condition is Dirichlet boundary condition, in which the values of $I(x,y)$ are specified for those points $(x,y)$ on the boundary of $I$.In your case, after you stretch your solution to another ...
I guess this is a straightforward non-linear optimization problem (to be solved with Newton variations, such as Trust-Region methods), where you don't even need to compute the Jacobian analytically. It appears to me that the optimization problem is written over $K_i$, and thus is the input to the cost function. To compute the cost, at each call to this ...
You say:I have a 128 point one dimensional k-space samples...The hanning window is the same size as the k-space vector (256)...Make sure that you have the appropriate sizes in your algorithm.Next, I convolved the k-space signal with the hanning window...Windowing is applied in $k$-space by multiplication - you simply have to multiply your window and ...
I would expect that you simply cut the outer parts of the image away. Certainly, there is always a hassle to figure out the proper pixel in the center, but that is a detail of the FT-algorithm employed.Remeber: Sampling denser than you need increases the FOV. Hence, the image you simulate should simply be larger with the original image (that you get by ...
Zero-padding data for a longer FFT is equivalent to interpolation by a (periodic) Sinc kernel. Interpolation by a (periodic) Sinc kernel can reconstruct points between samples of a signal that was strictly bandlimited (to below the Nyquist frequency) prior to sampling. See: https://ccrma.stanford.edu/~jos/resample/Theory_Ideal_Bandlimited_Interpolation....
Let us first assume you can produce estimates of the camera state (position and attitude) via sensors, a filter like a Kalman Filter, and a (simple) model for the camera itself. Using this information, you could then match features between a sequence of images and use those matches and camera state estimates to estimate relative 3D coordinates of those ...
No, you cannot reconstruct the original signal from the cyclic autocorrelation. The fundamental reason is that it results from an averaging operation. Like the autocorrelation and PSD, the cyclic autocorrelation is a many-to-one functional (you already noted that I think). For example, all BPSK signals with identical timing parameters (symbol clock phase and ... |
Two things to note here. First, subtracting inflation from the nominal interest rate
is an approximation to the real interest rate, but only in discrete time. Furthermore, the "true" relationship it's approximating isn't division of one rate by the other--you have to add 1 to all three of your quantities (inflation, real interest rate, and nominal interest rate) first to get the true relationship.
Here's a brief overview. Consider the Fisher equation of
$r = i - \pi$
where $r$ is the real interest rate, $i$ is the nominal interest rate, and $\pi$ is the inflation rate.
This equation is often introduced as a linear approximation to the true real rate of interest, given by the equation
$\frac{1 + i}{1 + \pi} = 1 + r$
Let's see how this holds in a discrete time model. Denote your nominal income as $Y$ and the price level as $P$. Your real income is $Y/P$. If all this income is invested in some interest-bearing asset in a discrete time model, your real income in the next time period becomes
$\frac{Y (1 + i)}{P (1 + \pi)}$
and if we want to find a real interest rate that summarizes this change in real income, we would need to write your real income in the next time period as
$\frac{Y}{P} (1 + r)$
which gives us the identity $\frac{1 + i}{1 + \pi} = 1 + r$ that the Fisher equation approximates. However, if we work in continuous time, this breaks down. First, the units don't work out--inflation rates and interest rates are measured in percentage change per year (or some other unit of time), so they cannot be added to the dimension-less number $1$. Second, it turns out that Fisher's approximation is actually completely correct in continuous time. Using derivatives, we define our quantities as follows:
$i = \frac{dY}{dt} \frac{1}{Y}$
$\pi = \frac{dP}{dt} \frac{1}{P}$
$r = \frac{d(Y/P)}{dt} \frac{1}{(Y/P)}$.
Using the quotient rule, we can rewrite $r$ as
$r = \frac{\frac{dY}{dt}P - \frac{P}{dt}Y}{P^{2}} \frac{1}{(Y/P)}$
which simplifies to
$r = (\frac{dY}{dt} \frac{1}{P} - \frac{dP}{dt} \frac{Y}{P^{2}}) \frac{P}{Y} = \frac{dY}{dt} \frac{1}{Y} - \frac{dP}{dt} \frac{1}{P} = i - \pi$
which gives us the Fisher equation, no approximations about it!
Note that this assumes continuous compounding from your nominal interest rate. If you instead have a compounding rate of $\tau$, we would define $i$ differently, and our equation becomes
$r = \tau \ln(1 + \frac{i}{\tau}) - \pi$.
However, for the purposes of the data you're working with, I am 90% sure that this modification is completely superfluous. If you want to use something more precise than the Fisher equation, you need to know exactly how your data was computed. What price index was used to calculate the inflation rate? Under what assumptions was the nominal interest rate calculated?
In short, while I can't speak to the reasons for your supervisor's recommendation, they're definitely correct that you should subtract inflation rather than divide by it. There's a reason why we use the Fisher equation. |
What identities are used to get $\sin x$ from $\tan x \operatorname{/} \sec x$? I was looking at an example in my textbook and the problem went from $\tan x \operatorname{/} \sec x$ to $\sin x$. I don't understand how.
Hints:
$\tan x=\frac{\sin x}{\cos x}$ and $\sec x=\frac{1}{\cos x}$
Here's an equivalent but more geometric answer. Let's say we have a right triangle with an angle $x$, the adjacent side with length $a$, the opposite side with length $b$ and the hypotenuse with length $c$.
Then $\tan x = \frac{b}{a}$, $\sec x = \frac{c}{a}$ and $\sin x = \frac{b}{c}$ (by definition of $\tan$, $\sec$, and $\sin$). That gives us
$$\frac{\tan x}{\sec x} = \frac{\frac{b}{a}}{\frac{c}{a}}=\frac{b}{a}\frac{a}{c}=\frac{b}{c}=\sin x$$
This method is also an easy way to derive the identity $\tan x = \frac{\sin x}{\cos x}$ that others are suggesting.
Just using the definitions: $$ \frac{\tan x}{\sec x}=\frac{\frac{\sin x}{\cos x}}{\frac1{\cos x}}=\sin x. $$
$$\tan x = \frac{\sin x}{\cos x} = \sin x \cdot \frac{1}{\cos x} = \sin x \sec x$$
Rearranging,
$$\sin x = \frac{\tan x}{\sec x}$$
Rewrite $\tan x$ as $\dfrac{\sin x}{\cos x}$. This makes it easier to see: $$\tan x = \dfrac{\sin x}{\cos x} = \sin x \cdot \dfrac{1}{\cos x}$$ $$\implies = \sin x \sec x$$
$$\sin x = \frac{\tan x}{\sec x}$$
$\tan(x)=\frac{\sin(x)}{\cos(x)}$ and $\sec(x)=\frac{1}{\cos(x)}$
Therefore, $\frac{\tan(x)}{\sec(x)}=\frac{\sin(x)}{\cos(x)}(\frac{1}{\cos(x)})^{-1}$.
To divide fractions, you must multiply by the reciprocal, so the answer is $\sin(x)$.
$\tan(\pi/2) = \infty$ and $\sec(\pi/2)$ is undefined whereas $\sin(\pi/2) = 1$ so the given equation is incorrect. |
I'm teaching an after school workshop for a few 7th graders. I was having them try to predict the next item in a complicated sequence. After some failed attempts, one of the kids started analyzing the ...
The Fibonacci numbers are one of the first sequences given as examples of sequences in many calculus textbooks as they have a definition that does not obviously have a closed form and they have many ...
Sequences and series are related concepts but differ extremely from one another. I feel that students in integral calculus frequently mix them up. Part of the problem is that:Sequences are usually ...
I'm writing (together with a colleague) a minicourse on mathematical analysis (currently we want to cover the Weierstrass theorem on functions on compact intervals, so the aim is to present only the ...
Let's say we study a reccurence relation such as$$a_{n+1}=a_{n}+n, n \geq 1$$I find many students are having difficulties when you ask them to find $$a_{n}$$ in terms of $$a_{n-1}$$ where they just ...
Basically the textbooks in my country are awful, so I searched on the web for a precalculus book and found this one: http://www.stitz-zeager.com/szprecalculus07042013.pdfHowever, it does not cover ...
In this article, I give counterexamples regarding real sequences. And in that one some others.In particular counterexamples answering questions like: "If for all $p \in \mathbb{N}$ $\lim\limits_{n \... |
The textbook Elements of Information Theory gives us an example:
For example, if we knew the true distribution p of the random
variable, we could construct a code with average description length
H(p). If, instead, we used the code for a distribution q, we would
need H(p) + D(p||q) bits on the average to describe the random
variable.
To paraphrase the above statement, we can say that if we change the information distribution(from q to p) we need D(p||q) extra bits on average to code the new distribution.
An illustration
Let me illustrate this using one application of it in natural language processing.
Consider that a large group of people, labelled B, are mediators and each of them is assigned a task to choose a noun from
turkey,
animal and
book and transmit it to C. There is a guy name A who may send each of them an email to give them some hints. If no one in the group received the email they may raise their eyebrows and hesitate for a while considering what C needs. And the probability of each option being chosen is 1/3. Toally uniform distribution(if not, it may relate to their own preference and we just ignore such cases).
But if they are given a verb, like
baste, 3/4 of them may choose
turkey and 3/16 choose
animal and 1/16 choose
book. Then how much information in bits each of the mediators on average has obtained once they know the verb? It is:
\begin{align*}D(p(nouns|baste)||p(nouns)) &= \sum_{x\in\{turkey, animal, book\}} p(x|baste) \log_2 \frac{p(x|baste)}{p(x)} \\&= \frac{3}{4} * \log_2 \frac{\frac{3}{4}}{\frac{1}{3}} + \frac{3}{16} * \log_2\frac{\frac{3}{16}}{\frac{1}{3}} + \frac{1}{16} * \log_2\frac{\frac{1}{16}}{\frac{1}{3}}\\&= 0.5709 \space \space bits\\\end{align*}
But what if the verb given is
read? We may imagine that all of them would choose
book with no hesitatation, then the average information gain for each mediator from the verb
read is:
\begin{align*}D(p(nouns|read)||p(nouns)) &= \sum_{x\in\{book\}} p(x|read) \log_2 \frac{p(x|read)}{p(x)} \\&= 1 * \log_2 \frac{1}{\frac{1}{3}} \\& =1.5849 \space \space bits \\\end{align*}We can see that the verb
read can give the mediators more information. And that's what relative entropy can measure.
Let's continue our story. If C suspects that the noun may be wrong because A told him that he might have made a mistake by sending the wrong verb to the mediators. Then how much information in bits can such a piece of bad news give C?
1) if the verb given by A was
baste:
\begin{align*}D(p(nouns)||p(nouns|baste)) &= \sum_{x\in\{turkey, animal, book\}} p(x) \log_2 \frac{p(x)}{p(x|baste)} \\&= \frac{1}{3} * \log_2 \frac{\frac{1}{3}}{\frac{3}{4}} + \frac{1}{3} * \log_2\frac{\frac{1}{3}}{\frac{3}{16}} + \frac{1}{3} * \log_2\frac{\frac{1}{3}}{\frac{1}{16}}\\&= 0.69172 \space \space bits\\\end{align*}
2) but what if the verb was
read?\begin{align*}D(p(nouns)||p(nouns|baste)) &= \sum_{x\in\{book, *, *\}} p(x) \log_2 \frac{p(x)}{p(x|baste)} \\&= \frac{1}{3} * \log_2 \frac{\frac{1}{3}}{1} + \frac{1}{3} * \log_2\frac{\frac{1}{3}}{0} + \frac{1}{3} * \log_2\frac{\frac{1}{3}}{0}\\&= \infty \space \space bits\\\end{align*}
Since C never know what would the other two nouns be and any word in the vocabulary would be possible.
We can see that the KL divergence is asymmetric.
I hope I am right, and if not please comment and help correct me. Thanks in advance. |
As usual, if you can put your hands on liquid quanto instruments' prices (e.g. some quanto futures like the dollar-quantoed Nikkei trading on the CME), then you could directly
imply the quanto adjustment $\rho\sigma\tilde{\sigma}$. The problem is that quanto option markets are not as developped as plain vanilla ones and you must resort to something else.
Assume you are pricing a quanto option of maturity $T$ and strike $K^Q$. The issue with doing what is mentioned in one of the comments, namely picking $$\sigma = \Sigma(K^Q,T),\quad \tilde{\sigma}=\tilde{\Sigma}(\text{atm},T)$$ where I used the notation $\Sigma(K,T)$ (resp. $\tilde{\Sigma}(K,T)$) to denote the full Black-Scholes volatility surface of the equity (resp. currency pair), is that this will make the quanto forward dependent on the strike of the quanto option to be priced, which represents an arbitrage opportunity (you can easily convince yourself using call put parity for quanto vanillas).
One remedy is to pick the ATM vol both for the currency and the equity pair $$\sigma = \Sigma(\text{atm},T),\quad \tilde{\sigma} = \tilde{\Sigma}(\text{atm},T)$$The problem in that case, is that in the limit as the equity/fx correlation $\rho$ tends towards zero, the prices of your quanto options will not be consistent plain vanilla ones, since you'll always use the $\text{atm}$ vol.
If you really want to be consistent with the smiles of the plain vanilla options (both in the equity and forex markets), then you need a more complex working modelling assumption to begin with (e.g. local or stochastic volatility model for both the equity and the currency pair).
If not you could also compute historical covariance and assimilate that to $\rho \sigma \tilde{\sigma}$. There is no perfect way of doing things here.
[Edit]
Let $S$ represent a risky asset denominated in a FOR(eign) currency. Let $\xi_0$ denote a constant FOR/DOM conversion rate agreed on at the quanto contract inception date (chosen equal to $1$ in most applications).
Do you agree that, in order to preclude arbitrage opportunities, $S_T$, or equivalently $\xi_0 S_T$, should have a unique distribution under some Equivalent Martingale Measure?
Stated differently, the distribution of $\xi_0 S_T$ shouldn't depend on some exogenous parameter, such as a strike level, since that would make it non unique (i.e. one distribution per exogenous parameter value).
Well in a BS world (equity $S$ and FX rate $X$ driven by correlated GBMs), one can show that the quanto forward computes as:$$ F^{\text{quanto}}(t,T) = F(t,T) e^{-\rho \sigma_S \sigma_X (T-t)} $$where $\sigma_S$ (resp. $\sigma_X$) are the constant volatilities of the individual GBMs; $\rho$ is their instantaneous correlation; and $F(t,T)$ the standard equity forward price at $T$ of the asset $S$ whose spot price is known at time $t$.
Now, if you pick $\sigma_S = f(K)$ then clearly the quanto forward becomes a function of $K$. This means that the first moment of the distribution of $\xi_0 S_T$ (hence $S_T$) is a function of $K$ which violates the former uniqueness assumption. |
A two-level system can be described by a density operator involving the Bloch vector
$$ \vec{r}; \quad r_x = Tr(\rho X); \quad r_y = Tr(\rho Y); \quad r_z = Tr(\rho Z) $$
as
$$ \rho = \frac{I + \vec{r}\cdot \vec{\sigma}}{2} $$
where $X$, $Y$, and $Z$ are the Pauli operators.
What is the physical idea behind defining the density operator for a two level system like this, and in particular what is $\vec{\sigma}$ here? |
So I had the privilege of seeing Donald Knuthspeak at UBC a few days ago. I was excited to see one of the eccentric Computing Science legends in real life.
The talk was in an interesting format. Knuth was in front of a relatively small theater, and simply answered any questions that were presented to him from the crowd. It was thus a spread of different questions and topics presented to him.
I took a few notes during the talk and I summarize the topics and Knuth’s responses here.
Knuth on writing:
translate to have less jargon
pays people to find errors in his books
write to have errors, that is, be precise and falsifiable (e.g. “13% improvement”, where this 13% can be verified)
have the history of discoveries with references
write about things that will be important 50 years from now
–> e.g. methods that are useful for many different applications
I remember this being pretty confusing the first time I looked at how to do this. The necessary files are in different zip packages and there’s no minimally working example. So here’s a simple example on how to compile your first \( \LaTeX \) document for Springer’s LNCS publications. Continue reading “How to use the Springer LNCS LaTeX template”
Scenario: You want to programmatically define where your figures are in your latex document without going through and manually editing all your paths. You have a folder called "figs" that contains all your figures, but this folder might move.
Here’s how to programmatically change the path of where your images are located
Or more specifically programmatically change where the "figs" folder is located
This beautiful app lets you draw any symbol into a box, then it will return the corresponding $\LaTeX$ command. So freakin’ handy! And it works. That obscure symbol you saw in some paper $\zeta$, and you can’t think of the Greek name of the top of your head? Simple! Just draw it in and bam, $\text{detextify}^2$ throws the command back (\zeta in case you are wondering).
I use this all the time when I’m writing anything sort of equations in LaTeX. A definite must know about.
Lately I’m writing a lot of papers in and every once and a while something comes up that drives me crazy trying to figure out.
Here’s how to easily switch between a bold vectorand an arrow vector .
% Minimal latex example:% Shows how to switch between bold and arrow vectors.% Specifies the type of document you have.\documentclass{article}% Used for the boldsymbol.\usepackage{amsmath}% Comment this out to represent vectors with an arrow on top.% Uncomment this to represent vectors as bold symbols.\renewcommand{\vec}[1]{\boldsymbol{#1}}% Start of the document.\begin{document}% Your content.My lovely vector: $\vec{x}$% End of the document.\end{document}
% Minimal latex example:% Shows how to switch between bold and arrow vectors. % Specifies the type of document you have.\documentclass{article} % Used for the boldsymbol.\usepackage{amsmath} % Comment this out to represent vectors with an arrow on top.% Uncomment this to represent vectors as bold symbols.\renewcommand{\vec}[1]{\boldsymbol{#1}}% Start of the document.\begin{document}% Your content.My lovely vector: $\vec{x}$ % End of the document.\end{document} |
Statistics in mathematics is a collection, interpretation, analysis, presentation, or organization of data. This is a popular technique for how data is sorted and summarized. It is also defined as the state of facts how they can be given in the form of numbers, tables, or any other classified arrangement.
Statistics is a branch of applied mathematics that is defined as numerical statements of facts in inquiry department placed together. It is also the study of the probability of occurrences of events based on qualitative data or collection of data. Statistics is generally used to inspect the heavy data collection in minimum budget only. Three popular types of statistical averages in mathematics are Mean, Median, Mode.
Mean is the average where the sum of total values is divided by the total count of numbers. Median is the middle value when numbers are arranged in increasing order. If there are two middle values then the average of both values is the final output. Last is median where the most repeated value is the final answer.
There is a list of basic statistics formulas whose values are related to statistical concepts or analyses. With the help of data collection, data analysis, and data summarization is easy.
Mean Statistics Formula
\[\large bar{x}=\frac{\sum x}{n}\]
Where, x = Items given n = Total number of items
Median Statistics Formula
If n is odd, then
M = \[\large (\frac{n+1}{2})^{th}term\] If n is even, then M = \[\large \frac{(\frac{n}{2})^{th}term+(\frac{n}{2}+1)^{th}term}{2}\]
where,
n = Total number of items
Mode Statistics Formula
The value which occurs most frequently
Variance Statistics Formula
\[\large \sigma ^{2} = \frac{\sum (x-bar{x})^{2}}{n}\]
Where, x = Items given barx = Mean n = Total number of items
Standard Deviation Statistics Formula
\[\large S = \sigma = \sqrt{\frac{\sum (x-bar{x})^{2}}{n}}\]
Where,
x = Items given barx = Mean n = Total number of items
Few terminologies related to Statistics include –
Here, is given a statistics problem with the solution to finding the favorite fruit of 100 people. The solution can be presented in the form of the pie chart.
The other popular forms that can be used to represent data include –
If you will see around then applications are just enormous like military, economics, population etc. This technique is also used for mathematical analysis like linear algebra, differential equation, or probability theory etc. Statistics are good for different sectors too like geology, psychology, weather forecasting, probability and many others. You just have to understand the data in deep or it can be named as Mathematical Science too. |
Everyone Dies
I assume the planets are on a "gentle" (shallow) approach to one another, which seems to match your description of "eventually takes it so close [that] they end up touching". There
will be panic as the planets draw nearer.
Everyone will die; it's just a question of when and how.
Tidal forces
As the planets approach, their mutual gravitational acceleration (doubled!) will pull them together and accelerate them to even higher relative speeds. The first problem is, the gravitational acceleration will not be uniform: the "near" pieces of the planets will feel a stronger pull than the "far" pieces, and this effect will be very pronounced.
It will cause great earthquakes and incredible ocean tides (and tsunamis), which will obliterate anything within a few hundred kilometers of a coastline. It will also destroy key infrastructure.
The atmospheres of both planets will be easily affected, causing weather patterns of a far greater magnitude than anything we know as both atmospheres will be pulled toward the center of mass of the two planets.
Roche Limit
If anyone is still alive after all of the above, this last bit should do them in.
Edit: Fixed math (and included steps!) Thanks to MadBender for the catch!
The Roche limit is the distance (radius) within which a celestial body (like a planet) can no longer hold itself together via its own gravity, and is then pulled apart by the gravitational tidal forces I introduced, above. The Roche limit (
d) for rigid bodies 1 $\, m$ and $M$ (your twin Earths), looks like this:
$$d = R_m \left( 2 \cdot \frac{\rho_M}{\rho_m} \right)^{1/3}$$
$\rho_M / \rho_m$ is the ratio of densities of both planets. Since they are identical, their density ratio will be 1/1, thus:
$$d = R_m \left( 2 \cdot \frac{1}{1} \right)^{1/3} \approx 1.26 R_m$$$$d \approx 1.26 \times 6\,371\text{ km} \approx 8\,027\text{ km}$$
As the two planets come within the Roche limit, the effects from the previous section will have already had catastrophic results, and started to elongate the planets. The difference is, that near the Roche boundary, gravity won't be enough to hold the planets together.
The overall mass stays the same, but the planets are literally torn to pieces. The atmospheres and oceans more or less go without a fight (see previous section), but the solid pieces will come bit by bit, and the (now very chaotic) motions will result in more impacts, which will continually pulverize the pieces until there isn't much left but a ring of debris around the star, almost certainly with no survivors.
What actually kills the remaining inhabitants is somewhat a matter of chance, but could be:
Direct impact or secondary impact forces Suffocation/decompression as the atmosphere is pulled towards the center of mass but your tiny planetoid carries on a different trajectory. Or, the atmosphere simply gets thinner as the mass of your planetoid is too weak to retain it at sufficient density to support human life. Other effects The magnetic fields of both planets will combine, quite likely in a way that would reduce the effectiveness of the magnetosphere, allowing cosmic rays to bombard the inhabitants, causing an increase in radiation sickness and cancers, however I don't think anyone will live long enough for that. Notes Of course the Earth isn't completely rigid. However, all of the liquid and gases would already have been pulled and squeezed into the gnarliest surf anyone has ever seen. |
Recovering two Lamé kernels in a viscoelastic system
1.
Dipartimento di Matematica “F. Enriques”, Universitá di Milano, via C. Saldini 50, 20133 Milano, Italy
2.
Sobolev Institute of Mathematics, Siberian branch of Russian Academy of Sciences, Acad. Koptyug prosp., 4, Novosibirsk, 630090, Russian Federation
sametemporal part, i.e. $\lambda_1(t,x)=k(t)p(x)$ and $\mu_1(t,x)=k(t)q(x)$. Furthermore, it is assumed that the spatial parts $p$ and $q$ of $\lambda_1$ and $\mu_1$ are unknownand the threeadditional measurements $\sum_{j=1}^3\sigma_{i,j}^0(t,x)$ n$_j(x) = g_i(t,x)$, $i=1,2,3$, are available on $(0,T)\times \partial \Omega$ for some (sufficiently large) subset $\Gamma\subset \partial \Omega$.
The fundamental task of this paper is to show the uniqueness of the pair $(p,q)$ as well as its continuous dependence on the boundary conditions, the initial data being kept fixed and the initial velocity being suitably related to the initial displacement.
Keywords:linear viscoelastic materials, Identification problems, hyperbolic second-order integrodifferential systems, recovering relaxation kernels, uniqueness, continuous dependence.. Mathematics Subject Classification:Primary: 45Q05, 45K05; Secondary: 35L20, 74H05, 74H45, 74J2. Citation:Alfredo Lorenzi, Vladimir G. Romanov. Recovering two Lamé kernels in a viscoelastic system. Inverse Problems & Imaging, 2011, 5 (2) : 431-464. doi: 10.3934/ipi.2011.5.431
References:
[1]
R. A. Adams, "Sobolev Spaces,",
[2]
A. L. Bukhgeim and M. V. Klibanov,
[3]
C. Cavaterra, A. Lorenzi and M. Yamamoto,
[4]
L. Hörmander, "Linear Partial Differential Operators,",
[5]
O. Yu. Imanuvilov,
[6]
O. Yu. Imanuvilov and M. Yamamoto,
[7]
O. Yu. Imanuvilov and M. Yamamoto,
[8]
O. Yu. Imanuvilov and M. Yamamoto,
[9]
O. Yu. Imanuvilov and M. Yamamoto,
[10]
V. Isakov, "Inverse Source Problems,",
[11] [12] [13]
V. Isakov, "Inverse Problems for Partial Differential Equations,",
[14]
V. Isakov and M. Yamamoto, "Carleman Estimate with the Neumann Boundary Condition and its Applications to the Observability Inquality and Inverse Hyperbolic Problems,",
[15] [16] [17]
M. V. Klibanov and A. Timonov, "Carleman Estimates for Coefficient Inverse Problems and Numerical Applications,",
[18]
M. M. Lavrent'ev, V. G. Romanov and S. P. Shishat'skiĭ, "Ill-posed Problems of Mathematics Physics and Analysis,",
64 (1986).
Google Scholar
[19] [20]
J. Nečas, "Les Methodes Directes en Theorie des Equations Elliptiques,",
[21]
J. Nečas and I. Hlaváček, "Mathematical Theory Of Elastic And Elasto-Plastic Bodies: An Introduction,",
[22] [23] [24] [25]
show all references
References:
[1]
R. A. Adams, "Sobolev Spaces,",
[2]
A. L. Bukhgeim and M. V. Klibanov,
[3]
C. Cavaterra, A. Lorenzi and M. Yamamoto,
[4]
L. Hörmander, "Linear Partial Differential Operators,",
[5]
O. Yu. Imanuvilov,
[6]
O. Yu. Imanuvilov and M. Yamamoto,
[7]
O. Yu. Imanuvilov and M. Yamamoto,
[8]
O. Yu. Imanuvilov and M. Yamamoto,
[9]
O. Yu. Imanuvilov and M. Yamamoto,
[10]
V. Isakov, "Inverse Source Problems,",
[11] [12] [13]
V. Isakov, "Inverse Problems for Partial Differential Equations,",
[14]
V. Isakov and M. Yamamoto, "Carleman Estimate with the Neumann Boundary Condition and its Applications to the Observability Inquality and Inverse Hyperbolic Problems,",
[15] [16] [17]
M. V. Klibanov and A. Timonov, "Carleman Estimates for Coefficient Inverse Problems and Numerical Applications,",
[18]
M. M. Lavrent'ev, V. G. Romanov and S. P. Shishat'skiĭ, "Ill-posed Problems of Mathematics Physics and Analysis,",
64 (1986).
Google Scholar
[19] [20]
J. Nečas, "Les Methodes Directes en Theorie des Equations Elliptiques,",
[21]
J. Nečas and I. Hlaváček, "Mathematical Theory Of Elastic And Elasto-Plastic Bodies: An Introduction,",
[22] [23] [24] [25]
[1]
Shitao Liu, Roberto Triggiani.
Recovering damping and potential coefficients for an inverse non-homogeneous second-order hyperbolic problem via a localized Neumann boundary trace.
[2]
Giuseppe Maria Coclite, Angelo Favini, Gisèle Ruiz Goldstein, Jerome A. Goldstein, Silvia Romanelli.
Continuous dependence in hyperbolic problems with Wentzell boundary conditions.
[3]
Davide Guidetti.
Some inverse problems of identification for integrodifferential parabolic systems with a boundary memory term.
[4]
Hongwei Lou.
Second-order necessary/sufficient conditions for optimal
control problems in the absence of linear structure.
[5]
Jaume Llibre, Amar Makhlouf.
Periodic solutions of some classes of continuous second-order differential equations.
[6]
Yi Zhang, Yong Jiang, Liwei Zhang, Jiangzhong Zhang.
A perturbation approach for an inverse linear second-order cone programming.
[7]
Leonardo Colombo, David Martín de Diego.
Second-order variational problems on Lie groupoids and optimal control applications.
[8]
Zhiqiang Yang, Junzhi Cui, Qiang Ma.
The second-order two-scale computation for integrated heat transfer problem with conduction, convection and radiation in periodic porous materials.
[9]
Alfredo Lorenzi, Eugenio Sinestrari.
Regularity and identification for an integrodifferential
one-dimensional hyperbolic equation.
[10]
Lassi Roininen, Petteri Piiroinen, Markku Lehtinen.
Constructing continuous stationary covariances as limits of the
second-order stochastic difference equations.
[11]
Guoshan Zhang, Peizhao Yu.
Lyapunov method for stability of descriptor second-order and high-order systems.
[12]
Gábor Kiss, Bernd Krauskopf.
Stability implications of delay distribution for first-order and
second-order systems.
[13]
David L. Russell.
Coefficient identification and fault detection in linear elastic systems; one dimensional problems.
[14]
Qi Hong, Jialing Wang, Yuezheng Gong.
Second-order linear structure-preserving modified finite volume schemes for the regularized long wave equation.
[15]
John R. Graef, Lingju Kong.
Uniqueness and parameter dependence of positive solutions of third order boundary value problems with $p$-laplacian.
[16]
Johnny Henderson, Rodica Luca.
Existence of positive solutions for a system of nonlinear second-order integral boundary value problems.
[17]
Yanhong Yuan, Hongwei Zhang, Liwei Zhang.
A smoothing Newton method for generalized Nash equilibrium problems with second-order cone constraints.
[18] [19]
Doyoon Kim, Seungjin Ryu.
The weak maximum principle for second-order elliptic and parabolic conormal derivative problems.
[20]
Luciano Pandolfi.
Joint identification via deconvolution of the flux and energy relaxation kernels of the Gurtin-Pipkin model in thermodynamics with memory.
2018 Impact Factor: 1.469
Tools Metrics Other articles
by authors
[Back to Top] |
Consider the stationary VAR process
$${\bf X}_t = \sum_{\tau = 1}^{L} A_\tau {\bf X}_{t-\tau} +{\bf \epsilon}_t$$
If the innovations $\epsilon_t \sim MVN({\bf 0},\Sigma)$ then is ${\bf X}_t$ a Gaussian stationary process?
Is it correct that due to the invertibility of the VAR into an MA and observing that ${\bf X}_t$ is the sum of zero mean MVN random variables the above is true, or is there a flaw somewhere in this. |
I'm interested in testing the relationship of public investment in human capital and infrastructure, and the rate of growth of the GDP for Colombia, during a 20 years period. Given that I'm only an undergraduate student and my knowledge is limited, I found that the closest theoretical model to test this hypothesis is Milbourne, Otto and Voss's "Public investment and economic growth" model, that has this production function:
$$ Y_{t}=A_{t}K^{\alpha }_{t}H^{1-\beta }_{t}\prod_{m }^{j=1}(G_{jt})^{\gamma _{j}}(A_{t}L_{t})^{1-\alpha -\beta-\gamma } $$
They use the following regression for economies that have not reached steady state output per capita levels:
$$ ln y_{t}-ln y_{0}=a_{0}+a_{1}ln s_{K}+ a_{2} ln s_{H} + a_{3} ln s_{G} + a_{4} ln(n-x+\delta )+ a_{5} ln y_{0} + \epsilon $$
Where y is output per capita, and $s_{i}$ represent the proportion of income dedicated to private capital investment (K), human capital investment (H) and government capital investment (G).
However, they use it on a cross-country study, and I don't know whether this is suitable for a country-specific time series study.
My econometric knowledge is at Gujarati's and Dinardo's level. I know how to estimate ARDL, VAR and VECM models. I would appreciate any advice regarding a time series estimation of this model. |
The best way to ensure a good grade is to make sure you
deeply understand the topics you are supposed to learn.It is of course important to remember the routine solution methods, but you should also be able to tell intuitively and at a glance why these methods work and where any given method is applicable.You of course need to remember some key results, but you should also be able to justify those results — or even better, give a (sketch of a) proof.
The point is that if you understand the topic well, you can quickly and reliably reconstruct all necessary information.If you remember the topic as a whole, it does not matter if you forget some little details.I have a PhD in mathematics and I still occasionally forget elementary things, but I can fill the gaps.For example, if you remember the differentiation rule of quotients but you are not sure about the signs, test it with some simple functions — the sign in the general case must be the same as in any example.
Teachers often focus on telling what is true (differentiation rules, ways to calculate limits), but I strongly recommend learning also what is not true.For example, if $\lim_{x\to\infty}f(x)=0$ and $\lim_{x\to\infty}g(x)=\infty$, do we necessarily have $0<\lim_{x\to\infty}f(x)g(x)<\infty$?If you are aware of some common "false rules" that are easy to believe, you can recognize when you have made a mistake.When solving a problem, try to make sure that you understand what you are doing at all times and test your claims in special cases if you are unsure.(The last sentence may sound trivial, but many students seem not to do this.)
You will make mistakes and you will forget things.We all do.If you want to make yourself good, try to make yourself robust — so that if you forget something, you can reconstruct it based on something else, and if you make a mistake, you can recognize it yourself.
So far I have answered a question like this: "What kind of a student will almost surely get perfect grades?"Another important question is: "How does one become such a student?"
For one thing, you should know what you want to become.If you really want to understand mathematics well, examine your own skills.Ask yourself what are the most important ideas, results and methods in higher order differentiation.If you cannot answer with confidence and give a couple of examples demonstrating these ideas, you need to work more.
For another thing, do not limit your scope to the present course if possible.The big picture you create for yourself shouldn't be only about the course at hand, but mathematics as a whole.I would even suggest not trying to remember which course a given topic was covered in and which course you are having at the moment.The borders between different courses are somewhat artificial and you don't need to respect them.
Also, if you have the extra time, look what is coming ahead: find a follow-up course that builds on your current course and take a look at its book.When I was in high school (or the closest equivalent in Finland), other students thought that I didn't have to work at all because I understood quickly and could solve problems quite intuitively.The reason was that I was working ahead of them: I had already read the book of the next course, and that gave me plenty of context and motivation for the present topic and I could focus on building a solid big picture.I was working hard, but I was working on something different than others.It often happens that you properly understand something only when you have applied it in something else; no one masters the last thing they have learned.
As JPBurke suggests, working in a group also helps.But a group is not strictly necessary if you can't find equally motivated friends or suitable ways to collaborate.What you do need is someone to ask from if you don't understand something on your own.It can be a fellow student, a teacher, an older sibling or anyone willing to help.
I realize that this answer gives somewhat grandiose goals.A perfect understanding is too much to ask for, but I do suggest putting goals in this direction.For me playful interest and idle curiosity in mathematics is what kept and still keeps me going; there is no need to be serious in order to become good.The most valuable thing you can have when trying to get good grades is a passion to understand. |
Illinois Journal of Mathematics Illinois J. Math. Volume 43, Issue 2 (1999), 391-402. An injectivity result for Hermitian forms over local orders Abstract
Let $\Lambda$ be a ring endowed with an involution $a \mapsto \tilde{a}$. We say that two units $a$ and $b$ of $\Lambda$ fixed under the involution are
congruent if there exists an element $u \in \Lambda^{\times}$ such that $a = ub\tilde{u}$. We denote by $\mathcal{H}(\Lambda)$ the set of congruence classes. In this paper we consider the case where $\Lambda$ is an order with involution in a semisimple algebra $A$ over a local field and study the question of whether the natural map $\mathcal{H}(\Lambda) \rightarrow \mathcal{H}(\Lambda)$ induced by inclusion is injective. We give sufficient conditions on the order $\Lambda$ for this map to be injective and give applications to hermitian forms over group rings. Article information Source Illinois J. Math., Volume 43, Issue 2 (1999), 391-402. Dates First available in Project Euclid: 19 October 2009 Permanent link to this document https://projecteuclid.org/euclid.ijm/1255985221 Digital Object Identifier doi:10.1215/ijm/1255985221 Mathematical Reviews number (MathSciNet) MR1703194 Zentralblatt MATH identifier 0939.11020 Subjects Primary: 11E39: Bilinear and Hermitian forms Secondary: 11E08: Quadratic forms over local rings and fields 11E70: $K$-theory of quadratic and Hermitian forms 19G38: Hermitian $K$-theory, relations with $K$-theory of rings Citation
Fainsilber, Laura; Morales, Jorge. An injectivity result for Hermitian forms over local orders. Illinois J. Math. 43 (1999), no. 2, 391--402. doi:10.1215/ijm/1255985221. https://projecteuclid.org/euclid.ijm/1255985221 |
Tagged: subspace Problem 709
Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
\[ \mathbf{v}_{1}= \begin{bmatrix} 1 \\ 2 \\ 2 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{2}= \begin{bmatrix} 1 \\ 3 \\ 1 \\ 1 \end{bmatrix} ,\;\mathbf{v}_{3}= \begin{bmatrix} 1 \\ 5 \\ -1 \\ 5 \end{bmatrix} ,\;\mathbf{v}_{4}= \begin{bmatrix} 1 \\ 1 \\ 4 \\ -1 \end{bmatrix} ,\;\mathbf{v}_{5}= \begin{bmatrix} 2 \\ 7 \\ 0 \\ 2 \end{bmatrix} .\] Find a basis for the span $\Span(S)$. Problem 706
Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set
\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\] still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample. Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
\[ \ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .\]
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.Add to solve later
Problem 659
Fix the row vector $\mathbf{b} = \begin{bmatrix} -1 & 3 & -1 \end{bmatrix}$, and let $\R^3$ be the vector space of $3 \times 1$ column vectors. Define
\[W = \{ \mathbf{v} \in \R^3 \mid \mathbf{b} \mathbf{v} = 0 \}.\] Prove that $W$ is a vector subspace of $\R^3$. Problem 658
Let $V$ be the vector space of $n \times n$ matrices with real coefficients, and define
\[ W = \{ \mathbf{v} \in V \mid \mathbf{v} \mathbf{w} = \mathbf{w} \mathbf{v} \mbox{ for all } \mathbf{w} \in V \}.\] The set $W$ is called the center of $V$.
Prove that $W$ is a subspace of $V$.Add to solve later
Problem 612
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.
Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$. (a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
Add to solve later
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$. Problem 611
An $n\times n$ matrix $A$ is called
orthogonal if $A^{\trans}A=I$. Let $V$ be the vector space of all real $2\times 2$ matrices.
Consider the subset
\[W:=\{A\in V \mid \text{$A$ is an orthogonal matrix}\}.\] Prove or disprove that $W$ is a subspace of $V$. Problem 604
Let
\[A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) Find a basis for the null space $\calN(A)$. (b) Find a basis of the range $\calR(A)$. (c) Find a basis of the row space for $A$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution Problem 601
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 $W$.
Add to solve later
(The Ohio State University, Linear Algebra Midterm) Read solution |
We do plenty of math, so I’d like to test out MathJax support.
Here is an example of MathJax inline rendering — \( 1/x^{2} \). And here is a block rendering:\[ r_{XY} = \frac{\mathrm{cov}(X,Y)}{\sqrt{\mathrm{var}(X)\mathrm{var}(Y)}} \]
Now, if we’d like to get serious, we’d do something involving multiline aligned equations, like \[\begin{align}\mathcal{N}(t, \mu, \sigma) &= \mathrm{normal} \newline &= \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{(t-\mu)^2}{2 \sigma^2}}\end{align}\]or even an inline formula like \( \sum_{t=0}^{\infty} \frac{x^t}{t!} = e^x\).
Or we could try defining a command, like this. \( \newcommand{\water}{\mathrm{H}_{2}\mathrm{O}} \)
Buffer slides off the sides of our tubes like \(\water\) off a duck’s back. |
That definition just does not work.
A first attempt to fix it would be to say that locally, the tangent line and the curve only intersect once. Even then, by that definition, two intersecting lines would be tangent to each other.
A more intuitive, but not rigorous, definition would be to say that the line, $\tau$ is tangent to the function $f$ at the point $(a, f(a))$ if
$\tau(a) = f(a)$ The more you zoom in on the point $(a, f(a))$ the better the line $y = \tau(x)$ approximates the curve $y = f(x)$.
The major idea is that the tangent line, $y = \tau(x)$ is, locally, the best linear approximation to the curve $y = f(x)$ at the point $a$. There is, in fact, a formal definition of derivative in more complicated spaces,
the total derivative , which says exaclty that.
In undergrad calculus, the definition $f'(a) = \lim_{x \to a} \dfrac{f(x) - f(a)}{x-a}$ does produce that theorem $f(a + h) = f(a) + hf'(a) + O(h^2)$ which can be interpreted as saying that the line $\tau(x) = f(a) + (x-a)f'(a)$ satisfies conditions $(1.)$ and $(2.)$ stated above. |
Reference documentation for deal.II version Git 840703ecf3 2019-10-14 17:15:20 -0400
In order to compute local contributions of an individual cell to the global matrix and right hand side, we usually employ two techniques:
\[ A^K_{ij} = \int_K \nabla \varphi_i(\bf x) \cdot \nabla \varphi_j(\bf x) \; dx \]into
\[ A^K_{ij} = \int_{\hat K} \left[ J^{-1}(\hat{\bf x}) \hat \nabla \varphi_i(\hat{\bf x}) \right] \cdot \left[ J^{-1}(\hat{\bf x}) \hat \nabla \varphi_j(\hat{\bf x}) \right] \; |\textrm{det}\; J(\hat{\bf x})| \;\; d\hat x, \]where a hat indicates reference coordinates, and \(J(\hat{\bf x}_q)\) is the Jacobian \(\frac{\partial \bf F_K(\hat{\bf x})}{\partial\bf \hat x}\) of the mapping \(\bf x = \bf F_K(\hat{\bf x})\).
\[ A^K_{ij} = \sum_{q}J^{-1}\left[(\hat{\bf x}_q) \hat \nabla \varphi_i(\hat{\bf x}_q)\right] \cdot \left[J^{-1}(\hat{\bf x}_q) \hat \nabla \varphi_j(\hat{\bf x}_q)\right]\ |\textrm{det}\ J(\hat{\bf x}_q)| w_q, \]where \(q\) indicates the index of the quadrature point, \(\hat{\bf x}_q\) its location on the reference cell, and \(w_q\) its weight.
In order to evaluate such an expression in an application code, we have to access three different kinds of objects: a quadrature object that describes locations \(\hat{\bf x}_q\) and weights \(w_q\) of quadrature points on the reference cell; a finite element object that describes the gradients \(\hat\nabla \varphi_i(\hat{\bf x}_q)\) of shape functions on the unit cell; and a mapping object that provides the Jacobian as well as its determinant. Dealing with all these objects would be cumbersome and error prone.
On the other hand, these three kinds of objects almost always appear together, and it is in fact very rare for deal.II application codes to do anything with quadrature, finite element, or mapping objects besides using them together. For this reason, deal.II uses the FEValues abstraction combining information on the shape functions, the geometry of the actual mesh cell and a quadrature rule on a reference cell. Upon construction it takes one object of each of the three mentioned categories. Later, it can be "re-initialized" for a concrete grid cell and then provides mapped quadrature points and weights, mapped shape function values and derivatives as well as some properties of the transformation from the reference cell to the actual mesh cell.
Since computation of any of these values is potentially expensive (for example when using high order mappings with high order elements), the FEValues class only computes what it is explicitly asked for. To this end, it takes a list of flags of type UpdateFlags at construction time specifying which quantities should be updated each time a cell is visited. In the case above, you want the gradients of the shape functions on the real cell, which is encoded by the flag
update_gradients, as well as the product of the determinant of the Jacobian times the quadrature weight, which is mnemonically encoded using the term
JxW and encoded in the flag
update_JxW_values. Because these flags are represented by single bits in integer numbers, producing a
set of flags amounts to setting multiple bits in an integer, which is facilitated using the operation
update_gradients | update_JxW_values (in other words, and maybe slightly confusingly so, the operation "this operation
and that operation" is represented by the expression "single-bit-in-an-integer-for-this-operation binary-or single-bit-in-an-integer-for-that-operation"). To make operations cheaper, FEValues and the mapping and finite element objects it depends on really only compute those pieces of information that you have specified in the update flags (plus some information necessary to compute what has been specified, see below), and not everything that could possibly be computed on a cell. This optimization makes it much cheaper to iterate over cells for assembly, but it also means that one should take care to provide the minimal set of flags possible.
In addition, once you pass a set of flags for what you want, the functions filling the data fields of FEValues are able to distinguish between values that have to be recomputed on each cell (for example mapped gradients) and quantities that do not change from cell to cell (for example the values of shape functions of the usual \(Q_p\) finite elements at the same quadrature points on different cells; this property does not hold for the shape functions of Raviart-Thomas elements, however, which must be rotated with the local cell). This allows further optimization of the computations underlying assembly.
Let's say you want to compute the Laplace matrix as shown above. In that case, you need to specify the
update_gradients flag (for \(\nabla\varphi_i(\bf x_q)\)) and the
update_JxW_values flag (for computing \(|\textrm{det}\; J(\bf x_q)|w_q\)). Internally, however, the finite element requires the computation of the inverse of the full Jacobian matrix, \(J^{-1}(\bf x_q)\) (and not just the determinant of the matrix), and to compute the inverse of the Jacobian, it is also necessary to compute the Jacobian matrix first.
Since these are requirements that are not important to the user, it is not necessary to specify this in user code. Rather, given a set of update flags, the FEValues object first asks the finite element object what information it needs to compute in order to satisfy the user's request provided in the update flags. The finite element object may therefore add other flags to the update flags (e.g., in the example above, an FE_Q object would add
update_covariant_transformation to the list, since that is the necessary transformation from \(\hat\nabla\hat\varphi_i(\hat{\bf x}_q)\) to \(\nabla\varphi_i(\bf x_q)\)). With these updated flags, FEValues then asks the mapping whether it also wants to add more flags to the list to satisfy the needs of both the user and the finite element object, by calling Mapping::requires_update_flags(). (This procedure of first asking the finite element and then the mapping does not have to be iterated because mappings never require information computed by the finite element classes, while finite element classes typically need information computed by mappings.) Using this final list, the FEValues object then asks both the finite element object and mapping object to create temporary structures into which they can store some temporary information that can be computed once and for all, and these flags will be used when re-computing data on each cell we will visit later on.
As outlined above, we have now determined the final set of things that are necessary to satisfy a user's desired pieces of information as conveyed by the update flags she provided. This information will then typically be queried on every cell the user code visits in a subsequent integration loop.
Given that many of the things mappings or finite element classes compute are potentially expensive, FEValues employs a system whereby mappings and finite element objects are encouraged to pre-compute information that can be computed once without reference to a concrete cell, and make use of this when asked to visit a particular cell of the mesh. An example is that the values of the shape functions of the common FE_Q element are defined on the reference cell, and the values on the actual cell are simply exactly the values on the reference cell – there is consequently no need to evaluate shape functions on every cell, but it is sufficient to do this once at the beginning, store the values somewhere, and when visiting a concrete cell simply copying these values from their temporary location to the output structure. (Note, however, that this is specific to the FE_Q element: this is not the case if we used a FE_RaviartThomas element, since there, computing the values of the shape functions on a cell involves knowing the Jacobian of the mapping which depends on the geometry of the cell we visit; thus, for this element, simply copying pre-computed information is not sufficient to evaluate the values of shape functions on a particular cell.)
To accommodate this structure, both mappings and finite element classes may internally split the update flags into two sets commonly referenced as
update_once and
update_each (though these names do not appear in any public interfaces). The former contains all those pieces of information that can be pre-computed once at the time the FEValues object starts to interact with a mapping or finite element, whereas the latter contains those flags corresponding to things that need to be computed on every cell. For example, if
update_flags=update_values, then the FE_Q class will set
update_once=update_values and
update_each=0, whereas the Raviart-Thomas element will do it the other way around.
These sets of flags are intended to be mutually exclusive. There is, on the other hand, nothing that ever provides this decomposition to anything outside the mapping or finite element classes – it is a purely internal decomposition.
As outlined above, data is computed at two different times: once at the beginning on the reference cell, and once whenever we move to an actual cell. The functions involved in each of these steps are discussed next:
Computing data on the reference cell before we even visit the first real cell is a two-step process. First, the constructor of FEValues, FEFaceValues and FESubfaceValues, respectively, need to allow the Mapping and FiniteElement objects to set up internal data structures. These structures are internal in the following sense: the FEValues object asks the finite element and mapping objects to create an object of type FiniteElement::InternalDataBase and Mapping::InternalDataBase each; the actual finite element and mapping class may in fact create objects of a derived type if they wish to store some data beyond what these base classes already provide. The functions involved in this are
The FEValues object then takes over ownership of these objects and will destroy them at the end of the FEValues object's lifetime. After this, the FEValues object asks the FiniteElement and Mapping objects to fill these InternalDataBase objects with the data that pertains to what can and needs to be computed on the reference cell. This is done in these functions:
Once initialization is over and we call FEValues::reinit, FEFaceValues::reinit or FESubfaceValues::reinit to move to a concrete cell or face, we need to calculate the "update_each" kinds of data. This is done in the following functions:
This is where the actual data fields for FEValues, stored in internal::FEValues::MappingRelatedData and internal::FEValues::FiniteElementRelatedData objects, are computed. These functions call the function in Mapping first, such that all the mapping data required by the finite element is available. Then, the FiniteElement function is called. |
Simple linear regression model
$$ y_i = \alpha + \beta x_i + \varepsilon $$
can be written in terms of probabilistic model behind it
$$\mu_i = \alpha + \beta x_i \\y_i \sim \mathcal{N}(\mu_i, \sigma)$$
i.e. dependent variable $Y$ follows normal distribution parametrized by mean $\mu_i$, that is a linear function of $X$ parametrized by $\alpha,\beta$, and by standard deviation $\sigma$. If you estimate such model using ordinary least squares, you do not have to bother about the probabilistic formulation, because you are searching for optimal values of $\alpha,\beta$ parameters by minimizing the squared errors of fitted values to predicted values. On another hand, you could estimate such model using maximum likelihood estimation, where you would be looking for optimal values of parameters by maximizing the likelihood function
$$ \DeclareMathOperator*{\argmax}{arg\,max} \argmax_{\alpha,\,\beta,\,\sigma} \prod_{i=1}^n \mathcal{N}(y_i; \alpha + \beta x_i, \sigma) $$
where $\mathcal{N}$ is a density function of normal distribution evaluated at $y_i$ points, parametrized by means $\alpha + \beta x_i$ and standard deviation $\sigma$.
In Bayesian approach instead of maximizing the likelihood function alone, we would assume
prior distributions for the parameters and use Bayes theorem
$$ \text{posterior} \propto \text{likelihood} \times \text{prior} $$
The likelihood function is the same as above, but what changes is that you assume some
prior distributions for the estimated parameters $\alpha,\beta,\sigma$ and include them into the equation
$$ \underbrace{f(\alpha,\beta,\sigma\mid Y,X)}_{\text{posterior}} \propto \underbrace{\prod_{i=1}^n \mathcal{N}(y_i\mid \alpha + \beta x_i, \sigma)}_{\text{likelihood}} \; \underbrace{f_{\alpha}(\alpha) \, f_{\beta}(\beta) \, f_{\sigma}(\sigma)}_{\text{priors}} $$
"What distributions?" is a different question, since there is unlimited number of choices. For $\alpha,\beta$ parameters you could, for example assume normal distributions parametrized by some hyperparameters, or $t$-distribution if you want to assume heavier tails, or uniform distribution if you do not want to make much assumptions, but you want to assume that the parameters can be a priori "anything in the given range", etc. For $\sigma$ you need to assume some prior distribution that is bounded to be greater then zero, since standard deviation needs to be positive. This may lead to the model formulation as illustrated below by John K. Kruschke.
(source: http://www.indiana.edu/~kruschke/BMLR/)
While in maximum likelihood you were looking for a single optimal value for each of the parameters, in Bayesian approach by applying Bayes theorem you obtain the
posterior distribution of the parameters. The final estimate will depend on the information that comes from your data and from your priors, but the more information is contained in your data, the less influential are priors.
Notice that when using uniform priors, they take form $f(\theta) \propto 1$ after dropping the normalizing constants. This makes Bayes theorem proportional to likelihood function alone, so the posterior distribution will reach it's maximum at exactly the same point as maximum likelihood estimate. What follows, the estimate under uniform priors will be the same as by using ordinary least squares since minimizing the squared errors corresponds to maximizing the normal likelihood.
To estimate a model in Bayesian approach in some cases you can use
conjugate priors, so the posterior distribution is directly available (see example here). However in vast majority of cases posterior distribution will not be directly available and you will have to use Markov Chain Monte Carlo methods for estimating the model (check this example of using Metropolis-Hastings algorithm to estimate parameters of linear regression). Finally, if you are only interested in point estimates of parameters, you could use maximum a posteriori estimation, i.e.
$$ \argmax_{\alpha,\,\beta,\,\sigma} f(\alpha,\beta,\sigma\mid Y,X) $$
For more detailed description of logistic regression you can check the Bayesian logit model - intuitive explanation? thread.
For learning more you could check the following books:
Kruschke, J. (2014). Doing Bayesian Data Analysis: A Tutorial with R,
JAGS, and Stan. Academic Press.
Gelman, A., Carlin, J. B., Stern, H. S., and Rubin, D. B. (2004).
Bayesian data analysis. Chapman & Hall/CRC. |
From doing some research on the site I have found that many people have posted about solutions of $\ce{HCl}$ where $[\ce{HCl}] = \pu{1e-8 mol dm-3}$
Here they are able to deduce
$$[\ce{H+}] = \frac{[\ce{HCl}]_0}{2} + \sqrt{\frac{[\ce{HCl}]_0^2}{4} + K_\mathrm{w}}$$
where $[\ce{HCl}]_0$ is the initial concentration of $\ce{HCl}$ dissolved.
Can we make a generalisation of this and say if $\ce{H_xA}$ is a strong acid that will dissociate fully when dissolved in water? Then if $[\ce{H_xA}]_0 < \sqrt{K_\mathrm{w}}$ then the $\mathrm{pH}$ of solution can be expressed by
$$\mathrm{pH} = -\log_{10} \left(\frac{x[\ce{H_xA}]_0}{2} + \sqrt{\frac{(x[\ce{H_xA}]_0)^2}{4} + K_\mathrm{w}}\right)$$ |
The following is a linear algebra problem I have been stuck for a while now.
Be $F_i=(a_{i1}, a_{i2}, \dots, a_{im}) \in M_{1\times n}(K)$ for $i=1,2,\dots,m.$ Let $W$ be the vector subespace of $M_{1\times n}(K)$ generated by $F_1,F_2,\dots F_m$. Let $A=(a_{ij}) \in M_{m \times n}(K).$ With Gauss-Jordan elimination using only row operations we convert the matrix $A$ to the matrix $A'=(a'_{ij})$ with exactly $r$ rows not full of zeros.
Prove that the $r$ rows not full of zeros of $A'$ form a basis of $W$.
I know these $r$ rows are linearly independent, but how can they form a basis of $W$ when $W$ is generated by $m$ vectors and $ r\leq m$? |
Subject: Machine Design -I Topic: Design against static Loads,Bolted and welded joints under eccentric loading. Power Screw Difficulty: Medium Subject: Machine Design -I Topic: Design against static Loads,Bolted and welded joints under eccentric loading. Power Screw Difficulty: Medium
The efficiency of a square threaded screw is calculated by,
$\eta=\frac{tan\alpha}{tan(\phi+\alpha} $ (a)
For a self-locking screw,
$\phi \geq \alpha$
Substituting the limiting value for $ \phi=\alpha $ in eqn (a)
$\eta \leq\ \frac{tan\alpha}{tan(\phi+\phi}$
$\eta \leq\ \frac{tan\alpha}{tan2 \phi}$
$tan2 \phi= \frac{2tan\phi}{1-tan^2\phi}$
$\eta \leq\ \frac{tan\phi(1-tan^2\phi)}{2tan\phi}$
$\eta \leq\ [\frac{1}{2}-tan^2{\phi}/{2}]$
Therefore, efficiency of a self-locking square threaded power screw is less than 1/2 i.e less than 50% |
I would like to know how to compute the statistics of the discrete Fourier transform of a noise signal. To illustrate what I mean, I will first explain in detail a computation I have managed to do myself.
Suppose we have a discrete time series of values $x_n$ with $n$ from 0 to $N-1$. Each $x_n$ is a random variable, uncorrelated with the others, and Gaussian distributed with width $\sigma$. If I define the discrete Fourier transform
$$X_k = \frac{1}{N}\sum_{n=0}^{N-1} x_n e^{-2 \pi i n k / N}$$
then I find that $X_k$ is a complex random variable with real and imaginary parts Gaussian distributed with width $\sigma/\sqrt{2 N}$. I did the computation by using the fact that the distribution of a sum is the convolution of the distributions, etc.
Now I want to know how to do this computation in the case that $x_n$ are correlated. How does one approach this problem? I can make the assumption that the process is Markovian.
I had originally asked this on the Computation Science site, but I think here is a better fit. |
I'm having trouble with little o notation.
Help me show that:
$2(n^2 + 100n)\log^5n = o(n^2\sqrt{n})$.
It is the last hwk on my sheet and I don't understand it, if someone can help me with little o notation, that would be great thank you kindly.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
I'm having trouble with little o notation.
Help me show that:
$2(n^2 + 100n)\log^5n = o(n^2\sqrt{n})$.
It is the last hwk on my sheet and I don't understand it, if someone can help me with little o notation, that would be great thank you kindly.
Hint: For positive functions $f(n)$ and $g(n)$, to show that $f(n)=o(g(n))$, one needs to show that $$\lim_{n\to\infty} \frac{f(n)}{g(n)}=0.$$In our particular case, after writing down what $\dfrac{f(n)}{g(n)}$ is, I would suggest dividing top and bottom by $n^2$. You will get something like $$2\left(1+\frac{100}{n}\right) \frac{\log^5n}{n^{1/2}}.$$ Finally, you need information on the speed of growth of the logarithm, and powers of it, in comparison with the speed of growth of positive powers of $n$. It might help to note that$$\frac{\log^5 n}{n^{1/2}}=\left(\frac{\log n}{n^{1/10}}\right)^5.$$ |
Statistics - Akaike information criterion (AIC) 1 - About
AIC stands for Akaike Information Criterion.
Akaike is the name of the guy who came up with this idea.
AIC is a quantity that we can calculate for many different model types, not just linear models, but also classification model such logistic regression and so on.
2 - Articles Related 3 - Definition
The AIC criterion is defined for a large class of models fit by maximum likelihood:
<MATH> AIC = -2 log L + 2 . d </MATH>
where:
L is the maximized value of the likelihood function for the estimated model. d is the total # of parameters used in the model (regression coefficients + intercept) 4 - Linear model
<MATH> 2 log L = \frac{\href{RSS}{RSS}}{\href{variance}{\hat{\sigma}}^2} </MATH>
where:
<math>\hat{\sigma}^2</math> is the variance estimate of the error associated with each response measurement (ie each error epsilon in the linear model)
Then by plugging it in the aboe formula, we can see that AIC and Mallow's Cp are actually proportional to each other. They are the same thing for linear models. |
What is the relation between estimator and estimate?
E. L. Lehmann, in his classic
Theory of Point Estimation, answers this question on pp 1-2.
The observations are now postulated to be the values taken on by random variables which are assumed to follow a joint probability distribution, $P$, belonging to some known class...
...let us now specialize to point estimation...suppose that $g$ is a real-valued function defined [on the stipulated class of distributions] and that we would like to know the value of $g$ [at whatever is the actual distribution in effect, $\theta$]. Unfortunately, $\theta$, and hence $g(\theta)$, is unknown. However, the data can be used to obtain an estimate of $g(\theta)$, a value that one hopes will be close to $g(\theta)$.
In words: an
estimator is a definite mathematical procedure that comes up with a number (the estimate) for any possible set of data that a particular problem could produce. That number is intended to represent some definite numerical property ($g(\theta)$) of the data-generation process; we might call this the "estimand."
The estimator itself is
not a random variable: it's just a mathematical function. However, the estimate it produces is based on data which themselves are modeled as random variables. This makes the estimate (thought of as depending on the data) into a random variable and a particular estimate for a particular set of data becomes a realization of that random variable.
In one (conventional) ordinary least squares formulation, the data consist of ordered pairs $(x_i, y_i)$. The $x_i$ have been determined by the experimenter (they can be amounts of a drug administered, for example). Each $y_i$ (a response to the drug, for instance) is assumed to come from a probability distribution that is Normal but with unknown mean $\mu_i$ and
common variance $\sigma^2$. Furthermore, it is assumed that the means are related to the $x_i$ via a formula $\mu_i = \beta_0 + \beta_1 x_i$. These three parameters--$\sigma$, $\beta_0$, and $\beta_1$--determine the underlying distribution of $y_i$ for any value of $x_i$. Therefore any property of that distribution can be thought of as a function of $(\sigma, \beta_0, \beta_1)$. Examples of such properties are the intercept $\beta_0$, the slope $\beta_1$, the value of $\cos(\sigma + \beta_0^2 - \beta_1)$, or even the mean at the value $x=2$, which (according to this formulation) must be $\beta_0 + 2 \beta_1$.
In this OLS context, a
non-example of an estimator would be a procedure to guess at the value of $y$ if $x$ were set equal to 2. This is not an estimator because this value of $y$ is random (in a way completely separate from the randomness of the data): it is not a (definite numerical) property of the distribution, even though it is related to that distribution. (As we just saw, though, the expectation of $y$ for $x=2$, equal to $\beta_0 + 2 \beta_1$, can be estimated.)
In Lehmann's formulation, almost any formula can be an estimator of almost any property.
There is no inherent mathematical link between an estimator and an estimand. However, we can assess--in advance--the chance that an estimator will be reasonably close to the quantity it is intended to estimate. Ways to do this, and how to exploit them, are the subject of estimation theory. In short: an estimator is a function and an estimate is a value that summarizes an observed sample.
An
estimator is a function that maps a random sample to the parameter estimate:
$$\hat{\Theta}=t(X_1,X_2,...,X_n)$$Note that an estimator of
n random variables $X_1,X_2,...,X_n$ is a random variable $\hat{\Theta}$. For instance, an estimator is the sample mean:$$\overline{X}=\frac{1}{n}\sum_{n=1}^nX_i$$An estimate $\hat{\theta}$ is the result of applying the estimator function to a lowercase observed sample $x_1,x_2,...,x_n$:
$$ \hat{\theta}=t(x_1,x_2,...,x_n) $$ For instance, an estimate of the observed sample $x_1,x_2,...,x_n$ is the sample mean: $$ \hat{\mu}=\overline{x}=\frac{1}{n}\sum_{n=1}^nx_i $$
It might be helpful to illustrate whuber's answer in the context of a linear regression model. Let's say you have some bivariate data and you use Ordinary Least Squares to come up with the following model:
Y = 6X + 1
At this point, you can take any value of X, plug it into the model and predict the outcome, Y. In this sense, you might think of the individual components of the generic form of the model (
mX + B) as estimators. The sample data (which you presumably plugged into the generic model to calculate the specific values for m and B above) provided a basis on which you could come up with estimates for m and B respectively.
Consistent with @whuber's points in our thread below, whatever values of
Y a particular set of estimators generate you for are, in the context of linear regression, thought of as predicted values. (edited -- a few times -- to reflect the comments below)
Suppose you received some data, and you had some observed variable called theta. Now your data can be from a distribution of data, for this distribution, there is a corresponding value of theta that you infer which is a random variable. You can use the MAP or mean for calculating the estimate of this random variable whenever the distribution of your data changes. So the random variable theta is known as an
estimate, a single value of the unobserved variable for a particular type of data.
While estimator is your data, which is also a random variable. For different types of distributions you have different types of data and thus you have a different estimate and thus this corresponding random variable is called the
estimator. |
An exponential equation is an expression where both sides can be presented in the form of same based and it can be solved with the help of a property. It is generally used to express a graph in many applications like Compound interest, radioactive decay, or growth of population etc. The general form of an exponential equation includes –
\[\large y=a^{x}\]
Where a are the constants and x, y are the variables.
The exponential function is a special type where the input variable works as the exponent. A function f(x) = bx + c or function f(x) = a, both are the exponential functions. It is used everywhere, if we talk about the C programming language then the exponential function is defined as the e raised to the power x. Here, x could be any real number. The syntax for exponential functions in C programming is given as –
The mean of the
Exponential (λ) Distribution is calculated using integration by parts as –
\[\large E(X) = \int_{0}^{\infty } x\lambda e^{-\lambda x} \; dx\]
\[\large = \lambda \left [ \frac{-x \; e^{-\lambda x}}{\lambda}|_{0}^{\infty } + \frac{1}{\lambda }\int_{0}^{\infty } e^{-\lambda x} dx \right ]\]
\[\large = \lambda \left [ 0 + \frac{1}{\lambda }\frac{-e^{-\lambda x}}{\lambda} |_{o}^{\infty }\right ]\]
\[\large = \lambda \frac{1}{\lambda ^{2} }\]
\[\large = \frac{1}{\lambda }\]
double exp (double x);
Parameters or Arguments – X
In case, the magnitude of the variable is too large then it may throwback an error. Understanding exponential functions are not easy but it is necessary when they are needed to use for the real-life applications.
The exponential distribution in probability is the distribution that explains the time among events in a Poisson process. Further, we will discuss the exponential growth and exponential decay formulas and how can you use them practically.
Exponential growth is the condition where the growth rate of the mathematical function is directly proportional to the current value of the function that results in growth with time being an exponential function. The Exponential growth formula in mathematics is given as –
Formula of Exponential Growth
\[\large P(t)=P_{0}e^{rt}\]
Where:
t = time (number of periods) P(t) = the amount of some quantity at time t P 0 = initial amount at time t = 0 r = the growth rate
Where t is the time (total number of periods), P(t) is the amount of a quantity at given time t, P0 is the initial among at the time t = 0, and r is taken as the growth rate.
Do you know the fact that most of the graphs have the same arcing shape? Any graph could not have a constant rate of change but it may constant ratios that grows by common factors over particular intervals of time.
There are two popular cases in case of Exponential equations. These are the exponential growth and the exponential decay. In the case of Exponential Growth, quantity will increase slowly at first then rapidly. In the case of exponential decay, the quantity will decrease faster at first then it will move slowly. Their formulas can be given as shown below:
\[\large y=ab^{x}\]
Here,
x and y are the variables a and b are constants. |
Let $T: \R^n \to \R^m$ be a linear transformation.Suppose that the nullity of $T$ is zero.
If $\{\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k\}$ is a linearly independent subset of $\R^n$, then show that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.
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.
Let $G, H, K$ be groups. Let $f:G\to K$ be a group homomorphism and let $\pi:G\to H$ be a surjective group homomorphism such that the kernel of $\pi$ is included in the kernel of $f$: $\ker(\pi) \subset \ker(f)$.
Define a map $\bar{f}:H\to K$ as follows.For each $h\in H$, there exists $g\in G$ such that $\pi(g)=h$ since $\pi:G\to H$ is surjective.Define $\bar{f}:H\to K$ by $\bar{f}(h)=f(g)$.
(a) Prove that the map $\bar{f}:H\to K$ is well-defined.
(b) Prove that $\bar{f}:H\to K$ is a group homomorphism.
Let $\calF[0, 2\pi]$ be the vector space of all real valued functions defined on the interval $[0, 2\pi]$.Define the map $f:\R^2 \to \calF[0, 2\pi]$ by\[\left(\, f\left(\, \begin{bmatrix}\alpha \\\beta\end{bmatrix} \,\right) \,\right)(x):=\alpha \cos x + \beta \sin x.\]We put\[V:=\im f=\{\alpha \cos x + \beta \sin x \in \calF[0, 2\pi] \mid \alpha, \beta \in \R\}.\]
(a) Prove that the map $f$ is a linear transformation.
(b) Prove that the set $\{\cos x, \sin x\}$ is a basis of the vector space $V$.
(c) Prove that the kernel is trivial, that is, $\ker f=\{\mathbf{0}\}$.(This yields an isomorphism of $\R^2$ and $V$.)
(d) Define a map $g:V \to V$ by\[g(\alpha \cos x + \beta \sin x):=\frac{d}{dx}(\alpha \cos x+ \beta \sin x)=\beta \cos x -\alpha \sin x.\]Prove that the map $g$ is a linear transformation.
(e) Find the matrix representation of the linear transformation $g$ with respect to the basis $\{\cos x, \sin x\}$.
Suppose that the vectors\[\mathbf{v}_1=\begin{bmatrix}-2 \\1 \\0 \\0 \\0\end{bmatrix}, \qquad \mathbf{v}_2=\begin{bmatrix}-4 \\0 \\-3 \\-2 \\1\end{bmatrix}\]are a basis vectors for the null space of a $4\times 5$ matrix $A$. Find a vector $\mathbf{x}$ such that\[\mathbf{x}\neq0, \quad \mathbf{x}\neq \mathbf{v}_1, \quad \mathbf{x}\neq \mathbf{v}_2,\]and\[A\mathbf{x}=\mathbf{0}.\]
(Stanford University, Linear Algebra Exam Problem)
Let $V$ be the subspace of $\R^4$ defined by the equation\[x_1-x_2+2x_3+6x_4=0.\]Find a linear transformation $T$ from $\R^3$ to $\R^4$ such that the null space $\calN(T)=\{\mathbf{0}\}$ and the range $\calR(T)=V$. Describe $T$ by its matrix $A$.
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 $n-1$.
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 $\R^n$.
(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 $\mathbf{v}\in \calN(T)$.
Let $A$ be the matrix for a linear transformation $T:\R^n \to \R^n$ with respect to the standard basis of $\R^n$.We assume that $A$ is idempotent, that is, $A^2=A$.Then prove that\[\R^n=\im(T) \oplus \ker(T).\]
(a) Let $A=\begin{bmatrix}1 & 2 & 1 \\3 &6 &4\end{bmatrix}$ and let\[\mathbf{a}=\begin{bmatrix}-3 \\1 \\1\end{bmatrix}, \qquad \mathbf{b}=\begin{bmatrix}-2 \\1 \\0\end{bmatrix}, \qquad \mathbf{c}=\begin{bmatrix}1 \\1\end{bmatrix}.\]For each of the vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$, determine whether the vector is in the null space $\calN(A)$. Do the same for the range $\calR(A)$.
(b) Find a basis of the null space of the matrix $B=\begin{bmatrix}1 & 1 & 2 \\-2 &-2 &-4\end{bmatrix}$.
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)
Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by\[\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,\]where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring homomorphism, called the augmentation map and the kernel of $\epsilon$ is called the augmentation ideal.
(a) Prove that the augmentation ideal in the group ring $RG$ is generated by $\{g-e \mid g\in G\}$.
(b) Prove that if $G=\langle g\rangle$ is a finite cyclic group generated by $g$, then the augmentation ideal is generated by $g-e$. |
Search
Now showing items 1-1 of 1
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 ... |
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
That seems like what I need to do, but I don't know how to actually implement it... how wide of a time window is needed for the Y_{t+\tau}? And how on earth do I load all that data at once without it taking forever?
And is there a better or other way to see if shear strain does cause temperature increase, potentially delayed in time
Link to the question: Learning roadmap for picking up enough mathematical know-how in order to model "shape", "form" and "material properties"?Alternatively, where could I go in order to have such a question answered?
@tpg2114 For reducing data point for calculating time correlation, you can run two exactly the simulation in parallel separated by the time lag dt. Then there is no need to store all snapshot and spatial points.
@DavidZ I wasn't trying to justify it's existence here, just merely pointing out that because there were some numerics questions posted here, some people might think it okay to post more. I still think marking it as a duplicate is a good idea, then probably an historical lock on the others (maybe with a warning that questions like these belong on Comp Sci?)
The x axis is the index in the array -- so I have 200 time series
Each one is equally spaced, 1e-9 seconds apart
The black line is \frac{d T}{d t} and doesn't have an axis -- I don't care what the values are
The solid blue line is the abs(shear strain) and is valued on the right axis
The dashed blue line is the result from scipy.signal.correlate
And is valued on the left axis
So what I don't understand: 1) Why is the correlation value negative when they look pretty positively correlated to me? 2) Why is the result from the correlation function 400 time steps long? 3) How do I find the lead/lag between the signals? Wikipedia says the argmin or argmax of the result will tell me that, but I don't know how
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
Because I don't know how the result is indexed in time
Related:Why don't we just ban homework altogether?Banning homework: vote and documentationWe're having some more recent discussions on the homework tag. A month ago, there was a flurry of activity involving a tightening up of the policy. Unfortunately, I was really busy after th...
So, things we need to decide (but not necessarily today): (1) do we implement John Rennie's suggestion of having the mods not close homework questions for a month (2) do we reword the homework policy, and how (3) do we get rid of the tag
I think (1) would be a decent option if we had >5 3k+ voters online at any one time to do the small-time moderating. Between the HW being posted and (finally) being closed, there's usually some <1k poster who answers the question
It'd be better if we could do it quick enough that no answers get posted until the question is clarified to satisfy the current HW policy
For the SHO, our teacher told us to scale$$p\rightarrow \sqrt{m\omega\hbar} ~p$$$$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$And then define the following$$K_1=\frac 14 (p^2-q^2)$$$$K_2=\frac 14 (pq+qp)$$$$J_3=\frac{H}{2\hbar\omega}=\frac 14(p^2+q^2)$$The first part is to show that$$Q \...
Okay. I guess we'll have to see what people say but my guess is the unclear part is what constitutes homework itself. We've had discussions where some people equate it to the level of the question and not the content, or where "where is my mistake in the math" is okay if it's advanced topics but not for mechanics
Part of my motivation for wanting to write a revised homework policy is to make explicit that any question asking "Where did I go wrong?" or "Is this the right equation to use?" (without further clarification) or "Any feedback would be appreciated" is not okay
@jinawee oh, that I don't think will happen.
In any case that would be an indication that homework is a meta tag, i.e. a tag that we shouldn't have.
So anyway, I think suggestions for things that need to be clarified -- what is homework and what is "conceptual." Ie. is it conceptual to be stuck when deriving the distribution of microstates cause somebody doesn't know what Stirling's Approximation is
Some have argued that is on topic even though there's nothing really physical about it just because it's 'graduate level'
Others would argue it's not on topic because it's not conceptual
How can one prove that$$ \operatorname{Tr} \log \cal{A} =\int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr}e^{-s \mathcal{A}},$$for a sufficiently well-behaved operator $\cal{A}?$How (mathematically) rigorous is the expression?I'm looking at the $d=2$ Euclidean case, as discuss...
I've noticed that there is a remarkable difference between me in a selfie and me in the mirror. Left-right reversal might be part of it, but I wonder what is the r-e-a-l reason. Too bad the question got closed.
And what about selfies in the mirror? (I didn't try yet.)
@KyleKanos @jinawee @DavidZ @tpg2114 So my take is that we should probably do the "mods only 5th vote"-- I've already been doing that for a while, except for that occasional time when I just wipe the queue clean.
Additionally, what we can do instead is go through the closed questions and delete the homework ones as quickly as possible, as mods.
Or maybe that can be a second step.
If we can reduce visibility of HW, then the tag becomes less of a bone of contention
@jinawee I think if someone asks, "How do I do Jackson 11.26," it certainly should be marked as homework. But if someone asks, say, "How is source theory different from qft?" it certainly shouldn't be marked as Homework
@Dilaton because that's talking about the tag. And like I said, everyone has a different meaning for the tag, so we'll have to phase it out. There's no need for it if we are able to swiftly handle the main page closeable homework clutter.
@Dilaton also, have a look at the topvoted answers on both.
Afternoon folks. I tend to ask questions about perturbation methods and asymptotic expansions that arise in my work over on Math.SE, but most of those folks aren't too interested in these kinds of approximate questions. Would posts like this be on topic at Physics.SE? (my initial feeling is no because its really a math question, but I figured I'd ask anyway)
@DavidZ Ya I figured as much. Thanks for the typo catch. Do you know of any other place for questions like this? I spend a lot of time at math.SE and they're really mostly interested in either high-level pure math or recreational math (limits, series, integrals, etc). There doesn't seem to be a good place for the approximate and applied techniques I tend to rely on.
hm... I guess you could check at Computational Science. I wouldn't necessarily expect it to be on topic there either, since that's mostly numerical methods and stuff about scientific software, but it's worth looking into at least.
Or... to be honest, if you were to rephrase your question in a way that makes clear how it's about physics, it might actually be okay on this site. There's a fine line between math and theoretical physics sometimes.
MO is for research-level mathematics, not "how do I compute X"
user54412
@KevinDriscoll You could maybe reword to push that question in the direction of another site, but imo as worded it falls squarely in the domain of math.SE - it's just a shame they don't give that kind of question as much attention as, say, explaining why 7 is the only prime followed by a cube
@ChrisWhite As I understand it, KITP wants big names in the field who will promote crazy ideas with the intent of getting someone else to develop their idea into a reasonable solution (c.f., Hawking's recent paper) |
I don't completely understand your notation; my apologies if I misunderstand. But I think MI is too weak for regression, because it only enforces dependence between variables, rather than any specific relationship (as regression requires). (In fact, it is this generality that makes MI better for detecting hidden statistical relationships.)
For example, in some latent variable generative models (e.g., VAEs, see
ELBO surgery: yet another way to carve up the variational evidence lower bound, Hoffman & Johnson), we maximize the mutual information between the data $x$ and its latent representation $z$. But we have to constrain this with other terms (reconstruction likelihood and matching the prior). Indeed, optimizing mutual information is a very active research area in modern machine learning (which I suppose could be considered research in the area of regression), especially in models like variational auto-encoders and generative adversarial networks (see the bottom of this answer).
Separately, something interesting to you might be Bayesian linear regression, which looks at regression from a more general probabilistic viewpoint.
On the other hand, it did make me think of the following simple idea, since mutual information is closely related to KL-divergence:$$ \mathcal{I}[A;B] = \mathbb{E}_{B}\left[ \mathfrak{D}_\text{KL}[P(A|B)\,||\,P(A)] \right] $$
Suppose we assume the following
data generation process:$$ x\sim\mathcal{U},\;\; y \,|\, x\sim\mathcal{N}(\alpha_r x +\beta_r, \sigma^2_r) $$We want to learn the parameters $\alpha,\beta$ of a stochastic regressor: given $x$,$$ \hat{y}\sim\mathcal{N}(\alpha x + \beta, \sigma^2) $$noting that the KL-divergence is asymmetric,what is the KL-divergence on the conditional distributions between the data generator and the regressor?\begin{align}\mathfrak{D}_\text{KL}[p(y|x)\,||\,p(\hat{y}|x)]&= \log(\sigma_\hat{y} / \sigma_y) + \frac{\sigma_y^2 + (\mu_y - \mu_{\hat{y}})^2}{2\sigma_{\hat{y}}^2} - \frac{1}{2}\\&= \log(\sigma) - \log(\sigma_r) + \frac{\sigma_r^2 + (\alpha_r x +\beta_r - \alpha x - \beta)^2}{2\sigma^2} - \frac{1}{2}\\\end{align}Suppose we have data $D=\{ (x_i,y_i) \}$. Then we want to minimize\begin{align} \sum_i \mathfrak{D}_\text{KL}[p(y_i|x_i)\,||\,p(\hat{y}_i|x_i)] &= c + \frac{1}{2\sigma^2} \sum_i (y_i - \alpha x_i - \beta)^2\end{align}wrt $\alpha$ and $\beta$,where $c$ is a constant (assuming $\sigma$ and $\sigma_r$ are).But this just amounts to minimizing the squared error loss here. |
I want to generate correlated random variables with a given correlation matrix, means, and variances. Does the Cholesky decomposition only work when the initial random variables are iids with the same mean and variance?
Let Z be uncorrelated random variables normally distributed with mean 0 and variance 1. This means $$Z\sim N\left(0,I\right)$$ If you make an affine transformation $$X\equiv A+BZ$$ then $X$ has a distribution $$X\sim N\left(A,BB'\right)$$ In our case, we want $BB'=\Sigma$, so applying the cholesky decomposition to $\Sigma$ is one way to find a suitable $B$. Thus, to simulate from $X\sim N\left(\mu,\Sigma\right)$, you would simulate $Z$, set $A=\mu$ and $B=chol(\Sigma)$, and apply the transformation above.
So to answer your question, uncorrelated variables of mean 0 and variance 1 can be transformed to generic multivariate normal distributions through the use of affine transformations, depending on the mean vector and cholesky decomposition of the covariance matrix. This is how multivariate normal random number generators generally work.
To address your other point in the comments, suppose you simulate $Z$ and make a transformation $$Y\equiv chol(C)Z$$ so that $Y\sim N(0, C)$. You can still get to $X$ from $Y$. As I discussed in the comments, you can easily do this by multiplying the univariate distributions by the respective standard deviations and adding the respective mean.
More formally, you can consider this another affine transformation $$X\equiv A+SY$$ However, since $Y$ is not uncorrelated, the affine transformation would give $$X\sim N\left(A,SCS'\right)$$ In this case, to get $SCS'=\Sigma$, you would need to set $S$ equal to a matrix with the standard deviations on the diagonal and zeros elsewhere. This is equivalent to applying the transformations on a univariate basis.
Additionnal items:
We need to decompose the convariance matrix Σ into orthogonal matrix product as mentionned above.
There are different ways to get the orthogonal matrix B, Cholesky is one of them. We can also use SVD (Singular Value Decomposition) to get B, equivalent to calculate the eigenvalues/vectors for PDM.
If you want to use the decomposition to simulate brownians over time, you can also use Brownian Bridge with Haar transformation. |
584 0
Can someone please help me out with the following question?
Q. A simple harmonic oscillator, of mass m and natural frequency w_0, experiences an oscillating driving force f(t) = macos(wt). Therefore its equation of motion is:
[tex]\frac{{d^2 x}}{{dt^2 }} + \omega _0 ^2 x = a\cos \left( {\omega t} \right)[/tex]
Given that at t = 0 we have x = dx/dt = 0, find the function x(t). Describe the solution if w is approximately, but not exactly, equal to w_0.
I got:
[tex]y\left( t \right) = \frac{a}{{\left( {\omega _0 ^2 - \omega ^2 } \right)}}\left( {\cos \left( {\omega t} \right) - \cos \left( {\omega _0 t} \right)} \right)[/tex]
The answer says a couple of things about the behaviour of the solution for w ~ w_0 but I can't figure out how they got it. For instance "for large t it shows beats of maximum amplitude 2((w_0)^2 - w^2)^-1." How is that deduced and how would I determine which are the main characterstics of motion that I need to note. Any help would be appreciated.
Q. A simple harmonic oscillator, of mass m and natural frequency w_0, experiences an oscillating driving force f(t) = macos(wt). Therefore its equation of motion is:
[tex]\frac{{d^2 x}}{{dt^2 }} + \omega _0 ^2 x = a\cos \left( {\omega t} \right)[/tex]
Given that at t = 0 we have x = dx/dt = 0, find the function x(t). Describe the solution if w is approximately, but not exactly, equal to w_0.
I got:
[tex]y\left( t \right) = \frac{a}{{\left( {\omega _0 ^2 - \omega ^2 } \right)}}\left( {\cos \left( {\omega t} \right) - \cos \left( {\omega _0 t} \right)} \right)[/tex]
The answer says a couple of things about the behaviour of the solution for w ~ w_0 but I can't figure out how they got it. For instance "for large t it shows beats of maximum amplitude 2((w_0)^2 - w^2)^-1." How is that deduced and how would I determine which are the main characterstics of motion that I need to note. Any help would be appreciated.
Last edited: |
The Force of Gravity and Gravitational Potential
The law of universal gravitation was formulated by Isaac Newton \(\left(1643-1727\right)\) and published in \(1687.\)
In accordance with this law, two point masses attract each other with a force that is directly proportional to the masses of these bodies \({m_1}\) and \({m_2},\) and inversely proportional to the square of the distance between them:
\[F = G\frac{{{m_1}{m_2}}}{{{r^2}}}.\]
Here, \(r\) is the distance between the centers of mass of the bodies, \(G\) is the gravitational constant, whose value found by experiment is \(G =\) \( 6,67 \times {10^{ – 11}}{\large\frac{{{\text{m}^3}}}{{\text{kg} \cdot {\text{s}^2}}}\normalsize}.\)
The force of gravitational attraction is a central force, that is directed along a line passing through the centers of the interacting bodies.
In a system of two bodies (Figure \(2\)), the attraction force \({\mathbf{F}_{12}}\) of the second body acts on the first body of mass \({m_1}.\)
Similarly, the attraction force \({\mathbf{F}_{21}}\) of the first body acts on the second body of mass \({m_2}.\) Both the forces \({\mathbf{F}_{12}}\) and \({\mathbf{F}_{21}}\) are equal and directed along \(\mathbf{r},\) where
\[\mathbf{r} = {\mathbf{r}_2} – {\mathbf{r}_1}.\]
Using the Newton’s second law we can write the following differential equations describing the motion of each body:
\[
{{m_1}\frac{{{d^2}{\mathbf{r}_1}}}{{d{t^2}}} = G\frac{{{m_1}{m_2}}}{{{r^3}}}\mathbf{r},\;\;\;}\kern-0.3pt {{m_2}\frac{{{d^2}{\mathbf{r}_2}}}{{d{t^2}}} = – G\frac{{{m_1}{m_2}}}{{{r^3}}}\mathbf{r}} \]
or
\[
{\frac{{{d^2}{\mathbf{r}_1}}}{{d{t^2}}} = G\frac{{{m_2}}}{{{r^3}}}\mathbf{r},\;\;\;}\kern-0.3pt {\frac{{{d^2}{\mathbf{r}_2}}}{{d{t^2}}} = – G\frac{{{m_1}}}{{{r^3}}}\mathbf{r}.} \]
It follows from the last two equations that
\[
{\frac{{{d^2}{\mathbf{r}_1}}}{{d{t^2}}} – \frac{{{d^2}{\mathbf{r}_2}}}{{d{t^2}}} }={ G\frac{{{m_2}}}{{{r^3}}}\mathbf{r} + G\frac{{{m_1}}}{{{r^3}}}\mathbf{r},\;\;}\Rightarrow {\frac{{{d^2}\mathbf{r}}}{{d{t^2}}} = -G\frac{{{m_1} + {m_2}}}{{{r^3}}}\mathbf{r}.} \]
This differential equation describes the change of the vector \(\mathbf{r}\left( t \right),\) i.e. the relative motion of two bodies under the force of gravitational attraction.
With a large difference in mass of the bodies, we can neglect the smaller body mass in the right side of this equation. For example, the mass of the Sun is \(333,000\) times greater than the mass of the Earth. In this case, the differential equation can be written in a simpler form:
\[\frac{{{d^2}\mathbf{r}}}{{d{t^2}}} = – G\frac{{{M_\text{C}}}}{{{r^3}}}\mathbf{r},\]
where \({M_\text{S}}\) is the mass of the Sun.
The gravitational interaction of bodies takes place through a gravitational field, which can be described by a scalar potential \(\varphi.\) The force acting on a body of mass \(m,\) placed in a field with potential \(\varphi,\) is equal to
\[{\mathbf{F} = m\mathbf{a} }={ – m\,\mathbf{\text{grad}}\,\varphi .}\]
In the case of a point mass \(M,\) the potential of the gravitational field is given by
\[\varphi = – \frac{{GM}}{r}.\]
The latter formula is also valid for distributed bodies with central symmetry, such as a planet or star.
Kepler’s Laws
The basic laws of planetary motion were established by Johannes Kepler \(\left(1571-1630\right)\) based on the analysis of astronomical observations of Tycho Brahe \(\left(1546-1601\right)\). In \(1609,\) Kepler formulated the first two laws. The third law was discovered in \(1619.\) Later, in the late \(17\)th century, Isaac Newton proved mathematically that all three laws of Kepler are a consequence of the law of universal gravitation.
Kepler’s First Law
The orbit of each planet in the solar system is an ellipse, one focus of which is the Sun (Figure \(3\)).
Kepler’s Second Law
The radius vector connecting the Sun and the planet describes equal areas in equal intervals of time. Figure \(4\) shows the two sectors of the ellipse corresponding to the same time intervals.
According to Kepler’s second law, the areas of these sectors are equal.
Kepler’s Third Law
The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit:
\[{T^2} \propto {a^3}.\]
The proportionality coefficient is the same for all planets in the solar system. Therefore, for any two planets, one can write the relationship
\[\frac{{T_2^2}}{{T_1^2}} = \frac{{a_2^3}}{{a_1^3}}.\]
Solved Problems
Click a problem to see the solution. |
Research talks;Algebraic and Complex Geometry;Topology
Let $\overline{M_{g,n}}$ be the moduli space of stable curves of genus $g$ with $n$ marked points. It is a classical problem in algebraic geometry to determine which of these spaces are rational over $\mathbb{C}$. In this talk, based on joint work with Mathieu Florence, I will address the rationality problem for twisted forms of $\overline{M_{g,n}}$ . Twisted forms of $\overline{M_{g,n}}$ are of interest because they shed light on the arithmetic geometry of $\overline{M_{g,n}}$, and because they are coarse moduli spaces for natural moduli problems in their own right. A classical result of Yu. I. Manin and P. Swinnerton-Dyer asserts that every form of $\overline{M_{0,5}}$ is rational. (Recall that the $F$-forms $\overline{M_{0,5}}$ are precisely the del Pezzo surfaces of degree 5 defined over $F$.) Mathieu Florence and I have proved the following generalization of this result.
Let $ n\geq 5$ is an integer, and $F$ is an infinite field of characteristic $\neq$ 2. (a) If $ n$ is odd, then every twisted $F$-form of $\overline{M_{0,n}}$ is rational over $F$. (b) If $n$ is even, there exists a field extension $F/k$ and a twisted $F$-form of $\overline{M_{0,n}}$ which is unirational but not retract rational over $F$. We also have similar results for forms of $\overline{M_{g,n}}$ , where $g \leq 5$ (for small $n$ ). In the talk, I will survey the geometric results we need about $\overline{M_{g,n}}$ , explain how our problem reduces to the Noether problem for certain twisted goups, and how this Noether problem can (sometimes) be solved. Keywords: rationality - moduli spaces of marked curves - Galois cohomology - Noether's problem
Let $\overline{M_{g,n}}$ be the moduli space of stable curves of genus $g$ with $n$ marked points. It is a classical problem in algebraic geometry to determine which of these spaces are rational over $\mathbb{C}$. In this talk, based on joint work with Mathieu Florence, I will address the rationality problem for twisted forms of $\overline{M_{g,n}}$ . Twisted forms of $\overline{M_{g,n}}$ are of interest because they shed light on the ...
14E08 ; 14H10 ; 20G15
... Lire [+] |
Thermodynamics Heat Engines, Refrigerators and Heat pumps A heat engine is a device which converts thermal energy in to other useful forms of energy such as Mechanical energy, Electrical energy Efficiency of heat engine \tt \eta=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}, where Q 1is the absorbed heat and Q 2is the rejected heat. The efficiency of an irreversible engine is always less than or equal to that of reversible engine when operated between the same temperature limits. A refrigerator is just reverse to heat engine in refrigerator the working substance extracts an amount of heat from cold reservoir. Coefficient of performance of a refrigerator \tt \beta=\frac{Q_2}{W}, where Q 2is amount of heat extracts from cold reservoir. W is the external work. View the Topic in this video From 46:53 To 1:02:13
Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers.
1. Thermal efficiency of a heat engine is given by \tt \eta = \frac{Work \ done / cycle}{Total \ amount \ of \ heat \ absorbed/cycle}
\tt \eta = 1 - \frac{Q_{2}}{Q_{1}} = 1 - \frac{T_{2}}{T_{1}}
2. A refrigerator or heat pump is a device used for cooling things. It absorb heat from sink at lower temperature and reject a large amount of heat to source at higher temperature.
Coefficient of performance of refrigerator is given by \beta = \frac{Q_{2}}{W} = \frac{Q_{2}}{Q_{1} - Q_{2}} = \frac{T_{2}}{T_{1} - T_{2}}
3. Relation between efficiency (η) and coefficient of performance (β)
\beta = \frac{1 - \eta}{\eta} |
The Discrete Hilbert Transform is an "ideal" (implying that this is not the practical implementation) linear time-invariant filter with input $x[n]$ having output
$$ \mathscr{H}\Big\{ x[n] \Big\} \triangleq \hat{x}[n] = \sum\limits_{i=-\infty}^{\infty} h[i] \, x[n-i] $$
Where the impulse response of a discrete-time Hilbert transformer is:
$$ h[n] = \begin{cases} \frac{\big(1 - (-1)^n\big)}{\pi n} \quad & n \ne 0 \\\\0 & n = 0\end{cases}$$
Because $1 - (-1)^n = 0$ for even $n$, this can be restated as
$$ h[n] = \begin{cases} \frac{2}{\pi n} \quad & n \text{ odd} \\\\0 & n \text{ even}\end{cases}$$
This is not a causal impulse response, nor is it finite in length. To make it finite in length, you would need to window it with a decent window:
$$ h[n] = \begin{cases} \frac{2}{\pi n} w[n] \quad & n \text{ odd} \\\\0 & n \text{ even}\end{cases}$$
where $w[n]$ is some window function of width $L+1$ samples (and $L$ is an even positive integer). If it were a Hamming Window, it would be:
$$ w[n] = \begin{cases} 0.54 \ + \ 0.46 \cdot \cos\left(\pi \frac{n}{L/2} \right) \qquad & |n| \le L/2 \\ 0 & |n| > L/2 \\\end{cases}$$
If it were a Kaiser window it would be
$$ w[n] = \begin{cases} \frac{1}{I_0(\beta)} \, I_0\left(\beta \sqrt{1 - \left(\frac{n}{L/2}\right)^2 } \right) \qquad & |n| \le L/2 \\ 0 & |n| > L/2 \\\end{cases}$$
where
$$ I_0(u) \triangleq \sum\limits_{k=0}^{\infty} \frac{(-1)^k \big( \tfrac{u}{2} \big)^{2k}}{(k!)^2} $$
$I_0(x)$ is the 0th-order modified Bessel function of the 1st kind. $L+1$ is the number of non-zero samples or FIR taps (the FIR filter order is $L$) and, in my centered and symmetrical case, must be even. $\beta$ is a
"shape parameter", maybe around 5 or 6, i dunno.
Now, to make this causal, your impulse response has to be delayed by $\frac{L}2$ samples to be $h[n-\frac{L}2]$, but then should also all other signals that this Hilbert output is compared to, they should also be delayed by $\frac{L}2$ samples to keep the phase relationship correct.
With a windowed finite-length impulse response (which is what we call an
"FIR"), the Hilbert transform output is
$$ \hat{x}[n] = \sum\limits_{i=-L/2}^{L/2} h[i] \, x[n-i] $$
and delaying the output so that the filter is causal you get
$$ \hat{x}[n-\tfrac{L}2] = \sum\limits_{i=0}^{L} h[i-\tfrac{L}2] \, x[n-i] $$
but with this delayed output $\hat{x}[n-\tfrac{L}2]$, you must compare that only to the like delayed input $x[n-\tfrac{L}2]$ in order for the two signals to have their 90° phase relationship (which is fundamentally what the Hilbert Transform is about). Note that every even-indexed sample of $h[n]$ is zero, so the number of taps is not really $L+1$ but is $\frac{L}2 - 1$ taps with non-zero coefficients. |
Correlation Coefficient is a popular term in mathematics that is used to measure the relationship between two variables. One of the popular categories of Correlation Coefficient is Pearson Correlation Coefficient that is denoted by the symbol R and commonly used in linear regression. If you wanted to start with statistics then Pearson Correlation Coefficient is the first thing you need to learn.
It would help in measuring the strength between two variables and their relationship. It is also named as Pearson Test in mathematics. When you conduct a statistical test among two variables, this would be an excellent choice using Pearson correlation coefficient formula to check how strong is the relationship between two variables.
You must be wondering what is the meaning of coefficient values here. The coefficient value lies between -1.00 and 1.00 in statistics. If the value is negative then it signifies that there is negative relation between two variables. At the same time, if value is positive then relationship is also positive. Greater the value of variable, stronger the relationship would be.
\[\large r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}\]
Where,
r = Pearson correlation coefficient x = Values in the first set of data y = Values in the second set of data n = Total number of values.
For example, if you wanted to check the relationship between age and reported income of participants then you should use Pearson correlation coefficient formula here. It will give you either positive or negative values as mentioned earlier.
Mostly, the relationship between two variables is stronger because as the age will grow, the income will also increase in the same ration. If you are interested in learning more about relationship strength then don’t stop here but practice more problems to get a better understanding of the concept. |
I need to find the periodicity of the following signal:
$$ x\left [ n \right ] = \cos (\frac{\pi n^{2}}{8}) $$ Now I understand that the basic procedure to determine the periodicity is to find a $ N $ such that $ x\left [ n \right ] = x\left [ n + N\right ] $. I applied the procedure to the aforementioned signal and got the following results: $$ x\left [ n \right ] = \cos (\frac{\pi n^{2}}{8}) \\ x\left [ n + N\right ]= \cos (\frac{\pi (n + N)^{2}}{8})\\ = \cos (\frac{\pi (n^{2} + 2nN + N^{2})}{8})\\ = \Re \left \{\exp (i \frac{\pi (n^{2} + 2nN + N^{2})}{8}) \right \}\\ = \Re \left \{\exp (i \frac{\pi n^{2}}{8})\exp (i \frac{\pi 2nN}{8})\exp (i \frac{\pi N^{2}}{8}) \right \} $$ Now for the signal to conform to $ x\left [ n \right ] = x\left [ n + N\right ]$: $$ \Re \left \{\exp (i \frac{\pi 2nN}{8})\exp (i \frac{\pi N^{2}}{8}) \right \} = 1 $$ $ \exp(i\omega) = 1 $ only when $ \omega = 2\pi k $ where $k$ is an integer. Hence: $$ \frac{\pi 2nN}{8} = \frac{\pi N^{2}}{8} = 2\pi k $$ Now logically I can see that if $ N = 8 $ the first term would be reduced to a multiple of $ 2\pi $ for all $ n $ and hence the fundamental period would be $ N = 8 $ but how would I go about proving this mathematically, that the minimum period is indeed 8? like what set of operations would I perform on: $$ \frac{\pi 2nN}{8} = \frac{\pi N^{2}}{8} = 2\pi k $$ to obtain $ N = 8 $?
I need to find the periodicity of the following signal:
You want to prove that $\frac{nN}{8}$ is an integer for any integer $n$. Consider $n=1$. Clearly, $N$ cannot be less than 8.
You also need to prove that $\frac{N^2}{16}$ is an integer. This means that $N$ is a multiple of 4. So, the smallest $N$ that meets both conditions is 8.
You have to prove that the period $N$ is the smallest number satisfying
$$\frac{\pi}{8}(n+N)^2=\frac{\pi n^2}{8}+2\pi k,\quad k\in\mathbb{N}\tag{1}$$
From (1) you get
$$\frac{\pi}{8}n^2+\frac{\pi}{8}2nN+\frac{\pi}{8}N^2=\frac{\pi n^2}{8}+2\pi k$$
which is equivalent to
$$\frac{\pi}{8}2nN+\frac{\pi}{8}N^2=2\pi k\tag{2}$$
or
$$2nN+N^2=16k\tag{3}$$
for any value of $n$. Clearly, the smallest number $N$ for which the left-hand side of (3) is a multiple of $16$, regardless of the value of $n$, is $N=8$:
$$16n+16\cdot 4=16k$$ |
Hexagon is the polygon that has six equal sides and the six edges. Hexa is a Greek word whose meaning is six. Naturally, when all six sides are equal then perimeter will be multiplied by 6 of one side of the hexagon. Abd each internal angle is measured as 120-degree. Take an example, if each side of the hexagon is marked as ‘a’ then the perimeter would be given as 6 * a.
The area is calculated as the total space within boundaries of a hexagon. Further, to make the calculation easy, you can divide the hexagon into six isosceles triangles. Calculate the area of each triangle one by one and multiply it by 6 to find the total area of a Hexagon. The perimeter of hexagon formula is written as –
\[\large P=6\times a\]
Where a is the side of the hexagon and all aides are equal in length.
Hexagon is a polygon or two-dimensional closed figure made up of straight line segments. Hexagon is a polygon with six sides and measurement of angle for each side is equal. It may be called as regular hexagon too whose sides are congruent to each other. There are a predefined set of hexagon formulas that are used collectively to find the area, perimeter, and the side of a hexagon. A Hexagon with the side length of a is given as:
\[\large Perimeter\;of\;an\;Hexagon = 6 \times a\]
\[\large Area\;of\;an\;Hexagon =\frac{3 \sqrt{3}}{ 2} \times a^{2}\]
With the help of these formulas, you can quickly calculate the area or perimeter of various hexagonal objects. Here are the few properties of Hexagon for a better understanding of the concept. These are –
A pyramid having a hexagonal base and six triangular lateral faces then it is named as the hexagonal pyramid. The other popular name for the hexagonal pyramid is heptahedron. The hexagonal pyramid formulas can be given as below
Base area of a hexagonal pyramid = 3ab Surface area of a hexagonal pyramid = 3ab+3bs Volume of a hexagonal pyramid = abh
Where,
a is apothem length. b is base length. s is slant height. h is height.
Regular hexagon fits together well like an equilateral triangle and they are commonly seen as tiles in the house. The exterior angle is measured as 60-degree in case of the regular hexagon and the interior angle is measured as 120-degree. The shape has six rotational symmetries that can be divided into six isosceles triangles too. This is the only regular polygon that can be divided into more regular polygons. The area of a hexagon formula in mathematics is given as –
\[\large A=3 \times s\times r\]
Where,
s is the side of a hexagon. r is the Radius of a Hexagon. |
Thank you for using the timer!We noticed you are actually not timing your practice. Click the START button first next time you use the timer.There are many benefits to timing your practice, including:
Does GMAT RC seem like an uphill battle? e-GMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT
So we need to calculate the angle made by Arc AXC at the centre of the Circle
Property-2: Angle made at the centre by any arc is always twice the angle made at the circumference by the same arc
i.e. Since Arc AXC makes an angle of 40 degrees at the circumference so i.e. Arc AXC will make an angle of 2*40 = 80 degrees at the Centre of the circle
i.e. Length of the Arc AXC = (80/360)*2*π*9 [Area = πr^2 = 81π i.e. Radius = 8]
i.e. Length of the Arc AXC = 4π
Answer: Option D_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html
NOw, as any angle subtended by an arc on the circumference is half the angle subtended by the same arc at the center of the circle, thus the angle subtended by arc AXC at the center of the circle is 2*40 deg.= 80 deg.
For the complete circumference, 2\(\pi\) or 360 is subtended by 2\(\pi\) r of the circumference. 80 deg will thus subtend = 80/360 * 2\(\pi\) r = 4\(\pi\). thus D is the answer.
Re: Consider a circle with area 81π . An arc on this circle is such that i[#permalink]
Show Tags
03 Jul 2015, 09:23
2
adityadon wrote:
does this mean ... whenever we are calculating length of arc we need to know angle subtended by it at centre ... not at circumference .. Right ? I calculated it as 2*pi
Yes, the angle always has to be the one that the arc subtends at the center of the circle.
the direct formula for calculating the length of any arc of a circle = l = \(r*\theta\) , where \(\theta\) is the angle measure at the center of the circle in RADIANS and not in degrees and r is the radius of the circle.
\(2\pi\) radians equal 360 degrees.
In the question above, 80 degrees will be equal to \((80/360) * 2\pi\) radians. Radius, r = 9.
Thus the length of the minor arc = \(r*\theta\) = \(9*(80/360) * 2\pi\) = \(4\pi\)
If the area of the circle is \(81\pi\), then the radius of the circle is 9 (\(A = \pi r^2\)). Therefore, the total circumference of the circle is \(18\pi\) (\(C = 2\pi r\)). Angle ABC, an inscribed angle of 40°, corresponds to a central angle of 80°. Thus, arc AXC is equal to 80/360 = 2/9 of the total circumference:
Consider a circle with area 81π . An arc on this circle is such that i[#permalink]
Show Tags
01 Apr 2016, 20:15
Chiragjordan wrote:
Consider a circle with area 81π.An arc on this circle is such that it makes a 40 degree inscribed angle with a point P on the circle. What is the length of this arc? [A] 2π 4π [C] 8π [D] 12π [E] Cannot be determined
Hi,
[b]A GOOD Q.. +1kudos Difficulty level should 600-700...
1) first, since we are talking of arc length, lets change the Area to Perimeter.. \(A=81π=π*r^2\) so r=9 and P= 18π
2) Next would be to find the angle that this arc makes at center.. Since the ARC makes 40 degree angle at a point on the circumference, ARC will make 2*40 at the center..
3) Finally lets correlate the angle and perimeter to find length of ARC.. 360 degree makes the perimeter or 18π.. so 1 degree will make \(\frac{18π}{360}\), and 2*40 or 80 degree will make\(\frac{18π}{360} * 80 = 4π\)_________________
Re: Consider a circle with area 81π . An arc on this circle is such that i[#permalink]
Show Tags
09 Mar 2019, 14:55
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email._________________ |
I'm trying to integrate
$$\int_0^{x<a} {\frac{e^x}{\sqrt{x(a-x)}}\,\mathrm dx}$$
I've tried various substitutions to no avail. Any ideas?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
$$\int_0^1 \frac{e^x}{\sqrt{x(1-x)}}\, dx = \sqrt{e} \, \pi \, I_0(1/2)$$
and
$$\int_0^2 \frac{e^x}{\sqrt{x(2-x)}}\, dx = e \, \pi \, I_0(1)$$
suggesting that the general solution might be
$$\int_0^a \frac{e^x}{\sqrt{x(a-x)}}\, dx = e^{a/2} \pi I_0(a/2).$$
where $I_0$ is a modified Bessel function of the first kind.
UPDATE:
Now that I have a second, I was able to do the substitutions suggested by the form of the expressions above and it indeed works out:
$$\int_0^a \frac{e^x \, dx}{\sqrt{x(a-x)}} = \int_0^{2b} \frac{e^x \, dx}{\sqrt{x(2b-x)}}$$
(where $a=2b$),
$$= \int_0^{2b} \frac{e^x \, dx}{\sqrt{b^2 - (x-b)^2}} = \int_{-1}^{1} \frac{b e^{b u + b} \, du}{\sqrt{b^2 - (b u)^2}}$$
(where $b u = x + b$),
$$= e^b \int_{-1}^{1} \frac{e^{b u} \, du}{\sqrt{1 - u^2}} = e^b \int_{-\pi/2}^{\pi/2} \frac{e^{b \sin\theta} \, \cos \theta \, d\theta}{\sqrt{1 - \sin^2 \theta}}$$
(where $u = \sin \theta$)
$$= e^b \int_{-\pi/2}^{\pi/2} e^{b \sin\theta} d\theta = e^b \pi I_0(b) = e^{a/2} \pi I_0\left(\frac{a}{2}\right)$$
Here $$I_0(x) = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} e^{x \sin\theta} \, d\theta$$
I am not sure that any closed form could be found and, may be, only series expansions could be the way to go.
First, let $x=ay$ to make $$I=\int_0^{t} {\frac{e^x}{\sqrt{x(a-x)}}\, dx}=\int_0^{\frac ta} {\frac{e^{ay}}{\sqrt{y(1-y)}}\, dy}$$ Expanding around $y=0$ would give $$\frac{1}{\sqrt{y(1-y) }}=\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}y^{n-\frac 12}$$ and $$\int e^{ay}\,y^{n-\frac 12}\,dy=-y^{n+\frac{1}{2}} E_{\frac{1}{2}-n}(-a y)$$ where appears the exponential integral function. |
2019-10-11 06:14
Implementation of CERN secondary beam lines T9 and T10 in BDSIM / D'Alessandro, Gian Luigi (CERN ; JAI, UK) ; Bernhard, Johannes (CERN) ; Boogert, Stewart (JAI, UK) ; Gerbershagen, Alexander (CERN) ; Gibson, Stephen (JAI, UK) ; Nevay, Laurence (JAI, UK) ; Rosenthal, Marcel (CERN) ; Shields, William (JAI, UK) CERN has a unique set of secondary beam lines, which deliver particle beams extracted from the PS and SPS accelerators after their interaction with a target, reaching energies up to 400 GeV. These beam lines provide a crucial contribution for test beam facilities and host several fixed target experiments. [...] 2019 - 3 p. - Published in : 10.18429/JACoW-IPAC2019-THPGW069 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPGW069 Record dettagliato - Record simili 2019-10-09 06:01
HiRadMat: A facility beyond the realms of materials testing / Harden, Fiona (CERN) ; Bouvard, Aymeric (CERN) ; Charitonidis, Nikolaos (CERN) ; Kadi, Yacine (CERN)/HiRadMat experiments and facility support teams The ever-expanding requirements of high-power targets and accelerator equipment has highlighted the need for facilities capable of accommodating experiments with a diverse range of objectives. HiRadMat, a High Radiation to Materials testing facility at CERN has, throughout operation, established itself as a global user facility capable of going beyond its initial design goals. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPRB085 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPRB085 Record dettagliato - Record simili 2019-10-09 06:01
Commissioning results of the tertiary beam lines for the CERN neutrino platform project / Rosenthal, Marcel (CERN) ; Booth, Alexander (U. Sussex (main) ; Fermilab) ; Charitonidis, Nikolaos (CERN) ; Chatzidaki, Panagiota (Natl. Tech. U., Athens ; Kirchhoff Inst. Phys. ; CERN) ; Karyotakis, Yannis (Annecy, LAPP) ; Nowak, Elzbieta (CERN ; AGH-UST, Cracow) ; Ortega Ruiz, Inaki (CERN) ; Sala, Paola (INFN, Milan ; CERN) For many decades the CERN North Area facility at the Super Proton Synchrotron (SPS) has delivered secondary beams to various fixed target experiments and test beams. In 2018, two new tertiary extensions of the existing beam lines, designated “H2-VLE” and “H4-VLE”, have been constructed and successfully commissioned. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPGW064 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPGW064 Record dettagliato - Record simili 2019-10-09 06:00
The "Physics Beyond Colliders" projects for the CERN M2 beam / Banerjee, Dipanwita (CERN ; Illinois U., Urbana (main)) ; Bernhard, Johannes (CERN) ; Brugger, Markus (CERN) ; Charitonidis, Nikolaos (CERN) ; Cholak, Serhii (Taras Shevchenko U.) ; D'Alessandro, Gian Luigi (Royal Holloway, U. of London) ; Gatignon, Laurent (CERN) ; Gerbershagen, Alexander (CERN) ; Montbarbon, Eva (CERN) ; Rae, Bastien (CERN) et al. Physics Beyond Colliders is an exploratory study aimed at exploiting the full scientific potential of CERN’s accelerator complex up to 2040 and its scientific infrastructure through projects complementary to the existing and possible future colliders. Within the Conventional Beam Working Group (CBWG), several projects for the M2 beam line in the CERN North Area were proposed, such as a successor for the COMPASS experiment, a muon programme for NA64 dark sector physics, and the MuonE proposal aiming at investigating the hadronic contribution to the vacuum polarisation. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPGW063 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPGW063 Record dettagliato - Record simili 2019-10-09 06:00
The K12 beamline for the KLEVER experiment / Van Dijk, Maarten (CERN) ; Banerjee, Dipanwita (CERN) ; Bernhard, Johannes (CERN) ; Brugger, Markus (CERN) ; Charitonidis, Nikolaos (CERN) ; D'Alessandro, Gian Luigi (CERN) ; Doble, Niels (CERN) ; Gatignon, Laurent (CERN) ; Gerbershagen, Alexander (CERN) ; Montbarbon, Eva (CERN) et al. The KLEVER experiment is proposed to run in the CERN ECN3 underground cavern from 2026 onward. The goal of the experiment is to measure ${\rm{BR}}(K_L \rightarrow \pi^0v\bar{v})$, which could yield information about potential new physics, by itself and in combination with the measurement of ${\rm{BR}}(K^+ \rightarrow \pi^+v\bar{v})$ of NA62. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPGW061 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPGW061 Record dettagliato - Record simili 2019-09-21 06:01
Beam impact experiment of 440 GeV/p protons on superconducting wires and tapes in a cryogenic environment / Will, Andreas (KIT, Karlsruhe ; CERN) ; Bastian, Yan (CERN) ; Bernhard, Axel (KIT, Karlsruhe) ; Bonura, Marco (U. Geneva (main)) ; Bordini, Bernardo (CERN) ; Bortot, Lorenzo (CERN) ; Favre, Mathieu (CERN) ; Lindstrom, Bjorn (CERN) ; Mentink, Matthijs (CERN) ; Monteuuis, Arnaud (CERN) et al. The superconducting magnets used in high energy particle accelerators such as CERN’s LHC can be impacted by the circulating beam in case of specific failure cases. This leads to interaction of the beam particles with the magnet components, like the superconducting coils, directly or via secondary particle showers. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPTS066 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPTS066 Record dettagliato - Record simili 2019-09-20 08:41
Shashlik calorimeters with embedded SiPMs for longitudinal segmentation / Berra, A (INFN, Milan Bicocca ; Insubria U., Varese) ; Brizzolari, C (INFN, Milan Bicocca ; Insubria U., Varese) ; Cecchini, S (INFN, Bologna) ; Chignoli, F (INFN, Milan Bicocca ; Milan Bicocca U.) ; Cindolo, F (INFN, Bologna) ; Collazuol, G (INFN, Padua) ; Delogu, C (INFN, Milan Bicocca ; Milan Bicocca U.) ; Gola, A (Fond. Bruno Kessler, Trento ; TIFPA-INFN, Trento) ; Jollet, C (Strasbourg, IPHC) ; Longhin, A (INFN, Padua) et al. Effective longitudinal segmentation of shashlik calorimeters can be achieved taking advantage of the compactness and reliability of silicon photomultipliers. These photosensors can be embedded in the bulk of the calorimeter and are employed to design very compact shashlik modules that sample electromagnetic and hadronic showers every few radiation lengths. [...] 2017 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 64 (2017) 1056-1061 Record dettagliato - Record simili 2019-09-20 08:41
Performance study for the photon measurements of the upgraded LHCf calorimeters with Gd$_2$SiO$_5$ (GSO) scintillators / Makino, Y (Nagoya U., ISEE) ; Tiberio, A (INFN, Florence ; U. Florence (main)) ; Adriani, O (INFN, Florence ; U. Florence (main)) ; Berti, E (INFN, Florence ; U. Florence (main)) ; Bonechi, L (INFN, Florence) ; Bongi, M (INFN, Florence ; U. Florence (main)) ; Caccia, Z (INFN, Catania) ; D'Alessandro, R (INFN, Florence ; U. Florence (main)) ; Del Prete, M (INFN, Florence ; U. Florence (main)) ; Detti, S (INFN, Florence) et al. The Large Hadron Collider forward (LHCf) experiment was motivated to understand the hadronic interaction processes relevant to cosmic-ray air shower development. We have developed radiation-hard detectors with the use of Gd$_2$SiO$_5$ (GSO) scintillators for proton-proton $\sqrt{s} = 13$ TeV collisions. [...] 2017 - 22 p. - Published in : JINST 12 (2017) P03023 Record dettagliato - Record simili 2019-04-26 08:32
Baby MIND: A magnetised spectrometer for the WAGASCI experiment / Hallsjö, Sven-Patrik (Glasgow U.)/Baby MIND The WAGASCI experiment being built at the J-PARC neutrino beam line will measure the ratio of cross sections from neutrinos interacting with a water and scintillator targets, in order to constrain neutrino cross sections, essential for the T2K neutrino oscillation measurements. A prototype Magnetised Iron Neutrino Detector (MIND), called Baby MIND, has been constructed at CERN and will act as a magnetic spectrometer behind the main WAGASCI target. [...] SISSA, 2018 - 7 p. - Published in : PoS NuFact2017 (2018) 078 Fulltext: PDF; External link: PoS server In : 19th International Workshop on Neutrinos from Accelerators, Uppsala, Sweden, 25 - 30 Sep 2017, pp.078 Record dettagliato - Record simili 2019-04-26 08:32 Record dettagliato - Record simili |
I will quote an answer of Matt L. from this post (I didn't comment there because I can't)
If you have a continuous-time signal $x(t)$, then the two signals you're talking about are
$$\begin{align} x_c(t) &=x(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT) \\ &=\sum_{n=-\infty}^{\infty}x(t)\delta(t-nT) \\ &=\sum_{n=-\infty}^{\infty}x(nT)\delta(t-nT) \\ \tag{1} \end{align}$$
and you define
$$x_d[n] \triangleq x(nT)\tag{2}$$
The first signal given by $(1)$ is technically a continuous-time signal, even though it is only non-zero at discrete times $t=nT$. The reason why it is considered a continuous-time signal is because it can and must be transformed using the continuous-time Fourier transform (CTFT). So $(1)$ is the continuous-time representation of a sampled signal. Eq. $(2)$ is the discrete-time representation of the
samesignal. Here the sampled signal is represented as a sequence of numbers. You can't apply the CTFT to $(2)$, but you must use the discrete-time Fourier transform (DTFT).
The nice thing is now that the CTFT of $x_c(t)$ given by $(1)$ and the DTFT of $x_d[n]$ given by $(2)$ are identical. So if the CTFT
$$\begin{align} X_c(j\Omega) &= \int_{-\infty}^{\infty}x_c(t)e^{-j\Omega t}dt \\ &= \int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}x(nT)\delta(t-nT) e^{-j\Omega t}dt \\ &= \sum_{n=-\infty}^{\infty}x(nT) \int_{-\infty}^{\infty}\delta(t-nT) e^{-j\Omega t}dt \\ &= \sum_{n=-\infty}^{\infty}x(nT) e^{-j\Omega nT} \\ &= \sum_{n=-\infty}^{\infty}x_d[n] e^{-j n(\Omega T)} \\ \end{align}$$
and the DTFT:
$$X_d(e^{j\omega})=\sum_{n=-\infty}^{\infty}x_d[n]e^{-jn\omega}$$
we have:
$$X_d(e^{j\omega})=X_c\left(\tfrac{j\omega}{T}\right)\tag{3}$$
In sum, the signals $(1)$ and $(2)$ are just two different representations of the same signal, and their spectra (one defined by the CTFT, the other defined by the DTFT) are identical.
My question arises due to a question I got when the period of the impulse train is not the same as the period of the sample:
$$x_c(t)=\sum_{n=-\infty}^{\infty}x(nT)\delta\big(t-n(\tfrac{2}{3}T)\big)\tag{1}$$
and
$$x_d[n]=x(nT)\tag{2}$$
does we still get the connection:
$$X_d(e^{j\omega})=X_c\left(\tfrac{j\omega}{T}\right)\tag{3}$$
if yes, does the $T$ in the connection imply to the impulse train period ($\frac{2}{3}T$) or to the sample time ($T$) and why?
When I set in $X_d(e^{j\omega})$ the value $\omega=\frac{2}{3}T\omega$ I got an answer which should be true. but I don't know if it's by luck or really true and I don't know why. |
Category: Ring theory Problem 624
Let $R$ and $R’$ be commutative rings and let $f:R\to R’$ be a ring homomorphism.
Let $I$ and $I’$ be ideals of $R$ and $R’$, respectively. (a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$. (b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$
Add to solve later
(c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$ Problem 618
Let $R$ be a commutative ring with $1$ such that every element $x$ in $R$ is idempotent, that is, $x^2=x$. (Such a ring is called a
Boolean ring.) (a) Prove that $x^n=x$ for any positive integer $n$.
Add to solve later
(b) Prove that $R$ does not have a nonzero nilpotent element. Problem 543
Let $R$ be a ring with $1$.
Suppose that $a, b$ are elements in $R$ such that \[ab=1 \text{ and } ba\neq 1.\] (a) Prove that $1-ba$ is idempotent. (b) Prove that $b^n(1-ba)$ is nilpotent for each positive integer $n$.
Add to solve later
(c) Prove that the ring $R$ has infinitely many nilpotent elements. |
This is a followup question to one I asked earlier based on the chat after the answer given by @hotpaw2, and cross-posted from stackoverflow since it was suggested it is more relevant to DSP. I have a signal which will be a single cosine wave, with a phase offset. My task is to extract (with very high accuracy required) the amplitude and phase of this single frequency component.
On paper, the following relations hold assuming a properly normalized fourier transform $T$:
$$T(\omega) = \mathcal{F}\left\{A \cos(\omega_0 t+ \phi) \right\} $$ $$A = 2|T(\omega_0)|= 2\sqrt{\mathrm{Re}(T(\omega_0))^2+\mathrm{Im}(T(\omega_0))^2} $$
$$ \begin{align} \phi & = \arg \left\{\mathrm{Re}(T(\omega_0)) + j \mathrm{Im}(T(\omega_0)) \right\} \\ & = \mathrm{atan2}\left( \mathrm{Im}(T(\omega_0)), \ \mathrm{Re}(T(\omega_0) \right) \\ \end{align} $$
Unsurprisingly, there is a bit more to the DFT than simply taking the transform and picking off the relevent components. In particular, the discussion suggested to me that I was not entirely clear on what the phase offset is being measured with respect to, and that there are significant edge effects that can destroy the accuracy of the result if data is not properly windowed.
I have been googling around but most of the discussion is fairly technical and light on example, so I was hoping that someone can shed some light on things. In particular, I came across one example suggesting that instead of doing a simple transform, I should be shifting it first.
I put together this code to test some ideas:
import numpy as npimport pylab as pldef flattop(n): return [1.0 - 1.93*np.cos(2*np.pi*k/(float(n-1))) + 1.29*np.cos(4*np.pi*k/(float(n-1)))- 0.388*np.cos(6*np.pi*k/(float(n-1))) + 0.028*np.cos(8*np.pi*k/(float(n-1))) for k in range(int(n))]f = 30.0w = 2.0*np.pi*fphase = np.pi/7num_t = 10*fwindow = flattop(num_t)t, dt = np.linspace(0, 1, num_t, endpoint=False, retstep=True)signal = np.cos(w*t+phase)+np.random.normal(0,0.05,len(t))pl.plot(t,window*signal)pl.show()amp = np.fft.rfft(window*signal)freqs = np.fft.rfftfreq(t.shape[-1],dt)index = np.argmax(np.abs(amp))print indexprint(np.arctan2(amp.imag,amp.real))[index]print (np.abs(amp))[index]*(2.0/len(t))pl.subplot(211)pl.plot(freqs,np.abs(amp))pl.subplot(212)pl.plot(freqs,(np.arctan2(amp.imag,amp.real)))pl.show()
Some observations:
When I use
num=10*f, I get perfect results for both phase and amplitude. However, if I use
num=10*f+1, the phase I get is completely wrong. I have tried using a window (in particular, a flat top window since it preserves amplitude) and I get the same thing: unless I control the number of samples to be an integer multiple (actually, 10.2, 10.4, 10.6, and 10.8*f also give good results for some reason) of the signal frequency, I get garbage back.
It was suggested that I could improve this by measuring the phase in the center of the window and that this could be achieved using fftshift. Googling around gave me a couple of examples which I have implemented in the code, but the results are the same.
So: first set of questions: can someone explain the point of using fftshift, and what exactly it is doing to the data? Why does using an inverse shift, transforming, and then shifting then require a frequency component set which is only shifted once, with no inverse opration? Is this approach the correct one (ignoring for the moment the issue of a window).
It also seems that using a window is required for finite time-domain signals, which I can understand. What's not clear to me is how to correct the amplitude and phase for the effect of the window in general. I have played with flat top windows to preserve the amplitude, and it seems to work, but I don't understand why the amplitude gain is not a function of frequency.
Second set of questions: if I window my data, presumably this will affect the amplitude and likely the phase(?) of the result. Is there an analytical way to correct for the amplitude change for a given window shape? I can find a few tables that list correction factors, but I haven't really seen a good explanation yet.
In the linked question, it was stated that phase should be measured near the center of the hump of the window function. But because the window function is a time-domain function and I want the phase for a specific frequency, I don't quite understand what that means.
In reality my sample signal will not be a generated perfect cosine wave and I won't have control over the number of samples, so I need to refine the method to give accurate information regardless of the exact number of samples.
EDIT: it appears that the first answer here might be a solution, but I still would like to understand the issues raised in the question in more depth if someone could provide insight.
EDIT 2: it seems that the problem is related to the fact that for accurate phase information we require that the frequency of interest be exactly centered in a DFT bin. How can that be ensured for an arbitrary number of samples? The question linked above provides a nice example that functions well for a single frequency, but there must be a general way to do this for more than one frequency. |
First, let's try and clean up the MTW expression a little bit by getting rid of the manifestly coordinate variant stuff, by introducing a proper contraction, a proper integral over the surface at infinity, and the unit normal $r^{a}$ to the surface at infinity:
$$M_{ADM}={16\pi}\oint\sqrt{\gamma}\,\,d^{2}x \gamma^{ab}r^{c}\left(\gamma_{ac,b}-\gamma_{ab,c}\right)$$
This is only valid, however, not just only for in asymptotically flat spaces, but actually only for asymptotically
Cartesian coordinates. You can prove this to yourself by calculating the ``ADM Mass'' of the flat 3-metric in polar coordinates:
$\begin{align*}16\pi M_{ADM}&=\lim_{r\rightarrow\infty}\oint r^{2}\sin\theta \gamma^{ab}\left(\gamma_{ra,b}-\gamma_{ab,r}\right)d\theta d\phi\\&=\lim_{r\rightarrow\infty}4\pi r^{2}\left(\gamma^{rr}\gamma_{rr,r}-\gamma^{ab}\gamma_{ab,r}\right)\\&=\lim_{r\rightarrow\infty}4\pi r^{2}\left(0-\frac{4}{r}\right)\\&=-\infty\end{align*}$
Obviously, this is wrong, and just an artifact of the way that spherical coordinates behave at infinity. To fix this, we need to always subtract the divergence from the ADM mass of flat spacetime from the ADM mass of the coordinate system in question. So, that's the origin of the $H$ and $H_{0}$ terms. The $H$ term is the extrinsic curvature of the space in question, while $H_{0}$ is the extrinsic curvature of flat spacetime. Now, it's just a matter of showing that the expression inside the integral is equal to $H$.
Typically, I define the extrinsic curvature as $\gamma^{ab}\nabla_{a}r_{b}$. So, let's go on an adventure:
$\begin{align*}H&=\gamma^{ab}\nabla_{a}r_{b}\\&=\gamma^{ab}\left(\partial_{a}r_{b}-\Gamma_{ab}{}^{c}r_{c}\right)\\&=\gamma^{ab}\partial_{a}r_{b}-\frac{1}{2}\gamma^{ab}\gamma^{cd}\left(2\gamma_{ad,b}-\gamma_{ab,d}\right)r_{c}\\&=\gamma^{ab}\partial_{a}\left(\gamma_{bc}r^{c}\right)-\gamma^{ab}r^{c}\left(\gamma_{ac,b}-\frac{1}{2}\gamma_{ab,c}\right)\\&=\delta^{a}{}_{c}\partial_{a}r^{c} + \gamma^{ab}r^{c}\gamma_{bc,a}-\gamma^{ab}r^{c}\left(\gamma_{ac,b}-\frac{1}{2}\gamma_{ab,c}\right)\\&=\partial_{a}r^{a}-\frac{1}{2}\gamma^{ab}r^{c}\left(\gamma_{ac,b}-\gamma_{ab,c}\right)\end{align*}$
Now, we note that in any coodinate system adapted so that the coordinate $r=constant$ that determines the surface is chosen for one of the coordinates, we have, necessarily $r^{a}=(1,0,0)$, we note that $\partial_{a}r^{a}=0$, and we thus conclude that the two expressions for ADM mass are equivalent. |
I can prove a lower bound of
$69$ tries.
The first step is to notice that the question is equivalent to
finding an $132$-node graph $G$ with no independent set of size $\ge 66$.
The reason is:
1. If we have the graph $G$, then for each edge $(x,y)$ in $G$, we put the $x$-th and $y$-th card into the slot. If we failed then no edge has two green cards as its endpoint, so $66$ green cards induce an independent set of $G$.
and
2. If we have a strategy to win the game, then the strategy must be nonadaptive, since we don't gain any information from each failure to choose two green cards. That is, the strategy of each step is already determined before the game gets played. If the $i$-th step is to test card $a_i$ and card $b_i$, we insert edge $(a_i,b_i)$ into $G$. If $G$ has a size-$66$ independent set, our strategy would be invalid when this set of cards are green.
Then we observe
$G$ must contain an odd cycle. Because otherwise $G$ is a bipartite graph which always have an independent set of size $\ge \frac{132}{2}=66$.
and
If $G$ has $C$ connected components, then $G$ has at least $133-C$ edges. For an arbitrary graph $G'$, if it has $C'$ components and $V'$ vertices, then it has at least $V'-C'$ edges, so we infer that $G$ has at least $132-C$ edges. However as we demonstrated above, $G$ has a cycle, and deleting an edge from that cycle doesn't change the number of its components. So we conclude $G$ has $\ge (132-C)+1$ edges.
So there are a few cases. If
$C\ge 66$, then we can take any vertex from a component to form an independent set of size $C$. This is invalid;
If
$C\le 64$, then $G$ has $\ge 133-64=69$ edges;
If
$C=65$, then every component must be a clique: if some component is not a clique, we can take two vertices from it and one vertex from every other component and still form an independent set (of size $C+1$). Now let $a_1,a_2,\dots,a_{65}$ be the number of vertices in each component, then $\sum_{i=1}^{65}a_i=132$ and the number of edges is $E=\sum_{i=1}^{65}\frac{a_i(a_i-1)}{2}$. Even if we allow $a_i$'s be real numbers (instead of integers), the minimum of $E$ is reached when each $a_i=a=132/65$, and $E=65\cdot \frac{a(a-1)}{2}\approx 68.03$. Therefore $69$ edges are needed in this case. (We note that this corresponds to @trolley813's answer, and optimal integer solution is that each $a_i$ is $2$ except two of them are $3$.) |
2019-10-14 17:21
Performance of VELO clustering and VELO pattern recognition on FPGA/LHCb Collaboration This document contains plots and tables showing the performance obtained on VELO clustering and VELO pattern recognition using algorithms implementable on FPGA. The data used are simulated with LHCb Upgrade conditions.. LHCB-FIGURE-2019-011.- Geneva : CERN, 2019 - 16. Detaljnije - Slični zapisi 2019-10-11 14:20
TURBO stream animation /LHCb Collaboration An animation illustrating the TURBO stream is provided. It shows events discarded by the trigger in quick sequence, followed by an event that is kept but stripped of all data except four tracks [...] LHCB-FIGURE-2019-010.- Geneva : CERN, 2019 - 3. Detaljnije - Slični zapisi 2019-10-10 15:48 Detaljnije - Slični zapisi 2019-09-12 16:43
Pending/LHCb Collaboration Pending LHCB-FIGURE-2019-008.- Geneva : CERN, 10 Detaljnije - Slični zapisi 2019-09-10 11:06
Smog2 Velo tracking efficiency/LHCb Collaboration LHCb fixed-target programme is facing a major upgrade (Smog2) for Run3 data taking consisting in the installation of a confinement cell for the gas covering $z \in [-500, -300] \, mm $. Such a displacement for the $pgas$ collisions with respect to the nominal $pp$ interaction point requires a detailed study of the reconstruction performances. [...] LHCB-FIGURE-2019-007.- Geneva : CERN, 10 - 4. Fulltext: LHCb-FIGURE-2019-007_2 - PDF; LHCb-FIGURE-2019-007 - PDF; Detaljnije - Slični zapisi 2019-09-09 14:37
Background rejection study in the search for $\Lambda^0 \rightarrow p^+ \mu^- \overline{\nu}$/LHCb Collaboration A background rejection study has been made using LHCb Simulation in order to investigate the capacity of the experiment to distinguish between $\Lambda^0 \rightarrow p^+ \mu^- \overline{\nu}$ and its main background $\Lambda^0 \rightarrow p^+ \pi^-$. Two variables were explored, and their rejection power was estimated applying a selection criteria. [...] LHCB-FIGURE-2019-006.- Geneva : CERN, 09 - 4. Fulltext: PDF; Detaljnije - Slični zapisi 2019-09-06 14:56
Tracking efficiencies prior to alignment corrections from 1st Data challenges/LHCb Collaboration These plots show the first outcoming results on tracking efficiencies, before appli- cation of alignment corrections, as obtained from the 1st data challenges tests. In this challenge, several tracking detectors (the VELO, SciFi and Muon) have been misaligned and the effects on the tracking efficiencies are studied. [...] LHCB-FIGURE-2019-005.- Geneva : CERN, 2019 - 5. Fulltext: PDF; Detaljnije - Slični zapisi 2019-09-06 11:34 Detaljnije - Slični zapisi 2019-09-02 15:30
First study of the VELO pixel 2 half alignment/LHCb Collaboration A first look into the 2 half alignment for the Run 3 Vertex Locator (VELO) has been made. The alignment procedure has been run on a minimum bias Monte Carlo Run 3 sample in order to investigate its functionality [...] LHCB-FIGURE-2019-003.- Geneva : CERN, 02 - 4. Fulltext: VP_alignment_approval - TAR; VELO_plot_approvals_VPAlignment_v3 - PDF; Detaljnije - Slični zapisi 2019-07-29 14:20 Detaljnije - Slični zapisi |
Let $X_m = \frac{1}{\sqrt{m}}\sum_{k=1}^m Z_k$ where $Z_k$ are iid equally likely on $\{\pm 1\}$. Then $X_m$ convergens to $X \sim \mathcal{N}(0,1)$ in distribution by CLT.
Let $f$ be a smooth bounded function on $\mathbb{R}$. Then $\mathbb{E}[f(X_m)] \to \mathbb{E}[f(X)]$. I wonder if there is any general method to give sharp asymptotic estimate of the error term $\mathbb{E}[f(X_m)] - \mathbb{E}[f(X)]$, which I expect to be $\Theta(1/m)$. The scaling constant should depend on $f$ (as well as the distribution of $Z_k$ if they are not binary).
For law of large number, this type of estimate can be done via the Delta method (e.g., to estimate $\mathbb{E}[f(\bar{Z})] - f(0)$). There must be a counterpart for CLT... I haven't found the Edgeworth expansion useful because it seems to work with distribution with densities.
Edited: To be clear, I am only interested in some specific nice function (e.g., $f(x) = x^2 e^{-x^2/4}$) and finding a sharp expansion for the error term of the form, say, $c/m + o(1/m)$, where $c$ will depend n $f$. As pointed by Mark, the worst-case rate of all bounded smooth function $f$ is $1/\sqrt{m}$, which agrees with the upper bound given by Stein's method. |
I am trying to understand connection between non determinism of grammar and LL(1) conflicts introduced by it.
As per my understanding non deterministic context free grammar is a context free grammar in which at some point in time during parsing, multiple actions are allowed. Also I believe non determinism occurs when we have multiple productions in grammar with same prefix of production (right hand side) bodies. Also I know we eliminate non determinism by left factoring grammar.
Wikipedia article says:
We do left factoring to eliminate FIRST-FIRST conflicts. ref Left recursion is a special case of FIRST-FIRST conflict. ref Substitution is used for removing FIRST-FOLLOW conflict. However that may introduce FIRST-FIRST conflict. ref Left recursion is eliminated using rule like below ref: $A\rightarrow A\alpha | \beta $ ios converted to $A\rightarrow \beta A'$ $A'\rightarrow \alpha A' | \epsilon$
I have following doubts:
Since both non determinism and FIRST-FIRST conflicts are removed with left factoring and since left recursion is special case of FIRST-FIRST conflict, is non determinism a reason behind FIRST-FIRST conflict and left recursion (and also for FIRST-FOLLOW conflict)? If yes, does non determinism cause anything else and there is nothing else apart from these three (FIRST-FIRST conflict, left recursion and FIRST-FOLLOW conflict) in which non determinism manifests?
Does the solutions (left factoring, substitution and left recursion removal rule) for removing above can also remove non determinism from all grammars?
I know LL grammar should be both non deterministic and unambiguous. Does ambiguity also manifests in some way in LL parsing table or grammar? I read ambiguity can be eliminated by enforcing association and precedence between grammar symbols and this resolves SHIFT-REDUCE conflicts in LR parsers. Is anything like this required for eliminating ambiguity to make grammar a valid LL grammar? |
There's a problem in the derivation for $C_\infty$.
Eventhough its mechanics is a quite simple limit process, the associated partially-true observation yields a misleading result; it assumes that when the bandwidth $B$ goes to infinity, so will be the total noise power $N$ (which is true), however the total signal power $S$ is assumed to be
fixed (which is not true) as which may also go to infinity and hence the SNR $\frac{S}{N}$ would remain fixed! Rather it's assuming that signal energy $S$ is fixed and therefore SNR goes to zero as $B$ goes to infinity from which it would deduce that $$C_\infty = B_\infty \times \log(1 + SNR_\infty) = \infty \times \log(1) = \infty \times 0$$
Then he finds the limiting value as $$C_\infty = 1.44 \frac{S}{\eta} $$ where $S$ is the
finite signal power that is transmitted through the infinite bandwidth channel and $\eta$ is the noise power spectral density.
You can easily see the fact that if you have infinite bandwidth $B$, then you could simultaneously transmit infinite different message signals $m_k(t)$ of each having a finite nonzero bitrate $R_k$ (for example by just using a simple FDM scheme) the total of which would add up to inifnite bits per second at once which make your information capacity to go infinity thereof.
Another consequence of SNR is the following observation: Given any analog channel with zero noise, it's information capacity is infinite. Proof: consider a system where you send an analog voltage level and the receiver converts it into a digital bitstream with N bits. If the channel has zero noise, then you can in principle send infinitely precise analog signal values. For example you can send the
exact value of $\pi$ Volts over the channel. Since there is neither noise nor distortion in the channel (of smallest possible bandwidth) you would be transmitting infinite digits of the number $\pi$ to the receiver which requires infinite many bits to store. Therefore you send a single analog voltage evalue which is equivalent to infinite number of bits to represent digitally. When there's nonzero noise however, you can only transmit analog values up to noise floor precision which yields for example SNR based dynamic range limits of ADC systems. |
A College Algebra teacher would like to find out whether the possession of a textbook enhances achievement score $X$. From an engineering freshmen class of 50 students, she took the scores in a 20 – item quiz of 5 randomly chosen students with textbook (1) and 5 without textbook (0). Compute the correlation between X and Y. My question, is $Y$ nominal and $X$ interval scale in this data set?What correlation should I use? Spearman Rho ; Pearson R ; Chi-Square? Thanks.
Your $X$ consists of achievement scores, so they are probably ratio scale, as I imagine they are numeric scores with a clear definition of $0$. Your $Y$ is a binary indicator variable; this one is tricky because although you may be tempted to label is as nominal, $1$ indicates the presence of a textbook and, in contrast, $0$ indicates the absence of one. In this situation, you should consider $Y$ to be ordinal.
For comparing quantitative data to a binary variable, I would start by trying a logit model:
$$ \log\left(\frac{\Pr(Y)}{1-\Pr(Y)}\right)=\beta_0+\beta_1X $$
You can then perform a t-test to test the hypothesis that $\beta_1 = 0$. If it the hypothesis is rejected, then $\beta_1$ tells us something about the direction and strength of the relationship between $X$ and $Y$. |
I am learning quantization of MCS theory. $$L_{MCS}=-\frac{1}{4}F^{\mu \nu}F_{\mu\nu}+\frac{g}{2} \epsilon^{\mu \nu \rho}A_\mu\partial_\nu A_\rho$$ I have reached the commutation relation $$[A_i(\vec x),\pi^j(\vec y)] = i \delta^j_i \delta^2(\vec x- \vec y).$$ Now I don't know how to reach the commutation relation between Electric fields and electric and magnetic fields. The relations are as follows. $$[E_i(\vec x),E_j(\vec y)] = - i g \epsilon_{ij} \delta^2(\vec x- \vec y)$$ $$[E_i(\vec x),B_j(\vec y)] = i \delta_{ij}\partial_j \delta^2(\vec x- \vec y)$$ Note: I have tried some conventional 3+1 QFT methods to get these relations but I get stuck on polarization sums and also this $\epsilon_{ij}$ term can not appear there. I got $$\pi^i = \dot A_i + \frac{\mu}{2}\epsilon^{ij}A_j$$
I have done this and here are main steps. $$\pi^i = \dot A_i + \frac{\mu}{2}\epsilon^{ij}A_j$$ In above expression we have $E^i = \dot A_i$ so we have $$\pi^i = E^i + \frac{\mu}{2}\epsilon^{ij}A_j$$ or $$E^i= \pi^i - \frac{\mu}{2}\epsilon^{ij}A_j.$$
we can compute $[E^i(\vec x),E^j(\vec y)]$ as $$[(\pi^i - \frac{\mu}{2}\epsilon^{ij}A_j),(\pi^i - \frac{\mu}{2}\epsilon^{ij}A_j)] $$expand above commutator and use following known commutation relations
$$[A_i(\vec x),\pi^j(\vec y)] = i \delta^j_i \delta^2(\vec x- \vec y)$$ $$[A_i(\vec x),A_j(\vec y)] =0$$ $$[\pi^i(\vec x),\pi^j(\vec y)] =0$$ and you will get your result. similarly we can use the relation $B=\epsilon^{ij}\partial_iA_j$ to compute the second commutation relation. |
Search
Now showing items 1-9 of 9
Production of $K*(892)^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$ =7 TeV
(Springer, 2012-10)
The production of K*(892)$^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$=7 TeV was measured by the ALICE experiment at the LHC. The yields and the transverse momentum spectra $d^2 N/dydp_T$ at midrapidity |y|<0.5 in ...
Transverse sphericity of primary charged particles in minimum bias proton-proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV
(Springer, 2012-09)
Measurements of the sphericity of primary charged particles in minimum bias proton--proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV with the ALICE detector at the LHC are presented. The observable is linearized to be ...
Pion, Kaon, and Proton Production in Central Pb--Pb Collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012-12)
In this Letter we report the first results on $\pi^\pm$, K$^\pm$, p and pbar production at mid-rapidity (|y|<0.5) in central Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV, measured by the ALICE experiment at the LHC. The ...
Measurement of prompt J/psi and beauty hadron production cross sections at mid-rapidity in pp collisions at root s=7 TeV
(Springer-verlag, 2012-11)
The ALICE experiment at the LHC has studied J/ψ production at mid-rapidity in pp collisions at s√=7 TeV through its electron pair decay on a data sample corresponding to an integrated luminosity Lint = 5.6 nb−1. The fraction ...
Suppression of high transverse momentum D mesons in central Pb--Pb collisions at $\sqrt{s_{NN}}=2.76$ TeV
(Springer, 2012-09)
The production of the prompt charm mesons $D^0$, $D^+$, $D^{*+}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at the LHC, at a centre-of-mass energy $\sqrt{s_{NN}}=2.76$ TeV per ...
J/$\psi$ suppression at forward rapidity in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012)
The ALICE experiment has measured the inclusive J/ψ production in Pb-Pb collisions at √sNN = 2.76 TeV down to pt = 0 in the rapidity range 2.5 < y < 4. A suppression of the inclusive J/ψ yield in Pb-Pb is observed with ...
Production of muons from heavy flavour decays at forward rapidity in pp and Pb-Pb collisions at $\sqrt {s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012)
The ALICE Collaboration has measured the inclusive production of muons from heavy flavour decays at forward rapidity, 2.5 < y < 4, in pp and Pb-Pb collisions at $\sqrt {s_{NN}}$ = 2.76 TeV. The pt-differential inclusive ...
Particle-yield modification in jet-like azimuthal dihadron correlations in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2012-03)
The yield of charged particles associated with high-pT trigger particles (8 < pT < 15 GeV/c) is measured with the ALICE detector in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV relative to proton-proton collisions at the ...
Measurement of the Cross Section for Electromagnetic Dissociation with Neutron Emission in Pb-Pb Collisions at √sNN = 2.76 TeV
(American Physical Society, 2012-12)
The first measurement of neutron emission in electromagnetic dissociation of 208Pb nuclei at the LHC is presented. The measurement is performed using the neutron Zero Degree Calorimeters of the ALICE experiment, which ... |
We have that by $x-y =t \to 0^+$ the given limit reduces to
$$\lim_{(x,y)\to(1,1)} \frac{(x-y)^{(x-y)}} {(x-y)}=\lim_{t\to 0^+} \frac{t^t} {t}\to \infty$$
indeed the points $(x-y)<0$ are not included within the domain of definition for the function that is $$D=\{(x,y)\in\mathbb{R^2}:x-y>0\}$$
Refer also to the related
Edit for a remark
As noticed by Jyrky Lahtonen in the comment "
some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$".
That important issue has been deeply discussed here
My observation is that not all sources have to be considered at the same level. Using the first definition is a reasonable way to deal with limits at a lower (high school) level whereas, in a more advanced context, the second more general definition (according to Rudin) should be adopted.
Moreover, note that, according to the first definition, the discussion of limits would reduce, as in the present case, to a (more or less) trivial determination on the domain of definition for functions which is of course a very different topic.
Therefore, when we deal with limits, I strongly suggest to refer to the more general definition excluding points outside the domain of definition for the function considered. |
I have a few doubts about my understanding of spontaneous symmetry breaking. To keep it simple, I will take a global U(1) transformation on a complex scalar field as an example. The questions are marked in bold and numbered, in order to encourage structured, easy-to-read answers that will hopefully be useful to other students as well.
The kinetic Lagrangian for complex scalar fields $\mathcal{L} = \partial_\mu \phi\ \partial^\mu \phi^*$ is invariant with respect to a global U(1) transformation $\phi(x) \rightarrow \phi'(x) = e^{i \alpha} \phi(x)$. Let's add a potential $V(|\phi|^2)$ that respects this symmetry, but makes the vacuum expectation value of $\phi$ non-vanishing:
$$V(|\phi|^2)=\frac{\lambda}{2}\left(|\phi|^2 - \frac{v^2}{2} \right)\tag{1}$$
However
(1) our usual machinery for the treatment of the Lagrangian is designed for fields with potentials taking their minimum value at $\phi = 0$. This bit comes from Griffiths. Why is that true?. Anyway this can be solved easily by shifting the field ($\tilde{\phi} = \phi - v/\sqrt{2}$) and rewriting the Lagrangian:
$$\mathcal{L} = \partial_\mu \tilde{\phi}\ \partial^\mu \tilde{\phi}^* - \frac{\lambda v^2}{4} \left( \tilde{\phi} + \tilde{\phi}^* \right)^2 + \mathcal{O} \left( \text{cubic terms in the fields} \right)\tag{2}$$
(2) In the following, I will ignore the cubic terms in the field, as done in this source for example. Do we ignore these terms, because they correspond to interactions and are not relevant to our present study? A complex field can be written as two real scalar fields ($\tilde{\phi} = \tilde{\phi}_1 + i \tilde{\phi}_2$), and in this form the Lagrangian becomes:
$$\mathcal{L} = \partial_\mu \tilde{\phi}_1 \partial^\mu \tilde{\phi}_1 + \partial_\mu \tilde{\phi}_2 \partial^\mu \tilde{\phi}_2 - \frac{\lambda v^2}{2} \tilde{\phi}_1^2\tag{3}$$
So in this picture we now have a massive, real scalar field $\tilde{\phi}_1$ with mass $m_{\tilde{\phi}_1}=\lambda v^2$, and a massless, real scalar field $\tilde{\phi}_2$ (corresponding to a Goldstone mode).
Let's go back to eq. 2 and perform the global U(1) transformation on $\tilde{\phi}$:
$$\mathcal{L}'= \partial_\mu \tilde{\phi}\ \partial^\mu \tilde{\phi}^* - \frac{\lambda v^2}{4} \left(e^{2 i \alpha} \tilde{\phi} \tilde{\phi}+ e^{-2 i\alpha} \tilde{\phi}^* \tilde{\phi}^* + 2\tilde{\phi} \tilde{\phi}^* \right) \tag{4}$$
(3) Is this strange-looking expression correct? Is this why we say that the symmetry "has been broken", since the Lagrangian is now not invariant under this transformation? (4) If that is correct, how is that a symmetry "breaking" at all? The original U(1) transformation was with respect to $\phi$, and is still valid. It seems to me more that you could say something like: the symmetry observed in the Lagrangian when expressed in terms of $\phi$ is a deceiving one, if the statement in question (1) is correct. Or said differently, it is a hidden one when the Lagrangian is expressed in terms of $\tilde{\phi}$.
Any other relevant comment is of course welcome. Thank you very much in advance for your answers!
Julien. |
π - Printable Version
+- HP Forums (
https://www.hpmuseum.org/forum)
+-- Forum: HP Calculators (and very old HP Computers) (
/forum-3.html)
+--- Forum: General Forum (
/forum-4.html)
+--- Thread: π (
/thread-3371.html)
Pages:
1 2 π - Massimo Gnerucci - 03-14-2015 09:44 AM
Happy Pi day, you MDY-ers!
RE: π - Thomas Radtke - 03-14-2015 09:48 AM
Thanks for the reminder, forgot it (being a DMY-er). Happy pi-day!
RE: π - Gerald H - 03-14-2015 09:57 AM (03-14-2015 09:48 AM)Thomas Radtke Wrote: Thanks for the reminder, forgot it (being a DMY-er). Happy pi-day!
As it's 20150314 I have a long time to wait.
RE: π - Tugdual - 03-14-2015 10:16 AM
What are you talking about? The real thing is Tau, I don't go for half a job.
Talk to you on June 28.
RE: π - Massimo Gnerucci - 03-14-2015 10:30 AM (03-14-2015 10:16 AM)Tugdual Wrote: What are you talking about? The real thing is Tau, I don't go for half a job.
What are you talking about? I only know of τ Ceti.
:D
RE: π - Massimo Gnerucci - 03-14-2015 10:34 AM
Pau is the answer!
xkcd
RE: π - John Smitherman - 03-14-2015 03:17 PM
In the US we get to celebrate twice since we tell time predominantly with a 12 hour a.m. / p.m. clock. ;-)
Regards,
John
RE: π - Jlouis - 03-14-2015 03:33 PM
[quote='Massimo Gnerucci' pid='30585' dateline='1426326276']
Happy Pi day, you MDY-ers!
Kate Bush is my favourite female singer since long time.
I'm glad that she is remembered this very especial day.
Happy Pi day to you all!
Live Long and Prosper!
RE: π - Gerson W. Barbosa - 03-14-2015 04:30 PM
FWIW,
\[\ln \left [ \frac{16\ln 878}{\ln \left ( 16\ln 878 \right )} \right ]=3.141 592 653 77\]
\[1000\left ( \sqrt{\frac{e}{\sqrt{5}}}-1 \right )-4\pi = 89.999 994 77\]
\[\frac{\sqrt{2}\left ( \pi ^{17}-80e^{\pi } \right )}{4}= 99 999 999.949\]
RE: π - Ángel Martin - 03-14-2015 04:45 PM (03-14-2015 03:33 PM)Jlouis Wrote:
Same here - And Aerial is bound to be her most subtle and intimate album ever..\
RE: π - Gerald H - 03-14-2015 06:10 PM
This is the chap who invented pi:
http://www.theguardian.com/profile/gareth-ffowc-roberts
RE: π - Tugdual - 03-14-2015 07:55 PM
In the US the Pi day is 2nd of March.
http://en.wikipedia.org/wiki/Indiana_Pi_Bill
RE: π - Jlouis - 03-14-2015 11:41 PM
Someone at YouTube coments of this Kate Bush song wrote that she made many mistakes singing the Pi number.
I tried to follow her singing but I can't understand many numbers she sings.
Could someone native english could check this out? Just for curiosity?
I know she is extremily perfectionist, some albuns took 2 or 3 years to be completed, then I doubt she could have made a mistake like that.
TIA
Edit: was the 35s available to Kate in 2005? Just kidding...
RE: π - Gerson W. Barbosa - 03-15-2015 12:01 AM (03-14-2015 11:41 PM)Jlouis Wrote: Someone at YouTube coments of this Kate Bush song wrote that she made many mistakes singing the Pi number.
You can check the lyrics (and the digits) here.
Gerson.
RE: π - Jlouis - 03-15-2015 12:20 AM (03-15-2015 12:01 AM)Gerson W. Barbosa Wrote:(03-14-2015 11:41 PM)Jlouis Wrote: Someone at YouTube coments of this Kate Bush song wrote that she made many mistakes singing the Pi number.
Thanks Gerson. I'm gonna check this out.
Abraços
RE: π - Gerson W. Barbosa - 03-15-2015 12:29 AM
There is also Liczba Pi (Number PI), a poem by Wislawa Szymborska (1996 Literature Nobel Prize Winner).
http://www.edulandia.pl/matura/1,117853,5854221,Liczby_Szymborskiej.html
I learned from my mother (a Polish Brazilian) how to count in Polish from 1 to 5 only, so I was able to check only the very first digits :-)
trzy koma jeden cztery jeden pięć
RE: π - Jlouis - 03-15-2015 12:38 AM (03-15-2015 12:20 AM)Jlouis Wrote:(03-15-2015 12:01 AM)Gerson W. Barbosa Wrote: You can check the lyrics (and the digits) here.
Yes, there are many mistakes.
Now I am certain that she used a 35s.
Kate Bush does not make mistakes.
RE: π - Gerson W. Barbosa - 03-15-2015 12:44 AM (03-15-2015 12:38 AM)Jlouis Wrote:
Neither does Katie. I haven't tried it yet, but I am quite sure Katie Wasserman's HP-32SII program gives only correct digits, even when run on the HP-35S :-)
http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/articles.cgi?read=899
[]s,
Gerson.
P.S.: I like Wuthering Heights better. I had just learned English then (or at least I thought I had :-) and I had also read the book (1977 I think).
RE: π - Jlouis - 03-15-2015 12:49 AM (03-15-2015 12:29 AM)Gerson W. Barbosa Wrote: There is also Liczba Pi (Number PI), a poem by Wislawa Szymborska (1996 Literature Nobel Prize Winner).
Now we understand where your genious comes from:
You have Jan Lukasiewicz DNA.
RE: π - patrice - 03-15-2015 07:29 AM
Hi all,
being a DMY-er I forgot this day, but you remained me , thanks.
did you noticed that for MDY-ers, yesterday was 4 decimals ? 3.1415 |
Prefix edit
Why not allow the double factorial? well, let's use it. For the record:
\begin{align}0!! &= 1 \\5!! &= 3 \cdot 5 = 15 \\6!! &= 2 \cdot 4 \cdot 6 = 48 \\\end{align}
Also the choose operator (also known as binomial coefficient).
Modular approach
I had some fun doing it the long way, but then I decided to jump into a more modular / exploitable strategy. I will build it for a little bit, so bear with me for now :)
First, let's make the following list with one four:
\begin{align}2 &= \sqrt 4 \\3 &= \left(\sqrt 4\right)? \\4 &= 4 \\6 &= \left(\sqrt 4\right)?! \\8 &= 4!! \\10 &= 4? \\21 &= \left(\sqrt 4\right)?!? \\24 &= 4! \\36 &= \left(4!!\right)? \\48 &= \left(\left(\sqrt 4\right)?!\right)!! = 6!!\\55 &= 4??\end{align}
I can also consider $ 4 = \sqrt 4 + \sqrt 4 $, so there is no need to consider the "extra fours".
From now on, I will use the fancy "one-four" substitution, and maybe some results will use less than four fours. But the translation from a compact equation to a "four-fours" equation is immediate.
Let's consider the following list of two-four numbers (I purposely omit the ones that can be obtained with a signel four):
\begin{align}0 &= 4 - 4 \\1 &= \frac{4}{4} \\5 &= 2 + 3 \\7 &= 4 + 3 \\9 &= 3 + 6 \\11 &= 21 - 10 \\12 &= 10 + 2 \\13 &= 10 + 3 \\14 &= 10 + 4 \\15 &= 21 - 6 \\16 &= 4 \cdot 4 \\17 &= 21 - 4 \\18 &= 21 - 3 \\19 &= 21 - 2 \\20 &= 24 - 4 \\22 &= 24 - 2 \\23 &= 21 + 2 \\25 &= 21 + 4 \\26 &= 24 + 2 \\27 &= 24 + 3 \\28 &= 24 + 4\end{align}
Up to this point we can get any integer below 28 with only two fours. My strategy will be obtaining numbers by combining the "high part" and "low part". So, given any two-four "high number" $n$ we can generate all integers between $n - 28$ and $n + 28$. And the resulting formula uses up to four fours.
We can trivially consider the following property:
$$\forall n < 55: \quad (n+1)? - n? < 56 $$
So we can have a "dense-enough" set of high numbers by simply using the $?$ operator to the list of "two-four numbers". The maximum number at the moment is:
$$ 434 = 28? + 28 = (4! + 4)? + 4! + 4 $$
To continue, we should pick integers with a maximum distance of 56 between them. The next integer should be at most 463, because $ 462 - 28 = 434 $.
Next hand-picked "high numbers" (credit to @f'' for most of them!):\begin{align}441 &= 21^2 \\465 &= (24 + 6)? \\504 &= 21*24 \\550 &= 55*10 \\600 &= 24?+24? \\630 &= \binom{36}{2} \\665 &= 6! - 55 \\720 &= 6! \\775 &= 6! + 55\end{align}
My original post contained a exhaustive list up to 132 and some odd holes up to 148, holes that user @f fixed in the comments, so credit for him for that. But now that I present the alternative strategy, the original post seems overweighted and slow to load :( |
Under the assumption that your background noise is
statistically independent of the signal you want to recover and that your second microphone picks up virtually nothing from the body noise, you have the following model for your measurements:$$ \begin{eqnarray} M_1(t) &=& L\{N(t)\} + B(t) \\ M_2(t) &=& N(t) \end{eqnarray}$$
Here, $M_1,M_2$ are the signals picked up by the microphone, $B(t)$ is the signal you want to recover, $N(t)$ is the environmental noise signal and $L$ is a linear time invariant system that maps the recorded environmental noise to what is picked up by the second microphone.
With $\langle\langle \cdot, \cdot \rangle\rangle$ as cross correlation, we also have the non-correlation constraint $\langle\langle N,B \rangle\rangle=0$ and $\langle\langle L\{N\},B \rangle\rangle=0$ as a consequence of statistical independence of $N$ and $B$.
These prerequisites guarantee a
cross correlation matching pursuit to converge against $B(t)$: Normalise $M_2$ with the standard inner product: $M_2(t)\leftarrow \frac{M_2}{\sqrt{\langle M_2(t),M_2(t) \rangle}}$ Find $c(t):=\langle\langle M_2,M_1\rangle\rangle$ Find the time associated with the maximum of the squared cross correlation: $\tau=\mathrm{argmax}\left( c^2(t) \right)$ Update $M_1$ to remove the contribution of the environmental noise: $M_1(t)\leftarrow M_1(t)-\langle M_2(t-\tau),M_2 \rangle\cdot M_2(t-\tau)$ If result not good enough, go back to 2.
In step 4 you may also use $c(\tau)$ instead of the inner product if the cross correlation uses the same inner product. Stopping the matching pursuit is a question of experience. Some of the assumptions you need for this algorithm are only approximately true, like the perfect uncorrelatedness of the environmental noise and the body signal. A good way to check if your matching pursuit runs into a non-productive state of signal modifications is to observe the natural square-norm of $M_1$. If it stops decreasing in step 4, terminate the loop.
After the algorithm ran, $M_1(t)$ will contain only your body signal $B(t)$. |
First, I assume finite dimensional operators: otherwise you need to check certain boundedness conditions on the operators. Because the CBH series is here truncated by the vanishing double commutators, the conditions for linear operators on
e.g. $\mathbf{L}^2(\mathbb{R})$ will be mild.
You need to practice operations with $\mathrm{Ad}$. Look up the following. In the Lie group $\mathfrak{G}$ with algebra $\mathfrak{g}$ the tangent vector to the path:
$$\sigma:\mathbb{R}\to\mathfrak{G};\;\sigma(\tau) = e^A\,e^{\tau\,B}\,e^{-A};\;A,\,B\in\mathfrak{g}\tag{1}$$
at the identity is $\mathrm{Ad}(e^A)\,B=\exp(\mathrm{ad}(A))\,B$. Here $\mathrm{Ad}:\mathfrak{G}\to GL(\mathfrak{g})$ is the Adjoint Representation. It is a Lie group homomorphism from the general Lie group $\mathfrak{G}$ to the matrix Lie group $GL(\mathfrak{g})$. Its kernel is the centre of $\mathfrak{G}$. Since it is a homomorphism, we have $\mathrm{Ad}(\gamma\,\zeta) = \mathrm{Ad}(\gamma)\,\mathrm{Ad}(\zeta);\,\forall \gamma,\,\zeta\in\mathfrak{G}$. Another useful identity is:
$$\begin{array}{lcl}\mathrm{Ad}(e^A)\,B &=& \exp(\mathrm{ad}(A))\,B \\&=&B + \mathrm{ad}(A) B + \frac{\mathrm{ad}(A)^2}{2!}\,B +\cdots \\&=& B+ [A,\,B] + \frac{1}{2!}\, [A,\,[A,\,B]] + \cdots\end{array}\tag{2}$$
and this series is
universally convergent if the operator $B\mapsto[A,\,B]$ is suitably bounded ( e.g. $\left\|[A,\,B]\right\| \leq K(A)\,\left\|B\right\|$ for some $K(A)\in\mathbb{R}$ - this is certainly true in finite dimensions).
Now, by (1) and the homomorphism property ($\mathrm{Ad}(e^{\lambda\,A}\,e^{\lambda\,B}) = \mathrm{Ad}(e^{\lambda\,A})\,\mathrm{Ad}(e^{\lambda\,B})$), you can find that:
$$\begin{array}{lcl}\mathrm{d}_\lambda f &=& A\,e^{\lambda\, A}\,e^{\lambda\,B}\,e^{-\lambda\,(A+B)} + e^{\lambda\, A}\,B\,e^{\lambda\,B}\,e^{-\lambda\,(A+B)} - e^{\lambda\, A}\,e^{\lambda\,B} \,(A+B)\,e^{-\lambda\,A+B)}\\&=& \left(A + e^{\lambda\,A}\,B\,e^{-\lambda\,A} - e^{\lambda\,A}\,e^{\lambda\,B}\,(A+B)\,e^{-\lambda\,B}\,e^{-\lambda\,A}\right)\,e^{\lambda\, A}\,e^{\lambda\,B}\,e^{-\lambda\,(A+B)}\\&=&\left(A+\mathrm{Ad}(e^{\lambda\,A})\left(B-\mathrm{Ad}(e^{\lambda\,B})\,(A+B)\right)\right)\,f\end{array}\tag{3}$$
All the above is perfectly general. You need to specialise it to your truncated case. So use the universally convergent (and here truncated to two terms) series (2) to expand $A+\mathrm{Ad}(e^{\lambda\,A})\left(B-\mathrm{Ad}(e^{\lambda\,B})\,(A+B)\right)$ and truncate it for your special case and I think you should make some headway.
A pedantic peeve: although both orders for the name are quite common, the order that accurately reflects the historical precedence is "Campbell-Baker-Hausdorff" as each of the authors made their contributions in 1897/1898 (Campbell), 1905 (Baker) and 1906 (Hausdorff), respectively. Each was aware of their forerunners' work, but, as stated in Fascicule 16 Ch 1 of Bourbaki (1960), "each found the demonstrations of his forerunners unconvincing(!)". That statement always makes me giggle and gives some comfort that I'm not the only one with about a 5% comprehension rate in reading technical literature (I reckon I need to read a paper about 20 times on average to "get" it). An amusing fact is that
none of these three actually worked out the series. Instead, they established the theorem that the series was convergent within some neigbourhood of $\mathbf{0}$ in the Lie algebra and comprises linear and Lie bracket operations only. The formula itself is due to Dynkin and was fully worked out in 1947! |
Search
Now showing items 1-1 of 1
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 ... |
Wikipedia definition
A sequence $X_n$ of random variables converges in probability towards the random variable $X$ if for all $\varepsilon > 0$:
\begin{equation} \lim_{n\to\infty}\Pr\big(|X_n-X| \geq \varepsilon\big) = 0 \end{equation}
Formally, pick any $\varepsilon > 0$ and any $\delta > 0$. Let $P_n$ be the probability that $X_n$ is outside the ball of radius $\varepsilon$ centered at $X$. Then for $X_n$ to converge in probability to $X$ there should exist a number $N$ (which will depend on $\varepsilon$ and $\delta$) such that for all $n \geq N$, $P_n < \delta$.
My question
When $X$ is a constant random variable, this definition is easy to understand. But what happens if $X$ is not constant ? Without the expression of the joint distribution $p(X_n, X)$, how can $Pr\big(|X_n - X| \geq \varepsilon\big)$ be computed ? |
Root mean square is the square root of a mean square or it can be defined as the arithmetic mean of squares of group values. This is a special case of quadratic mean whose exponent value is two. Root mean square is also taken as the variable function based on the integral of squares of values that are instantaneous in a cycle.
RMS is the square of function that defines the waveform in continuity. Also, the RMS of a periodic function is generally equal to RMS of a single function. RMS of a continuous function is generally calculated in approximate values by computing RMS of a sequence of equally spaced entities. Also, The RMS can be computed for different waveforms without using the calculus.
The other popular term associated with Root Mean Square is RMSE (Root Mean Square Error) predicted through an estimator or you could take help of a mode. It will tell the differences between the actual value and the calculated value.
\[\large X_{rms}=\sqrt{\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}….x_{n}^{2}}{n}}\]
Where,
$x_{1},\; x_{2},\; x_{3}$ are observations n is the total number of observations Question: Calculate the root mean square of the following observations; 5, 4, 8, 1 ? Solution:
Using the Root Menu Square formula: \[\large X_{rms}=\sqrt{\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}….x_{n}^{2}}{n}}\]
\[\large Root\;Mean\;Square=\sqrt{\frac{5^{2}+4^{2}+8^{2}+1^{2}}{4}}\]
Root mean square = 5.14
Chi-square is a common term in non-parametric statistics.it is used to analyze data consist of people distributed across various categories. It is necessary to check either distribution is performed as per the expected values or it is different. There is extremely small chi-square test in statistics that will check either observed data suits your expectations well or not. If the value is small then data fits your expectations well otherwise in case of large value you need to reject the null hypothesis.
Chi-square formula is the perfect way how can you show the relationship between two categorical variables. The two types of categoric variables are numerical variables and the non-numeric variables. And the final values will be the difference between observed frequency and the expected frequency. The common applications of chi square distribution are –
The chi-square test is associated with the non-parametric test and it can be utilized for nominal data too. This is used in advanced mathematics for tabular association and analysis. One major application is the hypothesis that will check either two populations are same or different in terms of characteristics and behavior.
Also, you can study two random samples in terms of different parameters. It can also be used to analyze a particular distribution by comparing the observed data with expected values. One of the drawbacks of chi-square test is that it does not allow the calculation of confidence intervals so the size of a sample is not available readily.
\[\LARGE X^{2}=\sum \frac{(O-E)^{2}}{E}\]
Where,
O = Observed frequency E = Expected frequency Σ = Summation X 2 = Chi Square value
In elementary mathematics, the completing a square is generally applied for the computation of quadrilateral polynomials. Completing the square formula is also used to derive the quadratic formulas.
\[\LARGE ax^{2}+bx+c\Rightarrow (x+p)^{2}+constant\]
Since the degree of this polynomial is two, so it has two roots or solutions. There are a plenty of methods to solve the roots of a quadratic equation and one of these popular techniques is completing the square method. |
The sample space $S$ is the set of all pairs $(i, j)$ with $1 \leq i, j \leq 6$, where $i$ and $j$ are the numbers corresponding to the first and the second die respectively. Hence $|S| = 36$. All of the $36$ possible outcomes are equally likely.
Solution (1)
Now we consider the outcomes that give rise to the event that the sum of the upturned faces is $5$. There are four such outcomes:\[(1, 4), (2, 3), (3, 2), (4, 1).\]Thus, the desired probability is $\frac{4}{36}=\frac{1}{9}$.
Solution (2)
The outcomes that give rise to this event are $(i, j)$ with $i < j$, where $i$ and $j$ are the numbers corresponding to the first and second die respectively.We count the number of such outcomes as follows.When $i=1$, possible values for $j$ are $j=2, 3, 4, 5, 6$. When $i=2$, possible values for $j$ are $j = 3, 4, 5, 6$.In general, $j$ can be $j = i+1, i+2, \dots, 6$. Note that $i$ cannot be $6$, otherwise $j$ is not strictly greater than $i$.Counting this way, we see that there are\[5+4+3+2+1 = 15\]outcomes, each of probability $1/36$.Hence the probability that the outcomes of the second die is strictly greater than the first die is\[15 \times \frac{1}{36} = \frac{15}{36} = \frac{5}{12}.\]
Conditional Probability Problems about Die RollingA fair six-sided die is rolled.(1) What is the conditional probability that the die lands on a prime number given the die lands on an odd number?(2) What is the conditional probability that the die lands on 1 given the die lands on a prime number?Solution.Let $E$ […]
Complement of Independent Events are IndependentLet $E$ and $F$ be independent events. Let $F^c$ be the complement of $F$.Prove that $E$ and $F^c$ are independent as well.Solution.Note that $E\cap F$ and $E \cap F^c$ are disjoint and $E = (E \cap F) \cup (E \cap F^c)$. It follows that\[P(E) = P(E \cap F) + P(E […]
Overall Fraction of Defective Smartphones of Three FactoriesA certain model of smartphone is manufactured by three factories A, B, and C. Factories A, B, and C produce $60\%$, $25\%$, and $15\%$ of the smartphones, respectively.Suppose that their defective rates are $5\%$, $2\%$, and $7\%$, respectively. Determine the overall fraction of […]
Independent and Dependent Events of Three Coins TossingSuppose that three fair coins are tossed. Let $H_1$ be the event that the first coin lands heads and let $H_2$ be the event that the second coin lands heads. Also, let $E$ be the event that exactly two coins lands heads in a row.For each pair of these events, determine whether […]
Independent Events of Playing CardsA card is chosen randomly from a deck of the standard 52 playing cards.Let $E$ be the event that the selected card is a king and let $F$ be the event that it is a heart.Prove or disprove that the events $E$ and $F$ are independent.Definition of IndependenceEvents […]
Jewelry Company Quality Test Failure ProbabilityA jewelry company requires for its products to pass three tests before they are sold at stores. For gold rings, 90 % passes the first test, 85 % passes the second test, and 80 % passes the third test. If a product fails any test, the product is thrown away and it will not take the […]
Pick Two Balls from a Box, What is the Probability Both are Red?There are three blue balls and two red balls in a box.When we randomly pick two balls out of the box without replacement, what is the probability that both of the balls are red?Solution.Let $R_1$ be the event that the first ball is red and $R_2$ be the event that the […] |
In Hoeffding's inequality $$ P(S_n-E[S_n]\geq t)\leq \exp(\frac{t^2}{\sum_{i=1}^n(a_i-b_i)^2}), $$ where $S_n=X_1+X_2+\dots+X_n$, with $X_i$ independent random variables satisfying $a_i<X_i<b_i$, is it allowed to let $a_i$ and $b_i$ depend on $n$?
Yes, it is allowed. Note that the inequality is a statement for
fixed $n$, hence there is no problem with $a_i$ and $b_i$ depending on $n$.
This is like asking if the $a_i$'s in the set $\{a_1, a_2, ..., a_n\}$ are allowed to depend on $n$. They can. Consider $a_i = i/n$ like in Riemann sums. Then
$$\{a_1, a_2, ..., a_n\} = \{1/n, 2/n, ..., n/n\}$$
We can have $a_i = (i-1)/n$
Then
$$\{a_1, a_2, ..., a_n\} = \{0/n, 1/n, ..., (n-1)/n\}$$ |
Real Analysis Exchange Real Anal. Exchange Volume 38, Number 1 (2012), 193-210. When is a Family of Generalized Means a Scale? Abstract
For a family \(\{k_\alpha \,\vert \,\alpha \in I\}\) of real \(\mathcal{C}^2\) functions defined on \(U\) (\(I\), \(U\) — open intervals) and satisfying some mild regularity conditions, we prove that the mapping \(I \ni \alpha \mapsto k_\alpha^{-1}\bigl(\sum_{i=1}^n w_i k_\alpha(a_i)\bigr)\) is a continuous bijection between \(I\) and \((\min\underline{a}, \,\max\underline{a})\), for every fixed non-constant sequence \(\underline{a} = \bigl(a_i\bigr)_{i=1}^n\) with values in \(U\) and every set, of the same cardinality, of positive weights \(\underline{w} = \bigl(w_i\bigr)_{i=1}^n\). In such a situation one says that the family of functions \(\{k_\alpha\}\) generates a
scale on \(U\). The precise assumptions in our result read (all indicated derivatives are with respect to \(x \in U\))
(i) \(k'_\alpha\) vanishes nowhere in \(U\) for every \(\alpha \in I\),
(ii) \(I \ni \alpha \mapsto \frac{k''_\alpha(x)}{k'_\alpha(x)}\) is increasing, 1-1 on a dense subset of \(U\) and onto the image \(\mathbb{R}\) for every \(x \in U\).
This result makes possible three things: 1) a new and extremely short proof of the classical fact that
power means generate a scale on \((0,+\infty)\), 2) a short proof of a fact, which is in a direct relation to two results established by Kolesárová in 2001, that, for every strictly increasing convex and \(\mathcal{C}^2\) function \(k \colon (0,\,1) \to (0,\,+\infty)\), the class \(\{\mathfrak{M}_{k_\alpha}\}_{\alpha \in (0,\,+\infty)}\) of quasi-arithmetic means (see Introduction for the definition) generated by functions \(k_\alpha\), \(k_\alpha(x) = k(x^\alpha)\), \(\alpha \in (0,\,+\infty)\), generates a scale on \((0,1)\) between the geometric mean and maximum (meaning that, for every \(\underline{a}\), \(\underline{w}\), if \(s \in \bigl(\prod_{i = 1}^n a_i^{\,w_i},\,\max(\underline{a})\bigr)\) then there exists exactly one \(\alpha\) such that \(\mathfrak{M}_{k_\alpha}(\underline{a},\underline{w}) = s\)). 3) a brief proof of one of the classical results of the Italian statistics' school from the 1910-20s that the so-called radical means generate a scale on \((0,\, +\infty)\). Article information Source Real Anal. Exchange, Volume 38, Number 1 (2012), 193-210. Dates First available in Project Euclid: 29 April 2013 Permanent link to this document https://projecteuclid.org/euclid.rae/1367265648 Mathematical Reviews number (MathSciNet) MR3083206 Zentralblatt MATH identifier 1277.26061 Citation
Pasteczka, Paweł. When is a Family of Generalized Means a Scale?. Real Anal. Exchange 38 (2012), no. 1, 193--210. https://projecteuclid.org/euclid.rae/1367265648 |
Revista Matemática Iberoamericana Rev. Mat. Iberoamericana Volume 19, Number 3 (2003), 873-917. The Pressure Equation in the Fast Diffusion Range Abstract
We consider the following degenerate parabolic equation $$ v_{t}=v\Delta v-\gamma|\nabla v|^{2}\quad\mbox{in $\mathbb{R}^{N} \times(0,\infty)$,} $$ whose behaviour depends strongly on the parameter $\gamma$. While the range $\gamma < 0$ is well understood, qualitative and analytical novelties appear for $\gamma>0$. Thus, the standard concepts of weak or viscosity solution do not produce uniqueness. Here we show that for $\gamma>\max\{N/2,1\}$ the initial value problem is well posed in a precisely defined setting: the solutions are chosen in a class $\mathcal{W}_s$ of local weak solutions with constant support; initial data can be any nonnegative measurable function $v_{0}$ (infinite values also accepted); uniqueness is only obtained using a special concept of initial trace, the $p$-trace with $p=-\gamma < 0$, since the standard concepts of initial trace do not produce uniqueness. Here are some additional properties: the solutions turn out to be classical for $t>0$, the support is constant in time, and not all of them can be obtained by the vanishing viscosity method. We also show that singular measures are not admissible as initial data, and study the asymptotic behaviour as $t\to \infty$.
Article information Source Rev. Mat. Iberoamericana, Volume 19, Number 3 (2003), 873-917. Dates First available in Project Euclid: 20 February 2004 Permanent link to this document https://projecteuclid.org/euclid.rmi/1077293809 Mathematical Reviews number (MathSciNet) MR2053567 Zentralblatt MATH identifier 1073.35128 Citation
Chasseigne, Emmanuel; Vázquez, Juan Luis. The Pressure Equation in the Fast Diffusion Range. Rev. Mat. Iberoamericana 19 (2003), no. 3, 873--917. https://projecteuclid.org/euclid.rmi/1077293809 |
Forewords
In a previous post, I wrote about the theoretical foundation of the CART algorithm with a particular focus on regression problems. I then put the theory into practice by conducting an experiment in R where some regression trees were fitted and interpreted. Continuing on the topic of CART, in this post I will write about how the algorithm can also be used to build classification trees. Since the two kinds of trees mainly differ in terms of the optimization criteria for splitting and pruning, these two aspects will be the focus of this post.
Optimization criteria for splitting
For both regression and classification problems, CART uses a greedy and binary recursive partitioning strategy to grow a tree. However, unlike regression trees which use mean squared error as the measure for node impurity, classification trees uses a measure called Gini index.
Assuming that the outcome has \(c\) classes, and the proportion of class \(i\) in a node is denoted as \(p_i\), then the Gini index for that particular node can be defined as:
\[ \begin{align} Gini \ index &= \sum_{i = 1}^{c}p_i(1 - p_i) \\ &= \sum_{i = 1}^{c}p_i - \sum_{i = 1}^{c}p_i^2 \\ &= 1 - \sum_{i = 1}^{c}p_i^2 \end{align} \]
The value of the Gini index has an inverse association with the variance of \(p\) (i.e., \(\sum_{i = 1}^{c}(p_i - 1 / c)^2\)), and is maximized when \(p_1 = p_2 = ... = p_c = \frac{1}{c}\), minimized when \(p_i = 1\) (\(i \in \{1, 2, ..., c\}\)).
The strategy of recursive partitioning is to find a best split for each node that has the minimum weighted sum of the Gini index of the two child nodes. Suppose the number of observations in a parent node \(R_k\) is denoted as \(N_k\); those in the two child nodes \(R_{kl}\) and \(R_{kr}\) are denoted as \(N_{kl}\) and \(N_{kr}\), respectively, then the measure to evaluate the split would be
1:
\[ Gini(R_{kl}) \times \frac{N_{kl}}{N_k} + Gini(R_{kr}) \times \frac{N_{kr}}{N_k} \]
where \(Gini(R_{kl})\) and \(Gini(R_{kr})\) respectively represents the Gini index for the two child nodes. Note that \(N_{kl} + N_{kr} = N_k\).
Prediction
The proportion of a class in a node is taken as the probability of that class in that specific node. Therefore, the outcome of any observation that falls into a leaf node will be predicted with the most frequent class in that node.
Tree pruning
After a full tree has been grown, it can be pruned using cost-complexity pruning. The Gini index can be used to guide the process, but typically the misclassification error rate is used instead. Let \(m\) denote the number of leaf nodes in a sub-tree \(T\), the pruning process aims to find the best sub-tree that has the minimal \(C_{\alpha}(T)\) defined as follows:
\[ C_{\alpha}(T) = \sum_{k = 1}^{m} ErrorRate(R_k) \times N_k + \alpha m \quad (3) \]
where \(\alpha\) is the hyperparameter that controls the process.
The pruning process starts with parent nodes of leave nodes and proceeds until only the root node is left. In effect, each time the algorithm chooses one node to collapse which leads to a minimal increase in \(\sum_{k = 1}^{m} ErrorRate(R_k) \times N_k\).
More on the optimization criteria
Why use the Gini index instead of the misclassification rate as the optimization criteria for splitting? One reason is that the Gini index is differentiable and is therefore more suitable for numerical optimization. Another reason is that the Gini index is more sensitive to changes in the node probabilities than the misclassification rate. Assume a two-class problem with 4 observations in each class, which can be denoted by \((4, 4)\). Suppose one split creates nodes \((3, 1)\) and \((1, 3)\), while another split creates nodes \((2, 4)\) and \((2, 0)\). Both splits produce a misclassification rate of 0.25, but the Gini index is lower for the second split. Since the second split produces a pure node and is therefore more preferable, the Gini index appears to be a better measure to evaluate the result of a split.
For comparison, the measure to evaluate a split in a regression tree is \(MSE(R_{kl}) \times \frac{N_{kl}}{N_k} + MSE(R_{kr}) \times \frac{N_{kr}}{N_k}\), where \(MSE\) stands for mean squared error.↩ |
In the case of a generalized flag manifold $G/P$, we have an explicit description of their cohomology groups due to Borel.(See herehere for a description.) I would like to know what the hard Lefschetz theorem looks like in this presentation.
I'll spell out Hard Lefschetz as an explicit combinatorial statement about the Grassmannian $G(d,n)$. I can definitely give you a version of this for the full flag manifold if you want it and probably for any $G/P$. I have no idea about a combinatorial proof.
$H^{\ast}(G(d,n))$ is entirely in even degrees, and has a standard basis given by the Schubert classes $X_{\lambda}$. The Schubert classes are indexed by partitions $(\lambda_1, \ldots, \lambda_d)$: Sequences of integers with $n-d \geq \lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_d \geq 0$. The class $X_{\lambda}$ lives in cohomological degree $2 \sum \lambda_i$. We set $\sum \lambda_i = |\lambda|$.
I'll take $G(2,5)$ as my running example. There are $10$ total partitions with $5-2 \geq \lambda_1 \geq \lambda_2 \geq 0$: namely $(0,0)$, $(1,0)$, $(2,0)$, $(1,1)$, $(3,0)$, $(2,1)$, $(3,1)$, $(2,2)$, $(3,2)$ and $(3,3)$. The groups $H^{2 \cdot 2}(G(2,5))$ and $H^{2 \cdot 4}(G(2,5))$ both have dimension $2$, with bases $([X_{2,0}], [X_{1,1}])$ and $([X_{3,1}], [X_{2,2}])$ respectively.
$H^2(G(d,n))$ is always one dimensional, with generator $\zeta = [X_{1000\ldots0}]$, and positive multiples of $\zeta$ are ample. Multiplication is given by the Pieri rule: $$\zeta \cdot [X_{\lambda}] = \sum_{|\mu| = |\lambda|+1,\ \mu_j \geq \lambda_j} [X_{\mu}].$$ For example, $\zeta \cdot X_{2,0} = X_{3,0} + X_{2,1}$. So the coefficient of $X_{\mu}$ in $\zeta^{|\mu| - |\lambda|} X_{\lambda}$ is the number of chains of partitions $(\lambda = \alpha^0, \alpha^1, \ldots, \alpha^N = \mu)$ with $|\alpha^{i+1}| = |\alpha^i|$ and $\alpha^{i+1}_j \geq \alpha^i_j$. In other words, the number of standard Young tableaux of shape $\mu \setminus \lambda$.
Taking our running example of the map $\zeta^2 : H^{2 \cdot 2}(G(2,5)) \to H^{2 \cdot 4}(G(2,5))$, we see that $$\zeta^2 [X_{2,0}] = 2 [X_{3,1}] + [X_{2,2}]$$ $$\zeta^2 [X_{1,1}] = [X_{3,1}] + [X_{2,2}].$$
So Hard Lefschetz in this case is the statement that $$\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$$ is invertible.
More generally, fix $(d,n,r)$. Let $A(d,n,r)$ be the set of partitions $n-d \geq \lambda_1 \geq \cdots \geq \lambda_d \geq 0$ with $\sum \lambda_i=r$. Hard Lefschetz says that, if I take a matrix whose rows are indexed by $A(d,n,r)$, and whose columns are indexed by $A(d,n,d(n-d)-r)$, and where the entry $L_{\lambda \mu}$ is the number of SYT of shape $\mu / \lambda$, then this matrix is invertible.
I think Harry Tamvakis has thought about whether this can be proved directly from symmetric function combinatorics.
Is this the sort of thing you are looking for? Is it worth it to you for me to give you the analogous versions for other $G/P$?
To add to David's answer a bit:
The Borel description of $H(G/P)$ is the $W_P$-invariant subring of the $W$ coinvariants, $S^{W_P}_W$, where $W$ is the associated Weyl group and $W_P\subseteq W$ is the parabolic subgroup associated to $P$. As David indicates above, we can choose a Schubert basis for $S^{W_P}_W$, and compute the matrices for the Lefschetz maps in that basis. The Pieri rule allows us to interpret these Lefschetz matrices as so-called ''weighted path matrices'' with respect to the Bruhat order restricted to the set of maximal coset representatives of $W/W_P$. A well known ''lemma'' of Gessel and Viennot is then helpful in computing these determinants (the original paper is available here). These ideas are used in this paper (now published in Journal of Algebra) to give a purely algebraic proof of the hard Lefschetz theorem for $S_W$ where $W$ is any finite Coxeter group, as well as for $S^{W_P}_W$ for certain maximal parabolics $W_P\subseteq W$. One could probably also use these ideas to prove that $S^{W_P}_W$ is hard Lefschetz for any parabolic subgroup, although the Bruhat order for $W/W_P$ is more complicated. |
$\DeclareMathOperator{\ex}{\mathbb E}\DeclareMathOperator{\Var}{Var}\DeclareMathOperator{\Cov}{Cov}$If there are two giant families, one with all the women in it and one with all the men in it, then men will have more brothers on average. So you cannot say anything
in general, even assuming that there are the same number of men and women.
So we will need to make some assumptions. In order to find out what the correct assumptions are, I shall define some notation. For convenience, I shall use probabilistic notation, but we are really just talking about counting.
Let $M$ be the random variable corresponding to the number of men in a family chosen at random from the set of families, and let $W$ be the quantity corresponding to the number of women in a randomly chosen family. Let $\mathcal M$ denote the total number of men, let $\mathcal W$ denote the total number of women and let $\mathcal F$ denote the total number of families.
Important Note: The families themselves should be treated as constants. They are not random samples drawn from some kind of
distribution or anything like that. When I use probabilistic
notation, it is purely for the sake of convenience - I am interpreting the question combinatorially, and it so happens that probabilistic constructs such as sample variance do a good job of capturing certain combinatorial quantities that are relevant in this question.
If you like, we are working over the discrete measure space $(F,
\mathcal P(F), \mathbb P)$, where $F$ is the set of families and
$\mathbb P(A)=|A|/|F|$ for any $A\subset F$. $M$ is then the random
variable defined by $M(f)=\textrm{number of men in $f$}$, while
$W(f)=\textrm{number of women in $f$}$. $\ex,\Var,\Cov$ will all take
their usual meanings as population mean, population variance and population
covariance.
We want to compute the average number of brothers that each man has. To do this, we shall double-count the set $A$ of pairs $(m_1,m_2)$ such that $m_1$ and $m_2$ are brothers.
The first way we count this set will be by family. For a family $f$, let $m_f$ denote the number of men in family $f$. Then we have\begin{align}|A|&=\sum_f m_f(m_f-1)\\&=\mathcal F\ex(M(M-1))\end{align}since in each family $f$, we have $m_f$ choices for the first brother and $m_f-1$ choices for the second brother.
The second way to count this set will be by man. For a man $m$, denote by $b_m$ the number of brothers that $m$ has. Then we have $$|A|=\sum_m b_m$$Therefore:$$\sum_m b_m=\mathcal F\ex(M(M-1))$$Then:\begin{align}\textrm{Average number of brothers a man has}&=\sum_m b_m/\mathcal M\\&=\frac{\mathcal F}{\mathcal M}\ex(M(M-1))\\&=\frac{\ex(M(M-1))}{\ex M}\end{align}
A similar argument gives us$$\textrm{Average number of brothers a woman has}=\frac{\ex(MW)}{\ex W}$$In this second case, if we write $w_f$ for the number of women in family $f$, then the number of pairs $(w,m)$ such that $w$ and $m$ are sister and brother is $m_fw_f$.
We would like to show that this second quantity is bigger than the first. Let's try and rewrite each quantity first.
\begin{align}\frac{\ex(M(M-1))}{\ex M}&=\frac1{\ex M}\left(\ex M^2-\ex M\right)\\&=\frac1{\ex M}\left(\Var M+(\ex M)^2-\ex M\right)\\&=\frac{\Var M}{\ex M} + \ex M - 1\end{align}
while
\begin{align}\frac{\ex(MW)}{\ex W}&=\frac{1}{\ex W}\left(\Cov(M,W)+\ex M\ex W\right)\\&=\frac{\Cov(M,W)}{\ex W} + \ex M\end{align}
Therefore, in order to ensure that the average number of brothers a woman has is greater than the average number of brothers a man has, we need to assume that:$$\frac{\Cov(M,W)}{\ex W}>\frac{\Var M}{\ex M}-1$$
We can turn this into a condition saying that the number of men per family has to have small variance$$\Var M<\ex M+\frac{\mathcal M}{\mathcal W}\Cov(M, W)$$
Is this a reasonable assumption to make? Assuming that the number of men in a family is independent of the number of women in that family, we should expect $\Cov(M, W)$ to be small. And if we assume that the total number of men is roughly equal to the to the number of women, then $\mathcal M/\mathcal W$ will be roughly equal to $1$. So women have more brothers if and only if the variance in the number of men per family is less than $\ex M$.
This is a surprising result, since there's no reason to suppose that the variance in the number of men per family should be less than $\ex M$. In fact, numerical experiments indicate that $\Var M$ is quite often larger than $\ex M$, which means that in fact it will be men who have more brothers than women.
So the answer is that it depends on how large the families are, but your intuition that women will have more brothers is
not true in general, even under fairly strong assumptions. If the number of men per family varies greatly from family to family, it is in fact men who have more brothers on average.
This surprising result can easily be confirmed with numerical experiment.
What's the explanation? Well, let's consider a situation in which the number of men per family varies greatly from family to family. Suppose we assume also that the number of men per family is uncorrelated with the number of women per family.
What this means is that there are going to be a significant number of families with lots of men and very few women, and a significant number of families with lots of women and very few men.
Now, within any given family, we know that the women will have more brothers than the men. But look at the overall contribution to the average:
The first type of family gives us lots of men with lots of brothers, and a small number of women with lots of brothers. The second type of family gives us lots of women with very few brothers, and a small number of men with very few brothers.
So on average, the men will tend to have lots of brothers, and the women will tend to have very few brothers, as long as the variance in the number of men is large enough to counteract the extra brother that each woman has. With large enough families, that one extra brother counts for less and less, and the variance effect takes over, giving you precisely the opposite effect from what you expected.
Let's take one last look at the result. We found that the important factor was that the variance in the number of men per family should not be too large. What if this value were equal to zero? That would mean that every family had the same number $a$ of men, so then it
would be true that women have more brothers, since every man would have $a-1$ brothers and every woman would have $a$ brothers.
The dependence on the covariance is interesting, too. By the Cauchy-Schwarz inequality, we have $\Cov(M,W)\le\sqrt{\Var M\Var W}$, and if we assume that $\Var M$ and $\Var W$ are roughly the same, then we have $\Cov(M,W)\le\Var M$. This extreme value occurs if $W=\lambda M$ for some positive constant $\lambda$. In that case, a simple counting argument shows us that women will have an average of one more brother than men. |
What would be the proof for $\langle q| p \rangle = e^{ipq}$?
Is it derived from canonical commutation relation?
($|q \rangle $ represents the position eigenstate, while $|p \rangle$ represents the momentum eigenstate.)
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.Sign up to join this community
From the canonical commutation relation in 1 dimension
$$[q,p] = i$$
you can choose an eigenbasis for q. Then note that any function can be expanded in terms of these eigenstates:
$$ |\psi\rangle = \int f(q) |q\rangle $$
Then consider the linear operator D which differentiates f:
$$ D |\psi\rangle = \int f'(q) |q\rangle $$
and note that
$$ D q - q D = 1 $$
So that -iD obeys the canonical commutation relations. This means that
$$ [ p + iD , q ] = 0 $$
so that if $p+iD$ is nonzero, it is a function of q, say
$$ p + i D = G(q) $$
Then a position dependent phase rotation of the wavefunction gets rid of G. If you redefine the eigenstates of position as follows:
$$ |q\rangle = e^{iH(x)}|q\rangle $$
Then the derivative operator gets an additional term
$$ iD \rightarrow iD + H'$$
and by choosing H to be an indefinite integral of G, you get rid of G, and
$$ p = -iD$$
The eigenstates of D are found by solving the differential equation
$$ - iD |\psi\rangle = k |\psi\rangle $$
and in terms of the expansion in eigenstates
$$ - i f'(q) = k $$
or
$$ f(q) = e^{ikq}$$
This is the overlap $\langle p | q \rangle $. This argument is directly lifted from Dirac's quantum mechanics book.
The rigorous version of this argument is due to Stone and Von Neumann, and it proceeds by rewriting the canonical commutation relations in exponetiated form:
$$ e^{iPx} q e^{-iPx} = q + x $$
In this form, you can see that, up to the arbitary choice of phase, the p operator is the generator of translations. I find the formal argument unilluminating, since it is essentially a repeat of Dirac.
What is interesting is Feynman's observation that this was essentially a 1d argument. If you have canonical commutation relations in multiple dimensions
$$ [p_i,q_j] = \delta_ij $$
Then you cannot just use this relation to figure out that the p's can be turned into differential operators. You also need the relation
$$ [p_i , p_j] = 0 $$
Without imposing the second condition, you cannot guarantee that the phases you get when you translate around an infinitesimal box is zero. If you drop the latter condition, you end up introducing a magnetic field vector potential to describe the failure of p commutativity, and this allows you to "derive Maxwell's equations from quantum mechanics". Feynman phrased it this way, but I think this is a poor phrasing. I think Feynman knew it was a poor phrasing, and this is why he did not publish, but people wanted to know the argument, so Dyson wrote it up in the American Journal of Physics in the mid 90s.
$$ \langle{q}|\hat{p}|\alpha\rangle = \frac{\hbar}{i}\frac{\partial}{\partial q}\langle q|\alpha \rangle $$ So, $$\langle{q}|\hat{p}|p\rangle = \frac{\hbar}{i}\frac{\partial}{\partial q}\langle q|p\rangle $$ This is equivalent to $$p\langle{q}|p\rangle = \frac{\hbar}{i}\frac{\partial}{\partial q}\langle q|p\rangle $$ then its just a first order linear differential equation, so $$\langle q|p\rangle = Ne^{\frac{i p q}{\hbar}}$$ where $N$ is some normalisation factor depends on the dimension of $q$.
Ref:
J.J. Sakurai, 1994, 'Modern Quantum Mechanic' |
A cylinder is a 3-dimensional structure with a circular base. It can also be taken as the set of circular disks which are stacked together one by one. Now take an example, where you are interested in painting the cylinder faces within the container. So, what you need to know exactly here? First of all, you need to calculate the quantity of paint for the container. For this purpose, you have to calculate the total surface area of the cone that is given as the sum of areas of a cylinder.
To make this concept little more interesting for you, let us take a cylinder with radius r and height h units. The circular base of the cylinder will be transformed to a rectangle with length ‘2𝝿r’ and height ‘h’ units. It can be given as shown in the figure below.
\[\large Surface\;Area\;of\;a\;Cylinder=2\pi r(r+h)\]
\[\large Curved\;Surface\;Area\;of\;a\;Cylinder=2\pi rh\]
Where,
r is the radius of the cylinder having circular base. h is the height of the cylinder with parallel faces.
As we have already mentioned that cylinder is simply set of circular disks which are stacked together one by one. So, we need to check the space occupied by these discs and this is possible by computing volume of a cylinder. For example, if circular discs are stacked up to the height h then cylinder volume formula could be written as –
\[\large Volume\;of\;a\;Cylinder=\pi r^{2}h\]
Where,
R is the radius of the cylinder having circular base and h is the height of the cylinder with parallel faces. In our day to day life too, we came across different types of situations where this is necessary to handle circular objects.
The formula is suitable to calculate the capacity of such type of cylindrical objects. The formulas are also helpful in designing cylindrical containers flasks, bottles, perfume bottles etc. This is easy to calculate the volume of a cylinder if you know the radius and the height of the cylinder.
To calculate the radius of a cylinder, you don’t have to put any extra efforts but rearrange the formula and put given values accordingly. In the end, you need to take the square root on the radius and this would be the final outcome.
\[\large Radius\;of\;a\;Cylinder=\sqrt{\frac{V}{\pi h}}\]
Where,
V is the Volume of the cylinder.
h is the height of the cylinder with parallel faces.
In most of the cases when you need to find the perimeter of a cylinder then diameter and height are given for that particular problem. The Perimeter of a cylinder is the outline of a two-dimensional shape. But cylinder is a 3-dimensional shape so how to calculate the perimeter here. This is possible by creating a projection on the base and reduce the shape to a rectangle. Now calculation of perimeter is possible in case of 3-dimensional cylinder too. The formula for the perimeter of a cylinder is given as –
\[\large Perimeter\;of\;a\;Cylinder=(2 \times d) + (2 \times h)\]
Where,
d is the diameter of the cylinder having circular base and h is the height of the cylinder with parallel faces.
Question 1: What will be the surface of the cylinder with height 10 cm and diameter of the base is 12 cm. Solution:
Given,
Height = 10 cm Diameter = 12 cm Radius = 6 cm
surface Area of Cylinder = 2πr(r+h)
= 2 X 3.14 X 6 (6+10) = 12 X 3.14 X 16 = 602.88 cm 2 Question 2: How many litres of water can a cylindrical water tank with base radius 20 cm and height 28 cm hold?
Solution: Given,
Base radius of the cylindrical water tank, r = 20 cm
Height of the cylindrical water tank, h = 28 cm
Volume of the cylindrical water tank = πr
2h
Volume of the cylindrical water tank = 3.14 X 20
228 = 35200
1 cubic centimeter = 0.001 litre =1 X 10
-3 litre
∴ 35200 cubic centimeter = 35200 X 10
-3 = 35.2 litres
The cylindrical water tank can hold 35.2 litres of water |
Can someone give me suggestions how can I construct a 2-tape Turing machine which simulates PDA ?
closed as unclear what you're asking by Evil, David Richerby, Rick Decker, Juho, hengxin Jul 9 '17 at 13:00
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
The first tape for input and the second for storing symbols (like on a stack). The first tape should be read-only so that the TM reads symbols one by one from left to right.
Every time your PDA writes a symbol on the stack, your TM moves the second head right and writes a symbol on the rightmost empty cell. When the PDA removes a symbol from the stack the TM replaces the rightmost symbol with the blank symbol (erases) and moves the second head left. On each state transition the first head should always move right.
In addition, your TM shouldn't be allowed to arbitrarily move heads right and left violating PDA rules.
Consider the following formal definition of moves for PDA (assume deterministic for simplicity):
UPDATE Definition of PDA moves
$\delta(q, a, Z) = (p, \alpha)$: the PDA in state $q$ with input $a$ and $Z$ on the top of the stack. The PDA enters the state $p$ and replaces the top symbol $Z$ with the symbols of the string $\alpha$. So, $\delta(q, a, Z) = (p, \epsilon)$ means remove the top stack symbol (pop). Advance the input head one symbol.
$\delta(q,\epsilon, Z) = (p, \alpha)$: the PDA in state $q$ with $Z$ on the top of the stack. Independent of the input symbol, PDA enters the state $p$ and replaces the top symbol $Z$ with the symbols of the strings $\alpha$. Input head does not advance in this case.
Examples of translation of PDA moves into TM moves Example 1.PDA move $\delta(q_1, 0, A) = (q_2, B)$ is translated as: TM in state $q_1$, head 1 (input head on the tape 1) reads $0$, and head 2 (stack head on the tape 2) reads $A$. Replace $A$ with $B$ (head 2 writes the symbol $B$ while it is on $A$). Advance head 1 one symbol. Enter state $q_2$. Example 2.$\delta(q_1, 0, A) = (q_2, BC)$ is translated as: TM in state $q_1$, head 1 reads $0$, and head 2 reads $A$. Head 2 writes $B$ (replaces $A$), advances head 2 one symbol, TM enters state $q_{21}$. Then, while TM in state $q_{21}$ and head 1 reads $0$, and head 2 reads the blank symbol: head 2 writes symbol $C$, enters state $q_2$. Advance head 1 one symbol.
The basic idea: in order to write $BC$ we introduced a new state $q_{21}$.
Analogously, $\delta(q,\epsilon, Z) = (p, \alpha)$ means head 2 just writes the blank symbol and moves left.
$\delta(q,\epsilon, Z) = (p, \alpha)$ means that you define the same transition
for every input symbol.
This is how a multitape Turing machine is defined. |
I'm studying from Goldstein's
Classical Mechanics, 3rd edition. In section 2.4, he discusses non-holonomic systems. We assume that the constraints can be put in the form $$f_\alpha(q, \dot{q}, t) =0, \qquad \alpha = 1 \dots m.\tag{2.20}$$ Then it also holds that $$\sum \lambda_\alpha f_\alpha = 0.\tag{2.21}$$ Using Hamilton's principle (i.e. that the action must be stationary), we get that
$$\delta \int_1^2 L\ dt = \int_1^2 dt\ \sum_{k=1}^n \left(\frac{\partial L}{\partial q_k} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q_k}}\right)\delta q_k = 0\tag{2.22}. $$
But we can't get Lagrange's equations from this because the $\delta q_k$ aren't independent. However, if we add this with $\sum \lambda_\alpha f_\alpha = 0$, it follows that
$$\delta \int_{t_1}^{t_2} \left(L +\sum_{\alpha=1}^m \lambda_\alpha f_\alpha\right)\ dt = 0.\tag{2.23}$$
And then Goldstein says that
The variation can now be performed with the $n\, \delta q_i$ and $m\, \lambda_\alpha$ for $m+n$ independent variables.
Why have the variables suddenly become independent? First we had $n$ dependent variables, why do we now have $m+n$ independent ones? |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.