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8.1 - One Sample Proportion
One sample proportion tests and confidence intervals are covered in Section 6.1 of the Lock
5 textbook.
In the last lesson you were introduced to the general concept of the Central Limit Theorem. The Central Limit Theorem states that if the sample size is sufficiently large then the sampling distribution will be approximately normally distributed for many frequently tested statistics, such as those that we have been working with in this course. When discussion proportions, we sometimes refer to this as the
Rule of Sample Proportions. According to the Rule of Sample Proportions, if \(np\geq 10\) and \(n(1-p) \geq 10\) then the sampling distributing will be approximately normal. When constructing a confidence interval \(p\) is not known but may be approximated using \(\widehat p\). When conducting a hypothesis test, we check this assumption using the hypothesized proportion (i.e., the proportion in the null hypothesis).
If assumptions are met, the sampling distribution will have a standard error equal to \(\sqrt{\frac{p(1-p)}{n}}\).
This method of constructing a sampling distribution is known as the
normal approximation method.
If the assumptions for the normal approximation method are not met (i.e., if \(np\) or \(n(1-p)\) is not at least 10), then the sampling distribution may be approximated using a binomial distribution. This is known as the
exact method. This course does not cover the exact method in detail, but you will see how these tests may be performed using Minitab Express.
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I've read the proof for why $\int_0^\infty P(X >x)dx=E[X]$ for nonnegative random variables (located here) and understand its mechanics, but I'm having trouble understanding the intuition behind this formula or why it should be the case at all. Does anyone have any insight on this? I bet I'm missing something obvious.
For the discrete case, and if $X$ is nonnegative, $E[X] = \sum_{x=0}^\infty x P(X = x)$. That means we're adding up $P(X = 0)$ zero times, $P(X = 1)$ once, $P(X = 2)$ twice, etc. This can be represented in array form, where we're adding column-by-column:
$$\begin{matrix} P(X=1) & P(X = 2) & P(X = 3) & P(X = 4) & P(X = 5) & \cdots \\ & P(X = 2) & P(X = 3) & P(X = 4) & P(X = 5) & \cdots \\ & & P(X = 3) & P(X = 4) & P(X = 5) & \cdots \\ & & & P(X = 4) & P(X = 5) & \cdots \\ & & & & P(X = 5) & \cdots\end{matrix}.$$
We could also add up these numbers row-by-row, though, and get the same result. The first row has everything but $P(X = 0)$ and so sums to $P(X > 0)$. The second row has everything but $P(X =0)$ and $P(X = 1)$ and so sums to $P(X > 1)$. In general, the sum of row $x+1$ is $P(X > x)$, and so adding the numbers row-by-row gives us $\sum_{x = 0}^{\infty} P(X > x)$, which thus must also be equal to $\sum_{x=0}^\infty x P(X = x) = E[X].$
The continuous case is analogous.
In general, switching the order of summation (as in the proof the OP links to) can always be interpreted as adding row-by-row vs. column-by-column.
Since the intuition behind the result is requested, let us consider a simple case of a discrete non-negative random variable taking on the three values $x_0 = 0$, $x_1$, and $x_2$ with probabilities $p_0$, $p_1$, and $p_2$. The cumulative distribution function (CDF) $F(x)$ is thus a staircase function $$F(x) = \begin{cases} 0, & x < 0, \\ p_0, & 0 \leq x < x_1,\\ p_0 + p_1, & x_1 \leq x < x_2,\\ 1, & x \geq x_2, \end{cases}$$ with jumps of $p_0$, $p_1$, and $p_2$ at $0$, $x_1$, and $x_2$ respectively. Note also that $$ E[X]= \sum_{i=1}^3 p_ix_i = p_1x_1 + p_2x_2. $$ Now, notice that $$\int_0^\infty P\{X > x\}\mathrm dx = \int_0^\infty [1 - F(x)]\mathrm dx$$ is the area of the region bounded by the curve $F(x)$, the vertical axis, and the line at height 1 above the horizontal axis. Standard Riemann integration techniques say that we should divide the region into narrow vertical strips, compute the area of each, take the sum, take limits etc. In our example, of course, all this can be bypassed since the region in question is the union of two adjoining non-overlapping rectangles: one of base $x_1$ and height $(1-p_0)$, and the other of base $x_2 - x_1$, and height $(1-p_0-p_1)$. BUT, suppose we divide the region under consideration into two different adjoining non-overlapping rectangles with the second lying above the first. The first rectangle has base $x_1$ and height $p_1$, while the second (lying above the first) has broader base $x_2$ and height $p_2$. The total area that we seek is easily seen to be $p_1x_1 + p_2x_2 = E[X]$.
Thus, for a non-negative random variable, $E[X]$ can be interpreted as the area of the region lying above its CDF $F(x)$ and below the line at height 1 to the right of the origin. The standard formula $$E[X] = \int_0^\infty x\mathrm dF(x)$$ can be thought of as computing this area by dividing it into thin horizontal strips of length $x$ and height $dF(x)$, while $$\int_0^\infty P\{X > x\}\mathrm dx = \int_0^\infty [1 - F(x)]\mathrm dx$$ (in the Riemann integral sense) can be thought of as computing the area by dividing it into thin vertical strips.
More generally, if $X$ takes on both positive and negative values, $$E[X] = \int_0^\infty [1 - F(x)]\mathrm dx - \int_{-\infty}^0 F(x) \mathrm dx$$ with similar interpretations.
A hint and a proof.
Hint: if $X=x$ with full probability, the integral is the integral of $1$ on $(0,x)$, hence the LHS and the RHS are both $x$. Proof: apply (Tonelli-)Fubini to the function $(\omega,x)\mapsto\mathbf 1_{X(\omega)>x}$ and to the sigma-finite measure $P\otimes\mathrm{Leb}$ on $\Omega\times\mathbb R_+$. One gets$$\int_\Omega\int_{\mathbb R_+}\mathbf 1_{X(\omega)>x}\mathrm dx\mathrm dP(\omega)=\int_\Omega\int_0^{X(\omega)}\mathrm dx\mathrm dP(\omega)=\int_\Omega X(\omega)\mathrm dP(\omega)=E(X),$$while, using the shorthand $A_x=\{\omega\in\Omega\mid X(\omega)>x\}$,$$\int_{\mathbb R_+}\int_\Omega\mathbf 1_{X(\omega)>x}\mathrm dP(\omega)\mathrm dx=\int_{\mathbb R_+}\int_\Omega\mathbf 1_{\omega\in A_x}\mathrm dP(\omega)\mathrm dx=\int_{\mathbb R_+}P(A_x)\mathrm dx=\int_{\mathbb R_+}P(X>x)\mathrm dx.$$
Perhaps by considering this question with a concrete physical example, it will provide some intuition.
Consider a beam of length $L = 10$ (you can pick your favorite units) attached to a wall. Now, at positions $1, 2, \ldots, 9$ hang weights $w_1,w_2,\ldots,w_9$. For simplicity, let's assume $\sum_{n=1}^9 w_n = 1$.
Then the
center of mass of the beam is $c = \sum_{n=1}^9 nw_n$. below is an example picture, with the weights in blue (heightsproportional to weight) and the center of mass in red.
In a probabilistic setting, our weights correspond to probabilities and $c = \mathbb E X$ where $X$ takes on the values $1,2,\ldots,9$ with probabilities $w_1,w_2,\ldots,w_9$, respectively.
Now, to explain how $c = \mathbb E X = \sum_{n=0}^9 \mathbb P(X > n) = \sum_{n = 0}^9 \sum_{k=n+1}^9 w_k$ comes about, expanding out the latter sum we have $$ c = (w_1 + \cdots + w_9) + (w_2 + \cdots + w_9) + \cdots + (w_9) \>, $$ so, $w_1$ appears once, $w_2$ appears twice, $w_3$ appears three times, etc. Hence $c = \sum_{n=1}^9 n w_n$.
In terms of the beam, we can think of the expression $\sum_{n=0}^9 \mathbb P(X > n)$ in the following way. Standing at zero, look out to the right and count up all the weights in front of you. Now, move one step to the right and repeat this process, adding the result to your initial sum. Continue this process until you get out to position 9, at which point there are no more weights in front of you.
The resulting sum is the center of mass, or, in probabilistic terms, the expectation $\mathbb E X$.
Extending this intuition to discrete random variables taking on non-integer values is straightforward. The extension to continuous variables is also not difficult.
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It seems that merely loading the
graphdrawing library can mess up drawing a tree using the standard TikZ syntax. Is there a way to avoid this unwanted effect?
For example:
\documentclass[tikz,border=10pt]{standalone}\usetikzlibrary{arrows.meta}\tikzset{ my tree/.style={ ->, nodes={draw, circle, minimum size = .5cm}, >=Stealth[], },}\begin{document}\tikz [ my tree ] \node {$\land$} child { node {$\lnot$} child { node {$a$} } } child { node {$\rightarrow$} child { node {$b$} } child { node {$c$} } };\end{document}
produces the following, expected result:
whereas
\documentclass[tikz,border=10pt]{standalone}\usetikzlibrary{arrows.meta,graphdrawing}\tikzset{ my tree/.style={ ->, nodes={draw, circle, minimum size = .5cm}, >=Stealth[], },}\begin{document}\tikz [ my tree ] \node {$\land$} child { node {$\lnot$} child { node {$a$} } } child { node {$\rightarrow$} child { node {$b$} } child { node {$c$} } };\end{document}
results in overlapping nodes and an overly compact tree:
Is there a way to confine the effects of
graphdrawing to pictures which are actually graphs?
The problem is caused by the following line of code in
tikzlibrarygraphdrawing.code.tex:
\tikzset{level distance=1cm, sibling distance=1cm}
If I comment this out, then the tree is rendered as expected. A comment for this line of code explains its inclusion thus:
% Patch the level and sibling distances so that gd and plain tikz are% in sync
But why is the
graphdrawing library insisting that plain TikZ must be in sync with it? If this is important, surely the library should configure itself to use the standard TikZ default level and sibling distances, rather than imposing its values on standard TikZ trees? What exactly is the motivation for 'sync'ing them anyway?
It just seems egregious: the library is
interfering with plain TikZ's defaults so that trees using the regular syntax will be deformed. Is this some sort of underhand marketing campaign for the graph drawing libraries? I break your code and then I offer you this great solution for those nasty trees you're having problems with which just so happens to require switching to my new-and-improved syntax?! Hopefully there is a less insidious explanation...? deliberately
I suspect this is must be a duplicate, but I can't find it.
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a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached).
b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct.
However,
your calculation of amplitude is not correct.
Peak-to-trough is
twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly.
CALCULATION OF AMPLITUDE OF OSCILLATIONS
1. Equation of Motion Method
First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision.
Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$
2. Energy Method
Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$.
At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$.
At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$.
By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1.
(c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it.
Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal.
Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is
always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result.
The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate.
Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$
We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off.
The total energy at P is the same as at U, so from above (
Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
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If the question is why the "$4\pi$" in the Coulomb constant (k=$\frac{1}{4\pi\epsilon_{0}}$), then an equally valid question could be why the "4$\pi$" in the magnetic permeability of vacuum, $\mu_{0}=4\pi\times10^{-7}H/m$ ?
Perhaps a clue can be found in the Maxwell's equation for the speed of electromagnetic wave (light) in a vacuum, $c=\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}$.
Of course, Maxwell derived this relationship much later than Coulomb.
Maxwell relates the electric permitivity to magnetic permeability in the vacuum, $\mu_{0}=\frac{1}{\epsilon_{0}c^{2}}$ which is given a value of $\mu_{0}=4\pi\times10^{-7}H/m$ in SI units.
The 'reason' for the "$4\pi$" appearing here and in Coulomb's constant (believe it or not) so that Maxwell's equations can be written without any $4\pi$' factors!
In order to understand this, consider how electrostatic phenomena are expressed in Coulombs law as "field intensity at a distance squared", compared to (the equivalent) Gauss' law, which describes the "flux through a closed surface enclosing the charge".
The total flux is the flux density multiplied by the surface area, which for a sphere of radius $r$ is given by $S=4\pi r^{2}$, so the ratio $S/r^{2}$ = $4\pi$ is simply the result of geometry of space and spherical symmetry.
The SI system of units (unlike the Gauss units) is said to be 'rationalized' because it allows the expression of Maxwell's equations without the $4\pi$ factors. To do this, the $4\pi$ factor has simply been "built into" the (SI unit) definition of the universal constant for permeability of the vacuum, $\mu_{0}=4\pi\times10^{-7}H/m$, from which we can express Coulomb's constant as k=$\frac{1}{4\pi\epsilon_{0}}$.
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a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached).
b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct.
However,
your calculation of amplitude is not correct.
Peak-to-trough is
twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly.
CALCULATION OF AMPLITUDE OF OSCILLATIONS
1. Equation of Motion Method
First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision.
Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$
2. Energy Method
Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$.
At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$.
At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$.
By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1.
(c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it.
Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal.
Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is
always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result.
The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate.
Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$
We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off.
The total energy at P is the same as at U, so from above (
Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
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a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached).
b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct.
However,
your calculation of amplitude is not correct.
Peak-to-trough is
twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly.
CALCULATION OF AMPLITUDE OF OSCILLATIONS
1. Equation of Motion Method
First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision.
Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$
2. Energy Method
Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$.
At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$.
At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$.
By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1.
(c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it.
Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal.
Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is
always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result.
The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate.
Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$
We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off.
The total energy at P is the same as at U, so from above (
Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
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Standard deviation formula is used to find the values of a particular data that is dispersed. In simple words, the standard deviation is defined as the deviation of the values or data from an average mean. Lower standard deviation concludes that the values are very close to their average. Whereas higher values mean the values are far from the mean value. It should be noted that the standard deviation value can never be negative.
Standard Deviation is of two types: Population Standard Deviation Sample Standard Deviation Formulas for Standard Deviation
Population Standard Deviation Formula \(\sigma= \sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}}{N}}\) Sample Standard Deviation Formula \(\sigma= \sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}}{N-1}}\) Notations for Standard Deviation σ = Standard Deviation x i= Terms Given in the Data x̄ = Mean N = Total number of Terms Standard Deviation Formula Based on Discrete Frequency Distribution
For discrete frequency distribution of the type:
x: x 1, x2, x3, … xn and f: f 1, f2, f3, … fn
The formula for standard deviation becomes:\(\sigma= \sqrt{\frac{1}{N}{\sum_{i=1}^{n}f_{i}\left(x_{i}-\bar{x}\right)^{2}}}\)
Here, N is given as:
N = n∑i=1 fi Another Formula for Standard Deviation
There is another standard deviation formula which is derived from the variance. This formula is given as:\(\sigma=\frac{1}{N}\sqrt{\sum_{i=i}^{n}f_{i}x_{i}^{2}-(\sum_{i=1}^{n}f_{i}x_{i})^{2}}\)
Example Question based on Standard Deviation Formula Question: During a survey, 6 students were asked how many hours per day they study on an average? Their answers were as follows: 2, 6, 5, 3, 2, 3. Evaluate the standard deviation. Solution:
Find the mean of the data:\(\frac{\left(2+6+5+3+2+3\right)}{6}\) = 3.5
Step 2: Construct the table:
x 1 x 1 − x̄ (x 1 − x̄) 2 2 -1.5 2.25 6 2.5 6.25 5 1.5 2.25 3 0.5 0.25 2 -1.5 2.25 3 -0.5 0.25 = 13.5 Step 3: Now, use the Standard Deviation formula
Sample Standard Deviation = \(\sqrt{\frac{\sum_{i=1}^{n}\left(x-x_{i}\right)^{2}}{n-1}}\) \(\sqrt{\frac{13.5}{6}}=\sqrt{2.25}\)
= 1.5
To check more maths formulas for different classes and for various concepts, stay tuned with BYJU’S. Also, register now to get access to various video lessons and get a more effective and engaging learning experience.
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You know that you have $\ell_1=\ell_2=1$, and the $m_{\ell_1}=+1$ while $m_{\ell_2}=-1$. Hence you must have (up to a factor of $1/\sqrt{2}$):$$\vert\psi_\pm\rangle = \vert 1 1\rangle_1\vert 1,-1\rangle_2 \pm \vert 1,-1\rangle_1\vert 1 1\rangle_2$$Now for the spin part, you can conclude likewise that$$\vert\chi_\pm\rangle = \vert 1/2,1/2\rangle_1\vert 1/2,-1/2\rangle_2\pm \vert 1/2,-1/2\rangle_1\vert 1/2,1/2\rangle_2\, .$$Note that, under the interchange of particle labels:$$P_{12}\vert\psi_\pm\rangle =\pm \vert\psi_\pm\rangle\, ,\qquad P_{12}\vert\chi_\pm \rangle= \pm \vert\chi_\pm\rangle\, .$$To have overall antisymmetry you should therefore have$$\vert\phi\rangle=\alpha \vert\psi_+\rangle\vert\chi_-\rangle +\beta \vert \psi_-\rangle \vert\chi_+\rangle$$with $\alpha$ and $\beta$ chosen so your states are properly normalized.
This way:\begin{align}P_{12}\vert\phi\rangle &= \alpha \left[P_{12}\vert\psi_+\rangle\right]\left[P_{12}\vert\chi_-\rangle\right]+ \beta \left[P_{12}\vert\psi_-\rangle\right]\left[P_{12}\vert\chi_+\rangle\right]\, ,\\&=- \alpha\vert\psi_+\rangle\vert\chi_-\rangle -\beta \vert\psi_-\rangle \vert\chi_+\rangle=-\vert\phi\rangle\, . \end{align}Basically, if you pick the symmetric combination for the spatial part, it must be multiplied by the antisymmetric combinations for the spin part so the product is overall antisymmetric, and
vice versa if you pick the antisymmetric combination for the spatial part you have to multiply it by the symmetric one for the spin part.
Note that the situation gets considerably more complicated with $3$ particles as the permutation group $S_3$ of three objects have a 2-dimensional representation of mixed symmetry which do not necessarily transform back to a $\pm 1$ multiple of themselves under permutation. The combination of such types of functions is a little more delicate than in the 2-particle case.
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a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached).
b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct.
However,
your calculation of amplitude is not correct.
Peak-to-trough is
twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly.
CALCULATION OF AMPLITUDE OF OSCILLATIONS
1. Equation of Motion Method
First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision.
Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$
2. Energy Method
Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$.
At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$.
At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$.
By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1.
(c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it.
Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal.
Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is
always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result.
The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate.
Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$
We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off.
The total energy at P is the same as at U, so from above (
Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
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A particle moves along the x-axis so that at time t its position is given by $x(t) = t^3-6t^2+9t+11$ during what time intervals is the particle moving to the left? so I know that we need the velocity for that and we can get that after taking the derivative but I don't know what to do after that the velocity would than be $v(t) = 3t^2-12t+9$ how could I find the intervals
Fix $c\in\{0,1,\dots\}$, let $K\geq c$ be an integer, and define $z_K=K^{-\alpha}$ for some $\alpha\in(0,2)$.I believe I have numerically discovered that$$\sum_{n=0}^{K-c}\binom{K}{n}\binom{K}{n+c}z_K^{n+c/2} \sim \sum_{n=0}^K \binom{K}{n}^2 z_K^n \quad \text{ as } K\to\infty$$but cannot ...
So, the whole discussion is about some polynomial $p(A)$, for $A$ an $n\times n$ matrix with entries in $\mathbf{C}$, and eigenvalues $\lambda_1,\ldots, \lambda_k$.
Anyways, part (a) is talking about proving that $p(\lambda_1),\ldots, p(\lambda_k)$ are eigenvalues of $p(A)$. That's basically routine computation. No problem there. The next bit is to compute the dimension of the eigenspaces $E(p(A), p(\lambda_i))$.
Seems like this bit follows from the same argument. An eigenvector for $A$ is an eigenvector for $p(A)$, so the rest seems to follow.
Finally, the last part is to find the characteristic polynomial of $p(A)$. I guess this means in terms of the characteristic polynomial of $A$.
Well, we do know what the eigenvalues are...
The so-called Spectral Mapping Theorem tells us that the eigenvalues of $p(A)$ are exactly the $p(\lambda_i)$.
Usually, by the time you start talking about complex numbers you consider the real numbers as a subset of them, since a and b are real in a + bi. But you could define it that way and call it a "standard form" like ax + by = c for linear equations :-) @Riker
"a + bi where a and b are integers" Complex numbers a + bi where a and b are integers are called Gaussian integers.
I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...
Does anyone know if $T: V \to R^n$ is an inner product space isomorphism if $T(v) = (v)_S$, where $S$ is a basis for $V$? My book isn't saying so explicitly, but there was a theorem saying that an inner product isomorphism exists, and another theorem kind of suggesting that it should work.
@TobiasKildetoft Sorry, I meant that they should be equal (accidently sent this before writing my answer. Writing it now)
Isn't there this theorem saying that if $v,w \in V$ ($V$ being an inner product space), then $||v|| = ||(v)_S||$? (where the left norm is defined as the norm in $V$ and the right norm is the euclidean norm) I thought that this would somehow result from isomorphism
@AlessandroCodenotti Actually, such a $f$ in fact needs to be surjective. Take any $y \in Y$; the maximal ideal of $k[Y]$ corresponding to that is $(Y_1 - y_1, \cdots, Y_n - y_n)$. The ideal corresponding to the subvariety $f^{-1}(y) \subset X$ in $k[X]$ is then nothing but $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$. If this is empty, weak Nullstellensatz kicks in to say that there are $g_1, \cdots, g_n \in k[X]$ such that $\sum_i (f^* Y_i - y_i)g_i = 1$.
Well, better to say that $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$ is the trivial ideal I guess. Hmm, I'm stuck again
O(n) acts transitively on S^(n-1) with stabilizer at a point O(n-1)
For any transitive G action on a set X with stabilizer H, G/H $\cong$ X set theoretically. In this case, as the action is a smooth action by a Lie group, you can prove this set-theoretic bijection gives a diffeomorphism
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Suppose $(f_n)_{n >0}$ is a sequence of bounded continuous functions that converges locally uniformly to $f$.Is the following statement true?
$$\lim_n \sup_{\gamma \in (0,2)}\int_0^\infty \frac{f_n(x+t)-f_n(x)}{t^\gamma}dt = \sup_{\gamma \in (0,2)}\int_0^\infty \frac{f(x+t)-f(x)}{t^\gamma}dt $$
Suppose $(f_n)_{n >0}$ is a sequence of bounded continuous functions that converges locally uniformly to $f$.Is the following statement true?
The statement is
not true.
Consider the bounded and continuous functions $$f_n(x)= \frac{x-n}n1_{[n,n+1)}(x)+\frac{1}{n}1_{[n+1,\infty)}(x).$$ Then, for $x<1$,
\begin{eqnarray*}\lim_n\sup_{\gamma\in(0,2)}\int_0^\infty\frac{f_n(x+t)-f_n(x)}{t^\gamma}dt&\geq& \lim_n \int_0^\infty\frac{f_n(x+t)-f_n(x)}{t}dt\\ &\geq&\lim_n \int_{n+1-x}^\infty\frac{1}{t\:n}dt=\infty. \end{eqnarray*} On the other hand, we have $f(x)=\lim_n f_n(x)=0$ uniformly, such that $$\sup_{\gamma\in(0,2)}\int_0^\infty\frac{f(x+t)-f(x)}{t^\gamma}dt=0.$$
Added later: There are many counterexamples. Take any smooth bounded function $g$ on $\mathbb R$ whose derivative is positive everywhere (for example, $g(t)=\arctan t$). Set $f_n(t) = g(t)/n.$ Then $f_n \to 0$ uniformly on $\mathbb R,$ and the hoped-for equality fails at every $x.$
Proof: Fix $x\in \mathbb R.$ Then there exists $\delta > 0$ such that
$$g(x+t)-g(x) > g'(x)t/2\,\,\text { for } 0<t<\delta.$$
Thus
$$\int_0^\infty \frac{f_n(x+t)-f_n(x)}{t^\gamma}dt \ge \frac{1}{n}\int_0^\delta \frac{g(x+t)-g(x)}{t^\gamma}dt \ge \frac{1}{n}\frac{g'(x)}{2}\int_0^\delta \frac{t}{t^\gamma}dt.$$
Now the last expression equals
$$\frac{1}{n}\frac{g'(x)}{2}\frac{\delta^{2-\gamma}}{2-\gamma}.$$
As $\gamma \to 2^-,$ this $\to \infty.$ Thus the supremum is $\infty$ for each $n.$ So as before, we have $\infty$ on the left and $0$ on the right, this time for every $x.$
Previous answer: It would appear that the result fails: Define $f_n(t) = \sqrt t/n.$ Then $f_n \to 0$ uniformly on compact subsets of $[0,\infty).$ Let $x=0.$ Then the integrals on the left are
$$ \frac{1}{n}\int_0^\infty \frac{\sqrt t}{t^\gamma}\, dt.$$
Now we take the $\sup$ over $\gamma \in (0,2).$ But notice $\gamma = 3/2$ makes each of these integrals $\infty.$ Thus the supremum is $\infty$ for each $n,$ and of course $\lim_{n\to \infty} \infty = \infty.$
On the right, each integral is just $0,$ so this appears to be a counterexample.
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I posted an algorithm yesterday, that purported to solve the co-NP Complete 'Boolean Tautology Problem' in polynomial time.
Link to the algorithm : polynomial time algorithm
In that post, I presented the algorithm's motivation, and implementation, but not proof of validity. Herein I present said proof :
Recap :
The algorithm accepts Exact $3-DNF$ expressions as input, and checks whether they are tautologies. The job of the algorithm is to derive, and mark clauses as true. Given some $n$ variable expression, the number of possible clauses are $8\times{n\choose 3}$ If all of them are derived, and marked true, then the algorithm claims that the expression is a tautology.
Given a $n$ variable expression, the algorithm first creates a graph in the following manner :
Start with all possible clauses in $n$ variables, as initial nodes. Create children nodes for each parent by adding one more literal to the parent clause. Such child clauses can have multiple parents. For each such child, there is a dual child which has the negation of that one extra literal. Repeat this process with the children as parent nodes, until you either exhaust your variables, or end up with children with $6$ literals in their clauses. All nodes are initially marked . false
For reference, a child is the $\land$ of the parent clause, and another literal. E.g. Parent : $(a\land b\land c)$, possible children : $(a\land b\land c\land d)$, $(a\land b\land c\land\lnot e)$, etc.
Dual children of a parent have opposite literals, e.g. $(a\land b\land c\land d)$, and $(a\land b\land c\land\lnot d)$ are dual children of $(a\land b\land c)$. So are $(a\land b\land c\land e)$, and $(a\land b\land c\land\lnot e)$, etc.
Then, the algorithm scans the input boolean expression to extract the clauses present, and works in the following manner :
Upon finding a clause in the expression, the corresponding clause node in the graph is marked true. If a parent node is true, all children nodes are marked true. If some dual couple children of a parent are true, then the parent is marked true. If every node is marked true, the original expression is a tautology, otherwise not.
This is a brief summary of the algorithm. I did not go into the reasoning behind these steps, as I've linked the original post. Now on to the proof of validity.
Rationale :
So, the question is, 'Why is the algorithm doing any of this?', and here's the answer :
We are given some $3-DNF$ boolean expression, and we can derive the presence of $3-DNF$ clauses from it. If the expression is a tautology, we can say that all possible clauses are present. Either they are directly present in the expression, or are a logical consequence of clauses within the expression. The job of the algorithm is to see if it can derive the presence of all possible clauses from the expression.
Obviously, the clauses present in the given expression are already present. The tricky part is to derive the remaining clauses. Each clause present in the expression carries with it some parts of the clauses not present directly. Each clause provides particular intersecting pieces of the missing, yet derivable clauses.
This is the reason why the algorithm stops making child nodes in the graph after they contain $6$ literals. This is because $6$ literals is the maximum number of possible literals in an intersection $(AND)$ of any two $3-DNF$ clauses.
Proof of Validity :
Herein, I go through the algorithm steps(after the graph is created) one by one, and justify them.
We find clauses in the input $3-DNF$ expression, by scanning through it. Every time we find a clause, we mark the corresponding node on the graph $true$. This implies that the particular clause is present , or has been derived to be present. 1) Any node on the graph represents a $\land$ of some literals. The propositional logic implication is that any assignment of variables that matches those of the literals present in the clause will result in the expression evaluating true. For example, if we consider the clause $(a\land b\land\lnot c)$ in $5$ variables, $\{a, b, c, d, e\}$, then any assignment of variables to the expression will result in the expression evaluating true, as long as we put $a=true$, $b=true$, $c=false$. 2)
Therefore, if any parent node is true, any child node of the parent must also be true, since it is just a $\land$ of the parent, and another literal, i.e. it falls within variable assignments of the parent. For example, if $(a\land b\land\lnot c)$ is true, then any child, like $(a\land b\land\lnot c\land y)$ must also be true.
Some dual couple children are true $\Rightarrow$ Parent is true 3) To prove :
Let us start with a parent $(x_1\land x_2\land\ldots\land x_k)$
If some dual couple children of this clause, with some variable $y$, are marked true, we get : $(x_1\land x_2\land\ldots\land x_k\land y)\lor (x_1\land x_2\land\ldots\land x_k\land\lnot y)$ $=(x_1\land x_2\land\ldots\land x_k)\land (y\lor\lnot y)$ $=(x_1\land x_2\land\ldots\land x_k)$ This proves : Some dual couple children are true $\Rightarrow$ Parent is true All nodes on graph marked true $\equiv$ Input $3-DNF$ expression is a tautology 4) To prove : Let us first prove : All nodes on graph marked true $\Rightarrow$ Input $3-DNF$ expression is a tautology
We assume that we have a case where all nodes are marked true, but the expression is not a tautology, and show it is impossible.
Firstly, any $3-DNF$ clauses present in the expression must have been marked true in the graph; same is true of all the children, and sub-children of those clauses.
Since the expression is known to not be a tautology, there must be some error in how we mark nodes true, i.e. we must be marking nodes true when they should not be marked. But the marking is done in only two ways.
The first way is directly marking from parent-to-child which happens if some parent is marked true. The second is marking a parent true if some dual couple children of its are true. Both methods are logically sound, as we have shown earlier. Hence there is no error in marking methods. So, it is impossible for all nodes to be marked true, if the expression is not a tautology.
Hence, All nodes on graph marked true $\Rightarrow$ Input $3-DNF$ expression is a tautology Now we prove : All nodes on graph marked true $\Leftarrow$ Input $3-DNF$ expression is a tautology.
We assume we have a case where we have a tautological expression, but some nodes on the graph are marked false. This implies that we have some error in our marking method such that we fail to mark some nodes as true, even when they are. Obviously this cannot be the case when we are marking top-to-bottom, as is the case when some parent node is marked true. So, the error must exist with the dual-couple-child method.
The claim now is that the dual-couple-child method sometimes fails to mark true nodes.
We know that the dual-couple-child method marks nodes true only when such nodes have some dual couple children both marked true.
If this fails to mark some true nodes, then there must be certain cases where a node is true, without any of its dual couple children being true(at most one of the dual couple may be true, not both). But this is trivially impossible since we can take the true children, in a disjunction to see if they add up to the parent. This is trivial for dual couple children, but impossible if they're not both true. For example, let us take our parent node to be $(a\land b\land c)$ in an expression of $6$ variables, $\{a,b,c,d,e,f\}$
So, possible duals of $(a\land b\land c)$ are : $(a\land b\land c\land d)true$, $(a\land b\land c\land\lnot d)false$ $(a\land b\land c\land e)false$, $(a\land b\land c\land\lnot e)false$ $(a\land b\land c\land f)false$, $(a\land b\land c\land\lnot f)true$
the $true/false$ next to them signifies their value. Take the disjunction of the true children :
$(a\land b\land c\land d)\lor(a\land b\land c\land\lnot f)$ $=(a\land b\land c)\land(d\lor\lnot f)$ Which trivially tells us that the true children have not added up to the parent, and that it is only possible when opposing literals are present, i.e. some dual couple children have to be true. Therefore, our dual-couple-child method is correctly assigning truth values to the nodes. Hence it is impossible to have a tautological expression, which results in some nodes on the graph marked false. Hence : All nodes on graph marked true $\Leftarrow$ Input $3-DNF$ expression is a tautology
Since we have proved both the left, and right implications, therefore :
All nodes on graph marked true $\equiv$ Input $3-DNF$ expression is a tautology Clarifications : Why is stopping at $6$ literals sufficient? a) We start with a $3-DNF$ boolean expression, in $n$ variables. Each clause in our expression is a $\land$ of $3$ literals, and the expression is a $\lor$ of the clauses. Let our expression have $m$ clauses ($m\leq8\times{n\choose3}$). The topmost nodes of the graph my algorithm creates are all possible clauses with $3$ literals. Ans :
It is trivially noted that given a tautological expression, we can check that every possible clause is present(truth table verification), and given a non-tautological expression, we cannot mark every possible clause as present.
The algorithm's job is to derive clauses from the given expression. The clauses present in the expression are trivially present, and marked $true$ on the graph.
Now, every such trivially present clause is basically a set of assignments to the variables of the expression, such that the clause is true. For example, if our expression is of $5$ variables, $\{a,b,c,d,e\}$, and contains the clause $(b\land\lnot d\land\lnot e)$, then it is equivalent to the following assignments being true :
$(\lnot a\land b\land\lnot c\land\lnot d\land\lnot e)$
$(\lnot a\land b\land c\land\lnot d\land\lnot e)$ $(a\land b\land\lnot c\land\lnot d\land\lnot e)$ $(a\land b\land c\land\lnot d\land\lnot e)$
Now, every such clause (set of assignments) contains a subset that is valid for other clauses. For example, let us take the clause $(a\land b\land c)$. The subset, of the above assignments, that is valid for this clause is :
$(a\land b\land c\land\lnot d\land\lnot e)$
which is equal to : $(b\land\lnot d\land\lnot e)\land(a\land b\land c)$
Any trivially present clause carries a subset of assignments that are valid for other clauses, which can be accessed by intersecting (taking $\land$ of) the trivially present clause, and the clause whose assignments we seek.
The algorithm does that, by marking all children of true nodes also true. Since the nodes are simply $\land$ of some literals, we're actually marking subsets of assignments true.
Now, the reason why we stop at $6$ literals is this :
Each trivially present clause carries pieces of clauses which may be true, but are absent from the expression. These pieces can be accessed by taking $\land$ of the present clauses(one at a time), and any missing clause, whose pieces we are seeking. If the missing clause is true, then the pieces will add up to it, and vice versa.
One can think of it this way - since the intersections are a subset of valid assignments to variables, then an intersection of $3$ clauses, say $(a\land b\land c)\land(b\land c\land\lnot d)\land(b\land\lnot d\land e)$ which is equal to $(a\land b\land c\land\lnot d\land e)$, will be contained in the intersection $(a\land b\land c)\land(b\land c\land\lnot d)$, or of $(b\land c\land\lnot d)\land(b\land\lnot d\land e)$, or $(a\land b\land c)\land(b\land\lnot d\land e)$
So, the intersection of more than $2$ clauses is automatically marked true within intersection of $2$ clauses, and we need not go seek intersections of arbitrary number of clauses.
We can do this efficiently because our expression is a $\lor$ of clauses, and each subset of assignments of missing clauses are also in a disjunction, so smaller subsets of assignments(gotten from $\land$ of more than $2$ clauses) maybe be marked true more than once(theoretically), for each intersection of $2$ clauses they belong to, but that is not a botheration, as they are invisible to the algorithm(practically).
Then the algorithm joins these subsets together to form any missing clauses, using the dual-couple-child method. The way we build missing clauses is by patching together the intersections of the $2$ clauses each, and checking to see if their respective (overlapping) sets of assignments to variables sum up to some missing clause. Summing is possible because they're in a disjunction, and overlapping, or redundant/repeated assignments are simply reduced (idempotence).
A simple example - say we are given this expression :
$(a\land b\land d)\lor(\lnot a\land b\land c)$
$(a\land b\land d)$ has a child $(a\land b\land c\land d)$
$(\lnot a\land b\land c)$ has a child $(\lnot a\land b\land c\land d)$ Those two children are dual couple children of $(b\land c\land d)$ Hence we have derived a missing clause from intersections of it, with other clauses. All in all, we stop at $6$ literals because that is the maximum number of literals that can be found in a $\land$ of $2$ clauses. Conclusion :
I have proved every step of the algorithm's verification process to be valid. Since the algorithm scans only up to clauses with $6$ variables each, no matter the number of variables in the expression, it scans over $8\times{n\choose3}+16\times{n\choose4}+32\times{n\choose5}+64\times{n\choose6}$ possible cases, which is in
. polynomial time
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Imagine I want to compute this
$$D^{\dagger2}D_{\alpha}(\Phi^*\Phi)$$
where the $D$-s are super-covariant derivatives and $\Phi$ is a chiral superfield. Following the notation of this review http://arxiv.org/abs/hep-ph/9709356 on supersymmetry, the chiral covariant derivatives are (page 33)
$$D_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}-i(\sigma^{\mu}\theta^{\dagger})_{\alpha}\partial_{\mu}\qquad{}D^{\alpha}=-\frac{\partial}{\partial\theta^{\alpha}}+i(\theta^{\dagger}\bar{\sigma}^{\mu})^{\alpha}\partial_{\mu}$$
$$D^{\dagger\dot{\alpha}}=\frac{\partial}{\partial\theta_{\alpha}^{\dagger}}-i(\bar{\sigma}^{\mu}\theta)^{\dot{\alpha}}\partial_{\mu}\qquad{}D_{\dot{\alpha}}^{\dagger}=-\frac{\partial}{\partial\theta^{\dagger\dot{\alpha}}}+i(\theta\sigma^{\mu})_{\dot{\alpha}}\partial_{\mu}$$
let's now consider the following coordinate change
$$x^{\mu}\to{}y^{\mu}=x^{\mu}+i\theta^{\dagger}\bar{\sigma}^{\mu}\theta$$
the review claims (page 34) that in these $y,\theta,\theta^{\dagger}$ coordinates the chiral covariant derivative are
$$D_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}-2i(\sigma^{\mu}\theta^{\dagger})_{\alpha}\frac{\partial}{\partial{}y^{\mu}}\qquad{}D^{\alpha}=-\frac{\partial}{\partial\theta^{\alpha}}+2i(\theta^{\dagger}\bar{\sigma}^{\mu})^{\alpha}\frac{\partial}{\partial{}y^{\mu}}$$
$$D^{\dagger\dot{\alpha}}=\frac{\partial}{\partial\theta_{\alpha}^{\dagger}}\qquad{}D_{\dot{\alpha}}^{\dagger}=-\frac{\partial}{\partial\theta^{\dagger\dot{\alpha}}}$$
we can as well go to coordinates
$$x^{\mu}\to{}\bar{y}^{\mu}=x^{\mu}-i\theta^{\dagger}\bar{\sigma}^{\mu}\theta$$
where in page 35 it is claimed that the chiral covariant derivatives take the form
$$D_{\alpha}=\frac{\partial}{\partial\theta^{\alpha}}\qquad{}D^{\alpha}=-\frac{\partial}{\partial\theta^{\alpha}}$$
$$D^{\dagger\dot{\alpha}}=\frac{\partial}{\partial\theta_{\alpha}^{\dagger}}-2i(\bar{\sigma}^{\mu}\theta)^{\dot{\alpha}}\frac{\partial}{\partial\bar{y}^{\mu}}\qquad{}D_{\dot{\alpha}}^{\dagger}=-\frac{\partial}{\partial\theta^{\dagger\dot{\alpha}}}+2i(\theta\sigma^{\mu})_{\dot{\alpha}}\frac{\partial}{\partial\bar{y}^{\mu}}$$
so far so fine. It is obvious that for some computations the choice of the clever coordinates will make computations considerably less cumbersome. This is where my problems arise. The computation I want to do contains a chiral superfield, which can be expressed most compactly using the $y,\theta$ coordinates, whilst the antichiral superfield is better written using $\bar{y},\theta^{\dagger}$. My question is, is itlegitimate to use the chiral covariant derivatives is the $\bar{y},\bar{\theta}$ form, even though what i wanna take the derivative of contains all $y\bar{y},\theta,\theta^{\dagger}$? even if the answer is no, what i the wisest way to perform my computation?
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a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached).
b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct.
However,
your calculation of amplitude is not correct.
Peak-to-trough is
twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly.
CALCULATION OF AMPLITUDE OF OSCILLATIONS
1. Equation of Motion Method
First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision.
Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$
2. Energy Method
Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$.
At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$.
At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$.
By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1.
(c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it.
Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal.
Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is
always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result.
The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate.
Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$
We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off.
The total energy at P is the same as at U, so from above (
Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
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Lets refactor and reword these statements for ease of thought.
Let $A(n)$ be $Θ(g(n))$.
Let $B(n)$ be $O(g(n))$.
Note that $A$ implies $B$ because $A$ is stronger than $B$.
This means for $A$ to be fulfilled we have to fulfill the criteria of $B$ and more. Therefore $A$ being fulfilled implies $B$ must also be.
Written set theoretically, we have the set of functions that fulfill the criteria of $A$ are a subset of the functions that fulfill the criteria of $B$.
Well this is not contradictory, but actually equivalent in a sense to the last statement which showed that $A$ has the criteria of $B$ and more.
From our understanding of this point we can determine that:
All the functions that fulfill the criteria of $A$ fulfill $B$, but not all functions that fulfill the criteria of $B$ fulfill the criteria of $A$.
This is really an equivalent statement to:
The functions that fulfill the criteria of $A$ are a subset of the functions that fulfill the criteria of $B$ because the second set contains the first.
And therefore $\Theta(g(n))\subseteq O (g(n))$ is true, and it is equally true that $f(n) = \Theta(g(n))$ implies $f(n) = O(g(n))$ without any contradiction.
Furthermore this makes sense with the definition for $\Theta(g(n))$ which states that:$f(n)\in \Theta(g(n))$ if and only if $f(n) \in O(g(n))$ and $f(n) \in \Omega(g(n))$. Clearly by this definition, $f(n) \in Θ(g(n))$ does imply $f(n) \in O(g(n))$, and $\Theta(g(n))$ is clearly a subset of $O(g(n))$ as well.
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I am currently reading Computability and Logic by Boolos Burgess Geoffrey for the proof on "undecidability of first order logic". however, I find the notations a bit confusing. Can anyone recommend any resource website link/ video lecture or a book perhaps which will help me understand the proof of undecidability of first order logic? I am a CS student so, I do not want a completely mathematical /philosophical proof. I came across plenty of those on the web.
Let me clarify one subtle point: first order logic is only undecidable for certain given
languages. In particular the language $\cal{L}$ that contains only monadic predicates, that is, predicates of the form $P(x)$ and no function symbols, is decidable.
If you allow function symbols or predicates with more than 1 argument, then $\cal{L}$ usually becomes undecidable. The proof involves encoding Turing machines and their computation sequences using the symbols of the logic. Then one adds a
finite series of axioms $\phi_1,\ldots, \phi_n$, and build a formula $\psi(x,y)$ such that$$ \phi_1,\ldots,\phi_n\vdash \psi(\overline{n},\overline{m})$$is provable if and only if
The Turing machine with index $n$ halts in at most $m$ steps on the empty input.
Undecidability of FOL follows, as proving
$$ \vdash \phi_1\wedge\ldots\wedge\phi_n\rightarrow \exists y,\psi(\overline{n},y)$$
is equivalent to determining wether the machine with index $n$ halts.
A detailed description of this proof is given in chapter 4 of Avigad's lecture notes on computability.
A Mathematical Introduction to Logic, Second Edition: Herbert B. Enderton
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Linear Classifiers are one of the most commonly used classifiers and Logistic Regression is one of the most commonly used linear classifiers. The concepts we are going to learn here will actually extend a lot of other classification methods beyond linear classifiers. We’re going to learn the fundamental concepts, but also the underlying algorithms that let us optimize the parameters of this model to fit your training data.
Motivation
Suppose, we want to make a spam filter which will filter spams from our email list automatically. So, for every email, we’re going to feed it to a classifier and the classifier will say if this is a spam or maybe we want to visit a restaurant for a specific food e.g. sushi but before visiting the restaurant we want to make sure that the sushi there is good. So, what we want to do is to check the review of others visiting this restaurant about sushi and see if it is positive. It is actually very difficult to manually check all the reviews. Instead, what we can do is to feed these reviews to a classifier and it is going to say, is this a positive sentiment or negative sentiment and then watching the number of positive sentiment and the negative sentiment we will decide whether the sushi there is good to go or not.
Linear Classifiers Intuition
A linear classifier will take some quantity x as input which in our case will be emails or reviews and is going to make a prediction \(\hat{y}\) that says is this a positive statement which means is this a no spam or a positive review in which case \(\hat{y} = +1\) or is this a negative statement which implies it is a spam or a negative review in which case \(\hat{y} = -1\) . We will use restaurant review classifier for the rest fo the tutorial as our use case. A linear classifier does a little bit more associating every word for weight or coefficient which says how positively or negatively influential this word is for a review. For example,
Word Coefficient good 1.0 delicious 1.5 wonderful 2.7 bad -1.0 terrific -2.1 awful -3.3 but, food, the, we, to, have, … 0.0 … …
Here good and delicious have a coefficient of \(1.0\) and \(1.5\) respectively. Wonderful is very positive and has a coefficient of \(2.7\). On the negative side, bad and terrific might have a coefficient of \(-1\) and \(-2.1\) respectively. But awful is just awful, so \(-3.3\). There are also some words that are not that relevant to the sentiment of the review, might have \(0\) coefficient.
Now how will we use these coefficient’s to make a prediction of whether a sentence is positive or negative? Let’s take a review, for example, $$\begin{aligned} &\text{Input } \textbf{x}_i: \\ &\text{ Sushi was} \textbf{ good} \text{, the food was } \textbf{delicious}\text{, but the service was } \textbf{terribfic.}\end{aligned} \\ \text{Score}(\textbf{x}_i) = 1.0 + 1.7 – 2.1 = 0. 6$$ here good, delicious and great have coefficients greater than \(0\) and rest of the terms having coefficients greater than \(0\) as they are not relevant. After calculating we get a positive score which implies the sentiment in the sentence is positive and \(\hat{y} = +1\). This is called a linear classifier because the output is the weighted sum of the inputs.
So more generally for a simple linear classifier, we are going to take a review and the coefficient associated with each word and feed that in our classifier and calculate the score for that input. If the score is greater than \(0\) the prediction is positive and \(\hat{y} = +1\) and if the score is less than \(0\) we say the prediction is \(\hat{y} = -1\). Now, what we need to do is to train the coefficients or weights of these linear classifiers from data.
So given some input training data that includes sentences of reviews labeled with either \(+1\) or \(-1\). We’re going to split those into some training set and some validation set. Then we’re going to feed that training set to some learning algorithm which is going to learn the weights associated with each word. And then after we learn this classifier, we’re going back and evaluate its accuracy on that validation set. Now, how do we learn this classifier from data?
Decision Boundary
The decision boundary is a boundary between positive and negative predictions. Let’s say that we have taken our data and trained our linear classifier and every word has zero weight except for two of them. Awesome has weight 1.0 and awful has weight -1.5. That means that the score of any sentence is 1.0 times the number of the word awesome minus 1.5 times the number of times the word awful shows up. We can plot that into a graph which depends on every sentence the number of awesome and the number of awful. So for example, for a sentence, $$\begin{aligned} \text{ Sushi was} \textbf{ awesome} \text{, the food was } \textbf{awesome}\text{, but the service was } \textbf{awful.}\end{aligned} $$ We’re going to plot that into a space where we’re going to have two awesome and one awful. So it gets plotted in the \((2,1)\) point. And then for every sentence that we might have in our training data set say, three awful and one awesome, three awesome and no awful and so on we will have different points and if we plot the dataset we find
The classifier that we’ve trained with the coefficients 1.0 and -1.5 will have a decision boundary that corresponds to a line plotted above, where 1.0 times awesome minus 1.5 times the number of awful is equal to zero. Everything below that line has a score greater than zero and any points above that line are going to have scored less than zero. So there’s that line, everything below the line is positive, everything above the line’s negative a linear decision boundary. That is what makes it a linear classifier.
Linear Classifiers Model
in our previous example that we had with just two features with no zero coefficients, awesome and awful and we have calculated the score as $$\text{Score(x)} = w_0 + w_1 \text{ #awesome} + w_2 \text{ #awful}$$ where \(w_0\) was \(0\) in our example. Now suppose that we had a third feature with no zero coefficient. Let’s say that the word great. So the score function then will be $$ \text{Score(x)} = w_0 + w_1 \text{ #awesome} + w_2 \text{ #awful} + w_3 \text{ #great}$$ So, the general form of our linear classifier model will be : $$\begin{aligned} &\text{Model: } \hat{y}_i = \text{sign(Score(}\textbf{x}_i)) \\ &\text{Score(}\textbf{x}_i) = w_0 + w_1 \textbf{x}_i[1]+ … + w_d \textbf{ x}_i[d] \end{aligned}$$ where $$\begin{aligned} feature \; 1 &= 1 \\ feature \; 2 &= \textbf{x}[1] \dots e.g. , \#awesome \\ feature \; 3 &= \textbf{x}[2] \dots e.g., \#awful \\ \vdots \\ feature \; d+1 &= \textbf{x}[d]\ \dots e.g., \#ramen \end{aligned}$$
Theses notations that we’ve used so far are without features associated with it. So we’re going to have these functions \(h_1\) through \(h_D\). They’ve defined some features we might extract from the data and we are going to encode the constant function that is \(h_0\). So the more generically our model will be looked like: $$\begin{aligned} \text{Model: } \hat{y}_i &= \text{sign(Score(}\textbf{x}_i)) \\ \text{Score(}\textbf{x}_i) &= w_0 h_0\textbf{x}_i + w_1 h_1\textbf{x}_i+ … + w_D h_D\textbf{x}_i \\ &= \textbf{w}^{T}h(\textbf{x}_i) \end{aligned}$$ where $$\begin{aligned} feature \; 1 &= h_0(\textbf{x}) \dots e.g., 1 \\ feature \; 2 &= h_1(\textbf{x}) \dots e.g. , \textbf{x}[1] = \#awesome \\ feature \; 3 &= h_2(\textbf{x}) \dots \textbf{x}[2] = \#awful \; or, \log(\textbf{x}[7]) \textbf{x}[2] = \log(\#bad) * \#awful \\ \vdots \\ feature \; D+1 &= h_D( \textbf{x}) \dots \text{some other function of} \; \textbf{x}[1], \dots, \textbf{x}[d]\end{aligned}$$ So this is our generic linear classifier model with multiple features.
Linear classifiers are a pretty abstract concept. Logistic regression is a specific case of that, where we use what’s called the logistic function to squeeze minus infinity to plus infinity into the interval \((0, 1)\) so we can predict probabilities for every class. Next, we will discuss Logistic Regression.
Also published on Medium.
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I'm trying to do a series of exercises from Spivak's Calculus, in chapter 8, Least Upper Bounds.
I'm trying to tackle these two exercises, $5.$ and $^*.6$
From $5.$ I have proven the first claim
$(a)$ Let $x-y>1$. Prove there is an integer $k$ such that $x<k<y$.
P Let $\ell$ be the greatest integer such that $\ell \leq x$. Then
$$y-x >1$$
$$y-\ell >1$$
$$y>1+\ell $$
Thus the integer $\ell +1$ is between $y$ and $x \text{ }\blacktriangle$.
$(b)$ Let $x<y$. Show there is a rational $r$ such that $x<r<y$.
P If $x<y$ then there is $\epsilon>0$ such that $x+\epsilon=y$. Thus $\epsilon=y-x$. But from T3, we have that there is an $n$ such that $1/n < \epsilon$, thus
$$\frac{1}{n} < y-x$$
$$1<ny-nx \text{ }\blacktriangle$$
and from the last theorem we have that there is a integer $k$ such that
$$nx<k<ny$$
$$x< \frac{k}{n}<y \text{ }\blacktriangle$$
$(c)$ Let $r<s$ be rational numbers. Prove there is an irrational number between $r$ and $s$. Hint: it is known there is an irrational number between $0$ and $1$.
Ok, this is a proof based on your answers.
P Since $\sqrt 3 $ is irrational and $\sqrt{3}<3$, then $\ell = \sqrt{3}/3<1$ and thus it is in $[0,1]$.
Now, $r<s \Rightarrow 0<s-r$. Then
$$0<\ell < 1$$
$$0<\ell(s-r) < s-r$$
$$r<r+\ell(s-r) < s$$
And since $\ell$ is irrational $r+\ell(s-r)$ is irrational. $\blacktriangle $
$(d)$ Show that if $x<y$ then there is an irrational number between $x$ and $y$: There is no need to work here, this is consequence of $(b)$ and $(c)$
This one was quite straightforward, but thanks anyways.
P
$$x<y$$
$$(b)\Rightarrow x<r<y,r<y \Rightarrow r<q<y \Rightarrow x<r<q<y$$
Then by $(c)$, there is an irrational $\ell$ such that
$$ x<r<\ell<q<y \text{ } \blacktriangle $$
This will let me conclude
$\mathbb Q$ is dense on any $[a,b]\subset \Bbb R$
$\mathbb I$ is dense on any $[a,b] \subset \Bbb R $
and will let me move on into $^*6.$ which is
$(a)$ Show that $f$ is continuous and $f(x)=0$ for all $x$ in a dense set $A$, then $f$ is $f(x)=0$ for all $x$.
$(b)$ Show that $f$ and $g$ are continuous and $f(x)=g(x)$ for all $x$ in a dense set $A$, then $f(x)=g(x)$ for all $x$.
$(c)$ If we suppose $f(x)\geq g(x)$ for all $x$ in $A$, then $f(x)\geq g(x)$ for all $x$. ¿Can $\geq$ be substituted with $>$ everywhere?
I'm not asking for solutions for this last problems (which will be asked separately), but for $(c)$ and $(d)$ in $5.$
The chapter has several important proofs, which might or might not be relevant here, but I think it is important you know what tools we have at hand:
THEOREM 7-1 If $f$ is continuous on $[a,b]$ and $f(a)<0<f(b)$, then there is $x \in [a,b]:f(x)=0$ THEOREM 1 If $f$ is continuous in $a$, then there exists a $\delta>0$ such that $f$ is bounded above in $(a-\delta,a+\delta)$. THEOREM 7-2 If $f$ is continuous on $[a,b]$ then $f$ is bounded on $[a,b]$. THEOREM 7-3 If $f$ is continuous on $[a,b]$ then there is an $y$ in $[a,b]$ such that $f(y)\geq f(x)$ for all $x$ in $[a,b]$. THEOREM 2 $\Bbb N$ is not bounded above. THEOREM 3 If $\epsilon >0$, there is an $n \in \Bbb N$ such that $1/n < \epsilon$.
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Let me give a more « conceptual» Galois answer. First , fix an integer
m and consider a field K of characteristic not dividing m, containing a primitive m-th root of unity. Let F/K be a finite Galois extension, and let $E = F (\sqrt[m]{\alpha})$, with $\alpha \in K*$ . By Kummer theory, E only depends on the class $[\alpha]$ of $\alpha$ modulo $F*^m$ , and E/F is cyclic of degree dividing $m$ . Question : when is E/K Galois ? Answer : iff $[\alpha]$ is invariant under the action of Gal(F/K), in other words, for any s in Gal(F/K), $s(\alpha)/\alpha$ $\in$ $F*^m$ (this is a classical exercise in Galois theory).
Coming back to your problem, take $m = 2$ , F/K quadratic ,
s a generator of its Galois group. According to the criterion above, for $E = F(\sqrt{\alpha})$ , E/K is Galois (of degree 4) iff $s(\alpha)/\alpha$ is a square in F*, equivalently iff $\alpha . s(\alpha)$ is a square in F* . It is important to stress « in F*, not necessarily in K* », because $\alpha . s(\alpha) = N(\alpha)$, where N is the norm map from F* to K*, and $N(\alpha)$ $\in$ K* . Suppose the criterion is met, then Gal(E/K) is a group of order 4, hence abelian, isomorphic to $C_4$ or $C_2 × C_2$ . Question : how to distinguish between the two cases, cyclic and biquadratic ? We must go back to the idea of proof in the above Galois exercise. In our situation, the original expression $s(\alpha) = \alpha . x^2$, $x$ $\in$ F*, implies that N(x) = $\pm$ 1. How to distinguish between the two signs ? By Hilbert 90 (or mere calculation), N(x) = 1 iff x is of the form $s(y)/y$, y$\in$ F*, hence $s(\alpha)/\alpha$ = $(s(y)/y)^2$, or equivalently, $\alpha = a. y^2$, a$\in$ K*. The latter equality means that E = F($\sqrt a$), or equivalently, writing F = K($\sqrt c$) in your original notation, that F is the biquadratic field K($\sqrt a, \sqrt c$), a and c $\in$ K* .
This solution may seem complicated, but could be useful in more elaborate associated problems, for instance : describe all the quadratic extensions L of K($\sqrt a, \sqrt c$) which are Galois but non abelian (i.e. which are diedral or quaternionic) over K .
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Interesting idea to use a grid. I doubt it will work for this question, since writing things in that grid format in a sense splits into arithmetic progressions modulo $p$, and usually saying things about the primes in a progression is harder.
However, I cannot help but mention that using a grid like that was applied brilliantly by Maier to prove a counter-intuitive result, (using the Prime Number Theorem for Arithmetic Progessions) and the idea is now called the Maier Matrix Method.
(A small digression, but it is interesting!)
The Maier Matrix Method
There are questions regarding primes in short intervals, and we can ask ourselves what does $$\pi(x+y)-\pi(x)$$ look like? To say anything meaningful, $y$ cannot be too small, but here lets suppose $y=\log^B(x)$ for some $B>2$. Selberg proved that under the Riemann Hypothesis, we have $$\pi(x+y)-\pi(x)\sim \frac{y}{\log x}$$ as $x\rightarrow \infty$ for almost all $x$. (A set with density $\rightarrow 1$) It was then conjectured that this asymptotic must hold for all $x$ which are sufficiently large. (This conjecture was made for several reasons, one of which is that it is true under Cramer's probabilistic model)
In a surprising turn of events, Maier showed it was false, and that there exists $\delta>0$ and arbitrarily large values of $x_1$ and $x_2$ such that both $$\pi(x_1 +\log^B (x_1))-\pi (x_1)> (1+\delta)\log^{B-1}(x_1)$$and $$\pi(x_2 +\log^B (x_2))-\pi (x_2)< (1-\delta)\log^{B-1}(x_2)$$hold, despite the fact that the asymptotic holds for a set of density $1$.
He proved this using a method which is now called the "Maier Matrix Method." Essentially, it is just drawing a grid which is similar to the one above, and then applying a clever combinatorial argument. The columns are arithmetic progressions, and by PNT4AP, we can easily say things about them to understand the number of primes in the grid. There is a little trick with oscillation of the Dickman Function, but then basically by the pigeon hole principle the question is solved.
I definitely think you might find this expository article by Dr.Andrew Granville to be interesting. (It is quite readable, and gives an a more in depth, and very clear explanation)
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The quantum harmonic oscillator $H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2$ has energies $E_n=\hbar\omega(n+1/2)$, so to achieve "classical" energies $n$ must be very large.
My question is: how does appear the classical behaviour of a point mass in this potential, as energy increases? Classically, a point mass has
exactly the position $x(t)=a\cos(\omega t)$, suggesting that the wavefuntion $\psi$ verifies that its square $|\psi(x,t)|^2\cong 0$ everywhere except near $x=a\cos(\omega t)$.
In this hyperphysics page appears that probability is well approximated. Taking the wave function $\psi_n$ corresponding to the energy $E_n$, with large $n$:
\begin{equation} \psi_n(x,t)=e^{-iE_nt/\hbar}\psi_n(x,0) \end{equation}
It yields:
\begin{equation} |\psi_n(x,t)|^2=|\psi_n(x,0)|^2 \end{equation}
For large energies, this quantum probability density approaches the classical probability density. But that is different than saying that $|\psi(x,t)|^2\cong 0$ everywhere except near $x=a\cos(\omega t)$!
I know that the mean value $\langle x\rangle$ does follow a sinusoidal "motion" when there is a linear combination of two or more energy eigenstates, but that is different because $\langle x\rangle$ is related to measuring $x$ a lot of times and then computing the mean value, while classically only one measure is needed. I mean, if $|\psi(x,t)|^2$
were $\cong 0$ everywhere except near $x=a\cos(\omega t)$, then, with only one measure of $x$, the result would be (in the vast majority of cases) $\cong a\cos(\omega t)$.
So, more realistically, let's consider a linear combination of states $\psi_n$ (a lot of states) all with high "classical" energies:
\begin{equation} \psi(x,t)=\sum_{n\ \text{large}}\psi_n(x,t) \end{equation}
Does $|\psi(x,t)|^2\cong 0$ everywhere except near $x=a\cos(\omega t)$, recovering the classical motion?
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I am looking at this paper : http://arxiv.org/pdf/1411.3419v1.pdf But somehow I am not being able to fish out a method to calculate this quantity called the "block-sensitivity".
Can someone kindly provide some discussion/examples or such?
I am wondering if there is a way to talk of "block sensitivity" of any function of the form $f:\Sigma ^n \rightarrow \{0,1\}$ where $\Sigma$ is a finite set. IF the above is possible then I guess given any such function one can Booleanize it to give another function $f' : \{0,1\}^{n \vert \vert \Sigma \vert \vert } \rightarrow \{0,1\}$ where $\vert \vert \Sigma \vert \vert$ is the number of bits required to encode $\Sigma$ in binary. Then is it true that I can always calculate the "block-sensitivity" of $f$ say $bs(f)$ and from there get $bs(f')$ as $bs(f) = \frac{1}{\vert \vert \Sigma \vert \vert} bs(f')$ ?
(the typical situation I have is where $\vert \vert \Sigma \vert \vert = poly(n)$ and I hope that this doesn't produce any extra complication!)
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Mount Everest is 8,848 metres (29,028 feet) above sea level and is the result of a continental plate smashing into another continental plate. Can a tectonic process build a mountain that's even higher? (Volcanoes excluded!)
Found a article that used a simple analytical modelling to determine how high a mountain can be. Reference
Based on simple physics, tallest a mountain will be on earth is ~10 km. This is based on:
Simple cone shape for the mountain. Vol ≈ $r^2 h$ Based on weight of the of the mountain: Weight W≈ $\rho g r^2 h$
Stress σ the mountain exerts on the ground underneath it is:
σ ≈ Weight/Area ≈ $(\rho g r^2 h)/r^2$ ≈ $\rho g h$
The limiting factor is the compressive strength of the rock: Assume granite with average density of granite is ρ = 3 g/cm3 Compressive strength is $\sigma_C$ = 200 MPa = $2 \times 10^8\, N/m^2$ Stress = Compressive strength of rock σ = $\sigma_C$ or $\rho g h_{max} = \sigma_C$.
Calculate max height:
$h_{max}$ ≈ $\sigma_C/(\rho g)$
$h_{max}$ ≈ $\frac{2 \times 10^8\, N/m^2}{3 \times 10^3 kg/m^3 \times 10\, m/s^2)}$≈ $10^4\, m$ = 10 km
The glacial buzzsaw hypothesis (summary; sample paper) is that mountains can't get much higher than the elevation at which glaciers form cirques. The upper walls of the cirques are steep and erode easily, which planes off the peaks above them, shortening the mountains. The evidence is, to summarize, that they
don't get much higher than the cirques. Cirques are lower in higher latitudes.
The highest possible mountain would therefore, I imagine, be near the equator and somewhere quite dry to minimize glaciation. But a very high mountain intrinsically alters global wind and weather patterns -- the Himalayas are sometimes called "the third pole". I don't know if you
could have a dry, tallest-in-the-world mountain near the equator, no matter what plate tectonics was trying to do.
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March 17th, 2017, 06:26 AM
# 1
Newbie
Joined: Mar 2017
From: Illinois
Posts: 3
Thanks: 0
Trying to derive a chem problem
I am trying to find a rate law for a chemical reaction through mechanisms.
I have the mechanisms set out below and I am left with the rate listed
1: 2A + 2B -> 2C (fast)
2: 2C + B -> D (slow)
3: D + E -> F (fast)
Step 2 is the rate determining step.
[A]^2[B]^2
--------------- = rate
[C]^2
My question is, How do I get the [C]^2 from the bottom to the top. In all of the class examples, the bottom was never squared, but in our examples, the prof took the inverse of the bottom and put it on the top.
March 17th, 2017, 11:26 AM
# 2
Senior Member
Joined: May 2016
From: USA
Posts: 1,310
Thanks: 551
We have no clue what the variables are, what the arrows mean, why you need to calculate a rate if you are given a formula for the rate, why that is a formula for a rate? Are those supposed to be chemical equations involves moles? Are they equations about times? About rates?
Please give the problem exactly and completely.
By the way, if you are talking about multiplicative inverses of exponentiated variables
$\dfrac{A^2 B^2}{C^2} = A^2 B^2 C^{-2}.$
Last edited by JeffM1; March 17th, 2017 at 11:31 AM.
March 18th, 2017, 04:44 AM
# 3
Senior Member
Joined: Jun 2015
From: England
Posts: 915
Thanks: 271
Hello austin and welcome.
First please let me ask you if you are confusing the reaction rate constant with the equilibrium constant?
This is a common misconception.
The rate of reaction is a differential equation (you should have three simultaneous ones)
not a fraction equal to a constant.
The reaction rate and its constant is determined purely empirically ( by experimental observation)
Are you sure that the reaction rates are second order as you suggest?
Second order reactions depend upon the initial concentrations / partial pressures which appear as the constants in the limits of integeration when the integration is performed to find the rate law.
Hopefully you also realise that just because the proposed rate law 'fits' the experimental data it is necessary but not sufficient for it to be correct.
Additional corroborating observational evidence is required for that.
The reactions themselves you are discussing are interesting.
Is there truly only a single product in each case?
You seem to be constructing quite a complex molecule.
@JeffM1 and other mathematicians
This is the Chemistry section!
The rate of a chemical reaction is defined as the rate of change of activity of some desired reaction component (reactant) with respect to time.
The activity may be measured by various properties depending so concentration or partial pressure are common.
The actual value is conventionally denoted by enclosing in square brackets.
Almost the simplest rate of reaction equation declares
$\displaystyle \frac{{d\left[ X \right]}}{{dt}} = {k_1}\left[ X \right]$
Which says that the rate of change of (say) concentration of X is proportional to the concentration of X.
With some reactions the indices are simple integer powers but some have fractional powers thus
$\displaystyle \frac{{d\left[ X \right]}}{{dt}} = {k_1}{\left[ X \right]^m}$
Last edited by studiot; March 18th, 2017 at 05:00 AM.
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I'm studying fluid mechanics in more depth during my Ph. D. and there is something related with the diffusive term that has been bothering me for a long time. Looking at the convection diffusion equation: $$ \frac{\partial v}{\partial t} + a\cdot\nabla v - \nabla(\nu \nabla v )=f, $$
and thinking in Fick's law, it's not hard for me thinking in diffusion as the process by which, for example, the particles of a solution with a concentration gradient get apart one from each other to "reduce" the energy of the system; being, each one of these particles, more "comfortable" inside the solvent, which is the same as saying "with the bigger free mean path possible".
But now when I look to the Navier Stokes equation (incompresible and viscous):
$$ \rho \left(\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}\right)=-\nabla p+\mu\nabla ^2\mathbf{v}+\mathbf{f} $$
I can easily see the viscous term as a diffusive one, but there's no way I can relate it with Fick's law. So, somebody can explain me how can I see viscosity as a diffusive process?
If you are interested in why I'm asking this, it's because in FEM there is a stabilization method called artificial viscosity that add some viscosity to increase diffusion and make the model more stable. So I understand a) why artificial viscosity increases the diffusion, b) why diffusion stabilizes the equation; but I don't understand why viscosity is diffusive (besides the fact that $\mu$ is multiplying $\nabla^2v$)
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Adjoint space of a topological vector space $E$
The vector space $E^{*}$ consisting of continuous linear functions on $E$. If $E$ is a locally convex space, then the functionals $f\in E^{*}$ separate the points of $E$ (the Hahn–Banach theorem). If $E$ is a normed space, then $E^{*}$ is a Banach space with respect to the norm \begin{equation*} \|f\| = \sup\limits_{x\ne0}\frac{|f(x)|}{\|x\|}. \end{equation*} There are two (usually different) natural topologies on $E^{*}$ which are often used: the strong topology determined by this norm and the weak-$*$-topology.
References
[1] D.A. Raikov, "Vector spaces" , Noordhoff (1965) (Translated from Russian) Comments
Instead of the term adjoint space one more often uses the term dual space. The weak-$*$-topology on $E^{*}$ is the weakest topology on $E^{*}$ for which all the evaluation mappings $f\mapsto f(x)$, $f\in E^{*}$, $x\in E$, are continuous.
References
[a1] H.H. Schaefer, "Topological vector spaces" , Macmillan (1966) How to Cite This Entry:
Adjoint space.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Adjoint_space&oldid=28995
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R V Anavekar
Articles written in Bulletin of Materials Science
Volume 6 Issue 6 December 1984 pp 1009-1012
The d.c. electrical conductivity of Na
2O-ZnO-B 2O 3 glass system has been measured as a function of temperature in the range of 350–600°K. The conductivity data show that the activation energy of Na + ions is dependent on ZnO concentration. The results have been discussed in the light of the cluster model of glasses.
Volume 40 Issue 3 June 2017 pp 523-526
Pr$^{3+}$ doped molybdenum lead-borate glasses with the chemical composition 75PbO−[25–($x+y$)B$_2$O$_3$]–$y$MoO$_3$–$x$Pr$_2$O$_3$ (where $x = 0.5$ and 1.0 mol% and y = 0 and 5 mol%) were prepared by conventional melt-quenching technique. Thermal, optical and structural analyses are carried out using DSC, UV and FTIR spectra. The physical parameters,like glass transition ($T_g$), stability factor ($\Delta T$), optical energy band gap ($E_{gopt}$), of these glasses have been determined as a function of dopant concentration. The $T_g$ and optical energy gaps of these glasses were found to be in the range of 290–350$^{\circ}$C and 2.45–2.7 eV, respectively. Stability of the glass doped with Pr$^{3+}$ is found to be moderate ($\sim$40). The results are discussed using the structural model of Mo–lead-borate glass
Current Issue
Volume 42 | Issue 6 December 2019
Click here for Editorial Note on CAP Mode
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Notes - Linear Equation in one variable
Category : 7th Class
LINEAR EQUATION IN ONE VARIABLE
FUNDAMENTALS
Symbols used to denote a constant are generally, 'c', 'k' etc...
e.g., 2x - 6; here, 2 is the coefficient of \['x';\,\,'x'\] is the variable and \[-6\]is the constant. Similarly in \[ay+b;\,\,a\] is the coefficient of \[y;\,\,'y'\] is the variable and \[(+b)\] is the constant.
e.g., (i) \[5x-1=6x+m\] (ii) \[3\left( x-4 \right)=5\]
(iii)\[2y+5=\frac{y}{6}-2\] (iv) \[\frac{t-1}{6}+\frac{2t}{7}=a\]
e.g., \[2x+6=3x-10\Rightarrow 6+10=3x-2x\Rightarrow 16=x\]
Verification
Substituting \[x=16\] we have LHS \[=2\times 16+6=38\] & RHS \[=\text{ }3\times 16-10=38\]
\[\therefore \]\[x=16\]is a solution of the above equation.
(a) Same number can be added to both sides of an equation.
(b) Same number can be subtracted from both sides of an equal.
(c) Both sides of an equation can be multiplied by the same non - zero number
(d) Both sides of an equation can be divided by the same non - zero number
(e) Cross multiplication: If\[\frac{ax+b}{cx+d}=\frac{p}{q}\], then \[q\left( ax+b \right)=p\left( cx+d \right).\]
This process is called cross multiplication.
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The time delay between neutrinos and photons does
not tell us directly about the absolute mass of the neutrinos. The time delay between a neutrino of mass $m$ and a massless neutrino does:$$\Delta t = \frac{d}{v} - \frac{d}{c} \approx 0.5 \left(\frac{mc^2}{E}\right)^2 d,$$where $\Delta t$ is the time delay in seconds, $v \approx c[1-\frac{1}{2}\left(\frac{mc^2}{E}\right)^2]$ from relativistic calculation, $d$ the distance to the supernova in 10 kpc (a "typical" value for Galactic supernova), neutrino rest mass $m$ in eV and neutrino energy in MeV. So it is a propagation-speed problem.
As for the reason of the delay between the neutrinos and the photons, several pieces of information should be helpful:
It is hot in the core of the supernova -- so hot that the photons scatter a lot with free electrons before escaping from the core (see comment for a more detailed version); (electromagnetic interaction) Yet it is not hot enough to influence two of the three neutrinos, $\nu_{\mu}$ and $\nu_{\tau}$, almost at all, and only slightly for another ($\nu_{e}$); they just fly out of the core. (weak interaction) So this is a different-interaction (or, cross-section) problem.
Take SN1987A for example, the neutrinos arrived 2-3 hours before the photons. On the other hand, $\Delta t$ calculated above, if detected, would be $\leq 1$ sec. With these two values and a bit more thinking, you might conclude that this "time of flight" method could not constrain very well the neutrinos with smaller-than-eV masses; and you would be right. There are better ways to do that. Here is a reference: http://arxiv.org/pdf/astro-ph/0701677v2.pdf.
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How can I prove that $$\int_0^1(1+\epsilon^{-1}e^{-\beta x/p\epsilon})dx\leq C,$$ where $\epsilon$ is a small parameter. In the limit $\epsilon^{-1}$ is unbounded so how is this integral bounded?
Just calculate:
$$\int_0^1 \! (1 + \epsilon^{-1}e^{-\beta x/p\epsilon}) \, dx$$ $$ = 1 + \epsilon^{-1}\int_0^1 \! e^{-\frac{\beta}{p \epsilon}x} \, dx$$ $$ = 1 + \frac{1}{\epsilon}\frac{-p\epsilon}{\beta}(e^{\frac{-\beta}{p\epsilon}} - 1)$$ $$ = 1 + \frac{p}{\beta}(1-e^{\frac{-\beta}{p\epsilon}})$$ This last expression is bounded in absolute value as long as $\epsilon$ does not become arbitrarily negative. Basically the reason the integral stays bounded is that the exponential term becomes small much faster than the $\epsilon^{-1}$ term becomes large as $\epsilon \to 0$ from the right. Notice also that you are only asking for an upper bound $$\int_0^1 \! (1 + \epsilon^{-1}e^{-\beta x/p\epsilon}) \, dx \le C$$ and such an expression is valid for any $\epsilon$ with $C = 1 + \frac{p}{\beta}$.
If I am not mistaken, your integral equals $$1+\frac{p}{\beta}(1-e^{-\frac{\beta}{p\varepsilon}})$$
Assuming that your parameters are positive, the result follows.
Change variables, $x=\epsilon u$ in your integral: $$ \begin{align} \int_0^1\left(1+\epsilon^{-1}e^{-\frac{\beta x}{p\epsilon}}\right)\,\mathrm{d}x &=\int_0^{1/\epsilon}\left(\epsilon+e^{-\frac{\beta u}{p}}\right)\,\mathrm{d}u\\ &\to1+\int_0^\infty e^{-\frac{\beta u}{p}}\,\mathrm{d}u\\ &=1+\frac{p}{\beta} \end{align} $$ as $\epsilon\to0$. So for $\epsilon$ near $0$, the integral is near $1+\frac{p}{\beta}$.
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Let $X_n(\Bbb{Z})$ be the simplicial complex whose vertex set is $\Bbb{Z}$ and such that the vertices $v_0,...,v_k$ span a $k$-simplex if and only if $|v_i-v_j| \le n$ for every $i,j$. Prove that $X_n(\Bbb{Z})$ is $n$-dimensional...
no kidding, my maths is foundations (basic logic but not pedantic), calc 1 which I'm pretty used to work with, analytic geometry and basic linear algebra (by basic I mean matrices and systems of equations only
Anyway, I would assume if removing $ fixes it, then you probably have an open math expression somewhere before it, meaning you didn't close it with $ earlier. What's the full expression you're trying to get? If it's just the frac, then your code should be fine
This is my first time chatting here in Math Stack Exchange. So I am not sure if this is frowned upon but just a quick question: I am trying to prove that a proper subgroup of $\mathbb{Z}^n$ is isomorphic to $\mathbb{Z}^k$, where $k \le n$. So we must have $rank(A) = rank(\mathbb{Z}^k)$ , right?
For four proper fractions $a, b, c, d$ X writes $a+ b + c >3(abc)^{1/3}$. Y also added that $a + b + c> 3(abcd)^{1/3}$. Z says that the above inequalities hold only if a, b,c are positive. (a) Both X and Y are right but not Z. (b) Only Z is right (c) Only X is right (d) Neither of them is absolutely right.
Yes, @TedShifrin the order of $GL(2,p)$ is $p(p+1)(p-1)^2$. But I found this on a classification of groups of order $p^2qr$. There order of $H$ should be $qr$ and it is present as $G = C_{p}^2 \rtimes H$. I want to know that whether we can know the structure of $H$ that can be present?
Like can we think $H=C_q \times C_r$ or something like that from the given data?
When we say it embeds into $GL(2,p)$ does that mean we can say $H=C_q \times C_r$? or $H=C_q \rtimes C_r$? or should we consider all possibilities?
When considering finite groups $G$ of order, $|G|=p^2qr$, where $p,q,r$ are distinct primes, let $F$ be a Fitting subgroup of $G$. Then $F$ and $G/F$ are both non-trivial and $G/F$ acts faithfully on $\bar{F}:=F/ \phi(F)$ so that no non-trivial normal subgroup of $G/F$ stabilizes a series through $\bar{F}$.
And when $|F|=pr$. In this case $\phi(F)=1$ and $Aut(F)=C_{p-1} \times C_{r-1}$. Thus $G/F$ is abelian and $G/F \cong C_{p} \times C_{q}$.
In this case how can I write G using notations/symbols?
Is it like $G \cong (C_{p} \times C_{r}) \rtimes (C_{p} \times C_{q})$?
First question: Then it is, $G= F \rtimes (C_p \times C_q)$. But how do we write $F$ ? Do we have to think of all the possibilities of $F$ of order $pr$ and write as $G= (C_p \times C_r) \rtimes (C_p \times C_q)$ or $G= (C_p \rtimes C_r) \rtimes (C_p \times C_q)$ etc.?
As a second case we can consider the case where $C_q$ acts trivially on $C_p$. So then how to write $G$ using notations?
There it is also mentioned that we can distinguish among 2 cases. First, suppose that the sylow $q$-subgroup of $G/F$ acts non trivially on the sylow $p$-subgroup of $F$. Then $q|(p-1) and $G$ splits over $F$. Thus the group has the form $F \rtimes G/F$.
A presentation $\langle S\mid R\rangle$ is a Dehn presentation if for some $n\in\Bbb N$ there are words $u_1,\cdots,u_n$ and $v_1,\cdots, v_n$ such that $R=\{u_iv_i^{-1}\}$, $|u_i|>|v_i|$ and for all words $w$ in $(S\cup S^{-1})^\ast$ representing the trivial element of the group one of the $u_i$ is a subword of $w$
If you have such a presentation there's a trivial algorithm to solve the word problem: Take a word $w$, check if it has $u_i$ as a subword, in that case replace it by $v_i$, keep doing so until you hit the trivial word or find no $u_i$ as a subword
There is good motivation for such a definition here
So I don't know how to do it precisely for hyperbolic groups, but if $S$ is a surface of genus $g \geq 2$, to get a geodesic representative for a class $[\alpha] \in \pi_1(S)$ where $\alpha$ is an embedded loop, one lifts it to $\widetilde{\alpha}$ in $\Bbb H^2$ by the locally isometric universal covering, and then the deck transformation corresponding to $[\alpha]$ is an isometry of $\Bbb H^2$ which preserves the embedded arc $\widetilde{\alpha}$
It has to be an isometry fixing a geodesic $\gamma$ with endpoints at the boundary being the same as the endpoints of $\widetilde{\alpha}$.
Consider the homotopy of $\widetilde{\alpha}$ to $\gamma$ by straightline homotopy, but straightlines being the hyperbolic geodesics. This is $\pi_1(S)$-equivariant, so projects to a homotopy of $\alpha$ and the image of $\gamma$ (which is a geodesic in $S$) downstairs, and you have your desired representative
I don't know how to interpret this coarsely in $\pi_1(S)$
@anakhro Well, they print in bulk, and on really cheap paper, almost transparent and very thin, and offset machine is really cheaper per page than a printer, you know, but you should be printing in bulk, its all economy of scale.
@ParasKhosla Yes, I am Indian, and trying to get in some good masters progam in math.
Algebraic graph theory is a branch of mathematics in which algebraic methods are applied to problems about graphs. This is in contrast to geometric, combinatoric, or algorithmic approaches. There are three main branches of algebraic graph theory, involving the use of linear algebra, the use of group theory, and the study of graph invariants.== Branches of algebraic graph theory ===== Using linear algebra ===The first branch of algebraic graph theory involves the study of graphs in connection with linear algebra. Especially, it studies the spectrum of the adjacency matrix, or the Lap...
I can probably guess that they are using symmetries and permutation groups on graphs in this course.
For example, orbits and studying the automorphism groups of graphs.
@anakhro I have heard really good thing about Palka. Also, if you do not worry about little sacrifice of rigor (e.g. counterclockwise orientation based on your intuition, rather than, on winding numbers, etc.), Howie's Complex analysis is good. It is teeming with typos here and there, but you will be fine, i think. Also, thisbook contains all the solutions in appendix!
Got a simple question: I gotta find kernel of linear transformation $F(P)=xP^{''}(x) + (x+1)P^{'''}(x)$ where $F: \mathbb{R}_3[x] \to \mathbb{R}_3[x]$, so I think it would be just $\ker (F) = \{ ax+b : a,b \in \mathbb{R} \}$ since only polynomials of degree at most 1 would give zero polynomial in this case
@chandx you're looking for all the $G = P''$ such that $xG + (x+1)G' = 0$; if $G \neq 0$ you can solve the DE to get $G'/G = -x/(x+1) = -1 + 1/(x+1) \implies \ln G = -x + \ln(1+x) \implies G = (1+x)e^(-x) + C$ which is obviously not a polyonomial, so $G = 0$ and thus $P = ax + b$
could you suppose that $\operatorname{deg} P \geq 2$ and show that you wouldn't have nonzero polynomials? Sure.
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(Sorry was asleep at that time but forgot to log out, hence the apparent lack of response)
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details
http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
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Carbon-14 has a half-life of 5,730 years. That means that after 5,730 years, half of that sample decays. After another 5,730 years, a quarter of the original sample decays (and the cycle goes on and on, and one could use virtually any radioactive isotope). Why is this so? Logically, shouldn't it take 2,865 years for the quarter to decay, rather than 5,730?
The right way to think about this is that, over 5,730 years,
each single carbon-14 atom has a 50% chance of decaying. Since a typical sample has a huge number of atoms 1, and since they decay more or less independently 2, we can statistically say, with a very high accuracy, that after 5,730 years half of all the original carbon-14 atoms will have decayed, while the rest still remain.
To answer your next natural question, no, this does not mean that the remaining carbon-14 atoms would be "just about to decay". Generally speaking,
atomic nuclei do not have a memory 3: as long as it has not decayed, a carbon-14 nucleus created yesterday is exactly identical to one created a year ago or 10,000 years ago or even a million years ago. All those nuclei, if they're still around today, have the same 50% probability of decaying within the next 5,730 years.
If you like, you could imagine each carbon-14 nucleus repeatedly tossing a
very biased imaginary coin very fast (faster than we could possibly measure): on each toss, with a very, very tiny chance, the coin comes up heads and the nucleus decays; otherwise, it comes up tails, and the nucleus stays together for now. Over a period of, say, a second or a day, the odds of any of the coin tosses coming up heads are still tiny — but, over 5,730 years, the many, many tiny odds gradually add up to a cumulative decay probability of about 50%. 1 A gram of carbon contains about 0.08 moles, or about 5 × 10 22 atoms. In a typical natural sample, about one in a trillion (1 / 10 12) of these will be carbon-14, giving us about 50 billion (5 × 10 10) carbon-14 atoms in each gram of carbon. 2 Induced radioactive decay does occur, most notably in fission chain reactions. Carbon-14, however, undergoes spontaneous β − decay, whose rate is not normally affected by external influences to any significant degree. 3 Nuclear isomers and other excited nuclear states do exist, so it's not quite right to say that all nuclei of a given isotope are always identical. Still, even these can, in practice, be effectively modeled as discrete states, with spontaneous transitions between different states occurring randomly with a fixed rate over time, just as nuclear decay events do.
I know exactly where you're coming from. If I can put it into my own words:
If it takes a sample some amount of time to decay, shouldn't a sample of half the size take half the time to decay? I have fallen into this seemingly sensical but somehow incorrect belief more than once.
Here's a graph that shows what I believe you're currently thinking.
The horizontal axis is time. On the vertical I graph amount of sample left. This graph would be true if half the sample took half the time to decay. (Can you see this in the graph? Look at $t=T/2$ where time $T$ is when the sample is gone.) I think this does make sense in a way, but this isn't how nature works.
Now here's a graph of what actually happens.
This graph is "exponentially decaying" It is a consequence of the following:
A sample half the size will decay at half the rate. This also makes sense in a way (thankfully): If you have half the sample size, you'll have half the decay rate. In contrast, note that the first graph has the constant rate of decay, no matter the size of the sample (that is, a constant slope).
So these two possibilities are mutually exclusive: Either the rate of decay is constant regardless of the size (first graph), or the rate of decay is proportional to the sample size (second graph). Observation shows the second graph is correct.
Logically, shouldn't it take 2,865 years for the quarter to decay, rather than 5,730?
Imagine that the quantity $q(n)$ of something decays as
$$q(n) = Q\cdot 2^{-n}$$
where $n$ is the number of half-lifes.
Initially, there is quantity $q(0) = Q\cdot 2^0 = Q$ of something.
After 1 half-life, there is $q(1) = Q \cdot 2^{-1} = \frac{Q}{2}$ remaining.
After 2 half-lifes, there is $q(2) = Q \cdot 2^{-2} = \frac{Q}{4}$ remaining.
After 3 half-lifes, there is $q(3) = Q \cdot 2^{-3} = \frac{Q}{8}$ remaining.
After 4 half-lifes...
Now, note that the quantity $\frac{q(n+1)}{q(n)} = \frac{1}{2}$ is
constant.
That is to say that, given a quantity
at any point in time (not just the "starting" point), one half-life later half of that quantity has decayed. This is the meaning of half-life.
From the Wikipedia article linked:
Half-life (t½) is the amount of time required for a quantity to fall to half its value as measured at the beginning of the time period.
Suppose you start with two kilograms of C-14. After 5730 years you have one kilogram left. Call that piece A. Now get another kilogram of C-14, call it piece B, and put it next to piece A.
You now have two identical pieces of C-14, and yet one of them (A) is supposed to half-decay in 2865 years and the other (B) is supposed to half-decay in 5730 years? Do you see how that doesn't make sense?
Hopefully this convinces you that the rate at which a radioactive element decays can only depend on how much of it there is at that moment, not on how much of the original sample is left.
As a crude analogy to give you some intuition, try the following: put 100 pennies in a shoebox, all with heads up. Shake the shoebox vigorously. Take out all the pennies that have changed to tails up. That's one half-life. Shake the box again, and again take out the pennies that are tails up. Repeat until there are no pennies left in the box.
The idea here is that heads up pennies represent carbon-14 atoms. The tails up pennies represent the atoms that have decayed. For any individual penny, each time you shake the box there is a 50/50 chance it will turn tails up, just as for every individual atom there is a 50-50 chance it will decay during one half-life period.
Half-life is used to describe
exponential decay. What you're describing would be linear decay.
In one half-life period,
on average, half of the C14 atoms would decay. So one would expect that if you start with four C14 atoms, you would after one half life have two, and after another half life only one would remain.
However, note that this process has a random component. You cannot predict exactly when an individual atom will decay. However once you have a larger number of atoms, you can make accurate predictions of how many will be left after a certain time period.
Basically, nuclear disintegration is probabilistic in its very nature. What it means is that one cannot say with conviction that, say, one atom kept on a table will disintegrate in, say, the next 1 minute. All one can say is that among a given sample of, say, 100 nuclei, 10% of it will disintegrate in the next 1 minute.
Nuclear disintegration follows what is known as first order kinetics which means that rate of reaction is directly proportional to the quantity of reactant present. In other words,
$$ d/dx(C) = -k C $$
where C is the current concentration of reactant and k is proportionality constant.
From this calculation, what one can get is a term called half-life, which means that after this time has elapsed, half of the concentration gets disintegrated (I'm using disintegrated and reacted interchangeably, since the reaction in nuclear disintegration is disintegration).
This means that a sample of 100 atoms after one half-life would remain 50 $=100 * (1/2)^1$, which after 2 half-lives would become 25 $=100 * (1/2)^2 = 50 * (1/2)^1$ and so on...
Suppose that decay worked the way you proposed, half as many atoms take half as long to decay. It sounds sort of plausible at first, but consider this: how does any one atom know when it's allowed to decay? It can't just roll a die and decay if it rolls a 1, it has to know how big the sample it's in is, and adjust its probability of decaying accordingly. If it didn't adjust its probability of decay, one would expect to exponential decay, because:
Thought experiment: roll a 20-sided die. If you roll a 1, it decays. How many rolls does it take to get to a 50% probability of decaying? (hint: it's not 10) How many rolls to get 100%?
Thought experiment 2: roll a hundred 20-sides dice. Any die that lands on 1 decays. How many do you likely have left after the first round? Should it take longer, on average, to make them all decay than if you had only 1?
Thought experiment 3: as many dice as there are atoms in a sizable chunk of material. How does it behave?
It should be clear that on average, you lose 5% of the dice
that you had left (not of how many you started out with, which is not something the system can remember) per round - exponential decay. The "half-life" of these dice is about 13.5 rounds, that's how long it takes before approximately half are decayed.
I think you're confused simply by the language. Remember that it's a quarter of the
original sample. So it's like compounding interest in the bank. You start with initial principal, once the interest is compounded, you might say that the percentage of that principal is ADDED TO the "principal", and then a percentage of THAT is calculated, and added to that second number. Similarly with nuclear decay, except you're subtracting, and you're subtracting an even half over a year instead of adding something like .05% every month (or whatever number banks use).
Half of that second sample is a quarter of the original. So you could express this fraction of the original as $\frac{1}{2^n}$ where $n = $the unit of time for your constant. In this case, a year. So for every year, $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}$, etc.
The mass of radioactive materials follows the ordinary differential equation: $$ m'(t)=-am(t), $$ where $m$ is the mass and $a$ a positive constant - i.e., constant relative rate of decay.
This implies $$m(t)=m(0)\mathrm{e}^{-at}. \tag{1}$$If $T_h$ is half life, then $$m(T_h)=m(0)\mathrm{e}^{-aT_h}=\frac{1}{2}m(0),$$which implies that$$T_h=\frac{\log 2}{a},$$and hence $(1)$ can be written also as$$m(t)=2^{-t/T_h}m(0).$$So the
quarter-life is $T_Q$, for which $m(T_Q)=\frac{1}{4}m(0)$ or$$m(T_Q)=2^{-T_Q/T_h}m(0)=\frac{1}{4}m(0),$$which holds only if $T_Q=2T_h$!
Imagine a sample of 1000 atoms with a half-life of 1 hour.
That means every hour, the sample is reduced to 50% of its size.
After one hour, you are left with 500 atoms. How much time for that new sample (500 atoms) to be reduced to 50% (250 atoms) ?
In your interpretation :
For the new sample to be reduced to 50%, it needs to lose 250 atoms. Since it lost 500 atoms in 1hour, it should take 30 minutes to lose 250 atoms. And that's where you're wrong. It still needs 1 hour for half of the atoms to decay.
You assume that the number of atoms decaying by time (500/hour) is constant, but it isn't.
What is constant is the probability for each atom decaying in an hour : 50% (In this exemple, that means we can expect 250 atoms to have decayed after 1 hour, and it gets way more precise with "real" atoms number on a longer period)
putting it simply: activity law states that :
dn/dt is proportional to n. which means
the rate of reaction of any substance depends on the amount of the substance itself. because there is a greater amount of carbon initially, the probability of some of it decaying is higher , than when there is less amount of it left.
Half life 1=5,730 years while the ratio is 1:1
Half life 2= 11,460 years while the ratio is 1:3
Half life 3= 17,190 years while the ratio is 1:7
Half life 4= 22,920 years while the ratio is 1:15
To get the 2-4 half lives, you just add the 5,730 each time.
So how I got the second half life you would do 5,730+5,730=11,460
protected by AccidentalFourierTransform Sep 19 at 0:27
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An object is dropped from a building of height $h\,\mathrm m$. The object falls half the total distance to the ground during the last $1.00\,\mathrm s$ of its fall. Find the height of the building. Take the downward direction to be negative and $9.80\,\frac{\mathrm m}{\mathrm s^2}$ to be the magnitude of the acceleration due to gravity.
Let $t,y,y_0,v_y,{v_y}_0$ respectively denote the time (in seconds), vertical displacement (in meters), the initial displacement, the vertical velocity (in meters per second), the initial velocity of the object. Let $a$ denote the acceleration's magnitude as above.
I've split up the drop into two time intervals, $0\le t\le T$ and $T\le t\le T+1$, where $T$ is the time it takes for the object to fall half the building's height from when it is dropped.
In the first time interval, I see that
$$\begin{cases}y_0=0\\{v_y}_0=0\end{cases}\implies\begin{cases}v_y=-at\\y=-\frac a2t^2\end{cases}$$
At the end of the first time interval, half the building's height is traversed, so that
$$-\dfrac h2=-\frac a2T^2\implies h=aT^2\quad(*)$$
In the second time interval, it seems clear to me that
$$\begin{cases}y_0=-\frac h2\\{v_y}_0=-aT\end{cases}\implies y=-\frac h2-aTt-\frac a2t^2$$
so that when the object hits the ground at $t=T+1$, I have
$$-h=-\frac h2-aT(T+1)-\frac a2(T+1)^2\implies h=a(3T^2+4T+1)\quad(**)$$
Equations $(*)$ and $(**)$ suggest that
$$3T^2+4T+1=T^2\implies T\approx-1.71,T\approx-0.293$$
which doesn't make sense to me. The time $T$ that elapses throughout the object's fall should be positive.
The solution is supposed to be
$h\approx57.1$ meters
which according to $(*)$ should correspond to $T\approx2.41$ seconds; on the other hand, $(**)$ would suggest $T\approx0.766$ seconds.
I'm fairly certain $(**)$ is incorrect, but I don't know what mistake I made in its derivation.
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Circular motion in special relativity is somewhat tricky: Note that for circular motion, the acceleration in the spaceship travelling in a circle is
not zero, so the spaceship is not in a single frame of inertia.
Here is an interesting thought: Distances
perpendicular to the direction of motion are not subject to contraction. Hence, if the observes on earth see the spaceship going on a circle with radius $R$, then in their frame of inertia the spaceship is always a distance $R$ away from the earth. Since the line from earth to spaceship is perpendicular to the direction of flight, the people on the spaceship will also believe that they are always a distance $R$ away from earth, so they will also fly on a circle.
They will, nonetheless, experience a different circumference! The best way to solve this is to consider a polygon with $N$ sides and then let $N$ go to infinity. If people on the earth measure each side of the polygon as $L_0/N$ where $L_0$ is the circumference of the polygon in the earth's frame of inertia, then the spaceship-people will measure each side to be $L_0/(N\gamma)$. Hence, for $N \rightarrow \infty$, the polygon becomes a circle. Measured from earth, it has circumference $L_0$, but for the spaceship it has circumference $L_0/\gamma$.
This suggests that the spaceship moves through non-Euclidean geometry, because it travels on a circle whose ratio between circumference and diameter is less than $2\pi$. This is a hint that accelerating frames have non-Euclidean geometry, which is excessively treated in General Relativity.
Reference: http://abacus.bates.edu/~msemon/WortelMalinSemon.pdf
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I am studying through the book Thermodynamics and Statistical Mechanics by Walter Greiner and I’ve got a couple of doubts when I was reading about the classical ensembles, specially the Canonical ensemble (chapter 7, pages 159-160, Springer, 2004).
In the case of the canonical ensemble it is considered that the system has a fixed temperature due to the contact with the termal reservoir and it can assume a large range of values of energy (I think, but I am not sure, that these different values of energy are attained by the system – in termal contact with the reservoir – until the final equilibrium state is attained.)
In the page 160, about the probability $p_{i}$ of finding the system $S$ in a microstate $i$ with an energy $E_{i}$, it is told:
“If $S$ is a closed system, $p_{i}$ will be proportional to the number of microstates ${\Omega}_{S}(E_{i})$. Analogously, $p_{i}$ is proprotional to the number of microstates in the total closed system for which $S$ lies in the microstate with the energy $E_{i}$. Obviously this is just equal to the number of microstates of the heat bath for the energy$E–E_{i}$, since $S$ assumes one microstate $i$ [...]”
What I don’t understand is: How “obviously” is that? Because to me it’s not obvious.
After that, I don’t understand too how to get the Equation (7.4), since we the expansion is “with respect to $E_{i}$”.
\begin{equation} k{\ln}{\Omega}_{R}(E-E_{i}){\approx}k{\ln}{\Omega}_{R}(E)-{\frac{\partial}{{\partial}E}}(k{\ln}{\Omega}_{R}(E))E_{i}+..., \end{equation} where $E$ is the total energy, $E_{R}$ is the energy of the reservoir, $E_{S}=E_{i}$ is the energy of the system in a given microstate and $E=E_{R}+E_{S}$.
If anybody could help, I really would appreciate.
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Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE
(Elsevier, 2017-11)
Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ...
Investigations of anisotropic collectivity using multi-particle correlations in pp, p-Pb and Pb-Pb collisions
(Elsevier, 2017-11)
Two- and multi-particle azimuthal correlations have proven to be an excellent tool to probe the properties of the strongly interacting matter created in heavy-ion collisions. Recently, the results obtained for multi-particle ...
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Computing the normal from vertex positions is quite simple using the vector cross product.
The cross product of two vectors $u$ and $v$ (noted $u \times v$, or sometimes $u \wedge v$) is a vector perpendicular to $u$ and $v$, of length $||u \times v|| = ||u|| \cdot ||v|| sin(\theta)$, with $\theta$ the angle between $u$ and $v$. The direction of the vector will depend on the order of the multiplication: $u \times v$ is the opposite of $v \times u$ (the two directions perpendicular to the plane).
If you are not familiar with the cross product, I invite you to read about it and get comfortable with it. Normals will then seem simple.
Flat shading normals
If you have a triangle $ABC$, $AB \times AC$ is a vector perpendicular to the triangle and with a length proportional to its area. Since the normal is the unit vector perpendicular to the triangle's plane, you can get the normal with:
$N = \dfrac{AB \times AC}{||AB \times AC||}$
In code, this would look like
n = normalize(cross(b-a, c-a)) for example. Just apply this over all your faces and you will have your normals per face.
For each triangle ABC
n := normalize(cross(B-A, C-A))
A.n := n
B.n := n
C.n := n
Note that this assumes vertices are not shared between triangles. I am not familiar with the Kinect API; it's quite possible that they are shared, in which case you would have to duplicate them, or move on to the next solution:
Smooth shading normals
After lighting with normals computed as above, you will notice that triangle edges are apparent. If this in not desirable, you can compute smooth normals instead, by taking into account all the faces that share a same vertex.
The idea is that if a same vertex is shared by three triangles $T1$, $T2$ and $T3$ for example, the normal $N$ will be the average of $N1$, $N2$ and $N3$. Moreover, if $T1$ is a big triangle and $T2$ is a tiny one, you probably want $N$ to be more influenced by $N1$ than by $N2$.
Remember how the cross product is proportional to the area? If you add up the cross products then normalize the sum, it will do exactly the weighted sum we want. So the algorithm becomes:
For each vertex
vertex.n := (0, 0, 0)
For each triangle ABC
// compute the cross product and add it to each vertex
p := cross(B-A, C-A)
A.n += p
B.n += p
C.n += p
For each vertex
vertex.n := normalize(vertex.n)
This technique is explained in longer detail this article by Iñigo Quilez: clever normalization of a mesh.
For more on normals, see also:
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If I have the hamiltonian of the simple harmonic oscillator
$$H = \frac{p^2}{2m} + \frac{1}{2} m \omega ^2 x^2 $$
Then it's partition function is:
$$Z = \frac{k_b T}{\hbar \omega} $$
You can get the average energy using $$U = \frac{\partial \ln{Z}}{\partial \beta} $$
where $\beta = k_b T $. in 1D , $$ U = kb * T $$
My question is, can you integrate U back up to obtain the original partition function, $Z = \frac{k_b T}{\hbar \omega} $? The closest I get is
$$\int U d\beta = \beta (T) - \beta(0) = k_b T - \hbar \omega $$
where $\hbar \omega $ is the zero point energy. The problem is that I then have to exponentiate to recover $Z$, and I get no where close to the original partition function. Any advice here?
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Now showing items 1-10 of 33
The ALICE Transition Radiation Detector: Construction, operation, and performance
(Elsevier, 2018-02)
The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ...
Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2018-02)
In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ...
First measurement of jet mass in Pb–Pb and p–Pb collisions at the LHC
(Elsevier, 2018-01)
This letter presents the first measurement of jet mass in Pb-Pb and p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV and 5.02 TeV, respectively. Both the jet energy and the jet mass are expected to be sensitive to jet ...
First measurement of $\Xi_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV
(Elsevier, 2018-06)
The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The ...
D-meson azimuthal anisotropy in mid-central Pb-Pb collisions at $\mathbf{\sqrt{s_{\rm NN}}=5.02}$ TeV
(American Physical Society, 2018-03)
The azimuthal anisotropy coefficient $v_2$ of prompt D$^0$, D$^+$, D$^{*+}$ and D$_s^+$ mesons was measured in mid-central (30-50% centrality class) Pb-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm ...
Search for collectivity with azimuthal J/$\psi$-hadron correlations in high multiplicity p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 and 8.16 TeV
(Elsevier, 2018-05)
We present a measurement of azimuthal correlations between inclusive J/$\psi$ and charged hadrons in p-Pb collisions recorded with the ALICE detector at the CERN LHC. The J/$\psi$ are reconstructed at forward (p-going, ...
Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(American Physical Society, 2018-02)
The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ...
$\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV
(Springer, 2018-03)
An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ...
J/$\psi$ production as a function of charged-particle pseudorapidity density in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2018-01)
We report measurements of the inclusive J/$\psi$ yield and average transverse momentum as a function of charged-particle pseudorapidity density ${\rm d}N_{\rm ch}/{\rm d}\eta$ in p-Pb collisions at $\sqrt{s_{\rm NN}}= 5.02$ ...
Energy dependence and fluctuations of anisotropic flow in Pb-Pb collisions at √sNN=5.02 and 2.76 TeV
(Springer Berlin Heidelberg, 2018-07-16)
Measurements of anisotropic flow coefficients with two- and multi-particle cumulants for inclusive charged particles in Pb-Pb collisions at 𝑠NN‾‾‾‾√=5.02 and 2.76 TeV are reported in the pseudorapidity range |η| < 0.8 ...
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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 ...
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that the electron and the proton are a particle pair
This may appear inconsistent with current theory, as it is widely assumed that particle pairs created in a single event, must have opposite charge and equal mass (matter/antimatter). This assumption turns out to be wrong.
We shall see that the difference in mass between the electron and the proton is due to mass defect, the same kind of mass defect as seen in heavy nuclei which is well understood. The electron is trapped in it's own negative electric potential and dispite having the same mass as the proton, appears to the observer as a smaller particle.
It was obvious from these assumptions that the observers potential (Ground Potential) had to lie somewhere between the electrical potential of the proton and the potential of the electron, and the reason is simply that we know of no charge more positive than the proton and likewise no charge more negative than the electron, ergo everything else must lie in-between.
This inspired me to go after the equation describing the relationship between electron-potential, ground-potential and proton-potential, it was tricky, but the solution presented itself to me within a couple of days of working on it.
\[ a (\gamma) = \frac{1}{2} (c-b) \]
Where 'a' is electon-potential, 'b' is ground-potential and 'c' is proton-potential.
The problem here is \(\gamma\) being the same kind of \(\gamma\) as in Einstein's relativity, being velocity dependent, so how does one define gamma?
After some further thinking it became apparent that four-velocity was a function of potential and since the proton represented the absolute maximum potential, it could be concidered as a physical constant in the same way as the speed of light, so the full equation now became;
\[ a = \frac{(c-b)}{2} \sqrt{1-\frac{b^2}{c^2}} \]
So here we have the first law of ground potential, a very simple and elegant statement for the first time defining ground potential.
Solving the above equation with known values for electron and proton potential gives a ground potential of 930 million volts.
I have chosen volts as the ideal unit to work in, because it is the SI unit which moves one elementary charge 1 meter in 1 second, so it is the perfect unit to define the potentials of electrons and protons.
A proton which has a mass of \(\frac{938 MeV}{c^2}\) contains 938 MeV of energy which divided by one elementary charge gives a potential of 938 million volts. Likewise the electron can be said to have 0.511 million volts potential.
This first law of GP shall be the law upon which all other laws shall rest.
Steven
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We look here at the continuity of a sequence of functions that converges pointwise and give some
counterexamples of what happens versus uniform convergence. Recalling the definition of pointwise convergence
We consider here real functions defined on a closed interval \([a,b]\). A sequence of functions \((f_n)\) defined on \([a,b]\) converges pointwise to the function \(f\) if and only if for all \(x \in [a,b]\) \(\displaystyle \lim\limits_{n \to +\infty} f_n(x) = f(x)\). Pointwise convergence is weaker than uniform convergence.
Pointwise convergence does not, in general, preserve continuity
Suppose that \(f_n \ : \ [0,1] \to \mathbb{R}\) is defined by \(f_n(x)=x^n\). For \(0 \le x <1\) then \(\displaystyle \lim\limits_{n \to +\infty} x^n = 0\), while if \(x = 1\) then \(\displaystyle \lim\limits_{n \to +\infty} x^n = 1\). Hence the sequence \(f_n\) converges to the function equal to \(0\) for \(0 \le x < 1\) and to \(1\) for \(x=1\).Although each \(f_n\) is a continuous function of \([0,1]\), their pointwise limit is not. \(f\) is discontinuous at \(1\).We notice that \((f_n)\) doesn't converge uniformly to \(f\) as for all \(n \in \mathbb{N}\), \(\displaystyle \sup\limits_{x \in [0,1]} \vert f_n(x) - f(x) \vert = 1\). That's reassuring as uniform convergence of a sequence of continuous functions implies that the limit is continuous! Continue reading Pointwise convergence and properties of the limit (part 1)
In that article, I described some properties of
Thomae’s function\(f\). Namely: The function is discontinuous on \(\mathbb{Q}\). Continuous on \(\mathbb{R} \setminus \mathbb{Q}\). Its right-sided and left-sided limits vanish at all points.
Let’s modify \(f\) to get function \(g\) defined as follow:
\[g: \left|\begin{array}{lrl} \mathbb{R} & \longrightarrow & \mathbb{R} \\ x & \longmapsto & 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \\ \frac{p}{q} & \longmapsto & q \text{ if } \frac{p}{q} \text{ in lowest terms and } q > 0 \end{array}\right.\] \(f\) and \(g\) both vanish on the set of irrational numbers, while on the set of rational numbers, \(g\) is equal to the reciprocal of \(f\). We now consider an open subset \(O \subset \mathbb{R}\) and \(x \in O\). As \(f\) right-sided and left-sided limits vanish at all points, we have \(\lim\limits_{n \to +\infty} f(x_n) = 0\) for all sequence \((x_n)\) of rational numbers converging to \(x\) (and such a sequence exists as the rational numbers are everywhere dense in the reals). Hence \(\lim\limits_{n \to +\infty} g(x_n) = + \infty\) as \(f\) is positive.
We can conclude that \(g\) is nowhere locally bounded.
The picture of the article is a plot of function \(g\) on the rational numbers \(r = \frac{p}{q}\) in lowest terms for \(0 < r < 1\) and \(q \le 50\).
Let’s discover the beauties of
Thomae’s function also named the popcorn function, the raindrop function or the modified Dirichlet function.
Thomae’s function is a real-valued function defined as: \[f: \left|\begin{array}{lrl} \mathbb{R} & \longrightarrow & \mathbb{R} \\ x & \longmapsto & 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \\ \frac{p}{q} & \longmapsto & \frac{1}{q} \text{ if } \frac{p}{q} \text{ in lowest terms and } q > 0 \end{array}\right.\] \(f\) is periodic with period \(1\)
This is easy to prove as for \(x \in \mathbb{R} \setminus \mathbb{Q}\) we also have \(x+1 \in \mathbb{R} \setminus \mathbb{Q}\) and therefore \(f(x+1)=f(x)=0\). While for \(y=\frac{p}{q} \in \mathbb{Q}\) in lowest terms, \(y+1=\frac{p+q}{q}\) is also in lowest terms, hence \(f(y+1)=f(y)=\frac{1}{q}\). Continue reading A function continuous at all irrationals and discontinuous at all rationals
From the mean value theorem, a real function whose
derivative is strictly positive at every point of an interval is strictly increasing. In particular, a continuously differentiable function \(f\) defined in a non-degenerate interval \(I\) with a strictly positive derivative at a point \(a\) of the interval is strictly increasing near that point. For the proof, we just have to notice that as \(f^\prime\) is continuous and \(f^\prime(a) > 0\), \(f^\prime\) is strictly positive within an interval \(J \subset I\) containing \(a\). By the mean value theorem, \(f\) is strictly increasing on \(J\).
We now suppose that \(f\) is differentiable on an interval \(I\) containing \(0\) with \(f^\prime(0)>0\). For \(x>0\) sufficiently close to zero we have \(\displaystyle \frac{f(x)-f(0)}{x-0} > \frac{f^\prime(0)}{2}>0\), hence \(f(x)>f(0)\). But that doesn’t imply that \(f\) is strictly increasing in a neighborhood of zero. Let’s prove it with a
counterexample. Continue reading A function whose derivative at 0 is one but which is not increasing near 0
In that article, I gave examples of real valued functions defined on \((0,+\infty)\) that converge to zero and whose derivatives diverge. But those functions were not monotonic. Here I give an example of a decreasing real valued function \(g\) converging to zero at \(+\infty\) and whose derivative is unbounded.
We first consider the polynomial map:
\[P(x)=(1+2x)(1-x)^2=1-3x^2+2x^3\] on the segment \(I=[0,1]\). \(P\) derivative equals \(P^\prime(x)=-6x(1-x)\). Therefore \(P\) is decreasing on \(I\). Moreover we have \(P(0)=1\), \(P(1)=P^\prime(0)=P^\prime(1)=0\) and \(P^\prime(1/2)=-3/2\). Continue reading A decreasing function converging to zero whose derivative diverges (part2)
In this article, I consider real valued functions \(f\) defined on \((0,+\infty)\) that converge to zero, i.e.:
\[\lim\limits_{x \to +\infty} f(x) = 0\] If \(f\) is differentiable what can be the behavior of its derivative as \(x\) approaches \(+\infty\)?
Let’s consider a first example:
\[\begin{array}{l|rcl} f_1 : & (0,+\infty) & \longrightarrow & \mathbb{R} \\ & x & \longmapsto & \frac{1}{x} \end{array}\] \(f_1\) derivative is \(f_1^\prime(x)=-\frac{1}{x^2}\) and we also have \(\lim\limits_{x \to +\infty} f_1^\prime(x) = 0\). Let’s consider more sophisticated cases! Continue reading Differentiable functions converging to zero whose derivatives diverge (part1)
In that article, I provide basic counterexamples on sequences convergence. I follow on here with some additional and more advanced examples. If \((u_n)\) converges then \((\vert u_n \vert )\) converge?
This is true and the proof is based on the
reverse triangle inequality: \(\bigl| \vert x \vert – \vert y \vert \bigr| \le \vert x – y \vert\). However the converse doesn’t hold. For example, the sequence \(u_n=(-1)^n\) is such that \(\lim \vert u_n \vert = 1\) while \((u_n)\) diverges. If for all \(p \in \mathbb{N}\) \(\lim\limits_{n \to +\infty} (u_{n+p} – u_n)=0\) then \((u_n)\) converges?
The assertion is wrong. A simple counterexample is \(u_n= \ln(n+1)\). It is well known that \((u_n)\) diverges. However for any \(p \in \mathbb{N}\) we have \(\lim\limits_{n \to +\infty} (u_{n+p} – u_n) =\ln(1+\frac{p}{n+1})=0\).
The converse proposition is true. Assume that \((u_n)\) is a converging sequence with limit \(l\) and \(p \ge 0\) is any integer. We have \(\vert u_{n+p}-u_n \vert = \vert (u_{n+p}-l)-(u_n-l) \vert \le \vert u_{n+p}-l \vert – \vert u_n-l \vert\) and both terms of the right hand side of the inequality are converging to zero. Continue reading Counterexamples on real sequences (part 2)
The purpose of this article is to provide some basic
counterexamples on real sequences. Counterexamples are provided as answers to questions.
I will come back later on with more complex cases. Unless otherwise stated, \((u_n)_{n \in \mathbb{N}}\) and \((v_n)_{n \in \mathbb{N}}\) are two real sequences. Continue reading Counterexamples on real sequences (part 1)
In this post, I mention that Peano existence theorem is valid for finite dimensional vector spaces, but not for Banach spaces of infinite dimension. I highlight here a second property of ordinary differential equations which is valid for finite dimensional vector spaces but not for infinite dimensional Banach spaces. Continue reading A solution of a differential equation not exploding in finite time
We build here a
continuous function of one real variable whose derivative exists on \(\mathbb{R} \setminus \mathbb{Q}\) and doesn’t have a left or right derivative on each point of \(\mathbb{Q}\).
As \(\mathbb{Q}\) is (infinitely)
countable, we can find a bijection \(n \mapsto r_n\) from \(\mathbb{N}\) to \(\mathbb{Q}\). We now reuse the function \(f\) defined here. Recall \(f\) main properties: Continue reading A continuous function not differentiable at the rationals but differentiable elsewhere
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"Abstract: We prove, once and for all, that people who don't use superspace are really out of it. This includes QCDers, who always either wave their hands or gamble with lettuce (Monte Zuma calculations). Besides, all nonsupersymmetric theories have divergences which lead to problems with things like renormalons, instantons, anomalons, and other phenomenons. Also, they can't hide from gravity forever."
Can a gravitational field possess momentum? A gravitational wave can certainly possess momentum just like a light wave has momentum, but we generally think of a gravitational field as a static object, like an electrostatic field.
"You have to understand that compared to other professions such as programming or engineering, ethical standards in academia are in the gutter. I have worked with many different kinds of people in my life, in the U.S. and in Japan. I have only encountered one group more corrupt than academic scientists: the mafia members who ran Las Vegas hotels where I used to install computer equipment. "
so Ive got a small bottle that I filled up with salt. I put it on the scale and it's mass is 83g. I've also got a jup of water that has 500g of water. I put the bottle in the jug and it sank to the bottom. I have to figure out how much salt to take out of the bottle such that the weight force of the bottle equals the buoyancy force.
For the buoyancy do I: density of water * volume of water displaced * gravity acceleration?
so: mass of bottle * gravity = volume of water displaced * density of water * gravity?
@EmilioPisanty The measurement operators than I suggested in the comments of the post are fine but I additionally would like to control the width of the Poisson Distribution (much like we can do for the normal distribution using variance). Do you know that this can be achieved while still maintaining the completeness condition $$\int A^{\dagger}_{C}A_CdC = 1$$?
As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including:ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room.An altern...
You're always welcome to ask. One of the reasons I hang around in the chat room is because I'm happy to answer this sort of question. Obviously I'm sometimes busy doing other stuff, but if I have the spare time I'm always happy to answer.
Though as it happens I have to go now - lunch time! :-)
@JohnRennie It's possible to do it using the energy method. Just we need to carefully write down the potential function which is $U(r)=\frac{1}{2}\frac{mg}{R}r^2$ with zero point at the center of the earth.
Anonymous
Also I don't particularly like this SHM problem because it causes a lot of misconceptions. The motion is SHM only under particular conditions :P
I see with concern the close queue has not shrunk considerably in the last week and is still at 73 items. This may be an effect of increased traffic but not increased reviewing or something else, I'm not sure
Not sure about that, but the converse is certainly false :P
Derrida has received a lot of criticism from the experts on the fields he tried to comment on
I personally do not know much about postmodernist philosophy, so I shall not comment on it myself
I do have strong affirmative opinions on textual interpretation, made disjoint from authoritorial intent, however, which is a central part of Deconstruction theory. But I think that dates back to Heidegger.
I can see why a man of that generation would be leaned towards that idea. I do too.
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Possible Duplicate: Showing that an inclusion is null homotopic
I asked this question here a while ago. Meanwhile I've come up with the following and was wondering if you could have a look and tell me if it's correct.
claim: $X$ homotopy equivalent to a point $\ast$ $\implies$ for all neighbourhoods $U$ of $\ast$ there exists a neighbourhood $V$ of $\ast$ such that the inclusion $i: V \hookrightarrow U$ is null homotopic. proof:
Let $h_t :id_X \simeq c$ where $c(x) = \ast$ denote the homotopy.
Let $U$ be a neighbourhood of $\ast$. This means there exists an open set $O$ such that $\ast \in O \subset U$.
claim: $i: O \hookrightarrow U$ is null homotopic.
proof: restrict $h_0$ to $O$: $h_0 |_O$ then $i = i \circ h_0|_O \simeq i \circ c = c$.
Thanks for your help.
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This is an old revision of the document!
If you need a pdf version of these notes you can get it here
The 2014 video was corrupted..again. I have made some changes on my computer and hopefully the problem will not happen again..
In a transformer two coils are coupled by an iron core so that the flux through the two coils is the same. The iron core is usually layered with insulating materials to prevent eddy currents. When an AC voltage is applied to the primary coil the magnetic flux passing through it is related to the applied field by
$V_{P}=N_{P}\frac{d\Phi_{B}}{dt}$
if we assume the coil has no resistance. The emf produced by the changing flux is $\mathcal{E}=-N_{P}\frac{d\Phi_{B}}{dt}$ and this means that the net potential drop in the coil is actually zero (as Kirchoff's rules say it must be if the resistance of the coil is zero). The voltage induced in the secondary coil will have magnitude
$V_{S}=N_{S}\frac{d\Phi_{B}}{dt}$
We can thus see that
$\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}}$
If we assume there is no power loss (which is fairly accurate) then $I_{P}V_{P}=I_{S}V_{S}$ and
$\frac{I_{S}}{I_{P}}=\frac{N_{P}}{N_{S}}$
In the example of a transformer we assumed that the flux induced in the secondary coil was equal to the flux in the primary coil. In general for two coils the relationship between the flux in one coil due to the current in another is described by a parameter called the mutual inductance.
$\Phi_{21}$ is the magnetic flux in each loop of coil 2 created by the current in coil 1. The total flux in the second coil is then $N_{2}\Phi_{21}$ and is related to the current in coil 1, $I_{1}$ by
$N_{2}\Phi_{21}=M_{21}I_{1}$
As, from Faraday's Law, the emf induced in coil 2 is $\mathcal{E}_{2}=-N_{2}\frac{d\Phi_{21}}{dt}$ so
$\mathcal{E}_{2}=-M_{21}\frac{dI_{1}}{dt}$
The mutual inductance of coil 2 with respect to coil 1, $M_{21}$ does not depend on $I_{1}$, but it does depend on factors such as the size, shape and number of turns in each coil, their position relative to each other and whether there is some ferromagnetic material in the vicinity.
In the reverse situation where a current flows in coil 2
$\mathcal{E}_{1}=-M_{12}\frac{dI_{2}}{dt}$
but in fact $M_{12}=M_{21}=M$
The mutual inductance is measured in Henrys ($\mathrm{H}$), $1\mathrm{H}=1\mathrm{\frac{Vs}{A}}=1\mathrm{\Omega s}$
To calculate the mutual inductance of the above situation we consider the magnetic field of a solenoid with cross-sectional area $A$
$B=\mu_{0}\frac{N_{1}}{l}I_{1}$
The magnetic flux through the loose coil due to the current in the solenoid is thus
$\Phi_{21}=BA=\mu_{0}\frac{N_{1}}{l}I_{1}A$
and the mutual inductance is then
$M=\frac{N_{2}\Phi_{21}}{I_{1}}=\frac{\mu_{0}N_{1}N_{2}A}{l}$
If we now consider a coil to which a time varying current is applied a changing magnetic field is produced, which induces an emf in the coil, which opposes the change in flux. The magnetic flux $\Phi_{B}$ passing through the coil is proportional to the current, and as we did for mutual inductance we can define a constant of proportionality between the current and the flux, the self-inductance $L$
$N\Phi_{B}=LI$
The emf $\mathcal{E}=-N\frac{d\Phi_{B}}{dt}=-L\frac{dI}{dt}$
The self-inductance is also measured in henrys.
A component in a circuit that has significant inductance is shown by the symbol.
When we draw this symbol it implies an inductor with negligible resistance (if the inductor has significant resistance we draw that as a resistor in series with the inductor). A large inductor does however reduce the AC current flowing through a circuit because the back emf generated opposes the applied potential so that the total potential across the inductor is small, and the current is also small. The degree to which an inductor opposes an AC current is called the reactance (more on that later!) .
We can can calculate the self-inductance of a solenoid from it's field
$B=\mu_{0}\frac{NI}{l}$
The flux in the solenoid is
$\Phi_{B}=BA=\mu_{0}\frac{N_{1}IA}{l}$
so
$L=\frac{N\Phi_{B}}{I}=\frac{\mu_{0}N^{2}A}{l}$
To find the inductance we need to know the total flux that is generated by the current.
$\Phi_{B}=\int\vec{B}\cdot d\vec{A}$
From Ampere's law ($\oint\vec{B}\cdot d\vec{l}=\mu_{0}I$)
$B=\frac{\mu_{0}I}{2\pi r}$
The magnetic flux through a rectangle of width $dr$ and length $l$ at a distance $r$ from the center
$d\Phi_{B}=B(l\,dr)=\frac{\mu_{0}I}{2\pi r}l\,dr$
The total flux is
$\Phi_{B}=\int d\Phi_{B}=\frac{\mu_{0}Il}{2\pi}\int_{r_{1}}^{r_{2}}\frac{dr}{r}=\frac{\mu_{0}Il}{2\pi}\ln\frac{r_{2}}{r_{1}}$
so the inductance is
$L=\frac{\Phi_{B}}{I}=\frac{\mu_{0}l}{2\pi}\ln\frac{r_{2}}{r_{1}}$
and the inductance per unit length is
$\frac{L}{l}=\frac{\mu_{0}}{2\pi}\ln\frac{r_{2}}{r_{1}}$
When a time varying current $I$ is being carried in an inductor $L$ the power being supplied to the inductor is
$P=I\mathcal{E}=LI\frac{dI}{dt}$
The amount of work done in a time $dt$ is
$dW=P\,dt=LI\,dI$
so the work done in increasing the current from zero to $I$ is
$W=\int dW=\int_{0}^{I}LI\,dI=\frac{1}{2}LI^{2}$
The work done in going from zero to $I$ is equivalent to the energy stored in the magnetic field
$U=\frac{1}{2}LI^{2}$
The formula $U=\frac{1}{2}LI^{2}$ can be applied to a solenoid ($L=\frac{\mu_{0}N^{2}A}{l}$ and $B=\mu_{0}\frac{NI}{l}$ )
$U=\frac{1}{2}\frac{\mu_{0}N^{2}A}{l}(\frac{Bl}{\mu_{0}N})^{2}=\frac{1}{2}\frac{B^{2}}{\mu_{0}}Al$
As the volume of the solenoid is $Al$ the energy density
$u=\frac{1}{2}\frac{B^{2}}{\mu_{0}}$
This formula is actually generally applicable to any magnetic field, although when we consider energy in a magnetic medium
$u=\frac{1}{2}\frac{B^{2}}{\mu}$
where $\mu$ is the permeability of the material.
This can be compared to the energy density of an electric field
$u=\frac{1}{2}\varepsilon_{0}E^{2}$
or
$u=\frac{1}{2}K\varepsilon_{0}E^{2}$
in a dielectric material with dielectric constant K.
When we take a resistor an inductor in series and connect it to a battery then Kirchoff's loop rule tells us that
$V_{0}-IR-L\frac{dI}{dt}=0$
which we can rearrange and integrate
$\int_{I=0}^{I}\frac{dI}{V_{0}-IR}=\int_{0}^{t}\frac{dt}{L}$
$-\frac{1}{R}\ln(\frac{V_{0}-IR}{V_{0}})=\frac{t}{L}$
$I=\frac{V_{0}}{R}(1-e^{-t/\tau})=I_{0}(1-e^{-t/\tau})$ where $\tau=\frac{L}{R}$
If we then switch back to the closed loop that does not include the battery then Kirchoff's loop rule gives us
$L\frac{dI}{dt}+RI=0$
$\int_{I_{0}}^{I}\frac{dI}{I}=-\int_{0}^{t}\frac{R}{L}dt$
$\ln\frac{I}{I_{0}}=-\frac{R}{L}t$
$I=I_{0}e^{-t/\tau}$
We can compare this to RC circuits and we see that the both have currents exponentially related to time, but the behavior is not quite the same, in our next lecture we will start to look at AC circuits where the time dependence of the current in inductors and capacitors is important in determining the AC behavior of a circuit.
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Suppose $W$ consists of all vectors $(x,y,z)$ such that $x + 2y - 3z = 0$. Which of the following is a basis for the orthogonal complement?
The answer is $(1,3,-2)$ but I don't understand that at all. How do I get to that answer?
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Suppose $W$ consists of all vectors $(x,y,z)$ such that $x + 2y - 3z = 0$. Which of the following is a basis for the orthogonal complement?
The answer is $(1,3,-2)$ but I don't understand that at all. How do I get to that answer?
This question appears to be off-topic. The users who voted to close gave this specific reason:
You have a plane $$ x + 2y - 3z = 0$$
which is a two dimensional subspace of your three dimensional space.
The orthogonal complement is a one dimensional subspace which is apanned by a vector perpendicular to the plane.
The normal vector to your plane $$N= (1,2,-3) $$is such a vector.
Thus the basis for yor the orthogonal complement is $$ B=\{ (1,2,-3)\}$$
Because linearity of inner product, orthogonal complement of a finite dimensional subespace can be obtained by its basis. A basis B for W is $B=(-2,1,0),(3,0,1)$
Then If $(x,y,z) \in W^{\perp}$ is orthogonal to every vector in B $$ ((x,y,z),(-2,1,0))=0$$ $$ ((x,y,z),(3,0,1))=0$$ Therefore $$ -2x+y=0 $$ $$ 3x+z=0 $$ It means a basis for orthogonal complement is $$ (1,2,-3) $$
A basis for this plane is any pair of linearly independent vectors that start at a point in it and end at another point in it. So, consider two points in the plane $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$, then a vector in the plane would be $(x_2-x_1,y_2-y_1,z_2-z_1).$
Since both points are the plane, we have that
$$z_1 = \frac{x_1+2y_1}{3}\quad\text{and}\quad z_2 = \frac{x_2+2y_2}{3}$$
so that
$$z_2-z_1 = \frac{x_2-x_1+2(y_2-y_1)}{3}.$$
This is the only restriction that the vectors must satisfy. We need to pick two linearly independent vectors that satisfy this restriction. A simple way to do that is to set $x_2-x_1=1$ and $y_2-y_1=0$ for the first one, so that
$$z_2-z_1 = \frac{1+2(0)}{3}=\frac{1}{3}.$$
and $x_2-x_1=0$ and $y_2-y_1=1$ for the second one, so that
$$z_2-z_1 = \frac{0+2(1)}{3}=\frac{2}{3}.$$
It follows that a basis for this plane is
$$ \left(1,0,\frac{1}{3}\right)\quad\text{and}\quad\left(0,1,\frac{2}{3}\right).$$
To make a vector, $(a,b,c)$, orthogonal to these two we can, then, use the fact that
$$(a,b,c)\times \left(1,0,\frac{1}{3}\right)=0\;\Rightarrow\; a+\frac{c}{3}=0\;\Rightarrow\; a=-\frac{c}{3},$$
and
$$(a,b,c)\times \left(0,1,\frac{2}{3}\right)=0\;\Rightarrow\; b+\frac{2c}{3}=0\;\Rightarrow\; b=-\frac{2c}{3}.$$
You have one degree of freedom so just set $c=-3$, then $a=1$ and $b=2$. A vector orthogonal to the plane is then
$$(a,b,c)=(1,2,-3).$$
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Differential Equations
An equation involving derivatives of one or more dependent variables with respect to one or more independent variables is called a
. For example: differential equation
\begin{equation} \frac{d^{2}y}{dx^2} + xy \left( \frac{dy}{dx}\right)^2 = 0 \end{equation} \begin{equation} \frac{d^{4}x}{dt^4} + 5 \frac{d^{2}x}{dt^2} + 3x = \sin t \end{equation} \begin{equation}\frac{\partial v}{\partial s} + \frac{\partial v}{\partial t} = v \end{equation} \begin{equation}\frac{\partial^{2}u}{\partial x^2} + \frac{\partial^{2}u}{\partial y^2} + \frac{\partial^{2}u}{\partial z^2} = 0 \end{equation}
Ordinary Differential Equations
A differential equation involving ordinary derivatives of one or more dependent variables with respect to a single independent variable is called an
. For example, +equation 1 and 2 are ordinary differential equations. In equation 1 the variable x is the single independent variable and y is a dependent variable and in equation 2 the independent variable is t, whereas x is dependent. ordinary differential equation Partial Differential Equations
A differential equation involving partial derivatives of one or more dependent variables with respect to one or more independent variables is called a
. Here, equation 3 and 4 are partial differential equations. In equation 3, the variables s and t are independent variables and v is the dependent variable. In equation 4 there are three independent variables: x, y, and z, in the equation u is dependent. partial differential equation Order
The
of the highest ordered derivative involved in a differential equation is called the order of the differential equation. The order of equation 1 is of second order since the highest derivative involved is a second derivative. Equation 2 is an ordinary differential equation of the fourth order. The partial differential 3 and 4 are of the first and second orders, respectfully. order
Linearity
A
of order n, in the dependent variable y and the independent variable x, is an equation that is in, or can be expressed in, the form linear ordinary differential equation
\begin{equation} a_0(x)\frac{d^{n}y}{dx^{n}} + a_1(x)\frac{d^{n-1}y}{dx^{n-1}} + … + a_{n-1}(x)\frac{dy}{dx} + a_n(x)y = b(x) \end{equation}
where \(a_0\) is not identically zero.
Following ordinary differential equations are both linear.
\begin{equation} \frac{d^{2}y}{dx^2} + 5 \frac{dy}{dx} + 6y=0\end{equation} \begin{equation} \frac{d^{4}y}{dx^4} + x^2 \frac{d^{3}y}{dx^3} + x^3\frac{dy}{dx}=x e^{x} \end{equation}
In each case, y is the dependent variable. Observe that y and its various derivatives occur to the first degree only and that no products of y and/or any of its derivatives are present.
A
is an ordinary differential equation that is not linear. The following ordinary differential equations are all nonlinear: nonlinear ordinary differential equation
\begin{equation}\begin{aligned}\frac{d^{2}y}{dx^2}+5\frac{dy}{dx}+6y^2 = 0 \\ \frac{d^{2}y}{dx^2}+5\left(\frac{dy}{dx}\right)^3+6y = 0 \\
\frac{d^{2}y}{dx^2}+5y\left(\frac{dy}{dx}\right)^3+6y = 0 \end{aligned}\end{equation} are further classified according to the nature of the coefficients of the dependent variables and their derivatives. For example, equation 6 is said to be Linear differential equations , while equation 6 is linear with constant coefficients . linear with variable coefficients Explicit vs Implicit
In mathematics, the
is a function in which the dependent variable has been given “explicitly” in terms of the independent variable. Or it is a function in which the dependent variable is expressed in terms of some independent variables. explicit function
Let’s define an nth-order ordinary differential equation
\begin{equation} F\left[ x,y, \frac{dy}{dx},…,\frac{d^{n}y}{dx^n}\right] = 0,\end{equation}
where F is a real function of its (n+2) arguments \(x,y,\frac{dy}{dx},…,\frac{d^{n}y}{dx^n}.\)
Let \(f\) be a real function defined for all x in a real interval I and having an nth derivative for all \(x \in I\). The function \(f\) is called an
of the differential equation (9) on I if it fulfills the following: explicit solution
$$F[x, f(x), f^{‘}(x),…,f^{(n)}(x)]$$ is defined for all \(x \in I,\) and
$$F[x, f(x), f^{‘}(x),…,f^{(n)}(x)] = 0$$ for all \(x \in I.\) For example
Let the function \(f\) defined for all real x by $$ f(x) = 2\sin x + 3 \cos x$$ is an explicit solution of the differential equation $$\frac{d^{2}y}{dx^2} + y = 0$$ for all real x.
Again, a relation \(g(x,y) = 0\) is called an
of (9) if this relation defines at least one real function \(f\) of the variable x on an interval I such that this function is an explicit solution of (9) on this interval. For example implicit solution
The relation \begin{equation}x^2 + y^2 – 25 = 0\end{equation} is an implicit solution of the differential equation \begin{equation}x+y\frac{dy}{dx} = 0\end{equation} on the interval I defined by \(-5<x<5.\) For the relation (10) defines the two real functions \(f_1\) and \(f_2\) given by $$f_1(x) = \sqrt{25 – x^2}$$ and $$f_2(x) = -\sqrt{25 – x^2}.$$
respectively, for all real \(x \in I,\) and both of these functions are explicit solutions of the differential equation (11) on \(I\).
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The original renewal process has independent and identically disributed (i.i.d.) inter-renewal times $\{T_1, T_2, T_3, ...\}$. Thus, starting from time 0, renewals occur at times $\{T_1, T_1+T_2, T_1+T_2+T_3, ...\}$. Now fix a probability $p >0$. Independently place each renewal time to $P_1$ with prob $p$. So we get new inter-renewal times $\{Z_1, Z_2, Z_3, ...\}$, where $Z_1 = \sum_{i=1}^GT_i$ and $G$ is an independent geometric random variable with success probability $p$, and $Z_2, Z_3, ...$ are i.i.d. That is because, when you generate them, you generate them independently but using the same probability law (so each is just a random sum of i.i.d. $T_i$ variables).
If the rate of $P$ was $\lim_{t\rightarrow\infty} \frac{N(t)}{t}= \frac{1}{E[T_1]}=\lambda$ (with prob 1), the rate of $P_1$ is $\lim_{t\rightarrow\infty} \frac{N_1(t)}{t} = p\lambda$ (with prob 1).
Here I am of course defining: \begin{align} &N(t) = \mbox{ Total number of renewals from $P$ during $[0,t]$}\\&N_1(t) = \mbox{ Total number of renewals from $P_1$ during $[0,t]$}\end{align}
A simple example is when $T_i=1/\lambda$ for all $i \in \{1, 2, 3, ...\}$, for a given constant $\lambda>0$. So inter-arrival times are constant (and hence trivially i.i.d.). Then $N(t)$ is a deterministic staircase function and $E[N(t)]=N(t)$ for all $t$, and indeed $$\lim_{t\rightarrow\infty} \frac{N(t)}{t} = \lim_{t\rightarrow\infty} \frac{E[N(t)]}{t} = \lambda $$Probabilistically splitting this deterministic process $N(t)$ (of rate $\lambda$ arrivals/time) using a probability $p$ gives a
random process $N_1(t)$ that has rate $p\lambda$ arrivals/time (and this new process indeed has i.i.d. inter-arrival times).
On visualizing the renewal times: I imagine renewal times as if they are things that arrive to a system, like job arrivals in a queueing system. So the original renewal process$P$ can be drawn over a timeline with spikes arising at the renewal times (the times $\{T_1, T_1+T_2, T_1+T_2+T_3, ...\}$). Then $N(t)$ counts the spikes up to time $t$. We can "probabilistically thin" this spikey process by independently including spikes with prob $p$, and throwing the others away. The thinned process $N_1(t)$ counts the number of included spikes, and so $N_1(t)\leq N(t)$ for all $t$.
As an interesting side note: Consider $\{T_1, T_2, T_3, ...\}$ as
any random sequence of inter-arrival times (not necessarily i.i.d.) and let $N(t)$ count the number of arrivals up to time $t$. Now probabilistically thin this process to a new process $N_1(t)$ by independently including each arrival with probability $p$. Then $E[N_1(t)] = pE[N(t)]$ for all $t\geq 0$ since: $$ E[N_1(t)] = E[E[N_1(t)|N(t)]] = E[pN(t)] = pE[N(t)] $$For example, if $E[N(t)]=\lambda t$ for all $t\geq 0$, then $E[N_1(t)]=p\lambda t$ for all $t\geq 0$. An example of such a process $N(t)$ that is not a Poisson process is this: Choose $T_1$ uniformly over $[0,1]$, then define $T_i=1$ for all $i\in\{2,3,4,...\}$.
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The most relative that I found on Google for
de morgan's 3 variable was: (ABC)' = A' + B' + C'.
I didn't find the answer for my question, therefore I'll ask here:
What is De-Morgan's theorem for
(A + B + C)'?
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
DeMorgan's Theorem applied to $(A + B + C)'$ is as follows:
$$(A + B + C)' = A'B'C'{}{}{}{}$$
We have $\;\;$NOT(A or B or C) $\;\equiv\;$ Not(A) and Not(B) and Not(C),
which in boolean-algebra equates to $A'B'C'$
Both these extensions from DeMorgan's defined for two variables can be justified precisely because we can apply DeMorgan's in a nested manner, and in so doing, reapply, etc, in the end, it is equivalent to an immediate extension of it's application to three variables (or more) variables, provided they are connected by the same connective, $\land, \lor$.
For example, we can write $(A+B+C)' \equiv \big(A + (B+C)\big)' \equiv \big(A' \cdot (B+C)'\big) \equiv A'\cdot (B'C') \equiv A'B'C'$.
Indeed, provided we have a negated series of multiple variables all connected by the
SAME connective (all and'ed or all or'ed), we can generalize DeMorgan's to even more than three variables, again, due to the associativity of AND and OR connectives. For any arbitrary finite number of connected variables:
So, $$(ABCDEFGHIJ)' = A' + B' + C' + \cdots + H' + I' + J'$$
And $$(A + B + C + \cdots + H + I + J)' = A'B'C'D'E'F'G'H'I'J'$$
This is one instance where introducing another variable provides some insight. Let $D = B\lor C$.
Then, we have: $$\begin{align} \neg(A\lor B\lor C) &= \neg(A\lor D)\\ &=\neg A \land \neg D \\ &=\neg A \land \neg(B\lor C) \\ &=\neg A \land \neg B \land \neg C \end{align}$$
Thus: $$\neg(A\lor B \lor C) = \neg A \land \neg B \land \neg C$$
The idea is effectively the same for even more terms. Thus we can have: $$\neg(P_1 \lor P_2 \lor \cdots \lor P_n) = \neg P_1 \land \neg P_2 \land \cdots \land \neg P_n$$ ...and... $$\neg(P_1 \land P_2 \land \cdots \land P_n) = \neg P_1 \lor \neg P_2 \lor \cdots \lor \neg P_n$$ (Note: I'm more familiar with this notation for logic, so I'm using it. $\lor$ is or, $\land$ is and, and $\neg$ is not.)
"+" is a closed operation on the truth set {T, F}. This means that for any two truth values "x" and "y", (x+y) can get assigned a truth value. Consequently, we know that for ((a+b)+c)' we can assign a truth value to (a+b) and c. (x+y)'=(x'+y') holds for anything to which we can assign truth values "x" and "y". Thus, ((a+b)+c)'=((a+b)'+c')=((a'+b')+c').
We can generalize this to any negated series of multiple variables connected by conjunction (AND) and disjunction (OR), so long as we switch the operations throughout. In other words, where X is a member of {AND, OR} and Y a member of {AND, OR}, (a X b Y...X c)'=(a' Y b' X...Y c'). This happens because, ' is a particular kind of isomorphism between the set of truth values under conjunction ({T, F}, AND) and the set of truth values under disjunction ({T, F}, OR). In particular we have that (x X y)'=(x' Y y'), and (x Y y)'=(x' X y') where X and Y indicate binary operations.
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Weierstrass Substitution Proof Technique
The
Weierstrass Substitution is an application of Integration by Substitution. The substitution is: $u \leftrightarrow \tan \dfrac \theta 2$
for $-\pi < \theta < \pi$, $u \in \R$.
It yields:
\(\displaystyle \sin \theta\) \(=\) \(\displaystyle \frac {2 u} {1 + u^2}\) \(\displaystyle \cos \theta\) \(=\) \(\displaystyle \frac {1 - u^2} {1 + u^2}\) \(\displaystyle \frac {\d \theta} {\d u}\) \(=\) \(\displaystyle \frac 2 {1 + u^2}\) This can be stated: $\displaystyle \int \map F {\sin \theta, \cos \theta} \rd \theta = 2 \int \map F {\frac {2 u} {1 + u^2}, \frac {1 - u^2} {1 + u^2} } \frac {d u} {1 + u^2}$
where $u = \tan \dfrac \theta 2$.
Proof
Let $u = \tan \dfrac \theta 2$ for $-\pi < \theta < \pi$.
Then:
\(\displaystyle u\) \(=\) \(\displaystyle \tan \dfrac \theta 2\) \(\displaystyle \leadsto \ \ \) \(\displaystyle \theta\) \(=\) \(\displaystyle 2 \tan^{-1} u\) Definition of Inverse Tangent \(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\d \theta} {\d u}\) \(=\) \(\displaystyle \dfrac 2 {1 + u^2}\) Derivative of Arctangent Function and Derivative of Constant Multiple \(\displaystyle \sin \theta\) \(=\) \(\displaystyle \dfrac {2 u} {1 + u^2}\) Tangent Half-Angle Substitution for Sine \(\displaystyle \cos \theta\) \(=\) \(\displaystyle \dfrac {1 - u^2} {1 + u^2}\) Tangent Half-Angle Substitution for Cosine The result follows from Integration by Substitution.
$\blacksquare$
Also known as
This technique is also known as
tangent half-angle subsitution. Also see Source of Name
This entry was named for Karl Theodor Wilhelm Weierstrass.
Sources 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables... (previous) ... (next): $\S 14$: Important Transformations: $14.58$ This article incorporates material from Weierstrass substitution formulas on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License. Weisstein, Eric W. "Weierstrass Substitution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/WeierstrassSubstitution.html
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Talk:Gamma-function $\Gamma$-function $ \newcommand{\abs}[1]{\left|#1\right|} \newcommand{\Re}{\mathop{\mathrm{Re}}} \newcommand{\Im}{\mathop{\mathrm{Im}}} $
A transcendental function $\Gamma(z)$ that extends the values of the factorial $z!$ to any complex number $z$. It was introduced in 1729 by L. Euler in a letter to Ch. Goldbach, using the infinite product $$ \Gamma(z) = \lim_{n\rightarrow\infty}\frac{n!n^z}{z(z+1)\ldots(z+n)} = \lim_{n\rightarrow\infty}\frac{n^z}{z(1+z/2)\ldots(1+z/n)}, $$ which was used by L. Euler to obtain the integral representation (Euler integral of the second kind, cf. Euler integrals) $$ \Gamma(z) = \int_0^\infty x^{z-1}e^{-x} \rd x, $$ which is valid for $\Re z > 0$. The multi-valuedness of the function $x^{z-1}$ is eliminated by the formula $x^{z-1}=e^{(z-1)\ln x}$ with a real $\ln x$. The symbol $\Gamma(z)$ and the name gamma-function were proposed in 1814 by A.M. Legendre.
If $\Re z < 0$ and $-k-1 < \Re z < -k$, $k=0,1,\ldots$, the gamma-function may be represented by the Cauchy–Saalschütz integral: $$ \Gamma(z) = \int_0^\infty x^{z-1} \left( e^{-x} - \sum_{m=0}^k (-1)^m \frac{x^m}{m!} \right) \rd x. $$ In the entire plane punctured at the points $z=0,-1,\ldots $, the gamma-function satisfies a Hankel integral representation: $$ \Gamma(z) = \frac{1}{e^{2\pi iz} - 1} \int_C s^{z-1}e^{-s} \rd s, $$ where $s^{z-1} = e^{(z-1)\ln s}$ and $\ln s$ is the branch of the logarithm for which $0 < \arg\ln s < 2\pi$; the contour $C$ is represented in Fig. a. [FIXME] It is seen from the Hankel representation that $\Gamma(z)$ is a meromorphic function. At the points $z_n = -n$, $n=0,1,\ldots$ it has simple poles with residues $(-1)^n/n!$.
Figure: g043310a
Contents Fundamental relations and properties of the gamma-function.
1) Euler's functional equation: $$ z\Gamma(z) = \Gamma(z+1), $$ or $$ \Gamma(z) = \frac{1}{z\ldots(z+n)}\Gamma(z+n+1); $$ $\Gamma(1)=1$, $\Gamma(n+1) = n!$ if $n$ is an integer; it is assumed that $0! = \Gamma(1) = 1$.
2) Euler's completion formula: $$ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z}. $$ In particular, $\Gamma(1/2)=\sqrt{\pi}$; $$ \Gamma\left(n+\frac{1}{2}\right) = \frac{1.3\ldots(2n-1)}{2^n}\sqrt{\pi} $$ if $n>0$ is an integer; $$ \abs{\Gamma\left(\frac{1}{2} + iy\right)}^2 = \frac{\pi}{\cosh y\pi}, $$ where $y$ is real.
3) Gauss' multiplication formula: $$ \prod_{k=0}^{m-1} \Gamma\left( z + \frac{k}{m} \right) = (2\pi)^{(m-1)/2}m^{(1/2)-mz}\Gamma(mz), \quad m = 2,3,\ldots $$ If $m=2$, this is the Legendre duplication formula.
4) If $\Re z \geq \delta > 0$ or $\abs{\Im z} \geq \delta > 0$, then $\ln\Gamma(z)$ can be asymptotically expanded into the Stirling series: $$ \ln\Gamma(z) = \left(z-\frac{1}{2}\right)\ln z - z + \frac{1}{2}\ln 2\pi + \sum_{n=1}^m \frac{B_{2n}}{2n(2n-1)z^{2n-1}} + O\bigl(z^{-2m-1}\bigr), \quad m = 1,2,\ldots, $$ where $B_{2n}$ are the Bernoulli numbers. It implies the equality $$ \Gamma(z) = \sqrt{2\pi} z^{z-1/2} z^{-z} \left( 1 + \frac{1}{12}z^{-1} + \frac{1}{288}z^{-2} - \frac{139}{51840}z^{-3} - \frac{571}{2488320}z^{-4} + O\bigl(z^{-5}\bigr) \right). $$ In particular, $$ \Gamma(1+x) = \sqrt{2\pi} x^{x+1/2} e^{-x + \theta/12x}, \quad 0 < \theta < 1. $$ More accurate is Sonin's formula [6]: $$ \Gamma(1+x) = \sqrt{2\pi} x^{x+1/2} e^{-x + 1/12(x+\theta)}, \quad 0 < \theta < 1/2. $$
5) In the real domain, $\Gamma(x) > 0$ for $x > 0$ and it assumes the sign $(-1)^{k+1}$ on the segments $-k-1 < x < -k$, $k = 0,1,\ldots$ (Fig. b).
Figure: g043310b
The graph of the function $ $.
For all real $x$ the inequality $$ \Gamma\Gamma^{\prime\prime} > \bigl(\Gamma^\prime\bigr)^2 \geq 0 $$ is valid, i.e. all branches of both $\abs{\Gamma(x)}$ and $\ln\abs{\Gamma(x)}$ are convex functions. The property of logarithmic convexity defines the gamma-function among all solutions of the functional equation $$ \Gamma(1+x) = x\Gamma(x) $$ up to a constant factor (see also the Bohr–Mollerup theorem).
For positive values of $x$ the gamma-function has a unique minimum at $x=1.4616321\ldots$ equal to $0.885603\ldots$. The local minima of the function $\abs{\Gamma(x)}$ form a sequence tending to zero as $x\rightarrow -\infty$.
Figure: g043310c
The graph of the function $ $.
6) In the complex domain, if $\Re z > 0$, the gamma-function rapidly decreases as $\abs{\Im z} \rightarrow \infty$, $$ \lim_{\abs{\Im z} \rightarrow \infty} \abs{\Gamma(z)}\abs{\Im z}^{(1/2)-\Re z}e^{\pi\abs{\Im z}/2} = \sqrt{2\pi}. $$
7) The function $ $ (Fig. c) is an entire function of order one and of maximal type; asymptotically, as $ $,
$ $
where
$ $
It can be represented by the infinite Weierstrass product:
$ $
which converges absolutely and uniformly on any compact set in the complex plane ($ $ is the Euler constant). A Hankel integral representation is valid:
$ $
where the contour $ $ is shown in Fig. d.
Figure: g043310d
$ $
G.F. Voronoi [7] obtained integral representations for powers of the gamma-function.
In applications, the so-called poly gamma-functions — $ $-th derivatives of $ $ — are of importance. The function (Gauss' $ $-function)
$ $
$ $
is meromorphic, has simple poles at the points $ $ and satisfies the functional equation
$ $
The representation of $ $ for $ $ yields the formula
$ $
where
$ $
This formula may be used to compute $ $ in a neighbourhood of the point $ $.
$ $
The functions $ $ and $ $ are transcendental functions which do not satisfy any linear differential equation with rational coefficients (Hölder's theorem).
The exceptional importance of the gamma-function in mathematical analysis is due to the fact that it can be used to express a large number of definite integrals, infinite products and sums of series (see, for example, Beta-function). In addition, it is widely used in the theory of special functions (thehypergeometric function, of which the gamma-function is a limit case, cylinder functions, etc.), in analytic number theory, etc.
References
[1] E.T. Whittaker, G.N. Watson, "A course of modern analysis" , Cambridge Univ. Press (1952) [2] H. Bateman (ed.) A. Erdélyi (ed.) , Higher transcendental functions , 1. The gamma function. The hypergeometric functions. Legendre functions , McGraw-Hill (1953) [3] N. Bourbaki, "Elements of mathematics. Functions of a real variable" , Addison-Wesley (1976) (Translated from French) [4] , Math. anal., functions, limits, series, continued fractions , Handbook Math. Libraries , Moscow (1961) (In Russian) [5] N. Nielsen, "Handbuch der Theorie der Gammafunktion" , Chelsea, reprint (1965) [6] N.Ya. Sonin, "Studies on cylinder functions and special polynomials" , Moscow (1954) (In Russian) [7] G.F. Voronoi, "Studies of primitive parallelotopes" , Collected works , 2 , Kiev (1952) pp. 239–368 (In Russian) [8] E. Jahnke, F. Emde, "Tables of functions with formulae and curves" , Dover, reprint (1945) (Translated from German) [9] A. Angot, "Compléments de mathématiques. A l'usage des ingénieurs de l'electrotechnique et des télécommunications" , C.N.E.T. (1957) Comments
The $ $-analogue of the gamma-function is given by
$ $
$ $ References
[a1] E. Artin, "The gamma function" , Holt, Rinehart & Winston (1964) [a2] R. Askey, "The $ $-Gamma and $ $-Beta functions" Appl. Anal. , 8 (1978) pp. 125–141 How to Cite This Entry:
Gamma-function.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Gamma-function&oldid=25595
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$$\frac {\Delta d_{t+1}}{d_t} = \frac {\frac {D_{t+1}}{Y_{t+1}}-\frac {D_{t}}{Y_{t}}}{\frac {D_{t}}{Y_{t}}} = \frac {D_{t+1}Y_{t}-D_{t}Y_{t+1}}{D_{t}Y_{t+1}}$$
$$=\frac {D_{t+1}Y_{t}}{{D_{t}Y_{t+1}}}-1 = \frac{Y_{t}}{Y_{t+1}}\cdot \left[{\frac {D_{t+1}}{D_{t}} -\frac {Y_{t+1}}{{Y_{t}}}}\right]$$
$$=\frac{Y_{t}}{Y_{t+1}}\cdot \left[{\frac {\Delta D_{t+1}}{D_{t}} -\frac {\Delta Y_{t+1}}{{Y_{t}}}}\right]$$
So the "rule of thumb" deviates from the above exact expression by (multiplicatively) $Y_{t}/Y_{t+1}$.
Now, we are talking about the growth
rate of the Debt/GDP ratio. Say the ratio was $120\text{%}$ and it went to what? in one year? Say it went to $D_{t+1} /Y_{t+1} = 130\text{%}$. Then its exact growth rate is $130/120 -1 = 8.33\text{%}$ Assume also that GDP grew at 5% yearly rate, so $Y_{t}/Y_{t+1} = 0.95238$.
In this numerical example, the "rule of thumb will give as Debt/GDP ratio growth rate $8.33\text{%} \times 0.95238 = 7.933 \text{%}$, or less than half percentage point underestimation. This is considered a small inaccuracy in the real world,
when one discusses broadly these magnitudes. And in general the above numerical example exaggerates, in most cases the growth rate of the GDP and of the Debt/GDP ratio are smaller than above, so the inaccuracy will be even smaller. Hence, the "rule of thumb" becomes acceptable in general also in theoretical models.
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a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached).
b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct.
However,
your calculation of amplitude is not correct.
Peak-to-trough is
twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly.
CALCULATION OF AMPLITUDE OF OSCILLATIONS
1. Equation of Motion Method
First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision.
Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$
2. Energy Method
Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$.
At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$.
At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$.
By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1.
(c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it.
Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal.
Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is
always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result.
The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate.
Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$
We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off.
The total energy at P is the same as at U, so from above (
Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
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What is the way to calculate the centroid of polygon? I have a concave polygon of 16 points, and I want know the centroid of that.
thanks
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I. CONSIDERATIONS. (In 2 parts. The formulas are at the end of each.) Case: Point List
I have sixteen marbles of equal mass. Let us agree they all lie in one plane.
Q1: What is the center of mass of the set of marbles?
The center of mass of a sphere is its center. So let the coordinates of the marble centers be:
$\;\;\;\;\;Pts = A_1, A_2.... A_{16}. $
To find the centroid C, average them:
$(1)\;\;\;\;\;C =(A_1+A_2+....+A_{16})/16. $
Case: Polygon
Now, using (
Pts) as the vertices, I cut out a 16-sided polygon, Jerome. Q2: What is the center of mass of Jerome? This is not the same question. How can I ask it correctly? Consider (1) above. What has changed? I see two problems:
There are many concave polygons through 16 given points. Which did I cut out? I must uniquely identify the region (area) I want to measure. Here are two ways:
a) Break up Jerome into distinct, convex regions. If a set of points are the vertices of a convex polygon, that polygon is unique. Each convex region of Jerome is uniquely determined by its vertices.
b) Arrange
Pts so that the order (A1 -> A2 -> A3 .... ->An) follows the perimeter in a single direction, without leaps or self-intersections. A unique perimeter is a unique polygon.
Both ways allow me to draw the figure. Note that the simple average (mean) does not distinguish order; it can't give the correct answer.
Jerome is a plane figure: Its mass is proportional to area. But "the mean of coordinates" is not the same thing as "proportional to area". Area
of what? Again, the mean can't, in general, give an answer.
I now have a unique definition of the task. I have also discarded the simple average where it can't apply, without knowing the answer or doing any math. Going a bit further, it seems that the mean could be right if the vertices are equally distributed about a common center (this hunch turns out to be correct; e.g. regular polygons).
Now to solve the problem...
(A correct formula is given here: Why doesn't a simple mean give the position of a centroid in a polygon? , but I want to know how to obtain that formula.)
II. SOLVE THE PROBLEM
Why doesn't a simple mean give the position of a centroid in a polygon? leads to formulas.
Frown. I want to know how to
set up and solve the problem. A solution can always be converted to a formula. The reverse --figuring out how a problem was solved by looking at the formula-- is very hard. So I will solve problem from scratch.
A working demonstration of the method below can be found here. The linked document is a Geogebra worksheet; feel free to download it, examine the code, and use it as you see fit.
A. Observation The centroid of triangle $\Delta ABC$ is the simple average of its vertices: $\;\;\;D = (A+ B + C)/3$. What? Point order does not matter here: take the vertices A,B,C, in any order, they are the same triangle. To prove the formula is correct, I might Integrate the triangle area; Find the balance point of figure ABC; or, with compass and straightedge, draw the intersection of the medians. I say, the point is the same in every case. B. Procedure I will convert the problem into one I know how to solve: Proposition: Find the centroid of a set of weighted points.
Chop Jerome up into mutually exclusive triangles.
Say there are
m triangles. For each triangle $\;\Delta_k, \;k =1, 2, \ldots, m,\;$ find the centroid $C_k$, and the area (weight) $w_k$. The set of weighted points is then $[\frac{w_k C_k}{J}],$ where J is the area of Jerome. Solution: Add 'em up.I say, the resulting point is the centroid of the polygon. I have a general solution (I can always carry out this procedure). But it's messy: I don't have a triangulation rule. Now what? A good way to generalize is to start with a case I can solve. C. Case: Jerome is Convex It is commonly known that, given a convex polygon $P$, we may choose any vertex, $V,$ draw segments from $V$ to every non-adjacent vertex, and $P$ is correctly triangulated.
I will also use the following formulas:
-Let any two sides of $\Delta ABC$ be vectors ${\rm u =(u_1, u_2),\;\; v= (v_1, v_2)}.\;$ Then $\;\;\;{\rm Area}_{\Delta ABC} = \tfrac{1}{2}|{\rm u \times v}|,\;\;\;\;\;$where (the determinant) ${\rm u \times v}={\rm u_1 v_2-u_2 v_1}$.
-Let sides $AB, AC$ of $\Delta ABC$ be vectors ${\rm u, v}.\;$ Then the centroid D= (A+B+C)/3 can be written
$\;\;\;D = A + \tfrac{1}{3}|{\rm u + v}|.$
I have gathered what I need.
D. Solution
Let the counterclockwise path of my
n-gon be given, in order, by $\;\;[A_i] = A_1, A_2, \ldots, A_n$ For covenience, I choose $V=A_1$.
Draw the
n-1 vectors from A_1 to the other vertices: $\;\;[{\rm a_i}] = (A_{k+1}-A_1),\;\; k = 1, 2, \ldots n\!-\!1$
I have
n-2 adjacent triangles with centroids $\;\;[C_i] = A_1 + \tfrac{1}{3}{\rm (a_k+a_{k+1})},\;\; k = 1, 2, \ldots n\!-\!2$ And areas (weights) $\;\;[w_i] = \tfrac{1}{2}{\rm (a_k\times a_{k+1})},\;\; k = 1, 2, \ldots n\!-\!2$ (I dropped the absolute value: counter-clockwise, ${\rm u \times v}$ is positive.) Then Total area = $\sum_{k=1}^{n-2} w_k$ and
which can be written
$$(2)\;\;C_J= A_1+{\large \frac{1}{3} \frac{\sum_{k=1}^{n-2} ({\rm a_k+ a_{k+1}})({\rm a_k \times a_{k+1}})}{\sum_{k=1}^{n-2} ({\rm a_k \times a_{k+1}})} }$$
I say, that this is in fact the complete solution: the determinants give
signed areas: +/- according as the direction of rotation from $A_k$ to $A_{k+1}$, about $A_1$ is positive or negative, preserving measure in either case. Being the thing to be done.
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Imagine we have a U-tube with constant diameter which is filled with water and oil in its equilibrium state (where water and oil are fully separated).
Then, we start inserting oil with constant speed at the end of the U-tube where the oil is.
My intuition says that if I take a control volume around the separation boundary between oil and water, the boundary will start moving towards the water and eventually my control volume will be full with oil. It feels just like the oil is a piston pressing the water out of the U-tube.
This implies that if I add $\frac{dV_{oil}}{dt}$ oil volume flow , I would get the same volume of water at other end of the U-tube, i.e., $\frac{dV_{oil}}{dt} = \frac{dV_{water}}{dt}$.
But this seems to contradict the mass conservation in my control volume because:
$$ \rho_{oil} \cdot \frac{dV_{oil}}{dt} = \rho_{water} \cdot \frac{dV_{oil}}{dt} \implies \rho_{oil} = \rho_{water} (WRONG) $$
Now imagine that the U-tube is full with water but the left half has very high temperature and the right half has very low temperature so that we have again the same densities as the previous example, i.e.: $\rho_{water left} = \rho_{oil}$ and $\rho_{water right} = \rho_{water previous}$. Now we insert water from the left side with volume rate $\frac{dV_{waterleft}}{dt}$.
In this case, I feel that the conservation of mass holds because the same material "get transformed" into water with different density. The mass conservation holds but with different speeds: $$ \rho_{waterleft} \cdot \frac{dV_{waterleft}}{dt} = \rho_{waterright} \cdot \frac{dV_{waterright}}{dt} \implies $$ $$ \rho_{waterleft} \cdot u_1 \cdot A = \rho_{waterright} \cdot u_2 \cdot A \implies $$ $$ u_2 = \frac{\rho_{waterleft}}{\rho_{waterright}} \cdot u_1 $$
Could anyone explain to me what I am doing wrong? Or, if I am not doing something wrong, how can I reconcile the first example with the mass conservation?
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In geometry, a tangent of a circle is a straight line that touches the circle at exactly one point, never entering the circle’s interior. It is a line through a pair of infinitely close points on the circle.
The tangent lines to circles form the subject of several theorems and play an important role in many geometrical constructions and proofs. Since the tangent line to a circle at a point
P is perpendicular to the radius to that point, theorems involving tangent lines often involve radial lines and orthogonal circles.
The tangent has two defining properties such as:
A Tangent touches a circle in exactly one place. The Tangent intersects the circle’s radius at $90^{\circ}$ angle.
The equation of the tangent is written as,
$\huge \left(y-y_{0}\right)=m_{tgt}\left(x-x_{0}\right)$
Tangents to two circles
Given two circles, there are lines that are tangents to both of them at the same time. If the circles are separate (do not intersect), there are four possible common tangents:
Two external tangents
Two internal tangents
If the two circles touch at just one point, there are three possible tangent lines that are common to both:
If the two circles touch at just one point, with one inside the other, there is just one line that is a tangent to both:
If the circles overlap – i.e. intersect at two points, there are two tangents that are common to both:
If the circles lie one inside the other, there are no tangents that are common to both. A tangent to the inner circle would be a secant of the outer circle.
Solved Examples of Tangent to a Circle Question: Determine the equation of the tangent to the circle:
$x^{2}+y^{2}-2y+6x-7=0\;at\;the\;point\;F(-2:5)$
Solution:
Write the equation of the circle in the form: $\left(x-a\right)^{2}+\left(y-b\right)^{2}+r^{2}$
Use the method of completing the square:
$x^{2}+y^{2}-2y+6x-7=0$
$x^{2}+6x+y^{2}-2y=7$
$\left(x^{2}+6x+9\right)-9+\left(y^{2}-2y+1\right)-1=7$
$\left(x+3\right)^{2}+\left(y-1\right)^{2}=17$
Draw a sketch
The centre of the circle is (−3;1) and the radius is $\sqrt{17}$ units.
Determine the gradient of the radius CF
$m_{CF}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$=\frac{5-1}{-2+3}=4$
Determine the gradient of the tangent
Let the gradient of the tangent line be
m.
$m_{CF}\times m=-1$
$4\times m=-1$
$\therefore m=-\frac{1}{4}$
Determine the equation of the tangent to the circle
Write down the gradient-point form of a straight line equation and substitute $m=-\frac{1}{4}\;and\;F(-2:5)$
So,
$y-y_{1}=m\left(x-x_{1}\right)$
$y-y_{1}=-\frac{1}{4}\left(x-x_{1}\right)$
$Substitute\;F\left(-2:5\right):\;y-5=-\frac{1}{4}\left(x-\left(-2\right)\right)$
$y-5=-\frac{1}{4}\left (x+2 \right )$
$y=-\frac{1}{4}x-\frac{1}{2}+5$
$=-\frac{1}{4}x+\frac{9}{2}$
The final answer
The equation of the tangent to the circle at $F\;is\;y=-\frac{1}{4}x+\frac{9}{2}$
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Yeah, this software cannot be too easy to install, my installer is very professional looking, currently not tied into that code, but directs the user how to search for their MikTeX and or install it and does a test LaTeX rendering
Some body like Zeta (on codereview) might be able to help a lot... I'm not sure if he does a lot of category theory, but he does a lot of Haskell (not that I'm trying to conflate the two)... so he would probably be one of the better bets for asking for revision of code.
he is usually on the 2nd monitor chat room. There are a lot of people on those chat rooms that help each other with projects.
i'm not sure how many of them are adept at category theory though... still, this chat tends to emphasize a lot of small problems and occasionally goes off tangent.
you're project is probably too large for an actual question on codereview, but there is a lot of github activity in the chat rooms. gl.
In mathematics, the Fabius function is an example of an infinitely differentiable function that is nowhere analytic, found by Jaap Fabius (1966). It was also written down as the Fourier transform off^(z)=∏m=1∞(cos...
Defined as the probability that $\sum_{n=1}^\infty2^{-n}\zeta_n$ will be less than $x$, where the $\zeta_n$ are chosen randomly and independently from the unit interval
@AkivaWeinberger are you familiar with the theory behind Fourier series?
anyway here's a food for thought
for $f : S^1 \to \Bbb C$ square-integrable, let $c_n := \displaystyle \int_{S^1} f(\theta) \exp(-i n \theta) (\mathrm d\theta/2\pi)$, and $f^\ast := \displaystyle \sum_{n \in \Bbb Z} c_n \exp(in\theta)$. It is known that $f^\ast = f$ almost surely.
(a) is $-^\ast$ idempotent? i.e. is it true that $f^{\ast \ast} = f^\ast$?
@AkivaWeinberger You need to use the definition of $F$ as the cumulative function of the random variables. $C^\infty$ was a simple step, but I don't have access to the paper right now so I don't recall it.
> In mathematics, a square-integrable function, also called a quadratically integrable function, is a real- or complex-valued measurable function for which the integral of the square of the absolute value is finite.
I am having some difficulties understanding the difference between simplicial and singular homology. I am aware of the fact that they are isomorphic, i.e. the homology groups are in fact the same (and maybe this doesnt't help my intuition), but I am having trouble seeing where in the setup they d...
Usually it is a great advantage to consult the notes, as they tell you exactly what has been done. A book will teach you the field, but not necessarily help you understand the style that the prof. (who creates the exam) creates questions.
@AkivaWeinberger having thought about it a little, I think the best way to approach the geometry problem is to argue that the relevant condition (centroid is on the incircle) is preserved by similarity transformations
hence you're free to rescale the sides, and therefore the (semi)perimeter as well
so one may (for instance) choose $s=(a+b+c)/2=1$ without loss of generality
that makes a lot of the formulas simpler, e.g. the inradius is identical to the area
It is asking how many terms of the Euler Maclaurin formula do we need in order to compute the Riemann zeta function in the complex plane?
$q$ is the upper summation index in the sum with the Bernoulli numbers.
This appears to answer it in the positive: "By repeating the above argument we see that we have analytically continued the Riemann zeta-function to the right-half plane σ > 1 − k, for all k = 1, 2, 3, . . .."
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I've been reading about the equation, and all of the sources I found state that the equation preserves the trace, self-adjointness, and positivity of the density matrix.
The first two properties are easily verified, but I can't seem to figure out how to prove the positivity ($\langle\alpha|\rho|\alpha\rangle\geq0$) is maintained. So far I wrote down:
$$ \langle\alpha|\frac{d\rho}{dt}|\alpha\rangle= \frac{d}{dt}\left(\langle\alpha|\rho|\alpha\rangle\right)= \frac{d\rho_{\alpha\alpha}}{dt}= -\frac{i}{\hbar}\langle\alpha|H\rho-\rho H|\alpha\rangle=-\frac{i}{\hbar} \sum_\mu \left(H_{\alpha\mu}\rho_{\mu\alpha} - \rho_{\alpha\mu}H_{\mu\alpha} \right) $$
Since both $H$ and $\rho$ are self-adjoint the last expression may be reduced to:
$$ \frac{d\rho_{\alpha\alpha}}{dt}= -\frac{i}{\hbar} \sum_\mu \left(H_{\alpha\mu}\rho_{\mu\alpha} - \left(\rho_{\mu\alpha}H_{\alpha\mu}\right)^* \right) =\frac{2}{\hbar}\text{Im}\left[ \sum_\mu H_{\alpha\mu}\rho_{\mu\alpha} \right] $$
I'm kind of stuck from here, any Help?
Edit:
This part of the question was edited for clarification.
I noticed another peculiar thing, when writing this down. My expressions don't relate to any particular basis. However, if I assume that the basis of choice is the eigenbasis of $\rho$ (or $H$), I trivially get $\frac{d\rho_{\alpha\alpha}}{dt}=0$ as:
$$ \frac{d\rho_{\alpha\alpha}}{dt}= -\frac{i}{\hbar}\langle\alpha|H\rho-\rho H|\alpha\rangle= -\frac{i}{\hbar}\left( \rho_{\alpha\alpha}\langle\alpha|H|\alpha\rangle - \rho_{\alpha\alpha}^*\langle\alpha|H|\alpha\rangle \right)= -\frac{i}{\hbar}\rho_{\alpha\alpha} \left( H_{\alpha\alpha} - H_{\alpha\alpha} \right)=0 $$
What is the meaning of this? As I didn't make any assumptions regarding the eigenbasis of $H$, it means the the diagonal terms of the commutator are zero, and therefore to populations are constant of motion.
This would make sense as for a statistical ensemble in equilibrium we claim $\rho\propto e^{-\beta H}$, which means $\left[ H,\rho\right]=0$, but if so what is the whole point of writing the equation to begin with?
This reasoning is of course flawed as it implies that for every system the diagonal of the density matrix is a constant of motion, which means we cannot alter populations of various states. Where does my reasoning fail? Is the source of the problem the fact that I wasn't careful enough ragarding the commutation of the Hamiltonian with itself at different times?
Edit 2:
I leave the question as it is for the sake others who might repeat my mistake. I found my error in the first step of the derivation. I wrote that:
$$ \langle\alpha|\frac{d\rho}{dt}|\alpha\rangle= \frac{d}{dt}\left(\langle\alpha|\rho|\alpha\rangle\right)= \frac{d\rho_{\alpha\alpha}}{dt} $$
This is of course wrong! Since the Liouville equation is written in the Schrödinger picture the states are time dependent and cannot be incorporated into the time derivative.
What I implicitly did by assuming the it could be incorporated into the time derivative, is I assumed that the state $|\alpha\rangle$, is a stationary state, i.e. that diagonalizes simultaneously $H$ and $\rho$, thus implicitly assuming that they commute, in which case indeed the populations are constants of motion.
In short the calculation of the RHS in the first equation I wrote is correct. However, it has nothing to do the the rate of change of the diagonal elements of $\rho$, and isn't related to population in any way.
However, instead I could write:
$$ \frac{d\rho_{\alpha\alpha}}{dt}= \frac{d}{dt}\left(\langle\alpha|\rho|\alpha\rangle\right)= \left(\frac{d\langle\alpha|}{dt}\right)\rho|\alpha\rangle + \langle\alpha| \left(\frac{d\rho}{dt}\right)|\alpha\rangle + \langle\alpha|\rho \left(\frac{d|\alpha\rangle}{dt}\right)= % \\ % =\frac{i}{\hbar}\left( \langle\alpha|H\rho|\alpha\rangle - \langle\alpha|\rho H|\alpha\rangle \right)+ \langle\alpha| \left(\frac{d\rho}{dt}\right)|\alpha\rangle= % \\ % =\frac{i}{\hbar} \langle\alpha|\left[H,\rho\right]|\alpha\rangle - \frac{i}{\hbar} \langle\alpha|\left[H,\rho\right]|\alpha\rangle=0 $$
Which essentially gives the same information as in the answer below. The last expression however shouldn't be interpreted the time evolution of any specific population. The time propogation of occupation probability is given by:
$$ P_{\alpha\alpha}(t) = |c_\alpha(t)|^2 = \langle\alpha(0)|\rho(t)|\alpha(0)\rangle= \left(G(t)\rho(0)G^\dagger(t)\right)_{\alpha\alpha} $$
where $c_\alpha(t)$ are the time dependent coefficients of some spectral decomposition: $|\psi(t)\rangle = \sum_\alpha c_\alpha(t)|\alpha\rangle$, and $G(t)$ is the time propagator between the states with time difference $t$.
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While performing my excavation activities on no-answer questions, I found this very sensible one, to which, I guess, by now the OP has found an answer.
But I realized that I had various questions of my own regarding the issue of perfect separation in logistic regression, and a (quick) search in the literature, did not seem to answer them. So I decided to start a little research project of my own (probably re-inventing the wheel), and with this answer I would want to share some of its preliminary results. I believe these results contribute towards an understanding of whether the issue of perfect separation is a purely "technical" one, or whether it can be given a more intuitive description/explanation.
My first concern was to understand the phenomenon in algorithmic terms, rather than the general theory behind it:
under which conditions the maximum likelihood estimation approach will "break-down" if fed with a data sample that contains a regressor for which the phenomenon of perfect separation exists?
Preliminary results (theoretical and simulated) indicate that: 1) It matters whether a constant term is included in the logit specification. 2) It matters whether the regressor in question is dichotomous (in the sample), or not. 3) If dichotomous, it may matter whether it takes the value $0$ or not. 4) It matters whether other regressors are present in the specification or not. 5) It matters how the above 4 issues are combined.
I will now present a set of sufficient conditions for perfect separation to make the MLE break-down. This is unrelated to whether the various statistical softwares give warning of the phenomenon -they may do so by scanning the data sample
prior to attempting to execute maximum likelihood estimation. I am concerned with the cases where the maximum likelihood estimation will begin -and when it will break down in the process.
Assume a "usual" binary-choice logistic regression model
$$P(Y_i \mid \beta_0, X_i, \mathbf z_i) = \Lambda (g(\beta_0,x_i, \mathbf z_i)), \;\; g(\beta_0,x_i, \mathbf z_i) = \beta_0 +\beta_1x_i + \mathbf z_i'\mathbf \gamma$$
$X$ is the regressor with perfect separation, while $\mathbf Z$ is a collection of other regressors that are not characterized by perfect separation. Also
$$\Lambda (g(\beta_0,x_i, \mathbf z_i)) = \frac 1{1+e^{-g(\beta_0,x_i, \mathbf z_i)}}\equiv \Lambda_i$$
The log-likelihood for a sample of size $n$ is
$$\ln L=\sum_{i=1}^{n}\left[y_i\ln(\Lambda_i)+(1-y_i)\ln(1-\Lambda_i)\right]$$
The MLE will be found by setting the derivatives equal to zero. In particular we want
$$ \sum_{i=1}^{n}(y_i-\Lambda_i) = 0 \tag{1}$$
$$\sum_{i=1}^{n}(y_i-\Lambda_i)x_i = 0 \tag{2}$$
The first equation comes from taking the derivative with respect to the constant term, the 2nd from taking the derivative with respect to $X$.
Assume now that in all cases where $y_1 =1$ we have $x_i = a_k$, and that $x_i$ never takes the value $a_k$ when $y_i=0$. This is the phenomenon of complete separation, or "perfect prediction": if we observe $x_i = a_k$ we know that $y_i=1$. If we observe $x_i \neq a_k$ we know that $y_i=0$. This holds irrespective of whether,
in theory or in the sample, $X$ is discrete or continuous, dichotomous or not. But also, this is a sample-specific phenomenon -we do not argue that it will hold over the population. But the specific sample is what we have in our hands to feed the MLE.
Now denote the abolute frequency of $y_i =1$ by $n_y$
$$n_y \equiv \sum_{i=1}^ny_i = \sum_{y_i=1}y_i \tag{3}$$
We can then re-write eq $(1)$ as
$$n_y = \sum_{i=1}^n\Lambda_i = \sum_{y_i=1}\Lambda_i+\sum_{y_i=0}\Lambda_i \Rightarrow n_y - \sum_{y_i=1}\Lambda_i = \sum_{y_i=0}\Lambda_i \tag{4}$$
Turning to eq. $(2)$ we have
$$\sum_{i=1}^{n}y_ix_i -\sum_{i=1}^{n}\Lambda_ix_i = 0 \Rightarrow \sum_{y_i=1}y_ia_k+\sum_{y_i=0}y_ix_i - \sum_{y_i=1}\Lambda_ia_k-\sum_{y_i=0}\Lambda_ix_i =0$$
using $(3)$ we have$$n_ya_k + 0 - a_k\sum_{y_i=1}\Lambda_i-\sum_{y_i=0}\Lambda_ix_i =0$$
$$\Rightarrow a_k\left(n_y-\sum_{y_i=1}\Lambda_i\right) -\sum_{y_i=0}\Lambda_ix_i =0$$
and using $(4)$ we obtain
$$a_k\sum_{y_i=0}\Lambda_ix_i -\sum_{y_i=0}\Lambda_ix_i =0 \Rightarrow \sum_{y_i=0}(a_k-x_i)\Lambda_i=0 \tag {5}$$
So :
if the specification contains a constant term and there is perfect separation with respect to regressor $X$, the MLE will attempt to satisfy, among others, eq $(5)$ also.
But note, that the summation is over the sub-sample where $y_i=0$ in which $x_i\neq a_k$ by assumption. This implies the following:
1) if $X$ is dichotomous in the sample, then $(a_k-x_i) \neq 0$ for all $i$ in the summation in $(5)$. 2) If $X$ is not dichotomous in the sample, but $a_k$ is either its minimum or its maximum value in the sample, then again $(a_k-x_i) \neq 0$ for all $i$ in the summation in $(5)$.
In these two cases, and since moreover $\Lambda_i$ is non-negative by construction, the only way that eq. $(5)$ can be satisfied is when $\Lambda_i=0$ for all $i$ in the summation. But
$$\Lambda_i = \frac 1{1+e^{-g(\beta_0,x_i, \mathbf z_i)}}$$
and so the only way that $\Lambda_i$ can become equal to $0$, is if the parameter estimates are such that $g(\beta_0,x_i, \mathbf z_i) \rightarrow -\infty$. And since $g()$ is linear in the parameters, this implies that at least one of the parameter estimates should be "infinity": this is what it means for the MLE to "break down": to not produce finite valued estimates. So cases 1) and 2) are sufficient conditions for a break-down of the MLE procedure.
But consider now the case where $X$ is not dichotomous, and $a_k$ is
not its minimum, or its maximum value in the sample. We still have complete separation, "perfect prediction", but now, in eq. $(5)$ some of the terms $(a_k-x_i)$ will be positive and some will be negative. This means that it is possible that the MLE will be able to satisfy eq. $(5)$ producing finite estimates for all parameters. And simulation results confirm that this is so.
I am not saying that such a sample does not create undesirable consequences for the properties of the estimator etc: I just note that in such a case, the estimation algorithm will run as usual.
Moreover, simulation results show that
if there is no constant term in the specification, $X$ is not dichotomous but $a_k$ is an extreme value, and there are other regressors present, again the MLE will run -indicating that the presence of the constant term (whose theoretical consequences we used in the previous results, namely the requirement for the MLE to satisfy eq. $(1)$), is important.
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Reduced Mass
\[\mu\].
The educed for masses
\[m, \: M\]is labelled
\[\mu\]and satisfies
\[\frac{1}= \frac{1}{m} + \frac{1}{M} \rightarrow \mu = \frac{mM}{m+M}\]
The reduced mass is less that the total mass
\[M=m\]and in fact less than each individual mass.
Or ital equations and equations using Newton's Second Law of motion can be rewritten in terms of the reduced mass.
Example: Suppose planets 1 and 2 with masses
\[M\]and m
\[m\]are in orbit.
1 will be attracted to 2 with a force
\[\mathbf{F}_{12}= \frac{GmM}{r^2} \mathbf{e}_{12} \]
1 will therefore experience an acceleration
\[\mathbf{a}_{12} = \frac{\mathbf{F}_{12}}{M}= \frac{Gm}{r^2} \mathbf{e}_{12} \]
2 will be attracted to 1 with a an equal and opposite force
\[- \mathbf{F}_{12}=- \frac{GmM}{r^2} \mathbf{e}_{12} \]
1 will therefore experience an acceleration
\[\mathbf{a}_{21} =- \frac{\mathbf{F}_{12}}{M}= \frac{Gm}{r^2} \mathbf{e}_{12} \]
Subtracting these gives a relative acceleration.
\[\mathbf{a} = \mathbf{a}_{21} - \mathbf{a}_{12}= -\frac{Gm}{r^2} \mathbf{e}_{12} - \frac{GM}{r^2} \mathbf{e}_{12} = - \frac{G \mu Mm}{r^2} \mathbf{e}_{12}\]
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Some NP-hard problems which are exponential on general graphs are subexponential on planar graphs because the treewidth is at most $4.9 \sqrt{|V(G)|}$ and they are exponential in the treewidth.
Basically I am interested if there are subexponential algorithms for PLANAR SAT which is NP-complete.
Let $\phi$ be a CNF formula on variables $x_i$ and the $i$-th clause is $c_i$.
The incidence graph p. 5 $G$ of $\phi$ is on vertices $V(G)=\{x_i\} \cup \{c_i\}$ and edges $(x_i,c_i)$ iff $x_i \in c_i$ or $\lnot x_i \in c_i$.
$\phi$ is in PLANAR SAT if the incidence graph is planar.
Are there subexponential algorithms for PLANAR SAT in terms of $\phi$?
I don't exclude the possibility reduction SAT to PLANAR SAT to make this possible, though SAT still to be exponential and $\phi$ is subexponential because of the increase in the size.
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When I click on the Wolfram output, I see $-\frac{2}{3} \sqrt{1 - \cos(3t)}$ only in the indefinite integral calculation: there are no bounds at all. (There are some specific, numerically computed definite integrals below the symbolic calculation, but this is not where $-\frac{2}{3} \sqrt{1 - \cos(3t)}$ is.)
This is important, because you cannot in general evaluate definite integrals by grabbing Wolfram Alpha's symbolically calculated output for an indefinite integral, and then plugging in bounds. The issue is that the output is not guaranteed to work for all possible choices of bounds. In general it can only be expected to work
when the bounds lie in a specific interval.
This is what that "restricted $t$ values" business is all about. Wolfram Alpha is telling you to look out, because it has detected that the function its algorithm produced may only work as an indefinite integral on an
interval of $t$ values--- not everywhere.
In slightly more precise language: you're putting in $f(t) = \sqrt{1 + \cos(t) \cos(2t) - \sin(t) \sin(2t)}$ and asking for an indefinite integral. An indefinite integral of $f$ is by definition an
antiderivative of $f$, that is, a function whose derivative is $f$. Its output $F(t) = -\frac{2}{3} \sqrt{1 - \cos(3t)}$ has the property that $F'(t) = f(t)$, but only for $t$ in a restricted interval. The equality $F'(t) = f(t)$ does not hold for all $t$. So you can't evaluate all definite integrals involving $f$, by plugging into $F$.
Recall the hypotheses of the theorem that lets you evaluate integrals with antiderivatives, namely, the fundamental theorem of calculus. It says: if you have a nice enough function $f(t)$ defined on $[a,b]$ and you have some function $F$ that satisfies $F'(t) = f(t)$ for
all $t$ in $[a,b]$, then$$
\int_a^b f(t) \, dt = F(b) - F(a).
$$If $F'(t) = f(t)$ does not hold for all $t$ in $[a,b]$, then the FTC doesn't apply, and you have no reason to expect $\int_a^b f(t) \, dt$ to be $F(b) - F(a)$. That's what's going on here.
To convince yourself that that's what's going on here, note that your $f(t)$ is nonnegative for all $t$ (graph it). If $F(t)$ were to satisfy $F'(t) = f(t)$ for all $t$, then because the derivative of $F$ is nonnegative, the function $F$ would have to be
nondecreasing, on the whole real line. But you can see (look at the plot of $F$) that it isn't.
Another way to convince yourself: ask Wolfram to plot the derivative of $F$, and also $f$, on the same axes. You'll see that the graphs match, but only on some intervals, and
not on the entire interval $[0, 2\pi]$ (in particular: $F'$ has jump discontinuities in this interval, and $f$ doesn't). So: you can't expect the integral to be $F(2\pi) - F(0)$.
Why is Wolfram bugging out? I don't know the precise algorithm it's using, but roughly, it's because the function $f$ you're asking to integrate on $[0,2\pi]$ is not differentiable on all of $[0,2\pi]$--- the graph has several sharp corners. Symbolic calculators aren't good at handling functions like this. (Humans aren't, either: doing calculus with such functions generally requires working in pieces, and paying very close attention to the hypotheses of each calculational step--- ie, you're not just manipulating symbols anymore.)
Here is another perspective. If you look up what "substitution" for definite integrals actually says--- as a statement that, under specific hypotheses, one can rewrite one definite integral as another--- you'll find hypotheses preventing the calculation suggested by Wolfram Alpha from working for your bounds. The familiar process of symbolic replacement we know for "doing a change of variables"--- where you replace symbols with other symbols and put in new numbers as the bounds--- is guaranteed to work only when the relationship between the old and new variable defines a one-to-one function. If you plot $s = \cos(u) + 1$ as a function of $u = 3t$ on the $u$-interval from $3 \cdot0$ to $3 \cdot 2\pi$ you see it's a wavelike graph, not a one-to-one function (it fails the horizontal line test).
I hope one of these perspectives helped.
[I should say: your issue isn't all that uncommon. You can expect it almost any time you are symbolically evaluating definite integrals using symbolic formulas found via "trigonometric substitutions". I don't think any software package is any better at it than Wolfram Alpha is.]
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Consider the Lagrangian density
\[ \mathcal{L} (\tilde{\phi},\phi) = \tilde{\phi}\left((\partial_t + D_A(r_A – \nabla^2)\right)\phi – u\tilde{\phi}(\tilde{\phi} – \phi)\phi + \tau \tilde{\phi}^2\phi^2. \]
Particle physicists of the 1970s would recognize this as the Lagrangian for a Reggeon field theory with triple- and quadruple-Pomeron interaction vertices. In the modern literature on theoretical ecology, it encodes the behaviour of a spatially distributed predator-prey system near the predator extinction threshold.
Such is the perplexing unity of mathematical science: formula
X appears in widely separated fields A and Z. Sometimes, this is a sign that a common effect is at work in the phenomena of A and those of Z; or, it could just mean that scientists couldn’t think of anything new and kept doing whatever worked the first time. Wisdom lies in knowing which is the case on any particular day.
[Reposted from the archives, in the light of John Baez’s recent writings.]
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Intertemporal choice is the study of how people make choices about what and how much to do at various points in time, when choices at one time influence the possibilities available at other points in time. These choices are influenced by the relative value people assign to two or more payoffs at different points in time. Most choices require decision-makers to trade off costs and benefits at different points in time. These decisions may be about savings, work effort, education, nutrition, exercise, health care and so forth. Since early in the twentieth century, economists have analyzed intertemporal decisions using the discounted utility model, which assumes that people evaluate the pleasures and pains resulting from a decision in much the same way that financial markets evaluate losses and gains, exponentially ‘discounting’ the value of outcomes according to how delayed they are in time. Discounted utility has been used to describe how people actually make intertemporal choices and it has been used as a tool for public policy. Policy decisions about how much to spend on research and development, health and education all depend on the discount rate used to analyze the decision.
[1]
The Keynesian consumption function was based on two major hypotheses. Firstly, marginal propensity to consume lies between 0 and 1. Secondly, average propensity to consume falls as income rises. Early empirical studies were consistent with these hypotheses. However, after World War II it was observed that savings did not rise as incomes rose. The Keynesian model therefore failed to explain the consumption phenomenon and thus emerged the theory of intertemporal choice. Intertemporal choice was introduced by John Rae in 1834 in the "Sociological Theory of Capital". Later, Eugen von Böhm-Bawerk in 1889 and Irving Fisher in 1930 elaborated on the model. A few other models based on intertemporal choice include the Life Cycle Income Hypothesis proposed by Modigiliani and the Permanent Income Hypothesis proposed by Milton Friedman. The concept of Walrasian Equilibrium may also be extended to incorporate intertemporal choice. The Walrasian analysis of such an equilibrium introduces two "new" concepts of prices: futures prices and spot prices.
Contents Fisher's Model of Intertemporal Consumption 1 Modigliani's Life Cycle Income Hypothesis 2 Friedman's Permanent Income Hypothesis 3 Hyperbolic discounting 4 See also 5 References 6 Fisher's Model of Intertemporal Consumption
Intertemporal budget constraint with consumption of period 1 and 2 on x-axis and y-axis respectively.
The figure depicts the intertemporal choice exercised by the consumer, given the utility preferences and the budget constraint.
Irving Fisher developed the theory of intertemporal choice in his book
Theory of interest (1930). Contrary to Keynes, who related consumption to current income, Fisher’s model showed how rational forward looking consumers choose consumption for the present and future to maximize their lifetime satisfaction.
According to Fisher, an individual's impatience depends on four characteristics of his income stream: the size, the time shape, the composition and risk. Besides this, foresight, self-control, habit, expectation of life, and bequest motive (or concern for lives of others) are the five personal factors that determine a person's impatience which in turn determines his time preference.
[2]
In order to understand the choice exercised by a consumer across different periods of time we take consumption in one period as a composite commodity. Suppose there is one consumer, N commodities, and two periods. Preferences are given by U(x_1, x_2) where x_t = (x_{t1},\dots,x_{tN}). Income in period t is Y_t. Savings in period 1 is S_1, spending in period t is C_t, and r is the interest rate. If the person is unable to borrow against future income in the first period, then he is subject to separate budget constraints in each period:
C_1 + S_1 \le Y_1, (1)
C_2 \le Y_2 + S_1(1 + r). (2)
On the other hand, if such borrowing is possible then the person is subject to a single intertemporal budget constraint:
C_1 + \frac{C_2}{1+r} = Y_1 + \frac{Y_2}{1+r}. (3)
The left hand side shows the present value of expenditure and right hand side depicts the present value of income. Multiplying the equation by (1+r) would give us the corresponding future values.
Now the consumer has to choose a C_1 and C_2 so as to
Maximize U(C_1,C_2) subject to C_1+C_2/(1+r) = Y_1 + Y_2/(1+r).
If the consumer is a net saver, an increase in interest rate will have an ambiguous effect on the current consumption.
If the consumer is a net borrower, an increase in interest rate will reduce his current consumption.
A consumer may be a net saver or a net borrower. If he's initially at a level of consumption where he's neither a net borrower nor a net saver, an increase in income may make him a net saver or a net borrower depending on his preferences. An increase in current income or future income will increase current and future consumption(consumption smoothing motives).
Now, let us consider a scenario where the interest rates are increased. If the consumer is a net saver, he will save more in the current period due to the substitution effect and consume more in the current period due to the income effect. The net effect thus becomes ambiguous. If the consumer is a net borrower, however, he will tend to consume less in the current period due to the substitution effect and income effect thereby reducing his overall current consumption.
[3] Modigliani's Life Cycle Income Hypothesis
The Life Cycle Hypothesis is based on the following model:
\max U_t = \sum_t U(C_t)(1+\delta)^{-t}
subject to
\sum_t C_t (1+r)^{-t}=\sum_t Y_t (1+r)^{-t} +W_0,
where
U( C ) is satisfaction received from consumption in time period t t, C is the level of consumption at time t t, Y is income at time t t, δ is the rate of time preference ( a measure of individual preference between present and future activity), W 0 is the initial level of income producing assets.
Life Cycle Hypothesis
Typically, a person’s MPC (marginal propensity to consume) is relatively high during young adulthood, decreases during the middle-age years, and increases when the person is near or in retirement. The Life Cycle Hypothesis(LCH) model defines individual behavior as an attempt to smooth out consumption patterns over one's lifetime somewhat independent of current levels of income. This model states that early in one's life consumption expenditure may very well exceed income as the individual may be making major purchases related to buying a new home, starting a family, and beginning a career. At this stage in life the individual will borrow from the future to support these expenditure needs. In mid-life however, these expenditure patterns begin to level off and are supported or perhaps exceeded by increases in income. At this stage the individual repays any past borrowings and begins to save for her or his retirement. Upon retirement, consumption expenditure may begin to decline however income usually declines dramatically. In this stage of life, the individual dis-saves or lives off past savings until death.
[4] [5] Friedman's Permanent Income Hypothesis
After the Second World War, it was noticed that a model in which current consumption was just a function of current income clearly too simplistic. It could not explain the fact that the long-run average propensity to consume seemed to be roughly constant despite the marginal propensity to consume being much lower. Thus Milton Friedman's Permanent Income Hypothesis is one of the models which seeks to explain this apparent contradiction.
According to the Permanent Income Hypothesis, permanent consumption, C
P, is proportional to permanent income, Y P. Permanent income is a subjective notion of likely medium-run future income. Permanent consumption is a similar notion of consumption.
Actual consumption, C, and actual income, Y, consist of these permanent components plus unanticipated transitory components, C
T and Y T, respectively.
C
P t =β 2Y P t
C
t = C P t + C T t
Y
t = Y P t + Y T t [6] Hyperbolic discounting
The article so far has considered cases where individuals make intertemporal choices by considering the present discounted value of their consumption and income. Every period in the future is exponentially discounted with the same interest rate. A different class of economists, however, argue that individuals are often affected by what is called the temporal myopia. The consumer's typical response to uncertainty in this case is to sharply reduce the importance of the future of their decision making.This effect is called hyperbolic discounting. In the common tongue it reflects the sentiment “Eat, drink and be merry, for tomorrow we may die.”
[7]
Mathematically, it may be represented as follows: f_H(D)=\frac{1}{1+kD}\,
where,
f( D):discount factor, D: delay in the reward, k:parameter governing the degree of discounting [8]
When choosing between $100 or $110 a day later,individuals may impatiently choose the immediate $100 rather than wait for tomorrow for an extra $10. Yet, when choosing between $100 in a month or $110 in a month and a day, many of these people will reverse their preferences and now patiently choose to wait the additional day for the extra $10.
[9] See also References ^ S. Berns, Gregory; Laibson, David; Loewenstein, George. "Intertemporal choice – Toward an Integrative Framework". ^ Thaler, Richard.H. (1997). "Irving Fisher: The Modern Behavioral Economist". The American Economic Review. ^ Varian, Hal (2006). Intermediate Micro Economics. ^ Mankiw, Gregory (5th edition). Macroeconomics. ^ Barro, Robert J (5th edition). Macroeconomics. ^ "Adaptive expectations: Friedman's permanent income hypothesis". ^ Bellows, Alan. "Hyperbolic Discounting". ^ Hyperbolic discounting ^ P. Redden, Joseph. "Hyperbolic Discounting".
(1) http://nordhaus.econ.yale.edu/documents/Disounting_fredericketal_000.pdf
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a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached).
b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct.
However,
your calculation of amplitude is not correct.
Peak-to-trough is
twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly.
CALCULATION OF AMPLITUDE OF OSCILLATIONS
1. Equation of Motion Method
First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision.
Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$
2. Energy Method
Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$.
At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$.
At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$.
By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1.
(c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it.
Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal.
Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is
always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result.
The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate.
Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$
We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off.
The total energy at P is the same as at U, so from above (
Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
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Let $\Sigma = \{0, 1\}$ and $$L = \{ w \in \Sigma^* | w = \mathrm{odd}(w) · \mathrm{even}(w) \},$$ where $\mathrm{odd}(w) = w_1, w_3, w_5 \ldots{}$ and $\mathrm{even}(w) = w_2, w_4, w_6, \ldots{}$ are the words constructed from the characters in all the odd and even positions of the string. I.e. $w = 0110110110 \in L$, because $\mathrm{odd}(w) = 01101, \mathrm{even}(w) = 10110$ and $w = 01101 \cdot{} 10110 = \mathrm{odd}(w) \cdot{} \mathrm{even}(w)$.
Is $L$ regular?
I've tried using the Pumping Lemma for strings of the sort $0^{2n+1}1$ but I can't fix the value of $v^i$ from $w = uvx$ to have an either even or odd length.
I also tried to construct the DFA but it seems to need an infinite number of states to "track" which characters of $w$ get mapped to $\mathrm{odd}(w) \cdot \mathrm{even}(w)$, so my best bet is it's irregular, which means I need to find a way to prove it with the Pumping-Lemma.
Any ideas how I can find a suitable string for the Pumping Lemma?
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This is related to my academic project
An extended finite state machine is a tuple $SM=(I,S,T)$ (simplified):
$I$ is the set of identifiers and it's divided into two sets Inputs and outputs, for simplification we will just consider boolean variables. Let $\sigma$ be the evaluation function $\sigma :I \to \{0,1\}$ which associate to very identifier its value, And let $\Sigma$ be the set of all evaluations. $S$ is the set of states
$T\subseteq S\times S \times G\times A$ the set of transitions, a transition $t=(s_1,s_2,g,a)$ means :
$s_1$ the source state $s_2$ the target state
$g$ the guard,which is usual expressed in the guard language and here we can just consider it as its semantic $g:\Sigma \to \{0,1\}$ where $g(\sigma)$ is an element of $\{0,1\}$.
$a$ is usually expressed in the action language, and we will here identify it with its semantic $a:\Sigma \to \Sigma $ where $f(\sigma)$ is another evaluation function.
Now a configuration of the state machine $SM$ is a tuple $(s_i,\sigma)$ where $s_i$ is the current state and the $\sigma$ the current evaluation function. A run of the machine is : $$(\sigma_0,s_0)\to(\sigma_1,s_{1})\to (\sigma_2,s_2)\to \cdots (\sigma_k,s_k)\to(\sigma_{k+1},s_{k+1})\to \cdots $$ where for every $k$ we have $(s_k,s_{k+1},g,a)$ is a transition and $g(\sigma_k)=1$ (meaning true) and $a(\sigma_k)=\sigma_{k+1}$
Example of action : we denote an action by $Rep(a)=[e_1,\cdots,e_n] $ where $e_i$ is a boolean expression over the variables in $I=[o_1,\cdots,o_n,i_1,\cdots ,i_k]$ where $i_j$ are the inputs of the system and are determined by the environment. And from this representation of the action we can have $a(\sigma)(x)=e_j(\sigma) $ if $x=o_j$ and $a(\sigma)(x)=i_j$ if $x=i_j$
The guards are represented by expressions over the set of the variables $I$
What I am able to do : I was able to generate some boolean expressions that are valid or satisfiable using either well known examples, or using the threshold density of boolean formulas. Hence I am able to generate the guards and the actions (which are random, and have some properties like satisfiable or valid)
Question :how could I generate a random Extended State machine with the property that: Every state is reachable and every transition can be executed
you can use that fact that I am able to generate a formula that evaluates to $1$ or $0$ and I can generate a satisfiable or unsatisfiable formula
Any other ideas to do tests with Extended State Machines? Thanks
Edit 29/04/2016
As my question is either not very clear or difficult; I have an alternative question:
Question 2:what are some well known random Extended State Machines for which every state is reachable.
The purpose of my project is to test the efficiency of an algorithm
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EDIT AT 10/12/06:
ok, this is pretty much the best construction I can get, see if any one come up with better ideas.
Theorem. For each $n$ There is an $(5n+12)$-state NFA $M$ over alphabets $\Sigma$ with $|\Sigma|=5$ such that the shortest string not in $L(M)$ is of length $(2^n-1)(n+1)+1$.
This will give us $f(n) = \Omega(2^{n/5})$.
The construction is pretty much the same with the one in Shallit's, except we construct an NFA directly instead of representing the language by a regular expression first. Let
$\Sigma = \{{0 \brack 0},{0 \brack 1},{1 \brack 0},{1 \brack 1},\sharp\}$.
For each $n$, we are going to construct an NFA recognizing language $\Sigma^*-\{s_n\}$, where $s_n$ is the following sequence (take $n=3$ for example):
$s_3 = \sharp{0 \brack 0}{0 \brack 0}{0 \brack 1}\sharp{0 \brack 0}{0 \brack 1}{1 \brack 0}\sharp \ldots \sharp{1 \brack 1}{1 \brack 1}{0 \brack 1}\sharp$.
The idea is that we can construct an NFA consists of five parts;
a starter, which ensures the string starts with $\sharp{0 \brack 0}{0 \brack 0}{0 \brack 1}\sharp$; a terminator, which ensures the string ends with $\sharp{1 \brack 1}{1 \brack 1}{0 \brack 1}\sharp$; a counter, which keeps the number of symbols between two $\sharp$'s as $n$; an add-one checker, which guarantees that only symbols with the form $\sharp{x \atop x+1}\sharp$ appears; finally, a consistent checker, which guarantees that only symbols with the form $\sharp{x \atop y}\sharp{y \atop z}\sharp$ can appear concurrently.
Note that we do want to accept $\Sigma^*-\{s_n\}$ instead of $\{s_n\}$, so once we find out that the input sequence is disobeying one of the above behaviors, we accept the sequence immediately. Otherwise after $|s_n|$ steps, the NFA will be in the only possible rejecting state. And if the sequence is longer than $|s_n|$, the NFA also accepts. So any NFA satisfies the above five conditions will only reject $s_n$.
It may be easy to check the following figure directly instead of a rigorous proof:
We start at the upper-left state. The first part is the starter, and the counter, then the consistent checker, the terminator, finally the add-one checker. All the arc with no terminal nodes point to the bottom-right state, which is an all time acceptor. Some of the edges are not labeled due to lack of spaces, but they can be recovered easily. A dash line represents a sequence of $n-1$ states with $n-2$ edges.
We can (painfully) verify that the NFA rejects $s_n$ only, since it follows all the five rules above. So a $(5n+12)$-state NFA with $|\Sigma|=5$ has been constructed, which satisfies the requirement of the theorem.
If there's any unclearliness/problem with the construction, please leave a comment and I'll try to explain/fix it.
This question has been studied by Jeffrey O. Shallit et al., and indeed the optimal value of $f(n)$ is still open for $|\Sigma|>1$. (As for unary language, see the comments in Tsuyoshi's answer)
In page 46-51 of his talk on universality, he provided a construction such that:
Theorem. For $n\geq N$ for some $N$ large enough, there is an $n$-state NFA $M$ over binary alphabets such that the shortest string not in $L(M)$ is of length $\Omega(2^{cn})$ for $c=1/75$.
Thus the optimal value for $f(n)$ is somewhere between $2^{n/75}$ and $2^n$. I'm not sure if the result by Shallit has been improved in recent years.
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For a homework , I struggled to solve the following question but couldn't go further:
endowment of person 1 = (30,0)endowment of person 2 = (0,20)
utility functions are such that:\begin{eqnarray*} U (a_1,b_1) & = & \min(a_1,b_1) \\\\ U (a_2,b_2) & = & \min(4a_2,b_2).\end{eqnarray*}
What I am doing is setting a1 equal to b1 and 4a2 equal to b2. After that I'm writing these; $$ p_1 a_1 + p_2 b_1 = 30p_1 \mbox{ and } p_1a_2 + p_24a_2 = 20p_2 $$ Finally, by looking at feasibility condition, I'm writing down: \begin{eqnarray*} \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} & = & 30 \\ \\ \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} & = & 20. \end{eqnarray*}
Here, if I do some calculations, they result in $p_1 = p_2$ and accordingly $a_1=b_1=15, a_2=4$ and $b_2=16.$ But then the problem is that there is excess demand for good2 and there isn't a WE.
Alternatively, I am considering to have $p_2 = 0$ so that there exist a Walrasian Equilibrium.
BUT, I'm stuck at this point and lack the correct intuition for solving next steps. Or, I might be in a completely incorrect way. Please, explain me what will be done with this excess demand?
Thank you in advance.
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Automatic Feature Extraction
Summary
"
Automatic Feature Extraction (AFE) is a technique by which approximate eigen-functions can be extracted from gram matrices constructed via kernel functions. These eigen-functions are features engineered by a particular kernel."" Some Mathematical Background¶ Definitions¶
Let X_k \ \in \ \mathbb{R}^d \ , \ k = 1, \cdots ,n be a random sample drawn from a distribution F(x). Let C \in \mathbb{R}^d be a compact set such that, \mathcal{H} = \mathcal{L}^2(C) be a Hilbert space of functions given by the inner product below.
Further let M(\mathcal{H}, \mathcal{H}) be a class of linear operators from \mathcal{H} to \mathcal{H}.
Nyström method¶
Automatic Feature Extraction (AFE) using the Nyström method aims at finding a finite dimensional approximation to the kernel eigenfunction expansion of Mercer kernels, as shown below.
It is well known that Mercer kernels form a
Reproducing Kernel Hilbert Space ( RHKS) of functions. Every Mercer kernel defines a unique RHKS of functions as shown by the Moore-Aronszajn theorem. For a more involved treatment of RHKS and their applications the reader may refer to the book written by Bertinet et.al.
Mercer's theorem states that the spectral decomposition of integral operator of K, \mathcal{T} \in M(\mathcal{H},\mathcal{H}) defined below yields the eigenfunctions which span the RHKS generated by K and having an inner product defined as above.
Equation above is more commonly also known as the Fredholm integral equation of the first kind. Nyström's method method approximates this integral using the quadrature constructed by considering a finite kernel matrix constructed out of a prototype set X_k \ k = 1, \cdots, m and calculating its spectral decomposition consisting of eigenvalues \lambda_k and eigen-vectors u_k. This yields an expression for the approximate non-linear feature map \hat{\phi} : \mathbb{R}^d \longrightarrow \mathbb{R}^m.
AFE in DynaML Kernels¶
The
SVMKernel[M] contains an implementation of AFE in the method
1 2 3 4 featureMapping( decomposition: (DenseVector[Double], DenseMatrix[Double]))( prototypes: List[DenseVector[Double]])( data: DenseVector[Double]): DenseVector[Double]
Note
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SVMKernel class is extended by all the implemented library kernels in DynaML thereby enabling the use of AFE in potentially any model employing kernels.
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$$\newcommand{\holonomy}{[\mathcal{H\mathbb{O} \ell}]}$$
This answer is an expansion of Qmechanic's comment.
Holonomy
Holonomy can be imagined as the integral, or global version, of the Riemann Curvature Tensor. The Riemann Curvature tensor, indeed is
$$R_{\mu\nu\rho}^\sigma=\mbox{d}\holonomy$$
Where $\mathcal{\holonomy}$ is the Holonomy.
Holonomy Groups
Now, this holonomy is the group action of the
of the manifold. So, in other words, the holonomy of the identity of the holonomy group (not doing any sort of a transport) doesn't do anything to a point on the manifold, and that holonomies are hand - wavily, sort - of "associative" (use this statement with caution!), i.e., instead of writerighteing $\phi(g,x)$ or something, if we choose to write something like, say, $g\dagger x$, then: Holonomy group
$$g\dagger\left(h\dagger x\right)=\left(gh\right)\dagger x$$
Oh, and the first statpement becomes, :
$$e\dagger x =x $$
Now, this is not as trivial as it looks. $e$ is the identity of the
, holonomy group of the manifold! . NOT So,
where does $G(2)$ come in?
Now, where in the world does $G(2)$ come from? $G(2)$ is a holonomy group of
, called $G(2)$ manifolds. This means that it is $\bf{\mathbf{\it{7}}}$-dimensional manifolds to use this as a compactification manifold for M-theory. M-theory has a supersymmetry of $\mathcal N=8$. But, if we waNt a supersymmetry of $\mathcal{N}=1$ (accessible at lower energies), n the compactificaqtion manifoldk must get rid of $\frac78$ of the supersymmetry, i.e. retain only $\frac18$. possible
It so happens to be that $G(2)$ manifolds do indeed satisfy this criterion.
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I have a circuit that consists of a voltage source in series with a resistor and a parallel of an inductor with a capacitor. I have to determine the resonance frequency of this circuit. I am very confused about this. First of all the impedance of the circuit is given by
$$Z=R+j(\frac{\omega L}{1-\omega^2 L C})$$
Now my thought was immediately to proceed as I did in the RLC series circuit i.e. to make the imaginary part of Z zero.
I obtain \$\omega=0\$. However the correct answer should be: $$\omega=\frac{1}{\sqrt{LC}}$$
But that would make the imaginary part of the impedance infinity and therefore the whole impedance infinity and the current zero. But isn't impedance when the current reaches its peak?
I'm so confused about all of this. I read that there is series impedance and parallel resonance but what should I peak and why are there 2 types of resonance. Shouldn't they be equivalent? Isn't resonance just the circuit behaving as a resistor?
Can someone explain me what is going on? Thanks!
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The Hubbard model is a model to describe electrons in a lattice. In general, the Hubbard model Hamiltonian $H$ contains two terms:
The kinetic term:$$T=-t\sum_{\langle ij\rangle\sigma} [c_{i\sigma}^\dagger c_{j\sigma} + h.c.] $$ The onsite Coulomb interaction term:
$$U=u\sum_{i=1}^N n_{i\uparrow}n_{i\downarrow}$$
So my question is: why can we not diagonalize the Hamiltonian: $H=T+U$? Some books attribute the reason that $T$ doesn't commute with $U$, therefore we need to formulate a perturbation theory. In particular, I want to know whether the space of solution of $T$ has the same dimension compared to the space of solution of $U$ due to $[T,U] \neq 0$?
Edit:
For $T$ operator we have the following eigenequation: $$T|n\rangle=T_n|n\rangle$$ For $U$ operator we have another one eigenequation: $$U|\alpha\rangle=U_\alpha|\alpha\rangle$$ Due to $[T,U] \neq 0$, I am wandering whether $|n\rangle$ has the same dimension compared to $|\alpha\rangle$? And why $$[T+U]|N\rangle \overset{?}{=} H_N|N\rangle.$$
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A particle moves along the x-axis so that at time t its position is given by $x(t) = t^3-6t^2+9t+11$ during what time intervals is the particle moving to the left? so I know that we need the velocity for that and we can get that after taking the derivative but I don't know what to do after that the velocity would than be $v(t) = 3t^2-12t+9$ how could I find the intervals
Fix $c\in\{0,1,\dots\}$, let $K\geq c$ be an integer, and define $z_K=K^{-\alpha}$ for some $\alpha\in(0,2)$.I believe I have numerically discovered that$$\sum_{n=0}^{K-c}\binom{K}{n}\binom{K}{n+c}z_K^{n+c/2} \sim \sum_{n=0}^K \binom{K}{n}^2 z_K^n \quad \text{ as } K\to\infty$$but cannot ...
So, the whole discussion is about some polynomial $p(A)$, for $A$ an $n\times n$ matrix with entries in $\mathbf{C}$, and eigenvalues $\lambda_1,\ldots, \lambda_k$.
Anyways, part (a) is talking about proving that $p(\lambda_1),\ldots, p(\lambda_k)$ are eigenvalues of $p(A)$. That's basically routine computation. No problem there. The next bit is to compute the dimension of the eigenspaces $E(p(A), p(\lambda_i))$.
Seems like this bit follows from the same argument. An eigenvector for $A$ is an eigenvector for $p(A)$, so the rest seems to follow.
Finally, the last part is to find the characteristic polynomial of $p(A)$. I guess this means in terms of the characteristic polynomial of $A$.
Well, we do know what the eigenvalues are...
The so-called Spectral Mapping Theorem tells us that the eigenvalues of $p(A)$ are exactly the $p(\lambda_i)$.
Usually, by the time you start talking about complex numbers you consider the real numbers as a subset of them, since a and b are real in a + bi. But you could define it that way and call it a "standard form" like ax + by = c for linear equations :-) @Riker
"a + bi where a and b are integers" Complex numbers a + bi where a and b are integers are called Gaussian integers.
I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...
Does anyone know if $T: V \to R^n$ is an inner product space isomorphism if $T(v) = (v)_S$, where $S$ is a basis for $V$? My book isn't saying so explicitly, but there was a theorem saying that an inner product isomorphism exists, and another theorem kind of suggesting that it should work.
@TobiasKildetoft Sorry, I meant that they should be equal (accidently sent this before writing my answer. Writing it now)
Isn't there this theorem saying that if $v,w \in V$ ($V$ being an inner product space), then $||v|| = ||(v)_S||$? (where the left norm is defined as the norm in $V$ and the right norm is the euclidean norm) I thought that this would somehow result from isomorphism
@AlessandroCodenotti Actually, such a $f$ in fact needs to be surjective. Take any $y \in Y$; the maximal ideal of $k[Y]$ corresponding to that is $(Y_1 - y_1, \cdots, Y_n - y_n)$. The ideal corresponding to the subvariety $f^{-1}(y) \subset X$ in $k[X]$ is then nothing but $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$. If this is empty, weak Nullstellensatz kicks in to say that there are $g_1, \cdots, g_n \in k[X]$ such that $\sum_i (f^* Y_i - y_i)g_i = 1$.
Well, better to say that $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$ is the trivial ideal I guess. Hmm, I'm stuck again
O(n) acts transitively on S^(n-1) with stabilizer at a point O(n-1)
For any transitive G action on a set X with stabilizer H, G/H $\cong$ X set theoretically. In this case, as the action is a smooth action by a Lie group, you can prove this set-theoretic bijection gives a diffeomorphism
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I have the following normal cone inclusion
$$-(A x + b) \in \mathcal{N}_\mathcal{C}(x) \qquad (1)$$
where $\mathcal{N}_\mathcal{C}$ denotes the normal cone to the convex set $\mathcal{C}$ at the point $x \in \mathcal{C}$. Matrix $A$ is non-symmetric.
Normally if $A$ would be symmetric, the convex optimization problem is
$$\min \frac{1}{2} x^\top A x + b^\top x + I_{\mathcal{C}}(x) \qquad (2)$$ where $I_{\mathcal{C}}(x)$ is the indicator function of the set $\mathcal{C}$. The inclusion (1) is the necessary and sufficient optimality condition of the convex optimization problem (2).
However, what are the implications of $A$ beeing non-symmetric and how does that relate to an optimization function? I guess there does not exist an optimization function?
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First of all, it is every integer (not every rational) that can be written as you have presented it as a sequence $a_n\in\{0,\ldots, p-1\}$. The number $1/p$ cannot be so represented.
To answer your cardinality question, there are at least two arguments that $\mathbb Z_p$ is uncountable, and thus that $\mathbb Q_p$ is uncountable.
From your presentation of the $p$-adics as the series $\sum a_n p^n$, we see that the $\mathbb Z_p$ has the cardinality of the set of maps $\mathbb N^+\rightarrow \{0,\ldots, p-1\}$, (i.e., $n\mapsto a_n$) and hence uncountable.
Another argument is to start with the fact that $\mathbb Z_p$ is infinite and (Hausdorff) compact, e.g., because $\mathbb Z_p= \lim \mathbb Z/p^{n}$, so a closed set of a compact set, hence compact (and infinite). By Baire's theorem, in a compact Hausdorff space, the countable union of nowhere dense closed sets cannot contain a non-empty open set. In particular, an infinite countable set cannot be (Hausdorff and) compact.
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It is relatively easy to derive the
potential energy stored into the magnetic field of an uniformly magnetized sphere of radius $R$ and total magnetic moment $\mu$ :\begin{equation}\tag{1}U_{\text{magn}} = \frac{\mu_0 \, \mu^2}{4 \pi R^3}.\end{equation}
The field is uniform inside the sphere, and dipolar on the exterior side. It is exerting a pressure that I want to calculate. In thermodynamics, the pressure can be defined as the
partial derivative of the "internal" energy relative to a change of volume :\begin{equation}\tag{2}P = -\, \frac{\partial U}{\partial V}.\end{equation}Since $V = 4 \pi R^3 / 3$, it is tempting to derive directly (1) (assuming $\mu = \text{constant}$), to get this relation :\begin{align}P_{\text{magn} \, 1} = -\, \frac{\partial \,}{\partial V} \Big( \frac{\mu_0 \, \mu^2}{3 V} \Big) &= \frac{\mu_0 \, \mu^2}{3 V^2} \\[12pt]&\equiv \frac{U_{\text{magn}}}{V}. \tag{3}\end{align}This recalls the stiff equation of state $p = \rho$.
I don't think that $\mu$ could be considered as an independant variable (changing the sphere radius may have an effect on the total dipolar moment, unless there's some kind of constraint). The polar field $B_{\text{pole}}$ is related to $\mu$ by this :\begin{equation}\tag{4}\mu(R, B_{\text{pole}}) = \frac{2 \pi B_{\text{pole}}^2 \, R^3}{\mu_0}.\end{equation}If I assume that $B_{\text{pole}}$ is the independant variable (it may be some constraint), then substituting (4) into (1) and doing the derivative gives this relation instead (a
negative pressure = tension !) :\begin{equation}\tag{5}P_{\text{magn} \, 2} = -\, \frac{\partial \,}{\partial V} \Big( \frac{3 B_{\text{pole}}^2}{4 \mu_0} \, V \Big) = -\, \frac{U_{\text{magn}}}{V}.\end{equation}This recalls the cosmological constant relation of state : $p = -\, \rho$.
There's a third possibility (are there others ?). I can consider the
magnetic flux $\Phi = B_{\text{inside}} \, \pi R^2$ as the independant variable (flux of the sphere's internal field passing through its own equator) :\begin{equation}\tag{6}\Phi = \frac{\mu_0 \, \mu}{2 R}.\end{equation}In this case, the pressure would be\begin{equation}\tag{7}P_{\text{magn} \, 3} = -\, \frac{\partial \,}{\partial V} \Big( \frac{\Phi^2}{\pi \mu_0 \, R} \Big) = \frac{U_{\text{magn}}}{3 V},\end{equation}which recalls the electromagnetic equation of state from relativistic physics : $p = \frac{1}{3} \, \rho$.
I suspect that $P_{\text{magn} \, 3}$ should be the proper pressure of the magnetic field. But how to justify this ?
Take note that $U_{\text{magn}}/V$ is NOT the field energy density, since it's varying from one place to another (the field outside the sphere is not uniform, since it's dipolar). So I'm not sure how to interpret the $P$ above correctly, since it's a constant (i.e not depending on position).
So the question is the following :
What is the total magnetic pressure felt by an outside agent that changes a bit the volume of a magnetized sphere ? I expect this : \begin{equation}\tag{8} P_{\text{magn} \, 3} = \frac{U_{\text{magn}}}{3 V} = \frac{\mu_0 \, \mu^2}{(4 \pi R^3)^2} \equiv \frac{B_{\text{int}}^2}{4 \mu_0} \equiv \frac{B_{\text{pole}}^2}{4 \mu_0}. \end{equation}
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$N_2,N_1$ are constant. $M_0$ is initial manifold, and evolving under volume preserving mean curvature flow. $A$ is the second fundamental form. $H$ is mean curvature. $h(t)$ is the average of mean curvature $$ h(t)=\frac{\int_{M_t} H d\mu}{\int_{M_t} d\mu} $$ And $$ H_T=\max_{t\in[0,T]} \max_{M_t}H $$ Then, how to get the red line 2 and 3 ?
About getting the red line 2, I fail to deal $2N_1(1+h)|A|^2$. Because $C_6$ depends on $N_1$ and $M_0$. I feel part of $C_6$ is from $2N_1(1+h)|A|^2$.
About the red line 3, I am unfamiliar about the $\forall \eta>0$ .... Seemly, there is some all to know but I don't know, what is it?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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Alright, I have this group $\langle x_i, i\in\mathbb{Z}\mid x_i^2=x_{i-1}x_{i+1}\rangle$ and I'm trying to determine whether $x_ix_j=x_jx_i$ or not. I'm unsure there is enough information to decide this, to be honest.
Nah, I have a pretty garbage question. Let me spell it out.
I have a fiber bundle $p : E \to M$ where $\dim M = m$ and $\dim E = m+k$. Usually a normal person defines $J^r E$ as follows: for any point $x \in M$ look at local sections of $p$ over $x$.
For two local sections $s_1, s_2$ defined on some nbhd of $x$ with $s_1(x) = s_2(x) = y$, say $J^r_p s_1 = J^r_p s_2$ if with respect to some choice of coordinates $(x_1, \cdots, x_m)$ near $x$ and $(x_1, \cdots, x_{m+k})$ near $y$ such that $p$ is projection to first $m$ variables in these coordinates, $D^I s_1(0) = D^I s_2(0)$ for all $|I| \leq r$.
This is a coordinate-independent (chain rule) equivalence relation on local sections of $p$ defined near $x$. So let the set of equivalence classes be $J^r_x E$ which inherits a natural topology after identifying it with $J^r_0(\Bbb R^m, \Bbb R^k)$ which is space of $r$-order Taylor expansions at $0$ of functions $\Bbb R^m \to \Bbb R^k$ preserving origin.
Then declare $J^r p : J^r E \to M$ is the bundle whose fiber over $x$ is $J^r_x E$, and you can set up the transition functions etc no problem so all topology is set. This becomes an affine bundle.
Define the $r$-jet sheaf $\mathscr{J}^r_E$ to be the sheaf which assigns to every open set $U \subset M$ an $(r+1)$-tuple $(s = s_0, s_1, s_2, \cdots, s_r)$ where $s$ is a section of $p : E \to M$ over $U$, $s_1$ is a section of $dp : TE \to TU$ over $U$, $\cdots$, $s_r$ is a section of $d^r p : T^r E \to T^r U$ where $T^k X$ is the iterated $k$-fold tangent bundle of $X$, and the tuple satisfies the following commutation relation for all $0 \leq k < r$
$$\require{AMScd}\begin{CD} T^{k+1} E @>>> T^k E\\ @AAA @AAA \\ T^{k+1} U @>>> T^k U \end{CD}$$
@user193319 It converges uniformly on $[0,r]$ for any $r\in(0,1)$, but not on $[0,1)$, cause deleting a measure zero set won't prevent you from getting arbitrarily close to $1$ (for a non-degenerate interval has positive measure).
The top and bottom maps are tangent bundle projections, and the left and right maps are $s_{k+1}$ and $s_k$.
@RyanUnger Well I am going to dispense with the bundle altogether and work with the sheaf, is the idea.
The presheaf is $U \mapsto \mathscr{J}^r_E(U)$ where $\mathscr{J}^r_E(U) \subset \prod_{k = 0}^r \Gamma_{T^k E}(T^k U)$ consists of all the $(r+1)$-tuples of the sort I described
It's easy to check that this is a sheaf, because basically sections of a bundle form a sheaf, and when you glue two of those $(r+1)$-tuples of the sort I describe, you still get an $(r+1)$-tuple that preserves the commutation relation
The stalk of $\mathscr{J}^r_E$ over a point $x \in M$ is clearly the same as $J^r_x E$, consisting of all possible $r$-order Taylor series expansions of sections of $E$ defined near $x$ possible.
Let $M \subset \mathbb{R}^d$ be a compact smooth $k$-dimensional manifold embedded in $\mathbb{R}^d$. Let $\mathcal{N}(\varepsilon)$ denote the minimal cardinal of an $\varepsilon$-cover $P$ of $M$; that is for every point $x \in M$ there exists a $p \in P$ such that $\| x - p\|_{2}<\varepsilon$....
The same result should be true for abstract Riemannian manifolds. Do you know how to prove it in that case?
I think there you really do need some kind of PDEs to construct good charts.
I might be way overcomplicating this.
If we define $\tilde{\mathcal H}^k_\delta$ to be the $\delta$-Hausdorff "measure" but instead of $diam(U_i)\le\delta$ we set $diam(U_i)=\delta$, does this converge to the usual Hausdorff measure as $\delta\searrow 0$?
I think so by the squeeze theorem or something.
this is a larger "measure" than $\mathcal H^k_\delta$ and that increases to $\mathcal H^k$
but then we can replace all of those $U_i$'s with balls, incurring some fixed error
In fractal geometry, the Minkowski–Bouligand dimension, also known as Minkowski dimension or box-counting dimension, is a way of determining the fractal dimension of a set S in a Euclidean space Rn, or more generally in a metric space (X, d). It is named after the German mathematician Hermann Minkowski and the French mathematician Georges Bouligand.To calculate this dimension for a fractal S, imagine this fractal lying on an evenly spaced grid, and count how many boxes are required to cover the set. The box-counting dimension is calculated by seeing how this number changes as we make the grid...
@BalarkaSen what is this
ok but this does confirm that what I'm trying to do is wrong haha
In mathematics, Hausdorff dimension (a.k.a. fractal dimension) is a measure of roughness and/or chaos that was first introduced in 1918 by mathematician Felix Hausdorff. Applying the mathematical formula, the Hausdorff dimension of a single point is zero, of a line segment is 1, of a square is 2, and of a cube is 3. That is, for sets of points that define a smooth shape or a shape that has a small number of corners—the shapes of traditional geometry and science—the Hausdorff dimension is an integer agreeing with the usual sense of dimension, also known as the topological dimension. However, formulas...
Let $a,b \in \Bbb{R}$ be fixed, and let $n \in \Bbb{Z}$. If $[\cdot]$ denotes the greatest integer function, is it possible to bound $|[abn] - [a[bn]|$ by a constant that is independent of $n$? Are there any nice inequalities with the greatest integer function?
I am trying to show that $n \mapsto [abn]$ and $n \mapsto [a[bn]]$ are equivalent quasi-isometries of $\Bbb{Z}$...that's the motivation.
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6. Maclaurin and Taylor series – Edexcel Further Pure Mathematics 2 (FP2) What students need to learn: Third and higher order derivatives.
Derivation and use of Maclaurin series.
The derivation of the series expansion of \(e^x , \sin x, \cos x, ln (1 + x)\) and other simple functions may be required.
Derivation and use of Taylor series.
The derivation, for example, of the expansion of \(\sin x\) in ascending powers of \((x − \pi)\) up to and including the term in \((x − \pi)^3\) .
Use of Taylor series method for series solutions of differential equations.
Students may, for example, be required to find the solution in powers of \(x\) as far as the term in \(x^4\) , of the differential equation \(\dfrac{d^2y}{dx^2} + x \dfrac{dy}{dx} + y = 0\) , such that \(y = 1\), \(\dfrac{dy}{d x} = 0\) at \(x = 0\). Premium Content :
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Here $T$ is a sufficiently strong, effective theory of arithmetic,$\text{Pvbl}$ is the formalized provability predicate of $T$, and $\text{Con}(T)$ is the formalized statement of $T$'s consistency.
The case of #1, $T \vdash \text{Pvbl}(\text{Con}(T))$, is interesting.
If $T$ is satisfied by the standard model of arithmetic, then we have for each $\phi$ that $T \vdash \text{Pvbl}(\phi)$ if and only if $T \vdash \phi$ (weaker assumptions on $T$ will also cause that equivalence to hold). By the second incompleteness theorem, it is impossible for $T \vdash \text{Con}(T)$. So if $T$ is sufficiently reasonable (e.g. if it is satisfied by the standard model), then we will have $T \not \vdash \text{Pvbl}(\text{Con}(T))$. The usual term for this property is "soundness", and in particular the $T$ here would be sound for $\Sigma_1$ formulas such as $\text{Pvbl}(\phi)$.
On the other hand, we know in general that $T \not \vdash \text{Con}(T)$. So $T + \lnot \text{Con}(T)$ is consistent. Let $R = T + \lnot \text{Con}(T)$. Then, for all $\phi$, we will have $R \vdash \text{Pvbl}_R(\phi)$, assuming that we take the "natural" effective axiomatization of $R$.
The proof is as follows. First, $R \vdash \lnot \text{Con}(T)$, which means that $R \vdash \text{Pvbl}_T(\phi)$ for all $\phi$, because $R$ can formalize the usual arguments showing that every sentence is provable from an inconsistent theory. Second, $R \vdash \text{Pvbl}_T(\phi) \to \text{Pvbl}_R(\phi)$, because $T$ is a subtheory of $R$ and because we chose the axiomatization of $R$ so that $R$ can prove $T$ is a subtheory of $R$.
So, if we take the $T$ in the question to be this $R$, we will have$R \vdash \text{Pvbl}_R(\text{Con}(R))$. Of course, we do not have $R \vdash \text{Con}_R$. In particular, the theory $R$ is not satisfied by the standard model, even though it is consistent. This is because $R$ proves the existence of numbers (such as a coded proof of $\text{Con}(R)$) which cannot be found in the standard model. Even more precisely, the theory $R$ is not sound for $\Sigma_1$ sentences;
We can also see that $R$ proves $\lnot \text{Con}(R)$. This does not contradict anything - it is possible for a theory to "prove it is inconsistent", even if the the theory is actually consistent, as this example shows.
So the answer to #1 is that more information about $T$ is needed to know whether $T \vdash \text{Pvbl}( \text{Con}(T))$.
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Covariance Matrix is a measure of how much two random variables gets change together. It is actually used for computing the covariance in between every column of data matrix.
The Covariance Matrix is also known as dispersion matrix and variance-covariance matrix. The covariance between two jointly distributed real-valued random variables X and Y with finite second moments is defined as.
$\LARGE Cov(X,Y)=\sum \frac{(X_{i}-\overline{X})(Y_{i}-\overline{Y})}{N}=\sum \frac{X_{i}Y_{i}}{N}$
Covariance Matrix Formula Solved Examples Question: Calculation of Covariance Matrix from Data Matrix:
Suppose the data matrix $y_{1}=5_{z1-z2}$ and $y_{1}$ = $2_{z2}$ with rows corresponding to subjects and columns are variables. Calculate a mean for each variable and replace the data matrix.
$X$ $N$ $Y$ $X-\overline{X}$ 1 2 -2 -4 2 8 -1 2 3 6 0 0 4 4 1 -2 5 10 2 4
Now the matrix of deviations from the mean is: $Y-\overline{Y}$
Therefore the covariance matrix of the observation is
$Z=\begin{pmatrix} -2 & -4 \\ -1 & 2 \\ 0 & 0 \\ 1 & -2\\ 2 & 4 \end{pmatrix}$
The diagonal elements of this matrix are the variances of the variables, and the off-diagonal elements are the covariances between the variables.
$\frac{1}{N-1}Z^{1}Z=\frac{1}{4}\begin{pmatrix} -2 &-1 &0 &1 & 2\\ -4 &2 &0 &-2 &4 \end{pmatrix}\begin{pmatrix} -2 &-4 \\ -1 &2 \\ 0 &0 \\ 1 &-2 \\ 2 &4 \end{pmatrix}$
$=\frac{1}{4}\begin{pmatrix} 10 &12 \\ 12 &40 \end{pmatrix}$
$=\begin{pmatrix} 2.5 &3.0 \\ 3.0 &10.0 \end{pmatrix}$
$=\begin{pmatrix} S_{x}^{2}&S_{xy}\\ S_{xy} & S_{x}^{2} \end{pmatrix}$
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As the previous answer states, your problem is NP-hard by reduction from Max-Cut. It actually also turns out to be Poly-APX-hard by reduction from Max Independent Set (which is Poly-APX-complete). The reduction is almost a strict reduction, where the optimal solution for the instance of $P$ has as its cost the negative of the cost of the optimal solution for the Max Independent Set instance. The same reduction can also be used to prove NP-hardness.
Suppose we are given a graph $G = (V, E)$ with the task of determining the size of the maximum Independent Set, call it $I$. We reduce this to an instance of problem $P$ as follows:
Let $G' = (V', E')$ where $V' = V \cup \{x\}$ and $E' = E \cup \{(x, v) | v \in V\}$ and let $w : E' \rightarrow \mathbb{Z}^+$ be a weighting of the edges such that whenever $e \in E$ we have $w(e) = |V|$ and whenever $e = (x, v)$ for some $v \in V$ we have $w(e) = -1$.
We claim that the measure for $(G', w)$ is $-|I|$.
Consider the coloring of $G'$ which assigns one color to the set $I \cup \{x\}$ and assigns every other vertex a unique color. Under that coloring, the only monochromatic edges are those with both endpoints in $I \cup \{x\}$. Since the subgraph in $G'$ induced by the vertex set $V$ is exactly the graph $G$, and since $I$ is an independent set in $G$, we know that there are no edges with both endpoints in $I$. Thus the only monochromatic edges in $G'$ are those of the form $(x, v)$ where $v \in I$. There are $|I|$ such edges, each with weight $-1$, so the total weight of all monochromatic edges under this coloring is $-|I|$. The measure for $(G', w)$ is then bounded above by the value $-|I|$.
Next suppose for the sake of contradiction that the measure for $(G, w)$ is less than $-|I|$.
Every negatively weighted edge in $G'$ has weight $-1$. Therefore, in order for the total weight of the monochromatic edges to be less than $-|I|$, there must be more than $|I|$ monochromatic edges of weight $-1$ in $G'$. Every edge of weight $-1$ has $x$ as one end-point. Then since these edges are monochromatic, the other endpoints all have the same color. Thus, we have accumulated a set of vertices $I' \in V$ with $|I'| > |I|$ that are all the same color. Since $I$ is the maximum independent set in $G$, we know that $I'$ is not an independent set, and so there exists an edge $(u, v) \in E$ with endpoints in $I'$. This edge is then monochromatic, and contributes a total of $w((u, v)) = |V|$ to the sum of the weights of the monochromatic edges. The total weight of all of the negatively weighted edges in all of $G'$ is $-|V|$, so already the total monochromatic weight under this coloring must be non-negative. This contradicts the fact that the total weight of the monochromatic edges is supposed to be less than $-|I|$.
By contradiction, the measure of $(G', w)$ must be exactly $-|I|$.
The above should also be enough to see how a solution to the problem $P$ instance can be used to solve the Max independent Set instance (by taking the set of vertices of the same color as $x$ to be the independent set) though we will not go through the details here.
As a result, the above is an approximation-preserving reduction that is sufficient to show that problem $P$ is Poly-APX-hard.
Now a few notes:
Your reasoning was a bit off in several places.
First of all, you described a reduction from $P$ to MEDKP (you solved an instance of $P$ by solving several instances of MEDKP) and concluded that $P$ is at least as hard as MEDKP. In fact, that is backwards. What you can actually conclude is that MEDKP is at least as hard as $P$. To make it clearer, consider the version of problem $P$ that you get if you restrict weights to be non-negative. This problem is trivial, since the measure of every instance $(G, w)$ will always by zero (by taking a coloring where each vertex is uniquely colored). However, your argument applies to this problem without modification (you could solve this problem using several instances of MEDKP). If you were to conclude that that problem is at least as hard as MEDKP, then you would have proved that MEDKP is trivial, which it is clearly not.
The second issue I saw was in the construction of weights $w'$. The issue is that modifying the weights leads to the problem changing. When you increase the weights as you did, the total weight of the monochromatic edges in a coloring is itself increased by a value of $C$ for each monochromatic edge. Therefore the total weight of the monochromatic edges changes by different amounts for different colorings, depending on the number of monochromatic edges in the coloring. The optimal coloring could then change! For example, if you take an instance of $P$ produced by the reduction I provided above, finding an optimal coloring is computationally difficult. On the other hand, if you move all the weights up even just by one, the optimal coloring is trivial to find (simply take a coloring with each vertex uniquely colored). Clearly, the problem has changed since the tractability of solving it has changed so much.
Next, I was wondering about whether you had a reason to generalize to real numbers and not just generalize to all integers. Dealing with integers makes things easier in several ways. For one thing, it is intuitively clear how to encode an integer, and not so much so for a real number; depending on your encoding of real numbers, it could be possibly to embed all sorts of difficult problems into $P$ in ways that don't actually shed any light on the nature of $P$ (i.e. you would be forced to solve the sum-of-square-roots problem (see online; cou) in a situation where you had to compare two solutions whose values were each sums of square roots). In addition, it is much more obvious how to discuss terms like "polynomial in the size of the input" when the size of the input is easy to determine from the size of the numbers in the input. In particular, the integer version of $P$ is clearly in NP (and therefore NP-complete) while the real number version is of some unknown difficulty.
I also want to note that I couldn't think of a polynomial factor approximation algorithm for $P$ (either the real number or integer version), and so just like with the NP-hardness result, $P$ is Poly-APX-hard but not necessarily Poly-APX-complete. I would be very interested in seeing further results in either direction.
Finally, I wanted to say that I really liked the problem and found it very interesting to think about; thanks!
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Subalgebras whose root system is not a subset of the root system of the original algebra are called special subalgebras. Therefore, the generators are not a subset of the original's group generators.
That is not quite right. A special subalgebra is one such that their step operators do not form a subset of the algebra step operators.That is what is meant by a root system not being a subset of another root system. It does not imply that the subalgebra generators are not a subset of the algebra generators.
Example: Consider the three dimensional representation of $\mathfrak g=\mathfrak{su}(3)$, whose generators $T_a=\lambda_a/2$, where $\lambda_a$ are the Gell-Mann matrices. The step operators are
\begin{align*} E_{\pm\alpha_1}=T_1\pm iT_2,\\ E_{\pm\alpha_2}=T_6\pm iT_7,\\ E_{\pm(\alpha_1+\alpha_2)}=T_4\pm iT_5, \end{align*}
Regular embedding: The generators $T_1$, $T_2$ and $T_3$ form a subalgebra $\mathfrak{su}(2)$. This subalgebra step operators are $t_\pm=T_1\pm iT_2=E_{\pm\alpha_1}$. The subalgebra step operators form a subset of the algebra step operators.
Special embedding: Another $\mathfrak{su}(2)$ subalgebra is given by $T_2$, $T_5$ and $T_7$. In this case the step operators are $t_\pm=T_5\pm iT_7$ which cannot be written as a subset of the step operators of $\mathfrak{su}(3)$.
Notice the branching rules for the first embedding is $3=2\oplus 1$ while for the second is $3=3$.
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Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider
(American Physical Society, 2016-02)
The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ...
Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Elsevier, 2016-02)
Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < |\eta| < 4.0$) and associated particles in the central range ($|\eta| < 1.0$) are measured with the ALICE detector in ...
Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Springer, 2016-08)
The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2016-03)
The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ...
Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2016-03)
Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ...
Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV
(Elsevier, 2016-07)
The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ...
$^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2016-03)
The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ...
Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV
(Elsevier, 2016-09)
The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ...
Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV
(Elsevier, 2016-12)
We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ...
Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV
(Springer, 2016-05)
Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ...
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I can’t actually take much credit for this. I made the initial conjecture that sparked this theorem, but my friend John Bytheway was the first person to codify the actual theorem and prove it. My proof of it is pretty independent of his, but I probably wouldn’t have come up with it unless I already knew the theorem was true. (The neccesary part of the theorem is stolen from him, but the sufficiency part is entirely mine).
Note: In this post I take the slightly unconventional approach that a Normal space needn’t be \(T_1\). i.e. a Normal topological space is one in which any two disjoint closed sets have disjoint neighbourhoods, but points needn’t be closed. I could invent a new term for this, like quasinormal, but I really can’t be bothered. Feel free to replace ‘normal’ with ‘quasinormal’ wherever you see it if this makes you more comfortable. It may be readily verified that the usual proof of Urysohn’s lemma in no way depends on the closedness of points, so Urysohn’s lemma holds for this new definition of normal.
First we need some preliminary definitions.
Definition 1:
Let \(X\) be a topological space and \(f : X \to \mathbb{R}\) a function, not neccesarily continuous. For \(x \in X\) we define the oscillation of \(f\) at \(x\) \to be
\(\omega_f(x) = \inf\limits_{U \ni x} \mathrm{diam} f(U)\)
where \(U\) ranges over open sets.
We define the total oscillation of \(f\) to be
\(\delta(f) = \sup\limits_{x \in X} \omega_f (x) \)
We now prove some preliminary results about \(\delta\)
Proposition 2:
Let \(X\) be an arbitrary topological space and consider \(B(X)\) the Banach space of bounded functions on \(X\) and \(C(X)\) the closed subspace consisting of the continuous bounded functions. Consider \(\delta : B(X) \to [0, \infty)\)
\(\delta(f) = 0\) iff \(f\) is continuous. \(\delta(f + g) \leq \delta(f) + \delta(g)\) \(\delta(tf) = |t| \delta(f)\) \(\delta(f) \leq 2 ||f||\) \(\delta(f) \leq 2 d(f, C(X))\)
These are all perfectly trivial to prove, so I’m not going to bother.
What I noticed is that in almost every case I considered the final inequality was in fact an equality. I could prove a weaker result for Compact hausdorff spaces (I showed that \(||f|| \leq \delta(f)\), but I couldn’t do any better, so I passed this over to John to see what he could come up with. He came up with the following theorem, which is the main theorem of this post:
\(\delta(f) = 2 d(f, C(X))\) for every \(f \in B(X)\) iff \(X\) is normal.
I’ll prove this in two parts. For the right to left implication I’ll in fact prove something stronger:
Theorem 4:
Let \(X\) be a normal topological space and let \(f \in B(X)\). Then there is a (not usually unique) continuous function \(h\) with \(||f – h|| = 2 \delta(f)\).
As is my wont, the proof of this will precede by a couple clever definitions and then drop out as practically a corollary of a Big Theorem.
The big theorem in question is the following, which is due to Tong. I conjectured it, tinkered around with proving it for a bit, went to look up something about semicontinuous functions in Engelking’s general topology book and saw the theorem staring out at me from one of the exercises.
Theorem 5:
Let \(X\) be a normal topological space. Let \(f, g : X \to \mathbb{R}\) be upper semicontinuous and lower semicontinuous respectively with \(f \leq g\). There is a continuous function \(h\) with \(f \leq h \leq g\).
Note the direction of the inequality and which are upper and lower semicontinuous respectively. If you reverse this then the theorem becomes false.
I’m not actually going to prove this here – I’ve not yet totally sorted out the details of the proof in my mind. It’s basically a modified version of the standard proof of Urysohn’s lemma, but with additional constraints on the sets constructed. (Update: See here for a proof.)
Anyway, we now make some more definitions:
Definition 6:
Let \(f \in B(X)\). Define
\(f^*(x) = \inf\limits_{U \ni x} \sup f(U)\)
\(f_*(x) = \sup\limits_{U \ni x} \inf f(U)\)
where U ranges over open sets.
Proposition 7:
These satisfying the following properties:
\(f_* \leq f \leq f^*\) \(f_*\) is lower semicontinuous. \(f^*\) is upper semicontinuous. If \(f\) is upper semicontinuous then \(f^* = f\). Similary if \(f\) is lower semicontinuous then \(f_* = f\). The maps \(f \to f^*\) and \(f \to f_*\) are monotone with respect to the pointwise ordering. If \(g, h\) are lower, upper semicontinuous respectively and \(g \leq f \leq h\) then \(g \leq f_* \leq f \leq f^* \leq h\) \(\omega_f(x) = f^*(x) – f_*(x)\)
Again, these are all really very easy to prove (assuming you prove them in order), so I’m not going to do it. I’ll actually not use most of these, but those that I don’t use are of independent interest. i.e. they’re cool. :-)
Now, we have: Proof of theorem 4:
Note that \(\delta(f) = \sup (f^*(x) – f_*(x)\). Consequently we have that
\( f^* – \frac{1}{2}\delta(f) \leq f_* + \frac{1}{2} \delta(f) \)
Now, the left hand side is upper semicontinuous and the right hand side is lower semicontinuous. Thus we have an interpolating continuous function \(h\).
So
\( f^* – \frac{1}{2}\delta(f) \leq h \leq f_* + \frac{1}{2} \delta(f) \)
But we have that \(f \leq f^*\) and \(f_* \leq f\).
So
\( f – \frac{1}{2}\delta(f) \leq h \leq f + \frac{1}{2} \delta(f) \)
i.e. \(||f – h|| \leq \frac{1}{2} \delta\)
But we know that \(||f – h|| \geq \frac{1}{2} \delta\) by our very first proposition. Hence we have equality.
QED
Lemma 8:
If for every \(f \in B(X)\) we have \(d(f, C(X)) = \frac{1}{2} \delta(f)\) then \(X\) is normal.
Proof:
This proof is entirely John’s.
Let \(A, B\) be disjoint closed sets. Define \(f : X \to \mathbb{R}\) by \(f|_A = -1\), \(f|_B = 1\) and \(f(x) = 0\) otherwise.
By considering appropriate neighbourhoods we may see that \(\omega_f(x) \leq 1\) for every \(x\). i.e. \(\delta(f) \leq 1\). Consequently we have \(d(f, C(X)) \leq \frac{1}{2}\) and may find a continuous function \(h\) with \(||f – h|| \leq \frac{3}{4}\).
Thus we have that \(h|_A \leq – \frac{1}{4}\) and \(h|_B \geq \frac{1}{4}\). Composing with some appropriate function from \([-1, 1] \to [0, 1]\) we get a continuous function \(g\) which separates the two closed sets. Then \(g^{-1}([0, \frac{1}{2}))\) and \(g^{-1}((\frac{1}{2}, 1])\) are appropriate disjoint neighbourhoods of \(A\) and \(B\).
QED
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This is a question that it is not homework but I would like clear.
I have this Proximal interpretation, that is: the solution of the problem is a fixed point of the following mapping: $$ x^{\ast} \in \ \ \arg \min_{x \in X_{\text{adm}}} \{\ \frac{1}{2}||x-x^{\ast}||^{2} + \gamma \ \ f(x) \} \ , \ \gamma >0$$
with $$X_{\text{adm}} = \{\ x \in \mathbb{R^n} | x \geq 0 \ , \ A_{\text{eq}}x=b_{\text{eq}}, A_{\text{ineq}}x=b_{\text{ineq}} \}$$ These $A$'s can be interpreted as constrains and it is compact and convex subset of $\mathbb{R^n}$.
Here I suppose that $f(x)$ is convex and differentiable with the gradient satisfying the Lipschitz condition.
Does this proximal method always converge? , i.e , does the proximal interpretation diverge?
I think that yes it divenges but maybe someone can hit me with some example?
I want to understand this because I'm going to study the Proximal Gradient method. Thanks for your help and time.
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While I was studying the Euler-Lagrange equation, Hamilton's principle of least action and geodesics, I started to wonder how to find the equations of motion of a particle restricted to a particular surface (e.g. the sphere). I know that the Lagrangian is defined as $L=K-U$ where $K$ is the kinetic energy and $U$ is the potential energy, and that the Euler-Lagrange equations give the stationary points of a functional (e.g the action). Is it true that if a particle was constrained to move on a particular surface, it would follow the path of a geodesic on that surface? For example, on a sphere, would it follow great circles in its motion?
The more general question I seek to answer is this: is it possible to model the path $\gamma(t)$ a particle will take as a geodesic on a manifold $M$(or simply surface)?
To be a little more concrete, take the examples of projectile motion without drag, where the particle is launched with initial velocity $v_0$ at an angle of $\theta$ from the horizontal at an initial height $H_0$. Here are the functions of time that describe their position on a Cartesian coordinate system: $$y(t)=H_0+v_0\sin(\theta)t-\frac{1}{2}gt^2$$ $$x(t)=v_0\cos(\theta)t.$$ Here, $g$ is the acceleration due to gravity that is approximated as constant throughout the trajectory of flight. Going back to my question, is there a surface $S\subset\mathbb{R}^3$ on which the path of projectile motion this particle takes is a geodesic? How should I find such a path?
Update:
Ok. Let's say that a particle is constrained a to move on a frictionless surface $S(x, y, z) \subset \mathbb{R}^3$ under the influence of gravity, where $F_g=-mg\hat{k}$ where $\hat{k}$ is the unit vector in the upwards z direction. Our Lagrangian is then $$L=\frac{1}{2}m\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)-mgz$$ We then have 3 Euler-Lagrange differential equations to solve to find the stationary points and minimize the action. However, this is not taking into account that the particle has to stay on the surface $S$. We can add the following restrictions (?): $$x(t)=S_x$$ $$y(t)=S_y$$ $$z(t)=S_z$$ where $x, y, z$ are function of time derived form the Euler-Lagrange equation that tell us where the particle will be located, and $S_i$ represents the $i^{th}$ coordinate (e.g x-coordinate). One idea that I had in order to find the functions of time that conform to these restrictions is the use of Lagrange multipliers. However, I am not sure how to apply this technique to this problem.
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15 minute read
In this summary, I assume you are familiar with Markov Decision Processes. In a Markov Decision Process (MDP), an agent interacts with the environment, by taking actions that induce a change in the state of the environment.
MDP: The robot knows the position
of the fuze bottle.
Image courtesy of the Personal Robotics Lab [6, 7]
An important assumption in MDPs is that the agent knows the true state of the world. So, our robotic arm would know exactly where the soda can is located. Partially Observable Markov Decision Processes relax this assumption, in that the agent does not know the true state, but can receive observations through its sensors that enable it to
estimatethat state.
POMDP: The robot retains a
probability distribution over
possible fuze bottle locations.
Image courtesy of the Personal Robotics Lab [6, 7]
For each state and action, the agent receives a reward $r_t$. As in the case of MDPs, the goal of the agent is to maximize the expected accumulated reward that it will receive over a time horizon $T$:
\begin{equation}
\mathbb{E}\left[ \sum_{t=0}^{T-1} r_t \right]
\end{equation}
Formally, we define a partially observable Markov decision process as a tuple $<S, A, \mathcal{T}, \Omega, O, R>$, where:
$S$: finite set of world states; in our example, that is the set of possible robot configurations and positions of the fuze bottle.
$A$: set of a robot actions; for our robotic arm, an action can be a displacement of the robot end-effector as well as a grasping action.
$\mathcal{T}$: $S \times A \rightarrow \Pi(S)$ is a state-transition function, mapping a world state and robot action to a probability distribution over states. It represents how the true world state changes given a robot action; in our example it encodes the uncertainty in the motion of the robot and of the fuze bottles.
$\Omega$ is a finite set of observations; for the object manipulation example, it can be a set of possible point clouds from a depth sensor or readings from the contact sensors in the robot's fingers.
$O: S \times A \rightarrow \Pi(\Omega)$ is the observation function, which maps a world state and robot action to a probability distribution over observations. It represents the uncertainty in the observations by the robot's sensors.
$R: S \times A \rightarrow \mathbb{R}$ is a reward function that the agent receives given an action at a particular world state, for instance a positive reward if the robot succeeds in picking up the object and a negative reward for robot failure.
\begin{align}
b'(s') &= P(s'|o, a, b) \\
&= \frac{P(o | s', a, b) P(s' | a, b)}{P(o | a,b)} &(\textit{from Bayes' rule}) \\
&= \frac{P(o|s',a,b) \sum_{s \in S} P(s' | a,b,s)P(s | a, b)}{P(o | a, b)} & (\textit{marginalization})\\
&= \frac{P(o|s',a) \sum_{s \in S} P(s' | a, s)P(s | b)}{P(o | a, b)} & (\textit{from conditional independence})\\
& = \frac{O(s',a, o) \sum_{s \in S} \mathcal{T}(s, a, s') b(s)}{P(o|a,b)} & (\textit{by definition of $~\mathcal{T},O,b$})
\end{align}
Now, how should the agent optimize its expected accumulated reward? The agent can control only the action $a$ that it will take, given its belief $b$. We call the policy of the agent a mapping from beliefs to actions:
$\pi: B \rightarrow A$
The optimal policy $\pi_t^*(b)$ is the policy that returns the optimal action, that is the action that maximizes the expected reward that the agent will receive over the next $t$ steps.
To understand the computation of the optimal policy, we need to first note the concept of a policy tree. A policy tree $p$ is a tree describing sequences of actions and observations. Each arc in the tree is labelled with an observation, and each node with an action to be performed when the observation leading to that node is received.
Now, to compute the optimal policy, we need a metric of how good a given policy tree p is. The value of executing a policy tree p in state s is the immediate reward by executing the action $a(p)$ at the root node of the tree, plus the expected value of the future.
\begin{align}
V_t^p(s)&= R(s,a(p)) +(\textit{Expected value of the future})
\end{align}
The value of the future depends (i) on which state the agent will end up, and (ii) on which observation $o_i$ the agent will make. The agent does not have this information in advance, so we will take the expectation over subsequent states and observations: \begin{align}
V_t^p(s)&= R(s,a(p)) + \sum_{s'\in S} P(s'|s, a(p)) \sum_{o_i \in \Omega} P(o_i|s',a(p)) V_{t-1}^{p,o_i}(s')
\end{align}
Based on our definitions of the transition function $\mathcal{T}$ and observation function $O$, we have:
\begin{align}
V_t^p(s)&= R(s,a(p)) + \sum_{s'\in S} \mathcal{T}(s, a(p),s') \sum_{o_i \in \Omega}O(s', a(p),o_i) V_{t-1}^{p,o_i}(s')
\end{align}
We see that the value of a state can be computed by one-step look-ahead using $V_{t-1}$. We can then compute the value for every state $s$ recursively, using dynamic programming [4] .
Sadly, the agents does not know the true state, but has a belief $b$, which is a probability distribution over states. Therefore, we need to compute the value of executing a policy tree $p$ given an agent belief $b$. We do this by marginalizing over the agent's belief $b$:
\begin{align}
V_t^p(b) &=\sum_{s \in S} b(s) V_t^p(s)
\end{align}
We can write the above equation more compactly by writing the sum as an inner product of the belief with a set of vectors $\alpha_t^p = <V_t^p(s_1), ... , V_t^p(s_n)>$
\begin{align}
V_t^p(b) = b \cdot \alpha_t^p
\end{align}
So, we see that a policy tree $p$ is associated with a set of vectors $\alpha_t^p$, which we call "alpha vectors." We have also computed the value of a given policy tree $p$ given a belief $b$.
So, given a set of $t$ step policy trees $P_t$, we can compute the value of the optimal policy as the value of starting in belief $b$ and executing the best policy tree in that belief:
\begin{align}
V_t(b) = \max_{p \in P_t} b \cdot \alpha_t^p \end{align}
This is equivalent to to executing the best root action at time $t$, and the best assignment of policy trees -- and their corresponding $\alpha$ vectors -- at $t-1$:
\begin{align} V_t(b) = \max_{a \in A}\left [ \sum_{s \in S}R(s,a)b(s) + \gamma \sum_{o_i \in \Omega} \max_{\alpha \in P_{t-1}} \sum_{s \in S} \sum_{s'\in S} P(o_i|s',a)P(s'|s,a)b(s) \alpha(s') \right ] \end{align}
The equation above can be solved with exact value iteration: start with a set of $t-1$ step policy trees, and iteratively use that to construct a superset of $t$ step policy trees. Unfortunately, the size of the policy trees grows exponentially at each step. We have |A| actions, and for every action we need to consider $|P_{t-1}|^{|\Omega|}$ new trees at each step. This makes solving it intractably large over time.
Instead, we can use sampling-based algorithms to find efficiently approximate solutions. Most such algorithms alternate between sampling in belief-space, and performing an one-step look-ahead, which propagates the information of the children of $b$ back to the point $b$.
Here is an example of how this is possible [3]: Let's assume a sampled set of belief points $b \in B$, and substitute $\alpha^{a,*}=R(s,a)$ and $\alpha^{a,o_i}= \sum_{s'\in S} P(o_i|s',a)P(s'|s,a)$. Then, we can rewrite the value function as:
\begin{align} V_t(b) =\max_{a \in A}\left [ \alpha^{a,*}\cdot b + \sum_{o_i \in \Omega} \max_{\alpha^{a, o_i} \in P_{t-1}} \alpha^{a, o_i}\cdot b \right] \end{align}
Then, for each sampled belief $b$ we can compute the best $\alpha^{a, o_i}$ and store it ahead of time. Then, each time we update the value at $t$ from $t-1$ policy trees, we only need to consider |A| new trees!
This however begs the question, what beliefs should we consider to sample? Recent algorithms [5] have achieved dramatic performance improvements, by attempting to sample from the belief space that is reachable by the optimal policy, since it is usually much smaller than the full belief space.
We thank Siddhartha Srinivasa and Vaibhav Unhelkar for their feedback. References
[1] L. P. Kaelbling, M. L. Littman, and A. R. Cassandra. "Planning and acting in partially observable stochastic domains." Artificial intelligence, 1998.
[2] N. Roy, 16420: Planning Under Uncertainty, Massachusetts Institute of Technology, delivered Fall 2010.
[3] J. Pineau, G. Gordon, and S. Thrun. "Point-based value iteration: An anytime algorithm for POMDPs." IJCAI, 2003.
[4] D. Bertsekas, Dynamic Programming and Optimal Control, Athena scientific, 1995.
[5] H. Kurniawati, D. Hsu, W. Sun Lee. "SARSOP: Efficient Point-Based POMDP Planning by Approximating Optimally Reachable Belief Spaces." Robotics: Science and Systems, 2008.
[6] Michael Koval. "Robust Manipulation via Contact Sensing." PhD Thesis. Carnegie Mellon University, 2016.
[7] Personal Robotics Lab. University of Washington, 2018.
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One of my data anaylsis pet peeves is false precision. Just because it is possible calculate a quantity to three decimal places doesn’t mean all of those decimal places are meaningful. This post explores how much precision is justified in the context of two common sports statistics: batting average in Major League Baseball and save percentage in the National Hockey League. Using Bayesian hierarchical models, we find that though these quantities are conventionally calculated to three decimal places, only the first two decimal places of precision are justified.
%matplotlib inline
import arviz as azfrom matplotlib import pyplot as pltfrom matplotlib.ticker import StrMethodFormatterimport numpy as npimport pandas as pdimport pymc3 as pmimport scipy as spimport seaborn as sns
sns.set(color_codes=True)svpct_formatter = ba_formatter = StrMethodFormatter("{x:.3f}")
We begin by loading hitting data for the 2018 MLB season from Baseball Reference.
def get_data_url(filename): return f"https://www.austinrochford.com/resources/sports_precision/{filename}"
def load_data(filepath, player_col, usecols): df = pd.read_csv(filepath, usecols=[player_col] + usecols) return (pd.concat((df[player_col] .str.split('\\', expand=True) .rename(columns={0: 'name', 1: 'player_id'}), df.drop(player_col, axis=1)), axis=1) .rename(columns=str.lower) .groupby(['player_id', 'name']) .first() # players that switched teams have their entire season stats listed once per team .reset_index('name'))
mlb_df = load_data(get_data_url('2018_batting.csv'), 'Name', ['AB', 'H'])
mlb_df.head()
name ab h player_id abreujo02 Jose Abreu 499 132 acunaro01 Ronald Acuna 433 127 adamewi01 Willy Adames 288 80 adamja01 Jason Adam 0 0 adamsau02 Austin L. Adams 0 0
batter_df = mlb_df[mlb_df['ab'] > 0]n_player, _ = batter_df.shape
This data set covers nearly 1,000 MLB players.
n_player
984
Batting average(https://en.wikipedia.org/wiki/Batting_average_(baseball%29) is the most basic summary of a player’s batting performance and is defined as their number of hits divided by their number of at bats. In order to assess the amount of precision that is justified when calculating batting average, we build a hierarchical logistic model. Let \(n_i\) be the number of at bats for the \(i\)-th player and let \(y_i\) be their number of hits. Our model is
\[ \begin{align*} \mu_{\eta} & \sim N(0, 5^2) \\ \sigma_{\eta} & \sim \textrm{Half-}N(2.5^2) \\ \eta_i & \sim N(\mu, \sigma_{\eta}^2) \\ \textrm{ba}_i & = \textrm{sigm}(\eta_i) \\ y_i\ |\ n_i & \sim \textrm{Binomial}(n_i, \textrm{ba}_i). \end{align*} \]
def hierarchical_normal(name, shape, μ=None): if μ is None: μ = pm.Normal(f"μ_{name}", 0., 5.) Δ = pm.Normal(f"Δ_{name}", shape=shape) σ = pm.HalfNormal(f"σ_{name}", 2.5) return pm.Deterministic(name, μ + Δ * σ)
with pm.Model() as mlb_model: η = hierarchical_normal("η", n_player) ba = pm.Deterministic("ba", pm.math.sigmoid(η)) hits = pm.Binomial("hits", batter_df['ab'], ba, observed=batter_df['h'])
We proceeed to sample from the model’s posterior distribution.
CHAINS = 3SEED = 88564 # from random.org, for reproducibilitySAMPLE_KWARGS = { 'draws': 1000, 'tune': 1000, 'chains': CHAINS, 'cores': CHAINS, 'random_seed': list(SEED + np.arange(CHAINS))}
with mlb_model: mlb_trace = pm.sample(**SAMPLE_KWARGS)
Auto-assigning NUTS sampler...Initializing NUTS using jitter+adapt_diag...Multiprocess sampling (3 chains in 3 jobs)NUTS: [σ_η, Δ_η, μ_η]Sampling 3 chains: 100%|██████████| 6000/6000 [00:44<00:00, 133.94draws/s]
Before drawing conclusions from the posterior samples, we use
arviz to verify that there are no obvious problems with the sampler diagnostics.
az.plot_energy(mlb_trace);
az.gelman_rubin(mlb_trace).max()
<xarray.Dataset>Dimensions: ()Data variables: μ_η float64 1.01 Δ_η float64 1.0 σ_η float64 1.0 η float64 1.0 ba float64 1.0
First we’ll examine the posterior distribution of Mike Trout’s batting average.
fig, ax = plt.subplots(figsize=(8, 6))trout_ix = (batter_df.index == 'troutmi01').argmax()ax.hist(mlb_trace['ba'][:, trout_ix], bins=30, alpha=0.5);ax.vlines(batter_df['h'] .div(batter_df['ab']) .loc['troutmi01'], 0, 275, linestyles='--', label="Actual batting average");ax.xaxis.set_major_formatter(ba_formatter);ax.set_xlabel("Batting average");ax.set_ylabel("Posterior density");ax.legend();ax.set_title("Mike Trout");
We see that the posterior places significant mass between .260 and .320, quite a wide range of batting averages. This range roughly corresponds to the 95% credible interval for his 2018 batting average.
np.percentile(mlb_trace['ba'][:, trout_ix], [2.5, 97.5])
array([ 0.25516468, 0.32704036])
We will use the width of the 95% credible interval for each player’s batting average to determine how many digits of precision are justified.
mlb_df = batter_df.assign( width_95=sp.stats.iqr(mlb_trace["ba"], axis=0, rng=(2.5, 97.5)))
The following plot shows the width of these intervals, grouped by the number of at bats the player had in 2018.
def plot_ci_width(grouped, width): fig, ax = plt.subplots(figsize=(8, 6)) low = grouped.quantile(0.025) high = grouped.quantile(0.975) ax.fill_between(low.index, low, high, alpha=0.25, label=f"{width:.0%} interval"); grouped.mean().plot(ax=ax, label="Average") ax.set_ylabel("Width of 95% credible interval"); ax.legend(loc=0); return ax
ax = plot_ci_width(mlb_df['width_95'].groupby(mlb_df['ab'].round(-2)), 0.95)ax.set_xlim(0, mlb_df['ab'].max());ax.set_xlabel("At bats");ax.set_ylim(bottom=0.);ax.yaxis.set_major_formatter(ba_formatter);ax.set_title("Batting average");
We see that, on average, about 100 at bats are required to justify a single digit of precision in a player’s batting average. Even in the limit of very many at bats (600 at bats corresponds to just under four at bats per game across a 162 game season) the 95% credible interval has an average width approaching 0.060. This limit indicates that batting average is at most meaningful to the second digit, and even the second digit has a fair bit of uncertainty. This result is not surprising; calculating batting average to three decimal places is a historical convention, but I don’t think many analysts rely on the third digit for their decisions/arguments. While intuitive, it is pleasant to have a somewhat rigorous justification for this practice.
We apply a similar analysis to save percentage in the NHL. First we load 2018 goaltending data from Hockey Reference.
nhl_df = load_data(get_data_url('2017_2018_goalies.csv'), 'Player', ['SA', 'SV'])
nhl_df.head()
name sa sv player_id allenja01 Jake Allen 1614 1462 andercr01 Craig Anderson 1768 1588 anderfr01 Frederik Andersen 2211 2029 appleke01 Ken Appleby 55 52 bernijo01 Jonathan Bernier 1092 997
n_goalie, _ = nhl_df.shape
This data set consists of the goaltending performance of just under 100 players.
n_goalie
95
Our save percentage model is almost identical to the batting average model. Let \(n_i\) be the number of at shots the \(i\)-th goalie faced and let \(y_i\) be the number of saves they made. The model is
\[ \begin{align*} \mu_{\eta} & \sim N(0, 5^2) \\ \sigma_{\eta} & \sim \textrm{Half-}N(2.5^2) \\ \eta_i & \sim N(\mu, \sigma_{\eta}^2) \\ \textrm{svp}_i & = \textrm{sigm}(\eta_i) \\ y_i\ |\ n_i & \sim \textrm{Binomial}(n_i, \textrm{svp}_i). \end{align*} \]
with pm.Model() as nhl_model: η = hierarchical_normal("η", n_goalie) svp = pm.Deterministic("svp", pm.math.sigmoid(η)) saves = pm.Binomial("saves", nhl_df['sa'], svp, observed=nhl_df['sv'])
with nhl_model: nhl_trace = pm.sample(nuts_kwargs={'target_accept': 0.9}, **SAMPLE_KWARGS)
Auto-assigning NUTS sampler...Initializing NUTS using jitter+adapt_diag...Multiprocess sampling (3 chains in 3 jobs)NUTS: [σ_η, Δ_η, μ_η]Sampling 3 chains: 100%|██████████| 6000/6000 [00:17<00:00, 338.38draws/s]
Once again, the convergence diagnostics show no cause for concern.
az.plot_energy(nhl_trace);
az.gelman_rubin(nhl_trace).max()
<xarray.Dataset>Dimensions: ()Data variables: μ_η float64 1.0 Δ_η float64 1.0 σ_η float64 1.0 η float64 1.0 svp float64 1.0
We examine the posterior distribution of Sergei Bobrovsky’s save percentage.
fig, ax = plt.subplots(figsize=(8, 6))bobs_ix = (nhl_df.index == 'bobrose01').argmax()ax.hist(nhl_trace['svp'][:, bobs_ix], bins=30, alpha=0.5);ax.vlines(nhl_df['sv'] .div(nhl_df['sa']) .loc['bobrose01'], 0, 325, linestyles='--', label="Actual save percentage");ax.xaxis.set_major_formatter(ba_formatter);ax.set_xlabel("Save percentage");ax.set_ylabel("Posterior density");ax.legend(loc=2);ax.set_title("Sergei Bobrovsky");
We see that the posterior places significant mass between .905 and .925. We see below that the best and worst save percentages (for goalies that faced at least 200 shots in 2018) are separated by about 0.070.
(nhl_df['sv'] .div(nhl_df['sa']) [nhl_df['sa'] > 200] .quantile([0., 1.]))
0.0 0.8669951.0 0.933712dtype: float64
Sergei Bobrovsky’s 0.020-wide credible interval is a significant proportion of this 0.070 total range.
np.percentile(nhl_trace['svp'][:, bobs_ix], [2.5, 97.5])
array([ 0.90683748, 0.92526507])
As with batting average, we plot the width of each goalie’s interval, grouped by the number of shots they faced.
nhl_df = nhl_df.assign( width_95=sp.stats.iqr(nhl_trace["svp"], axis=0, rng=(2.5, 97.5)))
ax = plot_ci_width(nhl_df['width_95'].groupby(nhl_df['sa'].round(-2)), 0.95)ax.set_xlim(0, nhl_df['sa'].max());ax.set_xlabel("Shots against");ax.set_ylim(bottom=0.);ax.yaxis.set_major_formatter(svpct_formatter);ax.set_title("Save percentage");
This plot shows that even goalies that face many (2000+) shots have credible intervals wider that 0.010, a signifcant proportion of the total variation between goalies.
This post is available as a Jupyter notebook here.
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Question is to check which of the following holds (only one option is correct) for a continuous bounded function $f:\mathbb{R}\rightarrow \mathbb{R}$.
$f$ has to be uniformly continuous. there exists a $x\in \mathbb{R}$ such that $f(x)=x$. $f$ can not be increasing. $\lim_{x\rightarrow \infty}f(x)$ exists.
What all i have done is :
$f(x)=\sin(x^3)$ is a continuous function which is bounded by $1$ which is not uniformly continuous. suppose $f$ is bounded by $M>0$ then restrict $f: [-M,M]\rightarrow [-M,M]$ this function is bounded ad continuous so has fixed point. I could not say much about the third option "$f$ can not be increasing". I think this is also true as for an increasing function $f$ can not be bounded but i am not sure. I also believe that $\lim_{x\rightarrow \infty}f(x)$ exists as $f$ is bounded it should have limit at infinity.But then I feel the function can be so fluctuating so limit need not exists. I am not so sure.
So, I am sure second option is correct and fourth option may probably wrong but i am not so sure about third option.
Please help me to clear this.
Thank You. :)
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The supply elasticity at a point is the willingness to produce more or less after the price has changed (due to whatever reason?).
$$\mu_{Q,P}=\frac{\partial Q_S/Q_S}{\partial P/P}=\frac{\partial Q_S}{\partial P}\cdot\frac{P}{Q_S}$$
Where $Q_S$ is the quantity supply function, $P$ is the price. $\frac{P}{Q_S}$ is the current ratio (the one at the point of measurement or at equilibrium) of price to quantity.
Imagine there are three different suppliers with different supply functions $Q_1,Q_2,Q_3$ For low prices some suppliers won't be able to afford to produce so there will only be one supplier for example. for a certain range $(p_0,p_1]$ we can have that two firms are producing, and for $P\in(p_1,\infty)$ all three firms are producing.
In this example it just so happens that the aggregate supply function is continuous.
Say the equilibrium price is $p_1$ and the aggregate supply is $Q_S(p_1)=Q_1(p_1)+Q_2(p_2)$ (only two out of three firms produce for this price). If the slope of $Q_S$ is some number $\alpha$ at that point, (on the left, because on the right supplier 3 also produces and so the slope is different, say $\beta\ne\alpha$), then the supply elasticity is $\alpha\cdot \frac{p_1}{Q_1(p_1)+Q_2(p_1)}$.
Since the slope on the right of $p_1$ is beta, and the aggregate supply function is continuous (that implies $Q_1(p_1)+Q_2(p_1)=Q_S(p_1)=\lim\limits_{p\rightarrow p_1\\p> p_1}Q_1(p)+Q_2(p)+Q_3(p)$) the supply elasticity is $$\beta\cdot \frac{p_1}{Q_1(p_1)+Q_2(p_1)+Q_3(p_1)}$$
Can the "right and left" elasticities be different in absolut value?
How should we understand a market where supply elasticity is not continuous at the equilibrium?
NB: if we had to impose that the elasticity has to be continuous that would be imposing $$Q_3(p_1)=\frac{(\beta-\alpha)(Q_1(p_1)+Q_2(p_1))}{\alpha}$$
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Response Surfaces¶
Response surfaces are inexpensive surroagates for expensive computational models.
Polynomial Ridge Function¶ class
psdr.
PolynomialRidgeFunction(
basis, coef, U)¶
A polynomial ridge function
derivative_roots()¶
Compute the roots of the derivative of the polynomial
profile_grad(
X)¶
gradient of the profile function g
roots()¶
Compute the roots of the polynomial
set_scale(
X, U=None)¶
Set the normalization map
Polynomial Ridge Approximation¶ class
psdr.
PolynomialRidgeApproximation(
degree, subspace_dimension, basis='legendre', norm=2, n_init=1, scale=True, keep_data=True, domain=None, bound=None)¶
Constructs a ridge approximation using a total degree approximation
Given a basis of total degree polynomials \(\lbrace \psi_j \rbrace_{j=1}^N\) on \(\mathbb{R}^n\), this class constructs a polynomial ridge function that minimizes the mismatch on a set of points \(\lbrace \mathbf{x}_i\rbrace_{i=1}^M \subset \mathbb{R}^m\) in a \(p\)-norm:\[\min_{\mathbf{U} \in \mathbb{R}^{m\times n}, \ \mathbf{U}^\top \mathbf{U} = \mathbf I, \ \mathbf{c}\in \mathbb{R}^N } \sqrt[p]{ \sum_{i=1}^M \left|f(\mathbf{x}_i) - \sum_{j=1}^N c_j \psi_j(\mathbf{U}^\top \mathbf{x}_i) \right|^p}\]
This approach assumes \(\mathbf{U}\) is an element of the Grassmann manifold obeying the orthogonality constraint.
For the 2-norm (\(p=2\)) this implementation uses Variable Projection following [HC18] to remove the solution of the linear coefficients \(\mathbf{c}\), leaving an optimization problem posed over the Grassmann manifold alone.
For both the 1-norm and the \(\infty\)-norm, this implementation uses a sequential linear program with a trust region coupled with a nonlinear trajectory through the search space.
Parameters: degree( int) – Degree of polynomial subspace_dimension( int) – Dimension of the low-dimensional subspace associated with the ridge approximation. basis( [ 'legendre' , 'monomial' , 'chebyshev' , 'laguerre' , 'hermite' ]) – Basis for polynomial representation norm( [ 1 , 2 , np.inf , 'inf' ]) – Norm in which to evaluate the mismatch between the ridge approximation and the data scale( bool ( default:True )) – Scale the coordinates along the ridge to ameliorate ill-conditioning bound( [ None , 'lower' , 'upper' ]) – If ‘lower’ or ‘upper’ construct a lower or upper bound
References
[HC18] J. M. Hokanson and Paul G. Constantine. Data-driven Polynomial Ridge Approximation Using Variable Projection. SIAM J. Sci. Comput. Vol 40, No 3, pp A1566–A1589, DOI:10.1137/17M1117690.
fit(
X, fX, U0=None, **kwargs)¶
Given samples, fit the polynomial ridge approximation.
Parameters: X( array-like ( M , m )) – Input coordinates fX( array-like ( M , )) – Evaluations of the function at the samples Polynomial Ridge Bound¶ Gaussian Process Model¶ class
psdr.
GaussianProcess(
structure='const', degree=None, nugget=None, Lfixed=None, n_init=1)¶
Fits a Gaussian Process by maximizing the marginal likelihood
Given \(M\) pairs of \(\mathbf{x}_i \in \mathbb{R}^m\) and \(y_i \in \mathbb{R}\), this fits a model\[g(\mathbf{x}) = \sum_{i=1}^M \alpha_i e^{-\| \mathbf{L}( \mathbf{x} - \mathbf{x}_i) \|_2^2} + \sum_{j} \beta_j \psi_j(x)\]
where \(\mathbf{L}\) is a lower triangular matrix, \(\lbrace \psi_j \rbrace_j\) is a polynomial basis (e.g., linear), and \(\boldsymbol{\alpha}, \boldsymbol{\beta}\) are vectors of weights. These parameters are choosen to maximize the log-marginal likelihood (see [RW06], Sec. 2.7), or equivalently minimize\[\begin{split}\min_{\mathbf{L}} & \ \frac12 \mathbf{y}^\top \boldsymbol{\alpha} + \frac12 \log \det (\mathbf{K} + \tau \mathbf{I}), & \quad [\mathbf{K}]_{i,j} &= e^{-\frac12 \| \mathbf{L}(\mathbf{x}_i -\mathbf{x}_j)\|_2^2} \\ \text{where} & \ \begin{bmatrix} \mathbf{K} + \tau \mathbf{I} & \mathbf{V} \\ \mathbf{V}^\top & \mathbf{0} \end{bmatrix} \begin{bmatrix} \boldsymbol{\alpha} \\ \boldsymbol{\beta} \end{bmatrix} = \begin{bmatrix} \mathbf{y} \\ \mathbf{0} \end{bmatrix}, & \quad [\mathbf{V}]_{i,j} &= \psi_j(\mathbf{x}_i).\end{split}\]
Here \(\tau\) denotes the nugget–a Tikhonov-like regularization term commonly used to combat the ill-conditioning of the kernel matrix \(\mathbf{K}\). The additional constraint for the polynomial basis is described by Jones ([Jon01] eq. (3) and (4)).
This class allows for four kinds of distance matrices \(\mathbf{L}\):
scalar \(\mathbf{L} = \ell \mathbf{I}\) scalar multiple \(\mathbf{L} = \ell \mathbf{L}_0\) diagonal \(\mathbf{L} = \text{diag}(\boldsymbol{\ell})\) lower triangular \(\mathbf{L}\) is lower triangular
Internally, we optimize with respect to the matrix log; specificially, we parameterize \(\mathbf{L}\) in terms of some scalar or vector \(\boldsymbol{\ell}\); i.e.,
scalar \(\mathbf{L}(\ell) = e^{\ell}\mathbf{I}\) scalar multiple \(\mathbf{L}(\ell) = e^{\ell} \mathbf{L}_0\) diagonal \(\mathbf{L}(\ell) = \text{diag}(\lbrace e^{\ell_i} \rbrace_{i})\) lower triangular \(\mathbf{L}(\ell) = e^{\text{tril}(\boldsymbol{\ell})}\)
In the constant and diagonal cases, this corresponds to the standard practice of optimizing with respect to the (scalar) log of the scaling parameters. Our experience suggests that the matrix log similarly increases the accuracy when working with the lower triangular parameterization. For the first three classes we have simple expressions for the derivatives, but for the lower triangular parameterization we use complex step approximation to compute the derivative [MN10]. With this derivative information, we solve the optimization problem using L-BFGS as implemented in scipy.
Parameters: structure( [ 'tril' , 'diag' , 'const' , 'scalar_mult' ]) –
Structure of the matrix L, either
const: constant * eye scalar_mult tril: lower triangular diag: diagonal Returns: L( np.ndarray(m,m)) – Distance matrix alpha( np.ndarray(M)) – Weights for the Gaussian process kernel beta( np.ndarray(N)) – Weights for the polynomial component in the LegendreTensorBasis obj( float) – Log-likelihood objective function value
References
[RW06] Gaussian Processes for Machine Learning, Carl Edward Rasmussen and Christopher K. I. Williams, 2006 MIT Press [Jon01] A Taxonomy of Global Optimization Methods Based on Response Surfaces, Donald R. Jones, Journal of Global Optimization, 21, pp. 345–383, 2001. [MN10] “The complex step approximation to the Frechet derivative of a matrix function”, Awad H. Al-Mohy and Nicholas J. Higham, Numerical Algorithms, 2010 (53), pp. 133–148.
fit(
X, y, L0=None)¶
Fit a Gaussian process model
Parameters: X( array-like ( M , m )) – M input coordinates of dimension m y( array-like ( M , )) – y[i] is the output at X[i]
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@egreg It does this "I just need to make use of the standard hyphenation function of LaTeX, except "behind the scenes", without actually typesetting anything." (if not typesetting includes typesetting in a hidden box) it doesn't address the use case that he said he wanted that for
@JosephWright ah yes, unlike the hyphenation near box question, I guess that makes sense, basically can't just rely on lccode anymore. I suppose you don't want the hyphenation code in my last answer by default?
@JosephWright anway if we rip out all the auto-testing (since mac/windows/linux come out the same anyway) but leave in the .cfg possibility, there is no actual loss of functionality if someone is still using a vms tex or whatever
I want to change the tracking (space between the characters) for a sans serif font. I found that I can use the microtype package to change the tracking of the smallcaps font (\textsc{foo}), but I can't figure out how to make \textsc{} a sans serif font.
@DavidCarlisle -- if you write it as "4 May 2016" you don't need a comma (or, in the u.s., want a comma).
@egreg (even if you're not here at the moment) -- tomorrow is international archaeology day: twitter.com/ArchaeologyDay , so there must be someplace near you that you could visit to demonstrate your firsthand knowledge.
@barbarabeeton I prefer May 4, 2016, for some reason (don't know why actually)
@barbarabeeton but I have another question maybe better suited for you please: If a member of a conference scientific committee writes a preface for the special issue, can the signature say John Doe \\ for the scientific committee or is there a better wording?
@barbarabeeton overrightarrow answer will have to wait, need time to debug \ialign :-) (it's not the \smash wat did it) on the other hand if we mention \ialign enough it may interest @egreg enough to debug it for us.
@DavidCarlisle -- okay. are you sure the \smash isn't involved? i thought it might also be the reason that the arrow is too close to the "M". (\smash[t] might have been more appropriate.) i haven't yet had a chance to try it out at "normal" size; after all, \Huge is magnified from a larger base for the alphabet, but always from 10pt for symbols, and that's bound to have an effect, not necessarily positive. (and yes, that is the sort of thing that seems to fascinate @egreg.)
@barbarabeeton yes I edited the arrow macros not to have relbar (ie just omit the extender entirely and just have a single arrowhead but it still overprinted when in the \ialign construct but I'd already spent too long on it at work so stopped, may try to look this weekend (but it's uktug tomorrow)
if the expression is put into an \fbox, it is clear all around. even with the \smash. so something else is going on. put it into a text block, with \newline after the preceding text, and directly following before another text line. i think the intention is to treat the "M" as a large operator (like \sum or \prod, but the submitter wasn't very specific about the intent.)
@egreg -- okay. i'll double check that with plain tex. but that doesn't explain why there's also an overlap of the arrow with the "M", at least in the output i got. personally, i think that that arrow is horrendously too large in that context, which is why i'd like to know what is intended.
@barbarabeeton the overlap below is much smaller, see the righthand box with the arrow in egreg's image, it just extends below and catches the serifs on the M, but th eoverlap above is pretty bad really
@DavidCarlisle -- i think other possible/probable contexts for the \over*arrows have to be looked at also. this example is way outside the contexts i would expect. and any change should work without adverse effect in the "normal" contexts.
@DavidCarlisle -- maybe better take a look at the latin modern math arrowheads ...
@DavidCarlisle I see no real way out. The CM arrows extend above the x-height, but the advertised height is 1ex (actually a bit less). If you add the strut, you end up with too big a space when using other fonts.
MagSafe is a series of proprietary magnetically attached power connectors, originally introduced by Apple Inc. on January 10, 2006, in conjunction with the MacBook Pro at the Macworld Expo in San Francisco, California. The connector is held in place magnetically so that if it is tugged — for example, by someone tripping over the cord — it will pull out of the socket without damaging the connector or the computer power socket, and without pulling the computer off the surface on which it is located.The concept of MagSafe is copied from the magnetic power connectors that are part of many deep fryers...
has anyone converted from LaTeX -> Word before? I have seen questions on the site but I'm wondering what the result is like... and whether the document is still completely editable etc after the conversion? I mean, if the doc is written in LaTeX, then converted to Word, is the word editable?
I'm not familiar with word, so I'm not sure if there are things there that would just get goofed up or something.
@baxx never use word (have a copy just because but I don't use it;-) but have helped enough people with things over the years, these days I'd probably convert to html latexml or tex4ht then import the html into word and see what come out
You should be able to cut and paste mathematics from your web browser to Word (or any of the Micorsoft Office suite). Unfortunately at present you have to make a small edit but any text editor will do for that.Givenx=\frac{-b\pm\sqrt{b^2-4ac}}{2a}Make a small html file that looks like<!...
@baxx all the convertors that I mention can deal with document \newcommand to a certain extent. if it is just \newcommand\z{\mathbb{Z}} that is no problem in any of them, if it's half a million lines of tex commands implementing tikz then it gets trickier.
@baxx yes but they are extremes but the thing is you just never know, you may see a simple article class document that uses no hard looking packages then get half way through and find \makeatletter several hundred lines of trick tex macros copied from this site that are over-writing latex format internals.
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I'm working on the mathematical theory of parabolic equations. The prototype of such equations is heat equation given as follows : Let $\Omega$ be a bounded region of the space and $T>0$ a fixed time. In $\Omega_T=(0,T)\times \Omega$ we consider the following equation $$u_t =\alpha\Delta u -a(x)u,$$ $$u(0,x)=f(x),$$ where $f$ is the initial condition, $a$ a bounded potential, $\alpha>0$ is a constant, and $\Delta$ is the Laplacian. I'm wondering to know the physical meaning of the coefficient $a$ (and may be $\alpha$) and it's role in the heat process? Any reference or suggestion would be helpful.
As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$k\Delta T(x,t)-h(x)T(x,t)=c\rho\dot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $\rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $c\rho$, then we have
$$\alpha \Delta u-\left(\frac{h(x)}{c\rho}\right)u=u_t,$$
with $\alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $\alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $\frac{\partial u}{\partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case,
$a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
The proper
physical name of your equation is diffusion equation with a source term. The equation can be rearranged to continuity equation - $u_t-\alpha u_{xx} = Q$. For $Q=0$ the time dependent solution can be shown to have time independent norm, which is manifestation of local mass conservation law. Source term means that particles can be created and destroyed locally, according to $Q(x)$ variation. Continuity equation is a restatement of Gauss law - at given infinitesimal volume, the change in the number of particles in the volume is exactly equal to the number of particles crossed the surface of it in/out of this volume. You may gain some physical intuition exploring compressible aspect of the Navier-Stockes equation. The compressability is exactly the violation of continuity equation.
In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
The equation describes the flow of heat in presence of sources or sinks. The first term on the right hand side is the normal diffusion term. The second term can be thought of as a source or sink term. For more details see: https://www.math.ubc.ca/~peirce/M257_316_2012_Lecture_19.pdf
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Definition:T3 1/2 Space Definition
Let $T = \left({S, \tau}\right)$ be a topological space.
$\left({S, \tau}\right)$ is a $T_{3 \frac 1 2}$ space if and only if: For any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an Urysohn function for $F$ and $\left\{{y}\right\}$. Variants of Name
From about 1970, treatments of this subject started to refer to this as a
completely regular space, and what is defined on $\mathsf{Pr} \infty \mathsf{fWiki}$ as a completely regular space as a $T_{3 \frac 1 2}$ space.
However, the names are to a fair extent arbitrary and a matter of taste, as there appears to be no completely satisfactory system for naming all these various Tychonoff separation axioms.
Also see Results about $T_{3 \frac 1 2}$ spacescan be found here.
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Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
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(Elsevier, 2013-04-10)
The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (|y| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ...
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(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Multiplicity dependence of the average transverse momentum in pp, p-Pb, and Pb-Pb collisions at the LHC
(Elsevier, 2013-12)
The average transverse momentum <$p_T$> versus the charged-particle multiplicity $N_{ch}$ was measured in p-Pb collisions at a collision energy per nucleon-nucleon pair $\sqrt{s_{NN}}$ = 5.02 TeV and in pp collisions at ...
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 ...
Directed flow of charged particles at mid-rapidity relative to the spectator plane in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(American Physical Society, 2013-12)
The directed flow of charged particles at midrapidity is measured in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV relative to the collision plane defined by the spectator nucleons. Both, the rapidity odd ($v_1^{odd}$) and ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
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