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M → Tip, 4 bytes
Ṅ×ịß
Try it online!
The TIO link adds a footer to call the function with the example Tip program shown on the Esolang page (M's "automatic wrapper" to call functions as though they were programs can't handle rational or fixed-point numbers, or at least I haven't figured out how to tell it how, so I need to make the function into a full program by hand to be able to run it.)
This actually prints useful debug output; the program can't be written in 3 bytes in M because a program consisting of exactly three dyads triggers a special case in the parser, so I had to add an extra command to avoid the special case. Making it
Ṅ (print with newline) at least gives it a useful purpose.
Function submission, taking two arguments: the initial IP on the left, the program on the right. The program is 1-indexed (i.e. command 1 is the first command; M uses 1-indexing by default); goto commands are represented as M rationals, and the halt command as
ı (i.e. the imaginary unit, \$i=\sqrt{-1}\$).
Does not implement I/O (other than halt/no-halt). I/O is an extension to Tip (not part of the language itself), and not required for Turing-completeness.
Explanation/background
Ṅ×ịß
Ṅ Print {the left argument} and a newline; also resolves a parser ambiguity
ị {The left argument}th element of {the right argument}, wrapping on OoB
× Multiply {the left argument} by {the chosen element}
ß Recursive call; arguments: {the product} and {the same right argument}
I was reading through the answers to this entry and realised that iterated Collatz functions, which were used in quintopia's earlier answer, would be fairly short to represent in golfing languages in which list indexing wraps by default (i.e. the fifth element of
[1,2,3] is 2, because the list is being treated as
[1,2,3,1,). So it's easy to extract a particular Collatz operation from a list in very few characters. Can we implement the Collatz operation easily? Well, a Collatz operation is \$rx+s\$, which is a polynomial, and the "base conversion" builtin that many golfing languages have is actually a general-purpose polynomial evaluator in disguise. So all we have to do is index into a list of lists of digits, base-convert them, and we're done, right?
2,3,1,2,3,…]
Unfortunately, it's not that simple. The first problem is that although Collatz functions can be defined entirely in terms of integers, that requires a divmod to extract the new value of \$x\$ (the definition where \$x\$ is the same value that's used to index into the list of Collatz operations requires rationals). Well, we just need a golfing language that supports rationals, right? M is a Jelly derivative that supports many types of arbitrary-precision arithmetic, and arithmetic on the rationals is part of its arsenal of mathematical operators.
Then we get to the second problem: M's base-conversion builtin
ḅ takes its arguments in the wrong order (it wants the list of digits to appear before the base). The problem with this is that M's default method of chaining together two binary operators given two arguments is \$x\oplus(x\otimes y)\$, and yet we'd want the Collatz operation (which can only fit the \$x\otimes y\$ part of this structure, as it's obtained by an index) to be on the
left of the \${\oplus}\$. Sure, we could override the chaining behaviour to pretty much anything we want, but that would cost a whole byte, and the golfing language entries to this question are getting so short that a byte is a lot.
So I looked back and re-evaluated a bit. Are there any operations we could use instead of polynomial evaluation? Ideally, ones that are commutative, so we don't have to worry about argument order? Soon after that, I realised that Collatz functions are more complex than they need to be.
As a result, I created Tip, a simplification/tarpit-ification of iterated Collatz functions in which \$s\$ is always 0, meaning that instead of a polynomial evaluation, we can perform the various operations via a simple multiplication. The language is more complex to prove Turing-complete than Collatz functions are, but it still has enough power to implement any program; there's a proof on the Esolang page.
And of course, unlike base conversion (
ḅ), multiplication (
×) is commutative, and thus it doesn't matter what order the arguments are placed in. So all we need to write is
×ị, and then place the program into an infinite recursion with
ß, and we have a Turing-complete language. Right?
Unfortunately, we run into a new problem. If a program starts with three binary operations, M engages in a special case that chains them as \$(x\odot y)\oplus(x\otimes y)\$ which is the worst possible structure for us, as it doesn't have the three nested function calls we'd need (index, multiply, and recursive call). So no matter what, we're going to need a fourth byte to disambiguate.
¹×ịß (adding the identity function
¹ as a no-op so that the program doesn't
start with three binary operators) does exactly what we'd need, causing them to nest inside each other in the way we want. We can use other operations in place of
¹;
Ṅ is a good choice because it produces useful debug output.
Is three bytes possible? Unless I'm missing something, not with this specific choice of implementing and implemented language, but at this point it surely seems like it'd be possible somehow, as there are so many ways to do it in four and so many Turing-complete languages you could implement. |
Since you are referring to
convolution operation and a matrix then you are probably dealing with 2D LSI (linear shift invariant) systems that's encountered in image processing etc. For such a system the input output, I/O, relationship is expressed as a convolution sum (or integral) like this:
$$ y[n_1,n_2] = \sum_{k_1=-\infty}^{\infty} \sum_{k_2=-\infty}^{\infty} h[k_1,k_2] x[n_1-k_1, n_2-k_2]$$
$$ y(t_1,t_2) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(\tau_1, \tau_2) x(t_1-\tau_1, t_2-\tau_2)d\tau_1 d\tau_2$$
where the input signal (a sequence or a function) $x$ is mapped into the output signal $y$ by the convolutional transform.
From a mathematical point of view, in general an integral transform can be defined as: $$ y(s) = \int \phi(s,t) x(t) dt $$Where the input function is transformed from $t$ domain into $s$ domain by the transform
kernel of $\phi(s,t)$ which is a cross domain function. The simple example in signal processing is the Laplace transform or Fourier transform whose kernels are $\phi(s,t) = e^{-st}$ and $\phi(w,t) = e^{-jwt}$ resepctively.
Similarly a disrete transform$$ y[k] = \sum_n \phi[k,n] x[n] $$ has the
kernel of $\phi[k,n] = e^{-j \frac{2\pi}{N}kn}$ for DFT (discrete Fourier transform) for example.
In essence the kernel acts on the input and produces the output by modifying it. Therefore the kernel defines the character of the mapping (th transform). Programmatically the operation is such that the kernel remains fixed inside a function, whereas the input and outputs change by every function call utilization. Hence the name kernel. (You can of course create functions which accept kernels as arguments in addition to the input etc.)
The mathematical definition of the convolution operation can be seen to be similar to the transforms considered above with its
kernel being the impulse response $h(x_1, x_2)$ (or $h[n_1, n_2]$) of the LSI system. For a discrete-time 2D LSI system whose impulse reponse $h[n_1, n_2]$ has a finite domain of support (FIR) it can be represented as a matrix as well.
Masks, in general, are not convolutional operators. Rather, they operate pixelwise (samplewise) on the image being operated. They are more programming concept than signal processing but somewhat finds applications such as in a window treated as a mask being applied to the input signal or a quantization mask applied on to the DCT coeffcicients for example. Masks find extensive usage for image processing effects. Their operating matrix can also be called as a kernel. |
Suppose we have a proof system for classical first-order logic, where $\vDash$ denotes model-theoretic consequence and $\vdash$ denotes proof-theoretic consequence.
We can distinguish two forms of completeness of the proof system. Call them weak and strong, respectively*:
$$\text{for all sentences $\phi$, if $\vDash\phi$, then $\vdash\phi$}\tag{weak}$$ $$\text{for all sets of sentences $\Gamma$ and sentences $\phi$, if $\Gamma\vDash\phi$, then $\Gamma\vdash\phi$}\tag{strong}$$
Both of these statements are true of classical first-order logic with any of the standard proof systems. It is also obvious that strong completeness implies weak completeness, by taking $\Gamma=\emptyset$.
The weak statement of completeness also comes in two other equivalent forms: (1) every sentence $\phi$ is either satisfiable or refutable, and (2) if $\phi$ is consistent, then $\phi$ is satisfiable. Similarly, the strong statement of completeness is equivalent to (3) if $\Gamma$ is consistent, then $\Gamma$ is satisfiable. (My understanding is that in his 1930 dissertation Gödel proved a statement even stronger than (3), namely (4): if $\Gamma$ is consistent, then $\Gamma$ is satisfiable in a countable domain.)
Does weak completeness imply strong completeness? I get stuck when I try to prove it does.
Here is how I tried to prove that (weak) implies (strong). My thought was to show (weak) implies (3). Suppose $\Gamma$ is consistent. Then for all sentences $\phi$, $\Gamma\nvdash\phi\land\lnot\phi$. What you'd
like to do at this point is use the syntactic deduction theorem** to say $\nvdash\Gamma\rightarrow(\phi\land\lnot\phi)$, and then use (weak) to conclude $\nvDash\Gamma\rightarrow(\phi\land\lnot\phi)$. From here it would follow that $\Gamma$ is satisfiable. But of course pushing $\Gamma$ into the antecedent of a conditional doesn't make sense for arbitrary sets of sentences $\Gamma$ (instead of a finite set, which can be pushed into the antecedent as a conjunction).
This suggests some sort of compactness trick would help, and that weak completeness does not imply strong completeness in general. Is this right?
*If there are more standard names for these properties, please tell me.
**Which says $\Gamma\cup\left\{\phi\right\}\vdash\psi$ if and only if $\Gamma\vdash\phi\rightarrow\psi$. |
I'm having a hard time to insert a table here when I'm asking a question. Any help would be appreciated. Also, is there any, not too expensive, math programs that allow me to enter math terms and equations kind of thing?
Thank you
Tom
I'm having a hard time to insert a table here when I'm asking a question. Any help would be appreciated. Also, is there any, not too expensive, math programs that allow me to enter math terms and equations kind of thing?
Thank you
Tom
This question came from our site for people studying math at any level and professionals in related fields.
Regarding how to write general math expressions on the site, my answer here covers that well I think.
Specifically for tables, you can use LaTeX's
\array command. The
tabular command, which only works in text mode, is not available here, so if you want to include text in your table, you will have to do some tinkering. Here is an example table:
$$\begin{array}{c|c|c|} & \text{Column A} & \text{Column B} \\ \hline\text{Row 1} & 5 & \oplus \\ \hline\text{Row 2} & \int & 8 \\ \hline\end{array}$$
produces
$$\begin{array}{c|c|c|} & \text{Column A} & \text{Column B} \\ \hline \text{Row 1} & 5 & \oplus \\ \hline \text{Row 2} & \int & 8 \\\hline \end{array}$$
Since my example given in a comment to Zev Chonoles's answer and adapted from it, due to some reason, does not display properly now, I repeat it here.
$$\begin{array}{|c|c|c|}\hline&\zeta (3)&\zeta (2)\\\hline\sigma&3+4\ln(1+4\sqrt{2})&2+5\ln\left(\dfrac{1+\sqrt{5}}{2}\right)\\\hline\tau&-3+4\ln(1+\sqrt{2})&-2+5\ln\left(\dfrac{1+\sqrt{5}}{2}\right)\\\hline\mu=1+\dfrac{\sigma}{\tau}&\dfrac{8\ln(1+\sqrt{2})}{4\ln(1+\sqrt{2})-3}&\dfrac{10\ln\left(\dfrac{1+\sqrt{5}}{2}\right)}{5\ln\left(\dfrac{1+\sqrt{5}}{2}\right)-2}\\\hline\end{array}$$
It is produced by
$$\begin{array}{|c|c|c|} \hline &\zeta (3)&\zeta (2)\\ \hline\sigma&3+4\ln(1+4\sqrt{2})&2+5\ln\left(\dfrac{1+\sqrt{5}}{2}\right)\\ \hline\tau&-3+4\ln(1+\sqrt{2})&-2+5\ln\left(\dfrac{1+\sqrt{5}}{2}\right)\\ \hline\mu=1+\dfrac{\sigma}{\tau}&\dfrac{8\ln(1+\sqrt{2})}{4\ln(1 \sqrt{2})-3}& \dfrac{10\ln\left(\dfrac{1+\sqrt{5}}{2}\right)} {5\ln\left(\dfrac{1+\sqrt{5}}{2}\right)-2}\\ \hline\end{array}$$ |
Let $p$ be an odd prime with a primitive root $g$. Prove that $$\prod_{x=1}^{\frac{p-1}{2}}x^2 \equiv (-1)^{\frac{p+1}{2}}\pmod{p}.$$
Remark: I intend to use the relationship $g^{\frac{p-1}{2}} \equiv -1 \pmod{p}.$
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Let $p$ be an odd prime with a primitive root $g$. Prove that $$\prod_{x=1}^{\frac{p-1}{2}}x^2 \equiv (-1)^{\frac{p+1}{2}}\pmod{p}.$$
Remark: I intend to use the relationship $g^{\frac{p-1}{2}} \equiv -1 \pmod{p}.$
$\prod_1^{p-1}x=\prod_1^{(p-1)/2}x\prod_{(p+1)/2}^{p-1}x\equiv\prod_1^{(p-1)/2}x\prod_1^{(p-1)/2}(-x)=(-1)^{(p-1)/2}\prod_1^{(p-1)/2}x^2$ but also by Wilson's Theorem $\prod_1^{p-1}x\equiv-1\pmod p$ and there you are.
If you really want to use a primitive root, note that your product is equal to the product of all the non-zero squares (the quadratic residues) modulo $p$. But these are congruent to the even powers of our primitive root $g$. So your product is congruent to $$g^2 g^4 g^6\cdots g^{2\frac{p-1}{2}}.$$ This is equal to $$g^{2\left(1+2+3+\cdots +\frac{p-1}{2}\right)}.$$ The arithmetic progression which is the exponent of $g$ has sum $$\frac{p-1}{2}\frac{p+1}{2}.$$ Finally, $$g^{\frac{p-1}{2}\frac{p+1}{2}}=\left(g^{\frac{p-1}{2}}\right)^{\frac{p+1}{2}}\equiv (-1)^{\frac{p+1}{2}}\pmod{p}.$$
$\prod_{1\le x\le p-1}x\equiv\prod_{1\le y\le p-1}g^y$ $=g^{1+2+\cdots +p-1}=g^{\frac{p(p-1)}2}=(g^\frac{p-1}2)^p\equiv (-1)^p=-1$
$\prod_{1\le x\le p-1}x=\prod_{1\le x\le \frac{p-1}2}x(p-x)$ as to avoid omission and repetition of terms $x\le p-x\implies x\le \frac p 2$ i.e., $x\le\frac{p-1}2$ as $p$ is odd
So, $\prod_{1\le x\le p-1}x\equiv(-1)^\frac{p-1}2\prod_{1\le x\le \frac{p-1}2}x^2$
So, $(-1)^\frac{p-1}2\prod_{1\le x\le \frac{p-1}2}x^2\equiv-1$
Multiplying either sides by $(-1)^{p-1},$ (which is legal as $(-1)^{p-1}=1$ as $p$ is odd)
$\prod_{1\le x\le \frac{p-1}2}x^2\equiv (-1)(-1)^\frac{p-1}2=(-1)^\frac{p+1}2$ |
To answer your first question, one way to define (co)homology with local coefficients is the following.
Let $X$ be a space, let $\Pi(X)$ be the fundamental groupoid of $X$, ie. a category with objects points of $X$ and morphisms $x \rightarrow y$ given by homotopy classes of paths. A
local coefficient system $M$ on $X$ is a functor $M: \Pi(X) \rightarrow \mathcal{A}b$ from the fundamental groupoid to the category of abelian groups. In particular, $M$ associates a "group of coefficients" $M(x)$ to every point of $x \in X$.
Associated to $M$ is the singular complex given by
$C _{n}(X, M)= \bigoplus _{\sigma \in Sing_{n}(X)} M(\sigma(1,0,\ldots,0))$,
where $Sing _{n}(X)$ is the set of all maps $\Delta ^{n} \rightarrow X$. The differential can be defined using the fact that the groups $M(x)$ are functorial with respect to paths in $X$. Observe that this is a very similar to the usual definition of singular complex with coefficients in an abelian group $A$, which would be
$C _{n}(X, A) = \bigoplus _{\sigma \in Sing_{n}(X)} A$,
except in the "non-local" case, we count the occurances of any $\sigma: \Delta^{n} \rightarrow X$ in a given chain using the same group $A$ and in the local case, we use $M(\Delta^{n}(1, 0, \ldots, ))$, which might be different for different $\sigma$.
This is the
locality ( localness?) in the name, which should be contrasted with globality of usual homology with coefficients, where the choice of the group $A$ is global and the same for all points.
The definition I have given above is enlightening but perhaps not suitable for computations. Luckily under rather weak assumptions one can use the definition you allude to. Let me explain. Let $X$ be path-connected, let $x \in X$ and consider $\pi = \pi_{1}(X, x)$ as a category with one object and morphisms the elements of the group.
The obvious inclusion $\pi \hookrightarrow \Pi(X)$ is an equivalence of categories and so the functor categories $[\pi, \mathcal{A}b], [\Pi(X), \mathcal{A}b]$ are equivalent, too. But the left functor category is exactly the category of $Z[\pi]$-modules! In particular, we have a bijection between isomorphism classes of local coefficient systems on $X$ and $Z[\pi]$-modules.
If $X$ is nice enough to admit a universal cover $\tilde{X}$, the above allows us to give another definiton of homology with local coefficients, the one you know. Let $M$ be a local coefficient system and let $M^\prime$ be the associated $\mathbb{Z}[\pi]$-module under the above equivalence (which is unique up to a unique isomorphism). Since $\pi$ acts on $\tilde{X}$, it also acts on $C _{\bullet}(X, \mathbb{Z})$ and so the latter is a chain complex of $\mathbb{Z}[\pi]$-modules. We the can define homology with local coefficients to be homology of the complex
$C_{n}(X, M ^\prime) = C_{n}(\tilde{X}, \mathbb{Z}) \otimes _{\mathbb{Z}[\pi]} M^\prime$
One can show that this two definitions I gave agree, that is, for $X$ like above we have an isomorphism $H_{n}(X, M) \simeq H_{n}(X, M^\prime)$. |
The Heaviside-Feynman formula (see Feynman Lectures vol I Ch.28, vol II Ch. 21) gives the electric and magnetic fields measured at an observation point $P$ due to an arbitrarily moving charge $q$
$$ \mathbf{E} = -\frac{q}{4 \pi \epsilon_0} \left\{ \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{ret} + \frac{[r]_{ret}}{c} \frac{\partial}{\partial t} \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{ret} + \frac{1}{c^2} \frac{\partial^2}{\partial t^2} [\mathbf{\hat r}]_{ret} \right\} $$ $$ \mathbf{B} = -\frac{1}{c} [\mathbf{\hat r}]_{ret} \times \mathbf{E} $$
where $[\mathbf{\hat r}]_{ret}$ and $[r]_{ret}$ is the unit vector and distance
from the observation point $P$ at time $t$ to the retarded position of charge $q$ at time $t - [r]_{ret}/c$ (hence the minus signs).
This formula is remarkable in that it is completely relational. It does not refer to any external reference frame. The fields at point $P$ only depend on the vector $\mathbf{r}$ from point $P$ to the retarded position of charge $q$ and its first and second order rates of change with respect to local time $t$.
Now one can imagine two ways in which the vector $\mathbf{r}$ from point $P$ to charge $q$ can change. One could move charge $q$ and keep point $P$ fixed or one could move point $P$ and keep charge $q$ fixed. The above formula is valid for the former situation giving the fields at a fixed point $P$ due to a moving charge $q$. This is the conventional retarded solution of Maxwell's equations.
But what about the latter situation in which the observation point $P$ moves and the charge $q$ is fixed. The relational nature of the formula implies to me that it should still apply in this situation. Perhaps this is the situation in which the
advanced Heaviside-Feynman formula is valid given by
$$ \mathbf{E} = -\frac{q}{4 \pi \epsilon_0} \left\{ \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{adv} - \frac{[r]_{adv}}{c} \frac{\partial}{\partial t} \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{adv} + \frac{1}{c^2} \frac{\partial^2}{\partial t^2} [\mathbf{\hat r}]_{adv} \right\} $$ $$ \mathbf{B} = -\frac{1}{c} [\mathbf{\hat r}]_{adv} \times \mathbf{E} $$
where $[\mathbf{\hat r}]_{adv}$ and $[r]_{adv}$ is the unit vector and distance from the observation point $P$ at time $t$ to the advanced position of charge $q$ at time $t + [r]_{adv}/c$. The advanced Heaviside-Feynman formula is the time-reverse of the conventional retarded formula.
This interpretation of the advanced formula, if valid, implies that an accelerated observer inside a fixed charged insulating spherical shell would measure an electric field whose strength is proportional to the acceleration. This implies that an accelerated charge feels a kind of electromagnetic inertial force and thus has an electromagnetic inertia due to the presence of the charged spherical shell.
For example imagine an electron with charge $-e$ inside such a fixed charged insulating spherical shell at potential $+V$ volts. Using the above advanced Heaviside-Feynman formula one can calculate that this electromagnetic inertia $m_{em}$ is given by $$m_{em} = \frac{2}{3} \frac{eV}{c^2}$$ For a shell charged to a high voltage $V=1000000$ volts this electromagnetic inertia would be of similar order to the electron's native mass and should therefore be easily observable. It would probably be important that the spherical shell is a charged insulator rather than a conductor because it is assumed that the charges inside the shell remain fixed.
Finally, there is a close analogy between Maxwell's equations and Einstein's field equations in the limit of weak gravitational fields. There is a clear gravitational analogue of the advanced Heaviside-Feynman formula. Thus one would expect that a mass accelerated inside a fixed spherical shell of mass should experience a gravitational inertial force in a manner analogous to the above electrical example (just substitute mass $m$, gravitational potential $\phi$ for charge $e$, electrical potential $V$ in above expression).
Perhaps this is the origin of inertia as hypothesised in Mach's Principle? |
Consider a database that has the relation schema CR (StudentName, CourseName).An instance of the schema CR is as given below.
The following query is made on the database.
$\style{font-family:'Times New Roman'}{\begin{array}{l}T1\leftarrow{\mathrm\pi}_{CourseName}\;({\mathrm\sigma}_{StudentName\mathit=\mathit'SA\mathit'}(\mathrm{CR}))\\T2\leftarrow CR\;\div T1\end{array}}$
The number of rows in T2 is__________
T1 result:
$T_2$ result : $ \mathrm{CR}\div T_1\mathit\Rightarrow $ student name for which every course name of CA, CB, CC is
This is not the official website of GATE.
It is our sincere effort to help you. |
The shear stress is part of the pressure tensor. However, here, and many parts of the book, it will be treated as a separate issue. In solid mechanics, the shear stress is considered as the ratio of the force acting on area in the direction of the forces perpendicular to area. Different from solid, fluid cannot pull directly but through a solid surface. Consider liquid that undergoes a shear stress between a short distance of two plates as shown in Figure 1.3.
Figure 1.3: Schematics to describe the shear stress in fluid mechanics.
The upper plate velocity generally will be
\[U = \ff(A, F, h)\tag{2}\]
Where
Equations qref{intro:eq:Upropto} can be rearranged to be
\[U h \propto \dfrac{F}{A} \tag{4}\]
Shear stress was defined as
\[T_{xy} = \dfrac{F}{A}\tag{5}\]
The index
\[T_{xy} = \mu \dfrac{U}{h}\tag{6}\]
Where \mu is called the absolute viscosity or dynamic viscosity which will be discussed later in this chapter in a great length.
Fig. 1.4 The deformation of fluid due to shear stress as progression of time.
In steady state, the distance the upper plate moves after small amount of time, \delta t is
\[dll = U \delta t \tag{7}\]
From Figure 1.4 it can be noticed that for a small angle, \(\delta\beta \cong \sin\beta\), the regular approximation provides
\[dll = U \delta t = h\delta\beta\tag{8}\]
From equation qref{intro:eq:dUdx} it follows that
\[U=h\dfrac{\delta\beta}{\delta t}\tag{9}\]
Combining equation qref{intro:eq:dbdt} with equation qref{intro:eq:shearS} yields
\[T_{xy} = \mu\dfrac{\delta\beta}{\delta t}\tag{10}\]
If the velocity profile is linear between the plate (it will be shown later that it is consistent with derivations of velocity), then it can be written for small a angel that
\[\dfrac{\delta \beta}{\delta t} = \dfrac{dU}{dy}\tag{11}\]
Materials which obey equation qref{intro:eq:tau_dbdt} referred to as Newtonian fluid. For this kind of substance
\[T_{xy} = \mu\dfrac{dU}{dy}\tag{12}\]
Newtonian fluids are fluids which the ratio is constant. Many fluids fall into this category such as air, water etc. This approximation is appropriate for many other fluids but only within some ranges. Equation qref{intro:eq:dbdt} can be interpreted as momentum in the \x direction transferred into the \y direction. Thus, the viscosity is the resistance to the flow (flux) or the movement. The property of viscosity, which is exhibited by all fluids, is due to the existence of cohesion and interaction between fluid molecules. These cohesion and interactions hamper the flux in y–direction. Some referred to shear stress as viscous flux of x–momentum in the y–direction. The units of shear stress are the same as flux per time as following
\[\dfrac{F}{A}\left\dfrac{kg m}{sec^2}\dfrac{1}{m^2}\right = \dfrac{mU}{A}\left\dfrac{kg}{sec}\dfrac{m}{sec}\dfrac{1}{m^2}\right\tag\]
Thus, the notation of \(T_{xy}\) is easier to understand and visualize. In fact, this interpretation is more suitable to explain the molecular mechanism of the viscosity. The units of absolute viscosity are \([N sec / m^2]\). |
I am stuck on this problem from Folland's
Real Analysis, Second Edition: For $j = 1, 2$, let $\mu_j, \nu_j$ be $\sigma$-finite measures on $(X_j, \mathcal{M}_j)$ such that $\nu_j <\!\!< \mu_j$. Then $\nu_1 \times \nu_2 <\!\!< \mu_1 \times \mu_2$.
Here is about where I am at.
(1) It is immediate that if $\mu_1 \times \mu_2(A \times B) = 0$, then $\nu_1 \times \nu_2(A \times B) = 0$.
(2) Because $\nu_1 \times \nu_2$ is $\sigma$-finite, it is enough to prove the result for when $\nu_1 \times \nu_2$ is finite. Since the result is quickly verified if either $\nu_1$ or $\nu_2$ is the zero measure, we may assume that both $\nu_1$ and $\nu_2$ are finite. We then have the following equivalent formulation for $\nu_1 <\!\!< \mu_1$:
For every $\epsilon > 0$, there exists $\delta > 0$ such that $\mu_1(E) < \delta$ implies $\nu_1(E) < \epsilon$.
And similarly for $\nu_2 <\!\!< \mu_2$.
I thought I might be able to prove the same condition for $\nu_1 \times \nu_2$ with respect to $\mu_1 \times \mu_2$, which would then imply $\nu_1 \times \nu_2 <\!\!< \mu_1 \times \mu_2$ since $\nu_1 \times \nu_2$ has been reduced to being finite. As a suggestion, following Folland's technique, it might be easier to argue by contradiction, assuming first the $\epsilon-\delta$ condition is false.
(3) If $\mu_1 \times \mu_2(E) = 0$, then, by definition,
$$0 = \inf \bigg\{ \sum_n \mu_1(A_n)\mu_2(B_n) \colon A_n \times B_n \text{ are rectangles such that } E \subset \bigcup_n A_n \times B_n\bigg\}.$$
To show $\nu_1 \times \nu_2(E) = 0$, we want to show that the analogous equation holds for $\nu_1 \times \nu_2$.
I can't seem to put the pieces together, despite some effort.
Any help would be greatly appreciated. Thanks. |
I'm trying to implement PCA mixtures (Tipping & Bishop 2006 Appendix C) on the Tobomovirus. I'll summarize the mathematical background and algorithm here:
For a single PCA model, we assume a latent variable model
$$ \mathbf{t} = \mathbf{Wx} + \boldsymbol\mu + \boldsymbol\epsilon. $$ This implies a probability distribution $$ p(\mathbf{t}|\mathbf{x}) = (2\pi\sigma^2)^{-d/2}\mathrm{exp}\{-\frac{1}{2\sigma^2}||\mathbf{t} - \mathbf{Wx}- \boldsymbol\mu||^2\} $$ since the noise term $\boldsymbol\epsilon$ is zero-mean Gaussian with variance $\sigma^2$.
We define the Gaussian prior
$$ p(\mathbf{x}) = (2\pi)^{-q/2}\mathrm{exp}\{-\frac{1}{2}\mathbf{x}^T\mathbf{x}\} $$ which allows us to derive the marginal distribution
$$ p(\mathbf{t}) = (2\pi)^{-d/2}|\mathbf{C}|^{-1/2}\mathrm{exp}\{-\frac{1}{2}(\mathbf{t} - \boldsymbol\mu)^T\mathbf{C}^-1(\mathbf{t} - \boldsymbol\mu)\} $$
where $\mathbf{C} = \sigma^2\mathbf{I} + \mathbf{WW}^T$.
The PCA mixture model means that we use a mixture of PCA models like this:
$$ p(\mathbf{t}) = \sum_{i=1}^M \pi_i p(\mathbf{t}|i) $$
The algorithm:
Initialize $\pi_i$, $\boldsymbol\mu_i$, $\mathbf{W}_i$, $\sigma_i^2$. These are the mixing coefficient, mean (of the marginal), weight matrix and variance of each model. Calculate the responsibilities,
$$ R_{ni} = \frac{p(\mathbf{t}_n|i)\pi_i}{p(\mathbf{t}_n)} $$ where $p(\mathbf{t}|i)$ is the probability for a single PCA model $i$.
Calculate mixing coefficients
$$ \tilde\pi_i = \frac{1}{N}\sum^N_{n=1} R_{ni}. $$
Calculate the means
$$ \tilde{\boldsymbol\mu}_i = \frac{\sum_{n=1}^N R_{ni}\mathbf{t}_n}{\sum_{n=1}^NR_{ni}} $$
Calculate the new weight matrices
$$ \mathbf{\tilde W}_i = \mathbf{S}_i \mathbf{W}_i(\sigma^2 \mathbf{I}+\mathbf{M}^{-1}\mathbf{W}_i^T\mathbf{S}_i\mathbf{W}_i)^{-1} $$
where $\mathbf{S}_i = \frac{1}{\tilde\pi_i N}\sum^N_{n=1}R_{ni}(\mathbf{t}_n - \tilde{\boldsymbol\mu})(\mathbf{t}_n - \tilde{\boldsymbol\mu})^T$.
Repeat from step 2.
The tilde operator means that it is a new parameter being calculated.
Here's my problem:When implementing this on the Tobomovirus dataset, the long 18x1 feature vectors and the resulting 18x18 $\mathbf{C}$ matrix lead to the probabilities $p(\mathbf{t})$ in step 2 becoming very very small, leading to underflow. I would like to work with the log probabilities instead, but I'm not sure how this will affect the algorithm and what exactly I need to change. |
We give here an alternative definition of the canonical spinor on 3-Sasakian 7-dimensional manifolds discovered by Agricola and Friedrich [1]. This approach makes use of the Riemannian cone construction which we now recall.
The Riemannian cone over \((M,g)\)
is the Riemannian manifold \((\bar{M},\bar{g}):=(\mathbb {R}^+\times M,dt^2+t^2g)\)
. The radial vector \(\xi :=t\tfrac{\partial }{\partial t}\)
satisfies the equation
$$\begin{aligned} \bar{\nabla }_X\xi =X,\qquad \forall \ X\in \mathrm {T}\bar{M}, \end{aligned}$$
(31)
where \(\bar{\nabla }\)
denotes the Levi–Civita covariant derivative of \(\bar{g}\)
. Assume now that \(M\)
is 3-Sasakian. It is well known (and is nowadays the standard definition of 3-Sasakian structures) that \(\bar{M}\)
has a hyperkähler structure \(J_1,J_2,J_3\)
, such that the vector fields \(\xi _i:=J_i(\xi )\)
on \(\bar{M}\)
are Killing and tangent to the hypersurfaces \(M_t:=\{t\}\times M\)
. When restricted to \(M=M_1\)
, \(\xi _i\)
are unit Killing vector fields satisfying the 3-Sasakian relations.
Suppose now that \(M\)
has dimension 7. The real spin bundle of \(M\)
is canonically identified with the positive spin bundle \(\Sigma ^+\bar{M}\)
restricted to \(M_1=M\)
. With respect to this identification, if \(\psi \in \Gamma (\Sigma M)\)
is the restriction to \(M\)
of a spinor \(\Psi \in \Gamma (\Sigma ^+\bar{M})\)
and \(X\)
is any vector field on \(M\)
identified with a vector field on \(\bar{M}\)
along \(M_1\)
, then
$$\begin{aligned} X\cdot \psi =X\cdot \xi \cdot \Psi \end{aligned}$$
(32)
and
$$\begin{aligned} \nabla _X\psi =\bar{\nabla }_X\Psi +\frac{1}{2} X\cdot \xi \cdot \Psi . \end{aligned}$$
(33)
Recall now that the restriction to \(\mathrm {Sp}(2)\)
of the half-spin representation \(\Sigma _8^+\)
has a 3-dimensional trivial summand. Correspondingly, on \(\bar{M}\)
there exist three linearly independent \(\bar{\nabla }\)
-parallel spinor fields on which every 2-form from \(\mathfrak {sp}(2)\)
(i.e. commuting with \(J_1\)
, \(J_2\)
, \(J_3\)
) acts trivially by Clifford multiplication. Moreover, there exists exactly one such unit spinor \(\Psi _1\)
(up to sign) on which the Clifford action of \(\Omega _1\)
(the Kähler form of \(J_1\)
) is also trivial (cf. [20
]).
Lemma 6.1
The spinor \(\Psi _0:=\tfrac{1}{|\xi |^2}\xi \cdot \xi _1\cdot \Psi _1\)
satisfies
$$\begin{aligned} \bar{\nabla }_X\Psi _0=\bar{A}(X)\cdot \xi \cdot \Psi _0, \end{aligned}$$
(34)
where
$$\begin{aligned} \bar{A}(X):=\left\{ \begin{array}{ll} \quad \ 0 &{}\quad if\, X\, \hbox {belongs }\,\hbox {to }\, \hbox { the }\, \hbox {distribution }\, D:=\langle \xi ,\xi _1,\xi _2,\xi _3\rangle \\ -2X &{}\quad if\, X\, \in D^\perp . \end{array}\right. \end{aligned}$$
Proof
Since \(J_i\)
are \(\bar{\nabla }\)
-parallel, (31
) yields \(\bar{\nabla }_X\xi _i=J_i(X)\)
for all \(X\in \mathrm {T}\bar{M}\)
. We thus have
$$\begin{aligned} \bar{\nabla }_X\Psi _0=\frac{1}{|\xi |^2}\left( X\cdot \xi _1\cdot \Psi _1+\xi \cdot J_1(X)\cdot \Psi _1 \right) -\frac{2}{|\xi |^4}\bar{g}(\xi ,X)\xi \cdot \xi _1\cdot \Psi _1. \end{aligned}$$
(35)
This relation gives immediately \(\bar{\nabla }_\xi \Psi _0=0\)
and \(\bar{\nabla }_{\xi _1}\Psi _0=0\)
. Moreover, since the 2-form \(\xi \wedge \xi _1-\xi _2\wedge \xi _3\)
commutes with \(J_1,J_2,J_3\)
, it belongs to \(\mathfrak {sp}(2)\)
and thus acts trivially by Clifford multiplication on \(\Psi _1\)
. We then obtain \(\xi \cdot \xi _1\cdot \Psi _0=\xi _2\cdot \xi _3\cdot \Psi _0\)
, which together with (35
) yields \(\bar{\nabla }_{\xi _2}\Psi _0=\bar{\nabla }_{\xi _3}\Psi _0=0\)
.
It remains to treat the case where \(X\)
is orthogonal to \(\langle \xi ,\xi _1,\xi _2,\xi _3\rangle \)
. Assume that \(X\)
is scaled to have unit norm. We consider the orthonormal basis of \(\mathrm {T}\bar{M}\)
at some point \(x\in M_1\)
given by \(e_1=\xi ,e_2=\xi _1,e_3=\xi _2,e_4=\xi _3,e_5=X,e_6=J_1(X),e_7=J_2(X),e_8=J_3(X)\)
. Since \(\Omega _1\cdot \Psi _1=0\)
where \(\Omega _1=e_1\cdot e_2+e_3\cdot e_4+e_5\cdot e_6+e_7\cdot e_8\)
, we obtain \(\Omega _1\cdot \Psi _0=0\)
. Now, the 2-form \(e_5\wedge e_6-e_7\wedge e_8\)
belongs to \(\mathfrak {sp}(2)\)
and its Clifford action commutes with \(e_1\cdot e_2\)
, thus \(e_5\cdot e_6\cdot \Psi _0=e_7\cdot e_8\cdot \Psi _0\)
. Together with the relation \(e_1\cdot e_2\cdot \Psi _0=e_3\cdot e_4\cdot \Psi _0\)
proved above and the fact that \(\Omega _1\cdot \Psi _0=0\)
, we get
$$\begin{aligned} (e_1\cdot e_2+e_5\cdot e_6)\cdot \Psi _0=0. \end{aligned}$$
(36)
Using (35
) we then compute at \(x\)
:
$$\begin{aligned} \bar{\nabla }_X\Psi _0&= \left( X\cdot \xi _1\cdot \Psi _1+\xi \cdot J_1(X)\cdot \Psi _1\right) =(e_5\cdot e_2+e_1\cdot e_6)\cdot (-e_1\cdot e_2\cdot \Psi _0)\\&= (e_1\cdot e_5+e_2\cdot e_6)\cdot \Psi _0=e_5\cdot e_2\cdot (e_1\cdot e_2+e_5\cdot e_6)\cdot \Psi _0+2e_1\cdot e_5\cdot \Psi _0\\&= 2e_1\cdot e_5\cdot \Psi _0=-2X\cdot \xi \cdot \Psi _0, \end{aligned}$$
thus proving the lemma.\(\square \)
As a direct consequence of this result, together with (32), (33), we obtain the following:
Corollary 6.2
([1
], Thm. 4.1) The spinor \(\psi _0:=\Psi _0\arrowvert _M\)
is a generalized Killing spinor on \(M\)
satisfying
$$\begin{aligned} \nabla _X\psi _0=A(X)\cdot \psi _0, \end{aligned}$$
(37)
where
$$\begin{aligned} A(X):=\left\{ \begin{array}{ll} \quad \frac{1}{2}X&{}\quad if\, X\, \hbox {belongs }\, \hbox {to }\, \hbox {the }\, \hbox {distribution }\, D:=\langle \xi _1,\xi _2,\xi _3\rangle \\ -\frac{3}{2}X&{}\quad if\, X\, \in \, D^\perp . \end{array}\right. \end{aligned}$$ |
Two bodies with similar mass orbiting around a common barycenter
with elliptic orbits
(left)
. Two bodies with a slight difference in mass
orbiting around a common barycenter. The sizes, and this particular type of orbit are similar to the Pluto
–Charon
system and also to Earth–Moon system in which the center of mass is inside the bigger body instead
(right)
.
In classical mechanics, the
two-body problem is to determine the motion of two point particles that interact only with each other. Common examples include a satellite orbiting a planet, a planet orbiting a star, two stars orbiting each other (a binary star), and a classical electron orbiting an atomic nucleus (although to solve the electron/nucleus 2-body system correctly a quantum mechanical approach must be used).
The two-body problem can be re-formulated as two
one-body problems, a trivial one and one that involves solving for the motion of one particle in an external potential. Since many one-body problems can be solved exactly, the corresponding two-body problem can also be solved. By contrast, the three-body problem (and, more generally, the n-body problem for n ≥ 3) cannot be solved in terms of first integrals, except in special cases.
Contents Reduction to two independent, one-body problems 1 Center of mass motion (1st one-body problem) 1.1 Two-body motion is planar 2 Laws of Conservation of Energy for each of two bodies for arbitrary potentials 3 Central forces 4 Work 5 See also 6 References 7 Bibliography 8 External links 9 Reduction to two independent, one-body problems Jacobi coordinates
for two-body problem; Jacobi coordinates are \boldsymbol{R}=\frac {m_1}{M} \boldsymbol{x}_1 + \frac {m_2}{M} \boldsymbol{x}_2
and \boldsymbol{r} = \boldsymbol{x}_1 - \boldsymbol{x}_2
with M = m_1+m_2 \
.
[1]
Let
x 1 and x 2 be the vector positions of the two bodies, and m 1 and m 2 be their masses. The goal is to determine the trajectories x 1( t) and x 2( t) for all times t, given the initial positions x 1( t = 0) and x 2( t = 0) and the initial velocities v 1( t = 0) and v 2( t = 0).
When applied to the two masses, Newton's second law states that
\mathbf{F}_{12}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{1} \ddot{\mathbf{x}}_{1} \quad \quad \quad (\mathrm{Equation} \ 1) \mathbf{F}_{21}(\mathbf{x}_{1},\mathbf{x}_{2}) = m_{2} \ddot{\mathbf{x}}_{2} \quad \quad \quad (\mathrm{Equation} \ 2)
where
F 12 is the force on mass 1 due to its interactions with mass 2, and F 21 is the force on mass 2 due to its interactions with mass 1. The two dots on top of the x position vectors denote their second derivative, or their acceleration vectors.
Adding and subtracting these two equations decouples them into two one-body problems, which can be solved independently.
Adding equations (1) and (2) results in an equation describing the center of mass (barycenter) motion. By contrast, subtracting equation (2) from equation (1) results in an equation that describes how the vector r = x 1 − x 2 between the masses changes with time. The solutions of these independent one-body problems can be combined to obtain the solutions for the trajectories x 1( t) and x 2( t). Center of mass motion (1st one-body problem)
Addition of the force equations (1) and (2) yields
m_{1}\ddot{\mathbf{x}}_1 + m_2 \ddot{\mathbf{x}}_2 = (m_1 + m_2)\ddot{\mathbf{R}} = \mathbf{F}_{12} + \mathbf{F}_{21} = 0
where we have used Newton's third law
F 12 = − F 21 and where \ddot{\mathbf{R}} \equiv \frac{m_{1}\ddot{\mathbf{x}}_{1} + m_{2}\ddot{\mathbf{x}}_{2}}{m_{1} + m_{2}} \mathbf{R} is the position of the center of mass (barycenter) of the system.
The resulting equation:
\ddot{\mathbf{R}} = 0
shows that the velocity
V = d R/ dt of the center of mass is constant, from which follows that the total momentum m 1 v 1 + m 2 v 2 is also constant (conservation of momentum). Hence, the position R ( t) of the center of mass can be determined at all times from the initial positions and velocities. Two-body motion is planar
The motion of two bodies with respect to each other always lies in a plane (in the center of mass frame). Defining the linear momentum
p and the angular momentum L by the equations (where μ is the reduced mass) \mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times \mu \frac{d\mathbf{r}}{dt}
the rate of change of the angular momentum
L equals the net torque N \mathbf{N} = \frac{d\mathbf{L}}{dt} = \dot{\mathbf{r}} \times \mu\dot{\mathbf{r}} + \mathbf{r} \times \mu\ddot{\mathbf{r}} \ ,
and using the property of the vector cross product that
v × w = 0 for any vectors and v pointing in the same direction, w \mathbf{N} \ = \ \frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F} \ ,
with
F = μ d 2 r / dt 2.
Introducing the assumption (true of most physical forces, as they obey Newton's strong third law of motion) that the force between two particles acts along the line between their positions, it follows that
r × F = 0 and the angular momentum vector L is constant (conserved). Therefore, the displacement vector r and its velocity v are always in the plane perpendicular to the constant vector L. Laws of Conservation of Energy for each of two bodies for arbitrary potentials
In system of the center of mass for arbitrary potentials
~U_{12} = U(\mathbf{r}_1 - \mathbf{r}_2) ~U_{21} = U(\mathbf{r}_2 - \mathbf{r}_1)
the value of energies of bodies do not change:
~E_1 = m_1 \frac{v_1^2}{2} + \frac{m_2} {m_1+m_2} U_{12} = \text{Const}_1(t) ~E_2 = m_2 \frac{v_2^2}{2} + \frac{m_1} {m_1+m_2} U_{21} = \text{Const}_2(t) Central forces
For many physical problems, the force
F( r) is a central force, i.e., it is of the form \mathbf{F}(\mathbf{r}) = F(r)\hat{\mathbf{r}}
where r = |
r| and r̂ = r/r is the corresponding unit vector. We now have: \mu \ddot{\mathbf{r}} = {F}(r) \hat{\mathbf{r}} \ ,
where
F(r) is negative in the case of an attractive force. Work
The total work done in a given time interval by the forces exerted by two bodies on each other is the same as the work done by one force applied to the total relative displacement.
See also References ^ David Betounes (2001). Differential Equations. Springer. p. 58; Figure 2.15. Bibliography External links
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Independence is a statistical concept. Two random variables $X$ and $Y$ are statistically independent if their joint distribution is the product of the marginal distributions, i.e.$$f(x, y) = f(x) f(y)$$if each variable has a density $f$, or more generally$$F(x, y) = F(x) F(y)$$where $F$ denotes each random variable's cumulative distribution function.
Correlation is a weaker but related statistical concept. The (Pearson) correlation of two random variables is the expectancy of the product of the standardized variables, i.e.$$\newcommand{\E}{\mathbf E}\rho = \E \left [\frac{X - \E[X]}{\sqrt{\E[(X - \E[X])^2]}}\frac{Y - \E[Y]}{\sqrt{\E[(Y - \E[Y])^2]}}\right ].$$The variables are uncorrelated if $\rho = 0$. It can be shown that two random variables that are independent are necessarily uncorrelated, but not vice versa.
Orthogonality is a concept that originated in geometry, and was generalized in linear algebra and related fields of mathematics. In linear algebra, orthogonality of two vectors $u$ and $v$ is defined in inner product spaces, i.e. vector spaces with an inner product $\langle u, v \rangle$, as the condition that$$\langle u, v \rangle = 0.$$The inner product can be defined in different ways (resulting in different inner product spaces). If the vectors are given in the form of sequences of numbers, $u = (u_1, u_2, \ldots u_n)$, then a typical choice is the dot product, $\langle u, v \rangle = \sum_{i = 1}^n u_i v_i$.
Orthogonality is therefore not a statistical concept per se, and the confusion you observe is likely due to different translations of the linear algebra concept to statistics:
a) Formally, a space of random variables can be considered as a vector space. It is then possible to define an inner product in that space, in different ways. One common choice is to define it as the covariance:$$\langle X, Y \rangle = \mathrm{cov} (X, Y)= \E [ (X - \E[X]) (Y - \E[Y]) ].$$Since the correlation of two random variables is zero exactly if the covariance is zero,
according to this definition uncorrelatedness is the same as orthogonality. (Another possibility is to define the inner product of random variables simply as the expectancy of the product.)
b) Not all the variables we consider in statistics are random variables. Especially in linear regression, we have independent variables which are not considered random but predefined. Independent variables are usually given as sequences of numbers, for which orthogonality is naturally defined by the dot product (see above). We can then investigate the statistical consequences of regression models where the independent variables are or are not orthogonal. In this context, orthogonality does not have a specifically statistical definition, and even more: it does not apply to random variables.
Addition responding to Silverfish's comment: Orthogonality is not only relevant with respect to the original regressors but also with respect to contrasts, because (sets of) simple contrasts (specified by contrast vectors) can be seen as transformations of the design matrix, i.e. the set of independent variables, into a new set of independent variables. Orthogonality for contrasts is defined via the dot product. If the original regressors are mutually orthogonal and one applies orthogonal contrasts, the new regressors are mutually orthogonal, too. This ensures that the set of contrasts can be seen as describing a decomposition of variance, e.g. into main effects and interactions, the idea underlying ANOVA.
Since according to variant a), uncorrelatedness and orthogonality are just different names for the same thing, in my opinion it is best to avoid using the term in that sense. If we want to talk about uncorrelatedness of random variables, let's just say so and not complicate matters by using another word with a different background and different implications. This also frees up the term orthogonality to be used according to variant b), which is highly useful especially in discussing multiple regression. And the other way around, we should avoid applying the term correlation to independent variables, since they are not random variables.
Rodgers et al.'s presentation is largely in line with this view, especially as they understand orthogonality to be distinct from uncorrelatedness. However, they do apply the term correlation to non-random variables (sequences of numbers). This only makes sense statistically with respect to the sample correlation coefficient $r$. I would still recommend to avoid this use of the term, unless the number sequence is considered as a sequence of realizations of a random variable.
I've scattered links to the answers to the two related questions throughout the above text, which should help you put them into the context of this answer. |
DrivAer: grille shutters
In this post we continue to explore road vehicle aerodynamics with aid of the DrivAer model. This time, considering the benefit of adding grille shutters to improve aerodynamic drag.
The active grille shutters know when the engine, brakes and other components need air and automatically open the air vents. If no further air intake is required, the shutters close improving aerodynamics and reducing fuel consumption.
The following video demonstrates the system’s operating principle.
Computational domain
For this case I re-used the same meshDict as for the external aerodynamics case, just adding additional refinements for the underhood area. Specially around the grid area in order to better capture the small features; which ends up with 9.4 million cells.
For the boundary conditions I’m keeping the same settings as for the external aerodynamics cases, with the exception of adding a porous region.
Porous region
The porosity properties are added to a certain region of the mesh, the heat exchanger. This region can be defined during the meshing process, or afterwards using the setSet command-line utility.
The most common approach in OpenFOAM is to define the Darcy-Forchheimer coefficients (alternatives include the power law and fixed coefficients). Those two coefficients are added to the sink term of the Navier-Stokes equations, which looks like
$$ S_i = -\left( \mu D_{ij} + \frac{1}{2}\rho |u_{kk}| F_{ij} \right) u_i $$
where $D_{ij}$ is the resistance of the medium (viscous loss term) and $F_{ij}$ is the inertial resistance (inertial loss term). In OpenFOAM we define the coefficients as two vectors,
d and f which are then added to the diagonal of the tensors $D_{ij}$ and $F_{ij}$, respectively. Here you can see a sample of the fvOption dictionary used to define the porosity source term of my simulations.
As I didn’t find experimental values for the pressure drop in the heat exchangers defined in the DrivAer model, I re-used the same inertial and viscous resistance as in the formula student aerodynamics workshop, that is,
d d [ 0 -2 0 0 0 0 0 ] ( 18463.81 -1000 -1000 ); f f [ 0 -1 0 0 0 0 0 ] ( 1.09 -1000 -1000 );
The code will automatically multiply negative values with the largest component of the vector and switch the sign to a positive value.
Results
As expected, considering the underhood in the simulation has increased the aerodynamic drag of the vehicle. Introducing the engine bay in the simulation pumped up drag in 13 counts (+5.25 %), while closing the upper grille improved the latter value in about 11 counts (-4.21 %).
Of course, by opening up the grilles, the stagnating region at the front of the vehicle is reduced and so is the pressure drag at the front of the vehicle. That’s why the orange line has a lower peak. With the active grille shutters we have a higher peak at the front; because on the front side of the shutters the flow stagnates —just as in the reference model—, and on the other side we have the engine bay low pressure region, which combined makes the resistance in the front to be greater. We can appreciate this in the following pictures.
If we consider only the results of simplified models like this one, we might be inclined towards installing this kind of devices on any car; but actually one must also consider heat exchanger performance. Obviously, when the grille shutters are closed there’s no fresh air going through the radiators; but when fully open, real systems include complex mechanisms and elements and reconfigure the flow path inside the engine bay usually restricting the amount of flow available for cooling. So, if the base design is already performing at the lower target boundary in terms of cooling performance, adding grille shutters to improve a few counts aerodynamic drag might not be the best option.
References
m.bmw.co.uk. (2017).
Lightweight and Aerodynamics | BMW EfficientDynamics | BMW UK. [online] Available at: https://m.bmw.co.uk/en_gb/discover-bmw/technology/efficientdynamics/lightweight-aerodynamics [Accessed 8 Dec. 2017].
Elvar, H. (2009).
Porous Media in OpenFOAM. Göteborg: Chalmers University of Technology.
Schütz, T. (Ed.). (2015).
Aerodynamics of road vehicles (5th ed.). Warrendale, Pennsylvania, USA: SAE International. |
Difference between revisions of "Lower attic"
From Cantor's Attic
(the Takeuti-Feferman-Buchholz ordinal)
(15 intermediate revisions by 5 users not shown) Line 1: Line 1:
{{DISPLAYTITLE: The lower attic}}
{{DISPLAYTITLE: The lower attic}}
[[File:SagradaSpiralByDavidNikonvscanon.jpg | thumb | Sagrada Spiral photo by David Nikonvscanon]]
[[File:SagradaSpiralByDavidNikonvscanon.jpg | thumb | Sagrada Spiral photo by David Nikonvscanon]]
+
Welcome to the lower attic, where the countably infinite ordinals climb ever higher, one upon another, in an eternal self-similar reflecting ascent.
Welcome to the lower attic, where the countably infinite ordinals climb ever higher, one upon another, in an eternal self-similar reflecting ascent.
Line 10: Line 11:
** [[infinite time Turing machines#zeta | $\zeta$]] = the supremum of the eventually writable ordinals
** [[infinite time Turing machines#zeta | $\zeta$]] = the supremum of the eventually writable ordinals
** [[infinite time Turing machines#lambda | $\lambda$]] = the supremum of the writable ordinals,
** [[infinite time Turing machines#lambda | $\lambda$]] = the supremum of the writable ordinals,
−
* [[admissible]] ordinals and [[Church-Kleene#relativized Church-Kleene ordinal | relativized Church-Kleene $\omega_1^x$]]
* [[admissible]] ordinals and [[Church-Kleene#relativized Church-Kleene ordinal | relativized Church-Kleene $\omega_1^x$]]
* [[Church-Kleene | Church-Kleene $\omega_1^{ck}$]], the supremum of the computable ordinals
* [[Church-Kleene | Church-Kleene $\omega_1^{ck}$]], the supremum of the computable ordinals
−
* the [[omega one chess | omega one of chess]]
+
* the [[omega one chess | omega one of chess]]
+
[[omega one chess|$\omega_1^{\chess
+
}$
+ + + + + +
]]
* the [[Feferman-Schütte]] ordinal [[Feferman-Schütte | $\Gamma_0$]]
* the [[Feferman-Schütte]] ordinal [[Feferman-Schütte | $\Gamma_0$]]
* [[epsilon naught | $\epsilon_0$]] and the hierarchy of [[epsilon naught#epsilon_numbers | $\epsilon_\alpha$ numbers]]
* [[epsilon naught | $\epsilon_0$]] and the hierarchy of [[epsilon naught#epsilon_numbers | $\epsilon_\alpha$ numbers]]
+
* the [[small countable ordinals]], such as [[small countable ordinals | $\omega,\omega+1,\ldots,\omega\cdot 2,\ldots,\omega^2,\ldots,\omega^\omega,\ldots,\omega^{\omega^\omega},\ldots$]] up to [[epsilon naught | $\epsilon_0$]]
* the [[small countable ordinals]], such as [[small countable ordinals | $\omega,\omega+1,\ldots,\omega\cdot 2,\ldots,\omega^2,\ldots,\omega^\omega,\ldots,\omega^{\omega^\omega},\ldots$]] up to [[epsilon naught | $\epsilon_0$]]
−
* [[
+
* [[| Hilbert's hotel]]
* [[omega | $\omega$]], the smallest infinity
* [[omega | $\omega$]], the smallest infinity
* down to the [[parlour]], where large finite numbers dream
* down to the [[parlour]], where large finite numbers dream
Latest revision as of 13:37, 27 May 2018
Welcome to the lower attic, where the countably infinite ordinals climb ever higher, one upon another, in an eternal self-similar reflecting ascent.
$\omega_1$, the first uncountable ordinal, and the other uncountable cardinals of the middle attic stable ordinals The ordinals of infinite time Turing machines, including admissible ordinals and relativized Church-Kleene $\omega_1^x$ Church-Kleene $\omega_1^{ck}$, the supremum of the computable ordinals the omega one of chess $\omega_1^{\mathfrak{Ch}_{\!\!\!\!\sim}}$ = the supremum of the game values for white of all positions in infinite chess $\omega_1^{\mathfrak{Ch},c}$ = the supremum of the game values for white of the computable positions in infinite chess $\omega_1^{\mathfrak{Ch}}$ = the supremum of the game values for white of the finite positions in infinite chess the Takeuti-Feferman-Buchholz ordinal the Bachmann-Howard ordinal the large Veblen ordinal the small Veblen ordinal the Extended Veblen function the Feferman-Schütte ordinal $\Gamma_0$ $\epsilon_0$ and the hierarchy of $\epsilon_\alpha$ numbers indecomposable ordinal the small countable ordinals, such as $\omega,\omega+1,\ldots,\omega\cdot 2,\ldots,\omega^2,\ldots,\omega^\omega,\ldots,\omega^{\omega^\omega},\ldots$ up to $\epsilon_0$ Hilbert's hotel and other toys in the playroom $\omega$, the smallest infinity down to the parlour, where large finite numbers dream |
📖 There's nothing I can say that's better than Songho's notes for transforms on OpenGL.
Reducing a continuous-time signal to a discrete-time signal. Think of it as recording sounds in a digital sound studio.
🤔 Some intuition: The continuous-time signal may be some sound picked up by the microphone, some image, some sine wave!?
🍿 If time permits: Demo differently sampled audio tracks (in Adobe Audition)
Nyquist frequency: Half the sampling frequency We get no aliasing from frequencies less than the Nyquist frequency.
As an example, for an image of stripes (black & white), if we're sampling at every
16px, stripes at cycle every
32px or more will result in no aliasing.
Also, if there are stripes that cycle at every
8px (
4px black, then
4px white), we'd want to sample at every
4px to avoid aliasing.
We're only concering $R^2$ for this section
🤔 Do we really need to describe a point with all $\alpha$, $\beta$ and $\gamma$? What if we drop $\alpha$?\begin{align} P & = \alpha A + \beta B + \gamma C \\ P & = (1 - \beta - \gamma) A + \beta B + \gamma C \\ P & = A - \beta A - \gamma A + \beta B + \gamma C \\ P - A & = \beta (B - A) + \gamma (C - A) \end{align}
Now we can see it with $A$ as the origin, with $(B - A)$ and $(C - A)$ as two independent vectors.
This can lead us to another point-in-triangle test (that
doesn't worry about winding). For any $P$, how can we tell its Barycentric coordinates? It's a change in frame of reference.
This may look rather troublesome, but it lends to an approach for finding ray-triangle intersection in $R^3$ (more later).
🍿 If time permits: Demo Suzanne model (in Blender)
🤔 Why would you want mipmaps? |
DrivAer: grille shutters
In this post we continue to explore road vehicle aerodynamics with aid of the DrivAer model. This time, considering the benefit of adding grille shutters to improve aerodynamic drag.
The active grille shutters know when the engine, brakes and other components need air and automatically open the air vents. If no further air intake is required, the shutters close improving aerodynamics and reducing fuel consumption.
The following video demonstrates the system’s operating principle.
Computational domain
For this case I re-used the same meshDict as for the external aerodynamics case, just adding additional refinements for the underhood area. Specially around the grid area in order to better capture the small features; which ends up with 9.4 million cells.
For the boundary conditions I’m keeping the same settings as for the external aerodynamics cases, with the exception of adding a porous region.
Porous region
The porosity properties are added to a certain region of the mesh, the heat exchanger. This region can be defined during the meshing process, or afterwards using the setSet command-line utility.
The most common approach in OpenFOAM is to define the Darcy-Forchheimer coefficients (alternatives include the power law and fixed coefficients). Those two coefficients are added to the sink term of the Navier-Stokes equations, which looks like
$$ S_i = -\left( \mu D_{ij} + \frac{1}{2}\rho |u_{kk}| F_{ij} \right) u_i $$
where $D_{ij}$ is the resistance of the medium (viscous loss term) and $F_{ij}$ is the inertial resistance (inertial loss term). In OpenFOAM we define the coefficients as two vectors,
d and f which are then added to the diagonal of the tensors $D_{ij}$ and $F_{ij}$, respectively. Here you can see a sample of the fvOption dictionary used to define the porosity source term of my simulations.
As I didn’t find experimental values for the pressure drop in the heat exchangers defined in the DrivAer model, I re-used the same inertial and viscous resistance as in the formula student aerodynamics workshop, that is,
d d [ 0 -2 0 0 0 0 0 ] ( 18463.81 -1000 -1000 ); f f [ 0 -1 0 0 0 0 0 ] ( 1.09 -1000 -1000 );
The code will automatically multiply negative values with the largest component of the vector and switch the sign to a positive value.
Results
As expected, considering the underhood in the simulation has increased the aerodynamic drag of the vehicle. Introducing the engine bay in the simulation pumped up drag in 13 counts (+5.25 %), while closing the upper grille improved the latter value in about 11 counts (-4.21 %).
Of course, by opening up the grilles, the stagnating region at the front of the vehicle is reduced and so is the pressure drag at the front of the vehicle. That’s why the orange line has a lower peak. With the active grille shutters we have a higher peak at the front; because on the front side of the shutters the flow stagnates —just as in the reference model—, and on the other side we have the engine bay low pressure region, which combined makes the resistance in the front to be greater. We can appreciate this in the following pictures.
If we consider only the results of simplified models like this one, we might be inclined towards installing this kind of devices on any car; but actually one must also consider heat exchanger performance. Obviously, when the grille shutters are closed there’s no fresh air going through the radiators; but when fully open, real systems include complex mechanisms and elements and reconfigure the flow path inside the engine bay usually restricting the amount of flow available for cooling. So, if the base design is already performing at the lower target boundary in terms of cooling performance, adding grille shutters to improve a few counts aerodynamic drag might not be the best option.
References
m.bmw.co.uk. (2017).
Lightweight and Aerodynamics | BMW EfficientDynamics | BMW UK. [online] Available at: https://m.bmw.co.uk/en_gb/discover-bmw/technology/efficientdynamics/lightweight-aerodynamics [Accessed 8 Dec. 2017].
Elvar, H. (2009).
Porous Media in OpenFOAM. Göteborg: Chalmers University of Technology.
Schütz, T. (Ed.). (2015).
Aerodynamics of road vehicles (5th ed.). Warrendale, Pennsylvania, USA: SAE International. |
Figure 1.7. The shear stress as a function of the sear rate.
In equation ___, the relationship between the velocity and the shear stress was assumed to be linear. Not all the materials obey this relationship. There is a large class of materials which shows a non-linear relationship with velocity for any shear stress. This class of materials can be approximated by a single polynomial term that is \(a = bx^n\). From the physical point of view, the coefficient depends on the velocity gradient. This relationship is referred to as power relationship and it can be written as \[\tau = \overbrace{K (\frac{dU}{dx})^{n-1}}^{viscosity} \left(\frac{dU}{dx}\right)\tag{16}\] The new coefficients \((n, K)\) in equation ___ are constant. When \(n = 1\) equation represent Newtonian fluid and \(K\) becomes the familiar \(\mu\). The viscosity coefficient is always positive. When \(n\) is above one, the liquid is dilettante. When \(n\) is below one, the fluid is pseudoplastic. The liquids which satisfy equation ___ are referred to as purely viscous fluids. Many fluids satisfy the above equation. Fluids that show increase in the viscosity (with increase of the shear) referred to as thixotropic and those that show decrease are called rheopectic fluids (see Figure 1.5).
Materials which behave up to a certain shear stress as a solid and above it as a liquid are referred as Bingham liquids. In the simple case, the ``liquid side'' is like Newtonian fluid for large shear stress. The general relationship for simple Bingham flow is \[\tau_{xy} = -\mu \cdot \pm \tau_{0} \quad if \lvert\tau_{yx}\rvert > \tau_{0}\tag{17}\] \[\frac{dU_{x}}{dy} = 0 \quad if\lvert\tau_{yx}\rvert < \tau_{0}\tag{18}\] There are materials that simple Bingham model does not provide adequate explanation and a more sophisticate model is required. The Newtonian part of the model has to be replaced by power liquid. For example, according to Ferraris at el concrete behaves as shown in Figure 1.7. However, for most practical purposes, this kind of figures isn't used in regular engineering practice.
Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
Let $L$ be the language generated by a context-free grammar whose start variable is $S$. Does adding $S \rightarrow SS$ in this grammar creating language $L^*$, why? What about grammars in Chomsky normal form?
As chi pointed out in the comment, since $\varepsilon\in L^*$ and $\varepsilon$ may not belong to the new grammar, so adding $S\rightarrow SS$ does not always generate $L^*$. It makes more sense to ask whether it generates $L^+$, so the following answer focuses on $L^+$.
General Form
Consider the following grammar:
\begin{align} S&\rightarrow aSb\\ S&\rightarrow \varepsilon \end{align}
It generates the language $L=\{a^nb^n\mid n\in\mathbb{N}\}$.
Now after adding $S\rightarrow SS$ in this grammar, the string $aababb$ can be generated by the grammar but it does not belong to $L^+$.
Chomsky Normal Form
Easy to see $L^+$ can be generated by the new grammar. Now for any string $s$ generated by the new grammar, we can prove $s\in L^+$ by mathematical induction on the depth of its parse tree. Consider the derivation process of $s$.
If the first derivation does not use the rule $S\rightarrow SS$, we can conclude $S$ does not show up in the following derivation since $S$ does not show up in the right hand side of a CFG of Chomsky normal form. As a result, $s\in L$.
If the first derivation uses the rule $S\rightarrow SS$, the strings generated respectively by the two $S$'s in the right hand side must belong to $L^+$ due to inductive assumption, thus $s$, which is formed by concatenating the two strings, also belongs to $L^+$. |
How does one go about proving that the free group $<a,b,a^{-1},b^{-1}>$ lacks any Følner sequence?
I'm going to sketch the standard proof, but it actually has a sticky technical point, so I'm not sure how satisfying you will find it.
First, a group $G$ is said to be amenable if it has a finitely additive, left-invariant probability measure. That is, there is a function $\mu\colon \mathcal P(G)\to [0,1]$ such that $\mu(G)=1$, $\mu(X\amalg Y)=\mu(X)+\mu(Y)$, and for all $g\in G$, $\mu(X)=\mu(gX)$. (Here $X \amalg Y$ is the union of $X,Y$ assuming they are disjoint.)
Now if a group has a Følner sequence $F_i$, you can define$$\mu(X)=\operatorname{ulim}_{i\to \infty} \frac{|X\cap F_i|}{|F_i|}.$$The fancy $\operatorname{ulim}_{i\to\infty}$ is the
ultrafilter limit, and is a way to force limits to exist even when they don't. (This is that sticky technical point.) You can verify that $\mu$ is a finitely additive probability measure. See Wikipedia's article on Følner sequences. So in other words, the existence of a Følner sequence implies amenability.
But now, we can also show that $\langle a,b\rangle$ has no such probabilty measure. Let $W(\alpha)$ be the set of reduced words beginning with $\alpha$. Then \begin{align*} \langle a,b\rangle &=\{1\}\amalg W(a)\amalg W(a^{-1})\amalg W(b)\amalg W(b^{-1})\\ &=W(a)\amalg aW(a^{-1})\\ &=W(b)\amalg bW(b^{-1}) \end{align*} Applying $\mu$ gives a contradiction, since the latter two equations give that $\mu(W(a)\amalg aW(a^{-1}))=\mu(W(b)\amalg bW(b^{-1}))=1$, and then the first equation gives $1=\mu(\{1\})+2$, which is a contradiction. |
Direct FSI Approach to Computing the Acoustic Radiation Force
In an earlier blog post, we considered the computation of acoustic radiation force using a perturbation approach. This method has the advantage of being both robust and fast; however, it relies heavily on the theoretical evaluation of correct perturbation terms. The idea behind the method presented here is to solve the problem by deducing the radiation force from the solution of the full nonlinear set of Navier-Stokes equations, interacting with a solid, elastic microparticle.
Fluid-Structure Interaction (FSI) Formulation
The problem that we want to solve involves the motion of a fluid due to the acoustic wave, which in turn exerts force on a solid particle. The particle responds to the applied force by deformation and net motion and also applies reaction forces on the fluid. This means that in addition to solving the equations describing the fluid dynamics, you would also need to account for the deformation of the particle and the resulting deformation of the space occupied by the fluid. The pre-built
Fluid-Structure Interaction physics interface in COMSOL Multiphysics® software allows you to solve this problem by coupling fluid flow, structural stress analysis, and mesh deformation.
The acoustic radiation force is a nonlinear effect, where the nonlinearity is inherent to the flow and stems from the convective term (\mathbf u \cdot \nabla )\mathbf u in the Navier-Stokes equation rather than from the material nonlinearity. To support acoustic waves, the fluid has to be compressible. The fluid compressibility can be introduced by modifying the constitutive relation of the fluid. We assume a linear elastic fluid where p = c_0^2(\rho-\rho_0) and, for water, we put c_0 = 1500 m/s and \rho_0 = 1000 kg/m
3. Because we initially want to compare the method with classical, analytical models that neglect the effect of viscosity, we assume an arbitrary, small viscosity value.
The acoustic radiation force is much higher in the standing wave fields, and most practical applications utilizing this effect involve standing waves. Let us examine this case.
Because the problem is nonlinear, it must be solved in the time domain. Since the solution can be quite time consuming, we will solve it in a 2D axisymmetric geometry. To create a standing wave in a time-dependent solution, we consider a resonant box two wavelengths high and initiate the standing wave by setting up the corresponding initial conditions. For example, in a box two wavelengths high, the standing wave solution would be p(r,z,t) = p_0 cos(k_0 z) cos(\omega t) and, at the time t= 0 , the initial conditions should read p(r,z,t=0) = p_0 cos(k_0 z) and \mathbf u(r,z,t=0) = 0. This is illustrated below for a simulation box that is 3 mm high and has a wavelength of \lambda = 1.5 mm (the corresponding frequency is 1 MHz).
As noted in the previous blog post, we expect the nonlinear force term to be a few orders of magnitude smaller than the linear force, such that the particle will appear to be bouncing up and down in the acoustic field. However, every time the particle moves, it will go a little further in one direction than the other. This is a result of the nonlinear force component that does not change its direction when the field changes its sign. Because the nonlinear force is so small, it is very hard to extract the value of the nonlinear force from the time-domain solution, unless a very simple trick is used.
The essence of this trick is to utilize the very fact that the nonlinear force does not change direction when the excitation changes sign. Let us denote x_\textrm p (t) as the average displacement of the particle obtained when p_0 is positive and x_\textrm m (t) when p_0 is negative. The nonlinear displacement component will be given by the combination of the two, x_\textrm{nl}(t) = 1/2 \left [ x_\textrm p (t) +x_\textrm m (t) \right]. This method was first used here. The solution corresponding to the pressure amplitude of 0.1 MPa and a nylon particle of 100-μm radius is depicted in the plot below. You can see that x_\textrm p (t) and x_\textrm m (t) have opposite signs and otherwise appear identical; however, their sum is non-zero, albeit very small in comparison.
Here, the particle is shown in motion. Notes on Implementation in COMSOL Multiphysics® Software
The method outlined above considers nonlinear acoustic phenomena. Therefore, the usual rules of acoustic modeling apply. They are:
The mesh has to be sufficiently fine to resolve the shortest wavelength. Here, the basic wavelength is that of the externally imposed standing wave. Shorter wavelengths, however, are produced by the particle vibrating at its resonant frequency. These wavelengths will be captured by the model because it is solved in the time domain. A suitable time progression method when resolving wave phenomena in COMSOL Multiphysics is the generalized alpha method. To make sure that the solution is stable, a Courant–Friedrichs–Lewy (CFL) criterion has to be manually imposed by setting a manual step size with \delta t < 0.5\ h_\textrm{min} / c_0.
Postprocessing
The last step of the analysis is computing the force from the average nonlinear displacement given by the finite element model. We can observe the effect of the acoustic radiation force as a small offset of the displacement from the otherwise perfect oscillatory motion of the particle. Judging by the difference in the computed linear and nonlinear components of the displacements, the difference in force components is about three orders of magnitude. So, the effect of the acoustic radiation force is negligible during one acoustic cycle and, to evaluate it correctly, around five to ten acoustic cycles must be computed.
With this data, we can export the results to Excel® spreadsheet software and find the average acceleration by fitting a second-degree polynomial to the displacement curve and computing the force as F = m \ddot x_\textrm{nl}. The graph below shows the results of such an analysis.
Here, the maximum force at a λ/8 distance from the pressure node of the standing wave is evaluated as a function of the particle’s radius. A good agreement is obtained in the limit of the small particle k_0R_0 \ll 1 for which the analytical model is valid, and the deviation appears to increase as we depart from this limit.
Conclusion
We have outlined a direct method for evaluating nonlinear acoustic radiation force. The same approach can be applied to other nonlinear acoustic effects such as acoustic streaming. The advantage of this method is that it does not rely on any theoretical models (e.g., the perturbation method) to express the nonlinear terms. Note that because the fluid-structure interaction approach requires the model to be solved in the time domain, it is much slower than the perturbation method.
Other Posts in This Series Excel is a registered trademark of Microsoft Corporation in the United States and/or other countries. Comments (2) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
General Solution
We have a Deconvolution problem with known operator.
One way to define the objective function is:
$$ \arg \min_{x} \frac{1}{2} {\left\| h \ast x - y \right\|}_{2}^{2} = \arg \min_{x} f \left( x \right) $$
The are 3 method to solve this:
Use Gradient Descent by deriving the gradient of $ f \left( \cdot \right) $. Write the problem in Matirx Form and either use direct solver (One could use Toeplitz System Solver) of this Least Squares problem or again use iterative solver in Matrix Form. Solve this in Fourier Frequency Domain.
I previously solved similar problems using the Matrix Form (See my answer to Deconvolution of 1D Signals Blurred by Gaussian Kernel) so this time we'll solve it in the convolution form.
The Gradient
Clearly the derivative of Inner Product is related to the adjoint of the operator.
In the case of Convolution the Adjopint is the Correlation Operator.
Let's define the Convolution Operation $ \ast $ with MATLAB Code.
So $ y = h \ast x $ will be in MATLAB:
vY = conv(vX, vH, 'valid');
Defining $ f \left( \cdot \right) $ in MATLAB would be:
hObjFun = @(vX) 0.5 * sum((conv(vX, vH, 'valid') - vY) .^ 2);
With the Derivative being:
vG = conv2((conv2(vX, vH, 'valid') - vY), vH(end:-1:1), 'full');
Where when we convolve with the flipped version of a vector it means we're basically doing correlation.
Solution in MATLAB
So, one must pay attention that since we use
valid convolution the size of $ y $ and $ x $ are different (Depends on the size of $ h $).
Usually we chose $ h $ to have odd number of elements to have its radius to be defined.
Then the MATLAB solution would be:
vX = zeros(size(vY, 1) + (2 * kernelRadius), 1);
vObjVal(1) = hObjFun(vX);
for ii = 1:numIterations
vG = conv2((conv2(vX, vH, 'valid') - vY), vH(end:-1:1), 'full');
vX = vX - (stepSize * vG);
vObjVal(ii + 1) = hObjFun(vX);
end
Specific Solution
In your case you asked for $ h $ to be Box Blur.
So I did the above with this model. Let's go through results.
First, let's see if the solution we got to is a real solution.
In order to examine that, we will convolve the result of the optimization with $ h $ and compare result to the input data. We expect it to be similar to the input.
As can be seen above it is, indeed, a perfect reconstruction of the input data.
We can see this by the objective value as well:
Yet the result is not good:
So, why is the result so bad?
There are 2 options: The SNR is not good enough to solve the inverse problem as is. It requires additional regularization like Wiener / Tikhonov (They are the same). I implemented it in the code so you have Wiener Filter to play with and understand its working. The model of Box Blur doesn't fit.
So in my code I implemented Wiener / Tikhonov Regularization (See
paramLambda in my code).
The full code is available on my StackExchange Signal Processing Q51460 GitHub Repository. |
The acceleration into the "ground" that is experienced at a radial distance $r$ from the axis of rotation (the center of the space station) is given as:
$$a= \omega^2 r$$
where $\omega$ is the angular velocity of the space station (i.e. how many times it rotates per second). This is called a "centrifugal force" (not to be confused with "centripetal force" though they are intimately related). You can see that acceleration is linear with respect to how far you are from the axis of rotation.
A good way to demonstrate that this equation makes sense intuitively is to tie a ball to the end of a rope. If you hold the rope very close to the ball and begin to swing it around in a circular path, it is relatively easy to hold on to. If you increase the length of the rope, you can feel that it becomes more difficult to hang on.
Back to your space station example. If we set the angular velocity of the space station so that at $r=20$ meters the acceleration is equivalent to the acceleration due to gravity on Earth, it is easy to solve for the accelerations at the other two locations:
Since ring A is $10$ meters $= \frac{1}{2}(20)$ meters away from the axis, the acceleration there will be half of what it is at B, or $4.9$ meters/second/second. Ring C is $30$ meters $= \frac{3}{2}(20)$ meters away from the axis, so the acceleration there will be $1.5$ times what it is at B, or $14.7$ meters/second/second.
If the concentric rings were allowed to rotate independently (ignoring the fact that this would create problems when trying to get from one ring to another) then from our equation above we can see that ring A would need to rotate $\sqrt{2} \approx 1.414$ times faster than ring B, and ring C would need to rotate $\sqrt{\frac{2}{3}} \approx 0.816$ times slower for them all to have the same centrifugal acceleration. |
<< ToK
ToK Warszawa meeting - Rough Notes Thu 15 Feb 2007 These are just rough notes - feel free to correct them, add links, etc. Hector Rubinstein - stockholm - magnetic fields
Magnetic fields on kpc scales exist. They may exist on intergalacticscales - it's unclear whether or not their origin is primordial.
CMB - Planck - may be able to detect magnetic fields present at theepochs not long after nucleosynthesis and recombination
it is well known that photon has a thermal mass - about 10^{-39} [units =eV?]which is extremely small - related to electron loops
Maxwell eqns -> Proca eqns WikipediaEn:Proca_action
m_photon < 10^-26 eV
\exists galactic mag fields at z \approx 3 making dynamo mechanisms difficult to explain them
Boehm - LDM hypothesis 511 keV detection
Leventhal(sp?) 199x ApJ
OSSE 3 components Purcell et al 1997
candidates
stars - SNe, SNII, WRcompact sources - pulsars, BH, low mass binaries
- most excluded because they would imply 511keV from the disk- SNIa - need large escape fraction and explosion rate to maintain a steady flux- low mass X-ray binaries - need electrons to escape from the disk to the bulge
dm + dm -> e^- + e^+ e+ loses energy -> positronium e+e- -> positronium decays
para-positronium 2 gamma - monochromatic wih 511keVortho-positronium 3 gamma - continuum
predictions
positron emission should be maximal with highest DM concentration (n^2 effect?)
cdm spectrum
does NOT produce CDM-like power spectrum???
- at 10^9 M_sun essentially CDM-like
- by 10^6 M_sun, the difference would be important
spectrum
Ascasibar et al 2005, 2006
model through F- 511keV through Z' relic density link with neutrino mass
interaction/decay diagrams ->
link between neutrino mass and DM cross section:
\sigma_\nu well-known for relic density \sim 10^{-26} cm^3/sMeV" class="mmpImage" src="/foswiki/pub/Cosmo/ToK070215RoughNotes/_MathModePlugin_731486621df83fabc7268c4eba1fd524.png" />
to fit neutrino data
BBN: 1MeV < m_N
low energy Beyond SM MeV
DM has definitely escaped all previous low energy experimentsdue to lack of luminosity
BABAR/BES II ... ?
summary
...
explains low value of neutrino masses detection at LHC may be possible but requires work back to SUSY -> snu-neutralino-nu ? Conlon - hierarchy problems in string theory: the power of large volume
planck scale 10^18 GeV
... cosm constant scale (10^-3eV)^4
- large-volume models can generate hierarchies thorugh a stabilisedexponentially large volume
- predicts cosmological constant (but about 50 orders of magnitude toolarge - solving this problem is left to the reader/audience)
G\"unther Stigl - high-energy c-rays, gamma-rays, neutrinos
HESS - correlation of observations at GC with molecular cloud distribution
KASCADE - has made observations
Southern Auger - 1500km^2 - in Chile/Argentina
Hillas plot
c-rays at highest energies could be protons, could be ions
- most interactions produce pions; pi^\pm decays to neutrinos pi^0 decays to photons (gamma-rays)
origin of very high energy c rays remains one of the fundamental unsolved questions of astroparticle physics - even galactic c ray origin is unclear acceleration and sky distribution of c rays are strongly linked to the strength and distribution of cosmic magnetic fields - which are poorly known sources probably lie in fields of \mu-Gauss HE c-rays, pion-production, gamma-ray/neutrinos - all three fields should be considered together; strong constraints arise from gamma-ray overproduction Khalil - DM - SUSY - brane cosmology
(British University in Egypt = BUE)
- friedmann eqn modified in 5D (brane model)
- dark matter relic abundance
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(see the transcript file for additional information)
Output written on X0jSvQ0MJM.dvi (1 page, 752 bytes).
Transcript written on X0jSvQ0MJM.log. |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
Your reasoning is along the right lines, but it is incomplete.
enol of B is more stable because of more substituted alkene.
The methyl group in acetone does stabilize the carbon-carbon double bond in the enol form for the reasons you suggest.
However the methyl group also stabilizes the carbonyl double bond in the keto form (see here for example, this is why a carbonyl in a ketone is slightly stronger than a carbonyl in an aldehyde). In fact, in simple carbonyl compounds, the methyl group stabilizes the carbonyl double bond a bit more than it stabilizes the enolic carbon-carbon double bond. As a result, simple aldehydes generally have a higher enol content than simple ketones. Hence, the enol content in acetaldehyde is greater than the enol content in acetone.
$$\ce{carbonyl <=> enol}$$$$\mathrm{K_{eq}=\frac{[enol]}{[carbonyl]}}$$
\begin{array}{|c|c|c|c|} \hline\text{compound} & \mathrm{K_\text{eq}} \\ \hline\text{acetaldehyde} & \mathrm{6 \times 10^{-7}} \\ \hline\text{acetone} & \mathrm{5 \times 10^{-9} } \\ \hline\end{array}
The two $\beta$-dicarbonyl compounds have a higher enol content than the two monocarbonyl compound because hydrogen bonding
and conjugation stabilize their enols. The enol content in C (a mono aldehyde) is higher than D because of the reasons outlined above.
Therefore, the overall order of increasing enol content is C > D >> A > B.
Unless you've studied this before, you might not know that a "methyl group stabilizes the carbonyl double bond a bit more than it stabilizes the enolic carbon-carbon double bond". So a
general rule that might come in handy in these situations is that the enol content increases with the acidity of the enolic hydrogen. A more acidic $\alpha$-hydrogen implies a weaker $\ce{C-H}$ bond. Since the position of a keto-enol equilibrium is dependent on the relative stabilities of the keto and enol forms (the compound with the highest bond strengths overall will be more stable and will predominate at equilibrium) a weak $\ce{C-H}$ (lower $\mathrm{p}K_\text{a}$) generally implies a higher enol content.
\begin{array}{|c|c|c|c|} \hline\text{compound} & \mathrm{p}K_\text{a} \\ \hline\text{acetone} & 19.3 \\ \hline\text{acetaldehyde} & 16.7 \\ \hline\text{2,4-pentanedione} & 13.3 \\ \hline\end{array}
Finally, this earlier answer provides a review of the key factors controlling keto-enol equilibria. |
Let $Q$ be some finite set with $n = |Q|$. Then suppose I want to show that for every nonempty subset $P \subseteq Q$ some property $A$ holds. One natural way to approach this is using induction, and suppose I have an inductive proof in the following sense, I succeeded showing it for the one-element subsets. Now I assume that for some subset $P$ the property holds for every proper nonempty subsets. And then supposing that $P = P' \cup \{q\}$ with $q\notin P'$ I can show that under the assumption $A$ for $P'$ the property $A$ also holds for $P$.
Looking closely, I remember that an induction arguments shows the property holds for all $n$, and that in the above case we do induction on the cardinality of the sets. But its not induction on $n = |Q|$, as this was arbitrary but fixed in advance, its induction on the cardinality of the subsets of $Q$. But the point is we do not have subsets of arbitrary cardinality in $Q$, just of cardinality $k \le n$. So property $A$ cannot hold for every $k$, as for $k > n$ we do not even have such subsets.
So, is it legitimate to call the above proof scheme a proof by induction? And if I want to formalize this more, making explicit what axioms and deduction rules are used, how would this be done?
One idea, we actually prove the statement $$ A'(k) :\Leftrightarrow (k \le n) \rightarrow \forall P \subseteq Q \mbox{ property $A$ holds}. $$ So that the base case is $A'(1)$, and in the inductive step if $k \le n$ we use the hypotheses as $k-1 \le n$ and otherwise the statement holds just by definition, without relying on the hypotheses. But I do not know if this is the way to go here. |
Background
The computation over real numbers are more complicated than computation over natural numbers, since real numbers are infinite objects and there are uncountably many real numbers, therefore real numbers can not be faithfully represented by finite strings over a finite alphabet.
Unlike the classical computability over finite strings where different models of computation like: lambda calculus, Turing machines, recursive functions, ... turn out to be equivalent (at least for computability over functions on strings), there are various proposed models for computation over real numbers which are not compatible. For example, in the TTE model (see also [Wei00]) which is the closest one to the classical Turing machine model, the real numbers are represented using infinite input tapes (like Turing's oracles) and it is not possible to decide the comparison and equality relations between two given real numbers (in finite amount of time). On the other hand in the BBS/real-RAM models which are similar to RAM machine model, we have variables that can store arbitrary real numbers, and comparison and equality are among the atomic operations of the model. For this and similar reasons many experts say that the BSS/real-RAM models are not realistic (cannot be implemented, at least not on current digital computers), and they prefer the TTE or other equivalent models to TTE like effective domain theoretic model, Ko-Friedman model, etc.
On the other hand, it seems to me that in the implementation of the algorithms in Computational Geometry (e.g. LEDA), we are only dealing with algebraic numbers and no higher-type infinite objects or computations are involved (is this correct?). So it appears to me (probably naively) that one can also use the classical model of computation over finite strings to deal with these numbers and use the usual model of computation (which is also used for implementation of the algorithms) to discuss correctness and complexity of algorithms.
Questions:
What are the reasons that researchers in Computational Geometry prefer to use the BSS/real-RAM model? (reasons specific Computational Geometry for using the BSS/real-RAM model)
What are the problems with the (probably naive) idea that I have mentioned in the previous paragraph? (using the classic model of computation and restricting the inputs to algebraic numbers in Computational Geometry)
Addendum:
There is also the complexity of algorithms issue, it is very easy to decide the following problem in the BSS/real-RAM model:
Given two sets $S$ and $T$ of positive integers,
is $\sum_{s\in S} \sqrt{s} > \sum_{t\in T}\sqrt{t}$?
While no efficient integer-RAM algorithm is known for solving it. Thanks to JeffE for the example.
References:
Lenore Blum, Felipe Cucker, Michael Shub, and Stephen Smale, "Complexity and Real Computation", 1998 Klaus Weihrauch, "Computable Analysis, An Introduction", 2000 |
I am working thru a derivation of the group velocity formula and I get to this stage: $$y=2A\cos(x\frac{\Delta K}{2} -t\frac{\Delta \omega}{2})\sin( \bar k x-\bar \omega t)$$ Then all the derivations I have seen say that $\frac{\Delta \omega}{\Delta K} $ is the group velocity. I know mathematically why this is a velocity but what I don't get is why do we know that this is the group velocity rather then the phase velocity and that $\frac{\bar \omega}{\bar k}$ is the phase velocity and not the group velocity?
Consider a wave
$$A = \int_{-\infty}^{\infty} a(k) e^{i(kx-\omega t)} \ dk,$$
where $a(k)$ is the amplitude of the kth wavenumber, and $\omega=\omega(k)$ is the frequency, related to $k$ via a dispersion relation. Note, if we wanted to track a wave, with wavenumber $k$, with constant phase, we would see that this occurs when $kx=\omega t$, i.e. $x/t = \omega/k = c$, with $c$ the $\textbf{phase}$ velocity.
We would like to know the speed at which the envelope $|A|$ is traveling.
For $\textbf{narrow banded}$ waves, the angular frequency $\omega$ can be approximated via the taylor expansion around a central wavenumber $k_o$, i.e.
$$\omega(k) = \omega(k_o) + \frac{\partial \omega}{\partial k} (k-k_o) + \mathcal{O}((k-k_o)^2),$$
where the scale of the bandwidth is quantified by the small parameter $(k-k_o)$. Therefore, we can rewrite $A$ as
$$A \approx e^{-i(\omega(k_o)t-k_o\frac{\partial \omega}{\partial k}t)} \int_{-\infty}^{\infty} a(k) e^{ik(x-\frac{\partial \omega}{\partial k} t)} \ dk.$$
Therefore
$$|A| = \int_{-\infty}^{\infty} a(k) e^{ik(x-\frac{\partial \omega}{\partial k} t)} \ dk,$$
which says that the envelope, $|A|$, travels at speed $\frac{\partial \omega}{\partial k}$, i.e. $$|A(x,t)| = |A(x-c_g t,0)|,$$ where we have defined $$c_g \equiv \frac{\partial \omega}{\partial k}.$$
The group velocity has dynamical significance, as it is the velocity at which the energy travels.
Definitions
Before we begin, we should define some terms and parameters/functions that will be used later:
Wave Number: $\equiv$ effectively the number of wave crests (i.e., anti-node of local maximum) per unit length $\leftrightharpoons$ ``density'' of waves $\rightarrow$ $\boldsymbol{\kappa}$ $=$ $\boldsymbol{\kappa}\left(\omega,\textbf{x},t\right)$ in general Wave Frequency: $\equiv$ effectively the number of wave crests crossing position $\mathbf{x}$ per unit time $\leftrightharpoons$ ``flux'' of waves $\rightarrow$ $\omega$ $=$ $\omega\left(\boldsymbol{\kappa},\textbf{x},t\right)$ in general Wave Phase: $\equiv$ position on a wave cycle between a crest and a trough (i.e., anti-node of local minimum) $\rightarrow$ $\phi$ $=$ $\phi\left(\textbf{x},t\right)$ in general Phase and Continuity
Then, we can define an elementary solution to periodic wave equations as:$$ \psi\left( \mathbf{x}, t \right) = \mathcal{A} \ e^{ {\displaystyle i\left( \boldsymbol{\kappa} \cdot \mathbf{x} - \omega t \right) } }$$where $\mathcal{A}$ is the wave amplitude and, in general, can be a function of $\boldsymbol{\kappa}$ and/or $\omega$, but we will assume constant for now. Let us assume that a
dispersion relation, $\omega$ $=$ $\mathcal{W}\left( \boldsymbol{\kappa}, \textbf{x}, t \right)$, exists and may be solved for positive real roots. In general, there will be multiple solutions to the dispersion relation, where each solution is referred to as different modes. The term in the exponent is known as the wave phase, given by:$$ \phi\left( \mathbf{x}, t \right) = \boldsymbol{\kappa}\left( \omega, \mathbf{x}, t \right) \cdot \mathbf{x} - \omega\left( \boldsymbol{\kappa}, \mathbf{x}, t \right) \ t + \phi{\scriptstyle_{o}}$$Because $\phi\left(\textbf{x},t\right)$ results from solutions of the wave equation, its derivatives must satisfy the dispersion relation through the following:$$ - \frac{ \partial \phi\left( \mathbf{x}, t \right) }{ \partial t } = \mathcal{W}\left( \frac{ \partial \phi\left( \mathbf{x}, t \right) }{ \partial \mathbf{x} }, \mathbf{x}, t \right)$$and we can see from the equation for $\phi\left(\textbf{x},t\right)$ that the following is true:$$\begin{align} \boldsymbol{\kappa} & = \frac{ \partial \phi\left( \mathbf{x}, t \right) }{ \partial \mathbf{x} } \\ \omega & = - \frac{ \partial \phi\left( \mathbf{x}, t \right) }{ \partial t }\end{align}$$We also know that $\partial^{2} \phi$/$\partial \mathbf{x} \partial t$ $=$ $\partial^{2} \phi$/$\partial t \partial \mathbf{x}$, therefore:$$\begin{align} \frac{ \partial^{2} \phi }{ \partial t \partial \mathbf{x} } - \frac{ \partial^{2} \phi }{ \partial \mathbf{x} \partial t } & = 0 \\ & = \frac{ \partial \boldsymbol{\kappa} }{ \partial t } - \frac{ - \partial \omega }{ \partial \mathbf{x} } = 0 \\ & = \frac{ \partial \boldsymbol{\kappa} }{ \partial t } + \frac{ \partial \omega }{ \partial \mathbf{x} } = 0 \\ & = \frac{ \partial \boldsymbol{\kappa} }{ \partial t } + \nabla \omega = 0\end{align}$$One can see that this final form looks similar to a continuity equation, so long as $\boldsymbol{\kappa}$ $\leftrightharpoons$ density of the waves, and $\omega$ $\leftrightharpoons$ flux of the waves. Phase Velocity
From the above relations, we can see that on
contours of constant $\phi\left(\textbf{x},t\right)$, we are sitting on local wave crests (i.e., phase fronts) where $\boldsymbol{\kappa}$ is orthogonal to these contours. These phase fronts move parallel to $\boldsymbol{\kappa}$ at a speed, $\mathbf{V}_{\phi}$, known as the phase velocity. The general form for this speed is given by:$$ \mathbf{V}_{\phi} \equiv \frac{ \mathcal{W}\left( \boldsymbol{\kappa}, \mathbf{x}, t \right) }{ \kappa } \boldsymbol{\hat{\kappa}}$$ Group Velocity
We can rearrange our continuity equation by multiplying by unity to get:$$\begin{align} \frac{ \partial \boldsymbol{\kappa} }{ \partial t } + \frac{ \partial \omega }{ \partial \mathbf{x} } \cdot \frac{ \partial \boldsymbol{\kappa} }{ \partial \boldsymbol{\kappa} } & = 0 \\ \frac{ \partial \boldsymbol{\kappa} }{ \partial t } + \frac{ \partial \omega }{ \partial \boldsymbol{\kappa} } \cdot \frac{ \partial \boldsymbol{\kappa} }{ \partial \mathbf{x} } & = 0 \\ \frac{ \partial \boldsymbol{\kappa} }{ \partial t } + \left( \mathbf{V}_{g} \cdot \nabla \right) \boldsymbol{\kappa} & = 0\end{align}$$where $\mathbf{V}_{g}$ is called the
group velocity, where we note that:$$\frac{ \partial \omega }{ \partial \mathbf{x} } = \frac{ \partial \mathcal{W}\left( \boldsymbol{\kappa}, \mathbf{x}, t \right) }{ \partial \boldsymbol{\kappa} } \cdot \frac{ \partial \boldsymbol{\kappa} }{ \partial \mathbf{x} } + \frac{ \partial \mathcal{W}\left( \boldsymbol{\kappa}, \mathbf{x}, t \right) }{ \partial \mathbf{x} }$$which shows that $\partial \mathcal{W}$/$\partial \boldsymbol{\kappa}$ $=$ $\left( \partial \omega / \partial \boldsymbol{\kappa} \right){\scriptstyle_{\textbf{x}}}$ $\Rightarrow$ different $\boldsymbol{\kappa}$'s propagate with velocity $\mathbf{V}_{g}$. In other words, $\mathbf{V}_{g}$ is the propagation velocity for $\kappa$ and $\mid$$\mathcal{A}$$\mid^{2}$ propagates with velocity $\mathbf{V}_{g}$.
Thus, an observer moving with the phase fronts (crests) moves at $\mathbf{V}_{\phi}$, but they observe the local wavenumber and frequency to change in time $\Rightarrow$ neighboring phase fronts (crests) move away from the observer in this frame. In contrast, for an observer moving with $\mathbf{V}_{g}$, they observe constant local wavenumber and frequency (with respect to time), but phase fronts (crests) continuously move past the observer in this frame.
References
Whitham, G. B. (1999),
Linear and Nonlinear Waves, New York, NY: John Wiley & Sons, Inc.; ISBN:0-471-35942-4.
The frequency $\omega$ can be hight in a wave packet, but the envelope motion may be slow. The latter is determined with the $\cos(...)$; that is why they call it a group velocity. It is a velocity of displacement of the packet as a whole. There muct be applets on internet to show how a wave packet moves.
protected by Qmechanic♦ May 3 '16 at 16:44
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Answer
$x^2-7x+\dfrac{49}{4} \Rightarrow\left( x-\dfrac{7}{2} \right)^2 $
Work Step by Step
The third term of a perfect square trinomial is equal to the square of half the coefficient of the middle term. Hence, to complete the square of the given expression $ x^2-7x ,$ the third term must be \begin{array}{l}\require{cancel}\left( \dfrac{-7}{2}\right)^2 \\\\= \dfrac{49}{4} .\end{array} Using $a^2\pm2ab+b^2=(a\pm b)^2$, then \begin{array}{l}\require{cancel} x^2-7x+\dfrac{49}{4} \Rightarrow\left( x-\dfrac{7}{2} \right)^2 .\end{array} |
This is an old revision of the document! Sets and Relations Sets
Sets are unordered collection of elements.
We denote a finite set containing only elements , and by . The order and number of occurrences does not matter: .
iff
Empty set: . For every we have .
To denote large or infinite sets we can use set comprehensions: is set of all objects with property . \[
y \in \{ x. P(x) \} \ \leftrightarrow\ P(y)
\]
Notation for set comprehension:
Sometimes the binder can be inferred from context so we write simply . In general there is ambiguity in which variables are bound. (Example: what does the in refer to in the expression: \[
\{a \} \cup \{ f(a,b) \mid P(a,b) \}
\] does it refer to the outerone as in or is it a newly bound variable? The notation with dot and bar resolves this ambiguity.
Subset: means
\[
A \cup B = \{ x. x \in A \lor x \in B \}
\] \[
A \cap B = \{ x. x \in A \land x \in B \}
\] \[
A \setminus B = \{ x. x \in A \land x \notin B \}
\]
Boolean algebra of subsets of some set (we define ):
are associative, commutative, idempotent neutral and zero elements: , absorption: , deMorgan laws: , complement as partition of universal set: , double complement:
Which axioms are sufficient?
Infinte Unions and Intersections
Note that sets can be nested. Consider, for example, the following set \[
S = \{ \{ p, \{q, r\} \}, r \}
\] This set has two elements. The first element is another set. We have . Note that it is not the case that
Suppose that we have a set that contains other sets. We define union of the sets contained in as follows: \[
\bigcup B = \{ x.\ \exists a. a \in B \land x \in a \}
\] As a special case, we have \[
\bigcup \{ a_1, a_2, a_3 \} = a_1 \cup a_2 \cup a_3
\] Often the elements of the set are computed by a set comprehension of the form . We then write \[
\bigcup_{i \in J} f(i)
\] and the meaning is \[
\bigcup \{ f(i).\ i \in J \}
\] Therefore, is equivalent to .
We analogously define intersection of elements in the set: \[
\bigcap B = \{ x. \forall a. a \in B \rightarrow x \in a \}
\] As a special case, we have \[
\bigcap \{ a_1, a_2, a_3 \} = a_1 \cap a_2 \cap a_3
\] We similarly define intersection of an infinite family \[
\bigcap_{i \in J} f(i)
\] and the meaning is \[
\bigcap \{ f(i).\ i \in J \}
\] Therefore, is equivalent to .
Relations
Pairs: \[
(a,b) = (u,v) \iff (a = u \land b = v)
\] Cartesian product: \[
A \times B = \{ (x,y) \mid x \in A \land y \in B \}
\]
Relations is simply a subset of , that is .
Note: \[
A \times (B \cap C) = (A \times B) \cap (A \times C)
\] \[
A \times (B \cup C) = (A \times B) \cup (A \times C)
\]
Diagonal relation
, is given by \[
\Delta_A = \{(x,x) \mid x \in A\}
\]
Set operations
Relations are sets of pairs, so operations apply.
Relation Inverse
\[
r^{-1} = \{(y,x) \mid (x,y) \in r \}
\]
Relation Composition
\[
r_1 \circ r_2 = \{ (x,z) \mid \exists y. (x,y) \in r_1 \land (y,z) \in r_2\}
\]
Note: relations on a set together with relation composition and form a
monoid structure:\[\begin{array}{l} r_1 \circ (r_2 \circ r_3) = (r_1 \circ r_2) \circ r_3 \\ r \circ \Delta_A = r = \Delta_A \circ r
\end{array} \]
Moreover, \[
\emptyset \circ r = \emptyset = r \circ \emptyset
\] \[
r_1 \subseteq r_2 \rightarrow r_1 \circ s \subseteq r_2 \circ s
\] \[
r_1 \subseteq r_2 \rightarrow s \circ r_1 \subseteq s \circ r_2
\]
Relation Image
When and we define
image of a set under relation as\[ S\bullet r = \{ y.\ \exists x. x \in S \land (x,y) \in r \}
\]
Transitive Closure
Iterated composition let . \[ \begin{array}{l}
r^0 = \Delta_A \\ r^{n+1} = r \circ r^n
\end{array} \] So, is n-fold composition of relation with itself.
Transitive closure: \[
r^* = \bigcup_{n \geq 0} r^n
\]
Equivalent statement: is equal to the least relation (with respect to ) that satisfies \[
\Delta_A\ \cup\ (s \circ r)\ \subseteq\ s
\] or, equivalently, the least relation (with respect to ) that satisfies \[
\Delta_A\ \cup\ (r \circ s)\ \subseteq\ s
\] or, equivalently, the least relation (with respect to ) that satisfies \[
\Delta_A\ \cup\ r \cup (s \circ s)\ \subseteq\ s
\]
Some Laws in Algebra of Relations
\[
(r_1 \circ r_2)^{-1} = r_2^{-1} \circ r_1^{-1}
\] \[
r_1 \circ (r_2 \cup r_3) = (r_1 \circ r_2) \cup (r_1 \circ r_3)
\] \[
(r^{-1})^{*} = (r^{*})^{-1}
\]
Binary relation can be represented as a directed graph with nodes and edges
Graphical representation of , , and
Equivalence relation is relation with these properties:
reflexive: symmetric: transitive:
Equivalence classes are defined by \[
x/r = \{y \mid (x,y) \in r
\] The set is a partition:
each set non-empty sets are disjoint their union is
Conversely: each collection of sets that is a partition defines equivalence class by \[
r = \{ (x,y) \mid \exists c \in P. x \in c \land y \in c \}
\]
Congruence: equivalence that agrees with some set of operations.
Partial orders:
reflexive antisymmetric: transitive Functions Example: an example function for , is\[ f = \{ (a,3), (b,2), (c,3) \}
\]
Definition of function, injectivity, surjectivity.
- the set of all functions from to . For it is a strictly bigger set than .
(think of exponentiation on numbers)
Note that is isomorphic to , they are two ways of representing functions with two arguments.
There is also isomorphism between
n-tuples and functions , where Function update
Function update operator takes a function and two values , and creates a new function that behaves like in all points except at , where it has value . Formally, \[ f[a_0 \mapsto b_0](x) = \left\{\begin{array}{l}
b_0, \mbox{ if } x=a_0 \\ f(x), \mbox{ if } x \neq a_0
\]
Domain and Range of Relations and Functions
For relation we define domain and range of : \[
dom(r) = \{ x.\ \exists y. (x,y) \in r \}
\] \[
ran(r) = \{ y.\ \exists x. (x,y) \in r \}
\] Clearly, and .
Partial Function
Notation: means .
Partial function is relation such that\[ \forall x \in A. \exists^{\le 1} y.\ (x,y)\in f
\]
Generalization of function update is override of partial functions,
Range, Image, and Composition
The following properties follow from the definitions: \[
(S \bullet r_1) \bullet r_2 = S \bullet (r_1 \circ r_2)
\] \[
S \bullet r = ran(\Delta_S \circ r)
\]
Further references Gallier Logic Book, Chapter 2 |
I want to know how $\mathbb{E}[|\chi_n^2 - n|]$ behaves as $n\to \infty$. Simulating this in R suggest that it grows at a rate of $\sqrt{n}$, but I am unable to prove it. Setting $\chi_n^2 =\sum_{i=1}^n Z_i^2$, where $Z_i$ are iid standard normal random variables, there is a trivial upper bound $$ \mathbb{E}[|\chi_n^2 - n]|] = \mathbb{E}[|\sum_{i=1}^n (Z_i^2-1)|]\\ \leq \sum_{i=1}^n \mathbb{E}[|Z_i^2 - 1|], $$ which grows at a rate of $n$ as $n\to\infty$. However, this bound is clearly too high. What can we say about the limiting behaviour of $\mathbb{E}[|\chi_n^2 - n|]$, or is there even an explicit expression for it?
Your simulation is correct. Let us observe that $$ \frac1{\sqrt n}\sum_{i=1}^n(Z_i^2-1)\to\mathcal N(0,2) $$ in distribution as $n\to\infty$ by the central limit theorem. We can show that $$ \operatorname E\biggl|\frac1{\sqrt n}\sum_{i=1}^n(Z_i^2-1)\biggr|\to\operatorname E|Y| $$ as $n\to\infty$, where $Z\sim\mathcal N(0,2)$ (see Theorem 25.12 and its Corollary on p. 338 of Billingsley's textbook). Hence, $$ \operatorname E|\chi_n^2-n|=\sqrt n\operatorname E\biggl|\frac1{\sqrt n}\sum_{i=1}^n(Z_i-1)^2\biggr|\sim\sqrt n\operatorname E|Y| $$ as $n\to\infty$.
By the central limit theorem we have $\frac{1}{\sqrt{n}}(\chi_n^2 - n) \xrightarrow{n \to \infty} \mathcal{N}(0, 2)$, which suggests that the scaling speed of $\sqrt{n}$ is correct.
If you just need an upper bound, we can use Jensen's inequality. If $X_n$ is chi-square distributed with $n$ degrees of freedom and $Z_i$ are independent standard Gaussians, then $$E[|X_n - n|]^2 \le E[(X_n - n)^2] = E\left[\left(\sum \limits_{i = 1}^n (Z_i^2 - 1)\right)^2\right] = \sum \limits_{i = 1}^n E[(Z_i^2 - 1)^2] + \sum \limits_{i < j} \operatorname{cov}(Z_i^2, Z_j^2) = 2n$$
This shows that actually $E[|X_n - n|] \le \sqrt{2n}$.
More precisely, if we actually had $E[|X_n - n|] \le \sqrt{n} a_n$ for some null sequence $a_n$, we would get $E\left[\frac{1}{\sqrt{n}}|X_n - n|\right] \le a_n$, which would imply that $\frac{1}{\sqrt{n}} |X_n - n|$ converges to zero in distribution (but this would contradict the central limit theorem). |
Observables correspond to particular things you can do in the lab (or observe in nature). So let's first talk about something you can do in the lab.
You can take a particle with spin and subject it to an inhomogeneous magnetic field. A particle with spin has a magnetic moment proportional to the spin and we know to Hamiltonian for a particle with a magnetic moment in an externalmagnetic field. So we get a term in the Hamiltonian proportional to $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z.$ At each point, this is an Hermitian operator. By setting up the magnetic field to point in the $\hat z$ direction and to be inhomogeneous in that direction we can send a beam in that is eigen to $\hat\sigma_z$ and it will be deflected up or down (depending on the eigenvalue and on how we made the field vary in the z direction) and this deflection literally happens because of the evolution determined by the Schrödinger equation when you have a term in the Hamiltonian proportional to $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z.$
And so now when you have a general state and again evolve it according to the Schrödinger equation when you have a term in the Hamiltonian proportional to $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z$ then the incoming spatial state splits its beam into two beams, one going up and one going down and the size of the two beams is such that the total current in each beam has a ratio equal to the ratio of the projection of the spin state onto the eigenstates of $\hat\sigma_z.$ Again, that fact is determined by the Schrödinger equation evolution for he actual state of the subject (the thing with spin) and the Stern-Gerlach device (the thing with the inhomogeneous magnetic field). And the spin state evolves so that the branch that is spatially deflected up becomes spin up and the branch that is deflected down becomes spin down. And again this cones out from the evolution determined by the Schrödinger equation when you have a term in the Hamiltonian proportional to $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z$ then the incoming spatial state splits its
So we know what causes measurements. Devices interacting with subjects according to the laws of physics. The Hamiltonian always has a term proportional to $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z$ when there is a particle with spin and an external magnetic field, there is no choice about whether it is there.
And we know the effects of measurements, they split states into a sun of states. Each each term in the sum has an eigenstate of the operator and that eigenstate is entangled with some other state. In this example the spin state become entangled with the position state of the particle. Deflected up entangled with spin up, and deflected down entangled with spin down.
So the only mystery is why we call it measurement. And that is because of that polarization, now if you put it on a similar device again it won't be split it will just be deflected. The real measurement effect happens later when these different states have a chance to affect the states of many other things so the two branches no longer can interact just because it is too hard to get them to ever overlap again.
So now to the question of operator. It isn't random operators, it is operators that have real eigenvalues and that have orthogonal eigenvectors. That is what is important. Why?
The real eigenvalues allow the continuous splitting of a state into the multiple eigenstates, which is what allows the Schrödinger equation to be able to split the state to entangle the projections with something else in a continuous manner (which is what the Schrödinger equation requires and we have to evolve according to the Schrödinger equation for the actual experimental setup there is no choice about that and no other options). To actually do it well, you have to find something else to couple with the eigenvectors to get the entanglement. And for that you can't actually just do anything, you are restricted to real terms in real Hamiltonians and nature only provides so many to choose from.
So why do we need the eigenstates to be orthogonal. That is key to getting that do it twice get the same answer. Otherwise it is just an interaction, not a measurement.
So to call something a measurement it basically has to be a real linear combination of orthogonal projections onto mutually orthogonal states. And so they do need to be operators, very particular operators. And you can only actually observer them if you can find something else to couple differently to those different states.
It's still not clear to me what the
direct meaning of $\hat A |\psi\rangle$ is.
The point is that the different eigenvalues you get from $\hat A |\psi\rangle$ tell the other thing that is becoming entangled with your object to continuous change at different rates depending on the eigenvalue.
For instance if you have a general spin state then for a fixed magnetic field the spin zero doesn't couple at all so isn't deflected at all and the lowest non zero spins deflect but deflect less than larger spins so deflect in opposite directions but at a power rate of motion in the deflected direction. So spin zero goes straight, spin $\pm 1/2$ get sent a little bit up and down and spin $\pm 1$ get sent more up and down. And I mean the z component t being 0, 1/2, or 1 which I why I included the minus sign to make that clear. So the eigenvalue tell you how separated the other thing (the thing used the measure the state) becomes. This is literally why you project onto an eigenspace. When two things have the same eigenvalue the thing you couple with doesn't move differently for the things with the same eigenvalue so they don't get separated at all so the whole projection onto the eigenspaces gets entangled with the exact same state of the thing used to measure it.
It is a wavefunction
The eigenfunctions are the ones that don't change in repeated measurements son get deflected in different ways as a parameterized by the real umber eigenvalue.
how does its corresponding quantum state relate (in physical terms) to the original state $\psi$ and the measurement $\hat A$?
The original state can be decomposed as eigenstates and they evolve into entangled states as they separate. So when separated you have the different eigenstates entangled with different states of the thing used to measure it. Spin is easiest since it entangles with its own position so we don't really have to bring in anything else other than the inhomogeneous magnetic field. |
I am trying to make the relation between linear modulation and nonlinear one by using orthonormal expansion. The purpose is to understand what is orthonormal set in each case and to understand the operation of projection.
Linear modulation
In digital linear modulation, the baseband signal can be written as
$$x(t) = \sum_n a_n p(t-nT)$$ where $a_n$ is data, $\lbrace p_n(t) = p(t-nT)\textrm{, }n\in \mathbb{Z}\rbrace$ is orthonormal set, i.e. $<p_n(t),p_m(t)> = \delta_{m-n}$ with $<x(t),y(t)> = \int x(t)y^*(t)dt$ is inner product defined for the signal space we are working on. To get $a_n$ back, we need project $x(t)$ to $p_n(t)$.
Use matched filter $q(t) = p^*(-t)$ thus $q(t-\tau) = p^*(\tau - t)$.
by defining $g(t) = p(t) \star p^*(-t)$ where $\star$ is convolution. $g(t=0) = 1$ and $g(t=nT) = 0$ because $\delta_{m-n} = <p_n(t),p_m(t)> = g((m-n)T)$.
\begin{align} y(t) &= x(t) \star p^*(-t)\\ &= \sum_n a_n p(t-nT)\star p^*(-t)\\ &= \sum_n a_ng(t-nT)\\ \implies y(t=mT) &= \sum_n a_ng((m-n)T) = a_m \end{align}
If we have white noise $n(t)$ : $y(t) = x(t) + n(t)$. The noise is processed as
\begin{align} z(t) &= n(t) \star q(t)\\ &= \int n(\tau)q(t-\tau) d\tau\\ &= \int n(\tau) p^*(\tau -t) d\tau\\ \implies z(t=mT) &= \int n(\tau) p^*(\tau - mT) d\tau\\ &= <n(t), p(t-mT)> \end{align}
Then we can say that for a given $y(t)$, each processing $m$ at the receiver which ends with sampling at $t=mT$, the noise process is projected to the vector $p_m(t)$ creating $z_m$ and the data part is projected to the same vector creating $a_m$. $\lbrace p_m(t) = p(t-mT)\textrm{, }m\in \mathbb{Z}\rbrace$ is orthonormal set, thus $z_m$ are uncorrelated, with Gaussian assumption, $z_m$ are independent.
This is famous process in classical technical books. Two conclusions I get
The set of orthonormal vectors $\lbrace p_m(t) = p(t-mT)\textrm{, }m\in \mathbb{Z}\rbrace$ has infinite number of elements as $m \in \mathbb{Z}$.
The orthonormal vectors $p_m(t)$ can be no-time-limited and the orthogonal property holds on the entier time support.
Non-linear modulation
Let's take a binary FSK modulation as example. $M$ basis functions, $k \in \mathbb{S} = \lbrace 1, ..., M \rbrace$:
$$\phi_k(t) = \begin{cases}\sqrt{\frac 2T}\sin\left(\frac{k\pi t}{T}\right)& \text{for}\quad 0 \leq t \leq T\\0 &\text{otherwise}\end{cases}$$
I write the waveform as $x(t) = \sum_n \gamma_n(t-nT)$ where $\gamma_n(t) = \phi_k(t) \textrm{ if } a_n = k$.
Questions
Question 1 : At the receiver, for the $n^\rm{th}$ symbol, we try projecting to $M$ functions $\phi_k(t-nT)$. If $k=k_0$ gives the largest power, we decode $a_n = k_0$. Is $\lbrace \phi_k(t-nT) \textrm{, } n \in \mathbb{Z}\textrm{, } k \in \mathbb{S}\rbrace$ the orthonormal set as $<\phi_k(t-nT),\phi_l(t-mT)> = \delta_{(m-n)\times(k-l)}$ ?
I mean I expect something equivalent to the two properties (infinite number of element and basis functions are not time-limited) of orthonormal set in the linear modulation case.
Question 2 : is it correct if I say the noise samples are uncorrelated because the projection are $<n(t),\phi_k(t-nT)>$ and $<n(t),\phi_l(t-mT)>$ and $<\phi_k(t-nT),\phi_l(t-mT)> = \delta_{(m-n)\times(k-l)}$ ?
Question 3 : The $M$ basis functions are interpreted as constellation of $M$ points. Is there any analogy with the linear modulation case ? If we consider M-QAM, we have also a constellation of $M$ complex points but if I write $a_n = a_{rn} + ja_{in}$, the two functions $p(t-nT)$ and $jp(t-nT) = e^{j\pi/2}p(t-nT)$ are not orthogonal. |
The situation with solids is considerably more complicated, with different speeds in different directions, in different kinds of geometries, and differences between transverse and longitudinal waves. Nevertheless, the speed of sound in solids is larger than in liquids and definitely larger than in gases. Young's Modulus for a representative value for the bulk modulus for steel is 160 \(10^9\) N /\(m^2\). A list of materials with their typical velocity can be found in the above book. Speed of sound in solid of steel, using a general tabulated value for the bulk modulus, gives a sound speed for structural steel of
\[
\nonumber c = \sqrt{ E \over \rho} = \sqrt{160 \times 10^{9} N/m^{2} \over 7860 Kg /m^3} = 4512 m/s \tag{19} \]
Compared to one tabulated value the example values for stainless steel lays between the speed for longitudinal and transverse waves.
Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
The most robust way of answering this is to benchmark it. Failingthat, there are several things to note (roughly in order ofimportance).
First, the most cheap floating-point operations on a modern CPU areaddition and multiplication (both are equally fast; same as fusedmultiply-adds when available). Division is much slower (by a factor of~20), trigonometric functions are also slower (~200), square roots andlogs similar. In fact, sometimes (depends on architecture) trigfunctions, roots, logs are implemented in a library. The bestreference for this is usually the optimization guide published by thewhichever company made your CPU. So as a rule of thumb you want tominimize special functions first, then divisions, then multiplicationsand additions.
So by this measure the second formula is much worse: the there is oneinverse trig function, one square root (it would be wasteful toimplement your formula as written, with three square roots), and onesincos (since they are expensive it is better to evaluate sincos onceand then write things like $\tan\frac\theta2$ in terms of those). Yourfirst formula involves the other special functions, apart from sincos,only once at the end.
Second, you should not stop at Newton's method. For example, if youcompare Newton's method with Halley's method,$$ \frac{e \sin (x)-e x \cos (x)+M}{1-e \cos (x)}, \qquadx-\frac{(e \cos (x)-1) (e \sin (x)+M-x)}{e^2+e (M-x) \sin (x)-2 e \cos (x)+1}, $$both evaluate the expensive trigonometric functions at the samearguments, but with a little more algebra Halley's method can convergefaster; the extra algebra might not outweigh the savings ofevaluating trigonometric fewer times. So Halley's method (and otheriterative methods) also need to be checked.
Third, you can precompute some things. For example, if you start byreducing the argument to the range $0<E<2\pi$, you can experimentally,in advance, find the maximum number of iterations taken by themethod. Since the number of iterations is fairly small (depending onprecision) and mispredicted branches can be expensive, it may alsomake sense to unroll the iteration loop by hand to a fixed sufficientnumber of iterations. Unrolled code can also more easily benefit fromvectorization.
Fourth, it is difficult to intuitively predict which optimizationswill do the best. And since this is just one fairly simple equation,it is probably best to benchmark many different approaches and findthe best. When doing this, also consider what optimization flags yourcompiler supports for floating-point arithmetic (some of them areinteresting and important, many people know about
-ffast-math, butit actually decomposes into different helpful or harmful optimizationflags; here is gcc'slist for example). Another thing to do is to look at the assemblyoutput of your compiler to see which CPU instructions it actually endsup using.
Fifth, if you need to solve this equation many times for the sameeccentricity $e$, it is possible to rewrite the problem. If youconsider the function $E = E(M)$, on the range $[0,2\pi]$, given bythe solutions of the equation for fixed known $e$, you can approximatethe function $E(M)$ using, for example, Chebyshev series, which takesonly a small number of evaluations of $E(M)$ (which can be done withany root-finding method). Once you have a sufficiently closeapproximation, which might be a 20-term Chebyshev series or somethinglike that, you can evaluate that later without needing to solve theequation again. |
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 |
I've heard a little recently about equivariant homotopy theory, and so I decided to try out some baby examples just to get a feel for it. I'm not even sure if these are the right thing to look at, and I'm sure I'm butchering the notation, but I've attempted to compute $\pi_2^{C_n}(S^2)$ and $\pi_3^{C_3}(S^2)$; here, $C_n$ acts by rotations on the plane and $C_3$ acts by the standard representation on 3-space.
For the first one, there's the obvious equivariant cell structure on $S^2$, which has fixed vertices at 0 and at the basepoint $\infty$, one orbit of $n$ edges, and one orbit of $n$ faces. My maps are based and I need to map fixed points to fixed points, so either both vertices go to $\infty$ or else my map fixes them.
Suppose both go to $\infty$. The image of any one edge is an arbitrary based loop, and of course the other edges' images are determined by this. Assuming my faces are going to able to be mapped in at all, the inclusion of an edge into a face is a cofibration, so I may as well just homotope the edges to $\infty$ now. From here, the map is determined by the image of one face, i.e. an element of (nonequivariant) $\pi_2(S^2)$. In the case where $0$ and $\infty$ are both fixed, the image of an edge is an arbitrary path between $0$ and $\infty$, and then we just get to decide where a face goes, which I think is just an element of the preimage of 1 in the connecting map $\pi_2(S^2,S^1)\rightarrow \pi_1(S^1)$. There's no extension problem in the lexseq, so this is just a projection onto one factor $\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}$, so this looks like $\mathbb{Z}\times \{1\}$.
So as a set, $\pi_2^{C_n}(S^2)=\mathbb{Z} \sqcup \mathbb{Z}$.
For the second one, I have a 1-cell of fixed points at $(x,y,z)=(t,t,t)$, my three 2-cells are the half-planes through that 1-cell and one of the three axes, and my three 3-cells fill in the rest. The 1-cell needs to be sent to fixed points, and these are still just $0$ and $\infty$, so by continuity it's all sent to $\infty$. So the image of a 2-cell is just an element of $\pi_2(S^2)$. Once we choose that (which determines the images of the other 2-cells, of course) we can always extend to the 3-cells, since these are just homotopies from a map to itself. Once we've chosen one, any other gives an element of $\pi_3(S^2)=\mathbb{Z}$. So as a set, $\pi_3^{C_3}(S^2)=\mathbb{Z} \times \mathbb{Z}$.
So, my first question is: Are these right? Also, I've learned to compute (usual) homotopy groups of spheres by making a Postnikov tower, and I'm wondering if there's a sufficiently easy example for the equivariant case where I can do the analogous calculations by hand without the full generality of slice cells or whatever's going on (those could be the wrong words -- I don't think I understand what these are well enough to know whether this is a decent request, either). In any case, I'd love suggestions of better/more instructive examples. Lastly, I'm wondering if there are actually group structures here. In the first example, it looks like I can reasonably hope to add guys that are both in the same copy of $\mathbb{Z}$, but not otherwise. I think it's easy to show that there's no $C_n$-equivariant coproduct on $S^2$. On the other hand, quotienting by the plane $x+y+z=0$ in $\mathbb{R}^3$ looks like it gives a $C_3$-equivariant coproduct on $S^3$. If two elements came from the same choice of $\pi_2(S^2)$, then there's an obvious origin for the $\pi_3(S^2)$-torsor, namely using the trivial homotopy to extend the map over the 3-cells. I think this agrees with my equivariant coproduct, just because it matches up with the usual picture you draw of how to add elements of $\pi_2$ (two squares sitting on top of each other). However, I can't tell whether this makes any sense when the elements came from different choices of $\pi_2(S^2)$. It seems like if it does work at all, there might be something funnier than the obvious group structure... |
Answer
Estimate 140 trips Exact 112 trips
Work Step by Step
$140 \div 1 \frac{1}{4}$ Estimate Round fractions to nearest whole number. $140 \div 1$ = 140 trips Exact Convert mixed numbers to improper fractions. $\frac{140}{1} \div \frac{5}{4}$ Multiply by the reciprocal of the divisor. $\frac{140}{1} \times \frac{4}{5}$ Reduce by the common factor of 5. $\frac{28}{1} \times \frac{4}{1}$ = $28 \times 4$ = 112 trips |
I updated my proof to a general version as follows: please share your thoughts & 2cent. Thanks
Show a monotone continuous complete preorder on $\mathbb{R^L_+}$ has $y\geq x\rightarrow y\succsim x$.
Point of Clarification $X=\mathbb{R^L_+}$ Preorder means the usual reflexivity and transitivity. Complete means for any $x,y\in X$, have $x\succsim y$ or $y\succsim x$ Continuous means the relation is preserved under limits. Monotone means for any $x,y\in X$, if $y\gg x$, then $y\succ x$. $\succ$ and $\sim$ are respectively asymmetic and symmetric parts of $\succsim$ Outline of Proof Go through two cases: (1) $y\gg x$. Easily get the result by definition. (2) Some components are equal while else y is strictly greater x. Use continuity where you add a sequence of small positive $\epsilon$ to y, making it a sequence $y^n_\epsilon$ where for every n, $y^n_\epsilon\gg x$ $\forall n$.
Proof Suppose $\succsim$ is a monotone, continuous, complete preorder on $X=\mathbb{R^L_L}$. Case (1) $y\gg x$ (i.e. $y_i>x_i$ $\forall i\in B=\{1,\dots,L\}$). By definition, $y\succ x$, which implies $y\succsim x$. Case (2) $y_j=x_j$ for some $j\in B$. For $\forall k\not=j,k\in B, y_k>x_k.$ For some $\epsilon>0$, let the sequence $\epsilon^n\in\mathbb{R^L_+}$ such that $\epsilon_j=\epsilon$, $\epsilon_k=0$. Denote the sequence $y^n_\epsilon=y+\epsilon^n$. Then, for any $\epsilon>0$ and $\forall n$, $y^n_\epsilon\succ x$, hence $y^n_\epsilon\succsim x$. By continuity of $\succsim$, $$\lim_{\epsilon \to 0} {y^n_\epsilon} = y$$ Hence, $y\succsim x$. $\blacksquare$ OLDER VERSION My question is what is the valid reasoning behind that continuous rational and monotone preference relation implies $x\succsim0$. I have put a proof below and would appreciate if you share your 2 cent on the validity/rigor of the proof. Thanks!
Suppose $x\in\mathbb{R^+_L}=\{x\in\mathbb{R^L}:x_l\geq0$ $\forall l=1,\dots,L\}$.
Claim: For every $x\in\mathbb{R^+_L}$, monotonicity implies $x\succsim0$. Proof:
(1) Suppose $x=(0,\dots,0)$. Then, $x\sim0$ is possible.
(2) Suppose $x\gg y$. Then, by Definition of monotone preference, $x\succ y$ is possible.
(3) Suppose $\exists$ some $j$ such that $x_j>0$ and $1\leq j\leq L$.Then, I have the following process of elimination:
$x\succsim0$ is possible. $x\succsim0$ and $0\succsim x$ $\iff x\sim0$ is possible. $x\succsim0$ but not $0\succsim x$ $\iff x\succ0$ is possible. $0\succsim x$ but not $x\succsim 0$ $\iff 0\succ x$ is impossible.
The single preference relation that is common in all three scenarios is $x\succsim0$. Hence, for every $x\in\mathbb{R^+_L}$, monotonicity implies $x\succsim0$. Q.E.D |
81st SSC CGL level Solution Set, 17th on Algebra
This is the 81st solution set of 10 practice problem exercise for SSC CGL exam and the 17th on topic Algebra. For maximum gains, the test should be taken first and then this solution set should be referred to.
If you have not yet taken the corresponding test yet you may take it by referring to the
before going through the solution. SSC CGL level question set 81 on Algebra 17
You may also watch the
video solutions in the two-part video. Part 1: Q1 to Q5 Part 2: Q6 to Q10 81st solution set - 10 problems for SSC CGL exam: 17th on topic Algebra - answering time 12 mins Q1. If $x+\displaystyle\frac{1}{x}=3$, then value of $x^8+\displaystyle\frac{1}{x^8}$ is, 2213 2207 2201 2203 Solution 1: Problem analysis and solution by applying principle of inverses: Solving in mind
Given,
$x+\displaystyle\frac{1}{x}=3$.
Raising the equation to its square will eliminate $x$ in the middle term 2 because of interaction of inverses $x$ and $\displaystyle\frac{1}{x}$. We will then easily take the 2 from LHS to RHS reducing $3^2=9$ in RHS to $7$.
This gives at the first squaring,
$x^2+\displaystyle\frac{1}{x^2}=7$.
Without writing any more, we can raise this equation again to its square to get,
$x^4+\displaystyle\frac{1}{x^4}=7^2-2=47$.
And lastly squaring for the third time,
$x^8+\displaystyle\frac{1}{x^8}=47^2-2=2207$.
Answer: Option b: 2207. Key concepts used: -- Problem analysis Key pattern identification . -- Principle of interaction of inverses -- Solving in mind
Knowing the power and use of the principle of interaction of inverses, we could evaluate the final result easily in mind without writing the steps. For accuracy we calculated $47^2$ by hand.
Q2. If $(x-4)(x^2+4x+16)=x^3-p$, then value of $p$ is, 27 0 8 64 Solution 2: Problem analysis and solving in mind by coefficient comparison and mathematical reasoning
$-p$ on the RHS will be equal to the product of the numeric terms in the two factors on the LHS and will not depend on anything else. So,
$-p=-4\times{16}=-64$,
Or, $p=64$.
Answer: Option d : 64. Key concepts used: Coefficient comparison technique -- Solving in mind . Q3. If $a+b+c=15$ and $\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}=\displaystyle\frac{71}{abc}$, then the value of $a^3+b^3+c^3-3abc$ is, 180 160 220 200 Solution 3: Problem analysis and solution using three-variable sum of cubes expression
By the three-variable sum of cubes expression we have,
$a^3+b^3+c^3-3abc$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=15(a^2+b^2+c^2-ab-bc-ca)$.
We will get the second set of three terms of the above expression from the second given expression,
$\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}=\displaystyle\frac{71}{abc}$.
Multiplying both sides by $abc$,
$ab+bc+ca=71$.
And with values of $a+b+c$ and $ab+bc+ca$ known it is easy to get value of $a^2+b^2+c^2$,
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=225$,
Or, $a^2+b^2+c^2=225-142=83$.
Finally going back to our previous result,
$a^3+b^3+c^3-3abc$
$=15(a^2+b^2+c^2-ab-bc-ca)$
$=15\times(83-71)$
$=180$
Answer: Option a: 180. Key concepts used: Problem analysis -- Three-variable Sum of cubes -- Three-variable square of sum -- Solving in mind.
This problem also could also be solved in mind as the steps are clear and calculations easy. Due to two stages of solution—first, getting the value of $ab+bc+ca$ and second the value of $a^2+b^2+c^2$—the time taken was a bit more, but it was still within a minute.
Q4. If $x=12$, and $y=4$, then the value of $(x+y)^{\displaystyle\frac{x}{y}}$ is, 570 48 1792 4096 Solution 4: Problem analysis and solving in mind by unit's digit behavior analysis
Simplifying the target expression to $(x+y)=16$ raised to the power of 3 by $\displaystyle\frac{x}{y}=\frac{12}{4}=3$ is easy.
At the next step, we didn't do any multiplication at all. We have used just the result of unit's digit behavior for powers of 6,
$\text{Unit's digit of } 6^n=6$.
The only choice value 4096 having unit's digit of 6, we knew, $16^3$ must be equal to 4096.
Answer: Option d: 4096. Key concepts used: Indices -- Unit's digit behavior analysis -- Free resource use principle—we have used the free resource of choice values -- S olving in mind . Unit's digit behavior analysis—two principles define unit's digit behavior
The first principle of unit's digit behavior is,
Unit's digit of the product of two integers is the unit's digit of the product of the unit's digits of the integers.
For example, unit's digit of $4398\times{79}$ will be 2, the unit's digit of $8\times{9}=72$.
The second principle is,
Unit's digit for powers of the digits 2 through 9 will follow a fixed pattern specific to the digit.
For example, unit's digit of any power of 5 and 6 will remain always locked on 5 and 6 respectively, whereas unit's digit for powers of 3 will cycle through the 4 digits 3, 9, 7, 1 and then back to 3.
You can easily form the pattern specific to a digit and save valuable calculation time using these two principles.
Unit's digit for a few powers of 6 are,
$6^2=36$=>$\text{unit's digit } 6$.
$6^3=216$=>$\text{unit's digit } 6$.
$6^4=1296$=>$\text{unit's digit } 6$, and so on.
However many times you multiply 6 with itself, the unit's digit remains locked at the value of 6.
Q5. If $xy+yz+zx=1$, then the value of $\displaystyle\frac{1+y^2}{(x+y)(y+z)}$ is, 1 4 2 3 Solution 5: Problem analysis, key pattern identification and solving in mind by targeted substitution
We use the given expression value that can be used in the only way—by expanding $(x+y)(y+z)=xy+yz+zx+y^2$, and substituting given value 1 of $xy+yz+zx$ to get the denominator same as the numerator.
Answer: Option a: 1. Key concepts used: Key pattern identification -- Substitution -- solving in mind. Q6. If $x+y+z=0$, then the value of $\displaystyle\frac{x^2}{3z}+\displaystyle\frac{y^3}{3xz}+\displaystyle\frac{z^2}{3x}$ is, $0$ $3y$ $y$ $xz$ Solution 6: Problem analysis, key pattern identification and solving in mind
The key pattern identified as—multiplying the target expression by $xz$ transforms the numerator to $(x^3+y^3+z^3)$, and we know by three variable sum of cubes expression,
If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.
Thus the numerator is simplified to $3xyz$ and the denominator is $3xz$. The answer turns out to $y$.
Let us show the deductions.
Target expression,
$\displaystyle\frac{x^2}{3z}+\displaystyle\frac{y^3}{3xz}+\displaystyle\frac{z^2}{3x}$
$=\displaystyle\frac{1}{3xz}(x^3+y^3+z^3)$,
$=\displaystyle\frac{3xyz}{3xz}$, as with $x+y+z=0$, $x^3+y^3+z^3=3xyz$,
$=y$.
Answer: Option c : $y$. Key concepts used: . Key pattern identification -- Target expression transformation -- Three-variable sum of cubes expression -- Solving in mind Q7. If the expression $(px^3-2x^2-qx+18)$ is completely divisible by $(x^2-9)$, then the ratio of $p$ to $q$ is, 9 : 1 1 : 9 3 : 1 1 : 3 Solution 7: Problem analysis, mathematical reasoning and solving in mind by coefficient comparison
By examining the two given expressions we can form the quotient factor $(ax-2)$,
$px^3-2x^2-qx+18=(x^2-9)(ax-2)$.
Here $a$ is an additional dummy variable and the numeric term of $(ax-2)$ must be $-2$ to make the product $(-9)\times{(-2)}=18$.
Analyzing the two expressions on the LHS and RHS while comparing the coefficients of $x$ and $x^3$ we can form the relations,
$p=a$, and
$-q=-9a$,
Or, $q=9$.
The ratio is, $p:q=1:9$.
Answer: Option b: 1 : 9. Key concepts used: -- Key pattern identification Quotient expression formation -- Coefficient comparison -- Basic ratio concepts. Q8. If $\displaystyle\frac{x}{3}+\displaystyle\frac{3}{x}=1$, then the value of $x^3$ is, $0$ $27$ $-27$ $1$ Solution 8: Problem analysis, key pattern identification and solving in mind by factor extraction technique
Simplifying the given expression,
$\displaystyle\frac{x}{3}+\displaystyle\frac{3}{x}=1$
Or, $x^2-3x+9=0$.
Multiplying by $x$ to form term $x^3$,
$x^3-3x^2+9x=0$,
Or, $x^3-3(x^2-3x+9) + 27=0$, extracting zero valued factor,
Or, $x^3=-27$.
Answer: Option c: $-27$. Key concepts used: -- Problem analysis Key pattern identification -- Factor extraction technique -- Solving in mind .
The problem is formed intending to test your problem skill and not math skills. In fact, the values of $x$ are imaginary from the quadratic equation, whereas $x=-3$ from $x^3$. Nevertheless, we are not to look more closely into the value of $x$, we are to find the value of $x^3$.
Q9. If $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}=0$ and $x+y+z=9$, then the value of $x^3+y^3+z^3-3xyz$ is, 81 6561 729 361 Solution 9: Problem analysis, key pattern identification and solving in mind by three-variable sum of cubes relation
The first given equation is transformed to,
$xy+yz+zx=0$.
The three-variable sum of cubes relation is,
$x^3+y^3+z^3-3xyz$
$=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
$=9(x^2+y^2+z^2)$.
Again by three-variable square of sum expansion,
$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$,
Or, $x^2+y^2+z^2=9^2=81$.
So the value of target expression is,
$9^3=729$.
Answer: Option c: 729. Key concepts used: Key pattern identification -- Input transformation -- Three-variable sum of cubes -- Three-variable square of sum -- Solving in mind. Q10. If $a(a+b+c)=45$, $b(a+b+c)=75$ and $c(a+b+c)=105$, then the value of $a^2+b^2+c^2$ is, 75 83 225 217 Solution 10: Problem analysis, key pattern identification and Collection of friendly terms: Solving in mind
Identifying the pattern of $(a+b+c)$ common to the three equations we sum up the three to get,
$(a+b+c)^2=225$,
Or, $a+b+c=15$.
This immediately gives values of $a$, $b$ and $c$ as,
$a=3$, from first equation,
$b=5$, from second equation and,
$c=7$, from third equation.
So target expression,
$a^2+b^2+c^2=9+25+49=83$.
Answer: Option b: 83. Key concepts used: -- Key pattern identification -- Principle of collection of friendly terms Solving in mind. Additional help on SSC CGL Algebra
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large number of question and solution sets and a valuable article on " " rich with concepts and links, you may refer to our other articles specifically on Algebra listed on latest shown first basis, 7 Steps for sure success on Tier 1 and Tier 2 of SSC CGL First to read tutorials on Basic and rich Algebra concepts and other related tutorials SSC CGL Tier II level Questions and Solutions on Algebra Efficient solutions for difficult SSC CGL problems on Algebra in a few steps SSC CGL level Question and Solution Sets on Algebra SSC CGL level Solution Set 81, Algebra 17 |
Difference between revisions of "Huge"
Line 26: Line 26:
The first-order definition of $n$-huge is somewhat similar to [[measurable|measurability]]. Specifically, $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete [[filter|ultrafilter]], $U$, over $\kappa$. A cardinal $\kappa$ is $n$-huge with target $\lambda$ iff there is a normal $\kappa$-complete ultrafilter, $U$, over $\mathcal{P}(\lambda)$, and cardinals $\kappa=\lambda_0<\lambda_1<\lambda_2...<\lambda_{n-1}<\lambda_n=\lambda$ such that:
The first-order definition of $n$-huge is somewhat similar to [[measurable|measurability]]. Specifically, $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete [[filter|ultrafilter]], $U$, over $\kappa$. A cardinal $\kappa$ is $n$-huge with target $\lambda$ iff there is a normal $\kappa$-complete ultrafilter, $U$, over $\mathcal{P}(\lambda)$, and cardinals $\kappa=\lambda_0<\lambda_1<\lambda_2...<\lambda_{n-1}<\lambda_n=\lambda$ such that:
−
$$\forall i<n(\{x\subseteq\lambda:\
+
$$\forall i<n(\{x\subseteq\lambda:\{}(x\cap\lambda_{i+1})=\lambda_i\}\in U)$$
−
Where $\
+
Where $\{}(X)$ is the [[Order-isomorphism|order-type]] of the poset $(X,\in)$. <cite>Kanamori2009:HigherInfinite</cite> This definition is, more intuitively, making $U$ very large, like most ultrafilter characterizations of large cardinals ([[supercompact]], [[strongly compact]], etc.).
$\kappa$ is then super $n$-huge if for all ordinals $\theta$ there is a $\lambda>\theta$ such that $\kappa$ is $n$-huge with target $\lambda$, i.e. $\lambda_n$ can be made arbitrarily large. <cite>Kanamori2009:HigherInfinite</cite>
$\kappa$ is then super $n$-huge if for all ordinals $\theta$ there is a $\lambda>\theta$ such that $\kappa$ is $n$-huge with target $\lambda$, i.e. $\lambda_n$ can be made arbitrarily large. <cite>Kanamori2009:HigherInfinite</cite>
Revision as of 13:59, 5 December 2017 Huge cardinals (and their variants) were introduced by Kenneth Kunen in 1972 as a very large cardinal axiom. Kenneth Kunen first used them to prove that the consistency of the existence of a huge cardinal implies the consistency of $\text{ZFC}$+"there is a $\aleph_2$-saturated $\sigma$-complete ideal on $\omega_1$". It is now known that only a Woodin cardinal is needed for this result. [1] Contents Definitions
Their formulation is similar to that of the formulation of superstrong cardinals. A huge cardinal is to a supercompact cardinal as a superstrong cardinal is to a strong cardinal, more precisely. The definition is part of a generalized phenomenon known as the "double helix", in which for some large cardinal properties $n$-$P_0$ and $n$-$P_1$, $n$-$P_0$ has less consistency strength than $n$-$P_1$, which has less consistency strength than $(n+1)$-$P_0$, and so on. This phenomenon is seen only around the $n$-fold variants as of modern set theoretic concerns. [2]
Although they are very large, there is a first-order definition which is equivalent to $n$-hugeness, so the $\theta$-th $n$-huge cardinal is first-order definable whenever $\theta$ is first-order definable. This definition can be seen as a (very strong) strengthening of the first-order definition of measurability.
Elementary embedding definitions $\kappa$ is almost $n$-huge with target $\lambda$iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length less than $\lambda$ (that is, $M^{<\lambda}\subseteq M$). $\kappa$ is $n$-huge with target $\lambda$iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length $\lambda$ ($M^\lambda\subseteq M$). $\kappa$ is almost $n$-hugeiff it is almost $n$-huge with target $\lambda$ for some $\lambda$. $\kappa$ is $n$-hugeiff it is $n$-huge with target $\lambda$ for some $\lambda$. $\kappa$ is super almost $n$-hugeiff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is almost $n$-huge with target $\lambda$ (that is, the target can be made arbitrarily large). $\kappa$ is super $n$-hugeiff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is $n$-huge with target $\lambda$. $\kappa$ is almost huge, huge, super almost huge, and superhugeiff it is almost 1-huge, 1-huge, etc. respectively. Ultrafilter definition
The first-order definition of $n$-huge is somewhat similar to measurability. Specifically, $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete ultrafilter, $U$, over $\kappa$. A cardinal $\kappa$ is $n$-huge with target $\lambda$ iff there is a normal $\kappa$-complete ultrafilter, $U$, over $\mathcal{P}(\lambda)$, and cardinals $\kappa=\lambda_0<\lambda_1<\lambda_2...<\lambda_{n-1}<\lambda_n=\lambda$ such that:
$$\forall i<n(\{x\subseteq\lambda:\text{order-type}(x\cap\lambda_{i+1})=\lambda_i\}\in U)$$
Where $\text{order-type}(X)$ is the order-type of the poset $(X,\in)$. [1] This definition is, more intuitively, making $U$ very large, like most ultrafilter characterizations of large cardinals (supercompact, strongly compact, etc.). $\kappa$ is then super $n$-huge if for all ordinals $\theta$ there is a $\lambda>\theta$ such that $\kappa$ is $n$-huge with target $\lambda$, i.e. $\lambda_n$ can be made arbitrarily large. [1]
If $j:V\to M$ is such that $M^{j^n(\kappa)}\subseteq M$ (i.e. $j$ witnesses $n$-hugeness) then there is a ultrafilter $U$ as above such that, for all $k\leq n$, $\lambda_k = j^k(\kappa)$, i.e. it is not only $\lambda=\lambda_n$ that is an iterate of $\kappa$ by $j$; all members of the $\lambda_k$ sequence are.
Consistency strength and size
Hugeness exhibits a phenomenon associated with similarly defined large cardinals (the $n$-fold variants) known as the
double helix. This phenomenon is when for one $n$-fold variant, letting a cardinal be called $n$-$P_0$ iff it has the property, and another variant, $n$-$P_1$, $n$-$P_0$ is weaker than $n$-$P_1$, which is weaker than $(n+1)$-$P_0$. [2] In the consistency strength hierarchy, here is where these lay (top being weakest): measurable = 0-superstrong = almost 0-huge = super almost 0-huge = 0-huge = super 0-huge $n$-superstrong $n$-fold supercompact $(n+1)$-fold strong, $n$-fold extendible $(n+1)$-fold Woodin, $n$-fold Vopěnka $(n+1)$-fold Shelah almost $n$-huge super almost $n$-huge $n$-huge super $n$-huge $(n+1)$-superstrong
All huge variants lay at the top of the double helix restricted to some natural number $n$, although each are bested by I3 cardinals (the critical points of the I3 elementary embeddings). In fact, every I3 is preceeded by a stationary set of $n$-huge cardinals, for all $n$. [1]
Similarly, every huge cardinal $\kappa$ is almost huge, and there is a normal measure over $\kappa$ which contains every almost huge cardinal $\lambda<\kappa$. Every superhuge cardinal $\kappa$ is extendible and there is a normal measure over $\kappa$ which contains every extendible cardinal $\lambda<\kappa$. Every $(n+1)$-huge cardinal $\kappa$ has a normal measure which contains every cardinal $\lambda$ such that $V_\kappa\models$"$\lambda$ is super $n$-huge". [1]
In terms of size, however, the least $n$-huge cardinal is smaller than the least supercompact cardinal (assuming both exist). [1] This is because $n$-huge cardinals have upward reflection properties, while supercompacts have downward reflection properties. Thus for any $\kappa$ which is supercompact and has an $n$-huge cardinal above it, $\kappa$ "reflects downward" that $n$-huge cardinal: there are $\kappa$-many $n$-huge cardinals below $\kappa$. On the other hand, the least super $n$-huge cardinals have
both upward and downward reflection properties, and are all much larger than the least supercompact cardinal. It is notable that, while almost 2-huge cardinals have higher consistency strength than superhuge cardinals, the least almost 2-huge is much smaller than the least super almost huge.
Every $n$-huge cardinal is $m$-huge for every $m\leq n$. Similarly with almost $n$-hugeness, super $n$-hugeness, and super almost $n$-hugeness. Every almost huge cardinal is Vopěnka (therefore the consistency of the existence of an almost-huge cardinal implies the consistency of Vopěnka's principle). [1]
References Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Kentaro, Sato. Double helix in large large cardinals and iteration ofelementary embeddings., 2007. www bibtex |
April 27th, 2018, 03:25 PM
# 1
Newbie
Joined: May 2017
From: California
Posts: 15
Thanks: 1
Upper darboux integral x^3
I'm trying to calculate the upper darboux integral of $f(x) = x^3$ on $[0, b]$ where $b$ is a positive integer.
Let $P$ be a partition defined as $\lbrace x_0 = 0, x_1, x_2, ... ,x_n = b \rbrace$, $x_i < x_{i+1}$.
Define $M(f; S) = \sup\lbrace x^3 : x \epsilon S \rbrace$
Define Upper Darboux sum as $U(f; P) = \sum_{i = 1}^{n} (M(x^3 ; [x_{i-1}, x_i]) \cdot (x_i - x_{i-1}))$
So $U(f; P) = \sum_{i = 1}^n x_i^3 \cdot(x_i - x_{i-1})$
In the example for $f(x) = x^2$, we have to choose some $x_i$ to make the algebra pan out... but why are we choosing an $x_i$ ? I don't get why the next step is to choose $x_i = ...$
Please no full solution, thank you
Last edited by Choboy11; April 27th, 2018 at 03:28 PM.
April 28th, 2018, 12:05 PM
# 2
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Joined: May 2007
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Thanks: 733
For $f(x)=x^2$, the approach is the same as for $f(x)=x^3$, replacing $x_i^3\ by \ x_i^2$ in the expression for U. I don't understand your last remark.
April 28th, 2018, 01:34 PM
# 3
Newbie
Joined: May 2017
From: California
Posts: 15
Thanks: 1
Sorry fro the confusion
The full question I am working on is "Find the upper and lower Darboux integrals for $f(x) = x^3$ on the interval $[0, b]$."
update:
$U(f, P) = \sum_{i=1}^n M(f(x), [x_{i-1}, x_i]) \cdot \triangle x$
$= \sum_{i=1}^n x_i^3(x_i - x_{i-1})$
Similar to the example in my book, let $x_i = i \cdot \frac bn$. I realize $\frac bn$ is the value of $x_i - x_{i-1}$ assuming the $x_i$'s are spaced out equally, but why do we do this step? Why are we specifying a partition(i think thats what we are doing here...)?
After that I get,
$U(f,P) = \sum_{i=1}^n i^3 (\frac bn) ^3 \cdot \frac bn$
$= (\frac bn)^4 \cdot (\sum_{i=1}^n i)^2$
$= (\frac bn)^4 \cdot (\frac{n(n+1)}{2})^2$
$= (\frac bn)^4 \cdot (\frac {n^4 + 2n^3 + n^2}{4}$.
So $U(f,P) \rightarrow \frac {b^4}{4}$ as $n \rightarrow \infty$.
By similar steps, we can show $L(f, P) = (\frac bn)^4 \cdot \frac{n^4 - 2n^3 +n^2}{4} \rightarrow \frac {b^4}{4}$ as $n \rightarrow \infty$.
We can also conclude $U(f, P) > L(f, P) \ge \frac{b^4}{4}$. Here I am stuck.
In the book example (for f(x) = x^2), they have $U(f,P) \le (\frac {b^3}{3})$ and $L(f,P) \ge (\frac {b^3}{3}$ and conclude that f is integrable. I don't understand this logic.
As i understand it, to show $f$ is integrable on $[a, b]$, we need to show
$\inf \lbrace U(f, P) : \text{P is a partition of }[a,b]) \rbrace = \sup \lbrace L(f, P) : \text{P is a partition of [a,b]}) \rbrace$.
Last edited by Choboy11; April 28th, 2018 at 01:43 PM.
April 28th, 2018, 02:43 PM
# 4
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Joined: Sep 2016
From: USA
Posts: 669
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Math Focus: Dynamical systems, analytic function theory, numerics
If I understand you correctly, you are asking exactly the right question. In general, you have to show this limit exists for
any partition. However, this is what makes Darboux sums so nice to work with. There is a theorem which says Darboux integrability is equivalent to Riemann integrability. With this in hand, you are free to choose a partition. The reason this works is sketched below.
1. Suppose you pick a sequence of partitions, $P_n$ whose mesh size goes to $0$, and you show that $U(f,P_n) - L(f,P_n) \to 0$.
2. Now, prove that if $P$ is a partition and $Q$ is any refinement of $P$ (i.e $P \subset Q$), then $L(f,P) \leq L(f,Q) \leq U(f,Q) \leq U(f,P)$.
3. Now, it follows that if $Q_n$ is any other sequence of partitions whose mesh goes to $0$, then $Q_n \cup P_n$ is a common refinement. Combine this with (1),(2), and the squeeze theorem to conclude the result.
May 1st, 2018, 09:46 PM
# 5
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From: California
Posts: 15
Thanks: 1
Sorry for the late reply... finals are coming and we started power series so i've put integration on the back burner... for now.
So far i've proved part 2(read it in my book... but this is it in my own words i guess?)
Suppose $P, Q$ are partitions of some interval $[a, b]$ such that $P \subset Q$ and $P = \lbrace x_0 < x_1 < ... < x_n\rbrace$ and $Q = \lbrace x_0 < x_1 < ... < x_{k-1} < x_k < x_{k+1} < ...x_n \rbrace$. (EDIT: I still need to explain why choosing these partitions is sufficient to show the statement holds for all $P, Q$ such that $P \subset Q$.)
Observe $$U(f, P) - U(f, Q) = M(f, [x_{k-1}, x_{k+1}])\cdot (x_{k+1} - x_{k-1}) - M(f, [x_{k-1}, x_k]) \cdot (x_k - x_{k-1}) - M(f, [x_k, x_{k+1}])\cdot(x_{k+1} - x_k) \ge 0$$. (The $\ge 0$ step seems intuitive, do I need to justify it?)
Thus we can conclude $U(f, P) - U(f, Q) \ge 0$ so $U(f, P \ge U(f, Q)$. By similar logic $L(f, P) \le L(f, Q)$. It is obvious.. i guess.. that $L(f, Q) \le U(f, Q)$. Combining these statements gives us $L(f, P) \le L(f, Q) \le U(f, Q) \le U(f, P)$.
I had a few questions in the proof about if I need further justification.
I'll post again when I get 1, 3. Thank you for your time
Last edited by Choboy11; May 1st, 2018 at 09:55 PM.
May 14th, 2018, 10:52 PM
# 6
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Since f(x) is continuous, the Darboux integral is the Riemann integral by definition.*
But I think the point here is to show the actual mechanics of the definition because in general f doesn't have to be continuous.
If P= {1,2,3), by definition
$\displaystyle U(f,P)= 2^{3}(2-1) + 3^{3}(3-2)$
The Upper Darboux integral is the infimum of U(f,P)
for all P, not just equal intervals. That sounds impossible.
For equal intervals, you can write an expression for the sum for any division
and you have to show the sum is a minimum as number of subdivisions approaches $\displaystyle \infty$.
Darboux Integral -- from Wolfram MathWorld
Interesting. I always wondered what a Darboux integral was.
* EDIT
Actually, the Riemann integral is based on the max interval approaching zero, and it's not obvious why that result should be the same as the Darboux integral, for continuous functions, but I believe you can prove RI is the same for all equal intervals approahing 0 and max interval approaching zero,
Last edited by zylo; May 14th, 2018 at 11:11 PM.
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Twice a year UGC Jointly with CSIR conducts the Net exam for PhD and Lecturership aspirants. For science subjects such as Life Science, Chemical Science etc, the question paper is divided into three parts - Part A, B and C. Part A is on Maths related topics and contains 20 questions any 15 of which have to be answered. The other two parts are on the specific subject chosen by the student.
The questions in Maths are tuned towards judging the problem solving ability of the student using basic concepts in maths rather than procedural competence in maths.
The fourth set of answer and structured solution of 15 Net level questions follow. To gain maximum benefits from this resource, the student must first answer the corresponding 15 question set and then only refer to this solution set.
This is a set of 15 questions for practicing UGC/CSIR Net exam: ANS Set 4
Answer all 15 questions. Each correct answer will add 2 marks to your score and each wrong answer will deduct 0.5 mark from your score. Total maximum score 30 marks. Time: (25 mins). Q1. The unit's digit of the product $(693\times{694}\times{695}\times{698})$ is, 2 8 0 4 Solution: In a product of numbers, the unit's digit is always determined by the unit's digit of the product of unit's digits of the numbers.
To further simplify, if we find a 5 and a multiple of 2 in the unit's digits of the numbers, the final unit's digit will always be 0, as once a 5 is multiplied with a power of 2, it results in unit's digit 0, and any further multiplication stays at 0 value of the unit's digit.
Answer: Option c: 0. Key concepts used: Unit's digit in product of numbers concept -- use of 0 multiplication concept to cut short solution time. Q2. In a city average age of men and women are 72.4 years and 67.4 years respectively, while the average of the total number of citizens is 69.4 years. The percentage of men in the city is, 40 50 60 66.7 Solution: If $m$ and $n$ are number of men and women, $m\times{72.4} + n\times{67.4} = (m + n)\times{69.4}$
Or, $3m = 2n$,
Or, $\displaystyle\frac{m}{n} = \frac{2}{3} = \frac{2x}{3x}$, where $m = 2x$, $n=3x$ and $x$ is the HCF of $m$ and $n$.
Thus, total number of citizens = $5x$ and the percentage of men is = $\displaystyle\frac{2x}{5x}\times{100}=40$.
Alternatively, $\displaystyle\frac{m}{n} = \frac{2}{3}$
Or, $\displaystyle\frac{n}{m} = \frac{3}{2}$. Adding 1 to both sides,
$\displaystyle\frac{m + n}{m} = \frac{5}{2}$
Or, $\displaystyle\frac{m}{m + n} = \frac{2}{5}$, that is, the percentage of men is 40.
Answer: Option a: 40. Key concepts used: Average concept -- concept that in a ratio, original quantities can be obtained by multiplying each of the ratio elements by the HCF of the original quantities and this enables to get the total. Conceptual solution: The positive contribution of men to the combined average age is 3 years and the negative contribution of women is 2 years. It follows, contribution to the total combined age by the men is, $3\times{Men}$ and same by women, $2\times{Women}$ which must nullify each other (as the average is between two values) and so are equal. This results in a ratio of Men : Women as 2 : 3.
With any such ratios, experience guides us to the final result, Men are 40%, Women 60%, ratio 2 : 3 and total 100%. This solution is conceptual, intuitive and hence can be reached quickly.
Q3. Two angles of a triangle are 40% and 60% of the largest angle. The largest angle is, $120^0$ $90^0$ $80^0$ $100^0$ Solution: Ratios of the two angles to the largest angle in a triangle are 2 : 5 and 3 : 5 giving a ratio of 2 : 5 : 3 between the three angles. If ratio of a : c is p : q and b : c is r : q, we have to equalize the middle value of the ratios to combine three ratios. Inverting second ratio, c : b is q : r, we get, a : c : b = p : q : r.
With this proportion 2 : 5 : 3 between the three angles let's assume the angles as 2x, 5x and 3x where x is the HCF between the values of the three angles. As total angle measure of a triangle is $180^0$, we have, $2x + 5x + 3x = 10x = 180^0$.
Or, $x = 18^0$ giving the largest angle as $5x = 5\times{18^0} = 90^0$.
Answer: Option b: $90^0$. Key concepts used: Conversion of percentage to ratios -- Combining two ratios -- total value of three angles of a triangle $180^0$ -- each quantity in a ratio can be multiplied by their HCF to get the individual quantities and their total.
This solution is the conventional mathematical solution. Let's see how we can proceed conceptually.
Conceptual solution:
One angle is 40% of the largest and the second 60% of the largest. Together these two angles are equal to the largest (100%), forming a second equal part of the total angle $180^0$. Largest angle then is, half of $180^0$ = $90^0$.
Q4. A bird sitting at the top of a pole spots a centipede 36m away from the base of the pole. If the bird catches the centipede at 16m from the base of the pole both moving at the same speed, the height of the pole is, 18m 15m 12m $12\sqrt{2}$m Solution: AB is the vertical side of the right angled triangle ABC with BC as 36m. At point D 16m away from the base, the bird catches the centipede. As the bird and the centipede moves at the same speed for the same time when they meet, AD = DC = 20m.
So, $AB^2 = AD^2 - BD^2 = 20^2 - 16^2 = 144$ and so, $AB = 12m$.
Answer: Option c: 12m. Key concepts used: Right angle triangle geometry -- time and speed concept -- Pythagoras theorem. Q5. In the circle with centre at $O$, $AB$ is a chord. $\angle{ACB} + \angle{OAB}$ =
$120^0$ $60^0$ $90^0$ $180^0$ Solution: In $\triangle{AOB}$,
$\angle{OAB} = 180^0 - (\angle{AOB} + \angle{ABO})$
Or, $ \angle{OAB} = 180^0 - (2\times{\angle{ACB}} + \angle{ABO})$
Or, $2\angle{ACB} + \angle{ABO} + \angle{OAB} = 180^0$
Or, $2(\angle{ACB} + \angle{OAB}) = 180^0$, as $\angle{ABO} = \angle{OAB}$ in isosceles $\triangle{AOB}$
Or, $\angle{ACB} + \angle{OAB} = 90^0$
Answer: Option c: $90^0$. Key concepts used: Total of angles in a triangle $180^0$ -- recognizing that two same length radii creates an isosceles triangle -- angle held by a chord at centre is double the angle held by the chord at any point on the periphery on the same side of the chord. Q6. A car travels uphill to a point at a speed of 20km/hr and returns back to its original position at a speed of 60km/hr. The average speed in which the car covered the whole distance is, 30km/hr 40km/hr 45km/hr 35km/hr Solution: Assuming one way distance as d, time taken downhill is, $T_{1} = \displaystyle\frac{d}{60}$ and time taken uphill $T_{2} = \displaystyle\frac{d}{20} = 3T_{1}$
Total time taken $T = 4T_{1} = \displaystyle\frac{d}{60} + \displaystyle\frac{d}{20} = \displaystyle\frac{4d}{60} = \displaystyle\frac{d}{15}$,
Or, $d=15\times{T}$.
Total distance covered being $2d$, the average speed is,
$\displaystyle\frac{2d}{T} = 30km/hr$.
Alternate method: without going into formulas, let's assume distance as the LCM of speeds, that is 60km. So 120km is covered in 3 + 1 = 4 hours at an average speed of 30km/hr.
Note that, in this approach, with distance as any multiple of the LCM 60km, the same result will hold.
Answer: Option a: 30 km/hr. Key concepts used: Speed time concept -- averaging concept. Q7. If $5^{\sqrt{x}} + 12^{\sqrt{x}} - 13^{\sqrt{x}} = 0$, then x is, $\displaystyle\frac{25}{4}$ 4 9 6 Solution: Initial examination of the equation reveals the third term higher than the first two and being in negative shifted to the right side of equation thus simplifying the equation giving,
$5^{\sqrt{x}} + 12^{\sqrt{x}} = 13^{\sqrt{x}}$
As there is no apparent relationship between the three terms, trial and error approach should be the best one.
The important point to make here is that the trial values should be for the power, that is $\sqrt{x}$ and not just $x$. Immediately at trial value 2 we reach the answer. Thus, $\sqrt{x} = 2$.
So at $x = 4$, the equation is satisfied.
Answer: Option b: 4. Key concepts used: Analysis of expression and simplification, wherever possible use the right side of equation to simplify an equation -- trial and error approach -- idea of squares of integers. Q8. The angle between the hour and minute hands of a clock at 10 past 10 is, $120^0$ $115^0$ $65^0$ $55^0$ Solution: In a clock the minute hand moves full circle, that is 360$^0$, in 1 hour and the hour hand moves more slowly and moves 360$^0$ in 12 hours.
In 1 hour or 60 minutes the hour hand moves $360^0\div{12}=30^0$. This movement is from one hour mark to the next hour mark.
At 10' O clock, the hour hand stood at 10 hour mark and the minute hand stood at 12 hour mark. As time progressed, in 10 minutes the minute hand stood at 2 hour mark (each hour mark represents 5 minutes and 30 degrees angular separation) and the hour hand moved forward by $10\times{30}\div{60}=5^0$ from the 10 hour mark.
So the angle between the two hands at 10 past 10 is, $120^0 - 5^0 = 115^0$.
Answer: Option b : $115^0$. Key concepts used: Clock hand movement with relation to speed -- relative movement of the hands -- angular movement -- unitary method.
Q9. In a river, a boat traveling at a still water speed of 6m/sec overtook a second boat traveling in the same direction and of same length at a still water speed of 4m/sec in 10secs. The length of each boat is, 100m 5m 10m Cannot be determined Solution: At first glance it may seem that the silence about the river water flow speed should make the problem unsolvable. But, as the boats are traveling in the same direction, relative speed will be the difference in actual speeds.
Actual speed of each boat will be its still water speed $\pm$ river flow speed. When subtracting the actual speed of the slower boat from that of the faster boat, the river flow speed will cancel out.
Thus the relative speed will turn out to be $6m/sec - 4m/sec = 2m/sec$.
To overtake the slower boat, the faster boat has to cover double its length (both lengths being same). At relative speed 2m/sec in 10secs the faster boat covers 20m. So length is half of 20m, that is, 10m.
Answer: Option c: 10m. Key concepts used: Boat in river concept -- overtaking concept -- in overtaking, be it train or boat or car, faster moving object has to cover the sum of the two lengths.
For a man or pole, length is taken as zero.
In same direction overtaking or passing, relative speed is the difference in speeds whereas in opposite direction overtaking, speeds are added up to form the relative speed.
Q10. 16 men can do a piece of work in 20 days. How many extra percentage of men will have to be brought in to complete the work in 40% of the original time? 50% 250% 80% 150% Solution: Total work amount is 320 man-days. This work is to be finished $\frac{2}{5}$th of 20 days, that is, in 8 days. To do it a total of $\frac{320}{8} = 40$ men will be required. This is an extra 24 men, that is, extra 150% men. Answer: Option c: d : 150%. Key concepts used: Man-days concept to measure amount of work in work-day problems -- careful estimation of extra percentage. Q11. A cube of ice of volume $100cm^3$ floats in water with $\frac{9}{10}$ths of its volme under water. If its portion above water melts at a rate of $10{\%}$ per minute, what will be its approximate volume above water after 3 minutes? $7cm^3$ $7.29cm^3$ $9.7cm^3$ $7.3cm^3$ Solution: By physical laws of buoyancy in water, whatever be the volume of an ice cube, $10{\%}$ of its volume will float above water. In this case, loss due to melting is $10{\%}$ of $10{\%}$, that is, $1{\%}$ of its volume every minute.
This is a case of compound loss (instead of growth) and the formula applicable is simply,
End volume = $100(1- .01)^3$. As the rate of loss is very small, compounding it will not increase the loss significantly compared to linear loss of $1{\%}$ per minute.
Thus in 3 minutes loss will about $3cm^3$, the total volume will be $97cm^3$ and volume above water will be $9.7cm^3$.
Answer: Option c: $9.7cm^3$. Key concepts used: Buoyancy -- linear and compound growth/loss -- approximation. Q12. The minimum value of $2p^2 + 3q^2$, where $p^2 + q^2=1$ and $-1 {\leq} p,q {\leq} 1$ is, 0 2 1 2.5 Solution: As $p^2 + q^2=1$, $2p^2 + 3q^2=1$ will be transformed to, $2(p^2 + q^2) + q^2 = 2 + q^2.$
As q is in square, its negative values also will increase the value of the expression. So the minimum value of the expression will arise only when $q^2$ = 0.
Answer: Option b: 2. Key concepts used: Expression simplification -- relation of value of expression with the values of its variables. Q13. From a bag containing 1 green, 4 blue and 5 red balls, if a ball is picked up blindfolded, what is the probability of picking up a blue ball? $\frac{1}{4}$ $\frac{2}{5}$ $\frac{1}{10}$ $\frac{2}{3}$ Solution: Probability of an event happening,
$P = \displaystyle\frac{\text{Number of favorable outcomes of the event}}{\text{Number of total possible outcomes of the event}}$.
For example in a coin toss total possible outcomes is 2 and so, probability of getting Head in a coin toss is $\frac{1}{2}$.
Here total possible outcomes is the total number of balls and total number of favorable outcomes is number of blue balls which is 4. Thus the desired probability is, $\frac{4}{10}=\frac{2}{5}$.
Answer: Option b: $\frac{2}{5}$. Key concepts used: Total and favorable outcomes -- randomness because of blindfolding -- probability concept. Q14. A rubber mat of thickness 1cm is rolled tightly with no gaps between layers into a solid cylindrical shape that stood on ground with a base area of $1m^2$ and height 1m. The total surface area of the sheet is, $2.02m^2$ $202.02m^2$ $101.01m^2$ $200.02m^2$ Solution: As the base surface area itself represents the lengthwise thin-edge surface area of the sheet and there are two such thin edges, lengthwise thin-edge areas together is, $2m^2$. Rolling out the sheet, the thin edge surface area is its length multiplied by thickness. So its length = $\frac{1}{.01}m = 100m$.
The height or breadth wise thin-edge surface area = $2\times{1\times{.01}}=.02m^2$ and its broad surface area is two times length by breadth or, $200m^2$. This gives a total surface area as $202.02m^2$.
Answer: Option b: $202.02m^2$. Key concepts used: Concept of rolling or spiraling and its effect on areas of shapes -- areas of rectangular shapes. Q15. In a queue of girls Lakshmi stood at the 5th position from the front and 15th position from the end. In another queue girls Veeny stood at the 18th position from the front and 7th from the end. The total number of girls in two queues together is, 20 38 40 43 Solution: If front and back position counts of a person in row or queue added up, the person is counted twice in the sum. Total number of persons in the queue then = Sum of positions from back and front - 1.
So total number of girls in two queues is, (5 + 15 - 1) + (18 + 7 -1) = 43.
Answer: Option d: 43. Key concepts used: Ranking with respect to a row or queue analysis. |
Proceedings of the 36th International Conference on Machine Learning, PMLR 97:7174-7183, 2019.
Abstract
For SGD based distributed stochastic optimization, computation complexity, measured by the convergence rate in terms of the number of stochastic gradient calls, and communication complexity, measured by the number of inter-node communication rounds, are two most important performance metrics. The classical data-parallel implementation of SGD over N workers can achieve linear speedup of its convergence rate but incurs an inter-node communication round at each batch. We study the benefit of using dynamically increasing batch sizes in parallel SGD for stochastic non-convex optimization by charactering the attained convergence rate and the required number of communication rounds. We show that for stochastic non-convex optimization under the P-L condition, the classical data-parallel SGD with exponentially increasing batch sizes can achieve the fastest known $O(1/(NT))$ convergence with linear speedup using only $\log(T)$ communication rounds. For general stochastic non-convex optimization, we propose a Catalyst-like algorithm to achieve the fastest known $O(1/\sqrt{NT})$ convergence with only $O(\sqrt{NT}\log(\frac{T}{N}))$ communication rounds.
@InProceedings{pmlr-v97-yu19c,title = {On the Computation and Communication Complexity of Parallel {SGD} with Dynamic Batch Sizes for Stochastic Non-Convex Optimization},author = {Yu, Hao and Jin, Rong},booktitle = {Proceedings of the 36th International Conference on Machine Learning},pages = {7174--7183},year = {2019},editor = {Chaudhuri, Kamalika and Salakhutdinov, Ruslan},volume = {97},series = {Proceedings of Machine Learning Research},address = {Long Beach, California, USA},month = {09--15 Jun},publisher = {PMLR},pdf = {http://proceedings.mlr.press/v97/yu19c/yu19c.pdf},url = {http://proceedings.mlr.press/v97/yu19c.html},abstract = {For SGD based distributed stochastic optimization, computation complexity, measured by the convergence rate in terms of the number of stochastic gradient calls, and communication complexity, measured by the number of inter-node communication rounds, are two most important performance metrics. The classical data-parallel implementation of SGD over N workers can achieve linear speedup of its convergence rate but incurs an inter-node communication round at each batch. We study the benefit of using dynamically increasing batch sizes in parallel SGD for stochastic non-convex optimization by charactering the attained convergence rate and the required number of communication rounds. We show that for stochastic non-convex optimization under the P-L condition, the classical data-parallel SGD with exponentially increasing batch sizes can achieve the fastest known $O(1/(NT))$ convergence with linear speedup using only $\log(T)$ communication rounds. For general stochastic non-convex optimization, we propose a Catalyst-like algorithm to achieve the fastest known $O(1/\sqrt{NT})$ convergence with only $O(\sqrt{NT}\log(\frac{T}{N}))$ communication rounds.}}
%0 Conference Paper%T On the Computation and Communication Complexity of Parallel SGD with Dynamic Batch Sizes for Stochastic Non-Convex Optimization%A Hao Yu%A Rong Jin%B Proceedings of the 36th International Conference on Machine Learning%C Proceedings of Machine Learning Research%D 2019%E Kamalika Chaudhuri%E Ruslan Salakhutdinov%F pmlr-v97-yu19c%I PMLR%J Proceedings of Machine Learning Research%P 7174--7183%U http://proceedings.mlr.press%V 97%W PMLR%X For SGD based distributed stochastic optimization, computation complexity, measured by the convergence rate in terms of the number of stochastic gradient calls, and communication complexity, measured by the number of inter-node communication rounds, are two most important performance metrics. The classical data-parallel implementation of SGD over N workers can achieve linear speedup of its convergence rate but incurs an inter-node communication round at each batch. We study the benefit of using dynamically increasing batch sizes in parallel SGD for stochastic non-convex optimization by charactering the attained convergence rate and the required number of communication rounds. We show that for stochastic non-convex optimization under the P-L condition, the classical data-parallel SGD with exponentially increasing batch sizes can achieve the fastest known $O(1/(NT))$ convergence with linear speedup using only $\log(T)$ communication rounds. For general stochastic non-convex optimization, we propose a Catalyst-like algorithm to achieve the fastest known $O(1/\sqrt{NT})$ convergence with only $O(\sqrt{NT}\log(\frac{T}{N}))$ communication rounds.
Yu, H. & Jin, R.. (2019). On the Computation and Communication Complexity of Parallel SGD with Dynamic Batch Sizes for Stochastic Non-Convex Optimization. Proceedings of the 36th International Conference on Machine Learning, in PMLR 97:7174-7183
This site last compiled Mon, 16 Sep 2019 16:05:04 +0000 |
Figure 1: Overview of the setting of unsupervised domain adaptation and its difference with the standard setting of supervised learning. In domain adaptation the source (training) domain is related to but different from the target (testing) domain. During training, the algorithm can only have access to labeled samples from source domain and unlabeled samples from target domain. The goal is to generalize on the target domain.
One of the backbone assumptions underpinning the generalization theory of supervised learning algorithms is that the test distribution should be the same as the training distribution. However in many real-world applications it is usually time-consuming or even infeasible to collect labeled data from all the possible scenarios where our learning system is going to be deployed. For example, consider a typical application of vehicle counting, where we would like to count how many cars are there in a given image captured by the camera. There are over 200 cameras with different calibrations, perspectives, lighting conditions, etc. at different locations in Manhattan. In this case, it is very costly to collect labeled data of images from all the cameras. Ideally, we would collect labeled images for a subset of the cameras and still be able to train a counting system that would work well for all cameras.
Domain adaptation deals with the setting where we only have access to labeled data from the training distribution (a.k.a., source domain) and unlabeled data from the testing distribution (a.k.a., target domain). The setting is complicated by the fact that the source domain can be different from the target domain — just like the above example where different images taken from different cameras usually have different pixel distributions due to different perspectives, lighting, calibrations, etc. The goal of an adaptation algorithm is then to generalize to the target domain without seeing labeled samples from it.
In this blog post, we will first review a common technique to achieve this goal based on the idea of finding a
domain-invariant representation. Then we will construct a simple example to show that this technique alone does not necessarily lead to good generalization on the target domain. To understand the failure mode, we give a generalization upper bound that decomposes into terms measuring the difference in input and label distributions between the source and target domains. Crucially, this bound allows us to provide a sufficient condition for good generalizations on the target domain. We also complement the generalization upper bound with an information-theoretic lower bound to characterize the trade-off in learning domain-invariant representations. Intuitively, this result says that when the marginal label distributions differ across domains, one cannot hope to simultaneously minimize both source and target errors by learning invariant representations; this provides a necessary condition for the success of methods based on learning invariant representations. All the material presented here is based on our recent work published at ICML 2019. Adaptation by Learning Invariant Representations
The central idea behind learning invariant representations is quite simple and intuitive: we want to find a representations that is insensitive to the domain shift while still capturing rich information for the target task. Such a representation would allow us to generalize to the target domain by only training with data from the source domain. The pipeline for learning domain invariant representations is illustrated in Figure 3.
Note that in the framework above we can use different transformation functions \(g_S/g_T\) on the source/target domain to align the distributions. This powerful framework is also very flexible: by using different measures to align the feature distributions, we recover several of the existing approaches, e.g., DANN (Ganin et al.’ 15), DAN (Long et al.’ 15) and WDGRL (Shen et al.’ 18).
A theoretical justification for the above framework is the following generalization bound by Ben-David et al.’ 10: Let \(\mathcal{H}\) be a hypothesis class and \(\mathcal{D}_S/\mathcal{D}_T\) be the marginal data distributions of source/target domains, respectively. For any \(h\in\mathcal{H}\), the following generalization bound holds: $$\varepsilon_T(h) \leq \varepsilon_S(h) + d(\mathcal{D}_S, \mathcal{D}_T) + \lambda^*,$$ where \(\lambda^* = \inf_{h\in\mathcal{H}} \varepsilon_S(h) + \varepsilon_T(h)\) is the optimal joint error achievable on both domains. At a colloquial level, the above generalization bound shows that the target risk could essentially be bounded by three terms:
The source risk (the first term in the bound). The distance between the marginal data distributions of the source and target domains (the second term in the bound). The optimal joint error achievable on both domains (the third term in the bound).
The interpretation of the bound is as follows. If there exists a hypothesis that works well on both domains, then in order to minimize the target risk, one should choose a hypothesis that minimizes the source risk while at the same time aligning the source and target data distributions.
A Counter-example
The above framework for domain adaptation has generated a surge of interest in recent years and we have seen many interesting variants and applications based on the general idea of learning domain-invariant representations. Yet it is not clear whether such methods are guaranteed to succeed when the following conditions are met:
The composition \(h\circ g\) achieves perfect classification/regression on the source domain. The transformation \(g: \mathcal{X}\to\mathcal{Z}\) perfectly aligns source and target domains in the feature space \(\mathcal{Z}\).
Since we can only train with labeled data from the source domain, ideally we would hope that when the above two conditions are met, the composition function \(h\circ g\) also achieves a small risk on the target domain because these two domains are close to each other in the feature space. Perhaps somewhat surprisingly, this is not the case as we demonstrate from the following simple example illustrated in Figure 4.
Consider an adaptation problem where we have input space and feature space \(\mathcal{X} = \mathcal{Z} = \mathbb{R}\) with source domain \(\mathcal{D}_S = U(-1,0)\) and target domain \(\mathcal{D}_T = U(1,2)\), respectively, where we use \(U(a,b)\) to mean a uniform distribution in the interval \((a, b)\). In this example, the two domains are so far away from each other that their supports are disjoint! Now, let’s try to align them so that they are closer to each other. We can do this by shifting the source domain to the right by one unit and then shifting the target domain to the left by one unit.
As shown in Figure 4, after adaptation both domains have distribution \(U(0, 1)\), i.e., they are perfectly aligned by our simple translation transformation. However, due to our construction, now the labels are flipped between the two domains: for every \(x\in (0, 1)\), exactly one of the domains has label 1 and the other has label 0. This implies that if a hypothesis achieves perfect classification on the source domain, it will also incur the maximum risk of 1 on the target domain! In fact, in this case we have \(\varepsilon_S(h) + \varepsilon_T(h) = 1\) after adaptation for any classifier \(h\). As a comparison, before adaptation, a simple interval hypothesis \(h^*(x) = 1\) iff \(x\in (-1/2, 3/2)\) attains perfect classification on both domains.
A Generalization Upper Bound on the Target Error
So what insights can we gain from the previous counter-example? Why do we incur a large target error despite perfectly aligning the marginal distributions of the two domains and minimizing the source error? Does this contradict Ben-David et al.’s generalization bound?
The caveat here is that while the distance between the two domains becomes 0 after the adaptation, the optimal joint error on both domains becomes large. In the counter-example above, this means that after adaptation \(\lambda^{*} = 1\), which further implies \(\varepsilon_T(h) = 1\) if \(\varepsilon_S(h) = 0\). Intuitively, from Figure 4 we can see that the labeling functions of the two domains are “maximally different” from each other after adaptation, but during adaptation we are only aligning the marginal distributions in the feature space. Since the optimal joint error \(\lambda^*\) is often unknown and intractable to compute, could we construct a generalization upper bound that is free of the constant \(\lambda^*\) and takes into account the conditional shift?
Here is an informal description of what we show in our paper: Let \(f_S\) and \(f_T\) be the labeling functions of the source and target domains. Then for any hypothesis class \(\mathcal{H}\) and any \(h\in\mathcal{H}\), the following inequality holds: $$\varepsilon_T(h) \leq \varepsilon_S(h) + d(\mathcal{D}_S, \mathcal{D}_T) + \min\{\mathbb{E}_S[|f_S – f_T|], \mathbb{E}_T[|f_S – f_T|]\}.$$
Roughly speaking, the above bound gives a decomposition of the difference of errors between source and target domains. Again, the second term on the RHS measures the difference between the marginal data distributions. But, in place of the optimal joint error term, the third term now measures the discrepancy between the labeling functions of these two domains. Hence, this bound says that just aligning the marginal data distributions is not sufficient for adaptation, we also need to ensure that the label functions (conditional distributions) are close to each other after adaptation.
An Information-Theoretic Lower Bound on the Joint Error
In the counter-example above, we demonstrated that aligning the marginal distributions and achieving a small source error is not sufficient to guarantee a small target error. But in this example, it is actually possible to find another feature transformation that jointly aligns both the marginal data distributions and the labeling functions. Specifically, let the feature transformation \(g(x) = \mathbb{I}_{x\leq 0}(x)(x+1) + \mathbb{I}_{x > 0}(x)(2-x)\). Then, it is straightforward to verify that the source and target domains perfectly align with each other after adaptation. Furthermore, we also have \(\varepsilon_T(h) = 0\) if \(\varepsilon_S(h) = 0\).
Consequently, it is natural to wonder whether it is always possible to find a feature transformation and a hypothesis to align the marginal data distributions and minimize the source error so that the composite function of these two also achieves a small target error? Quite surprisingly, we show that this is not always possible. In fact, finding a feature transformation to align the marginal distributions can provably increase the joint error on both domains. With this transformation, minimizing the source error will only lead to increasing the target error!
More formally, let \(\mathcal{D}_S^Y/\mathcal{D}_T^Y\) be the marginal label distribution of the source/target domain. For any feature transformation \(g: X\to Z\), let \(\mathcal{D}_S^Z/\mathcal{D}_T^Z\) be the resulting feature distribution by applying \(g(\cdot)\) to \(\mathcal{D}_S/\mathcal{D}_T\) respectively. Furthermore, define \(d_{\text{JS}}(\cdot, \cdot)\) to be the Jensen-Shannon distance between a pair of distributions. Then, for any hypothesis \(h: Z\to\{0, 1\}\), if \(d_{\text{JS}}(\mathcal{D}_S^Y, \mathcal{D}_T^Y) \geq d_{\text{JS}}(\mathcal{D}_S^Z, \mathcal{D}_T^Z)\), the following inequality holds: $$\varepsilon_S(h\circ g) + \varepsilon_T(h\circ g)\geq \frac{1}{2}\left(d_{\text{JS}}(\mathcal{D}_S^Y, \mathcal{D}_T^Y) – d_{\text{JS}}(\mathcal{D}_S^Z, \mathcal{D}_T^Z)\right)^2.$$
Let’s parse the above lower bound step by step. The LHS corresponds to the joint error achievable by the composite function \(h\circ g\) on both the source and the target domains. The RHS contains the distance between the marginal label distributions and the distance between the feature distributions. Hence, when the marginal label distributions \(\mathcal{D}_S^Y/\mathcal{D}_T^Y\) differ between two domains, i.e., \(d_{\text{JS}}(\mathcal{D}_S^Y, \mathcal{D}_T^Y) > 0\), aligning the marginal data distributions by learning \(g(\cdot)\) will only increase the lower bound. In particular, for domain-invariant representations where \(d_{\text{JS}}(\mathcal{D}_S^Z, \mathcal{D}_T^Z) = 0\), the lower bound attains its maximum value of \(\frac{1}{2}d^2_{\text{JS}}(\mathcal{D}_S^Y, \mathcal{D}_T^Y)\). Since in domain adaptation we only have access to labeled data from the source domain, minimizing the source error will only lead to an increase of the target error. In a nutshell, this lower bound can be understood as an uncertainty principle: when the marginal label distributions differ across domains, one has to incur large error in either the source domain or the target domain when using domain-invariant representations.
Empirical Validations
One implication made by our lower bound is that when two domains have different marginal label distributions, minimizing the source error while aligning the two domains can lead to increased target error. To verify this, let us consider the task of digit classification on the MNIST, SVHN and USPS datasets. The label distributions of these three datasets are shown in Figure 5.
From Figure 5, it is clear to see that these three datasets have quite different label distributions. Now let’s use DANN (Ganin et al., 2015) to classify on the target domain by learning a domain invariant representation while training to minimize error on the source domain.
We plot four adaptation trajectories for DANN in Figure 6. Across the four adaptation tasks, we can observe the following pattern: the test domain accuracy rapidly grows within the first 10 iterations before gradually decreasing from its peak, despite consistently increasing source training accuracy. These phase transitions can be verified from the negative slopes of the least squares fit of the adaptation curves (dashed lines in Figure 6). The above experimental results are consistent with our theoretical findings: over-training on the source task can indeed hurt generalization to the target domain when the label distributions differ.
Future Work
Note that the failure mode in the above counter-example is due to the increase of the distance between the labeling functions during adaptation. One interesting direction for future work is then to characterize what properties the feature transformation function should have in order to decrease the shift between labeling functions. Of course domain adaptation would not be possible without proper assumptions on the underlying source/target domains. It would be nice to establish some realistic assumptions under which we can develop effective adaptation algorithms that align both the marginal distributions and the labeling functions. Feel free to get in touch if you’d like to talk more!
DISCLAIMER: All opinions expressed in this post are those of the author and do not represent the views of CMU. |
I was doing this question:
If a parallel plate capacitor made of square plates of side L and separation D is connected to voltage source V. And then a dielectric slab of width D is introduced into the space between the plates to a distance x, then the change in its potential energy with respect to x is
$${\displaystyle dU = + (K-1)\epsilon_{0} \frac{V^{2}L}{2D}dx}$$
and since this is stored in capacitor as the work done by the force due to the charges on the plate on the slab, so the force
$${\displaystyle \vec{F} = - \frac{dU}{dx} \hat{x} }$$
Which gives,
$${\displaystyle \vec{F} = - (K-1)\epsilon_{0} \frac{V^{2}L}{2D}\hat{x} }$$
But this is incorrect because when dielectric is partially inserted in capacitor the charges on the plates will try to pull it in the space of capacitor and not outward.
However if the voltage source is disconnected, then the change in potential energy with respect to x will be
$${\displaystyle dU = - (K-1)\epsilon_{0} \frac{V^{2}L}{2D}dx}$$
Which gives,
$${\displaystyle \vec{F} = (K-1)\epsilon_{0} \frac{V^{2}L}{2D}\hat{x} }$$
Which corresponds to correct result.
It's written in the question that in the first case potential difference remains constant across the plates and in the second case, the charges on the plates remain constant.
But I can't understand why is it so? And even if I agree on that then why the two cases gave completely different results? |
I thought of this weird reduction (chances that it is wrong are high :-). Idea: reduction from Hamiltonian path on grid graphs with degree $\leq 3$; each node of the planar graph can be shifted in such a way that every "row" ($y$ value) and every "column" ($x$ value) contains at most one node. The graph can be scaled and each node can be replaced by a square gadget with many points; horizontal links between the gadgets (the edges of the original graph) are made using pairs of points on distinct rows, vertical links using pair of points on distinct columns. Node traversals are forced using the "many points" of the square gadgets.
The node gadget is represented in the following figure:
It has 3 "interface points" $[W,N,E]$ (on distinct columns/rows), and an inner border of $C \times C$ points. A polyline that traverses the gadget from one interface point to another can have a number of corners that is proportional to $C$ (three traversals of the gadget are shown in the figure), in particular the number of corner points is between $2C$ and $2C+2$ (the total number of points of the gadget is $C \times C - 4 + 6$).The gadget can be rotated to get other interface point combinations ($[N,E,S]$, $[E,S,W]$, $[S,W,N]$).
Now we can shift the planar grid graph in such a way that for every pair of nodes $(x_1,y_1), (x_2,y_2)$, $x_1 \neq x_2$ and $y1 \neq y_2$. See the following figure of a simple $4 \times 3$ grid. Next, we can scale the graph and replace each node with the gadget above. At this stage each gadget is "isolated": a polyline cannot go from one gadget to another.
Now we can simulate the edges of the original graph using pairs of points on the bottom or on the right, each pair on a separate row or on a separate column; see the following figure for two adjacent nodes linked horizontally (in a new bottom row two points are added one on the same column of interface point $E$ of the first gadget, the other on the same column of interface point $W$ of the second gadget).
On every gadget there can be max $4 + 2C$ corner points (1 generated by the enter interface point, 1 generated by the exit interface point, 2 generated by the extra turn on straight traversals and 2C on the inner zig-zag), the points used for the edges can generate max $2e$ corner points.
Suppose that the original graph has $n$ nodes and $e$ edges, if we pick $C > (4n+2e)$, and $k = 2Cn$ as the number of corner points that must be used, then we force the "hidden" polygon of the puzzle to traverse every gadget; but every gadget can be entered/exited exactly once (through a pair of interface cells); so the problem has a solution if and only if the original grid graph has an Hamiltonian path. |
Context:
Griffith's book on Quantum Mechanics (QM), in Section 2.3.1, tries to solve for the stationary states $\psi(x)$ of a harmonic oscillator by solving the Time-Independent Schrodinger Equation (TISE),
$$\frac{1}{2m}[p^{2}+(m\omega x)^{2}]\psi=E\psi,$$
using the method of ladder operators. The ladder operators’ method starts from a
postulated definition of $a_{+}$ and $a_{-}$ (eq. 2.47):
$$ a_{\pm}\equiv \frac{1}{\sqrt{2\hbar\omega m}} (\mp i p+m\omega x),$$
which then was shown to work in terms of factorizing the Hamiltonian $H=[p^{2}+(m\omega x)^{2}]/(2m)$, and thus used to raise/lower energy in discrete steps. The discussion comes to a head in p.46 and footnote 21, where it is concluded that “
we can construct all the stationary states” by repeated application of this operator $a_{+}$ starting from the lowest energy level (rung) on the ladder, $E_{0}$, and that such ladder is unique because two ladders with the same step size ($\hbar \omega$) and common first rungs will completely overlap and therefore be identical.
However, from a logical point of you, there might be a little issue for the reader up to this point as it was not proven (or discussed) that such operators ($a_{+}$,$a_{-}$) and their ensuing ladder (with $\pm \hbar \omega$ step size) were the only possible ones that could represent the Hamiltonian (i.e. no uniqueness was discussed), and therefore one might start to imagine an example of another set of possible operators that produce steps of half the size of the ones discussed (so, $\hbar \omega/2$, instead of $\hbar \omega$), and produce a different ladder, that will still overlap with the original ladder, even for common bottom rung ($E_{0}$), but with twice the number of rungs. By extension, an infinite number of such ladders of step size equal to $\hbar \omega/n$, where $n$ is integer, might be imagined in this sense, and they would still not conflict each other. Such uniqueness is not rigorously discussed.
The issue:
I think that the key to concluding that we only have one unique ladder is to:
first prove that all operators would produce ladders with the same bottom rung; andthen to show that the original operators ($a_{\pm}$) actually give the smallestenergy step size (resolution) among all admissible operators/ladders (and thus be our only choice for operators, because other operators would be simply just integer multiples of them). But there is an issue here, as described below. Point (1) is easy to prove: any new operators like $a^{2}_{-}$, $a^{2}_{+}$, or indeed any generalization thereof ($a^{m}_{-}$, $a^{m}_{+}$, for $m\in \mathbb{Z}$), share the same bottom rung with the original ladder operators ($a_{-}$ and $a_{+}$). I can prove this rigorously as follows: say that we have $m=2$, and we wish to find its lowest rung state (call it $\bar{\psi_{0}}$ for this case, to distinguish it from the original case $\psi_{0}$), then we find it by putting:
\begin{eqnarray} a_{-}a_{-}\bar{\psi_{0}}&=&0 \nonumber\\ \Rightarrow a_{+}a_{-}(a_{-}\bar{\psi_{0}})&=&0 \nonumber\\ \Rightarrow (a_{-}a_{+}-1)(a_{-}\bar{\psi_{0}})&=&0 \nonumber\\ \Rightarrow a_{-}\bar{\psi_{0}}&=&a_{-}\underbrace{(a_{+}a_{-}\bar{\psi_{0}})}_{=(0)\bar{\psi_{0}}=0} \nonumber\\ \Rightarrow a_{-}\bar{\psi_{0}}&=& 0 \nonumber \end{eqnarray} \begin{equation} \Rightarrow \boxed{\bar{\psi_{0}} \equiv \psi_{0}}, \nonumber \end{equation}
which proves that all such ladders definitely share the same rung (I have used the fact that $a_{+}a_{-}\psi_{n}=n\psi_{n}$ in the above reduction to zero). Similar proofs can be done for higher $m$.
However,
point (2) is proving to be more subtle and addresses the possibility of constructing new operators that are not of the form $a^{m}_{-}$, $a^{m}_{+}$, for $m\in \mathbb{Z}$. We note that the entire choice of operators $a_{-}$, $a_{+}$ as equal to: $(\text{constant})[\pm\hbar D + m\omega x]$, where $D\equiv\frac{d}{dx}$, was originally based simply on them producing a second-order derivative ($D^{2}\equiv\frac{d^{2}}{dx^{2}}$) when multiplied together, to match the Time-Independent Schrodinger Equation (TISE). Since TISE is second-order, the choice of $a_{\pm}$ was made to have each operator with a first-order derivative operator ($D$), which is a natural choice. Then it was proved that their product is of the form:
$$b_{-}b_{+}\ \ \ \varpropto \ \ \ \left[\underbrace{-\hbar D^{2}+(m\omega x)^{2}}_{=2 m H} + \text{some constant} \right],$$
(here I chose letter $b_{\pm}$ to generalize the discussion and to later distinguish it from the original operator $a_{\pm}$) which is later written in $H$ in the following form
$$ H= \hbar\omega\left( b_{-}b_{+} - \phi \right)\ \ \ ;\ \ \ H= \hbar\omega\left( b_{+}b_{-} + \phi \right),$$
where $\phi$ is some constant. And this will give us later
$$ \text{Commutator}[b_{-},b_{+}]=2\phi, $$
which later gives us the key conclusions that energy jumps in steps as
$$ \boxed{H(b_{-}\psi)=[E-(2\phi)\hbar\omega]\ \psi \ \ \ ;\ \ \ H(b_{+}\psi)=[E+(2\phi)\hbar\omega]\ \psi}. $$
Now, yes, clearly, if we choose operators $b_{\pm}$ like before to be some higher integer order of the original operators $a_{\pm}$ (such as $a^{m}_{\pm}$, with $m\in\mathbb{Z}$), then clearly they will have LARGER steps on the ladder, and therefore $a_{\pm}$ are the operators with the finest allowed resolution of steps (smallest allowed step). And since we have proved (above) that the first rung is shared by all such operators ($a^{m}_{\pm}$, with $m\in\mathbb{Z}$), then the original $a_{\pm}$ operators are unquestionably brilliant and unique.
But, what if we chose operators $b_{\pm}$ that are not of the form $a^{m}_{\pm}$, with $m\in\mathbb{Z}$ ? What if I chose fractional derivatives, for example, such as $D^{1/2}$ or $D^{3/2}$ (which are also formal operators in applied mathematical analysis--- e.g. see wiki page), which when multiplied will still give second-order ($D^{2}$), and therefore may factorize $H$ and represent the TISE equation? In fact, their application to this problem might be especially convenient because we have the $x$ dependence of the form $x^{k}$, which lends itself relatively easily to fractional derivative operators.
Griffith's text doesn't discuss whether this is doable or not, and therefore leaves the door open to the reader's imagination (or unease) about uniqueness here. For example, what if we chose say:
$$ b_{-} \ \varpropto \ \ (\hbar D)^{1/2} + (m\omega x)^{3/2}\ \ \ ;\ \ \ b_{+} \ \varpropto\ \ - (\hbar D)^{3/2} + (m\omega x)^{1/2}$$
or some other similar definitions that, when multiplied, could lead us (with the help of Gamma function identities that usually result from fractional derivatives) again to the sought form of: $$b_{-}b_{+}\ \ \ \varpropto \ \ \ \left[\underbrace{-\hbar D^{2}+(m\omega x)^{2}}_{= 2 m H} + \text{some constant} \right],$$ (assuming we could make wise algebraic choices to produce the constant in this expression), and we could then find $$ H= \hbar\omega\left( b_{-}b_{+} - \phi \right)\ \ \ ; \ \ \ H= \hbar\omega\left( b_{+}b_{-} + \phi \right),$$
with some new $\phi$ that is less than $1/2$ (that is $\phi<0.5$), and therefore produce a new ladder with "legal" energy steps that are SMALLER than $\hbar\omega$ (namely, energy step size $\boxed{2\phi\hbar\omega}$) ?
Any help in settling this idea of uniqueness would be appreciated. |
The rule of thumb I know for the error on a count for random sampling is that it goes as the square root of the count. For instance, if I observe a radioactive source for $1$ minute and measure $100$ decay events, a reasonable uncertainty on the count is $100\pm\sqrt{100} = 100 \pm 10$.
I have a different sampling problem, though. To put it simply, imagine I have a bag filled with a large number of balls which are each either red or blue. I'm interested in measuring the fraction of red balls in the bag. I draw an unbiased sample of $N_{\rm balls}$ balls and count the red ones, finding $N_{\rm red}$. My estimate for the red fraction $f_{\rm red}=N_{\rm red}/N_{\rm balls}$, but what uncertainty should I associate with this? I'm tempted to say $f_{\rm red}=\frac{N_{\rm red}}{N_{\rm balls}}\pm\frac{\sqrt{N_{\rm red}}}{N_{\rm balls}}$, but I find this unsatisfying. This is because it makes no distinction in the case where I draw $N_{\rm red}=0$; if this occurs with $N_{\rm balls}=1$, then my measurement is not especially constraining and should have a large uncertainty, but if I found $N_{\rm red}=0$ with $N_{\rm balls}=10^5$ I would have a high confidence that $f_{\rm red}\sim0$. In the particular experiment I have in mind, repeated draws are not an option.
Could someone point me to a correct handling of the statistics for such an experiment?
I appreciate that the physics content here is fairly minimal, so if the community prefers that this move to CV.SE or M.SE that's fair enough; I put it here because the context is a physical experiment and I would prefer answers in terms familiar to a physicist. |
We now proceed to an obvious extension of the one-sample procedures: paired samples. This chapter is going to be much shorter as the procedures are very similar to the previously discussed one-sample tests.
Paired observations are pretty common in applications as each unit acts as its own control. For instance, before treatment vs. after, left side vs. right side, identical twin A vs. identical twin B, etc. Another way this comes about is via
matching (a common experimental design technique): the researcher matches each unit with another unit that shares relevant (or maybe irrelevant) characteristics - age, education, height, etc. and the two units differ on the treatment. Then, use (some of) our one-sample methods on the
differences between the paired observation values. By analogy, consider the paired t-test, which looks very much like a one-sample t-test. Wilcoxon Signed Rank Test
Suppose $12$ sets of identical twins were given a psychological test to measure the amount of aggressiveness in each person's personality. We're interested in whether the firstborn tends to be more aggressive than the other. A higher score is indicative of higher aggressiveness. We hypothesize
\[\begin{aligned} &H_0: \text{the firstborn does not tend to be more aggressive (no difference is also okay)} \\ \text{vs. } &H_1: \text{the firstborn twin tends to be more aggressive} \end{aligned}\]
Here's the data we need for the signed rank test on the differences $d_i$:
Firstborn ($x_i)$ Secondborn ($y_i$) Difference ($d_i$) Rank Signed rank 86 88 2 4 4 71 77 6 8 8 77 76 -1 2.5 -2.5 68 64 -4 5 -5 91 96 5 6.5 6.5 72 72 0 (1) $-$ 77 65 -12 11 -11 91 90 -1 2.5 -2.5 70 65 -5 6.5 -6.5 71 80 9 10 10 88 81 -7 9 -9 87 72 -15 12 -12
We use mid-ranks for the ties, then use Sprent and Smeeton "device" for handling the deviation of $0$. Our test statistic:
\[S_+ = 4 + 8 + 6.5 + 10 = 28.5\]
Now we can get the normal approximation and score representation of the test statistic:
\[\text{Normal approximation} = \frac{S_+ - \frac{1}{4}n(n+1)}{\sqrt{\frac{1}{24}n(n+1)(2n+1)}} = \frac{28.5 - \frac{1}{4} \times 12 \times 13}{\sqrt{\frac{1}{24} \times 12 \times 13 \times 25}} = -0.824\]
\[\text{Score representation} = \frac{S_+ - \frac{1}{2} \sum\limits_{i=1}^n {|S_i|}}{\frac{1}{2}\sqrt{\sum\limits_{i=1}^n {S_i^2}}} = \frac{28.5 - \frac{1}{2} \times 77}{\frac{1}{2}\sqrt{648}} = -0.786\]
For the score representation, we see a slight difference when we compare to: $\sum\limits_{i=1}^{12} i = 78$, and $\sum\limits_{i=1}^{12} i^2 = 650$. The difference is caused by the $-$ rank in the case where $d_i = 0$. In either case, there's really no evidence that the firstborn twin is more aggressive.
Notes Within a pair, observations may not be independent (obviously, as almost by construction sometimes they won't be), but the pairsthemselves should be. The typical $H_0$ here would be that the median of the differencesis $0$. If the differences are assumed to have a symmetric distribution(about $0$), then the Wilcoxon approach is suitable. Note then that the individual unit measurements do notneed to be assumed symmetric. You need to think about whether this is realistic for your given situation. The alternative therefore refers to a shift in centralityof the differences. You need to think about whether this is an interesting, relevant, important question. McNemar's Test
The test is a less obvious use of, or a modification to the sign test. Here's example 5.4 in the book. We have records on all attempts of two rock climbs, successful or not. For the $108$ who tried both:
First climb success First climb failure Second climb success 73 14* Second climb failure 9* 12
Is there evidence that one climb is harder? The only ones that had information
for this question are the ones that succeeded in one but failed in another. The people who succeeded / failed in both are essentially "ties".
We can frame this as a Binomial setup. Think of success $\rightarrow$ failure as a "+" for the first climb, and failure $\rightarrow$ success as a "-" for the first climb. This puts us under a sign test situation. Under $H_0:$ climbs are equally difficult, and the probability of a "+" is the same as the probability of a "-" $\Rightarrow Bin(23, 0.5)$ with a "+". The p-value is 0.405, so there seems to be no difference in the difficulty of the climbs.
This version of the sign test is called
McNemar's test. In general, we have pairs of data $(x_i, y_i)$ $i=1 \cdots m$ where $X_i$ and $Y_i$ take values $0$ and $1$ only. There are $4$ patterns of outcome: $(0, 0), (0, 1), (1, 0), (1, 1)$. The paired data are summarized in a $2 \times 2$
contingency table showing the classifications of $Y_i$ and $X_i$:
$Y_i = 0$ $Y_i = 1$ $X_i = 0$ $a\text{ (\# of (0, 0) pairs)}$ $b\text{ (\# of (0, 1) pairs)}$ $X_i = 1$ $c\text{ (\# of (1, 0) pairs)}$ $d\text{ (\# of (1, 1) pairs)}$ Assumptions Pairs are mutually independent. Two categories for each outcome ($X_i = 0, 1,/,Y_i = 0, 1$) The difference $P(X_i = 0, Y_i = 1) - P(X_i = 1, Y_i = 0)$ is negative for all $i$, 0 for all $i$, or positive for all $i$.
The null hypothesis can be formulated in various equivalent ways:
\[\begin{aligned} &H_0: P(X_i = 0, Y_i = 1) = P(X_i = 1, Y_i = 0) && \text{for all } i \\ \text{vs. } &H_1: P(X_i = 0, Y_i = 1) \neq P(X_i = 1, Y_i = 0) && \text{for all } i \end{aligned}\]
or
\[\begin{aligned} &H_0: P(X_i = 0) = P(Y_i = 0) && \text{for all } i \\ \text{vs. } &H_1: P(X_i = 0) \neq P(Y_i = 0) && \text{for all } i \end{aligned}\]
or
\[\begin{aligned} &H_0: P(X_i = 1) = P(Y_i = 1) && \text{for all } i \\ \text{vs. } &H_1: P(X_i = 1) \neq P(Y_i = 1) && \text{for all } i \end{aligned}\]
The first one, in terms of our example, is saying $P(\text{success} \rightarrow \text{failure}) = P(\text{failure} \rightarrow \text{success})$. The second one is saying $P(\text{failure on first climb}) = P(\text{failure on second climb})$.
Our test statistic - if $b$ and $c$ are reasonably small, take $T_2 = b$. Under $H_0$, $b \sim Bin(b+c, 0.5)$, which is our usual sign test. Or for larger $b, c$:
\[T_1 = \frac{(b-c)^2}{b+c}\]
Note that it does
not depend on $a$ and $d$. Under $H_0$, $T_1$ is approximately $\chi^2_{(1)}$. To derive $T_1$, let's call $b + c = n$. If $n$ is large enough, we can use a normal approximation to the Binomial:
\[\frac{T_2 - \frac{1}{2}n}{\sqrt{\frac{1}{2} \cdot \frac{1}{2}n}} = \frac{b - \frac{1}{2}(b+c)}{\frac{1}{4}(b+c)} = \frac{b-c}{\sqrt{b+c}} = \mathbf{Z} \approx N(0, 1)\]
\[T_1 = \mathbf{Z}^2 \Rightarrow T_1 \approx \chi^2_{(1)}\]
And that's it! Paired-sample tests really aren't that different from one-sample tests. Now we're ready for some real problems: two independent samples for example. |
Nowadays most machine learning (ML) models predict labels from features. In classification tasks, an ML model predicts a categorical value and in regression tasks, an ML model predicts a real value. These ML models thus require a large amount of feature-label pairs. While in practice it is not hard to obtain features, it is often costly to obtain labels because this requires human labor.
Can we do more? Can we learn a model
without too many feature-label pairs? Think of human learning: as humans, we do not need 1,000 cat images and labels “cat” to learn what is a cat or to differentiate cats from dogs. We can also learn the concept through comparisons. When we see a cat/dog, we can compare it with cats we have seen to decide whether we should label it “cat”.
Our recent papers (1,2) focus on using comparisons to build ML models. The idea of using comparisons is based on a classical psychological observation: It is easier for people to compare between items than evaluate each item alone. For example, what is the age of the man in the image?
Not very easy, right? Is he 20, 30 or 40? We can probably say he is not very old, but it is just hard to be very accurate on the exact age. Now, which person in the two images is older?
Now based on the wrinkles and silver hair, you can probably quickly judge that the second man is older.
This phenomenon is not only present for this task, but also in many other real-world applications. For example, to diagnose patients, it is usually more difficult to directly label each patient with a kind of disease by experimental tests, but easier to compare the physical conditions of two patients. In material synthesis, measuring the characteristics of a material usually requires expensive tests, but comparisons are relatively easy through simulations. For movie ratings, it is often hard for us to give scores for a specific movie, but easier to pick our favorite among a list of movies.
So how can we build ML models using comparisons? Here we describe an approach that uses comparisons to do inferences on the unlabeled samples and feed inferred labels into existing models. Below we will look at two ways for such inference, for classification and regression respectively.
The Setup
As described above, our setup starts with a set of unlabeled features \(x_1, x_2,…, x_n\), drawn independently and identically distributed (i.i.d.) from a feature distribution \(X\sim P_X\). Let the data dimension be \(d\). Our goal is to learn a function \(f: \mathbb{R}^d \rightarrow \mathcal{Y}\), where \(\mathcal{Y}\) is the label space. For example, for binary classification \(\mathcal{Y}=\{1, -1\}\), and for regression \(\mathcal{Y}=\mathbb{R}\).
We assume we can query either direct labels or pairwise comparisons. The direct label \(Y(x)\) is a (possibly noisy) version of \(f(x)\). The comparison \(Z\) is based on a pair of samples \(x,x’\) and indicates which one of \(x,x’\) can possibly have a larger \(f\) value. For binary classification, this means \(Z\) indicates the more positive sample; for regression, \(Z\) indicates the larger target (e.g., the older people of the pair of images). Our goal is to use as few direct label queries as possible.
Our high-level strategy is to obtain a fully labeled sample pool \(\hat{y}_1,…,\hat{y}_n\), where \(\hat{y}_i\) are either inferred or directly labeled, to feed into a supervised learning algorithm. We will show how such inference can happen, and how the querying process can neatly combine with the learning algorithm for a better performance.
Our Building Block: Ranking
Before we go to the algorithms, we first introduce our workhorse: Ranking from pairwise comparisons. We organize the comparisons to induce a ranking over all the samples. After that, we can do efficient inference with a very small amount of direct labels.
There is a vast amount of literature on ranking from pairwise comparisons, based on different assumptions on the comparison matrix and desired properties. If we have perfect and consistent comparisons, we can use QuickSort (or HeapSort, InsertSort) to rank all \(n\) samples with \(O(n\log n)\) comparisons. If comparisons are noisy and inconsistent, things will be more complicated, but we can still obtain some meaningful rankings. We will not go into more details about ranking since it is out of the scope of this post; we refer interested readers to this survey for more papers on this topic.
Now let’s suppose we have a ranking over all items. We denote it as \(x_1\prec x_2\prec \cdots\prec x_n\), where \(x_i\prec x_j\) means we think \(f(x_i)\leq f(x_j)\). Note that the actual ranking induced by \(f\) might be different from \(x_1\prec x_2\prec \cdots\prec x_n\), as we can have errors in our comparisons.
Classification: Binary Search on Ranking
Now we consider the binary classification problem. If we have a perfect ranking with \begin{align*}f(x_1)\leq f(x_2)\leq \cdots\leq f(x_n),\end{align*} this means the first few samples have labels -1, and then the remaining samples have label +1. Given this specific structure, we would want to find the
changing point between negative and positive samples. How are we going to find it?
Binary search! Since the ranking is in order, we just need \(\log n\) direct label queries to figure out the changing point. Note that this has a specific meaning in the context of classification: in the standard supervised learning setting, we need at least \(d\) labels to learn a classifier in \(d\) dimension. Now with this ranking information at hand, we only need to find a threshold in a sequence, which is equivalent to learning a classifier in one dimension. Note that in general the comparison queries are cheaper, so our algorithm can save a lot of cost.
There are a few more things to note for classification. First is about ties: Suppose our task is to differentiate between cats and dogs. If we are given two cat images, it doesn’t really matter how we rank them since we only care about the threshold between positive and negative samples.
Secondly, we can combine our algorithm with active learning to save even more label cost. Many active learning algorithms ask about a batch of samples in each round, and we can use our binary search to label each batch. In more detail, we show in our paper the following theorem:
Theorem (Informal). Suppose each label is correct with probability \(1/2+c\), for a constant \(c\). Then an active learning algorithm would require \(\Omega(d\log(1/\varepsilon))\) direct labels to achieve an error rate of \(\varepsilon\). On the other hand, using binary search on ranking will require \(O(\log(\varepsilon))\) direct labels, and \(O(d\log(d/\varepsilon))\) comparisons. Regression: Isotonic Regression
If we are doing regression, we cannot hope to find a threshold in the ranking, since we need to predict a real number for each label. However, ranking can still help regression through isotonic regression. Given a ranked sequence\begin{align*} f(x_1)\leq f(x_2)\leq\cdots \leq f(x_n) \text{ and } y_i=f(x_i)+\varepsilon_i, \varepsilon_i\sim \mathcal{N}(0,1),\end{align*} the isotonic regression aims to find the solution of
\begin{align*} \min_{\hat{y}_i} & \sum_{i=1}^n (\hat{y}_i-y_i)^2\\ s.t.& \hat{y}_i\leq \hat{y}_{i+1}, \forall i=1,2,…,n-1. \end{align*} If we use \(y_i\) as our labels, the mean-squared error \(\frac{1}{n}\sum_{i=1}^m (y_i-f(x_i))^2\) will have an expectation of 1, since \(\varepsilon_i\sim \mathcal{N}(0,1)\). Isotonic regression enjoys \(m^{-2/3}\) statistical rate, which is diminishing as \(n\rightarrow \infty\). For a reference, see (Zhang, 2002).
The \(m^{-2/3}\) decays faster than the optimal rates of many non-parametric regression problems because it is dimension-independent. Non-parametric methods typically have an error rate of \(m^{-\frac{2}{d+2}}\) given \(m\) labels, the so-called curse of dimensionality (see Tsybakov’s book for an introduction to non-parametric regression). Since the rate of isotonic regression decays much faster than the non-parametric regression problems, we only need a fraction of labels for good accuracy. We leverage this property to design the following algorithm: suppose we only directly query \(m\) labels. While having a ranking over \(n\) points, we can infer the unlabeled samples by just using their nearest labeled points. That is, we query \(y_{t_1},…,y_{t_m}\) and get refined values \(\hat{y}_{t_1},…,\hat{y}_{t_m}\) using the above isotonic regression formulation, we label each point as \(\hat{y}_i=\hat{y}_{t_j}\), where \(i \in 1,…,n\) and \(t_j\) is \(i\)’s nearest neighbor in \(\{t_1,…,t_m\}\).
In our paper, we analyze this algorithm under the non-parametric regression setting. We have the following theorem:
Theorem (Informal). Suppose the underlying function \(f\) is Lipschitz. If we use \(m\) direct labels, any algorithm will incur an error of at least \(\Omega\left(m^{-\frac{2}{d+2}}\right)\). If we use isotonic regression with nearest neighbors, the error will be \(m^{-\frac{2}{3}}+n^{-\frac{2}{d}}\), where \(m\) is the number of direct labels, and \(n\) is the number of ranked points. This rate is optimal for any algorithm using \(m\) direct labels and \(n\) ranked points.
Note the MSE of non-parametric regression using only the labeled samples is \(\Theta(m^{-\frac{2}{d+2}})\) which is exponential in \(d\) and makes non-parametric regression impractical in high-dimensions. Focusing on the dependence on \(m\), our result improves the rate to \(m^{-2/3}\), which is no longer exponential. Therefore, using the ranking information we can avoid the curse of dimensionality.
Comparison in Practice
Now let’s test our algorithm in practice. Our task is to predict the ages of people in images, as aforementioned. We use the APPA-REAL dataset, with 7,113 images and associated ages. The dataset is suitable for comparisons because it contains both the biological age, as well as the apparent age estimated from human labelers. Suppose our goal is to predict the biological age, and we can simulate comparisons by comparing the apparent ages.
Our classification task is to judge whether a person is under or over 30 years old. We compare our method with a base-line active learning method which only uses label queries. Both methods use a linear SVM classifier ( features are extracted from the 128-dimension top layer of FaceNet, an unsupervised method to extract features from faces). The shades represent standard variation over 20 repeats of experiments. The plots show comparisons indeed reduce the number of label queries.
Our regression task is to predict the actual ages, and we compute the mean squared error (MSE) to evaluate different methods. Our label-only baselines are nearest neighbors(NN) methods with 5 or 10 neighbors(5-NN and 10-NN), and support vector regression(SVR). Our methods use 5-NN or 10-NN after we have inferred the labels via isotonic regression. We thus name our methods R\(^2\) 5-NN and R\(^2\) 10-NN. Again, the experiment shows comparisons can reduce the number of label queries.
Future Works
Of course, binary classification and regression are not the only settings where using comparison information can have a big impact. Using the rank-and-infer approach, we hope to extend these results to multi-class classification, optimization, and reinforcement learning. Feel free to get in touch if you want to learn more!
DISCLAIMER: All opinions expressed in this post are those of the author and do not represent the views of Carnegie Mellon University. |
Let $M$ be a smooth manifold of dimension $n$. Consider the space $V_k(M)$ of pairs $(x,\alpha)$ where $x \in M$ and $\alpha$ is a linear embedding $\mathbb{R}^k \hookrightarrow T_x M$ (or equivalently, $k$ independent vectors in $T_xM$). This is topologized as a subspace of $M \times \operatorname{Map}(\mathbb{R}^k, TM)$ and it fibers over $M$ with $(x,\alpha) \mapsto x$. The fiber is (homotopy equivalent to) the Stiefel manifold $V_k(\mathbb{R}^n) = O(n) / O(n-k)$.
A slight variation of this requires an oriented manifold $M$ and asks gives a subspace $V'_k(M) \subset V_k(M)$ where the $\alpha$ are required to preserve orientation. The fiber is then $SO(n) / SO(n-k)$. I'm most interested in the case $k = n$, if it makes any difference.
How is this construction called? What do we know about the resulting manifold (the total space of the bundle)?
I know that if $M$ is parallelized, then both bundles are trivial, respectively $V_k(M) \cong M \times V_k(\mathbb{R}^n)$ and $V_k(M) \cong M \times V'_k(\mathbb{R}^n)$. Simply take a trivialization $\phi : M \times \mathbb{R}^n \to TM$, then there is an obvious map $M \times V_k(\mathbb{R}^n) \to V_k(M)$, $(x,v_1,\dots,v_k) \mapsto (x,\phi(v_1),\dots,\phi(v_k))$. More generally like all $SO(n)$ bundles $V'_k(M)$
is characterized by the Euler class and the Pontrjagin classes has an Euler class and Pontrjagin classes – what is the precise relationship? Are they related to the classes of the tangent bundle of $TM$ itself? I was not even able to say what $V'_2(S^2)$ looks like 😕. |
Difference between revisions of "Lower attic"
From Cantor's Attic
(the Takeuti-Feferman-Buchholz ordinal)
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{{DISPLAYTITLE: The lower attic}}
{{DISPLAYTITLE: The lower attic}}
[[File:SagradaSpiralByDavidNikonvscanon.jpg | thumb | Sagrada Spiral photo by David Nikonvscanon]]
[[File:SagradaSpiralByDavidNikonvscanon.jpg | thumb | Sagrada Spiral photo by David Nikonvscanon]]
+
Welcome to the lower attic, where the countably infinite ordinals climb ever higher, one upon another, in an eternal self-similar reflecting ascent.
Welcome to the lower attic, where the countably infinite ordinals climb ever higher, one upon another, in an eternal self-similar reflecting ascent.
Line 10: Line 11:
** [[infinite time Turing machines#zeta | $\zeta$]] = the supremum of the eventually writable ordinals
** [[infinite time Turing machines#zeta | $\zeta$]] = the supremum of the eventually writable ordinals
** [[infinite time Turing machines#lambda | $\lambda$]] = the supremum of the writable ordinals,
** [[infinite time Turing machines#lambda | $\lambda$]] = the supremum of the writable ordinals,
−
* [[admissible]] ordinals and [[Church-Kleene#relativized Church-Kleene ordinal | relativized Church-Kleene $\omega_1^x$]]
* [[admissible]] ordinals and [[Church-Kleene#relativized Church-Kleene ordinal | relativized Church-Kleene $\omega_1^x$]]
* [[Church-Kleene | Church-Kleene $\omega_1^{ck}$]], the supremum of the computable ordinals
* [[Church-Kleene | Church-Kleene $\omega_1^{ck}$]], the supremum of the computable ordinals
−
* the [[omega one chess | omega one of chess]]
+
* the [[omega one chess | omega one of chess]]
+
[[omega one chess|$\omega_1^{\chess
+
}$
+ + + + + +
]]
* the [[Feferman-Schütte]] ordinal [[Feferman-Schütte | $\Gamma_0$]]
* the [[Feferman-Schütte]] ordinal [[Feferman-Schütte | $\Gamma_0$]]
* [[epsilon naught | $\epsilon_0$]] and the hierarchy of [[epsilon naught#epsilon_numbers | $\epsilon_\alpha$ numbers]]
* [[epsilon naught | $\epsilon_0$]] and the hierarchy of [[epsilon naught#epsilon_numbers | $\epsilon_\alpha$ numbers]]
* [[indecomposable]] ordinal
* [[indecomposable]] ordinal
* the [[small countable ordinals]], such as [[small countable ordinals | $\omega,\omega+1,\ldots,\omega\cdot 2,\ldots,\omega^2,\ldots,\omega^\omega,\ldots,\omega^{\omega^\omega},\ldots$]] up to [[epsilon naught | $\epsilon_0$]]
* the [[small countable ordinals]], such as [[small countable ordinals | $\omega,\omega+1,\ldots,\omega\cdot 2,\ldots,\omega^2,\ldots,\omega^\omega,\ldots,\omega^{\omega^\omega},\ldots$]] up to [[epsilon naught | $\epsilon_0$]]
−
* [[
+
* [[| Hilbert's hotel]] and [[]]
* [[omega | $\omega$]], the smallest infinity
* [[omega | $\omega$]], the smallest infinity
* down to the [[parlour]], where large finite numbers dream
* down to the [[parlour]], where large finite numbers dream
Latest revision as of 13:37, 27 May 2018
Welcome to the lower attic, where the countably infinite ordinals climb ever higher, one upon another, in an eternal self-similar reflecting ascent.
$\omega_1$, the first uncountable ordinal, and the other uncountable cardinals of the middle attic stable ordinals The ordinals of infinite time Turing machines, including admissible ordinals and relativized Church-Kleene $\omega_1^x$ Church-Kleene $\omega_1^{ck}$, the supremum of the computable ordinals the omega one of chess $\omega_1^{\mathfrak{Ch}_{\!\!\!\!\sim}}$ = the supremum of the game values for white of all positions in infinite chess $\omega_1^{\mathfrak{Ch},c}$ = the supremum of the game values for white of the computable positions in infinite chess $\omega_1^{\mathfrak{Ch}}$ = the supremum of the game values for white of the finite positions in infinite chess the Takeuti-Feferman-Buchholz ordinal the Bachmann-Howard ordinal the large Veblen ordinal the small Veblen ordinal the Extended Veblen function the Feferman-Schütte ordinal $\Gamma_0$ $\epsilon_0$ and the hierarchy of $\epsilon_\alpha$ numbers indecomposable ordinal the small countable ordinals, such as $\omega,\omega+1,\ldots,\omega\cdot 2,\ldots,\omega^2,\ldots,\omega^\omega,\ldots,\omega^{\omega^\omega},\ldots$ up to $\epsilon_0$ Hilbert's hotel and other toys in the playroom $\omega$, the smallest infinity down to the parlour, where large finite numbers dream |
The inverse square law for an electric field is:
$$ E = \frac{Q}{4\pi\varepsilon_{0}r^2} $$ Here: $$\frac{Q}{\varepsilon_{0}}$$ is the source strength of the charge. It is the point charge divided by the vacuum permittivity or electric constant, I would like very much to know what is meant by source strength as I can't find it anywhere on the internet. Coming to the point an electric field is also described as: $$Ed = \frac{Fd}{Q} = \Delta V$$ This would mean that an electric field can act only over a certain distance. But according to the Inverse Square Law, the denominator is the surface area of a sphere and we can extend this radius to infinity and still have a value for the electric field. Does this mean that any electric field extends to infinity but its intensity diminishes with increasing length? If that is so, then an electric field is capable of applying infinite energy on any charged particle since from the above mentioned equation, if the distance over which the electric field acts is infinite, then the work done on any charged particle by the field is infinite, therefore the energy supplied by an electric field is infinite. This clashes directly with energy-mass conservation laws. Maybe I don't understand this concept properly, I was hoping someone would help me understand this better.
The inverse square law for an electric field is:
It goes out forever, but the total energy it imparts is finite. The reason is that when things fall off as the square of the distance, the sum is finite. For example:
$$ \sum_n {1\over n^2} = {1\over 1} + {1\over 4} + {1\over 9} + {1\over 16} + {1\over 25} + ... = {\pi^2\over 6} $$
This sum has a finite limit. Likewise the total energy you gain from moving a positive charge away from another positive charge from position R to infinity is the finite quantity
$$\int_r^{\infty} {Qq\over r^2} dr = {Qq\over r}$$
So there is no infinity. In two dimensions (or in one), the electric field falls off only like ${1\over r}$ so the potential energy
is infinite, and objects thrown apart get infinite speed in the analogous two-dimensional situation.
I just want to add something besides Ron's answer, which in my opinion you should accept as "the answer".
The second formula which you quote does not apply to the field produced by a point charges. It is only true for a
constant electric field. In general, the change in the potential energy when going from a point $\mathbf{r}_0$ in the space to infinity is
$$\Delta V = -\int_{\mathbf{r}_0}^\infty \mathbf{E}(\mathbf{r}) \cdot \text{d} \mathbf{r}$$.
which comes from the relation $\mathbf{E}(\mathbf{r})=-\nabla V(\mathbf{r})$ and integration in the space.
For your point charge in $\mathbb{R}^3$, you can integrate easily [using the the electric field is isotropic] to get
$$\Delta V = V(\infty)-V(\mathbf{r}_0) = \dfrac{Q}{4\pi\varepsilon_0 r_{0}}$$
where $r_0=|\mathbf{r}_0|=\sqrt{x_0^2+y_0^2+z_0^2}$ and $V(\infty)=0$.
Hence, even if the range of the electric field is infinite, the energy is always finite. Note that when $|\mathbf{r}_0| \rightarrow 0$ the potential energy blows up even though physically does not make any sense. To solve this you need to make use of quantum electrodynamics, the "quantum version" of electromagnetism.
The Landau Pole is not a problem for QED because at scales much smaller than it (the Planck scale, which is smaller than the Landau pole by 260 orders of magnitude) the (negative) gravitational self-energy of the particle will more than cancel out its electromagnetic self-energy. So string theory is not necessary in this case, just gravity. |
I was trying to include two equations side-by-side in a document, while retaining the labels following this answer.
However the spacing between the
minipage and the text is too small.How can I increase it to match the "default" vertical spacing used between other equations and text?
Here is the code
Let us now suppose that the sequence of variables follows a stochastic Markov evolution model defined by \begin{minipage}[t]{0.45\columnwidth}% \begin{equation} \nu_{t}=\frac{\beta\nu_{t-1}}{\gamma_{t}}\label{eq:v_evolution} \end{equation} % \end{minipage}% \begin{minipage}[t]{0.45\columnwidth}% \begin{equation} \phi_{t}=\frac{\phi_{t-1}\gamma_{t}}{\beta}\label{eq:v_precision_evolution} \end{equation} % \end{minipage} for the variance and the precision, respectively, where $\gamma_{t}$ is a time $t$ random impulse, with distribution
And here is the result:
Thank you! |
2018-09-11 04:29
Proprieties of FBK UFSDs after neutron and proton irradiation up to $6*10^{15}$ neq/cm$^2$ / Mazza, S.M. (UC, Santa Cruz, Inst. Part. Phys.) ; Estrada, E. (UC, Santa Cruz, Inst. Part. Phys.) ; Galloway, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; Gee, C. (UC, Santa Cruz, Inst. Part. Phys.) ; Goto, A. (UC, Santa Cruz, Inst. Part. Phys.) ; Luce, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; McKinney-Martinez, F. (UC, Santa Cruz, Inst. Part. Phys.) ; Rodriguez, R. (UC, Santa Cruz, Inst. Part. Phys.) ; Sadrozinski, H.F.-W. (UC, Santa Cruz, Inst. Part. Phys.) ; Seiden, A. (UC, Santa Cruz, Inst. Part. Phys.) et al. The properties of 60-{\mu}m thick Ultra-Fast Silicon Detectors (UFSD) detectors manufactured by Fondazione Bruno Kessler (FBK), Trento (Italy) were tested before and after irradiation with minimum ionizing particles (MIPs) from a 90Sr \b{eta}-source . [...] arXiv:1804.05449. - 13 p. Preprint - Full text დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-25 06:58
Charge-collection efficiency of heavily irradiated silicon diodes operated with an increased free-carrier concentration and under forward bias / Mandić, I (Ljubljana U. ; Stefan Inst., Ljubljana) ; Cindro, V (Ljubljana U. ; Stefan Inst., Ljubljana) ; Kramberger, G (Ljubljana U. ; Stefan Inst., Ljubljana) ; Mikuž, M (Ljubljana U. ; Stefan Inst., Ljubljana) ; Zavrtanik, M (Ljubljana U. ; Stefan Inst., Ljubljana) The charge-collection efficiency of Si pad diodes irradiated with neutrons up to $8 \times 10^{15} \ \rm{n} \ cm^{-2}$ was measured using a $^{90}$Sr source at temperatures from -180 to -30°C. The measurements were made with diodes under forward and reverse bias. [...] 2004 - 12 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 533 (2004) 442-453 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-23 11:31 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-23 11:31
Effect of electron injection on defect reactions in irradiated silicon containing boron, carbon, and oxygen / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Yakushevich, H S (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) Comparative studies employing Deep Level Transient Spectroscopy and C-V measurements have been performed on recombination-enhanced reactions between defects of interstitial type in boron doped silicon diodes irradiated with alpha-particles. It has been shown that self-interstitial related defects which are immobile even at room temperatures can be activated by very low forward currents at liquid nitrogen temperatures. [...] 2018 - 7 p. - Published in : J. Appl. Phys. 123 (2018) 161576 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-23 11:31 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-23 11:31
Characterization of magnetic Czochralski silicon radiation detectors / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) Silicon wafers grown by the Magnetic Czochralski (MCZ) method have been processed in form of pad diodes at Instituto de Microelectrònica de Barcelona (IMB-CNM) facilities. The n-type MCZ wafers were manufactured by Okmetic OYJ and they have a nominal resistivity of $1 \rm{k} \Omega cm$. [...] 2005 - 9 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 548 (2005) 355-363 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-23 11:31
Silicon detectors: From radiation hard devices operating beyond LHC conditions to characterization of primary fourfold coordinated vacancy defects / Lazanu, I (Bucharest U.) ; Lazanu, S (Bucharest, Nat. Inst. Mat. Sci.) The physics potential at future hadron colliders as LHC and its upgrades in energy and luminosity Super-LHC and Very-LHC respectively, as well as the requirements for detectors in the conditions of possible scenarios for radiation environments are discussed in this contribution.Silicon detectors will be used extensively in experiments at these new facilities where they will be exposed to high fluences of fast hadrons. The principal obstacle to long-time operation arises from bulk displacement damage in silicon, which acts as an irreversible process in the in the material and conduces to the increase of the leakage current of the detector, decreases the satisfactory Signal/Noise ratio, and increases the effective carrier concentration. [...] 2005 - 9 p. - Published in : Rom. Rep. Phys.: 57 (2005) , no. 3, pp. 342-348 External link: RORPE დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-22 06:27
Numerical simulation of radiation damage effects in p-type and n-type FZ silicon detectors / Petasecca, M (Perugia U. ; INFN, Perugia) ; Moscatelli, F (Perugia U. ; INFN, Perugia ; IMM, Bologna) ; Passeri, D (Perugia U. ; INFN, Perugia) ; Pignatel, G U (Perugia U. ; INFN, Perugia) In the framework of the CERN-RD50 Collaboration, the adoption of p-type substrates has been proposed as a suitable mean to improve the radiation hardness of silicon detectors up to fluencies of $1 \times 10^{16} \rm{n}/cm^2$. In this work two numerical simulation models will be presented for p-type and n-type silicon detectors, respectively. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 2971-2976 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-22 06:27
Technology development of p-type microstrip detectors with radiation hard p-spray isolation / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) ; Díez, S (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) A technology for the fabrication of p-type microstrip silicon radiation detectors using p-spray implant isolation has been developed at CNM-IMB. The p-spray isolation has been optimized in order to withstand a gamma irradiation dose up to 50 Mrad (Si), which represents the ionization radiation dose expected in the middle region of the SCT-Atlas detector of the future Super-LHC during 10 years of operation. [...] 2006 - 6 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 566 (2006) 360-365 დეტალური ჩანაწერი - მსგავსი ჩანაწერები 2018-08-22 06:27
Defect characterization in silicon particle detectors irradiated with Li ions / Scaringella, M (INFN, Florence ; U. Florence (main)) ; Menichelli, D (INFN, Florence ; U. Florence (main)) ; Candelori, A (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Bruzzi, M (INFN, Florence ; U. Florence (main)) High Energy Physics experiments at future very high luminosity colliders will require ultra radiation-hard silicon detectors that can withstand fast hadron fluences up to $10^{16}$ cm$^{-2}$. In order to test the detectors radiation hardness in this fluence range, long irradiation times are required at the currently available proton irradiation facilities. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 589-594 დეტალური ჩანაწერი - მსგავსი ჩანაწერები |
I am tring to prove the NP-completeness of the following Binary Integer Linear Program (BILP):
$$ \text{min}\sum_i\sum_j x_{ij}\\ \text{subject to} \sum_jr_{ij}x_{ij} \geq R,\ \ \forall \ \ i \in [L]\\ \sum_i x_{ij} \leq 1,\ \ \forall \ \ j \in [N]\\ x_{ij} \in \{0,1\}.$$
This a resource allocation problem in which I am trying to minimize the number of resources utilized while guaranteeing a certain minimum rate for all entities.
$x_{ij}$ is an indicator random variable that indicates whether or not a resource $j$ is allocated to an entity $i$. The first set of constraints impose a minimum amount of resource guaranteed for each entity. $r_{ij}$ is the rate that an entity $i$ can get in a resource element $j$. The second set of constraints ensure that a resource is not allocated to multiple users as sharing of resources is not possible in this problem. $[L]$ is the set of user entities and $[N]$ is the set of resources here.
I have tried looking at several covering problems like set cover that look like this but the second set of constraints seem to be falling out of place every time. Can anyone suggest an existing NP complete problem that can be reduced to this problem to prove it's NP-completeness?
Thanks |
I have a simple question, but I'm having trouble figuring it out. I'm trying to make the limits on a sum in math-mode appear as they do on the right hand side below.
I know that in-line displays the limits like on the right hand side by default, but writing
\textstyle before the sum in math mode of course makes the sum symbol smaller, much unlike the equations around it.
I did find a way to take the index variable out from underneath and put it near the bottom left. I have so far
\documentclass[12pt]{article}\begin{document}\begin{equation}\sum_{n=1}^\infty = \mathop{}_{\mkern+5mu n}\!\sum_1^\infty\end{equation}\end{document}
Which looks like
Now the limits just have to move beside the sum and be brought slightly closer together. I thought of using
\sum\ _1^\infty, which might technically work if the height of the blank vertical was higher (How could you make it higher as an aside?)(I'm new to LaTeX), but perhaps it would not look as nice as other ways that exist.
Question
How could the limits on the sum be brought from where they are in math-mode to where they would be with in-line, but preserving the size of the sum symbol as it is in math-mode?
Quick Secondary Question
Can you increase the height of the blank vertical,
\ (with a space after the backslash)? |
I have discovered that the following two tags are too similar to each other:
riemann-zeta-function, with 299 questions tagged, has the description
The Riemann zeta function is the function of one complex variable $s$ defined by the series $\zeta(s) = \sum_{n \geq 1} \frac{1}{n^s}$ when $\operatorname{Re}(s) > 1$. It admits a meromorphic continuation to $\mathbb{C}$ with only a simple pole at $1$. This function satisfies a functional equation relating the values at $s$ and $1-s$. This is the most simple example of an $L$-function and a central object of number theory.
and no tag wiki;
zeta-functions, with 195 questions tagged, has the description
The Riemann zeta function is defined as the analytic continuation of the function defined for $\sigma > 1$ by the sum of the preceding series.
and tag wiki
The Riemann zeta function, $\zeta(s)$, is a function of a complex variable $s$ that analytically continues the sum of the infinite series
$$\zeta(s) =\sum_{n=1}^\infty\frac{1}{n^s}$$
which converges when the real part of $s$ is greater than $1$.
There are two problems here:
does it make sense to have one tag dedicated to just the Riemann $\zeta$, and a single other one for the rest of the $\zeta$ functions? A single tag for all these functions should suffice.
if we still want tags to distinguish between Riemann and "non-Riemann" $\zeta$ functions, then the latter class of functions should be correctly described in the tag description and tag wiki - a thing that does not currently happen with the latter tag.
My suggestion is to just melt the former tag into the latter, and automatically retag all the questions. |
On the Number of Rich Words Abstract
Any finite word
w of length n contains at most \(n+1\) distinct palindromic factors. If the bound \(n+1\) is reached, the word w is called rich. The number of rich words of length n over an alphabet of cardinality q is denoted \(R_q(n)\). For binary alphabet, Rubinchik and Shur deduced that \({R_2(n)}\le c 1.605^n \) for some constant c. In addition, Guo, Shallit and Shur conjectured that the number of rich words grows slightly slower than \(n^{\sqrt{n}}\). We prove that \(\lim \limits _{n\rightarrow \infty }\root n \of {R_q(n)}=1\) for any q, i.e. \(R_q(n)\) has a subexponential growth on any alphabet. KeywordsRich words Enumeration Palindromes Palindromic factorization Notes Acknowledgments
The author wishes to thank Edita Pelantová and Štěpán Starosta for their useful comments. The author acknowledges support by the Czech Science Foundation grant GAČR 13-03538S and by the Grant Agency of the Czech Technical University in Prague, grant No. SGS14/205/OHK4/3T/14.
References 1.Balková, L.: Beta-integers and Quasicrystals, Ph. D. thesis, Czech Technical University in Prague and Université Paris Diderot-Paris 7 (2008)Google Scholar 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. |
50th SSC CGL level Solution Set, 1st on Simple Interest and Compound Interest problems
This is the 50th solution set of 10 practice problem exercise for SSC CGL exam and 1st on Simple interest and Compound interest problems. Students must complete the
in prescribed time first and then only refer to the solution set for extracting maximum benefits from this resource. corresponding question set
In MCQ test, you need to deduce the answer in shortest possible time and select the right choice.
Based on our analysis and experience we have seen that, for accurate and quick answering, the student
must have complete understanding of the basic concepts in the topic area is adequately fast in mental math calculation should try to solve each problem using the basic and rich concepts in the specific topic area and does most of the deductive reasoning and calculation in his or her head rather than on paper.
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Before going through the solutions you should go through our sessions on
for detailed explanation of the efficient Simple Interest and Compound Interest problem solving techniques and related concepts that we will use for this present set of 10 problems. Basic and rich concepts on Simple Interest and Compound Interest 50th solution set- 10 problems for SSC CGL exam: 1st on Simple interest and Compound interest problems - time 15 mins Problem 1.
A man borrowed Rs.16000 from two persons. He paid 6% interest per annum to one and 10% interest per annum to the other. In the first year he paid a total interest of Rs.1120. How much did he borrow at each rate?
Rs.12500 : Rs.3500 Rs.11000 : Rs.5000 Rs.12000 : Rs.4000 Rs.10000 : Rs.6000 Solution 1 : Problem analysis
As this problem involves interest on the first 1 year only, the type of interest, that is, simple or compound, does not matter. We can assume the problem as a problem on simple interest only.
Here Rs.16000 is the total sum of money borrowed. If we assume Rs. $x$ as the sum borrowed from the first person at the interest rate of 6%, then sum borrowed from the second person becomes $\text{Rs.}(16000-x)$, which he borrowed at the interest rate of 10%.
Applying the interest rates on the amounts and adding the two amounts of interest we will get the total interest which is Rs.1120. From the single linear equation in one variable it will be a short way off to the solution.
A small but important point is, the percentages need to be converted to equivalent decimals for efficient deduction of the value of $x$.
Solution 1 : Problem solving execution
Interest in 1 year for the money $\text{Rs.}x$ borrowed from the first person at the interest rate 6% is,
$I_1=0.06x$.
Interest in 1 year for the money $\text{Rs.}(16000-x)$ borrowed from the second person at the interest rate 10% is,
$I_2=0.1(16000-x)$.
Sum of these two interests is, Rs.1120. So,
$I_1+I_2=1600-0.04x=1120$,
Or, $0.04x=480$,
Or, $x=12000$, and
$16000-x=4000$.
Answer: Option c: Rs.12000 : Rs.4000 Key concepts used: -- Basic simple interest concepts . Basic algebraic concepts Problem 2.
A certain sum of money amounts to Rs.1680 in 3 years and to Rs.1800 in 5 years. Find the sum and rate of simple interest.
Rs.1200 : 4% Rs.1500 : 4% Rs.1800 : 5% Rs.1600 : 5% Solution 2 : Problem analysis and execution
In simple interest calculations, as the interest remains fixed for each year, the difference in amounts of the two years, Rs.120 is the interest for 2 years. So per year interest is,
$Ar=\text{Rs.}60$, where $A$ is the amount and $r$ the interest rate.
Taking the first amount after 3 years,
$1680=A +3Ar=A +180$,
Or, Amount $A=\text{Rs.}1500$, and,
Interest $r=\displaystyle\frac{60}{1500}=4$%.
Answer: Option b : Rs.1500 : 4%. Key concepts used: -- Basic simple interest concepts Change analysis technique -- Efficient simplification . Problem 3.
A sum of money at compound interest amounts to Rs.650 at the end of the first year and Rs.676 at the end of the second year. The sum of money is,
Rs.1300 Rs.650 Rs.1250 Rs.625 Solution 3 : Problem analysis and execution
In compounding of interest, interest accrued in a year will be rate of interest applied on the amount in the beginning of the year. So in this case of our problem, interest accrued in the second year is,
$676-650=26=650r$, where $r$ is the rate of interest,
Or, $r=\displaystyle\frac{26}{650}=\frac{1}{25}=4$%.
As amount at the end of first year is 650, if $A$ is the starting amount,
$650=A(1+r)=\displaystyle\frac{26}{25}A$,
Or, $A=25\times{25}=625$.
Answer: Option d: Rs.625. Key concepts used: -- Basic compound interest concepts -- Efficient simplification. Change analysis technique Note: We have kept the interest in the form of fraction for ease of calculation. That is efficient simplification. Problem 4.
A sum of money placed at compound interest doubles itself in 5 years. In how many years it would be 8 times of itself at the same rate of interest?
10 years 15 years 20 years 7 years Solution 4 : Problem analysis
The interest rate here is such that a sum of money doubles itself in 5 years. If this doubled sum is invested again in the beginning of the sixth year it will again be doubled, that is, will be 4 times the original sum after another 5 years, that is, a total of 10 years. So after another 5 years, that is, after a total period of 15 years, the original sum will be 8 times of itself.
Alternately, by the principle of compounding, in a period of every 5 years any invested sum is doubled at the rate of interest as stated. So to make itself 8 times, that is, three times doubling, a total of 3 times 5, that is, 15 years will be required.
Mathematically,
$A_5=A(1+r)^5=2A$,
Or, $(1+r)^5=2$.
Assuming that after $x$ years the original amount will make itself 8 times,
$A_x=8A=A(1+r)^x$,
Or, $2^3=(1+r)^x$, substituting for 2,
Or, $(1+r)^{15}=(1+r)^x$.
So $x=15$ years.
Answer: Option b: 15 years. Key concepts used: -- Basic compound interest concepts -- conceptual reasoning Base equalization technique. Problem 5.
The compound interest on Rs.2000 in 2 years, if the rate of interest is 4% per annum for the first year and 3% per annum for the second year, will be,
Rs.143.40 Rs.141.40 Rs.140.40 Rs.142.40 Solution 5 : Problem analysis
As the rate of interest is different for the first and second year, we need to enumerate the actual problem state of each year using basic compound interest concepts. This process of enumeration of each year status is a rich compound interest concept, namely,
Time breakdown technique. Solution 5 : Problem solving execution
The interest for the first year will be,
$I_1=Ar=2000\times{0.04}=80$.
And the amount after 1 year,
$A_1=A +Ar=2080$.
Interest at 3% for the second year will be on this amount,
$I_2=A_1r=2080\times{0.03}=62.40$.
So total interest is,
$80+62.40=142.40$.
Answer: Option d: Rs.142.40. Key concepts used: -- Basic compound interest concepts -- Rich compound interest concept . Time breakdown technique -- Enumeration technique Problem 6.
An amount of money at compound interest grows up to Rs.3840 in 4 years and up to Rs.3936 in 5 years. Find the rate of interest.
2% 2.05% 3.5% 2.5% Solution 6 : Problem analysis and execution
For the 5th year the interest is,
$A_4r=3936-3840=96$, where $A$ is the original amount, $r$ is the rate of interest, and $A_4r$ is the amount after 4 years.
Or, $3840r=96$,
Or, $r=\displaystyle\frac{1}{40}=2.5$%.
Answer: Option d : 2.5%. Key concepts used: -- Basic compound interest concepts Change analysis technique . Problem 7.
The time in which Rs.80000 amounts to Rs.92610 at 10% per annum compound interest, interest being compounded semi-annually, is
1.5 years 2 years 3 years 2.5 years Solution 7 : Problem analysis and execution
As the annual interest rate is 10% and the compounding is done semi-annually, that is, every 6 months, we can assume the period unit for deductions as 6 months and the interest rate for 1 period unit as 5%.
For the first 6 month period the interest will be,
$I_1=80000\times{0.05}=4000$.
As the final amount of Rs 92610 differs from the initial amount of Rs.80000 by Rs.12610, which is very near to, $3\times{4000}$, the interest that would have accrued if the interest would have been of simple type without compounding. Even at this point examining the choice values we can choose 3 periods of 6 months or 1.5 years as the answer.
Mathematically,
Interest for the first period is Rs. 4000, and the amount at the end of 1st period, Rs.84000. Interest for the second period will be on this amount,
$I_2=84000\times{0.05}=4200$.
The amount at the end of the second period will be, $84000+4200=88200$.
Interest for the 3rd period will be,
$I_3=88200\times{0.05}=4410$, and the total amount, $88200+4410=92610$.
Answer: Option a: 1.5 years. Key concepts used: -- Basic compound interest concepts Compounding period concept -- Estimation technique -- Free resource use principle -- Time breakdown technique -- Enumeration technique. Problem 8.
A man borrows Rs.2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly installments. How much will each installment be?
Rs.1352 Rs.1283 Rs.1377 Rs.1275 Solution 8 : Problem analysis and execution
Let us visualize how two equal installments will wipe out the debt at the same time ensure the interest accrual.
At the end of the first year the total amount due will be,
$A_1=2550\left(1+\displaystyle\frac{1}{25}\right)=102\times{26}$, dividing 2550 by 25 was instantaneous and so we did it but left the factor of 26 for later use.
Assuming $x$ is returned back as the first instalment, the amount on which the second year interest accrual would take place will be,
$A_{1n}=(102\times{26})-x$.
The amount at the end of second year will then be,
$A_2=\left(102\times{26}-x\right)\left(1+\displaystyle\frac{1}{25}\right)=x$, as this is the second instalment that has to complete the transaction.
So,
$\left(102\times{26}-x\right)\left(\displaystyle\frac{26}{25}\right)=x$,
Or, $\displaystyle\frac{51}{25}x=26\times{102}\times{\frac{26}{25}}$,
Or, $51x=26\times{102}\times{26}$,
Or, $x=2\times{676}=1352$
Answer: Option a: Rs.1352. Key concepts used: -- Installment concept -- Basic compound interest concept -- Delayed evaluation Efficient simplification. Note: As decimal calculation usually gives a bit of trouble, we have kept the interest in fraction form, and did the larger multiplication as late as possible. Though the steps shown are many, actual deduction is inherently simple and quick. Problem 9.
The difference between the compound interest and the simple interest at the same rate of interest, for the amount Rs.5000 in 2 years, is Rs.32. The rate of interest is,
5% 10% 12% 8% Solution 9 : Problem analysis and execution:
By the given condition,
$I_{compound}-I_{simple}=32$,
Or, $(Ar^2+2Ar)-2Ar=32$, where $A$ is the original amount and $r$ is the same rate of interest over 2 years,
Or, $Ar^2=32$,
Or, $5000\times{r^2}=32$,
Or, $r^2=\displaystyle\frac{32}{5000}=\frac{64}{10000}=\left(\frac{8}{100}\right)^2$,
Or, $r=8$%.
Answer: Option d: 8%. Key concepts used: Basic compound interest concept -- Basic simple interest concept -- Basic algebraic concepts. Problem 10.
If the compound interest on a sum for 2 years at 12.5% per annum is Rs.510, the simple interest for the same sum at the same rate for the same period of time is,
Rs.480 Rs.450 Rs.400 Rs.460 Solution 10 : Problem analysis and execution:
If $A$ is the amount invested for 2 years and $r$ be the interest, the compound interest is,
$I_{compound}=A(1+r^2)-A=Ar(r+2)$,
Or, $Ar\left(\displaystyle\frac{1}{8}+2\right)=510$, interest 12.5% is equivalent to the fraction $\frac{1}{8}$,
Or, $Ar=\displaystyle\frac{510\times{8}}{17}=240$.
So the simple interest for the same period on the same amount at the same interest rate is,
$I_{simple}=2Ar=480$.
Answer: Option a: Rs.480. Key concepts used: -- Basic compound interest concept -- Basic simple interest concept Efficient simplification . Note: To ease calculation we have kept the interest as fraction equivalent to 12.5%. Also knowing beforehand that simple interest per year is $Ar$, we have evaluated $Ar$ itself, not the components, avoiding evaluation of important component $A$ and the final multiplication with $r$. This is efficient simplification. You need to evaluate compound variables consisting of more than one variable, if that is directly used in the final solution. This approach speeds up the solution process considerably. |
I am currently learning viscosity solutions. In the lecture notes by N. Katzourakis, he introduced the theories of continuous viscosity solution to PDE with continuous nonlinearitiy. And in Chapter 9, he mentioned that neither the solutions nor the coefficients is actually needed to be continuous. He pointed out that
all the results hold true if the solution and the coefficient are replaced by their semi-continuous envelopes.
For example, for degenerate elliptic PDE $F(\cdot, u,Du,D^2u)=0$ with $$ F(x,r,p,X)\le F(x,r,p,Y),\quad \forall X\le Y\in \mathbb{S}(n), $$ a function $u:\Omega\to \mathbb{R}$ is a viscosity subsolution on $\Omega$ if for all $x\in \Omega$, and $(p,X)\in \mathcal{J}^{2,+}u^*(x)$, we have $$ F^*(x,u^*(x),p,X)\ge 0. $$ Here $u^*$ and $F^*$ denote the upper-semicontinuous envelops of $u$ and $F$, respectively.
I was wondering what does he mean by "all the results"? Does he mean all the properties of Jets, the stability results and well-posedness holds for this definition? Can you recommend a reference which discuss the properties of viscosity solutions in this sense? |
I have a problem in proving invalid the following argument:
Horses and cows are mammals.Some animals are mammals.Some animals are not mammals.Therefore all horses are animals.
If we translate it into the logical notation then we have :
The premises are :
$(\forall x)(Hx \lor Cx \rightarrow Mx)$.
$(\exists x)(Ax \land Mx)$
$(\exists x)(Ax \land \sim Mx)$
The conclusion is :
$(\forall x)(Hx \rightarrow Ax)$.
Where $Ax$ is means $x$ is an animal, $Hx$ means $x$ is a horse, $Cx$ means $x$ is a cow, $Mx$ means $x$ is a Mammal.
I have to show it is an invalid argument.But
the 2nd and 3rd premises are like contradictory to the conclusion part.
As per I know if I can show a truthvalue assignment for which the conclusion is false still the premises are true,then the argument will be invalid.But I am unable to find such a truthvalue assignment,
looking for a help. Thanks. |
Author Message Lonely-Star Tux's lil' helper Joined: 12 Jul 2003 Posts: 82
Posted: Wed Apr 27, 2005 9:12 am Post subject: Mathematical Symbols in Xfig and gnuplot Hi everybody,
Studying physics I am starting to use Xfig and gnuplot.
Now I wonder How I can use symbols like omega or phi in the text in xfig or labels in gnuplot.
Any help?
Thanks! furanku l33t Joined: 08 May 2003 Posts: 902 Location: Hamburg, Germany
Posted: Wed Apr 27, 2005 2:40 pm Post subject: Hi!
The good old "make-my-graphs-pretty" question
You have several possibilties.
gnuplot
1) The enhanced postscript driver
Start gnuplot. Try
Code: gnuplot> plot sin(x) title "{/Symbol F}(x)"
You will get a window with your graph labled verbose as "{/Symbol F}(x)". But now try
Code: gnuplot> set term post enh
Terminal type set to 'postscript'
Options are 'landscape enhanced monochrome blacktext \
dashed dashlength 1.0 linewidth 1.0 defaultplex \
palfuncparam 2000,0.003 \
butt "Helvetica" 14'
gnuplot> set output "test.ps"
gnuplot> plot sin(x) title "{/Symbol F}(x)
gnuplot> exit
View the resulting file "test.ps" in your favorite postscript viewer: Now you have a greek upper Phi as label. To learn more about the enhanced possibilities of the postscrip driver read the file "/usr/share/doc/gnuplot-4.0-r1/psdoc/ps_guide.ps" (or whatever gnuplot version you use).
Advantages: Easy to use, output file easy included in almost every document
Disadvantages: Limited possibilities, looks ugly, wrong fonts when included in other documents (esp. LaTeX)
2) The LaTeX drivers
I guess you want to include your graph in a LaTeX File (I hope you have learned LaTeX, if not do so, quickly, it's essential for all physical publications!)
Again several possibilities...
2a) The "latex" driver, which uses the pictex environment
Code: gnuplot> set term latex
Options are '(document specific font)'
gnuplot> set output "test.tex"
gnuplot> plot sin(x) title "$\Phi(x)$"
gnuplot> exit
Now gnuplot produced a file "test.tex" which you can include in your LaTeX Document with
Process your LaTeX Document and you'll see the graph labled with the TeX Fonts and all the glory you can use to typeset formulas in LaTeX: fractions, integrals, ... all you can do in LaTeX can be used.
Advantage: beautifull output, fonts fitting to the rest of your document
Disadvantage: more complicated to use, limited capabilities of the LaTeX picture environment
2b) Combined LaTeX and Postscript. Almost like 2a) but now the graph is in Postscript, just the lables are set by LaTeX:
Code: gnuplot> set term pslatex
Terminal type set to 'pslatex'
Options are 'monochrome dashed rotate'
gnuplot> set output "test.tex"
gnuplot> plot sin(x) title "$\Phi (x)$"
gnuplot> exit
Now the file "test.tex" will contain postscript special to draw the graph, the label is still set by LaTeX. Use it in your LaTeX Document like before.
Advantage: almost unlimited graphics possibilities due to postscript
Disadvantage: usage of poststcript needs converting the whole document to postscript afterwards (but that's normal anway), pdflatex isn't able to process postscript (well, VTeX's version can, but it's not open source)
3) The fig driver
You export your graph in gnuplot into xfigs "fig" file format, which can be usefull if you want to add modify your graph afterward, for example add some text and arrows (which can be also done in gnuplot but is a pain in the ass...)
Code: gnuplot> set term fig textspecial
Terminal type set to 'fig'
Options are 'monochrome small pointsmax 1000 landscape inches dashed textspecial fontsize 10 thickness 1 depth 10 version 3.2'
gnuplot> set output "test.fig"
gnuplot> plot sin(x) title "$\Phi (x)$"
gnuplot> exit
Open your file "test.fig" in xfig and go on as described below.
XFig
Xfig offers you, almost like gnuplot the possibility to add greek symbols as postscript fonts, and has also a "special flag" which is meant for using LaTeX code in your illustration, which is set in LaTeX later when compiling your document with latex.
4) The symbol postscript font
Click in xfig on the large "T" to get the text tool. Click on "Text font (Default)" in the lower right corner and select "Symbol (Greek)". Click somewhere in the image. Now you can type greek letters. Unlike in gnuplot they will appear on the screen. Export your file to postscript and you can use it in your documents like the files generated by gnuplot described in 1) above.
5) The special text flag
Note the option "textspecial" in the "set term fig textspecial" command in 3) above. This tells xfig that text set with this flag has a special meaning in some exported formats. You can set it manually in xfig with the button "Text flags" in the lower bar. Set "Special flag" to "Special" in the appearing dialog. Now click somewhere and type somthing like "$\int{-\infty}^\infty e^{-x^2} dx$. Now go to "File -> Export" and select one of "Latex picture" (which is like gnuplots "latex" terminal described above) or "Combined PS/LaTeX (both parts)" (which is like gnuplots "pslatex" driver, with the only exception that the LaTeX and Postscript code are stored in two separate files. Don't worry, you will just have to include the file ending with "_t" into your LaTeX document, this will automaticall include the other file).
[Edit:] It may necessary to set the "hidden" flag in newer versions of xfig to avoid getting both labels, the one set by xfig and the one from LaTeX on top of each other.
You will see that there is also an "Combined PDF/LaTeX (both parts)" export options which is usefull if you want to generate pdf files from your LaTeX sources directly using pdflatex, since that can't include postscript graphics. On the other hand you still can make a dvi file from your LaTeX sources and convert that to pdf using dvipdf, or convert your postscript files to pdf by epstopdf, or ...
You see there are a lot of possibilities, and I just mentioned the ones I used, and did a good job for me during my diploma thesis in physics, and still do.
Fell free to ask if you still have questions,
Frank Last edited by furanku on Thu Feb 14, 2008 10:31 am; edited 2 times in total Lonely-Star Tux's lil' helper Joined: 12 Jul 2003 Posts: 82
Posted: Wed Apr 27, 2005 5:14 pm Post subject: Thanks a lot for your help!
(it worked) incognito n00b Joined: 15 Jan 2004 Posts: 3
Posted: Wed Apr 27, 2005 11:30 pm Post subject: lurkers thank you furanku,
Great post - hopefully the moderators would consider putting it in the Documents, Tips, and Tricks section.
incognito adsmith Veteran Joined: 26 Sep 2004 Posts: 1386 Location: NC, USA
Posted: Thu Apr 28, 2005 1:11 am Post subject: There;s a script which does all that very nicely and automatically... google for "texfig" furanku l33t Joined: 08 May 2003 Posts: 902 Location: Hamburg, Germany
Posted: Thu Apr 28, 2005 8:29 am Post subject: Thanks, incognito, but I guess it's not gentoo related enough for the tips and trick section. But almost all of our new students in our workgroup come up with this question after a while, so I thought that's a nice occasion to write down what I learned about that and give them simply the URL (I guess, at least I have to spellcheck it on the weekend, sorry, I'm not a native english speaker and wrote it in a hurry yesterday...)
adsmith, that's a nice little script. It includes your exported fig file in a skeleton LaTeX document and processes and previews that. For larger documents I prefer the method using a make file which does the neccessary conversions (I didn't mention that xfig comes with a seperate program called "fig2dev" which can do all the exports xfig can do on the commandline). That combined with the preview-latex (screenshot) mode (which displays all your math and graphics inline in the [x]emacs, it's now part of auctex), gives me the for my taste most effective document writing environment. But as far as I can see new users are more attracted by kile (a KDE TeX environment, screenshots) or lyx (screenshots), and in that case texfig is a good help to get the LaTeX labels in your figures right. Thanks for your tip! nixnut Bodhisattva Joined: 09 Apr 2004 Posts: 10974 Location: the dutch mountains
Posted: Thu Feb 14, 2008 7:59 pm Post subject: Moved from Other Things Gentoo to Documentation, Tips & Tricks.
Tip/trick, so moved here _________________ Please add [solved] to the initial post's subject line if you feel your problem is resolved. Help answer the unanswered
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Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE
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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 ...
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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 ... |
So I've been doing regular languages a while and still need a better understanding of why all finite languages A ⊆ Σ* are regular? Is there a formal proof of it or is it just because a DFA can represent any finite language since the states would be finite as well?
The proof goes something like this:
If $A$ is a finite language, then it contains a finite number of strings $a_0, a_1, \cdots , a_n$. The language $\{a_i\}$ consisting of a single literal string $a_i$ is regular. The union of a finite number of regular languages is also regular. Therefore, $A = \{a_0\} \cup \{a_1\} \cup \cdots \cup \{a_n\}$ is regular.
There is also a direct proof. Let $P$ be the set of all prefixes of words in $A$. Since $A$ is finite, so is $P$. We construct a DFA whose states are $\{ q_p : p \in P \} \cup \{ q' \}$. The initial state is $q_\epsilon$. A state $q_p$ is final iff $p \in P$. When at state $q_p$ and reading $\sigma$, if $p\sigma \in P$ then we move to $q_{p\sigma}$, otherwise we move to $q'$. When at state $q'$, we always stay in $q'$.
What I described above is the minimal DFA. You can get a somewhat simpler DFA by taking an arbitrary superset of $P$, such as the set of all words of length at most $n$, where $n$ is the length of the longest word in $A$. |
Let $f:\Bbb R\to\Bbb R$ be a convex function. Then $f$ is differentiable at all but countably many points.
It is clear that a convex function can be non-differentiable at countably many points, for example $f(x)=\int\lfloor x\rfloor\,dx$.
I just made this theorem up, but my heuristic for why it should be true is that the only possible non-differentiable singularities I can imagine in a convex function are corners, and these involve a jump discontinuity in the derivative, so since the derivative is increasing (where it is defined), you get an inequality like $f'(y)-f'(x)\ge \sum_{t\in(x,y)}\omega_t(f')$, where $\omega_t(f')$ is the oscillation at $t$ (limit from right minus limit from left) and the sum is over all real numbers between $x$ and $y$. Since the sum is convergent (assuming that $x\le y$ are points such that $f$ is differentiable at $x$ and $y$ so that this makes sense), there can only be countably many values in the sum which are non-zero, and at all other points the oscillation is zero and so the derivative exists. Thus there are only countably many non-differentiable points in the interval $(x,y)$, so as long as there is a suitable sequence $(x_n)\to-\infty$, $(y_n)\to\infty$ of differentiable points, the total number of non-differentiable points is a countable union of countable sets, which is countable.
Furthermore, I would conjecture that the set of non-differentiable points has empty interior-of-closure, i.e. you can't make a function that is non-differentiable at the rational numbers, but as the above discussion shows there are still a lot of holes in the proof (and I'm making a lot of unjustified assumptions regarding the derivative already being somewhat well-defined). Does anyone know how to approach such a statement? |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
This problem is from the book [1]. In case of being closed as a duplication of that in [2], I first make a defense:
The accepted answer at [2] is still in dispute. The proof given by
@eh9is based on Kruskal's algorithm.
I am seeking for a proof independent of any MST algorithms.
Let $T$ be an MST of graph $G$. Given a connected subgraph $H$ of $G$, show that $T \cap H$ is contained in some MST of $H$. Problem:
My partial trial is
by contradiction:
Suppose that $T \cap H$ is not contained in any MST of $H$. That is to say, for any MST of $H$ (denoted $MST_{H}$), there exists an edge $e$ such that $e \in T \cap H$, and however, $e \notin MST_{H}$.
Now we can add $e$ to $MST_{H}$ to get $MST_{H} + {e}$ which contains a cycle (denoted $C$). Because $MST_{H}$ is a minimum spanning tree of $H$ and $e$ is not in $MST_{H}$, we have that every other edge $e'$ than $e$ in the cycle $C$ has weight no greater than that of $e$ (i.e., $\forall e' \in C, e' \neq e. w(e') \le w(e)$). There exists at lease one edge (denoted $e''$) in $C$ other than $e$ which is not in $T$. Otherwise, $T$ contains the cycle $C$.
Now we have $w(e'') \le w(e)$ and $e \in T \land e'' \notin T$, $\ldots$
I failed to continue... |
Some differentiate fluid from solid by the reaction to shear stress. The fluid continuously and permanently deformed under shear stress while the solid exhibits a finite deformation which does not change with time. It is also said that fluid cannot return to their original state after the deformation. This differentiation leads to three groups of materials: solids and liquids and all material between them. This test creates a new material group that shows dual behaviors; under certain limits; it behaves like solid and under others it behaves like fluid (see Figure 1.1). The study of this kind of material called rheology and it will (almost) not be discussed in this book. It is evident from this discussion that when a fluid is at rest, no shear stress is applied.
The fluid is mainly divided into two categories: liquids and gases. The main difference between the liquids and gases state is that gas will occupy the whole volume while liquids has an almost fix volume. This difference can be, for most practical purposes considered, sharp even though in reality this difference isn't sharp. The difference between a gas phase to a liquid phase above the critical point are practically minor. But below the critical point, the change of water pressure by 1000% only change the volume by less than 1 percent. For example, a change in the volume by more 5% will required tens of thousands percent change of the pressure. So, if the change of pressure is significantly less than that, Hence, the pressure will not affect the volume. In gaseous phase, any change in pressure directly affects the volume. The gas fills the volume and liquid cannot. Gas has no free interface/surface (since it does fill the entire volume).
There are several quantities that have to be addressed in this discussion. The first is force which was reviewed in physics. The unit used to measure is
The traditional quantity, which is force per area has a new meaning. This is a result of division of a vector by a vector and it is referred to as tensor. In this book, the emphasis is on the physics, so at this stage the tensor will have to be broken into its components. Later, the discussion on the mathematical meaning is presented (later version). For the discussion here, the pressure has three components, one in the area direction and two perpendicular to the area. The pressure component in the area direction is called pressure (great way to confuse, isn't it?). The other two components are referred as the shear stresses. The units used for the pressure components is \([N/m^2]\).
Fig. 1.2 Density as a function of the size of sample.
The density is a property which requires that liquid to be continuous. The density can be changed and it is a function of time and space (location) but must have a continues property. It doesn't mean that a sharp and abrupt change in the density cannot occur. It referred to the fact that density is independent of the sampling size. Figure 1.2 shows the density as a function of the sample size. After certain sample size, the density remains constant. Thus, the density is defined as \[\rho = \lim_{\Delta V\to \epsilon} \frac{\Delta m}{\Delta V}\tag{1}\]
It must be noted that
Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
I recently asked this on Biology.SE, and need to start with the master equation for the reaction: $$\ce{E + S -> ES}$$ Where $\ce{E}$ is our enzyme and $\ce{S}$ the substrate to get a probability density function for the binding time of a substrate with one specific enzyme to be: $$f{\left(t\right)}=k\left[\ce{S}\right]\mathrm e^{-k\left[\ce{S}\right]t}$$ I have read up on master equations, but really have no idea where to start in this derivation, any help would be appreciated?
The way to approach this is not how you have mentioned in your answer.
The representation of the master equation is correct and the generalized form of the chemical master equation (CME)
[1] would be:
$$\frac{\partial P(x,t)}{\partial t}=\displaystyle\sum _{\mu=1}^M a_\mu(x-v_\mu)\ P(x-v_\mu,t) - a_\mu(x)\ P(x,t)$$
Where:
$x$ denotes the state of the chemical system (number of molecules of each species) $\mu$ denotes a type of reaction $a_\mu$ denotes the probability of happening of reaction-$\mu$ in a small interval (also called reaction propensity) $v_\mu$ denotes the change in the state due to the reaction-$\mu$ $M$ denotes the total number of possible reactions
As far as I can understand (also from your question on Biology.SE), you are interested in knowing the distribution of the time interval when no reaction happens. A chemical reaction can be modelled as a Poisson process which is based on 3 main postulates:
$\lim\limits_{h\to0+}\dfrac{P(N_h=1)}h=\lambda$
i.e. the probability of occurrence of one event in a very small interval of time is equal to the macroscopic rate or intensity ($\lambda\,$).
$P(N_h\geqslant2)=o(h)$
i.e. the probability of occurrence of more than one events in an infinitesimal interval is essentially zero.
Consecutive events are independent of each other.
A chemical reaction seems to follow all these postulates. The detailed derivation of the CME can be found in [1]; I am trying to present a short derivation based on Poisson postulates.
The probability that system is in state $x$ at time $t+\Delta t$ would be the probability that system is in state $x-v_\mu$ at time $t$ times the probability that reaction $\mu$ happens in the interval $\Delta t$,
orthe probability that system is in state $x$ at time $t$ times probability that no reaction happens in the interval $\Delta t$.
$$P(x,t+\Delta t)=\sum_\mu P(x-v_\mu,t)\ P(\mu) + P(x,t)\left(1-\sum_\mu P(\mu) \right)$$
when you rearrange this equation, divide by $\Delta t$ and take the limit of $\Delta t \to 0$, you will end up with the CME. Note that, as per the first postulate: $$\lim\limits_{\Delta t\to 0}\frac{P(\mu)}{\Delta t}=a_\mu$$
Now, the probability that $k$ reactions would have happened in a time interval $\tau$ follows the Poisson distribution (see
[2] for the derivation).
$$P(N=k)=\frac{(\lambda\tau)^k e^{-\lambda \tau}}{k!}$$
This is a well known property of the Poisson distribution that the time interval between reactions will follow exponential distribution. The probability that no reaction has happened in time interval $\tau$ would be:
$$ P(N=0)=\frac{(\lambda\tau)^0 e^{-\lambda \tau}}{0!}=e^{-\lambda\tau}$$
in this case, $\lambda=\displaystyle\sum _{\mu=1}^M a_\mu$ i.e. probability of any reaction.
Conversely, the probability that
some reaction would happen in time $\tau$ would be $P(r\leq \tau) = 1-e^{-\lambda\tau}$. Where $r$ denotes the first reaction time. This is the cumulative distribution function for the exponential distribution.
In other words, the reaction time between intervals would follow an exponential distribution such that the probability that no reaction would happen will exponentially reduce with the size of the interval $\tau$.
In your specific case there are two reactions — association and dissociation, the propensities of whose will be:
$$a_f=\frac{k_f}{V}n_E.n_S\\[1em] a_r=k_r n_{ES}=k_r(n_{E_0}-n_E)$$
Where $k_f$ is the rate constant of association, $k_r$ is the rate constant for dissociation (as per mass action kinetics), $n_i$ denotes the number of molecules of species-$i$ and $V$ is the volume (of the reaction vessel). See
[3] for details. Note that the amounts of the species change with time, and so will the reaction propensities. You can calculate the exact reaction time distribution when the system is at the deterministic steady state (using the steady state concentrations).
I think this answers your question on [biology.se] as well.
References: Gillespie, Daniel T. "A rigorous derivation of the chemical master equation". Physica A: Statistical Mechanics and its Applications, 188 (1992): 404-425. Hogg, Robert V., and Allen T. Craig. Introduction to mathematical statistics. New York: Macmillan, 1978. Gillespie, Daniel T. "A general method for numerically simulating the stochastic time evolution of coupled chemical reactions." Journal of computational physics22.4 (1976): 403-434.
(Note I am the OP)
Master equation
The master equation can be written as: $$\frac{\mathrm d p_n{\left(t\right)}}{\mathrm dt}=\sum_{n'}\left\{W_{nn'}p_{n'}{\left(t\right)}-W_{n'n}p_n{\left(t\right)}\right\}$$ Where $p_n$ denotes the probability that the system is in state $n$ at time $t$. And $W_{nn'}\Delta t$ denotes the probability of the system changing from state $n'$ to the state $n$ in the time interval $\Delta t$.
Specific case
Let us say we have $n=1$ free enzyme, then: $$ \frac{\mathrm dp_0{\left(t\right)}}{\mathrm dt}=W_{01}p_1{\left(t\right)}$$ So what is $W_{01}$? This is the probability that the one enzyme is used up, let us call this $\mu$. $$\frac{\mathrm dp_0{\left(t\right)}}{\mathrm dt}=\mu p_1{\left(t\right)}$$ The average number of 'free enzymes' is given by: $$\left\langle n \right\rangle=p_1{\left(t\right)}$$ Differentiating this gives: $$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=\frac{\mathrm dp_1{\left(t\right)}}{\mathrm dt}$$ But in our case, $$ \frac{\mathrm dp_1{\left(t\right)}}{\mathrm dt}=-\mu p_1{\left(t\right)}$$ Thus $$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=-\mu p_1{\left(t\right)}$$ $$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=-\mu\left\langle n \right\rangle$$ From here with the assumption that $\mu\propto [\ce{S}]$ you can follow through my derivation on Biology.SE
The reason for the assumption $\mu\propto [\ce{S}]$ can be explained by the fact that $\mu$ is related to the rate of the reaction, which increases (approximately) linearly with substrate concentration. If this assumption is not valid, please can you comment.
Sources
1.'Stochastic processes in physics and chemistry' by N.G. Van Kampen chapter V |
That probably depends on what you consider to be the "form" of the Schrödinger equation.
As Todd succinctly said, the Schrödinger equation[s]
$$\mathrm i\hbar \frac{\mathrm d|\psi\rangle}{\mathrm dt} = \hat{H}|\psi\rangle \quad \text{(time-dependent)} \tag{1}$$
or
$$\hat{H}|\psi\rangle = E|\psi\rangle \quad\text{(time-independent)} \tag{2}$$
don't change. The Hamiltonian will, of course, be different, but the question is how different? The general form of the (non-relativistic) Hamiltonian for an atom has three components
$$\hat{H} = \color{red}{\underbrace{-\frac{\hbar^2}{2m_N}\nabla_N^2}_{\text{KE of nucleus}}} \,\,\,\, \color{blue}{\underbrace{- \sum_i \frac{\hbar^2}{2m_e}\nabla_i^2}_{\text{KE of electrons}}} \,\,\,\, \color{green}{\underbrace{-\sum_i\frac{Ze^2}{4\pi\varepsilon_0r_i}}_{\text{nucleus-electron PE}}}\,\,\,\, \color{purple}{\underbrace{+ \sum_i\sum_{j>i} \frac{e^2}{4\pi\varepsilon_0r_{ij}}}_{\text{electron-electron PE}}} \tag{3}$$
where the nucleus has mass $m_N$ and charge $Ze$; and $r_i$ denotes the distance between electron $i$ and the nucleus, and $r_{ij}$ denotes the distance between electrons $i$ and $j$.
Whether you are dealing with a hydrogen atom or a silver atom, this form remains valid. The only special thing about the hydrogen atom (or in a hydrogenic ion) is that there is only one electron, so the first two sums only have one term, and the last sum has no terms. So, the Hamiltonian for the hydrogen atom can be simplified to
$$\hat{H}_\text{H atom} = \color{red}{-\frac{\hbar^2}{2m_p}\nabla_p^2} \color{blue}{- \frac{\hbar^2}{2m_e}\nabla_e^2} \color{green}{- \frac{e^2}{4\pi\varepsilon_0r}} \tag{4}$$
but I'm of the opinion that the
form of the Hamiltonian
$$\hat{H} = \color{red}{\hat{T}_n} + \color{blue}{\hat{T}_e} + \color{green}{\hat{V}_{en}} + \color{purple}{\hat{V}_{ee}} \tag{5}$$
doesn't change, it's just that there are fewer terms in each sum. You might disagree, but I don't really care, as long as you write the correct Hamiltonian out. |
We are given a binary implementing some cryptographic scheme, a public key, and an encrypted flag. After some reverse engineering in IDA, we could restore the scheme, which can be described as follows:
First of all, we operate in the field $\mathbb{Z}_p$, where $p = 2^{32} - 5$.
Key generation:
Generate random matrices $X_0 \in \mathbb{Z}_p^{8\times8},\,A^\prime, B^\prime \in \mathbb{Z}_p^{4\times8}$. Obtain $A, B \in \mathbb{Z}_p^{8\times8}$ as rows of $A^\prime, B^\prime$ plus some their linear combinations. Obtain $C = -(A X_0^2 + B X_0)$. Finally, $X_0$ becomes a private key, and $\langle A, B, C\rangle$ becomes a public key.
Encryption:
Encode the plaintext message as $M \in \mathbb{Z}_p^{8\times8}$. Generate a random matrix $R \in \mathbb{Z}_p^{8\times8}$. Obtain $A_e = R A,\, B_e = R B,\, C_e = R C + M$. $\langle A_e, B_e, C_e\rangle$ becomes an encrypted message.
Decryption:
Obtain $M = A_e X_0^2 + B_e X_0 + C_e$. Decode $M$ as a plaintext message.
The key idea of this cryptographic scheme is to consider a quadratic matrix equation
$$AX^2 + BX + C = 0.$$
The decryption procedure works because we can cancel out the random $R$ using the property that the private $X_0$ satisfies this quadratic equation by construction of $C$:
$$
\begin{aligned} A_e X_0^2 + B_e X_0 + C_e &= \left(R A\right) X_0^2 + \left(R B\right) X_0 + \left(R C + M\right) \\ &= R \cdot \left(A X_0^2 + B X_0 + C\right) + M \\ &= R \cdot 0 + M \\ &= M. \end{aligned} $$
The matrices $A, B$ are generated the way that almost surely $\operatorname{rank} A = \operatorname{rank} B = 4$, so $A$ is not invertible, which prevents us from completing the square and scaling through by $A^{-1}$ to solve the quadratic equation effectively.
However, we can actually cancel out the $R$ with
another equation which we can solve effectively, for example, the linear one:
$$\left(A + B\right) Y + C = 0.$$
Since $\left(A + B\right)$ is invertible, $Y_0 = -\left(A + B\right)^{-1} C$ is obviously the solution of this linear equation, hence
$$
\begin{aligned} \left(A_e + B_e\right) Y_0 + C_e &= \left(R A + R B\right) Y_0 + \left(R C + M\right) \\ &= R \cdot \left(\left(A + B\right) Y_0 + C\right) + M \\ &= R \cdot 0 + M \\ &= M. \end{aligned} $$
Another way to look at this trick is to notice that
$$
\begin{aligned} R &= R \cdot I \\ &= R \cdot (A + B)(A + B)^{-1} \\ &= (RA + RB)(A + B)^{-1} \\ &= (A_e + B_e)(A + B)^{-1},\\ M &= C_e - RC \\ &= C_e - (A_e + B_e)(A + B)^{-1}C. \end{aligned} $$
Anyway, we can decrypt the flag using only the public key $\langle A, B, C\rangle$ and the encrypted message $\langle A_e, B_e, C_e\rangle$:
TWCTF{pa+h_t0_tomorr0w} |
Before we start with calculations we should think about
which values of $r$ we should take into account. $r=1,2$: We can exclude the trivial cases $r=1$ and $r=2$ which give a probability of zero, since we never reach a configuration with three people having the same birthday. $2<r\leq 365$: The main part where we put the focus of our analysis. Fortunately the reasoning can be easily extended to $r\leq 730$. $365<r\leq 2\cdot 365$: Assuming the year has $365$ days, the situation becomes slightly different if the number of people is greater than $365$. In these cases it is guaranteed that there are people with the same birthday and this needs some additional thoughts. $r>2\cdot 365$: According to the pigeonhole principle groups with more than $730$ people have at least three equal birthdays with probability $p=1$.
Starter:
In order to get a first impression and to have a verification for small numbers at hand we look at a smaller example. It should be small enough to make calculations easy and large enough to be representative for the problem.
We take $r=5$ people and instead of $365$ days we use $6$ days. If we take a fair die with six faces instead, this smaller version of the problem can then be stated as:
Small Problem: What is the probability that in a group of $5$ people, each of them throwing a fair die once, at least three people roll the same number.
Five people can throw a total of $6^5=7776$ different configuration of $5$-tuples from $(1,1,1,1,1)$ to $(6,6,6,6,6)$.
We answer the question by counting the $5$-tuples which have
no more than two equal faces.
Let $N_{(11111)}$ be the number of $5$-tuples with pairwise different entries. We see that there are $6\cdot5\cdot\ldots\cdot1=6!$ possibilities and get
\begin{align*}
N_{(11111)}=6!=720
\end{align*}
The index $11111$ indicates that there are
five pairwise different values counted.
Now we consider the number of $5$-tuples with one or more pairs of equal numbers, but with no occurrence of three or more equal numbers. There are two different constellations:
One pair and three other numbers which are pairwise different, denoted by $N_{(1112)}$.
The second constellation are two different pairs and another number different to each of them, denoted by $N_{(122)}$.
We obtain
\begin{align*}
N_{(1112)}&=\frac{1}{1!}\binom{5}{2}6\cdot5\cdot4\cdot3=3600\tag{1}\\
N_{(122)}&=\frac{1}{2!}\binom{5}{2}6\binom{3}{2}5\cdot4=1800\tag{2}
\end{align*}
Comment:
In (1) there are $\binom{5}{2}$ pairs which can take the values $1,\ldots,6$. The other three people have $5,4$ and $3$ possibilities to throw a number pairwise different to all the others.
In (2) there are $\binom{5}{2}\cdot6$ possibilities for one pair, leaving $\binom{5-2}{2}\cdot5$ possibilities for the second pair. Since we don't want to count different pair constellations more than once, we have to divide them by $2!$.
The distribution table of all different $5$-tuples can be easily calculated
\begin{array}{cccccccc}
\text{Total}&N_{(11111)}&N_{(1112)}&N_{(113)}&N_{(122)}&N_{(14)}&N_{(23)}&N_{(5)}\\
7776&\color{blue}{720}&\color{blue}{3600}& 1200& \color{blue}{1800} & 150 & 300 & 6
\end{array}
We are ready to calculate the wanted probability $p$ for this small example:
\begin{align*}
p&=1-\frac{N_{(11111)}}{6^5}-\left(\frac{N_{(1112)}}{6^5}+\frac{N_{(122)}}{6^5}\right)\\
&=1-\frac{720}{6^5}-\frac{5400}{6^5}\\
&=1-\frac{6120}{7776}\\
&=0.212963
\end{align*}
The real problem:
Most of the work has been done by the small example. If we now consider $365$ days and $r$ people we can go on straight forward.
At first we put the focus at
\begin{align*}
3\leq r \leq 365
\end{align*}
Let's denote with $N_{(1^{r})}$ the number of $r$-tuples with all entries pairwise different. The exponent in the index $1^{r}$ denotes the
multiplicity of $1$.
We obtain
\begin{align*}
N_{(1^{r})}=\frac{365!}{(365-r)!}\tag{3}
\end{align*}
Next we have to consider $r$-tuples containing $k$ pairs of tuples, with $1\leq k\leq \lfloor\frac{r}{2}\rfloor$ together with $r-2k$ single numbers pairwise different to all other single numbers and numbers within tuples. We denote them with $N_{(1^{r-2k}2^k)}$.
We obtain analogously to (2)
\begin{align*}
N_{(1^{r-2k}2^k)}&=\frac{1}{k!}\binom{r}{2}365\binom{r-2}{2}364\cdots\binom{r-2(k-1)}{2}\left(365-(k-1)\right)\\
&\qquad\cdot(365-k)\cdots((365-k)-(r-2k-1))\\
&=\frac{1}{k!}\binom{r}{2}\binom{r-2}{2}\cdots\binom{r-2(k-1)}{2}\frac{365!}{(365-(r-k))!}\\
&=\frac{1}{k!}\frac{r!}{2^k(r-2k)!}\frac{365!}{(365-(r-k))!}\tag{4}
\end{align*}
Plausibility check: If we compare the expression (4) with the small example above we could do a simple check. If we set\begin{align*}r=5,k=2,\text{ and }365\rightarrow 6\end{align*}we obtain\begin{align*}N_{(1^{5-4}2^2)}=N_{(12^2)}=\frac{1}{2!}\frac{5!}{2^2(5-4)!}\frac{6!}{(6-(5-2))!}=1800\end{align*}in accordance with the table entry $N_{(122)}$ above.
The range $365<r\leq 2\cdot 365$
If $r>365$ it is guaranteed that at least two people have the same birthday. It follows
$$N_{(1^r)}=0$$
We see after short consideration, the approach (4) for a group of people with $1\leq k\leq \lfloor r\rfloor$ distinct pairs is also valid up to $r\leq 730$.
Now it's time to harvest. In order to find the number of all $r$-tuples we have to exclude, we sum up $N_{(1^r)}$ and $N_{(1^{r-2k}2^k)}$ for $1\leq k \leq \lfloor\frac{r}{2}\rfloor$.
In fact, inspired by the comment of @robjohn, we can also respect
all other values of $r$ in the same formula, if we take $\frac{1}{n!}=0$ for $n<0$ and finally obtain
The probability $p$ that at least three people from a group of $r\geq 1$ people have the same birthday is according to (3) and (4)
\begin{align*}
p&=1-\frac{1}{365^r}\left(N_{(1^r)}+\sum_{k=1}^{\lfloor\frac{r}{2}\rfloor}N_{(1^{r-2k}2^k)}\right)\\
&=1-\frac{365!}{365^r}\left(\frac{1}{(365-r)!}
+r!\sum_{k=1}^{\lfloor\frac{r}{2}\rfloor}\frac{1}{2^kk!(r-2k)!(365-(r-k))!}\right)\\
&=1-r!\frac{365!}{365^r}\sum_{k=0}^{\lfloor\frac{r}{2}\rfloor}\frac{1}{2^kk!(r-2k)!(365-(r-k))!}
\end{align*} |
Under the auspices of the Computational Complexity Foundation (CCF)
We develop an analytic framework based on linear approximation and point out how a number of complexity related questions -- on circuit and communication complexity lower bounds, as well as pseudorandomness, learnability, and general combinatorics of Boolean functions -- fit neatly into this framework. ... more >>>
We establish hardness versus randomness trade-offs for a
broad class of randomized procedures. In particular, we create efficient nondeterministic simulations of bounded round Arthur-Merlin games using a language in exponential time that cannot be decided by polynomial size oracle circuits with access to satisfiability. We show that every language with ... more >>>
A new notion of predictive complexity and corresponding amount of
information are considered. Predictive complexity is a generalization of Kolmogorov complexity which bounds the ability of any algorithm to predict elements of a sequence of outcomes. We consider predictive complexity for a wide class of bounded loss functions which ... more >>>
For circuit classes R, the fundamental computational problem, Min-R,
asks for the minimum R-size of a boolean function presented as a truth table. Prominent examples of this problem include Min-DNF, and Min-Circuit (also called MCSP). We begin by presenting a new reduction proving that Min-DNF is NP-complete. It is significantly ... more >>>
The sign-rank of a matrix A=[A_{ij}] with +/-1 entries
is the least rank of a real matrix B=[B_{ij}] with A_{ij}B_{ij}>0 for all i,j. We obtain the first exponential lower bound on the sign-rank of a function in AC^0. Namely, let f(x,y)=\bigwedge_{i=1}^m\bigvee_{j=1}^{m^2} (x_{ij}\wedge y_{ij}). We show that the matrix [f(x,y)]_{x,y} has ... more >>>
We establish a generic form of hardness amplification for the approximability of constant-depth Boolean circuits by polynomials. Specifically, we show that if a Boolean circuit cannot be pointwise approximated by low-degree polynomials to within constant error in a certain one-sided sense, then an OR of disjoint copies of that circuit ... more >>>
We study the maximum possible sign rank of $N \times N$ sign matrices with a given VC dimension $d$. For $d=1$, this maximum is $3$. For $d=2$, this maximum is $\tilde{\Theta}(N^{1/2})$. Similar (slightly less accurate) statements hold for $d>2$ as well. We discuss the tightness of our methods, and describe ... more >>>
In the first part of this work, we introduce a new type of pseudo-random function for which ``aggregate queries'' over exponential-sized sets can be efficiently answered. An example of an aggregate query may be the product of all function values belonging to an exponential-sized interval, or the sum of all ... more >>>
The sign-rank of a matrix $A$ with entries in $\{-1, +1\}$ is the least rank of a real matrix $B$ with $A_{ij} \cdot B_{ij} > 0$ for all $i, j$. Razborov and Sherstov (2008) gave the first exponential lower bounds on the sign-rank of a function in AC$^0$, answering an ... more >>>
Threshold weight, margin complexity, and Majority-of-Threshold circuit size are basic complexity measures of Boolean functions that arise in learning theory, communication complexity, and circuit complexity. Each of these measures might exhibit a chasm at depth three: namely, all polynomial size Boolean circuits of depth two have polynomial complexity under the ... more >>>
The communication class $UPP^{cc}$ is a communication analog of the Turing Machine complexity class $PP$. It is characterized by a matrix-analytic complexity measure called sign-rank (also called dimension complexity), and is essentially the most powerful communication class against which we know how to prove lower bounds.
For a communication problem ... more >>>
In this paper we study the quantum learnability of constant-depth classical circuits under the uniform distribution and in the distribution-independent framework of PAC learning. In order to attain our results, we establish connections between quantum learning and quantum-secure cryptosystems. We then achieve the following results.
1) Hardness of learning ... more >>> |
Consider the Correlated Random Effects model $y_{it} = \alpha + x_{it}\beta + \bar x \gamma + w_i + \epsilon_{it} $ where $x_{it}$ is a scalar explanatory variable.
The correlated random effects GLS estimator $ \hat \beta_{CRE} $ is the OLS estimator of $\beta$ in the quasi-demeaned regression
$\tilde y_{it} = \delta + \tilde x_{it} \beta + \bar x_i \rho + u_{it} $,
where $\tilde y_{it} = y_{it} - \theta \bar y_i , \tilde x_{it} = x_{it} - \theta \bar x_i $ and $\theta = 1 - (\sigma^2_\epsilon/ (\sigma^2_\epsilon + T\sigma^2_w))^{1/2} $
Question: I need to show that the residuals from the regression of $x_{it} - \bar x_i $ on a constant and $\bar x_i $ is just $x_{it} - \bar x_i $ itself.
Attempt: Regress $x_{it} - \bar x_i = \alpha + \bar x_{i} + \tilde r_{it} $ rearrange to get the residuals, $\tilde r_{it} = (x_{it} - \bar x_i) - (\alpha + \bar x_{i})$ I'm not sure how to proceed. |
A number field $K$ is said to have a
power basis if there is an $\alpha \in K$ such that the full ring of integers $O_K$ is the $\mathbb{Z}$-linear span of $1,\alpha,\alpha^2,\ldots,\alpha^{\deg{K}-1}$. In other words, $O_K = \mathbb{Z}[\alpha]$; another term for this is monogenic. This happens for example for the quadratic and the cyclotomic fields. The monogenic number fields are presumably very rare; results of Bhargava imply that they are a negligible fraction (zero density) among the number fields of degree $3, 4$, or $5$, and this is conjectured to hold in every fixed degree $d > 2$.
It is easily seen that totally $p$-adic number fields of degree $d$ are never monogenic for $d \gg p$. (To illustrate this, consider that the split primes in the cyclotomic field of level $N$ are the ones $\equiv 1 \mod{N}$; so they are in particular $> N > \phi(N)$.)
Question. Are there monogenic totally real number fields of arbitrarily high degree?
To put it differently: for totally real algebraic integers of arbitrarily high degree, may the ring $\mathbb{Z}[\alpha]$ be integrally closed?
Dummit and Kisilevsky [
Indices in cyclic cubic fields, in "Number Theory and Algebra," 1977] have shown that infinitely many totally real cubic fields have a power basis. There exist totally real monogenic sextic fields; an example, taken at will from page 116 of [I. Gaal, Diophantine Equations and Power Integral Bases: New Computational Methods] is the field generated by a root of$$X^6 - 5X^5 +2X^4 +18X^3 - 11X^2 -19X+1.$$Are there examples of higher degree? |
Answer
Please see the work below.
Work Step by Step
We know that $v=\sqrt{\frac{F}{\mu}}$ This simplifies to: $F=\mu v^2$ We plug in the known values to obtain: $F=(0.170)(6.7)^2$ $F=7.6N$
You can help us out by revising, improving and updating this answer.Update this answer
After you claim an answer you’ll have
24 hours to send in a draft. An editorwill review the submission and either publish your submission or provide feedback. |
Francesco Polizzi
Mathematician, researching in Algebraic Geometry and related topics.
Italy
Member for 9 years, 3 months
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HP 17bII+ Silver solver - Printable Version
+- HP Forums (
https://www.hpmuseum.org/forum)
+-- Forum: HP Calculators (and very old HP Computers) (
/forum-3.html)
+--- Forum: General Forum (
/forum-4.html)
+--- Thread: HP 17bII+ Silver solver (
/thread-11398.html) HP 17bII+ Silver solver - pinkman - 09-14-2018 09:20 PM
I'm very satisfied with my HP 17bII+ Silver, which I find very powerful, but also nice and not looking too complicated (no [f] and [g] functions on keys like the HP 12c or HP 35s that I used to use at work every day, but a really complete calculator except for trigs and complex numbers calculations).
So the 17BII+ is my new everyday calculator. I don't come back to arguments where a good (HP) calculator is a perfect complement or subsitute to Excel.
I chose the 17BII+ after having carefully studied the programs I use in my day to day work:
- price and costs calculations: can be modeled in the solver, with 4 or 5 equations and share vars
- margin calculations: built-in functions
- time value of money: built-in functions
- time functions: built-in functions
For the few moments I need trigs or complex calculations, I always have Free42 or a real 35s / 15c not really far from me.
After having rebuilt my work environment in the (so powerful) solver, I started to study the behavior of the solver, looking at loops (Σ function), conditional branchings (IF function), menu selection (S function), and the Get and Let functions (G(), L()).
I googled a lot and found an interesting pdf file about the Solvers of the 19B, 17B, 17BII and - according to the author, the 17BII+ Silver.
The file is here : http://www.mh-aerotools.de/hp/documents/Solver%20GET-LET.pdf
I tried a few equations in the "Using New and Old Values" chapter, pages 4 and following.
There I found lots of differences with my actual 17BII+ Silver, which I would like to share here with you.
For instance, the following equation found page 4:
Code:
In the next example found page 5:
Code:
Then the equation :
Code:
I don't understand the first 2 cases. In the first one, A should be set to -B, not B, if not using the "old" value of A. In the second one, the solver does not use the "old" value, but it does not also solve the equation, as there is no defined solution.
In the last case, I understand that the equation is evaluated twice before finding an answer. So the old value is used there, but not the way I could expect.
I finally found one - and only one - way to use iterations in the solver, with the equation :
Code:
Note that neither A=G(A)+1 or A=1+G(A) or G(A)+2=A works.
I'm not disappointed, as the solver is a really interesting and useful feature of the calculator, but I'm just surprised not having found more working cases of iterations, or a clear understanding of how the solver works.
Comments are welcomed.
Regards,
Thibault
RE: HP 17bII+ Silver solver - rprosperi - 09-14-2018 10:14 PM
The 17BII+ (Silver Edition) is an excellent machine, in fact it has the best keyboard of any machine made today by HP, but the solver does have a bug, and there are also a few other smaller issues making the solver slightly inferior to the 17B/17BII/19B/19BII/27S version.
See these 3 articles for details:
http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv021.cgi?read=242551
http://www.hpmuseum.org/forum/thread-6573-post-58685.html#pid58685
http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv018.cgi?read=134189
Overall these are not dramatic issues, and once you understand the solver bug, you likely can create equations that can avoid the issue.
I have most HP machines but the 17BII and 17BII+ are the ones I use most often for real work (vs. playing, exploring or following along interesting threads here).
Edit: added 3rd link
RE: HP 17bII+ Silver solver - Don Shepherd - 09-15-2018 12:05 AM (09-14-2018 09:20 PM)pinkman Wrote: I'm not disappointed, as the solver is a really interesting and useful feature of the calculator, but I'm just surprised not having found more working cases of iterations, or a clear understanding of how the solver works.
Thibault, when Kinpo built the 17bii+ calculator years ago (both the gold one and the silver one), they basically goofed the solver implementation. This has been discussed at length in the HPMuseum forum. The solvers on the 17b and 17bii work fine, just as you would expect them to. If you plan on making significant use of the solver and its incredible capabilities, forget the + and get an original 17b or 17bii. You won't be sorry.
Also, get the manual (it's on the Museum DVD) Technical Applications for the HP-27s and HP-19b. It applies to the 17b as well.
The Sigma function also works fine on the 17b and 17bii.
RE: HP 17bII+ Silver solver - pinkman - 09-15-2018 04:36 AM
Thanks to both of you for the details, links and advice.
I've read the threads carefully, I did not find them by myself first. It's the end of the night now, I'll try to make few testing later.
RE: HP 17bII+ Silver solver - rprosperi - 09-15-2018 11:55 PM
If you want to really explore the capabilities of the awesome Pioneer Solver, and confidently try to push it without worrying about using L() this way or that, I agree with Don, buy a 17BII and use that for the Solver stuff, but continue to use the 17BII+ for every day stuff.
Here's a very nice 17BII for only $25 (shipping included, in US) and you can even find them cheaper if you're willing to wait:
https://www.ebay.com/itm/HP-17Bll-Financial-Calculator/113252533550
The 17BII is bug-free for solver use, while the 17BII+ has a much better LCD, readable in a wider range of lighting & use conditions.
In case you haven't seen this yet, here's an example of what can be done with the solver:
http://www.hpmuseum.org/forum/thread-2630.html
RE: HP 17bII+ Silver solver - pinkman - 09-16-2018 09:59 PM
Well I'll try to find one, even if I'm not in the US.
I also want to continue using my actual 17bII+ at work, as the solver is powerful enough for me (for the moment), and it looks really good.
Don and you Bob have done a lot to help understand what the solver can do, that's pretty good stuff.
Regards,
Thibault
RE: HP 17bII+ Silver solver - rprosperi - 09-16-2018 10:23 PM
Nah, Don and Gerson are the real Pioneer Solver Masters, I just have a good collection of links.
Of all the various tools built-in to the various calculator models I've explored, the Pioneer Solver is easily the one that most exceeds it's initial apparent capability. This awesome tool must have been incredibly well-tested by the QA team. The fact that the sheer size (and audacity!) of Gerson's Trig formulas still return amazingly accurate results says more about the underlying design and code than any comments I could add.
Enjoy exploring it, and when you've mastered it (or at least tamed it a bit), come back here and share some interesting Solver formulas. There are numerous folks here that enjoy entering the formulas and running some test cases. (well, I suppose "enjoy" is not really the right word about entering the formulas - I guess feel good about accomplishing it successfully is more accurate).
When I see some of these long Solver equations, it reminds of TECO commands back in the PDP-11 days.
RE: HP 17bII+ Silver solver - Thomas Klemm - 09-16-2018 11:05 PM
If you feel like entering long formulas you can solve the 8-queens problem.
):0)
Cheers
Thomas
RE: HP 17bII+ Silver solver - Don Shepherd - 09-17-2018 06:30 AM
How about a nifty, elegant, simple number base conversion Solver equation for the 17b/17bii, courtesy Thomas Klemm:
BC:ANS=N+(FROM-TO)\(\times \Sigma\)(I:0:LOG(N)\(\div\)LOG(TO):1:L(N:IDIV(N:TO))\(\times\)FROM^I)
Note: either FROM or TO must be 10 unless you are doing HEX conversions
RE: HP 17bII+ Silver solver - Thomas Klemm - 09-17-2018 07:49 AM (06-19-2014 02:06 PM)Thomas Klemm Wrote: You could set either FROM or TO to 100. This allows conversions between HEX and DEC.
Best regards
Thomas
RE: HP 17bII+ Silver solver - Don Shepherd - 09-17-2018 09:17 AM (09-17-2018 07:49 AM)Thomas Klemm Wrote:
Wow, I didn't realize that Thomas. Stunning.
Don
RE: HP 17bII+ Silver solver - Martin Hepperle - 09-17-2018 10:24 AM
As has already been written, unfortunately the solver in this calculator is flawed.
Part of the problems comes from the fact that it evaluates the equation twice which leads to incrementing by two instead of one etc. But there seem to be more quirks.
This is too bad, as the solver is a very capable tool (with G() and L()), while still easy and versatile to use (with the menu buttons).
Accomplishing something similar (solve for one variable today, for another tomorrow) with modern tools like Excel is more complicated.
Martin
RE: HP 17bII+ Silver solver - rprosperi - 09-17-2018 12:48 PM (09-16-2018 10:23 PM)rprosperi Wrote: Nah, Don and Gerson are the real Pioneer Solver Masters, I just have a good collection of links.
An obvious correction is in order here:
Don, Gerson and Thomas Klemm are the real Pioneer Solver Masters...
No disrespect intended by the omission. A review of past posts of significant solver formulas quickly reveals just how often all three of these three guys were the authors.
RE: HP 17bII+ Silver solver - pinkman - 09-28-2018 01:33 PM
Be sure I have great respect of everyone mentioned above
Thanks for the advices, my new old 17bii is now in my hands ($29 on French TAS), I'm just waiting for the next rainy Sunday for pushing the solver to it's limits (mmh, to MY limits I guess).
RE: HP 17bII+ Silver solver - Jlouis - 09-28-2018 07:47 PM
Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?
TIA
RE: HP 17bII+ Silver solver - mfleming - 09-29-2018 01:37 AM (09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?
Check out the document by Martin Hepperle in this thread which lists the functions available for each calculator. The document is an excellent introduction to the Solver application.
Solver GET-LET
~Mark
RE: HP 17bII+ Silver solver - Jlouis - 09-29-2018 02:25 AM (09-29-2018 01:37 AM)mfleming Wrote:(09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?
Thanks mfleming, I should have read the beginning of the manual. It is the same solver for the calculators of my question. I prefer to use the 19bII due the dedicated alpha keyboard, but the 27s is a pleasure to use too.
Cheers
RE: HP 17bII+ Silver solver - Don Shepherd - 09-29-2018 11:13 AM (09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?The 27S solver has a random number function; the 17b/bii do not. I'm not sure about the 18c and 19b. RE: HP 17bII+ Silver solver - Don Shepherd - 09-29-2018 11:53 AM (09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?Here is a 3-page cross-reference I created a few years ago of Solver functions present in the 19bii, 27s, and 17bii.
[attachment=6374]
[attachment=6375]
[attachment=6376]
RE: HP 17bII+ Silver solver - Jlouis - 09-29-2018 01:36 PM (09-29-2018 11:53 AM)Don Shepherd Wrote:
Really thanks Dom for taking time to make these documents.
This is what makes MoHC a fantastic community.
Cheers
JL
Edited: Looks like the 19BII is a little more complete than the 27s and that the 17BII is the weaker one. |
https://doi.org/10.1351/goldbook.F02514
Source of @C01130@ in which the @A00106@ is an electron beam moving at speeds close to the speed of light in the spatially periodic magnetic field produced by an array of magnets (the wiggler). The emitted @W06659@, \(\lambda _{\text{L}}\), is approximately given by \[\frac{\lambda _{\omega }}{4\ E^{2}}\] with \(\lambda _{\omega }\) being the wiggler period and \(E\) the electron's energy in \(\mathrm{MeV}\).
See:
laser |
I'm making a poster (using baposter) where I wish to include some microscopy images. To make the fluorescent signals stand out better, I changed the background of that cell to "pitchblack", which I have defined as
\definecolor{pitchblack}{cmyk}{0,0,0,1}
I've also tried this with
\definecolor{pitchblack}{rgb}{0,0,0}
However, with my microscopy images in the foreground, I find out that "black" isn't really black:
The
baposterclass brings
xcolor and
tikz along for the ride, so I'm assuming my color issue is with those packages. So my question is, how I can make the "black" of the background frame black enough to match the background of these images?
Full code of the frame:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\headerbox{Preliminary Images}{name=prelim,column=2,row=0,boxColorOne=pitchblack}{%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\begin{center}\renewcommand{\arraystretch}{0}\begin{tabular}{@{}c@{}c@{}c@{}}\raisebox{-2ex}{\parbox[t]{0.25\textwidth}{\scriptsize\color{white} Representative volume renderings of HeLa cells imaged using TALEsagainst the $\beta$-globin locus. Signals are approximately 0.3$\mu$min diameter.}} &\imagetop{\includegraphics[width=0.3\textwidth]{dataimages/5_02_volume_view.jpg}} & \imagetop{\includegraphics[width=0.3\textwidth]{dataimages/5_04_volume_view.jpg}} \\ \imagetop{\includegraphics[width=0.3\textwidth]{dataimages/5_05_volume_view.jpg}} & \imagetop{\includegraphics[width=0.3\textwidth]{dataimages/5_07_volume_view.jpg}} & \imagetop{\includegraphics[width=0.3\textwidth]{dataimages/5_10_volume_view.jpg}}\end{tabular}\end{center}}
As a final note, the problem appears to be independent of the pdf viewer (i.e. it looks this way in evince,acroread,and gimp). |
Can you guys help me solve this integral? $$ \int \frac{x^3}{x^4 + 2x^2 - 6}dx $$
closed as off-topic by Watson, Shailesh, Claude Leibovici, Davide Giraudo, R_D Jul 25 '16 at 14:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
" This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Watson, Shailesh, Claude Leibovici, Davide Giraudo, R_D
Let $t=x^2+1$ then your integral becomes $$\frac{1}{2}\int \frac{t-1}{t^2-7}\,dt=\frac{1}{2}\int \frac{tdt}{t^2-7}- \frac{1}{2}\int \frac{dt}{t^2-7}.$$ Now $$\int \frac{tdt}{t^2-7}=\frac{1}{2}\ln(t^2-7)$$ and $$\int \frac{dt}{t^2-7}=-\frac{1}{\sqrt{7}}\mbox{arctanh}(t/\sqrt{7})$$
HINT:
$$x^4+2x^2-6=(x^2+1)^2-7$$
Let $x^2+1=y\implies2x\ dx=dy$ and $x^2=y-1$
$$\dfrac{2Ay+B}{y^2-7}=A\cdot\dfrac{2y}{y^2-7}+\dfrac B{y^2-7}$$
Use $\dfrac{2a\ du}{u^2-a^2}=\dfrac{u+a-(u-a)}{(u+a)(u-a)}=?$
$$I= \int \frac{x^3}{x^4 + 2x^2 - 6}dx$$
As Arthur suggests, let $u = x^2,du = 2x$
$$I = \frac{1}{2}\int \frac{u}{u^2 + 2u - 6}du$$ $$ = \frac{1}{4\sqrt{7}} (\int \frac{u}{u+1-\sqrt{7}}du -\int \frac{u}{u+1+\sqrt{7}}du ) $$
Let $z = u+1-\sqrt{7},w = u+1+\sqrt{7}$
$$I = \frac{1}{4\sqrt{7}} (\int \frac{z-1+\sqrt{7}}{z}dz -\int \frac{w-1-\sqrt{7}}{w}dw) = \frac{1}{4\sqrt{7}} (z +(\sqrt{7}-1)\ln{z} - w+(1+\sqrt{7}) \ln w)\\ = \frac{1}{4\sqrt{7}} (u+1-\sqrt{7} +(\sqrt{7}-1)\ln(u+1-\sqrt{7}) - u+1+\sqrt{7} + (1+\sqrt{7})\ln (u+1+\sqrt{7}) ) \\ = \frac{1}{4\sqrt{7}}(2+(\sqrt{7}-1)\ln(u+1-\sqrt{7})+(\sqrt{7}-1)\ln(u+1-\sqrt{7})) = \frac{1}{4\sqrt{7}}(2+(\sqrt{7}-1)\ln(x^2+1-\sqrt{7})+(\sqrt{7}-1)\ln(x^2+1-\sqrt{7}))$$ |
$$\int _0^{\frac{\pi }{2}}\sin^2x\cos^2x\,dx$$
I have been trying to solve this integral using the various trig identities but none worked. When I went on one of the online integral calculators it said that "hyperbolic identities" needed to be used. I have never encountered these before. Symbolab said it was an invalid input.
In the solutions manual, this was the first step:
$$\int _0^{\frac{\pi }{2}}\sin^2x\cos^2x\,dx=\int _0^{\frac{\pi }{2}}\frac{1}{4}\left(4\sin^2x\cos^2x\right)\,dx$$
I am already confused as to how this was done. It does not seem to be related to any usual trig identity. Can it be done without these "hyperbolic identities"?
Any help? |
Like you said Cook’s Distance measures the change in the regression by removing each individual point. If things change quite a bit by the omission of a single point, than that point was having a lot of influence on your model. Define $\hat{Y}_{j(i)}$ to be the fitted value for the jth observation when the ith observation is deleted from the data set. Cook’s Distance measures how much $i$ changes all the predictions.
$$D_i = \frac{\sum_{j=1}^{n}\hat{Y}_j - \hat{Y}_{j(i)})^2}{pMSE}$$$$= \frac{e_i^2}{pMSE}[\frac{h_{ii}}{(1-h_{ii})^2}]$$
If $D_i \geq 1$ it is extreme (for small to medium datasets).
Cook’s Distance shows the effect of the ith case on all the fitted values. Note that the ith case can be influenced by
big $e_i$ and moderate $h_{ii}$
moderate $e_i$ and big $h_{ii}$
big $e_i$ and big $h_{ii}$
In R, use the
influence.measures package with
cooks.distance(model) |
wiki-latex plugin
Hi,
this plugin provides a wiki macro to render images from latex source. It was inspired by the graphviz-macro.
The requirements are: latex, dvips, convert (from ImageMagick).
You can find it here: http://github.com/nisrael/redmine-wiki_latex_plugin/tree/master
Nils
Replies (40) RE: wiki-latex plugin - Added by Jerome Vanthournout over 10 years ago
Do you have a description of your plugin (snapshot, syntax, ...)?
RE: wiki-latex plugin - Added by Nils Israel over 10 years ago
Here a screenshot:
The syntax is as follows:
{{latex($\sum_i c_i$)}}
But I would like to use it with more comlpex formulas, like:
{{latex($\sum_{i=1}^k c_i$)}}
or
{{latex( \begin{equation} \sum_{i=1}^k c_i \end{equation} )}}
The problem here are the curly braces und newlines. All newlines are replaced by br or p tags
before the params are passed to the macro. And since the regex to match the macro arguments in formatter.rb, Line 105 matches everything, but not the closing curly brace I have no idea how to solve this without modifying the
formatter.rb.
RE: wiki-latex plugin - Added by Yoann Rousseau about 10 years ago
Hi !
I have some problems with this plugin.
When I write a formula like this : {{latex($\sum_{i=1}^k c_i$)}}
I don't obtain the desired result. The formula is simply printed without being interpreted.
The problem seems to come with curly braces. When I remove them, image of the formula is displayed.
So I can't use complex formulas.
Do you have any solution to resolve this problem ?
Thanks !
P.S. : The use \begin{equation} syntax doesn't change the problem.
RE: wiki-latex plugin - Added by Nils Israel about 10 years ago
Loot at this: http://www.redmine.org/issues/3061
or use this MACROS_RE in ./lib/redmine/wiki_formatting/textile/formatter.rb to
MACROS_RE = / (!)? # escaping ( \{\{ # opening tag ([\w]+) # macro name (\(((.(?!\}\}))*)\))? # optional arguments \}\} # closing tag ) /mx unless const_defined?(:MACROS_RE) RE: wiki-latex plugin - Added by Matt Ueckermann about 10 years ago
Hello,
I foolishly added the folder as "nisrael-redmine-wiki_latex_plugin-28000b9bf5a044f3a84c3cdd3cc0d562513a98db" instead of just "wiki_latex_plugin" and I got an error, that the exact plugin name (wiki_latex_plugin) should be used. I subsequently changed the folder name, and re-built the db, but then got an error the "wiki_latexes" table already exists. I no longer get an error for the plugin name usage in redmine, but I also do not get the rendered latex formula, I just get $\sum_i c_i$.
Foolish because I didn't back up my db before proceeding. Any suggestions?
Thanks!
~Matt RE: wiki-latex plugin - Added by Matt Ueckermann about 10 years ago
Also I'm running on Windows... unfortunately.
Matt Ueckermann wrote:
Hello,
I foolishly added the folder as "nisrael-redmine-wiki_latex_plugin-28000b9bf5a044f3a84c3cdd3cc0d562513a98db" instead of just "wiki_latex_plugin" and I got an error, that the exact plugin name (wiki_latex_plugin) should be used. I subsequently changed the folder name, and re-built the db, but then got an error the "wiki_latexes" table already exists. I no longer get an error for the plugin name usage in redmine, but I also do not get the rendered latex formula, I just get $sum_i c_i$.
Foolish because I didn't back up my db before proceeding. Any suggestions?
Thanks!
~Matt RE: wiki-latex plugin - Added by Matt Ueckermann about 10 years ago
Sorry, now I'm suspecting Windows to be the culprit. I use TeXnicCenter with MikTeX, and I have ImageMagick installed, but I'm not sure your plugin knows where to look for these in Windows. Anyway, that's a suspicion, I have no idea really. It would be nice to get it working, so any suggestions are very much appreciated.
Thanks! (Sorry for spamming the board. Every time after I post, I get new ideas).
RE: wiki-latex plugin - Added by Matt Ueckermann about 10 years ago
Sorry, now I'm suspecting Windows to be the culprit. I use TeXnicCenter with MikTeX, and I have ImageMagick installed, but I'm not sure your plugin knows where to look for these in Windows. Anyway, that's a suspicion, I have no idea really. It would be nice to get it working, so any suggestions are very much appreciated.
Thanks! (Sorry for spamming the board. Every time after I post, I get new ideas).
RE: wiki-latex plugin - Added by Yoann Rousseau about 10 years ago
Hi !
I don't really know the Windows environment, but on Linux this plugin needs 2 commands to run correctly. The first one is latex, the second one is dvipng.
If you look quickly the code of the plugin, you can see these two lines :
/app/controllers/wiki_latex_controller.rb, line 45 :
fork_exec(dir, "usr/bin/latex --interaction=nonstopmode "+@name+".tex 2> /dev/null > /dev/null")
/app/controllers/wiki_latex_controller.rb, line 46 :
fork_exec(dir, "/usr/bin/dvipng "+@name+".dvi -o "+@name+".png")
I don't know if this plugin can run on Windows, but you can try to check the two previous path in order to indicate to the plugin where to look for.
I hope it could help you.
Matt Ueckermann wrote:
Sorry, now I'm suspecting Windows to be the culprit. I use TeXnicCenter with MikTeX, and I have ImageMagick installed, but I'm not sure your plugin knows where to look for these in Windows. Anyway, that's a suspicion, I have no idea really. It would be nice to get it working, so any suggestions are very much appreciated.
Thanks! (Sorry for spamming the board. Every time after I post, I get new ideas).
RE: wiki-latex plugin - Added by Matt Ueckermann about 10 years ago
Thanks for pointing to the controller file... gives me something to play with.
Thus far I have tried:
fork_exec(dir, "C:\Progra~1\MiKTeX~1.7\miktex\bin\latex --src --interaction=nonstopmode "+@name+".tex")
and
fork_exec(dir, "C:\Progra~1\MiKTeX~1.7\miktex\bin\dvipng "+@name+".dvi -o "+@name+".png")
but no luck. I have two guesses, either I am suffering from my original folder problem, or Windows does not like the "+@name+" syntax. I tried replacing "+@name+" with a test .tex file and running the commands from the command window in Windows, and I ended up with a nice .png image of my test file.
Is there any way to make the temporary directory stick around so I can look at it? Perhaps figure out where in the chain things fail? I have not written any Redmine scripts, nor do I have to time right now to delve into the details, so any help would be appreciated. The problem might be that the controller file is no longer executed... the temporary directory never shows up when I check, so either it is removed each time, or it never gets created.
Thanks!
RE: wiki-latex plugin - Added by Matt Ueckermann about 10 years ago
I just fixed my folder naming issue... I changed a line in:
app\views\wiki_latex\macro_inline.html.erb
from
<%= stylesheet_link_tag "wiki_latex.css", :plugin => "wiki_latex_plugin", :media => :all %>
to (the rather ugly)
<%= stylesheet_link_tag "wiki_latex.css", :plugin => "nisrael-redmine-wiki_latex_plugin-28000b9bf5a044f3a84c3cdd3cc0d562513a98db", :media => :all %>
And I no longer get the error. Still, no nice latex formulas though, I just get $\sum$ displayed for instance. Strange, at some point the temporary directory was created, but now no longer... the mystery continues.
RE: wiki-latex plugin - Added by Yoann Rousseau about 10 years ago
Your problem could be caused by the used LaTeX packages. If your look for in the controller, you can see the following lines :
temp_latex.puts('\usepackage{amsmath}') temp_latex.puts('\usepackage{amsfonts}') temp_latex.puts('\usepackage{amssymb}') temp_latex.puts('\usepackage[active,displaymath,textmath,graphics]{preview}')
If you don't have installed these packages, it is normal that a problem occures. So check your LaTeX installation for each of these packages.
Think to have a look in the logs of latex, the package(s) in question should be referred. For that, don't forget to put the following line into comment because as you say that, temporary files are deleted for each execution : File.unlink(basefilename+"."+ext) RE: wiki-latex plugin - Added by Matt Ueckermann about 10 years ago
Hello, commenting that line helped a lot, thanks Yoann!
I have all the package for latex, but! no luck yet. A few steps closer: windows doesn't like fork(), so I had to try and do without. I wrote a small batch script, latex.bat containing the following:
C:\Progra~1\MiKTeX~1.7\miktex\bin\latex --src --interaction=nonstopmode %1.tex
C:\Progra~1\MiKTeX~1.7\miktex\bin\dvipng %1.dvi -o %1.png
I've also modified the controller.rb file, and it now reads:
Dir.chdir(dir)
exec("latex "+@name+"")
Strange thing, running the batch file from cmd I get a nice .png image from a .tex file. However, when redmine creates the .tex file and runs it, there is no .png file. Not sure why. My only guess is that redmine doesn't wait for exec to finish its task. I've also reversed the order of the commands in latex.bat (after the .dvi was created) and then the .png images are created. However, they still do not get displayed, I get a:
The image "imagename.png" cannot be displayed because it contains errors
I should just really give this up... but at this point it's personal. Any suggestions appreciated.
RE: wiki-latex plugin - Added by Carlos de la Torre almost 10 years ago
I've just installed the LaTeX plugin but I got the following error trying to use it:
I wrote
{{latex($\sum_i c_i$)}}
in a Wiki, and I get the following error (in a red box):
Error executing the latex macro (Mysql::Error: #42S02Table 'redmine_private.wiki_latexes' doesn't exist: SHOW FIELDS FROM `wiki_latexes`) RE: wiki-latex plugin - Added by Matt R almost 10 years ago
Hello, first of all the plugin is really cool!
I passed all the steps in order to set it up and it works fine. With one exception:
Whenever I put a comma inside the latex formula, it disapears in the output image. It doesn't even get parsed. Probably it gets devoured somewhere in the regexp guts, but not sure. How can I fix it?
Thanks in advance!
RE: wiki-latex plugin - Added by Nils Israel almost 10 years ago
Hi,
very quick hack:
Replace in
./lib/redmine/wiki_formatting/textile/formatter.rb (in
def inline_macros(text))
args = ($5 || '').split(',').each(&:strip)
with
if macro == 'latex' args = $5 else args = ($5 || '').split(',').each(&:strip) end RE: wiki-latex plugin - Added by Steve G over 9 years ago
Hi,
Thanks for your great plugin!Just a few remarks:
in 'vendor/plugins/wiki_latex_plugin/app/controllers/wiki_latex_controller.rb': latex and dvipng paths must be fixed if they are not '/usr/bin' the '&' character can't be used yet in LaTeX formulas. It is transcribed to ' x % x % ' (without the spaces), which seems to be equivalent to '&' in both sides (specific to Redmine or inherent to Ruby?)
Any ideas to manage the '&' issue?
RE: wiki-latex plugin - Added by Steve G over 9 years ago
Steve G wrote:
Hi,
Thanks for your great plugin!Just a few remarks:
in 'vendor/plugins/wiki_latex_plugin/app/controllers/wiki_latex_controller.rb': latex and dvipng paths must be fixed if they are not '/usr/bin' the '&' character can't be used yet in LaTeX formulas. It is transcribed to ' x % x % ' (without the spaces), which seems to be equivalent to '&' in both sides (specific to Redmine or inherent to Ruby?)
Any ideas to manage the '&' issue?
A quick and dirty fix could be to replace ' x % x % ' by '&' in wiki_latex_controller.rb:
40: temp_latex.puts @latex.source.gsub("x%_remove_x%","&")
where the '_remove_' must be deleted (added in the comment to prevent ' x % x % ' to be converted to '&')
Then
{{latex($\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$)}} leads to a nice
RE: wiki-latex plugin - Added by Thimios Dimopulos over 9 years ago
thank you very much for the great plugin.
I just installed it on Redmine 0.9.1.stable.3369 and did the basic test with {{latex($\sum_i c_i$)}} but it does not seem to work, i just get $\sum_i c_i$ on the wiki page.
Has anyone tried to install the plugin on Redmine 0.9.1.stable.3369?
The plugin code on github is one year old.
Are there any plans for reviving it?
thanks again.
RE: wiki-latex plugin - Added by Paul H over 9 years ago
I have it working on redmine 0.9.3 :)
The 'File.unlink' hint was key for me getting it working - I didn't have all the latex packages going.
Note that if you see $\sum_i c_i$ as described in previous post, try right-clicking it. You'll find its an image, and the text is the image's alt tag. In my case, latex wasn't able to generate the image, so only the alt tag was shown.
RE: wiki-latex plugin - Added by João Lopes over 9 years ago
I'm using wiki-latex plugin in redmine 0.9.3 and I'm experiencing the following dicultulty.
I made the change in order to support the & in latex as described bellow. Puting the code:
{{latex(
\large \begin{align*} \frac{\partial \varepsilon}{\partial t} {+} \overline{U}_\alpha \frac{\partial \varepsilon }{\partial x_\alpha} = & {-}2\nu \left( \overline{ \frac{\partial u_i}{\partial x_\gamma} \frac{\partial u_i}{\partial x_\alpha } } \frac{\partial \overline{U}_\alpha}{\partial x_\gamma} {+}\overline{ \frac{\partial u_i}{\partial x_\gamma} \frac{\partial u_\alpha}{\partial x_\gamma } } \frac{\partial \overline{U}_i}{\partial x_\alpha} {+} \overline{u_\alpha \frac{\partial u_i}{\partial x_\gamma} } \frac{\partial^2 \overline{U}_i}{\partial x_\gamma \partial x_\alpha} \right) \\ & {-} 2 \nu \overline{\frac{\partial u_i}{\partial x_\gamma} \frac{\partial u_i}{\partial x_\alpha} \frac{\partial u_\alpha}{x_\gamma} } {-} \nu \frac{\partial}{\partial x_\alpha} \overline{u_\alpha \frac{\partial u_i}{\partial x_\gamma } \frac{\partial u_i}{\partial x_\gamma } }\\ & {-} 2 \nu \frac{ \partial}{\partial x_i} \overline{\frac{\partial u_i }{\partial x_\gamma} \frac{\partial p}{ \partial x_\gamma} } \\ & {-} 2\nu \overline{ \frac{\partial u_i}{\partial u_\gamma} \frac{\partial^2 \tau^{sub}_{i\alpha} }{\partial x_\alpha \partial x_\gamma} }\\ & {+} \nu^2 \frac{\partial^2 \varepsilon}{\partial x_\alpha \partial x_\alpha}\\ & 2 \nu \end{align*} )}}
I have an error whem I save. If I remove the last \nu everything goes Ok.
I also had several problems when using the signs + and - but replacing by {+} and {-} the latex is correctly generated. RE: wiki-latex plugin - Added by João Lopes over 9 years ago
Related with the problem posted yesterday,
The display of the document generates:
Parameters: {"action"=>"show", "id"=>"14", "controller"=>"documents"}
Rendering template within layouts/base Rendering documents/show
ActionView::TemplateError (Stack overflow in regexp matcher: /
(!)? # escaping ( \{\{ # opening tag ([\w]+) # macro name (\(((.(?!\}\}))*)\))? # optional arguments \}\} # closing tag ) /mx) on line #11 of app/views/documents/show.rhtml: 8: <p><em><%=h @document.category.name ><br /> 9: <= format_date @document.created_on ></em></p> 10: <div class="wiki"> 11: <= textilizable @document.description, :attachments => @document.attachments %> 12: </div>
in the production.log
RE: wiki-latex plugin - Added by João Lopes over 9 years ago
I Solve solve the problem related in the last two posts changing the
redmine/lib/redmine/wiki_formatting/textile/formatter.rb
to
MACROS_RE = / (!)? # escaping ( \{\{ # opening tag ([\w]+) # macro name (\((.*?)\))? # optional arguments \}\} # closing tag ) /xm unless const_defined?(:MACROS_RE)
Instead of the solution posted here! |
Interested in the following function:$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}, $$where $\pi(n)$ is the prime counting function.When $s=2$ the sum becomes the following:$$ \Psi(2)=\sum_{n=2}^\infty \frac{1}{\pi(n)^2}=1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1...
Consider a random binary string where each bit can be set to 1 with probability $p$.Let $Z[x,y]$ denote the number of arrangements of a binary string of length $x$ and the $x$-th bit is set to 1. Moreover, $y$ bits are set 1 including the $x$-th bit and there are no runs of $k$ consecutive zer...
The field $\overline F$ is called an algebraic closure of $F$ if $\overline F$ is algebraic over $F$ and if every polynomial $f(x)\in F[x]$ splits completely over $\overline F$.
Why in def of algebraic closure, do we need $\overline F$ is algebraic over $F$? That is, if we remove '$\overline F$ is algebraic over $F$' condition from def of algebraic closure, do we get a different result?
Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$).Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outwa...
@AlessandroCodenotti That is a poor example, as the algebraic closure of the latter is just $\mathbb{C}$ again (assuming choice). But starting with $\overline{\mathbb{Q}}$ instead and comparing to $\mathbb{C}$ works.
Seems like everyone is posting character formulas for simple modules of algebraic groups in positive characteristic on arXiv these days. At least 3 papers with that theme the past 2 months.
Also, I have a definition that says that a ring is a UFD if every element can be written as a product of irreducibles which is unique up units and reordering. It doesn't say anything about this factorization being finite in length. Is that often part of the definition or attained from the definition (I don't see how it could be the latter).
Well, that then becomes a chicken and the egg question. Did we have the reals first and simplify from them to more abstract concepts or did we have the abstract concepts first and build them up to the idea of the reals.
I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, ...
I was watching this lecture, and in reference to above screenshot, the professor there says: $\frac1{1+x^2}$ has a singularity at $i$ and at $-i$, and power series expansions are limits of polynomials, and limits of polynomials can never give us a singularity and then keep going on the other side.
On page 149 Hatcher introduces the Mayer-Vietoris sequence, along with two maps $\Phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$ and $\Psi : H_n(A) \oplus H_n(B) \to H_n(X)$. I've searched through the book, but I couldn't find the definitions of these two maps. Does anyone know how to define them or where there definition appears in Hatcher's book?
suppose $\sum a_n z_0^n = L$, so $a_n z_0^n \to 0$, so $|a_n z_0^n| < \dfrac12$ for sufficiently large $n$, so $|a_n z^n| = |a_n z_0^n| \left(\left|\dfrac{z_0}{z}\right|\right)^n < \dfrac12 \left(\left|\dfrac{z_0}{z}\right|\right)^n$, so $a_n z^n$ is absolutely summable, so $a_n z^n$ is summable
Let $g : [0,\frac{ 1} {2} ] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $lim_{n→∞} n!g_n(t) = 0,$ for all $t ∈ [0,\frac{1}{2}]$ .
Can you give some hint?
My attempt:- $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$
If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero.
I have a bilinear functional that is bounded from below
I try to approximate the minimum by a ansatz-function that is a linear combination
of any independent functions of the proper function space
I now obtain an expression that is bilinear in the coeffcients
using the stationarity condition (all derivaties of the functional w.r.t the coefficients = 0)
I get a set of $n$ equations with the $n$ the number of coefficients
a set of n linear homogeneus equations in the $n$ coefficients
Now instead of "directly attempting to solve" the equations for the coefficients I rather look at the secular determinant that should be zero, otherwise no non trivial solution exists
This "characteristic polynomial" directly yields all permissible approximation values for the functional from my linear ansatz.
Avoiding the neccessity to solve for the coefficients.
I have problems now to formulated the question. But it strikes me that a direct solution of the equation can be circumvented and instead the values of the functional are directly obtained by using the condition that the derminant is zero.
I wonder if there is something deeper in the background, or so to say a more very general principle.
If $x$ is a prime number and a number $y$ exists which is the digit reverse of $x$ and is also a prime number, then there must exist an integer z in the mid way of $x, y$ , which is a palindrome and digitsum(z)=digitsum(x).
> Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert. Herr H. A. Schwarz gab dann ein einfacheres Beispiel.
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
It's discussed very carefully (but no formula explicitly given) in my favorite introductory book on Fourier analysis. Körner's Fourier Analysis. See pp. 67-73. Right after that is Kolmogoroff's result that you can have an $L^1$ function whose Fourier series diverges everywhere!! |
I've just started learning about limits. Why can we say $$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} = 0 $$ even though $\lim_{x\rightarrow \infty} \sin x$ does not exist?
It seems like the fact that sin is bounded could cause this, but I'd like to see it algebraically.
$$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} = \frac{\lim_{x\rightarrow \infty} \sin x} {\lim_{x\rightarrow \infty} x} = ? $$
L'Hopital's rule gives a fraction whose numerator doesn't converge. What is a simple way to proceed here? |
I have a second order ordinary differential equation of the form :
$ a \frac{d^2p(x)}{dx^2} + x \frac{dp(x)}{dx} + p(x) = 0 $
Can anyone tell me how I can solve it?
Thanks in advance .
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I have a second order ordinary differential equation of the form :
$ a \frac{d^2p(x)}{dx^2} + x \frac{dp(x)}{dx} + p(x) = 0 $
Can anyone tell me how I can solve it?
Thanks in advance .
Assuming that $a$ is a constant:
$$ap''(x)+xp'(x)+p(x)=0\Longleftrightarrow$$
Substitute $\frac{\text{d}}{\text{d}x}\left(x\right)=1$:
$$ap''(x)+xp'(x)+\frac{\text{d}}{\text{d}x}\left(x\right)p(x)=0\Longleftrightarrow$$
Apply the reverse product rule:
$$\frac{\text{d}}{\text{d}x}\left(ap'(x)\right)+\frac{\text{d}}{\text{d}x}\left(xp(x)\right)=0\Longleftrightarrow$$ $$\frac{\text{d}}{\text{d}x}\left(ap'(x)+xp(x)\right)=0\Longleftrightarrow$$ $$\int\frac{\text{d}}{\text{d}x}\left(ap'(x)+xp(x)\right)\space\text{d}x=\int0\space\text{d}x\Longleftrightarrow$$ $$ap'(x)+xp(x)=\text{C}\Longleftrightarrow$$ $$p'(x)+\frac{xp(x)}{a}=\text{C}\Longleftrightarrow$$
Let $v(x)=\exp\left[\int\frac{x}{a}\space\text{d}x\right]=e^{\frac{x^2}{2a}}$.
Multiply both sides by $v(x)$:
$$e^{\frac{x^2}{2a}}p'(x)+\frac{xe^{\frac{x^2}{2a}}p(x)}{a}=\text{C}e^{\frac{x^2}{2a}}\Longleftrightarrow$$
Substitute $\frac{xe^{\frac{x^2}{2a}}}{a}=\frac{\text{d}}{\text{d}x}\left(e^{\frac{x^2}{2a}}\right)$:
$$e^{\frac{x^2}{2a}}p'(x)+\frac{\text{d}}{\text{d}x}\left(e^{\frac{x^2}{2a}}\right)p(x)=\text{C}e^{\frac{x^2}{2a}}\Longleftrightarrow$$
Apply the reverse product rule:
$$\frac{\text{d}}{\text{d}x}\left(p(x)e^{\frac{x^2}{2a}}\right)=\text{C}e^{\frac{x^2}{2a}}\Longleftrightarrow$$ $$\int\frac{\text{d}}{\text{d}x}\left(p(x)e^{\frac{x^2}{2a}}\right)\space\text{d}x=\int\text{C}e^{\frac{x^2}{2a}}\space\text{d}x\Longleftrightarrow$$ $$p(x)e^{\frac{x^2}{2a}}=\int\text{C}e^{\frac{x^2}{2a}}\space\text{d}x\Longleftrightarrow$$ $$p(x)=\frac{\int\text{C}e^{\frac{x^2}{2a}}\space\text{d}x}{e^{\frac{x^2}{2a}}}$$
This can also be solved as an Eigenvalue problem. With some manipulation, you can turn it into Hermite's Differential Equation having Hermite polynomials as solutions. One benefit, this allows the use of Rodrique's Formulas to generate new solutions. |
I need to show that the above sequence is a Cauchy sequence. However, when using the definition of a Cauchy sequence, I get that $s(n) - s(m)$ is equal to some complicated sigmal notation expression, which I need to show to be less than $\epsilon$ for every $\epsilon>0$ for $n,m > N$. Any help would be appreciated.
We don't need an exact value of $a_n-a_m$; we just need an estimate. Assuming WLOG that $n>m$, we get $a_n-a_m=(a_{m+1}-a_m)+(a_{m+2}-a_{m+1})+\cdots+(a_n-a_{n-1}) = \frac1{3^m}+\frac1{3^{m+1}}+\cdots+\frac1{3^{n-1}}$.
For fixed $m$, how big can that get? Well, clearly, the worst case comes as $n\to\infty$. What is the infinite sum $\frac1{3^m}+\frac1{3^{m+1}}+\cdots+\frac1{3^n}+\cdots$?
This series (a geometric series) is one of the ones you should know exactly. In most cases, you'll want to compare the series to something - but here, the geometric series is one of the standard things to compare to.
Alternatively, you can find a formula for $a_n$ and this makes your problem easier.Let's write the recurrence relation for $a_n$,$a_{n-1}$,...,$a_2$ :
$a_n=a_{n-1}+\frac{1}{3^{n-1}}$ $a_{n-1}=a_{n-2}+\frac{1}{3^{n-2}}$ ................................. $a_2=a_1+\frac{1}{3}$ After we add the above lines we get that $a_n=1+\frac{1}{3}+...\frac{1}{3^{n-1}}$,which is a geometric progression.Hence,$a_n=\frac{3-\left(\frac{1}{3}\right)^{n-1}}{2}$. Hint That complicated sigma notation expression is a geometric sum and it is easy to calculate.
If $n <m$ then $$|s(m)-s(n)|=\sum_{k=n}^{m-1}\frac{1}{3^k}=\frac{1}{3^n} \sum_{k=n}^{m-1}\frac{1}{3^{k-n}}=\frac{1}{3^n} \sum_{j=0}^{m-n-1}\frac{1}{3^{j}}=\frac{1}{3^n} \frac{1-\frac{1}{3^{m-n}}}{1-\frac{1}{3}}$$ |
Question:
Using the reduction formula, find the integral of {eq}\displaystyle \int x^n \cos (x)\ dx {/eq} in terms of an integral involving {eq}x^{n - 1} {/eq}.
Integration by Reduction Formula.
Reduction formula reduces an integral from higher to lower power with the same form.
Also reduction formula repeated continuously until it forms an easier integration.
The formula is:
{eq}\displaystyle\int cosx\ dx=sinx+c\\\\ \displaystyle\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+c\\\\ \displaystyle\int sinx\ dx=-cosx+c\\\\ {/eq}
Answer and Explanation:
We have to integrate the given equation by using reduction formula:
{eq}\displaystyle\int x^ncos(x)dx\\\\ I_n=\displaystyle\int x^ncos(x)dx\\\\ {/eq}
Applyin formula of integration by parts:
{eq}\displaystyle\int (uv)dx=u\displaystyle\int v\ dx-\displaystyle\int \left [ \dfrac{d}{dx}(u)\displaystyle\int v\ dx \right ]dx\\\\ =x^n\displaystyle\int cosx\ dx-\displaystyle\int \left [ \dfrac{d}{dx}(x^n)\displaystyle\int cosx\ dx \right ]dx\\\\ =x^n\ sinx-\displaystyle\int \left [ nx^{n-1}\ sinx \right ]dx\\\\ =x^n\ sinx-n\displaystyle\int x^{n-1}sinx\ dx\\\\ {/eq}
Here the equation is going towards non ending loop.
Again integrating:
{eq}=x^n\ sinx-n\left [ x^{n-1}\displaystyle\int sinx\ dx-\displaystyle\int \left [ \dfrac{d}{dx}(x^{n-1})\displaystyle\int sinx\ dx \right ] \right ]dx\\\\ =x^n\ sinx-n\left [ x^{n-1}(-cosx)-\displaystyle\int \left [ (n-1)x^{n-2}(-cosx) \right ]dx \right ]\\\\ =x^n\ sinx+nx^{n-1}cosx-n(n-1)\displaystyle\int x^{n-2}cosx\ dx\\\\ =x^n\ sinx+nx^{n-1}cosx-n(n-1)\ I_{n-2}\\\\ {/eq}
This is the required reduction formula.
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from AP Calculus AB & BC: Homework Help ResourceChapter 13 / Lesson 13 |
(Sorry for the fuzzy title, could not think of something more informative. Feel free to suggest improvements)
This question is somewhat of a generalization of "Osborne, Nash equilibria and the correctness of beliefs". Consider the normal form game
$G = \langle P, S, U \rangle$ with
$P = \{1,\dots, m\}$ the set of players,
$S =\{S_1,\dots,S_m\}$ the $m$-tuple of pure strategies for players in $P$
$U = \{u_1,\dots,u_m\}$ the $m$-tuple of payoff functions for players in $P$
The structure $G$ is rich enough to define the notion of a Nash equilibrium (NE).
However, authors sometimes describe NE based on richer models. For instance, the description by Osborne discussed in "Osborne, Nash equilibria and the correctness of beliefs" hinges on a structure
$\hat{G} = \langle P, S, B, U \rangle$ with
$P = \{1,\dots, m\}$ the set of players,
$S =\{S_1,\dots,S_m\}$ the $m$-tuple of pure strategies for players in $P$
$B = \{B_1,\dots,B_m\}$ the $m$-tuple of possible beliefs about each other's actions for players in $P$
$U = \{u_1,\dots,u_m\}$ the $m$-tuple of payoff functions for players in $P$
In this setting a NE is equivalent to
A strategy profile $s^*$ and a belief profile $b^*$ in which for all $i\in P$
$$ u_i ( s^*_i ~|~ s_{-i} = b^*_{i}) \geq u_i ( s' ~|~ s_{-i} = b^*_{i}) \text{ for all } s' \in S_i$$ and $$b^*_i = s^*_{-i}$$
Based on this equivalence, Osborne describes a NE as a situation in which
First, each player chooses her action according to the model of rational choice, given her beliefs about the other players' actions. Second, every player's belief about the other players' actions is correct.
Now in other settings, I have seen justifications of NE based on other structures richer than $G$. For instance, one could have a game with incomplete information
$\tilde{G} = \langle P, \Theta, p , S, U \rangle$ with
$P = \{1,\dots, m\}$ the set of players,
$\Theta = \{\Theta_1,\dots,\Theta_m\}$ the $m$-tuple of possible types for players in $P$
$p$, a joint probability distribution over types
$S =\{S_1,\dots,S_m\}$ the $m$-tuple of pure strategies for players in $P$
$U = \{u_1,\dots,u_m\}$ the $m$-tuple of payoff functions for players in $P$
In $\tilde{G}$, a NE is equivalent to
A Bayesian-Nash equilibrium in which agent know each other types for sure, that is $p$ is degenerate.
In this case, one could therefore describe a NE as a situation in which
Players know each other's type for sure and play according to a Bayesian-Nash equilibrium
(See for instance "Featherstone, C., & Niederle, M. (2008). Ex ante efficiency in school choice mechanisms: an experimental investigation")
Now these two examples puzzle me. One seems to recommend to view NE as a situation in which
beliefs about actions are correct, whereas the other suggests that NE be viewed as situations in which agent have perfect information about each other's preferences.
So my question is :
Is any of these two argument better than the another in any meaningful way? Is any of the two generalization of $G$ (to $\tilde{G}$ or $\hat{G}$) any more helpful than the other in terms of understanding the necessary conditions for a NE to prevail?
My impression is that the concept of NE stands alone, independently of these generalizations and that there is no "correct" way of enriching the NE framework. As I understand it, "a NE is a situation in which agents are rational and beliefs are correct" is not more or less true than "a NE is a situation in which agents have perfect information on each other's payoffs". But as I see such somewhat conflicting claims showing up again and again, I am worried that I might be missing something. |
Depth Estimation
This post is gonna be a bit long and mathematically more involved and covers the following topics.
Depth Estimation using triangulation Depth Estimation using sequential Bayesian updates Interesting papers for further reading on above topics References
There are multiple methods to estimate depth but what I am most interested in is estimating depth in a robust way. The ability to detect outliers arising from dynamic objects in the scene, occlusions etc is crucial in any depth estimator. Simple triangulation unfortunately cannot do this. [1] presents a statistical analysis of depth and derives a probabilistic approach to explicitly handle outliers.
Robust depth estimation is crucial for SLAM systems. Approaches such as feature based ORB-SLAM [3] systems estimate depth by first estimating frame-to-frame egomotion and then triangulating and using the triangulated map points for egomotion estimation, loop closing etc.
On the other hand, SVO [2] approaches the SLAM problem in an incremental fashion. First estimate current frame pose relative to the last frame(not keyframe). The pose is estimated by defining photometric loss over the existing points projected into the current frame. Since this frame-to-frame motion has very small disparity, the patches corresponding to each point only move in a small amount. Next they do feature alignment by violating epipolar constraints to get better cross correlation across patches. And as a final step, motion-only BA and structure-only BA steps are carried out. What impressed me most is the fast, accurate, robust depth estimator in the backend which will be discussed in great detail in the following sections. Since each pixel is treated independently; this can be parallelized and dense reconstruction can be obtained as shown in [4].
Depth estimation using triangulation
Simplest of all the methods; this involves solving an overdetermined system of equations. Let \(T_1\) and \(T_2\) be the camera matrices from the above two views. Then for the point \(p_1\);
From the above;
where \(\left[u_1\right]_{\text{x}}\) is skew symmetric matrix representation of \(u_1\). Each of the above two equations provides two linearly independent equations. Using SVD, one can find \(p_1\).
Depth estimation using sequential Bayesian updates
How do we represent depth of a pixel in a parametric way? How do we parameterize the depth in the first place? Read along to find out!
Fig 1: Setup
From Figure 1, let \({\bf u}=\left(u, v\right)^T\) be a pixel in image \(\mathcal{I}\). For a given depth \(z\), let \({\bf p} = \left(x, y, z\right)^T\) be the corresponding 3D point along the optical ray through pixel \({\bf u}\). Let \(\mathcal{I’}\) be another image that has \({\bf p}\) in it’s field-of-view with \({\bf u’}=\left(u’, v’\right)^T\) being back-projected point in the image. Let \(W\) and \(W’\) be patches around \({\bf u}\) and \({\bf u’}\) respectively. Figure 2(a) shows the Normalized Cross Correlation score for patches \(W\) and \(W’\) as \(z\) is varied along the optic ray through \({\bf u}\). Figure 2(b) shows on the other hand shows histogram of depth \(z\) for 60 neighbouring images measured along the optic ray through \({\bf u}\).
Fig 2 Image source [1]
From the above plots it is statistically clear that true depth measurement is concentrated around single depth value (good measurement) and there is a lot of noise (bad measurement) albeit uniformly distributed along different depth values.
So a probability distribution that is combination of Gaussian (for good measurement) and uniform (for bad measurement) makes sense. The two are weighted by the inlier ratio \(\rho\) which indicates the probability of the measurement being inlier. Mathematically, given the inlier ratio \(\rho\) and the true depth \(\hat d\), the depth after \(k^{th}\) measurement takes the form;
\([d_{min}, d_{max}]\) is the interval from which noisy depth measurements are sampled. These can be set based on the near and far plane of the camera. \({\tau}_k^2\) is the variance of depth by assuming 1 pixel measurement noise in the image. Section outlines how to compute \(\tau_k^2\).
Say we have sequence of \(n\) frames \(k = \{r, r+1, \dots, r+n \}\) that observe the point \(\bf p\) first observed in the reference frame \(r\). Using triangulation method described above we can generate sequence of depth hypothesis \(\mathcal{D} = \{d_{r+1}, \dots, d_{r+n}\}\) where \(d_k^{th}\) hypothesis is obtained by triangulating views \(r\) and \(k\).
Now that we have parameterized the depth and have sequence of depth hypothesis, how do we compute the parameters of the above density model?
Assuming independent observations, likelihood function is given by;
The parameters of the above likelihood can be obtained in a maximum likelihood framework using Expectation Maximization. But as noted in [1], the experiments seemed to be trapped in a local optimum. Other way round? Sequential Bayesian updates to the rescue!
Posterior takes the form;
where \(p\left(\hat d, \rho \right)\) is the prior on depth and inlier ratio which is assumed to be uniform and modeled using a dense 2d histogram. Authors of [1] show that above posterior can be modelled using product of a Gaussian distribution for the depth and a Beta distribution for the inlier ratio.
where \(a_k\) and \(b_k\) are parameters of Beta distribution. But the choice of the posterior as product of a Gaussian and a Beta distribution could be surprising right? Section gives a proof of why. In fact this is the approximating distribution that minimizes KL divergence from the true posterior. Updates in the above equation are implemented in a sequential manner where posterior after \((k-1)^{th}\) observation is used as prior for \(k^{th}\) observation. Now, for \(k^{th}\) observation, the new posterior takes the form;
But the new posterior is no longer of the form \(Gaussian \times Beta\) but this can be approximated using moment matching. The new parameters \(a_k, b_k, \mu_k, \sigma_k^2\) are computed so that the posterior \(p \left(\hat d, \rho \mid d_{r+1}, \dots, d_k\right)\) and the approximation \( q \left(\hat d, \rho \mid a_k, b_k, \mu_k, \sigma_k^2 \right) \) have same first and second order moments wrt \(\rho\) and \(\hat d\). The updates are derived in the following section.
Derivation of parameters for recursive Bayesian updates
Here I shall derive the expression for parameters in the approximated posterior form. Specifically we match the moments wrt \(\rho\) and \(\hat d\) in the posterior approximation;
and the updated posterior;
after \(k^{th}\) observation. To simplify notation subscripts are dropped. \(a, b, \mu, \sigma^2\) and \(a’, b’, \mu’, \sigma’^2\) are the prior and posterior parameters respectively.
Moments of approximated posterior
Mean and variance wrt \(\hat d\) in equation 1 are \(\mu’\) and \(\mu’^2 + \sigma’^2\) respectively.
Similarly; mean and variance wrt \(\rho\) in equation 1 are \(\frac{a’}{a’+b’}\) and \(\frac{a’(a’+1)}{(1+a’+b’)(a’+b’)}\) respectively.
Moments of actual posterior
Substitute the density model in (2) to get;
Now, using the properties of \(\text{Beta}\) distribution,
From the above;
and similarly;
Substituting (4) and (5) in (3), (2) takes the form;
The above can further be simplified to;
where
First to make things simpler;
First integrating out (6) w.r.t \(\rho\) we obtain the marginal distribution over \(\hat d\). Mean wrt \(\hat d\) in (6) is given by \(C_1 m + C_2\mu\) and variance by \(C_1 (s^2+m^2) + C_2(\mu^2+\sigma^2)\)
Similarly integrating out w.r.t \(\hat d\) we obtain the marginal distribution over \(\rho\). Mean and variance w.r.t \(\rho\) are given by \(C_1\frac{a+1}{a+b+1}+C_2\frac{a}{a+b+1}\) and \(C_1\frac{(a+1)(a+2)}{(a+b+1)(a+b+2)} + C_2\frac{a(a+1)}{(a+b+1)(a+b+2)}\) respectively.
Equating the above mean and variances we get the following 4 equations;
Solving the above 4 equations we get the updated parameters \(a’, b’, \mu’, \sigma’^2\).
Variational Reasoning for approximate posterior
Here, the derivation for the approximate posterior for equation (2) using factorization is provided. The Dynamic Bayesian network corresponding to the density mixture model is given in the following figure. Quantities in the red box denote \(N\) i.i.d data. The observed data \((d_k)\) is highlighted in orange.
The corresponding joint probability distribution is give by;
where \(D = \{d_1, d_2, \dots, d_N\}\) are \(N\) depth hypothesis and \(Z = \{z_1, z_2, \dots, z_N\}\) are the latent variables with \(z_k = 1\) indicating that \(k^{th}\) depth hypothesis is inlier.
Restating the mixture model;
Assuming independent initial prior, we have \(p\left(\rho, \hat d\right) = p(\rho)p(\hat d)\). Introducing the latent variable \(z_k\); we have the following;
and
By marginilizing the latent variables from (9) and (10) we get the original mixture model (8) i.e,
We now aim to approximate the posterior \(p(\hat d, \rho, Z \mid D)\) with \(q(\hat d, \rho, Z)\). Factorizing the approximate posterior as follows;
Now the factorized distributions are given by;
What we are most interested in is equation (12).
Taking exponential of the above gives;
where \(\alpha\) is a constant, \(r_k = \mathbf{E}\left[z_k\right]\) and \(S = \sum_{k}\mathbf{E}\left[z_k\right]\). The above equation can further be factorized into two indepenent distributions over \(\rho\) and \(\hat d\) i.e,
where;
Now by choosing the priors over \(\rho\) and \(\hat d\) to follow uniform and Gaussian respectively; equation (14) reduces to Beta distribution and with equation (15) reducing to Gaussian.
Computing Variance in triangulation
Variance is \(\tau_k^2 = \left(\vert \vert {\bf {}_r p}\vert \vert - \vert \vert {\bf {}_r p^+} \vert \vert \right)^2 \).
Let \(\bf t\) be the translation component of the transformation \({}^k T_r\). \({\bf {}_r p}\) is a point \(z\) units away from \(C_r\). Let \(\bf f\) be the unit vector from \(C_r\) through the pixel \(u\).
where the term added to \(\beta\) in equation (4) is the angle generated by the ray for 1 pixel in the image as shown in the following figure (\(f\) is the focal length of camera). Equation (5) ensures sum of all angles in a triangle is \(\pi\) and equation (6) is obtained by applying law of sines
Interesting papers Neural RGB->D Sensing: Depth and Uncertainty from a Video Camera Depth from Videos in the Wild: Unsupervised Monocular Depth Learning from Unknown Cameras Learning the Depths of Moving People by Watching Frozen People References
[1] Video-based, Real-Time Multi View Stereo
[2] SVO: Fast Semi-Direct Monocular Visual Odometry
[3] ORB-SLAM: A Versatile and Accurate Monocular SLAM System
[4] REMODE: Probabilistic, Monocular Dense Reconstruction in Real Time |
R/data.R
evo_rates.Rd
From Table 1 in Sung et al. (2016).
evo_rates
A data frame with 15 rows and 8 variables:
domain
Either
Bacteria or
Eukarya for what type of organismthe species is.
species
Species name.
Ge
Effective genome size using only coding DNA.
Gc_Gnc
Effective genome size using coding DNA and non-coding DNA that is under purifying selection.
indels
Rate of insertions and deletions (\(10^{-10}\) events per site per generation).
subs
Base-substitution mutation rate (\(10^{-10}\) events per site per generation).
Ne
Effective population size (\(\times 10^{6}\)).
theta_s
Population mutation rate estimated using \(\theta_s\).
pi_s
Population mutation rate estimated using \(\pi_s\).
Sung, W., M. S. Ackerman, M. M. Dillon, T. G. Platt, C. Fuqua, V. S. Cooper, andM. Lynch. 2016. Evolution of the insertion-deletion mutation rate across thetree of life.
G3: Genes | Genomes | Genetics 6:2583–2591. |
I will outline how the problem can be approached and statewhat I think the end result will be for the special casewhen the shape parameters are integers, but not fill in thedetails.
First, note that $X-Y$ takes on values in $(-\infty,\infty)$and so $f_{X-Y}(z)$ has support $(-\infty,\infty)$.
Second, from the standard results that the density of the sum of two independent continuous random variables is theconvolution of their densities, that is,$$f_{X+Y}(z) = \int_{-\infty}^\infty f_X(x)f_Y(z-x)\,\mathrm dx$$and that the density of the random variable $-Y$ is$f_{-Y}(\alpha) = f_Y(-\alpha)$, deduce that$$f_{X-Y}(z) = f_{X+(-Y)}(z) = \int_{-\infty}^\infty f_X(x)f_{-Y}(z-x)\,\mathrm dx= \int_{-\infty}^\infty f_X(x)f_Y(x-z)\,\mathrm dx.$$
Third, for non-negative random variables $X$ and $Y$, note that theabove expression simplifies to$$f_{X-Y}(z) = \begin{cases}\int_0^\infty f_X(x)f_Y(x-z)\,\mathrm dx, & z < 0,\\\int_{0}^\infty f_X(y+z)f_Y(y)\,\mathrm dy, & z > 0.\end{cases}$$
Finally, using parametrization $\Gamma(s,\lambda)$ to mean arandom variable with density $\lambda\frac{(\lambda x)^{s-1}}{\Gamma(s)}\exp(-\lambda x)\mathbf 1_{x>0}(x)$,and with$X \sim \Gamma(s,\lambda)$ and $Y \sim \Gamma(t,\mu)$ random variables, we have for $z > 0$ that$$\begin{align*}f_{X-Y}(z) &= \int_{0}^\infty \lambda\frac{(\lambda (y+z))^{s-1}}{\Gamma(s)}\exp(-\lambda (y+z))\mu\frac{(\mu y)^{t-1}}{\Gamma(t)}\exp(-\mu y)\,\mathrm dy\\&= \exp(-\lambda z) \int_0^\infty p(y,z)\exp(-(\lambda+\mu)y)\,\mathrm dy.\tag{1}\end{align*}$$Similarly, for $z < 0$,$$\begin{align*}f_{X-Y}(z) &= \int_{0}^\infty \lambda\frac{(\lambda x)^{s-1}}{\Gamma(s)}\exp(-\lambda x)\mu\frac{(\mu (x-z))^{t-1}}{\Gamma(t)}\exp(-\mu (x-z))\,\mathrm dx\\&= \exp(\mu z) \int_0^\infty q(x,z)\exp(-(\lambda+\mu)x)\,\mathrm dx.\tag{2}\end{align*}$$
These integrals are not easy to evaluate but for the special case$s = t$, Gradshteyn and Ryzhik,
Tables of Integrals, Series, and Products,Section 3.383, lists the value of$$\int_0^\infty x^{s-1}(x+\beta)^{s-1}\exp(-\nu x)\,\mathrm dx$$in terms of polynomial, exponential and Bessel functions of $\beta$and this can be used to write down explicit expressions for $f_{X-Y}(z)$.
From here on, we assume that $s$ and $t$ are integers sothat $p(y,z)$ is a polynomial in $y$ and $z$ of degree $(s+t-2, s-1)$and $q(x,z)$ is a polynomial in $x$ and $z$ of degree $(s+t-2,t-1)$.
For $z > 0$, the integral $(1)$ is the sum of $s$ Gamma integrals with respect to $y$ with coefficients$1, z, z^2, \ldots z^{s-1}$. It follows that the density of$X-Y$ is proportional to a
mixture density of $\Gamma(1,\lambda), \Gamma(2,\lambda), \cdots, \Gamma(s,\lambda)$random variables for $z > 0$. Note that this resultwill hold even if $t$ is not an integer.
Similarly, for $z < 0$,the density of$X-Y$ is proportional to a
mixture density of $\Gamma(1,\mu), \Gamma(2,\mu), \cdots, \Gamma(t,\mu)$random variables flipped over, that is,it will have terms such as $(\mu|z|)^{k-1}\exp(\mu z)$instead of the usual $(\mu z)^{k-1}\exp(-\mu z)$.Also, this result will hold even if $s$ is not an integer. |
One of the application of this concept is the idea of measuring the atmospheric pressure. Consider a situation described in Figure 4.3. The liquid is filling the tube and is brought into a steady state. The pressure above the liquid on the right side is the vapor pressure. Using liquid with a very low vapor pressure like mercury, will result in a device that can measure the pressure without additional information (the temperature).
Example 4.4
Calculate the atmospheric pressure at \(20^{\circ}C\). The high of the Mercury is 0.76 [m] and the gravity acceleration is 9.82[\(m/sec\)]. Assume that the mercury vapor pressure is 0.000179264[kPa]. The description of the height is given in Figure 4.3. The mercury density is 13545.85[\(kg/m^3\)].
Solution 4.4
The pressure is uniform or constant plane perpendicular to the gravity. Hence, knowing any point on this plane provides the pressure anywhere on the plane. The atmospheric pressure at point \(a\) is the same as the pressure on the right hand side of the tube. Equation (15) can be utilized and it can be noticed that pressure at point \(a\) is
\[ P_{a} = \rho \, g \, h + P_{vapor} \label{static:eq:vaporP} \tag{31} \] The density of the mercury is given along with the gravity and therefore, \[ P_a = 13545.85 \times 9.82 \times 0.76 \sim 101095.39 [Pa] \sim 1.01[Bar] \tag{32} \] The vapor pressure is about \(1 \times 10^{-4}\) percent of the total results.
The main reason the mercury is used because of its large density and the fact that it is in a liquid phase in most of the measurement range. The third reason is the low vapor (partial) pressure of the mercury. The partial pressure of mercury is in the range of the 0.000001793[Bar] which is insignificant compared to the total measurement as can be observed from the above example.
Example 4.5
A liquid \(a\) in amount \(H_a\) and a liquid \(b\) in amount \(H_b\) in to an U tube. The ratio of the liquid densities is \(\alpha = \rho_1/\rho_2\). The width of the U tube is \(L\). Locate the liquids surfaces.
Solution 4.5
The question is to find the equilibrium point where two liquids balance each other. If the width of the U tube is equal or larger than total length of the two liquids then the whole liquid will be in bottom part. For smaller width, \(L\), the ratio between two sides will be as
\[ \rho_1\,h_1 = \rho_2\,h_2 \rightarrow h_2 = \alpha \, h_1 \tag{33} \] The mass conservation results in \[ H_a + H_b = L + h_1 + h_2 \tag{34} \] Thus two equations and two unknowns provide the solution which is \[ h_1 = \dfrac{H_a + H_b - L}{1+\alpha} \tag{35} \] When \(H_a > L\) and \(\rho_a\left( H_a -L\right) \geq \rho _b\) (or the opposite) the liquid \(\mathbf{a}\) will be on the two sides of the U tube. Thus, the balance is \[ h_1 \, \rho_b + h_2\, \rho_a = h_3\, \rho_a \tag{36} \] where \(h_1\) is the height of liquid \(\mathbf{b}\) where \(h_2\) is the height of "extra'' liquid \(\mathbf{a}\) and same side as liquid \(\mathbf{b}\) and where \(h_3\) is the height of liquid \(\mathbf{b}\) on the other side. When in this case \(h_1\) is equal to \(H_b\). The additional equation is the mass conservation as \[ H_a = h_2+L+h_3 \tag{37} \] The solution is \[ h_2 = \dfrac{(H_a-L)\,\rho_a - H_b\rho_b}{2\,\rho_a} \tag{38} \] Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
R1C1R2C2: The Two-Pole Passive RC Filter
I keep running into this circuit every year or two, and need to do the same old calculations, which are kind of tiring. So I figured I’d just write up an article and then I can look it up the next time.
This is a two-pole passive RC filter. Doesn’t work as well as an LC filter or an active filter, but it is cheap. We’re going to find out a couple of things about its transfer function.
First let’s find out the transfer function of this circuit:
Not very difficult:
$$ \frac{V_{out}}{V_{in}} = \frac{Z_4}{Z_2 + Z_4} \cdot \frac{Z_A}{Z_1 + Z_A}$$
where \( Z_A = Z_3 \parallel (Z_2 + Z_4). \)
After a bunch of what I call Grungy Algebra, we get
$$ \frac{V_{out}}{V_{in}} = \frac{Z_3Z_4}{Z_1Z_2 + Z_1Z_3 + Z_1Z_4 + Z_2Z_3 + Z_3Z_4} $$
The problem with Grungy Algebra is that it sows the seeds of Doubt. And we all know that Doubt is bad in engineering. So a couple of ways of double-checking here:
as \( Z_3 \to 0 \) or \( Z_4 \to 0 \), \( \frac{V_{out}}{V_{in}} \to 0 \) as \( Z_3 \to \infty, \frac{V_{out}}{V_{in}} \to Z_4/(Z_1+Z_2+Z_4) \) as \( Z_4 \to \infty, \frac{V_{out}}{V_{in}} \to Z_3/(Z_1+Z_3) \)
Numerically:
for Z1, Z2, Z3, Z4 in [(1.0,2.0,3.0,4.0), (4.0,3.0,2.0,1.0), (1.0,10.0,2.0,20.0)]: ZA = 1/(1/Z3 + 1/(Z2+Z4)) print "Z1=%f, Z2=%f, Z3=%f, Z4=%f" % (Z1,Z2,Z3,Z4) print "Vo/Vi = Z4ZA/(Z2+Z4)(Z1+ZA) = %f" % (Z4/(Z2+Z4)*ZA/(Z1+ZA)) print " = (direct formula) %f" % (Z3*Z4/(Z1*Z2+Z1*Z3+Z1*Z4+Z2*Z3+Z3*Z4)) Z1=1.000000, Z2=2.000000, Z3=3.000000, Z4=4.000000 Vo/Vi = Z4ZA/(Z2+Z4)(Z1+ZA) = 0.444444 = (direct formula) 0.444444 Z1=4.000000, Z2=3.000000, Z3=2.000000, Z4=1.000000 Vo/Vi = Z4ZA/(Z2+Z4)(Z1+ZA) = 0.062500 = (direct formula) 0.062500 Z1=1.000000, Z2=10.000000, Z3=2.000000, Z4=20.000000 Vo/Vi = Z4ZA/(Z2+Z4)(Z1+ZA) = 0.434783 = (direct formula) 0.434783
All looks good!
Now we just need to handle our original circuit.
In this case, \( Z_1=R_1, Z_2=R_2, Z_3=1/sC_1, Z_4=1/sC_2. \) And… we have more Grungy Algebra. After which we get
$$ H(s) = \frac{V_o}{V_i} = \frac{1}{R_1R_2C_1C_2s^2 + R_1C_1s + R_1C_2s + R_2C_2s + 1}$$
More double-checking: if \( C_1=0 \) then we get \( H(s) = \frac{1}{(R_1+R_2)C_2s + 1} \); if \( C_2 = 0 \) then \( H(s) = \frac{1}{R_1C_2s + 1}. \)
A couple of things about this:
For low frequencies, the \( s^2 \) term is very small and \( H(s) \approx \frac{1}{\tau s+1} \) where \( \tau = R_1C_1 + R_1C_2 + R_2C_2 \).
The standard form of the second order system is
$$H(s) = \frac{1}{\tau^2s^2 + 2\zeta\tau s + 1} = \frac{{\omega_n}^2}{s^2 + 2\zeta\omega_n s + {\omega_n}^2}$$
so here we have \( \tau^2 = R_1R_2C_1C_2 \) and \( \zeta = \frac{R_1C_1 + R_1C_2 + R_2C_2}{2\sqrt{R_1R_2C_1C_2}} \)
If \( R_1 = R_2 = R \) and \( C_1 = C_2 = C \), then \( \tau = RC \) and \( \zeta = \frac{3RC}{2RC} = 1.5 \), which is a fairly high damping ratio.
We can get close to a damping ratio of 1 by making the second stage impedances larger; for example if \( R_1 = R, R_2 = kR, C_1 = C, C_2 = \frac{1}{k}C \) then we have \( \tau = RC \) and \( \zeta = \frac{RC + \frac{RC}{k} + RC}{2RC} = 1 + \frac{1}{2k}, \) which means if \( k=10 \) then we get \( \zeta = 1.05. \) We can get close to 1 but we can’t reach it or get below it with this circuit; we’d need inductors or an active filter using transistors or op-amps.
We can also simulate this system using a state-space representation; since currents \( I_1, I_2 \) into the capacitors are \( I_1 = C\frac{dV_1}{dt} = \frac{V_{in} - V_1}{R_1} + \frac{V_2 - V_1}{R_2} \) and \( I_2 = C\frac{dV_2}{dt} = \frac{V_1 - V_2}{R_2}, \) then:
\( \frac{d}{dt}\begin{bmatrix}C_1V_1 \cr C_2V_2\end{bmatrix} = \begin{bmatrix}-\frac{1}{R_1}-\frac{1}{R_2} & \frac{1}{R_2} \cr \frac{1}{R_2} & -\frac{1}{R_2} \end{bmatrix} \begin{bmatrix}V_1 \cr V_2 \end{bmatrix} + \begin{bmatrix}\frac{1}{R_1} \cr 0 \end{bmatrix} V_{in} \)
or
\( \frac{d}{dt}\begin{bmatrix}V_1 \cr V_2\end{bmatrix} = \begin{bmatrix}-\frac{1}{R_1C_1}-\frac{1}{R_2C_1} & \frac{1}{R_2C_1} \cr \frac{1}{R_2C_2} & -\frac{1}{R_2C_2} \end{bmatrix} \begin{bmatrix}V_1 \cr V_2 \end{bmatrix} + \begin{bmatrix}\frac{1}{R_1C_1} \cr 0 \end{bmatrix} V_{in} \)
which is in the canonical state-space form \( \frac{dV}{dt} = AV + Bu \), and we can simulate it with
scipy.signal.StateSpace:
import scipy.signal import matplotlib.pyplot as plt import numpy as np %matplotlib inline C1 = C2 = 0.1e-6 R1 = 1e3 R2 = 10e3 A = np.array([[-1.0/R1/C1-1.0/R2/C1, 1.0/R2/C1], [1.0/R2/C2, -1.0/R2/C2]]) B = np.array([[1.0/R1/C1], [0]]) C = np.array([0,1]) D = 0 ss = scipy.signal.StateSpace(A,B,C,D) t = np.arange(5000)*1e-6 t, yout, xout = scipy.signal.lsim(ss, t>1e-6, t) plt.plot(t,yout,'b',label='2nd order system') tau3 = R1*C1+R1*C2+R2*C2 ss_approx = scipy.signal.StateSpace(-1.0/tau3, 1.0/tau3, 1, 0) t, yout_approx, xout_approx = scipy.signal.lsim(ss_approx, t>1e-6, t) plt.plot(t,yout_approx,'r',label='1st order approximation') plt.xlabel('t') plt.ylabel('y') plt.legend(loc='lower right');
For practical circuit implementations of filters:
Use NP0/C0G capacitors — these are class 1 ceramic capacitors and are much more ideal than X5R/X7R/Y5V/Z5U capacitors, with lower temperature coefficient, lower tolerances, lower parasitic losses, and less unwanted effects like microphony. Don’t use resistors that are too small or too large. Too small of a resistor can load down the input; too large of a resistor can make the output high enough impedance to allow noise to couple into the circuit. The range 1KΩ - 1MΩ is what you will find most often. Don’t use capacitors that are too small or too large. Too small of a capacitor may be too high impedance; too large of a capacitor will cost a lot. The range 10pF - 1μF is what you will find most often. Low-pass filters with cutoff frequencies of 10Hz or less are difficult to design in analog, and you are probably better off using DSP techniques to filter out any lower frequency content.
And that’s about all there is to say on the subject.
© 2018 Jason M. Sachs, all rights reserved.
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Dear Mr.Jason Sachs;
I am new in coding for MCU like(dsPIC), I designed few controllers for power electronics , the controllers are working fine say with the MATLAb, simulink, in other simulation platforms., now that i like to write the c code for them, i am getting some trouble with the output limitation, or understanding of prpoer fixed point math, some of the C codes are working well , but some of them are not., cuz probably I did not understand the adc dac , fixed point or the output-limitting technique in C .
If, you could Please, help. in whivh case I will send a complete c code and the Pi controolers parameters with the output limiters value with the circuit diagram. I am using dsPICj32mc204, and first I simulate always in Proteus then I usually bread board.
I was looking at this post of yours: https://www.embeddedrelated.com/showarticle/123.ph...
my email is: atalhaque@gmail.com
Thanks, Ahmad
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I would like to uncover a set of equations line by line with and the covered equations should be transparent. The problem occurs when I want to align those equations: somehow the transparency stops at the
&-sign.
I attached a minimum working example of the problem, where you can see that only the arrow is transparent instead of the whole line.
I've read the
beamer User's Guide but I couldn't find any detailed information about the
\uncover-command and known complications, but maybe somebody here is able to help me.
I appreciate any help.
\documentclass[xcolor=dvipsnames]{beamer}\mode<presentation>\usetheme{Boadilla}\setbeamercovered{transparent}\usepackage[ngerman]{babel}\usepackage[utf8]{inputenc}\usepackage{lmodern,amsfonts,amsmath,amssymb,amsthm}\title{Dummy-Titel}\subtitle{Dummy-Untertitel}\author{Autor}\date{\today}\begin{document}\begin{frame} \begin{align*} \uncover<1->{& a = b \wedge b = c \\} \uncover<2->{\Rightarrow & a = c \\} \end{align*}\end{frame}\end{document} |
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