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I am reading about the 2nd order transfer function of a 2nd order system (like the mass-spring-damper system). I am constantly seeing the following form as the standard one:
\begin{equation} H(s) = \frac{\omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2} \end{equation}
But my questions are:
Why there are not zeros in this standard form? From what I am reading the general 2nd order transfer function can have zeros but in that form there are none and it is said to be the standard one.
Why there is no gain? I hardly found out sources that even mentioned the gain. Their standard form was:
\begin{equation} H(s) = \frac{K \cdot \omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2} \end{equation}
From what I understand the general form, the one that should be called standard, should be derived from the general form of the 2nd order differential equation:
\begin{equation} a_{2}\frac{d^{2}y(t)}{dt^{2}}+a_{1}\frac{dy(t)}{dt}+a_{0}y(t)=b_{m}\frac{d^{m}x(t)}{dt^{m}}+b_{m-1}\frac{d^{m-1}x(t)}{dt^{n-1}}+...+b_{0}x(t) \end{equation}
But from what I see it is derived from the not so general form:
\begin{equation} \frac{d^{2}y(t)}{dt^{2}}+a_{1}\frac{dy(t)}{dt}+a_{0}y(t)=x(t) \end{equation}
Why is that? |
It breaks up into three simple sections that are each relatively easy to explain:
simulate this circuit – Schematic created using CircuitLab
The first part is the diode that provides reverse voltage protection. If for some reason the polarity of the input voltage is wired opposite to what it is supposed to be, then \$D_1\$ will block it and the output will also be essentially off. Only if the polarity is correct, with the rest of the circuit be operational. The price of including this added protection is a voltage drop of perhaps \$700\:\text{mV}\$. (I exaggerated this voltage drop a little in the diagram. But it gets the point across.)
The next section is below that. It's a zener regulator. The resistor is there to limit the current. The zener tends to have the same voltage across it, when reverse-biased with sufficient voltage (and \$11-13\:\text{V}\$ is more than sufficient.) With \$R_1\$ as given, you'd expect the current to be somewhere from about \$5\:\text{mA}\$ to \$10\:\text{mA}\$. This is a "normal" operating current for many zeners. (You could go look up the datasheet and find out, exactly. I didn't bother here.) So the voltage at the top of the zener should be close to \$9.1\:\text{V}\$. The exact current through the zener will have a slight impact on this. But not much. (The capacitor, \$C_1\$, is there to "average out" or "smooth out" the zener noise. It's not critical. But it is helpful.)
The final section on the right is there to "boost up" the current compliance. Since the zener only has a few milliamps to work with, if you didn't include this added section your load could only draw a very small few milliamps, at most, without messing up the zener's regulated voltage. So to get more than that, you need a current boosting section. This is composed of what is often called an "emitter follower" BJT. This BJT's emitter will "follow" the voltage at the base. Since the base is at \$9.1\:\text{V}\$, and since the base-emitter voltage drop will be about \$600-700\:\text{mV}\$, you can expect the emitter to "follow," but here with a slightly lower voltage (as indicated in the schematic.) This BJT doesn't require much base current in order to allow a lot of collector current. So the BJT here may "draw" current from its collector, by also drawing a much smaller, tiny base current ("stolen" from the zener, so it can't be allowed to be very much), and then this sum of the two becomes the total emitter current. This emitter current can be as much as several hundred times the base current. So here, the BJT might draw \$1\:\text{mA}\$ of base current (which is okay, because there is several times that much available due to \$R_1\$) in order to handle perhaps as much as \$200\:\text{mA}\$ of emitter current. In keeping with the idea of "being conservative" the specification only says \$100\:\text{mA}\$ -- and that's very much the right way to go when telling someone what this is capable of. Be conservative.
\$R_2\$ is there as a bit of a short-circuit current limit. It doesn't serve much else. But if the load tries to pull too much current via the emitter then there will be an increasingly larger voltage drop across \$R_2\$ and this will cause the collector to have access to lower remaining voltage. At some point, the emitter will be "cramped." In this case, a drop of more than \$2\:\text{V}\$ (perhaps a little more) will probably begin the process of cramping the output. This means the limit is somewhere above \$\frac{2\:\text{V}}{22\:\Omega}\approx 100\:\text{mA}\$. Overall, \$R_2\$ is a very cheap way to add some modest protection to help make the whole thing just a little more bullet-proof, so to speak.
Note: \$C_2\$ is an output capacitor providing some added current compliance if there's a momentary, short-term demand by the load. I'd also normally want to include an output resistor across \$C_2\$ (not shown) of perhaps \$4.7\:\text{k}\Omega\$ as a
bleed resistor to provide a DC path to ground from the output and to discharge \$C_2\$ after a few seconds, when the input source of power is removed. |
I am trying to prove the convergence/divergence of the series´ $$\sum_{i=1}^\infty \frac{1}{F_n} = 1+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{8}+...$$${F_n}$ being the Fibonacci sequence.
The Fibonacci sequence is defined without recursion by: $${F_n}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}} \quad\land\quad\phi=\frac{1+\sqrt{5}}{2} $$ I have tried to prove its convergence with the Root Test and the Ratio Test because of the $n$ exponent but can't manage to do it because of the difference in the fraction. Can anyone help me? Thank you
I am trying to prove the convergence/divergence of the series´ $$\sum_{i=1}^\infty \frac{1}{F_n} = 1+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{8}+...$$${F_n}$ being the Fibonacci sequence.
Since $$F_{n}=\frac{\phi^{n}-\left(-\phi\right)^{-n}}{\sqrt{5}}$$ we have $$F_{n}\sim\frac{\phi^{n}}{\sqrt{5}}$$ as $n\rightarrow\infty$ so $$\sum_{n\ge1}\frac{1}{F_{n}}\sim\sqrt{5}\sum_{n\ge1}\frac{1}{\phi^{n}}=\frac{\sqrt{5}}{\phi-1}.$$
You may prove by induction that for any $n\geq 5$ we have $F_{n+5}\geq 11 F_n$. That is enough to deduce convergence by comparison with a geometric series and further get that:
$$ S = \sum_{n\geq 1}\frac{1}{F_n} =\frac{17}{6}+\sum_{n\geq 5}\frac{1}{F_n} = \frac{17}{6}+\sum_{n=5}^{9}\frac{1}{F_n}+\sum_{n\geq 10}\frac{1}{F_n}\leq \frac{17}{6}+\frac{88913}{185640}+\frac{1}{11}\sum_{n\geq 5}\frac{1}{F_n}$$ such that: $$ \frac{10}{11}\sum_{n\geq 5}\frac{1}{F_n}\leq \frac{88913}{185640},\qquad S\leq \frac{17}{6}+\frac{11}{10}\cdot \frac{88913}{185640}=\frac{2079281}{618800}. $$
For $n \ge 3$, we have
$$\frac{1}{F_n} \le \frac{F_{n-1}}{F_nF_{n-2}} = \frac{F_{n} - F_{n-2}}{F_nF_{n-2}} = \frac{1}{F_{n-2}} - \frac{1}{F_n} = \left(\frac{1}{F_{n-2}} + \frac{1}{F_{n-1}}\right) - \left(\frac{1}{F_{n-1}} + \frac{1}{F_n}\right)\\ = \frac{F_{n}}{F_{n-2}F_{n-1}} - \frac{F_{n+1}}{F_{n-1}F_n} $$
The partial sums is monotonic increasing and bounded from above. $$\sum_{n=1}^N \frac{1}{F_n} = 2 + \sum_{n=3}^N \frac{1}{F_n} \le 2 + \frac{F_3}{F_1F_2} - \frac{F_{N+1}}{F_{N-1}F_N} \le 2 + \frac{2}{1\cdot 1} = 4$$ As a result, the series converges.
$$\frac{F_{n+1}}{F_n}=\frac{\phi^{n+1}-(-\phi)^{-n-1}}{\phi^n-(-\phi)^{-n}}\ge\phi\frac{1-\phi^{-2n-2}}{1+\phi^{-2n}}.$$
The expression on the right is an increasing function of $n$ that exceeds $1$ as of $n=2$ (and quickly tends to $\phi$).
Let $F_n$ be the fibonacci sequence
We know $F_{2n+2}=F_{2n+1}+F_{2n}\geq 2F_{2n}$
Similarly $F_{2n+1}=F_{2n}+F_{2n-1}\geq 2F_{2n-1}$
So $1/F_2+1/F_4+1/F_6\dots<1/F_2+1/2F_2+1/4F_2\dots=1/F_2(1/2+1/4+1/8\dots)$
$1/F_1+1/F_3+1/F_5\dots<1/F_1+1/2F_1+1/4F_1\dots=1/F_1(1/2+1/4+1/8\dots)$
Now I think it's clearly evident that why the sum of reciprocals of the Fibonacci sequence is convergent, only the definition of the Fibonacci sequence is enough!
Suppose $2 \ge \frac{F_{n+1}}{F_n} \ge \frac32 $ for $n$ and $n+1$.
Then
$\frac{F_{n+2}}{F_{n+1}} =\frac{F_{n+1}}{F_{n+1}}+\frac{F_{n}}{F_{n+1}} =1+\frac{F_{n}}{F_{n+1}} \ge 1 + \frac12 =\frac32 $
and
$\frac{F_{n+2}}{F_{n+1}} =\frac{F_{n+1}}{F_{n+1}}+\frac{F_{n}}{F_{n+1}} =1+\frac{F_{n}}{F_{n+1}} \lt 1 + 1 =2 $.
I think I have an interesting proof for the summability of the reciprocals of the FN. It is based on the Kummer's test and the fact that $F_{n-1}\geq 2 F_{n}$, $n\geq 2$. First note that $(1/F_n)$ is summable, if and only if, there exists a sequence $(u_n)$ of positive numbers such that $$u_n F_{n+1}-u_{n+1}F_{n}\geq F_{n},$$ for all $n$ sufficiently large. (This is the Kummer's test applied to the sequence $(1/F_n)$). Now, from the definition of FN: $F_{n+1}-F_{n}=F_{n-1}$, for all $n\geq 2$.
That is, taking $u_n = 2$, for all $n\geq 1$, we have that $2 F_{n+1}- 2 F_{n}= 2 F_{n-1}\geq F_{n}$, for all $n\geq 2$. As so, by the Kummer's test, the sequence $(1/F_n)$ is summable. |
My intuitive way of thinking about it is that it is $2/2/2$ or $2/2^2$, So why then is it $1/2^2$? what is the flaw in my thinking?
Repeated multiplication can be seen as $$ \overset{\text{m terms}}{a \cdots a} = a^m.$$
Dividing this term repeatedly can then be seen as subtracting from $a^m$, because $$ \frac{a^m}{a} = \frac{\overset{\text{m terms}}{a \cdots a}}{a} = \overset{\text{m - 1 terms}}{a \cdots a} = a^{m-1}.$$
So $$\frac{1}{2^2} = \frac{2^0}{2^2} = 2^{-2}.$$
Well, $2^{-n} := {1 \over 2^n} (n \geq 0)$ is a definition.
Why this definition?
Because the usual rules (e.g. $2^{n+m}=2^n2^m$) which were established for ($n,m\in \mathbb{N}$) are now true for all $n,m \in \mathbb{Z}$. Which is nice, and shortens a lot of proofs.
$2^2 = 2\cdot 2 = 4$, $2^1 = 2$, $2^0 = 1$, $2^{-1} = 1/2$, $2^{-2} = 1/4$.
I guess the problem in your way of thinking is that when you're thinking about multiplication, you should "start" from $1$, whereas you're "starting" from $2$. The first $2$ in your expression $2/2/2$ is playing a different role than the other two $2$s - it should really be a $1$. |
Based on the existing answers, which opened my eyes for what's going on here, I would like to present yet another very simple expression for the solution, which is just slightly different from the one in AlexTP's answer (and which turned out to be equivalent to the one given in Jason R's answer, as shown below in the EDIT-part).
[EDIT: now that AlexTP has edited his answer, our expressions for the PDF are identical; so all three answers finally agree with each other].
Let the complex random variable $Z=X+jY$ be defined as
$$Z=\rho e^{j\theta}\tag{1}$$
where the radius $\rho$ is
deterministic and given, whereas the angle $\theta$ is random and uniformly distributed on $[0,2\pi)$. I state without further proof that $Z$ is circularly symmetrical, from which it follows that its probability density function (PDF) must satisfy
$$f_Z(z)=f_Z(x+jy)=f_Z(r),\qquad\textrm{with}\quad r=\sqrt{x^2+y^2}\tag{2}$$
i.e., it can be written as a function of the radius (magnitude) $r$.
Since the PDF must be zero everywhere except for $r=\rho$, and since it must integrate to unity (when integrated over the 2-dimensional plane), the only possible PDF is
$$f_Z(r)=\frac{1}{2\pi}\delta(r-\rho)\tag{3}$$
It can be shown that $(3)$ leads to the correct marginal densities for the random variables $X$ and $Y$.
EDIT:
After some very useful discussion in the comments it appears that we've managed to agree on one solution to the problem. I will show in the following that the unassuming formula $(3)$ is actually equivalent to the more involved looking formula in Jason R's answer. Note that I use $r$ for the magnitude (radius) of the complex RV $Z$, whereas in Jason's answer $r$ denotes the real part of $Z$. I will use $x$ and $y$ for the real and imaginary parts, respectively. Here we go:
$$f_Z(r)=\frac{1}{2\pi}\delta(r-\rho)=\frac{1}{2\pi}\delta\left(\sqrt{x^2+y^2}-\rho\right)\tag{4}$$
We know that $\delta\big(g(x)\big)$ is given by
$$\delta\big(g(x)\big)=\sum_i\frac{\delta(x-x_i)}{|g'(x_i)|}\tag{5}$$
where $x_i$ are the (simple) roots of $g(x)$. We have
$$g(x)=\sqrt{x^2+y^2}-\rho\quad\textrm{and}\quad g'(x)=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}\tag{6}$$
The two roots $x_i$ are
$$x_{1,2}=\pm\sqrt{\rho^2-y^2}\tag{7}$$
Consequently,
$$|g'(x_1)|=|g'(x_2)|=\frac{\sqrt{\rho^2-y^2}}{\rho}=\sqrt{1-\left(\frac{y}{\rho}\right)^2}\tag{8}$$
With $(5)$-$(8)$, Eq. $(4)$ can be written as
$$f_{X,Y}(x,y)=\frac{1}{2\pi\sqrt{1-\left(\frac{y}{\rho}\right)^2}}\left[\delta\left(x-\sqrt{\rho^2-y^2}\right)+\delta\left(x+\sqrt{\rho^2-y^2}\right)\right]\tag{9}$$
For $\rho=1$, Eq. $(9)$ is identical to the expression given in Jason R's answer.
I think we can now agree that Eq. $(3)$ is a correct (and very simple) expression for the PDF of the complex RV $Z=\rho e^{j\theta}$ with deterministic $\rho$ and uniformly distributed $\theta$. |
I have a certain need of rigorously justify some steps of proofs that use limits (maybe because I don't want to rely on intuition about limits that I have from first courses of calculus or somewhat), but I'm not sure what to think when I read "let $x\to a$".
I not sure if I'll make myself clear. I have done basic courses of calculus and I'm actually studying undergraduate level Real Analysis. I know basic properties of limits and know how to prove limits using epsilons and deltas. I'm interested in whether I can use the term "tending to" as it is used, for example, in the following proof of Fermat's Theorem (this proof is given on theorem 5.8 of Rudin's Principles of Mathematical Analysis):
Theorem:Let $f$ be defined on $[a,b]$; if $f$ has a local maximum at a point $x\in(a,b)$, and if $f'(x)$ exists, then $f'(x)=0$. Beginning of the proof: Choose $\delta>0$ so that $f(x)$ is a local maximum, that is, $f(x)\ge f(t)$ for all $t\in (x-\delta, x+\delta)$. Then take $x-\delta<t<x$ then $$\frac{f(t)-f(x)}{t-x}\ge 0.$$ Letting $t\to x$, we have $f'(x)\ge 0$.
I do know I can assume that $f'(x)$ is negative and choose $\epsilon = -f'(x)$ so that the quotient will also be negative, getting a contradiction hence we must have $f'(x)\ge 0.$ But this is quite non-practical as the term "tending to" suggests. Any answer that helps me "translate" the term into rigorous thought would be appreciated. |
A regular unit regarding assess utilized with revolving mechanical products can be rotations each and every minute, abbreviated
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Once we begin our own basic harmonic motion having absolutely no rate in addition to highest possible displacement (
x=X), then a entire electrical power is definitely: This resource efficiency of your energy just for this program around situation style is definitely so: Try to remember the fact that possibilities vitality ( PE), kept in your early spring which employs Hooke’s Law will be: the time period in this instance can be 12 moments 60/5.. uniform sale paper motion: Movements about some sort of rounded route along with frequent quickness. When a golf swings ( amplitudes ) tend to be small, lower than in relation to 15?, a pendulum gives simple harmonic oscillator having period of time [latex]\text=2\pi \sqrt where L will be the whole chain and gary the gadget guy may be the acceleration and speed caused by gravity. Uniform rounded motions is also sinusoidal as the projection with this action functions just like a very simple harmonic oscillator. Solving your differential formula previously mentioned normally produces methods that happen to be sinusoidal as the name indicated. As an example, by(testosterone levels), 5(testosterone levels), some sort of(testosterone), Okay(to), plus You actually(t) have the ability to sinusoidal alternatives for simple harmonic action. When out of place, a pendulum will probably oscillate all over the equilibrium level on account of traction within steadiness with all the restoring push of the law of gravity. angular frequency: A angular displacement a component period.
We are able to take advantage of this system to indicate the actual power performing inwards within the item employing Newton’s minute law,
F=ma. where m will be the huge with the rotaing system, x is it is displacement from the steadiness placement, and k is definitely the spring season frequent. Inside x-direction, really the only compel there exists could be the frictional push. Identify factors that affect the period on the straightforward pendulum Whenever we carry on and boost the rotation fee of your turn table, and thus increasing the velocity of any target located on them, eventually the particular frictional push defintely won’t be sufficient to maintain the article driving some sort of circle, plus the target will switch into the outside of the turntable along with leave. The following paying attention is viewed from the concept to get v max; it’s proportional for the rectangular reason this push frequent k. Velocity, centripetal acceleration in addition to power [ edit ]
The situation of twisting offers: A drive frequent okay is related to your solidity (and also solidity) of any system-the much larger this compel constant, the larger the repairing compel, plus the more rigid it. If
? is definitely indicated inside radians, the particular arc period inside of a group is related to it is radius ( L in this situation) by means of: Identify parameters affecting the time scale of an actual physical pendulum Somebody who is concept is moving on the right and remaining, then it need to have the leftward compel on there if it’s on the correct area, and a rightward pressure only when it’s for the still left area. It is fairly totally obvious that the arc span is definitely specifically proportionate for the angle : but, is there a continuous with proportionality? Properly, a good perspective with corresponds to a arc time .
Since the target begins to proceed, the variable prospective electrical power is definitely changed into kinetic strength, starting to be totally kinetic electricity within the steadiness posture. The probable electricity
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1 And 1.09425 Implies times Or 1
x Implies A person Per 2.09425 = 10.61 Volume is normally denoted using a Latin notification f and also by way of a Ancient greek mail ? (nu). An average physics science lab being active is to measure reestablishing makes made by spgs, see whether they follow Hooke’s legislation, and calculate their particular compel constants if they do. Additionally differentiation leads to the constituents of the acceleration and speed (that happen to be merely the amount of alter in the swiftness factors): The larger the bulk from the concept is usually, the better the phase T. frequency: Your quotient of the range of times m some sort of routine phenomenon happens across the time frame testosterone where it arises: y Equals m And t. oscillator: A design of which profits for its unique condition, inside the very same orientation and situation, from a limited range of a long time. The scenario of motion of which details basic harmonic movements can be acquired by simply merging Newton’s Secondly Rules along with Hooke’s Law in a second-order straight line regular differential scenario: [latex]\text_ way to get the particular drive is usually to be aware that Newton’s primary regulation lets us know: Using this specific formula along with the picture discovering a linear pace sooner, we will connect the actual angular rate of a twisting resist the straight line rate associated with some time in distance ron that object: There are in the shape which the world wide web compel for the joe is usually tangent towards arc and compatible ? mgsin ?. Gravity serves through the middle regarding huge in the firm entire body. The actual bike might be powered around this quickness whether or not it were ridden. Hence, within the small-angle approximation sin\theta \approx \theta, The posture of any concept in the plane might be altered coming from polar to be able to cartesian matches in the equations
In the event you adjust the predatory instincts so that you can count on
large frequenciesto get coupled with short periods, along with the other way around, you could possibly prevent quite a few humiliating problems about physics checks. The development inside uniform rounded movement is obviously led inside and it is offered by: where m will be the size on the shifting target. Solving the particular differential equation previously, a simple solution the industry sinusoidal purpose is usually purchased. Here, a duplication regularity , , from the action is actually scored in rounds per next –otherwise referred to as hertz (symbolic representation Hertz). We is able to use differential calculus and find the rate as well as acceleration being a use of occasion:
A equivalent calculations for the uncomplicated pendulum provides a comparable consequence, specifically: From this, a number of handy interaction stick to: Using Newton’s Second Legislations, Hooke’s Regulation, as well as some differential Calculus, we’re capable to gain the time scale as well as consistency in the huge oscillating for a early spring that we experienced over the last area! Note that the and also frequency are completely in addition to the plethora.
Example: A motor vehicle travels close to some sort of group of friends connected with distance Thirty t at a acceleration involving 10 m/s. At that level, it can be useful to outline a whole new angular model known as the radian (image rad.). Be aware the particular dependence of Ton g. Velocity, centripetal acceleration in addition to power [ edit ]
This change regularly rate ensures that the object is quickening (
centripetal acceleration). The essential thing to remember concerning this relationship is the fact that time remains in addition to the size of the inflexible body. If there’s not any general activity, you may have interferance chaffing. For instance, within an plenitude associated with ?= 23° it truly is 1% larger. Distance Equates to pace * moment 0 188.Your five = 10 * time Time Equals 188.5 ? 10 Is equal to In search of.425 seconds This is termed the centripetal speeding; sixth v Two Versus s is definitely the special constitute the speeding takes when i am handling items experiencing uniform circular motions. (w) The net drive is actually absolutely nothing within the sense of balance place, although the leader has strength along with carries on proceed to the ideal. Angular Frequency
Gravity operates over the middle with bulk of the strict human body. In case you adjust your own intuition so you count on
large frequenciesto be paired with short periods, as well as the opposite way round, chances are you’ll keep away from several awkward problems upon science tests. Within this interval, the thing transforms through a small angle along with records available this short arc of time-span , exactly where
Someone suggests an individual this inquiry, “For what period of your time were you running around circle keep track of within the playground?” You reply, “I has been walking around the actual round trail 5 minutes”. The actual bike will be powered only at that rate whether it have been ridden.
Define the time of round movements and provide it has the rate of recurrence?
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The possible power
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I’ve avoided doing any manifold (regretting it somewhat) courses, however do have some understanding. Let $p$ be a point on a surface $S:U\to \Bbb{R}^3$, we define:
The tangent space to $S$ at $p$, $T_p(S)=\{k\in\Bbb{R}^3\mid\exists\textrm{ a curve }\gamma:(-ε,ε)\to S\textrm{ with }\gamma(0)=p,\gamma'(0)=k\}$.
The tangent plane to $S$ at $p$ as the plane $p+T_p(S)\subseteq\Bbb{R}^3$.
My current understanding is, in the diagram below the tangent plane is the plane shown, whilst the tangent space would be p minus each element of the plane, hence the corresponding plane passing through the origin. Is this correct or is it incorrect? I’m doing a course called geometry of curves and surfaces and being unsure about this is making understanding later topics difficult.
Edit - can't post images, here's a link instead! http://standards.sedris.org/18026/text/ISOIEC_18026E_SRF/image022.jpg
Thanks! |
Question:
A sample of an unknown compound is vaporized at 150degree C . The gas produced has a volume of 960.mL at a pressure of 1.00atm , and it weighs 0.941g . Assuming the gas behaves as an ideal gas under these conditions, calculate the molar mass of the compound. Round your answer to 3 significant digits.
Ideal Gas Equation
The ideal gas equation applies to ideal gases which do have any intermolecular forces. This equation can also be used to calculate the density of gases under standard conditions.
Answer and Explanation: Given Data: The mass of compound is 0.941 g. The pressure is 1 atm. The temperature is {eq}150\;^\circ {\rm{C}} {/eq}. The volume of solution is 960 mL(0.960 L).
The molar mass can be calculated using the ideal gas equation, which is shown below.
{eq}\begin{align*} PV &= nRT\\ PV &= \dfrac{{{\rm{Given}}\;{\rm{mass}}}}{{{\rm{Molecular}}\;{\rm{mass}}}} \times RT \end{align*} {/eq}
Where,
{eq}V {/eq} is the volume of {eq}{\rm{C}}{{\rm{O}}_{\rm{2}}} {/eq}. {eq}n {/eq} is the moles of {eq}{\rm{C}}{{\rm{O}}_{\rm{2}}} {/eq}. {eq}P {/eq} is the pressure. {eq}T {/eq} is the temperature. {eq}R {/eq} is the gas constant (0.0821 L-atm/mol-K).
The temperature is in degrees
The conversion of degrees into Kelvin is shown below.
{eq}0^\circ {\rm{C}} = 273\;{\rm{K}} {/eq}
The conversion of {eq}150^\circ {\rm{C}} {/eq} into Kelvin is shown below
{eq}\begin{align*} 150^\circ {\rm{C}} &= \left( {150 + 273} \right)\;{\rm{K}}\\ &= {\rm{423}}\;{\rm{K}} \end{align*} {/eq}
Substitute the values in the above formula to calculate the molar mass.
{eq}\begin{align*} {\rm{1}}\;{\rm{atm}} \times {\rm{0}}{\rm{.960}}\;{\rm{L}} &= \dfrac{{0.941\;{\rm{g}} \times 0.0821\;\dfrac{{{\rm{L}} \cdot {\rm{atm}}}}{{{\rm{mol}} \cdot {\rm{K}}}} \times 423\;{\rm{K}}}}{{{\rm{Molecular}}\;{\rm{weight}}\left( {{\rm{g/mol}}} \right)}}\\ &= 34.04\;{\rm{g}} \end{align*} {/eq}
The molar mass of the unknown compound up to 3 significant figures is {eq}\underline {34.0\;{\rm{g}}} {/eq}.
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from General Studies Science: Help & ReviewChapter 27 / Lesson 24 |
For a Levy process $(X_t)_{t\geq 0}$, we have $\mathbb{E}[X_t]=t\mathbb{E}[X_t^1]$ and $\text{Var}(X_t)=t\text{Var}(X_t^1)$. Does the same hold for the first absolute moment, i.e. does $\mathbb{E}[|X_t|]=O(t)$ hold? Assume there is no continuous part, and that the process has bounded jumps (the absolute moment of the compound Poisson part is O(t)), so: \begin{align*} X_t = \lim_{\epsilon\to 0}\int_{\epsilon<x<1,s\leq t}x\bar J(ds,dx) \end{align*} where $\bar J(ds,dx)=J(ds,dx)-\nu(dx)ds$ is the compensated jump measure.
For a symmetric $\alpha$-stable process $(X_t)_{t \geq 0}$, the scaling property $X_t \sim t^{1/\alpha} X_1$ implies
$$\mathbb{E}(|X_t|) = t^{1/\alpha} \mathbb{E}(|X_1|)$$
for any $\alpha>1$. This shows that we cannot expect $\mathbb{E}(|X_t|) = O(t)$ for
any Lévy process $(X_t)_{t \geq 0}$. Even if we assume additionally that the process has no drift and diffusion part and that the process has bounded jumps, this is, in general, not true (see the counterexample below). There is the following statement on fractional moments:
Theorem:Let $(X_t)_{t \geq 0}$ be a Lévy process with Lévy triplet $(0,0,\nu)$. Suppose that there exist $\alpha,\beta>0$ such that $$\int_{|y| \leq 1} |y|^{\alpha} \, \nu(dy) <\infty \quad \text{and} \quad \int_{|y|>1} |y|^{\beta} \, \nu(dy)<\infty.$$ Then $$\mathbb{E}(|X_t|^{\kappa}) \leq C t^{\min\{\kappa/\alpha,1\}} \qquad \text{for all $t \leq 1$}$$ where $\kappa \in (0,\beta)$, $\kappa \neq \alpha$.
Roughly speaking, the behaviour of the Lévy measure $\nu$ at $\infty$ gives the existence of (fractional) moments whereas the behaviour of $\nu$ at $0$ determines the asymptotic of the fractional moments (for small $t$).
So, for a Lévy process $(X_t)_{t \geq 0}$ with bounded jumps (without drift and diffusion part), we have $\mathbb{E}(|X_t|) = O(t)$ as $t \to 0$ if $\int_{|y| \leq 1} |y|^{\alpha} \, \nu(dy)<\infty$ for some $\alpha \in (0,1)$.
The following lemma gives an example for a Lévy process with bounded jumps which does
not satisfy $\mathbb{E}(|X_t|) = O(t)$.
Example:Let $(X_t)_{t \geq 0}$ be a truncated $\alpha$-stable process for some $\alpha \in (1,2)$, i.e. a Lévy process with Lévy measure $$\nu(dy) = \frac{1}{|y|^{1+\alpha}} 1_{(0,1)}(|y|) \, dy.$$ Then $\mathbb{E}(|X_t|) = O(t)$ does not hold. Proof: Let $(Y_t)_{t \geq 0}$ be an independent Lévy process with Lévy measure $\frac{1}{|y|^{1+\alpha}} 1_{[1,\infty)}(|y|) \, dy$. Then $Z_t := X_t + Y_t$ defines a symmetric $\alpha$-stable process. By the scaling property, we have
$$\mathbb{E}(|Z_t|) = t^{1/\alpha} \mathbb{E}(|Z_1|).$$
Moreover, by the above theorem, we find
$$\mathbb{E}(|Y_t|) \leq C t$$
for all $t \leq 1$ for some constant $C>0$. Consequently, by the reverse triangle inequality,
$$\begin{align*} \mathbb{E}(|X_t|) &= \mathbb{E}(|Z_t-Y_t|) \geq \mathbb{E}(|Z_t|) - \mathbb{E}(|Y_t|) \\ &\geq t^{1/\alpha} \mathbb{E}(|Z_1|) - C t. \end{align*}$$
As $\alpha>1$ this clearly shows that $\mathbb{E}(|X_t|) = O(t)$ does not hold true for small $t$.
Further reading:
Luschgy, H., Pagés, G.: Moment Estimates for Lévy processes. Sato: Lévy processes. (General results on moments, e.g. characterization of existence of moments in terms of Lévy triplet.) Kühn, F.: Existence and Estimates of Moments for Lévy-type Processes. (This article provides moment estimates for a larger class of processes, so-called Lévy-type (or Feller) processes.) |
I am looking for the relations and analogies between the Perelman's entropy functional,$\mathcal{W}(g,f,\tau)=\int_M [\tau(|\nabla f|^2+R)+f-n] (4\pi\tau)^{-\frac{n}{2}}e^{-f}dV$, and notions of entropy from statistical mechanics. Would you please explain it in details?
For metrics on $S^{2}$ with positive curvature, Hamilton introduced the entropy $N\left( g\right) =-\int\ln(R\operatorname{Area})Rd\mu.$ If the initial metric has $R>0,$ he proved that this is nondecreasing under the Ricci flow on surfaces; note that $Rd\mu$ satisfies $(\frac{\partial}{\partial t}-\Delta )(Rd\mu)=0.$ Let $T$ be the singular time; then $$ \frac{d}{dt}N\left( g\left( t\right) \right) =2\int\left\vert \operatorname{Ric}+\nabla^{2}f-\frac{1}{2\tau}g\right\vert ^{2}d\mu+4\int% \frac{\left\vert \operatorname{div}(\operatorname{Ric}+\nabla^{2}f-\frac {1}{2\tau}g)\right\vert ^{2}}{R}d\mu, $$ where $\tau=T-t$ and $\Delta f=r-R.$ ($f$ satisfies $\frac{\partial f}{\partial t}=\Delta f+rf$; since $n=2$, $\operatorname{Ric}=\frac{1}{2}Rg$)
Perelman's entropy has the main term: $\int fe^{-f}d\mu,$ which is the classical entropy with $u=e^{-f}$ as Deane Yang wrote. (Besides Section 5 of Perelman, further discussion of entropy appeared later in some of Lei Ni's papers as well as elsewhere.) Even though this term is lower order (in terms of derivatives), geometrically it is the most significant as can be seen by taking the test function to be the characteristic function of a ball (multiplied by a constant for it to satisfy the constraint); technically, one chooses a cutoff function. Thus Perelman proved finite time no local collapsing below any given scale only assuming a local upper bound for $R,$ since the local lower Ricci curvature bound (control of volume growth is needed to handle the cutoff function) can be removed by passing to the appropriate smaller scale.
Heuristically (ignoring the cutoff issue), since the constraint is $\int(4\pi\tau)^{-n/2}e^{-f}d\mu=1,$ if we take $\tau=r^{2}$ and $e^{-f}=c\chi_{B_{r}},$ then $c\approx\frac{r^{n}% }{\operatorname{Vol}B_r}.$ So, if the time and scale are bounded from above, by Perelman's monotonicity, we have $$-C\leq\mathcal{W}(g,f,r^{2})\lessapprox r^{2}% \max_{B_{r}}R+\ln\frac{\operatorname{Vol}B_r}{r^{n}},$$ yielding the volume ratio lower bound.
Added December 12, 2013:For all of the following, see Perelman, Ni, Topping, etal. Let $\mathcal{N}=\int fe^{-f}d\mu$ be the classical entropy. Then, under $\frac{\partial}{\partial t}g=-2(\operatorname{Ric}+\nabla^{2}f)$ and $\frac{\partialf}{\partial t}=-\Delta f-R$, we have $-\frac{d\mathcal{N}}{dt}=\mathcal{F}(g,f)\doteqdot\int(R+\left\vert \nabla f\right\vert ^{2})e^{-f}d\mu$(Perelman's energy). If the solutions are on $[0,T)$, then $\mathcal{F}(t)\leq\frac{n}{2\left( T-t\right) }\int e^{-f}d\mu$, which implies that$\frac{d}{dt}(\mathcal{N}-(\frac{n}{2}\int e^{-f}d\mu)\log(T-t))\geq0$. Let$\mathcal{W}(g,f,\tau)=(4\pi\tau)^{-n/2}\left( \tau\mathcal{F}+\mathcal{N}\right) -n\int(4\pi\tau)^{-n/2}e^{-f}d\mu$ (Perelman's entropy). Under$\frac{\partial}{\partial t}g=-2\operatorname{Ric}$ and $\frac{\partialf}{\partial t}=-\Delta f+|\nabla f|^{2}-R+\frac{n}{2\tau}$, we have have$\frac{d\mathcal{F}}{dt}=2\int|\operatorname{Ric}+\nabla^{2}f|^{2}e^{-f}d\mu-\frac{n}{2\tau}\mathcal{F}$ and $\frac{d\mathcal{N}}{dt}=-\mathcal{F}+\frac{n}{2\tau}\int e^{-f}d\mu-\frac{n}{2\tau}\mathcal{N}$. So, by couplingwith $\frac{d\tau}{dt}=-1$, we obtain (Perelman's entropy formula)\begin{align*}(4\pi\tau)^{n/2}\frac{d\mathcal{W}}{dt} & =\frac{n}{2\tau}\left(\tau\mathcal{F}+\mathcal{N}\right) -\mathcal{F}+\tau\frac{d\mathcal{F}}{dt}+\frac{d\mathcal{N}}{dt}\\& =2\tau\int|\operatorname{Ric}+\nabla^{2}f-\frac{1}{2\tau}g|^{2}e^{-f}d\mu.\end{align*}
Perelman himself wrote about his entropy formula for the Ricci flow that
"The interplay of statistical physics and (pseudo)-riemannian geometry occurs in the subject of Black Hole Thermodynamics, developed by Hawking et al. Unfortunately, this subject is beyond my understanding at the moment."
Subsequently, this connection has been explored in some detail in two papers by Samuel and Chowdury: "Geometric flows and black hole entropy" and "Energy, entropy and the Ricci flow". For a discussion of both these papers, see chapter 6 of "On the Emergence Theme of Physics" by Robert Carroll.
Perelman has given a gradient formulation for the Ricci flow, introducing an "entropy function" which increases monotonically along the flow. We pursue a thermodynamic analogy and apply Ricci flow ideas to general relativity. We investigate whether Perelman's entropy is related to (Bekenstein-Hawking) geometric entropy as familiar from black hole thermodynamics. From a study of the fixed points of the flow we conclude that Perelman entropy is not connected to geometric entropy. However, we notice that there is a very similar flow which
doesappear to be connected to geometric entropy. The new flow may find applications in black hole physics suggesting for instance, new approaches to the Penrose inequality.
There is a nice interpretation of Perelman's monotonicity formulas in terms of optimal transportation, see e.g. these lecture notes by Peter Topping
It seems helpful to look at the elliptic case first. As discovered by Lott-Villani and Sturm nonnegative Ricci curvature can be characterized by the property that the Boltzmann-entropy is convex along optimal transportation. This is very intuitive, imagine e.g. a pile of sand being transported from the south to the north-pole on the sphere.
The idea for the Ricci flow is similar (being a (super)Ricci flow can be viewed as parabolic version of having nonnegative Ricci curvature), but the details are a bit more complicated. The $W$-functional can be written as derivative of a suitable Boltzmann-entropy (see Section 5 in Perelman's first paper) and the monotonicity of $W$ can be interpreted as convexity of this entropy, see the above lecture notes for details. |
This post comes as a result of a comment a friend made about what he thought was a good idea for a /r/dataisbeautiful post: “Some kind of graph showing the correlation between the population of a country and the number of people that answered it in the countries of the world sporcle quiz.” he said in a WhatsApp message. I think he knew exactly what would happen once I saw that suggestion…
My grandfather used to work as a painter in a car garage. I remember him telling me that the sorts of people he worked with, mechanics and engineery types, had this quirk where if you were stuck with something, you simply had to stand in the middle of the shop looking puzzled. “What’s that you’ve got there? Oh let me have a look. Ah yeah, you see, what you need there…” And solutions to your problem would be offered up by these inquisitive minds.
Apparently, I suffer from the same problem, except with data.
Anyway…
Sporcle gives you access to the global results at the end of the quiz (I got 124 out of 197, managing to forget that Afghanistan and Colombia exist along the way). I’m well aware that this isn’t a representative sample of the global population. Judging by the Alexa results, Sporcle’s users are mainly in English-speaking countries like the UK, USA and Australia, so if anything this is a look at how well the English-speaking world knows the rest.
Next, population data: Wikipedia was my starting point. I headed on over to the List of countries by population (United Nations) article, and then converted the table to a CSV with this handy tool. There’s a couple of different articles with similar titles, but this one also includes labels of UN continental region and UN statistical region which will be useful later on when looking for regional effects on the “memorability” of a country.
After a bit of manual cleaning up of the data - things like removing British Overseas Territories and Crown Dependencies, adding Kosovo, renaming “Côte d’Ivoire” and “Macedonia” to “Ivory Coast” and “Republic of Macedonia” for consistency etc. - it was a simple case of joining the two tables together in R and outputting to a CSV file.
Here’s a sample:
Country Percentage Answered UN continental region UN statistical region Population2017 Afghanistan 79.5% Asia Southern Asia 35530081 Albania 64.5% Europe Southern Europe 2930187 Algeria 71.6% Africa Northern Africa 41318142 Andorra 66.2% Europe Southern Europe 76965 Angola 58.9% Africa Middle Africa 29784193
Full table is available here.
And finally some plots:
These are the same plots with differing labels.
The grey lines are a linear-regression of “Memorability” against population: countries above the line are more memorable than their population would predict, countries below the line, less memorable.
No real surprises here, the upper right region is dominated by the G20 countries with the bizarre outlier that is the Vatican over on the top left, and then the African states that apparently people forget exist.
I posted these to /r/dataisbeautiful, where another user suggested that using GDP in place of population might yield some interesting results. I tried it out. Here’s the same plots as above but with Nominal GDP on the X-axis:
There’s a lot less spread here which suggets that GDP makes for a better predictor than population. Another user suggested that the reason for this is that Nominal GDP is relevant for international trade.
Finally, I did a bit of simple modelling in Stan to determine what the continent-level effects were, just a fixed-effects model with a Student-T residual:
$$ \text{Percentage Answered} \sim \text{Student}(\nu, \beta^\text{Continent} + \beta^\text{Population} \cdot log(\text{Population}), \sigma) $$
And similarly, using Nominal GDP:
$$ \text{Percentage Answered} \sim \text{Student}(\nu, \beta^\text{Continent} + \beta^\text{GDP} \cdot log(\text{Nominal GDP}), \sigma) $$
The Stan model file is here for interested parties, it’s the same for both models but with differing input data.
What’s important is the results. First, with population:
Here general geographic effects are evident, negative effects for Africa, positive effects for Europe and North America.
And then with GDP:
Arguably these results are more interesting: once the effect of GDP is removed, there’s not an awful lot of distance between geographic regions.
I originally did this all back in June, but only just bothered to write it up. |
There is (at least) one way to prove unambiguity of a grammar $G = (N,T,\delta,S)$ for language $L$. It consists of two steps:
Prove $L \subseteq \mathcal{L}(G)$. Prove $[z^n]S_G(z) = |L_n|$.
The first step is pretty clear: show that the grammar generates (at least) the words you want, that is correctness.
The second step shows that $G$ has as many syntax trees for words of length $n$ as $L$ has words of length $n$ -- with 1. this implies unambiguity. It uses the structure function of $G$ which goes back to Chomsky and Schützenberger [1], namely
$\qquad \displaystyle S_G(z) = \sum_{n=0}^\infty t_nz^n$
with $t_n = [z^n]S_G(z)$ the number of syntax trees $G$ has for words of length $n$. Of course you need to have $|L_n|$ for this to work.
The nice thing is that $S_G$ is (usually) easy to obtain for context-free languages, although finding a closed form for $t_n$ can be difficult. Transform $G$ into an equation system of functions with one variable per nonterminal:
$\qquad \displaystyle \left[ A(z) = \sum\limits_{(A, a_0 \dots a_k) \in \delta} \ \prod\limits_{i=0}^{k} \ \tau(a_i)\ : A \in N \right] \text{ with } \tau(a) = \begin{cases} a(z) &, a \in N \\ z &, a \in T \\ \end{cases}.$
This may look daunting but is really only a syntactical transformation as will become clear in the example. The idea is that generated terminal symbols are counted in the exponent of $z$ and because the system has the same form as $G$, $z^n$ occurs as often in the sum as $n$ terminals can be generated by $G$. Check Kuich [2] for details.
Solving this equation system (computer algebra!) yields $S(z) = S_G(z)$; now you "only" have to pull the coefficient (in closed, general form). The TCS Cheat Sheet and computer algebra can often do so.
Example
Consider the simple grammar $G$ with rules
$\qquad \displaystyle S \to aSa \mid bSb \mid \varepsilon$.
It is clear that $\mathcal{L}(G) = \{ww^R \mid w \in \{a,b\}^*\}$ (step 1, proof by induction). There are $2^{\frac{n}{2}}$ palindromes of length $n$ if $n$ is even, $0$ otherwise.
Setting up the equation system yields
$\qquad \displaystyle S(z) = 2z^2S(z) + 1$
whose solution is
$\qquad \displaystyle S_G(z) = \frac{1}{1-2z^2}$.
The coefficients of $S_G$ coincide with the numbers of palindromes, so $G$ is unambiguous.
The Algebraic Theory of Context-Free Languages by Chomsky, Schützenberger (1963) On the entropy of context-free languages by Kuich (1970) |
I am looking for a datastructure to represent the complex relationships between a bunch of
abstract sets. The "abstract" means that these sets are not defined by their elements, but by their relationship to each other. This means that for me a set is nothing more than a named thing with a relations to other named things in the same space. For two sets $A,B$ the relation can be they represent the same set, i.e. $A=B$, they are complementary, i.e. $\bar A=B$, they are disjoint $A\cap B=\varnothing$ but not complementary, they are overlapping, i.e. they are not equivalent but have a non-empty intersection, one is a propersubset of the other set, e.g. $A\subset B$ essentially unknown, but maybe some of the former cases can be excluded.
Besides storing these relationships, I want to ask questions on the datastructure like
Are the sets $A$ and $B$ disjoint? Name me a subset of $A\cap\bar B\cap \bar C$. What is the relation between the sets $X$ and $\bar Y\cap Z$?
Also the datastructure should be updatable. There will be inserted new sets all the time which must be integrated in the structure correctly. There can be
lots of sets and it is no option to explicitely store the relation between any pair of sets. Instead, some relations should be deducable by the query algorithms. For example, if we have $A\subset B\subset C$, then we will not store $A\subset C$ as this follows logically from the other two relations.
I thought about lots of tree like structures or cycle-free directed graphs. But maybe someone already developed exactly what I am looking for.
One more thing: I would prefer when the datastructure does note introduce big differences between sets and their complement. This means, asking whether $A\cap B$ is empty should not invoke vastly different algorithms than asking this for $A\cap \bar B$.
Update
To answer some comments, here are some more details on my intend.
You can think about this as a big (hyper-)graph problem, where the whole graph is too big to be stored but has much structure, so maybe can be stored very efficiently in another way. The nodes of the graph are sets. The (hyper-)edges are relations between the sets. The edges are colored to represent the six possible relation types from above. In this sense, the graph is an edge-colored complete graph. Note that the edge type
unknown is very important to me!
To describe the queries in general, let me introduce the following notation: given a set $A$, we write $A^+=A$ and $A^-=\bar A$. The following queries should be supported as efficient as possible:
Given a sequence of nodes $A_i$ and a sequence of signs $s_i\in\{+,-\}$. Which of the following three options is true for $\bigcap_i A_i^{s_i}$: it is empty, there is a node representing a subset of this intersection, we do not know anything. Given two sequences $A_i$, $B_i$ of nodes and two sequences $s_i$ and $r_i$ of signs. What kind of edge is between $\bigcap_i A_i^{s_i}$ and $\bigcap_i B_i^{r_i}$?
The last query seems trivial for the graph stored with all its edges. But as I said, the graph is probably too big to be stored in this naive way.
Of course, I never stated that for sets $A_i$ and signs $s_i$ my data structure must contain the set $\bigcap_i A_i^{s_i}$,
but information can be deduced in other ways. For example, assume $A$ and $B$ are sets, but $A\cap B$ is not stored. But we have a set $C\subset A,B$. So we know that $A\cap B$ $-$ despite being not stored explicitely $-$ is not empty. If we lack any of these information, then we may conclude that "we do not know anything about $A\cap B$". The fact that $A\cap B$ is empty is stored by an edge between $A$ and $B$ signalling that these sets are disjoint.
Why I wrote
hyper-edge instead of just edge? It might be necessary to store that $A\cap B\cap C=\varnothing$ but no two of these sets are disjoint. This cannot be deduced from any two set relation. So we may need edges involving more than just two sets. |
Look at this simplified itnernal layout (from its datasheet):
You can see there's an opamp in there driving a transistor (NPN), whose collector is considered the output (the anode is grounded). This means that, by taking a quantity from the output (cathode), and bringing it to the input, you create a negative feedback (because the collector inverts the phase), so RF1 and CF1, together with RH1 and RL1, are there to provide a zero in the overall transfer function: \$f_Z=\frac{1}{2\pi C(R_{F1}+R_{H1}||R_{L1})}\$. That zero compensates for the loss of phase of the converter at around the switching frequency, ensuring it does not oscillate. Here's a quick proof:
The zero frequency is at \$\frac{1}{2\pi 1e-9[10000 + (1000||1000)]}\approx15.16\text{kHz}\$. Looking at the readings, you see the phase is 45
o at around 13.2kHz, which is different than the math because of the gain-bandwidth product of the opamp, plus the parasitic capacitances of the transistor itself (both thrown in there for exemplification, only, no other reason), which influence the magnitude and the phase, both.
In short, the TL431, together with the feedback network, acts like an error amplifier, frequency compensated. |
Differences
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Line 293: Line 293: Generalization of function update is override of partial functions, $f \oplus g$ Generalization of function update is override of partial functions, $f \oplus g$ + Line 299: Line 300: ==== Range, Image, and Composition ==== ==== Range, Image, and Composition ==== + The following properties follow from the definitions: \[ \[ - S \bullet r = ran(\ Delta_S \circ r) + (S \bullet r_1) \bullet r_2 = S \ bullet (r_1 \circ r_2) \] \] \[ \[ - (S \bullet r_1) \bullet r_2 = S \ bullet (r_1 \circ r_2) + S \bullet r = ran(\ Delta_S \circ r) \] \] Line 309: Line 311: * [[:Gallier Logic Book]], Chapter 2 * [[:Gallier Logic Book]], Chapter 2 + * [[sav08:discrete_mathematics_by_rosen|Discrete Mathematics by Rosen]] |
This question already has an answer here:
$f:\mathbb{N}\rightarrow \mathbb{N}$ is a one-to-one function such that $f(mn)=f(m)f(n).$ Find the lowest possible value of $f(999)$.
The answer is given as $24$ but I never get that.
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This question already has an answer here:
$f:\mathbb{N}\rightarrow \mathbb{N}$ is a one-to-one function such that $f(mn)=f(m)f(n).$ Find the lowest possible value of $f(999)$.
The answer is given as $24$ but I never get that.
$f(999)=f(37)\cdot f(3)\cdot f(3)\cdot f(3)$
Now, $f(3)$ or $f(37)$ cannot be $1.$ Because, if, for example, $f(3)=1,$ then $f(999)=f(37)$ which implies $999=37$ (because the function is one-to-one).
Therefore, $f(3)$ can have the lowest value $2.$ And hence $f(37)$ can have the lowest value $3.$ Thus, $f(999)=3\cdot 2\cdot 2\cdot 2=24.$
$$m=n=1 \Rightarrow f(1)=1$$ so we have $f(3)\ge 2,f(37)\ge 2$ since $f(3)\neq f(37)$
$$f(999)=f(27)\cdot f(37)=(f(3))^3 \cdot f(37)\ge 2^3\cdot 3=24$$
Let us define $f(2)=37,f(3)=2,f(37)=3$ and $f(p)=p$ for all primes $p\neq 2,p\neq 3,p\neq 37$
Let us show the function is injective.
The condition implies $f(\prod_{i=1}^k p_i^{e_i})=\prod_{i=1}^k f(p_i)^{e_i}$
let $m=2^{a}\cdot 3^{b}\cdot 37^{c}\cdot\prod_{i=1}^k p_i^{e_i},n=2^{p}\cdot 3^{q}\cdot 37^{r}\cdot\prod_{i=1}^l p_i^{e_i}, $ be prime factorizations of $m$ and $n$ and $f(m)=f(n)$ then
$$f(m)=f(n) \Leftrightarrow 37^{a}\cdot2^{b}\cdot 3^{c}\cdot \prod_{i=1}^k p_i^{e_i}=37^{p}\cdot2^{q}\cdot 3^{r}\cdot \prod_{i=1}^l p_i^{e_i} \Rightarrow$$
$$a=p,b=q,c=r,\land \prod_{i=1}^k p_i^{e_i}=\prod_{i=1}^l p_i^{e_i} \Rightarrow m=n$$
so the funciton is injective |
I have a digital option that pays out \$1M at time $T$ if the price of the underlying stock is higher than \$1300 (with current price ~\$1000) and, obviously, zero otherwise. I am in the Black-Scholes setting and there are no dividends up to date $T$.
I have used the following to calculate the price of the digital option at $t=0$:
$C(0)=e^{-rT}N(d_2)$
where
$d_2 = \frac{\ln\big(\frac{S(0)}{Ke^{-rT}}\big) - \frac{\sigma^2}{2}}{\sigma\sqrt{T}}$
I am not given any of the parameters (apart from $K$ and $T$) and have been asked to make an educated guess as to what the call price should be. I have checked this using an online calculator and the price I get (~\$0.18) seems to be correct given the parameters I have fed into this model.
I have a couple of questions on this as it is not explained anywhere in my lecture notes and I cannot seem to find what I need online.
Q1: Why is the price of the call option not dependent on the pay-out? If this option was paying \$1 or \$0 then a price of \$0.18 seems reasonable, but not for \$1M. I would place that bet every week as it is cheaper and surely better odds than the lottery. I'm obviously missing some understanding here.
Q2: How would the trader who sold this digital option hedge against the potential losses? I have read online that this is possible using a call-spread but I'm not sure I understand how this would work. I understand that you need to go short on the call with the higher strike price and long on the other but I don't see how to apply this to this situation? |
Research
I study the distribution of algebraic numbers, mathematical statistical physics and roots/eigenvalues of random polynomials/matrices.
Projects in Progress
1
The distribution of values of the non-archimedean absolute Vandermonde determinant and the non-archimedean Selberg integral (with Jeff Vaaler). The Mellin transform of the distribution function of the non-archimedean absolute Vandermonde (on the ring of integers of a local field) is related to a non-archimedean analog of the Selberg/Mehta integral. A recursion for this integral allows us to find an analytic continuation to a rational function on a cylindrical Riemann surface. Information about the poles of this rational function allow us to draw conclusions about the range of values of the non-archimedean absolute Vandermonde.
2
Non-archimedean electrostatics. The study of charged particles in a non-archimedean local field whose interaction energy is proportional to the log of the distance between particles, at fixed coldness $\beta$. The microcanonical, canonical and grand canonical ensembles are considered, and the partition function is related to the non-archimedean Selberg integral considered in 1. Probabilities of cylinder sets are explicitly computable in both the canonical and grand canonical ensembles.
3
Adèlic electrostatics and global zeta functions (with Joe Webster). The non-archimedean Selberg integral/canonical partition function are examples of Igusa zeta functions, and as such local Euler factors in a global zeta function. This global zeta function (the exact definition of which is yet to be determined) is also the partition function for a canonical electrostatic ensemble defined on the adèles of a number field. The archimedean local factors relate to the ordinary Selberg integral, the Mehta integral, and the partition function for the complex asymmetric $\beta$ ensemble. The dream would be a functional equation for the global zeta function via Fourier analysis on the idèles, though any analytic continuation would tell us something about the distribution of energies in the adèlic ensemble.
4
Pair correlation in circular ensembles when $\beta$ is an even square integer (with Nate Wells and Elisha Hulbert). This can be expressed in terms of a form in a grading of an exterior algebra, the coefficients of which are products of Vandermonde determinants in integers. Hopefully an understanding of the asymptotics of these coefficients will lead to scaling limits for the pair correlation function for an infinite family of coldnesses via hyperpfaffian/Berezin integral techniques. This would partially generalize the Pfaffian point process arising in COE and CSE. There is a lot of work to do, but there is hope.
5
Martingales in the Weil height Banach space (with Nathan Hunter). Allcock and Vaaler produce a Banach space in which $\overline{\mathbb Q}^{\times}/\mathrm{Tor}$ embeds densely in a co-dimension 1 subspace, the (Banach space) norm of which extends the logarithmic Weil height. Field extensions of the maximal abelian extension of $\mathbb Q$ correspond to $\sigma$-algebras, and towers of fields to filtrations. Elements in the Banach space (including those from $\overline{\mathbb Q}^{\times}/\mathrm{Tor}$) represent random variables, and the set up is ready for someone to come along and use martingale techniques—including the optional stopping time theorem—to tell us something about algebraic numbers. Instruction
I have three current PhD students and one current departmental Honors student. I have supervised two completed PhDs and six completed honors theses. You can find a list of current and completed PhD and honors students on my CV.
My teaching load has been reduced for the last five years (or so) due to an FTE release for serving on the Executive Council of United Academics. As President of United Academics, and Immediate Past President of the University Senate I am not teaching in the 2018 academic year. In AY2019, I am scheduled to teach a two-quarter sequence on mathematical statistical physics.
I take my teaching seriously. I prepare detailed lecture notes for most courses (exceptions being introductory courses, where my notes are better characterized as well-organized outlines). When practical and appropriate I use active learning techniques, mostly through supervised group work. I am a tough, but fair grader.
Service
Service encompasses pretty much everything that an academic does outside of teaching and research. This includes advising, serving on university and departmental committees, reviewing papers, writing letters of recommendation, organizing seminars and conferences, serving on professional boards, etc. The impossibility of doing it all allows academics to decide what types of service they are going specialize based on their interests and abilities.
I have spent the last three years heavily engaged in university level service. I currently serve as the president of United Academics of the University of Oregon, and I am the immediate-past president of the University Senate. Before that I was the Vice President of the Senate and the chair of the Committee on Committees. All of these roles are difficult and require a large investment of thought and energy. The reward for this hard work is a good understanding of how the university works, who to go to when issues need resolution, and who can be safely ignored.
I know what academic initiatives are underway, being involved in several of them. I am spearheading, with the new Core Education Council, the reform of general education at UO. I am working on the New Faculty Success Program—an onboarding program for new faculty—with the Office of the Provost and United Academics. I am currently on the Faculty Salary Equity Committee and its Executive Committee. I have been a bit player in many other projects and initiatives including student evaluation reform, the re-envisioning of the undergraduate multicultural requirement, and the creation of an expedited tenure process to allow the institution alacrity when recruiting imminent scholars. This list is incomplete.
Next year, with high probability, I will be the chair of the bargaining committee for the next collective bargaining agreement between United Academics and the University of Oregon (this assumes I am elected UA president). I will also be working with the Core Ed Council to potentially redefine the BA/BS distinction, with a personal focus on ensuring quantitative/data/information literacy is distributed throughout our undergraduate curriculum. I will also be working to help pilot (and hopefully scale) the Core Ed “Runways” (themed, cohorted clusters of gen ed courses) with the aspirational goal of having 100% of traditional undergraduates in a high-support, high-engagement, uniquely-Oregon first-year experience within the next 3-5 years.
As important as the service I
am doing, is the service I am not doing. I do little to no departmental service (though part of this derives from the CAS dean’s interpretation of the CBA) and I avoid non-required departmental functions (for reasons). I do routinely serve on academic committees for graduate/honors students, etc. I decline most requests to referee papers/grants applications, and serve on no editorial boards. The national organizations for which I am an officer are not mathematical organizations, but rather organizations dedicated to shared governance. Diversity & Equity
The two principles which drive my professional work are
truth and fairness.
I remember after a particularly troubling departmental vote, a senior colleague attempted to assuage my anger at the department by explaining that “the world is not fair.” I hate this argument because it removes responsibility from those participating in such decisions, and places blame instead on a stochastic universe. And, while there
is stochasticity in the universe, we should be working toward ameliorating inequities caused by chance, and in instances where we have agency, making decisions which do not compound them.
I do not think the department does a very good job at recognizing nor ameliorating inequities. Indeed, there are individuals, policies and procedures that negatively impact diversity. See my recent post Women & Men in Mathematics for examples.
My work on diversity and equity issues has been primarily through the University Senate and United Academics. As Vice-president of the UO Senate, I sat on the committee which vetted the Diversity Action Plans of academic units. I also worked on, or presided over several motions put forth by the University Senate which address equity, diversity and inclusion. Obviously, the work of the Senate involves many people, and in many instances I played only a bit part, but nonetheless I am proud to have supported/negotiated/presided over the following motions which have addressed diversity and equity issues on campus:
Implementing A System for the Continuous Improvement and Evaluation of Teaching Proposed Changes to Multicultural Requirement Resolution denouncing White Supremacy & Hate Speech on Campus Proposed Change to Admissions Policies Requiring Disclosure of Criminal and Disciplinary Hearing A Resolution in support of LGBTQAI Student Rights Declaring UO a Sanctuary Campus Reaffirming our Shared Values of Respect for Diversity, Equity, and Inclusion Student Sexual and gender-Based Harassment and Violence Complaint and Response Policy
Besides my work with the Senate, I have also participated in diversity activities through my role(s) with United Academics of the University of Oregon. United Academics supports both a Faculty of Color and LGBTQ* Caucus which help identify barriers and propose solutions to problems affecting those communities on campus. United Academics bargained a tenure-track faculty equity study, and I am currently serving on a university committee identifying salary inequities based on protected class and proposing remedies for them.
I have attended in innumerable rallies supporting social justice, and marched in countless marches. I flew to Washington D.C. to attend the March for Science. I’ve participated in workshops and trainings on diversity provided by the American Federation of Teachers, and the American Association of University Professors.
I recognize that I am not perfect. I cannot represent all communities nor emulate the diversity of thought on campus. I have occasionally used out-moded words and am generally terrible at using preferred pronouns (though I try). I recognize my short-comings and continually work to address them.
There are different tactics for turning advocacy into action, and individuals may disagree on their appropriateness and if/when escalation is called for. My general outlook is to work within a system to address inequities until it becomes clear that change is impossible from within. In such instances, if the moral imperative for change is sufficient then I work for change from without. This is my current strategy when tackling departmental diversity issues; I work with administrative units, the Senate and the union to put forth/support policies which minimize bias, discrimination and caprice in departmental decisions. I ensure that appropriate administrators know when I feel the department has fallen down on our institutional commitment to diversity, and I report incidents of bias, discrimination and harassment to the appropriate institutional offices (subject to the policy on Student Directed Reporters).
Fairness is as important to me as truth, and I look forward to the day where I can focus more of my time uncovering the latter instead of continually battling for the former. |
As Mohan points out in the comments, there is a canonical $K$-vector space isomorphism $\Omega^1_{X/k,\eta}\simeq\Omega_{K/k}^1$, where $\eta$ is the generic point of $X$.
(It seems like you're assuming $X$ is regular, or equivalently smooth over $k$, since otherwise $\Omega_{X/k}^1$ need not be locally free.) What you're talking about with divisors associated to "rational differentials" is a particular case of a somewhat more general construction. Let $X$ be any integral scheme with generic point $\eta$ and function field $K$. If $\mathscr{L}$ is an invertible $\mathscr{O}_X$-module, then (non-zero) elements of the $1$-dimensional $K$-vector space $\mathscr{L}_\eta$ are often called rational sections of $\mathscr{L}$ (maybe $0$ deserves to be a rational section too, but let's focus on non-zero rational sections). By the definition of $\mathscr{L}_\eta$ as the colimit over $\mathscr{L}(U)$ for $U$ non-empty and open in $X$, such rational sections can also be described as equivalence classes of pairs $(U,s)$, where $U$ is a non-empty open of $X$ and $s$ a section in $\mathscr{L}(U)$, and the equivalence relation is equality on intersections (
a priori the equivalence relation looks weaker than this, but the restriction maps of $\mathscr{L}$ are injective due to integrality of $X$ and $\mathscr{L}$ being finite locally free, so the apparently stronger relation of equality on intersections holds). Now assume, in addition, that $X$ is Noetherian and regular at all codimension-$1$ points, that is, the stalks of $\mathscr{O}_X$ at points of codimension $1$ are discrete valuation rings. Then we can associate to any non-zero rational section $s$ as above a Weil divisor $\mathrm{div}^\mathscr{L}(s)$ by looking at the image of $s$ under trivializations of $\mathscr{L}$ in neighborhoods of each codimension -$1$ point and taking valuations of these images. So one can write $\mathrm{div}^\mathscr{L}(s)=\sum_x\mathrm{ord}_x(s)\cdot Z_x$, where the sum is over the points of codimension $1$ and $Z_x=\overline{\{x\}}$ is the corresponding integral closed subscheme of codimension $1$ (i.e. prime Weil divisor). (The Noetherianity ensures that $\mathrm{ord}_x(s)=0$ for all but finitely many $x$ in the sum.)
Suppose now that $s$ and $t$ are non-zero rational sections of $\mathscr{L}$. Then because $\mathscr{L}_\eta$ is $1$-dimensional over $K$, there is a non-zero element of $K$, i.e. a rational function $f$ on $X$ (which is incidentally by our definition a rational section of the trivial sheaf $\mathscr{O}_X$) for which $s=ft$ (the equality taking place in $\mathscr{L}_\eta$). One can verify from the definition that for each codimension-$1$ point $x$, $\mathrm{ord}_x(s)=\mathrm{ord}_x(ft)=\mathrm{ord}_x(f)+\mathrm{ord}_x(t)$. This means that for the associated Weil divisors we have $\mathrm{div}^\mathscr{L}(s)=\mathrm{div}(f)+\mathrm{div}^\mathscr{L}(t)$, where $\mathrm{div}(f)$ is the usual principal divisor associated to a non-zero element of $K$, and coincides with $\mathrm{div}^{\mathscr{O}_X}(f)$ as we have defined it. At any rate, we find that any two non-zero rational sections of $\mathscr{L}$ give rise to linearly equivalent Weil divisors. Thus to every invertible $\mathscr{O}_X$-module we have a well-defined associated Weil divisor class.
This can be applied in particular to a smooth projective curve $X$ over $k$ and the invertible $\mathscr{O}_X$-module $\Omega_{X/k}^1$, recovering the equality you ask about, and giving a well-defined canonical
divisor class. Note the rational differentials have to be non-zero. This makes sense because for such an $X$, one can define an invertible $\mathscr{O}_X$-module $\mathscr{O}_X(\mathrm{div}^{\Omega_{X/k}^1}(\omega))$ for any non-zero rational differential $\omega$, and this will be isomorphic to the invertible sheaf from which $\omega$ came, namely $\Omega_{X/k}^1$, the canonical sheaf of $X$. |
On Monday, Celestalon kicked off the official Alpha Theorycrafting season by posting a Theorycrafting Discussion thread on the forums. And he was kind enough to toss a meaty chunk of information our way about
Resolve, the replacement for Vengeance.
Resolve: Increases your healing and absorption done to yourself, based on Stamina and damage taken (before avoidance and mitigation) in the last 10 sec.
In today’s post, I want to go over the mathy details about how Resolve works, how it differs from Vengeance, and how it may (or may not) fix some of the problems we’ve discussed in previous blog posts.
Mathemagic
Celestalon broke the formula up into two components: one from stamina and one from damage taken. But for completeness, I’m going to bolt them together into one formula for resolve $R$:
$$ R =\frac{\rm Stamina}{250~\alpha} + 0.25\sum_i \frac{D_i}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t_i )}{10} \right )$$
where $D_i$ is an individual damage event that occurred $\Delta t_i$ seconds ago, and $\alpha$ is a level-dependent constant, with $\alpha(100)=261$. The sum is carried out over all damaging events that have happened in the last 10 seconds.
The first term in the equation is the stamina-based contribution, which is always active, even when out of combat. There’s a helpful buff in-game to alert you to this:
My premade character has 1294 character sheet stamina, which after dividing by 250 and $\alpha(90)=67$, gives me 0.07725, or about 7.725% Resolve. It’s not clear at this point whether the tooltip is misleadingly rounding down to 7% (i.e. using floor instead of round) or whether Resolve is only affected by the stamina from gear. The Alpha servers went down as I was attempting to test this, so we’ll have to revisit it later. We’ve already been told that this will update dynamically with stamina buffs, so having Power Word: Fortitude buffed on you mid-combat will raise your Resolve.
Once you’re in combat and taking damage, the second term makes a contribution:
I’ve left this term in roughly the form Celestalon gave, even though it can obviously be simplified considerably by combining all of the constants, because this form does a better job of illustrating the behavior of the mechanic. Let’s ignore the sum for now, and just consider an isolated damage event that does $D$ damage:
$$0.25\times\frac{D}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t )}{10} \right )$$
The 0.25 just moderates the amount of Resolve you get from damaging attacks. It’s a constant multiplicative factor that they will likely tweak to achieve the desired balance between baseline (stamina-based) Resolve and dynamic (damage-based) Resolve.
The factor of $D/{\rm MaxHealth}$ means that we’re normalizing the damage by our max health. So if we have 1000 health and take an attack that deals 1000 damage (remember, this is before mitigation), this term gives us a factor of 1. Avoided auto-attacks also count here, though instead of performing an actual damage roll the game just uses the mean value of the boss’s auto-attack damage. Again, nothing particularly complicated here, it just makes Resolve depend on the percentage of your health the attack would have removed rather than the raw damage amount. Also note that we’ve been told that dynamic health effects from temporary multipliers (e.g. Last Stand) aren’t included here, so we’re not punished for using temporary health-increasing cooldowns.
The term in parentheses is the most important part, though. In the instant the attack lands, $\Delta t=0$ and the term in parentheses evaluates to $2(10-0)/10 = 2.$ So that attack dealing 1000 damage to our 1000-health tank would give $0.25\times 1 \times 2 = 0.5,$ or 50% Resolve.
However, one second later, $\Delta t = 1$, so the term in parentheses is only $2(10-1)/10 = 1.8$, and the amount of resolve it grants is reduced to 45%. The amount of Resolve granted continues to linearly decrease as time passes, and by the time ten seconds have elapsed it’s reduced to zero. Each attack is treated independently, so to get our total Resolve from all damage taken we just have to add up the Resolve granted by every attack we’ve taken, hence the sum in my equation.
You may note that the time-average of the term in parentheses is 1, which is how we get the advertised “averages to ~Damage/MaxHealth” that Celestalon mentioned in the post. In that regard, he’s specifically referring to just the part within the sum, not the constant factor of 0.25 outside of it. So in total, your average Resolve contribution from damage is 25% of Damage/MaxHealth.
Comparing to Vengeance
Mathematically speaking, there’s a world of difference between Resolve and Vengeance. First and foremost is the part we already knew: Resolve doesn’t grant any offensive benefit. We’ve talked about that a lot before, though, so it’s not territory worth re-treading.
Even in the defensive component though, there are major differences. Vengeance’s difference equation, if solved analytically, gives solutions that are exponentials. In other words, provided you were continuously taking damage (such that it didn’t fall off entirely), Vengeance would decay and adjust to your new damage intake rather smoothly. It also meant that damage taken at the very beginning of an encounter was still contributing some amount of Vengeance at the very end, again, assuming there was no interruption. And since it was only recalculated on a damage event, you could play some tricks with it, like taking a giant attack that gave you millions of Vengeance and then riding that wave for 20 seconds while your co-tank takes the boss.
Resolve does away with all of that. It flat-out says “look, the only thing that matters is the last 10 seconds.” The calculation doesn’t rely on a difference equation, meaning that when recalculating, it doesn’t care what your Resolve was at any time previously. And it forces a recalculation at fixed intervals, not just when you take damage. As a result, it’s much harder to game than Vengeance was.
Celestalon’s post also outlines a few other significant differences:
No more ramp-up mechanism No taunt-transfer mechanism Resolve persists through shapeshifts Resolve only affects self-healing and self-absorbs
The lack of ramp-up and taunt-transfer mechanisms may at first seem like a problem. But in practice, I don’t think we’ll miss either of them. Both of these effects served offensive (i.e. threat) and defensive purposes, and it’s pretty clear that the offensive purposes are made irrelevant by definition here since Resolve won’t affect DPS/threat. The defensive purpose they served was to make sure you had
some Vengeance to counter the boss’s first few hits, since Vengeance had a relatively slow ramp-up time but the boss’s attacks did not.
However, Resolve ramps up a
lot faster than Vengeance does. Again, this is in part thanks to the fact that it isn’t governed by a difference equation. The other part is because it only cares about the last ten seconds.
To give you a visual representation of that, here’s a plot showing both Vengeance and Resolve for a player being attacked by a boss. The tank has 100 health and the boss swings for 30 raw damage every 1.5 seconds. Vengeance is shown in arbitrary units here since we’re not interested in the exact magnitude of the effect, just in its dynamic properties. I’ve also ignored the baseline (stamina-based) contribution to Resolve for the same reason.
As a final note, while the blog post says that Resolve is recalculated every second, it seemed like it was updating closer to every half-second when I fooled with it on alpha, so these plots use 0.5-second update intervals. Changing to 1-second intervals doesn’t significantly change the results (they just look a little more fragmented).
The plot very clearly shows the 50% ramp-up mechanism and slow decay-like behavior of Vengeance. Note that while the ramp-up mechanism gets you to 50% of Vengeance’s overall value at the first hit (at t=2.5 seconds), Resolve hits this mark as soon as the second hit lands (at 4.0 seconds) despite not having
any ramp-up mechanism.
Resolve also hits its steady-state value much more quickly than Vengeance does. By definition, Resolve gets there after about 10 seconds of combat (t=12.5 seconds). But with Vengeance, it takes upwards of 30-40 seconds to even approach the steady-state value thanks to the decay effect (again, a result of the difference equation used to calculate Vengeance). Since most fights involve tank swaps more frequently than this, it meant that you were consistently getting stronger the longer you tanked a boss. This in turn helped encourage the sort of “solo-tank things that should not be solo-tanked” behavior we saw in Mists.
This plot assumes a boss who does exactly 30 damage per swing, but in real encounters the boss’s damage varies. Both Vengeance and Resolve adapt to mimic that change in the tank’s damage intake, but as you could guess, Resolve adapts much more quickly. If we allow the boss to hit for a random amount between 20 and 40 damage:
You can certainly see the similar changes in both curves, but Resolve reacts quickly to each change while Vengeance changes rather slowly.
One thing you’ve probably noticed by now is that the Resolve plot looks very jagged (in physics, we might call this a “sawtooth wave”). This happens because of the linear decay built into the formula. It peaks in the instant you take the attack – or more accurately, in the instant that Resolve is recalculated after that attack. But then every time it’s recalculated it linearly decreases by a fixed percent. If the boss swings in 1.5-second intervals, then Resolve will zig-zag between its max value and 85% of its max value in the manner shown.
The more frequently the boss attacks, the smoother that zig-zag becomes; conversely, a boss with a long swing timer will cause a larger variation in Resolve. This is apparent if we adjust the boss’s swing timer in either direction:
It’s worth noting that every plot here has a new randomly-generated sequence of attacks, so don’t be surprised that the plots don’t have the same profile as the original. The key difference is the size of the zig-zag on the Resolve curve.
I’ve also run simulations where the boss’ base damage is 50 rather than 30, but apart from the y-axis having large numbers there’s no real difference:
Note that even a raw damage of 50% is pretty conservative for a boss – heroic bosses in Siege have frequently had raw damages that were larger than the player’s health. But it’s not clear if that will still be the case with the new tanking and healing paradigm that’s been unveiled for Warlords.
If we make the assumption that raw damage will be lower, then these rough estimates give us an idea of how large an effect Resolve will be. If we guess at a 5%-10% baseline value from stamina, these plots suggest that Resolve will end up being anywhere from a 50% to 200% modifier on our healing. In other words, it has the potential to double or triple our healing output with the current tuning numbers. Of course, it’s anyone’s guess as to whether those numbers are even remotely close to what they’ll end up being by the end of beta.
Is It Fixed Yet?
If you look back over our old blog posts, the vast majority of our criticisms of Vengeance had to do with its tie-in to damage output. Those have obviously been addressed, which leaves me worrying that I’ll have nothing to rant about for the next two or three years.
But regarding everything else, I think Resolve stands a fair chance of addressing our concerns. One of the major issues with Vengeance was the sheer magnitude of the effect – you could go from having 50k AP to 600k AP on certain bosses, which meant your abilities got up to 10x more effective. Even though that’s an extreme case, I regularly noted having over 300k AP during progression bosses, a factor of around 6x improvement. Resolve looks like it’ll tamp down on that some. Reasonable bosses are unlikely to grant a multiplier larger than 2x, which will be easier to balance around.
It hasn’t been mentioned specifically in Celestalon’s post, but I think it’s a reasonable guess that they will continue to disable Resolve gains from damage that could be avoided through better play (i.e. intentionally “standing in the bad”). If so, there will be little (if any) incentive to take excess damage to get more Resolve. Our sheer AP scaling on certain effects created situations where this was a net survivability gain with Vengeance, but the lower multiplier should make that impossible with Resolve.
While I still don’t think it needs to affect anything other than active mitigation abilities, the fact that it’s a multiplier affecting everything equally rather than a flat AP boost should make it easier to keep talents with different AP coefficients balanced (Eternal Flame and Sacred Shield, specifically). And we already know that Eternal Flame is losing its Bastion of Glory interaction, another change which will facilitate making both talents acceptable choices.
All in all, I think it’s a really good system, if slightly less transparent. It’s too soon to tell whether we’ll see any unexpected problems, of course, but the mechanic doesn’t have any glaring issues that stand out upon first examination (unlike Vengeance). I still have a few lingering concerns about steady-state threat stability between tanks (ironically, due to the
removal of Vengeance), but that is the sort of thing which will become apparent fairly quickly during beta testing, and at any rate shouldn’t reflect on the performance of Resolve. |
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
Let $A, B, C$ be $n\times n$ invertible matrices. When you simplify the expression\[C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2,\]which matrix do you get?(a) $A$(b) $C^{-1}A^{-1}BC^{-1}AC^2$(c) $B$(d) $C^2$(e) $C^{-1}BC$(f) $C$
Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.Let\[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\]where\begin{align*}p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3.\end{align*}
(a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.
(b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be a vector space and $B$ be a basis for $V$.Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$.Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form\[\begin{bmatrix}1 & 0 & 2 & 1 & 0 \\0 & 1 & 3 & 0 & 1 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0\end{bmatrix}.\]
(a) What is the dimension of $V$?
(b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.Let\[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix}a & b\\c& -a\end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\]
(a) Show that $W$ is a subspace of $V$.
(b) Find a basis of $W$.
(c) Find the dimension of $W$.
(The Ohio State University, Linear Algebra Midterm)
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 2 and contains Problem 4, 5, and 6.Check out Part 1 and Part 3 for the rest of the exam problems.
Problem 4. Let\[\mathbf{a}_1=\begin{bmatrix}1 \\2 \\3\end{bmatrix}, \mathbf{a}_2=\begin{bmatrix}2 \\-1 \\4\end{bmatrix}, \mathbf{b}=\begin{bmatrix}0 \\a \\2\end{bmatrix}.\]
Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.
Problem 5.Find the inverse matrix of\[A=\begin{bmatrix}0 & 0 & 2 & 0 \\0 &1 & 0 & 0 \\1 & 0 & 0 & 0 \\1 & 0 & 0 & 1\end{bmatrix}\]if it exists. If you think there is no inverse matrix of $A$, then give a reason.
Problem 6.Consider the system of linear equations\begin{align*}3x_1+2x_2&=1\\5x_1+3x_2&=2.\end{align*}
(a) Find the coefficient matrix $A$ of the system.
(b) Find the inverse matrix of the coefficient matrix $A$.
(c) Using the inverse matrix of $A$, find the solution of the system.
(Linear Algebra Midterm Exam 1, the Ohio State University) |
Multiple solutions for a fractional nonlinear Schrödinger equation with local potential
1.
School of Science, Jiangxi University of Science and Technology, Ganzhou, Jiangxi 341000, China
2.
School of Mathematical Sciences, Dalian University of Technology, Dalian, 116024, China
$\left\{ \begin{align} &{{\varepsilon }^{2\alpha }}{{\left( -\Delta \right)}^{a}}u+V\left( x \right)u=f\left( u \right),\ \ x\in {{\mathbb{R}}^{N}}, \\ &u\in {{H}^{a}}\left( {{\mathbb{R}}^{N}} \right),u>0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\in {{\mathbb{R}}^{N}}, \\ \end{align} \right.$
$0<α<1$
$N>2α$
$\varepsilon>0$
$V$
$f$
$\text{cat}_{M_{δ}}(M)$ Keywords:Penalization techniques, Ljusternik-Schnirelmann theory, fractional Schrödinger equation, multiple solutions, single spike solutions. Mathematics Subject Classification:Primary: 35A15, 35J60; Secondary: 35S15. Citation:Wulong Liu, Guowei Dai. Multiple solutions for a fractional nonlinear Schrödinger equation with local potential. Communications on Pure & Applied Analysis, 2017, 16 (6) : 2105-2123. doi: 10.3934/cpaa.2017104
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V. Benci and G. Cerami,
The effect of the domain topology on the number of positive solutions of nonlinear elliptic problems,
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C. Bucur and E. Valdinoci,
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Nonlinear equations for fractional Laplacians Ⅰ: Regularity, maximum principles, and Hamiltonian estimates,
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X. Cabré and Y. Sire,
Nonlinear equations for fractional Laplacians Ⅱ: Existence, uniqueness and qualitative properties of solutions,
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S. Cingolani and M. Lazzo,
Multiple positive solutions to nonlinear Schrödinger equations with competing potential functions,
[15]
J. Dávila, M. del Pino, S. Dipierro and E. Valdinoci,
Concentration phenomena for the nonlocal Schrödinger equation with Dirichlet datum,
[16]
J. Dávila, M. Del Pino and J. C. Wei,
Concentrating standing waves for the fractional nonlinear Schrödinger equation,
[17]
M. Del Pino and P. Felmer,
Local mountain passes for semilinear elliptic problems in unbounded domains,
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S. Dipierro, M. Medina, I. Peral and E. Valdinoci, Bifurcation results for a fractional elliptic equation with critical exponent in $\mathbb{R}^n$
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S. Dipierro, M. Medina and E. Valdinoci, Fractional elliptic problems with critical growth in the whole of $\mathbb{R}^n$ in
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M. M. Fall and E. Valdinoci,
Uniqueness and nondegeneracy of positive solutions of $(Δ)^s u+u=u^p$ in $\mathbb{R}^N$ when $s$ is close to $1$,
[22]
M. M. Fall, F. Mahmoudi and E. Valdinoci,
Ground states and concentration phenomena for the fractional Schrödinger equation,
[23]
P. Felmer, A. Quaas and J. Tan,
Positive solutions of the nonlinear Schrödinger equation with the fractional Laplacian,
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G. M. Figueiredo and G. Siciliano, A multiplicity result via Ljusternick-Schnirelmann category and Morse theory for a fractional Schrdinger equation in $\mathbb{R}^N$
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W. Liu, Multiple solutions for a fractional nonlinear Schrödinger equation with a general nonlinearity, prepared.Google Scholar
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Lewy-Stampacchia type estimates for variational inequalities driven by (non)local operators,
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X. Shang and J. Zhang,
Concentrating solutions of nonlinear fractional Schrödinger equation with potentials,
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[1]
C. O. Alves, G. M. Figueiredo and M. F. Furtado,
Multiple solutions for a nonlinear Schrödinger equation with magnetic fields,
[2] [3]
B. Barrios, E. Colorado, R. Servadei and F. Soria,
A critical fractional equation with concave-convex power nonlinearities,
[4]
V. Benci and G. Cerami,
The effect of the domain topology on the number of positive solutions of nonlinear elliptic problems,
[5] [6]
C. Bucur and E. Valdinoci,
[7] [8]
X. Cabré and Y. Sire,
Nonlinear equations for fractional Laplacians Ⅰ: Regularity, maximum principles, and Hamiltonian estimates,
[9]
X. Cabré and Y. Sire,
Nonlinear equations for fractional Laplacians Ⅱ: Existence, uniqueness and qualitative properties of solutions,
[10] [11] [12] [13] [14]
S. Cingolani and M. Lazzo,
Multiple positive solutions to nonlinear Schrödinger equations with competing potential functions,
[15]
J. Dávila, M. del Pino, S. Dipierro and E. Valdinoci,
Concentration phenomena for the nonlocal Schrödinger equation with Dirichlet datum,
[16]
J. Dávila, M. Del Pino and J. C. Wei,
Concentrating standing waves for the fractional nonlinear Schrödinger equation,
[17]
M. Del Pino and P. Felmer,
Local mountain passes for semilinear elliptic problems in unbounded domains,
[18] [19]
S. Dipierro, M. Medina, I. Peral and E. Valdinoci, Bifurcation results for a fractional elliptic equation with critical exponent in $\mathbb{R}^n$
[20]
S. Dipierro, M. Medina and E. Valdinoci, Fractional elliptic problems with critical growth in the whole of $\mathbb{R}^n$ in
[21]
M. M. Fall and E. Valdinoci,
Uniqueness and nondegeneracy of positive solutions of $(Δ)^s u+u=u^p$ in $\mathbb{R}^N$ when $s$ is close to $1$,
[22]
M. M. Fall, F. Mahmoudi and E. Valdinoci,
Ground states and concentration phenomena for the fractional Schrödinger equation,
[23]
P. Felmer, A. Quaas and J. Tan,
Positive solutions of the nonlinear Schrödinger equation with the fractional Laplacian,
[24]
G. M. Figueiredo and G. Siciliano, A multiplicity result via Ljusternick-Schnirelmann category and Morse theory for a fractional Schrdinger equation in $\mathbb{R}^N$
[25] [26] [27] [28] [29]
W. Liu, Multiple solutions for a fractional nonlinear Schrödinger equation with a general nonlinearity, prepared.Google Scholar
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R. Servadei and E. Valdinoci,
Lewy-Stampacchia type estimates for variational inequalities driven by (non)local operators,
[32] [33] [34] [35] [36] [37]
X. Shang and J. Zhang,
Concentrating solutions of nonlinear fractional Schrödinger equation with potentials,
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Walter Dambrosio, Duccio Papini.
Multiple homoclinic solutions for a one-dimensional Schrödinger equation.
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Shuangjie Peng, Huirong Pi.
Spike vector solutions for some coupled nonlinear Schrödinger equations.
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Xudong Shang, Jihui Zhang.
Multiplicity and concentration of positive solutions for fractional nonlinear Schrödinger equation.
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Weichung Wang, Tsung-Fang Wu, Chien-Hsiang Liu.
On the multiple spike solutions for singularly perturbed elliptic
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Weiming Liu, Lu Gan.
Multi-bump positive solutions of a fractional nonlinear Schrödinger
equation in $\mathbb{R}^N$.
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On the blow-up solutions for the fractional nonlinear Schrödinger equation with combined power-type nonlinearities.
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César E. Torres Ledesma.
Existence and concentration of solutions for a non-linear fractional Schrödinger equation with steep potential well.
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Miaomiao Niu, Zhongwei Tang.
Least energy solutions for nonlinear Schrödinger equation involving the fractional Laplacian and critical growth.
[12]
Van Duong Dinh.
On blow-up solutions to the focusing mass-critical nonlinear fractional Schrödinger equation.
[13]
Rossella Bartolo, Anna Maria Candela, Addolorata Salvatore.
Infinitely many solutions
for a perturbed Schrödinger equation.
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Kaimin Teng, Xiumei He.
Ground state solutions for fractional Schrödinger equations with critical Sobolev exponent.
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Ran Zhuo, Yan Li.
Nonexistence and symmetry of solutions for Schrödinger systems involving fractional Laplacian.
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John Baxley, Mary E. Cunningham, M. Kathryn McKinnon.
Higher order boundary value problems with multiple solutions: examples and techniques.
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Wentao Huang, Jianlin Xiang.
Soliton solutions for a quasilinear Schrödinger equation with critical exponent.
[20]
Kun Cheng, Yinbin Deng.
Nodal solutions for a generalized quasilinear Schrödinger equation with critical exponents.
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The electric scalar potential field \(V({\bf r})\), defined in Section 5.12, is useful for a number of reasons including the ability to conveniently compute potential differences (i.e., \(V_{21} = V({\bf r}_2) - V({\bf r}_1)\)) and the ability to conveniently determine the electric field by taking the gradient (i.e., \({\bf E}=-\nabla V\)). One way to obtain \(V({\bf r})\) is by integration over the source charge distribution, as described in Section 5.13. This method is awkward in the presence of material interfaces, which impose boundary conditions on the solutions that must be satisfied simultaneously. For example, the electric potential on a perfectly conducting surface is constant
1 – a constraint which is not taken into account in any of the expressions in Section 5.13 (this fact is probably already known to the reader from past study of elementary circuit theory; however, this is established in the context of electromagnetics in Section 5.19.)
In this section, we develop an alternative approach to calculating \(V({\bf r})\) that accommodates these boundary conditions, and thereby facilitates the analysis of the scalar potential field in the vicinity of structures and spatially-varying material properties. This alternative approach is based on
Poisson’s Equation, which we now derive.
We begin with the differential form of Gauss’ Law (Section 5.7):
\[\nabla \cdot {\bf D} = \rho_v\]
Using the relationship \({\bf D}=\epsilon{\bf E}\) (and keeping in mind our standard assumptions about material properties, summarized in Section 2.8) we obtain
\[\nabla \cdot {\bf E} = \frac{\rho_v}{\epsilon}\]
Next, we apply the relationship (Section 5.14):
\[{\bf E} = -\nabla V\] yielding \[\nabla \cdot \nabla V = - \frac{\rho_v}{\epsilon}\]
This is Poisson’s Equation, but it is not in the form in which it is commonly employed. To obtain the alternative form, consider the operator \(\nabla \cdot \nabla\) in Cartesian coordinates:
\[\begin{aligned} \nabla \cdot \nabla & = \left[ \frac { \partial } { \partial x } \hat { \mathbf { x } } + \frac { \partial } { \partial y } \hat { \mathbf { y } } + \frac { \partial } { \partial z } \hat { \mathbf { z } } \right] \cdot \left[ \frac { \partial } { \partial x } \hat { \mathbf { x } } + \frac { \partial } { \partial y } \hat { \mathbf { y } } + \frac { \partial } { \partial z } \hat { \mathbf { z } } \right] \\ & = \frac { \partial ^ { 2 } } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } } { \partial y ^ { 2 } } + \frac { \partial ^ { 2 } } { \partial z ^ { 2 } } \\ &= \nabla ^ { 2 } \end{aligned}\]
i.e., the operator \(\nabla \cdot \nabla\) is identically the Laplacian operator \(\nabla^2\) (Section 4.10). Furthermore, this is true regardless of the coordinate system employed. Thus, we obtain the following form of Poisson’s Equation:
\[\boxed{ \nabla^2 V = - \frac{\rho_v}{\epsilon} } \label{m0067_ePoisson}\]
Poisson’s Equation (Equation \ref{m0067_ePoisson}) states that the Laplacian of the electric potential field is equal to the volume charge density divided by the permittivity, with a change of sign.
Note that Poisson’s Equation is a partial differential equation, and therefore can be solved using well-known techniques already established for such equations. In fact, Poisson’s Equation is an
inhomogeneous differential equation, with the inhomogeneous part \(-\rho_v/\epsilon\) representing the source of the field. In the presence of material structure, we identify the relevant boundary conditions at the interfaces between materials, and the task of finding \(V({\bf r})\) is reduced to the purely mathematical task of solving the associated boundary value problem (see “Additional Reading” at the end of this section). This approach is particularly effective when one of the materials is a perfect conductor or can be modeled as such a material. This is because – as noted at the beginning of this section – the electric potential at all points on the surface of a perfect conductor must be equal, resulting in a particularly simple boundary condition.
In many other applications, the charge responsible for the electric field lies outside the domain of the problem; i.e., we have non-zero electric field (hence, potentially non-zero electric potential) in a region that is free of charge. In this case, Poisson’s Equation simplifies to
Laplace’s Equation:
\[\boxed{ \nabla^2 V = 0 ~~\mbox{(source-free region)} } \label{m0067_eLaplace}\]
Laplace’s Equation (Equation \ref{m0067_eLaplace}) states that the Laplacian of the electric potential field is zero in a source-free region.
Like Poisson’s Equation, Laplace’s Equation, combined with the relevant boundary conditions, can be used to solve for \(V({\bf r})\), but only in regions that contain no charge.
Contributors
Ellingson, Steven W. (2018) Electromagnetics, Vol. 1. Blacksburg, VA: VT Publishing. https://doi.org/10.21061/electromagnetics-vol-1 Licensed with CC BY-SA 4.0 https://creativecommons.org/licenses/by-sa/4.0. Report adoption of this book here. If you are a professor reviewing, adopting, or adapting this textbook please help us understand a little more about your use by filling out this form. |
This is tricky (Edit: And so tricky that this answer -six upvotes- is wrong. )
WARNING: This answer is
wrong. Please don't upvote (or accept!) it. I've decided against deleting it because it still might be helpful.
The issue is this: the traditional-elementrary $\epsilon,\delta$ definition of the limit of a function of a real variable around $x=a$ involves a "deleted neighbourhood" $ 0 < | x − a | < \delta$, i.e., a punctured open interval: we must find some $\delta$ such that the function evaluated inside that neighbourhood falls near the limit. The question is: do we require that the function is defined in all that (real) interval? Actually
we don't (that was my mistake). (If that were the case, then a function defined only on the rationals would not have any limits.) All that we require is the condition is fullfiled for all points of the domain that are inside that neighbourhood. (Actually, if the function is not defined in some deleted neighbourhood, we need to assert at least that the $x=a$ is a limit point of the domain. Without this, $\lim_{x\to 0} \sqrt{x-1}=3$ would be vacuously true)
(WRONG answer begins)
First, notice that the function is defined for all reals except for $x=0$, and for the points where the denominator is zero: $$\sin(1/x)=0 \iff 1/x= k\pi \iff x = \frac{1}{k\pi}$$
for any integer $k$.
Outside these prohibited points, the function equals $1$.
Now, as you correctly guessed, that the function is not defined at $x=0$ does not matter for computing the limit.
But what matters is that the other prohibited points get arbitrarily close to $x=0$, hence you cannot find any neighborhood around $x=0$ where the funcion is defined. Then, the limit does not exist. (
WRONG)
(The graph of this function would consist of an horizontal line ($y=1$) with "holes" at $x=0$, and $x=1/k\pi$ . These holes get more and more concentrated around $x=0$... You cannot trust a computer generated graph for this kind of function.) |
Let $$S_a(N)=\sum_{n\le N}\frac{\varphi(an)}{n^2}.$$ The usual machinery gives an asymptotic formula $$S_a(N)=\frac1{\zeta(2)}\cdot\frac{a^2}{\varphi_+(a)}\log N+C(a)+O(N^{-1+\varepsilon}a^{1+\varepsilon}),$$ where $C(a)$ some complicated function and $$\varphi_+(a)=a\prod_{p\mid a}\left(1+\frac1p\right).$$ Is it possible to give a reference on this asymptotic formula? (My proof is rather long.)
I have found a proof of more general formula in the book
Postnikov, A. G. Introduction to analytic number theory American Mathematical Society, 1988, (section 4.2). This proof is simple but it has a small mistake inside. (For arithmetic progression starting from $0$ this mistake vanishes.)
Please give more references if you know ones. |
Statistics - (Linear spline|Piecewise linear function) 1 - About linear in the left and the right. forced to be continuous at the knot. 2 - Articles Related 3 - Definition / Representation
This model can be represented as:
<MATH> y_i = \beta_0 + \beta_1.b_1(x_i) + \beta_2.b_2(x_i) + \dots + \beta_{K+3}b_{K+3}(x_i) + \epsilon_i </MATH>
where the <math>b_k</math> are basis functions and are:
the variable itself. One of these basis functions is just the variable itself
<MATH> \begin{array}{rll} b_1(x_i) & = & x_i \\ \end{array} </MATH>
and additional variables that are a collection of truncated basis transformation functions at each of the knots. They are called truncated power functions, but this one is to power 1.
<MATH> \begin{array}{rll} b_{k+1}(xi) & = & (xi - \xi_k)_+ \text{ where } k = 1, \dots, K \\ \text{where the ()+ means positive part. i.e. } (xi - \xi_k) & = & \left\{ \begin{matrix} xi - \xi_k & \text{ if } x_i > \xi_k & \\ 0 & \text{Otherwise} & \text{It's not allowed to go negative.} \end{matrix} \right. \end{array} </MATH>
4 - Plot
There is one knot in these pictures, but we can put down multiple knots.
When we fit a linear model with a global linear function plus one of the basis function and their respective coefficient, we get a function that's allowed to change its slope at the knot. |
Tagged: invertible matrix Problem 583
Consider the $2\times 2$ complex matrix
\[A=\begin{bmatrix} a & b-a\\ 0& b \end{bmatrix}.\] (a) Find the eigenvalues of $A$. (b) For each eigenvalue of $A$, determine the eigenvectors. (c) Diagonalize the matrix $A$.
Add to solve later
(d) Using the result of the diagonalization, compute and simplify $A^k$ for each positive integer $k$. Problem 582
A square matrix $A$ is called
nilpotent if some power of $A$ is the zero matrix. Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.
Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$.
Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample. Problem 562
An $n\times n$ matrix $A$ is called
nonsingular if the only vector $\mathbf{x}\in \R^n$ satisfying the equation $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$. Using the definition of a nonsingular matrix, prove the following statements. (a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular. (b) Let $A$ and $B$ be $n\times n$ matrices and suppose that the product $AB$ is nonsingular. Then: The matrix $B$ is nonsingular. The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.) Problem 552
For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix.
Add to solve later
(a) $A=\begin{bmatrix} 1 & 3 & -2 \\ 2 &3 &0 \\ 0 & 1 & -1 \end{bmatrix}$ (b) $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$. Problem 548
An $n\times n$ matrix $A$ is said to be
invertible if there exists an $n\times n$ matrix $B$ such that $AB=I$, and $BA=I$,
where $I$ is the $n\times n$ identity matrix.
If such a matrix $B$ exists, then it is known to be unique and called the
inverse matrix of $A$, denoted by $A^{-1}$.
In this problem, we prove that if $B$ satisfies the first condition, then it automatically satisfies the second condition.
So if we know $AB=I$, then we can conclude that $B=A^{-1}$.
Let $A$ and $B$ be $n\times n$ matrices.
Suppose that we have $AB=I$, where $I$ is the $n \times n$ identity matrix.
Prove that $BA=I$, and hence $A^{-1}=B$.Add to solve later
Problem 546
Let $A$ be an $n\times n$ matrix.
The $(i, j)$
cofactor $C_{ij}$ of $A$ is defined to be \[C_{ij}=(-1)^{ij}\det(M_{ij}),\] where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column.
Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$.
The matrix $\Adj(A)$ is called the adjoint matrix of $A$.
When $A$ is invertible, then its inverse can be obtained by the formula
For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula.
(a) $A=\begin{bmatrix} 1 & 5 & 2 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}$. (b) $B=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &4 \\ 3 & 0 & 1 \end{bmatrix}$. Problem 506
Let $A$ be an $n\times n$ invertible matrix. Then prove the transpose $A^{\trans}$ is also invertible and that the inverse matrix of the transpose $A^{\trans}$ is the transpose of the inverse matrix $A^{-1}$.
Namely, show that \[(A^{\trans})^{-1}=(A^{-1})^{\trans}.\] Problem 500
10 questions about nonsingular matrices, invertible matrices, and linearly independent vectors.
The quiz is designed to test your understanding of the basic properties of these topics.
You can take the quiz as many times as you like.
The solutions will be given after completing all the 10 problems.
Click the View question button to see the solutions. Problem 452
Let $A$ be an $n\times n$ complex matrix.
Let $S$ be an invertible matrix. (a) If $SAS^{-1}=\lambda A$ for some complex number $\lambda$, then prove that either $\lambda^n=1$ or $A$ is a singular matrix. (b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$.
Add to solve later
(c) Suppose that all the eigenvalues of $A$ are integers and $\det(A) > 0$. If $n$ is odd and $SAS^{-1}=A^{-1}$, then prove that $1$ is an eigenvalue of $A$. Problem 438
Determine whether each of the following statements is True or False.
(a) If $A$ and $B$ are $n \times n$ matrices, and $P$ is an invertible $n \times n$ matrix such that $A=PBP^{-1}$, then $\det(A)=\det(B)$. (b) If the characteristic polynomial of an $n \times n$ matrix $A$ is \[p(\lambda)=(\lambda-1)^n+2,\] then $A$ is invertible. (c) If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible. (d) If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$. (e) If $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ with corresponding eigenvalue $\lambda_1$, and if $\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_2$, then $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$.
(
Stanford University, Linear Algebra Exam Problem) Read solution |
Suppose $n \neq 6,p,2^n$ for $n \geq 1$. Let $D = \{k \leq n, (k,n) = 1\}$. Suppose that $D = (1 , 1 + d,..., k_0 + rd)$ for some $r,d$.
Clearly, $1+d$ is the smallest non-prime factor of $n$. It follows that if $d = 2$ then $n$ has to be a power of two. Similarly, $d=1$ would imply that $n$ has to be a prime. For the number $6$, we see that there are only two terms, so there is not really an arithmetic progression.
Now, we have $d = p-1$ where $p$ is the smallest prime non-factor of $n$. Let $q$ be the largest prime smaller than $p$. This means that $q$ divides $n$. Let us look at the sequence $1,p,2p-1,...,n-1$. How many terms does it contain? If $n -1 = rd+1$, then the sequence has $r+1$ terms, so we conclude it has $\frac{n-2}{p-1} + 1$ terms.
Claim : If $\frac{n-2}{p-1} + 1 > q$, then we are done.
Why is this true? Well, note that $1 , 1 + d , ..., n-1$ is a set of integers, whose size is greater than $q$. Let us make the size exactly $q$, by choosing only the numbers $1, 1 + d, ..., 1 + (q-1)d$.
I claim that each of these leave different remainders when divided by $q$, since if say $1 + fd$ and $1+gd$ leave the same remainder(where $0 \leq f,g < q$), then $q | (f-g)d$, but then $q | d$ or $q | f-g$. $q | f-g$ is not possible, since $|f-g| < q$.
The fact that $q|p-1$ is not possible stems from the fact that $q$ is the biggest prime smaller than $p$, and by Bertrand's postulate we see that $q > \frac {p-1}2$, since there is a prime strictly between $\frac{p-1}{2}$ and $p-1$, and $q$ must be greater than that prime, so cannot possibly divide $p-1$.
Therefore, by the pigeonhole principle, one of the $1+md$ is a multiple of $q$ (leaves the remainder $0$), a contradiction as then it wouldd have non-trivial gcd with $n$ (both are multiples of $q$), but is a member of $D$.
Claim : Indeed, $\frac{n-2}{p-1} + 1 > q$.
Proof : We will rearrange this, to get $n > (p-1)(q-1) + 2$. However, note that $n$ is a multiple of every prime number until $p$, so $n \geq \prod_{p' < p} p'$, where $p'$ are prime. Furthermore, note that $(p-1)(q-1) + 2 < pq$, so this really reduces to the wonderful result : $\prod_{p' < q} p' > p$.
We can see this by induction, roughly, since each time the right hand side can be multiplied by atmost $2$ (Bertrand's postulate!) while the right hand side is multiplied by a prime, which is larger than or equal to $2$, each time.
This proves the second claim. Therefore, for large enough $d = p-1$, indeed
no number exists whose set of relatively prime numbers are in arithmetic progression. I think the proof above (the induction, Bertrand's postulate etc.) starts working for small enough $d$ (I think $p=7$ would do). $p=3,5$ can be checked individually (it is actually quite clear: I leave it as an exercise).
Therefore, the only numbers with the stated property are prime, powers of two, and six.
EDIT : I would then continue to think as follows : if not geometric progression, then what more regularity can we demand from the co-prime set? I was thinking, possibly this condition : there exist a pair of consecutive numbers in this set i.e. for some $k$, both $k,k+1 \in D$. Can we assert , that for non-prime $n$, and non-powers of $2$, this would happen? |
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
Let $A, B, C$ be $n\times n$ invertible matrices. When you simplify the expression\[C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2,\]which matrix do you get?(a) $A$(b) $C^{-1}A^{-1}BC^{-1}AC^2$(c) $B$(d) $C^2$(e) $C^{-1}BC$(f) $C$
Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.Let\[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\]where\begin{align*}p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3.\end{align*}
(a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.
(b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be a vector space and $B$ be a basis for $V$.Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$.Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form\[\begin{bmatrix}1 & 0 & 2 & 1 & 0 \\0 & 1 & 3 & 0 & 1 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0\end{bmatrix}.\]
(a) What is the dimension of $V$?
(b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.Let\[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix}a & b\\c& -a\end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\]
(a) Show that $W$ is a subspace of $V$.
(b) Find a basis of $W$.
(c) Find the dimension of $W$.
(The Ohio State University, Linear Algebra Midterm)
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 2 and contains Problem 4, 5, and 6.Check out Part 1 and Part 3 for the rest of the exam problems.
Problem 4. Let\[\mathbf{a}_1=\begin{bmatrix}1 \\2 \\3\end{bmatrix}, \mathbf{a}_2=\begin{bmatrix}2 \\-1 \\4\end{bmatrix}, \mathbf{b}=\begin{bmatrix}0 \\a \\2\end{bmatrix}.\]
Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.
Problem 5.Find the inverse matrix of\[A=\begin{bmatrix}0 & 0 & 2 & 0 \\0 &1 & 0 & 0 \\1 & 0 & 0 & 0 \\1 & 0 & 0 & 1\end{bmatrix}\]if it exists. If you think there is no inverse matrix of $A$, then give a reason.
Problem 6.Consider the system of linear equations\begin{align*}3x_1+2x_2&=1\\5x_1+3x_2&=2.\end{align*}
(a) Find the coefficient matrix $A$ of the system.
(b) Find the inverse matrix of the coefficient matrix $A$.
(c) Using the inverse matrix of $A$, find the solution of the system.
(Linear Algebra Midterm Exam 1, the Ohio State University) |
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
(a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular?
(b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular?
(c) Let $A$ be a $4\times 4$ matrix and let\[\mathbf{v}=\begin{bmatrix}1 \\2 \\3 \\4\end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix}4 \\3 \\2 \\1\end{bmatrix}.\]Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular?
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.
Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
Prove that the matrix\[A=\begin{bmatrix}0 & 1\\-1& 0\end{bmatrix}\]is diagonalizable.Prove, however, that $A$ cannot be diagonalized by a real nonsingular matrix.That is, there is no real nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix.
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
An $n\times n$ matrix $A$ is called nonsingular if the only vector $\mathbf{x}\in \R^n$ satisfying the equation $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$.Using the definition of a nonsingular matrix, prove the following statements.
(a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular.
(b) Let $A$ and $B$ be $n\times n$ matrices and suppose that the product $AB$ is nonsingular. Then:
The matrix $B$ is nonsingular.
The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.)
Let $A$ be an $n\times (n-1)$ matrix and let $\mathbf{b}$ be an $(n-1)$-dimensional vector.Then the product $A\mathbf{b}$ is an $n$-dimensional vector.Set the $n\times n$ matrix $B=[A_1, A_2, \dots, A_{n-1}, A\mathbf{b}]$, where $A_i$ is the $i$-th column vector of $A$.
Prove that $B$ is a singular matrix for any choice of $\mathbf{b}$.
For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix. |
Addition is fast because CPU designers have put in the circuitry needed to make it fast. It does take significantly more gates than bitwise operations, but it is frequent enough that CPU designers have judged it to be worth it. See https://en.wikipedia.org/wiki/Adder_(electronics).Both can be made fast enough to execute within a single CPU cycle. They'...
I assume that the task is to compute $mul(10, a)= 10a$. You don't need to do multiplication. A single binary adder is enough since $$10a = 2^3a + 2a$$meaning you add one-time left-shifted $a$ to 3-time left-shifted $a$.For general multiplication $mul(x,y)$ please see this article.
This seems a very basic question to me, so excuse me if I lecture you a bit. The most important point for you to learn here is that a number is not its digit representation. A number is an abstract mathematical object, whereas its digit representation is a concrete thing, namely a sequence of symbols on a paper (or a sequence of bits in compute memory, or a ...
Short version: it doesn't know. There's no way to tell.If 1111 represents -7, then you have a sign-magnitude representation, where the first bit is the sign and the rest of the bits are the magnitude. In this case, arithmetic is somewhat complicated, since an unsigned add and a signed add use different logic. So you'd probably have a SADD and a UADD opcode,...
The best algorithm that is known is to express the factorial as a product of prime powers. One can quickly determine the primes as well as the right power for each prime using a sieve approach. Computing each power can be done efficiently using repeated squaring, and then the factors are multiplied together. This was described by Peter B. Borwein, On the ...
There are several aspects.The relative cost of a bitwise operation and an addition. A naive adder will have a gate-depth which depend linearly of the width of the word. There are alternative approaches, more costly in terms of gates, which reduce the depth (IIRC the depth then depend logarithmically of the width of the word). Others have given ...
To compute the exact mean (no confidence interval or estimate) of each exam, you must at least observe every student's exam score. This takes $\Omega(r)$ per exam. There are $c$ exams you must do this for, this problem should take at least $\Omega(c \cdot r)$ time.
CPUs operate in cycles. At each cycle, something happens. Usually, an instruction takes more cycles to execute, but multiple instructions are executed at the same time, in different states.For example, a simple processor might have 3 steps for each instruction: fetch, execute and store. At any time, 3 instructions are being processed: one is being fetched, ...
There is no way to represent all real numbers without errors if each number is to have a finite representation. There are uncountably many real numbers but only countably many finite strings of 1's and 0's that you could use to represent them with.
Your two algorithms are equivalent (at least for positive integers, what happens with negative integers in the imperative version depends on Java's semantics for % which I don't know by heart). In the recursive version, let $a_i$ and $b_i$ be the argument of the $i$th recursive call:$$\begin{gather*}a_{i+1} = b_i \\b_{i+1} = a_i \mathbin{\mathrm{mod}...
It all depends what you want to do.For example, what you show is a great way of representing rational numbers. But it still can't represent something like $\pi$ or $e$ perfectly.In fact, many languages such as Haskell and Scheme have built in support for rational numbers, storing them in the form $\frac{a}{b}$ where $a,b$ are integers.The main reason ...
In typical floating point implementations, the result of a single operation is produced as if the operation was performed with infinite precision, and then rounded to the nearest floating-point number.Compare $a+b$ and $b+a$: The result of each operation performed with infinite precision is the same, therefore these identical infinite precision results ...
In 1's complement you just invert all the bits.Consider these 2 examples (assuming 8 bits):$4 = 00000100$, so $-4= 11111011$$0 = 00000000$, so $-0=11111111$.So you have 2 ways to represent the number 0In 2's complement you add 1 to the 1's complement representation of the negative number$-4$ that in 1's complement was $11111011$ becomes $...
Keep in mind that the factorial function grows so fast that you'll need arbitrary-sized integers to get any benefit of more efficient techniques than the naive approach. The factorial of 21 is already too big to fit in a 64-bit unsigned long long int.As far as I know, there is no algorithm to compute $n!$ (factorial of $n$) which is faster than doing the ...
If your algorithm uses asymptotically less than $n$ time, then it does not have enough time to read all the digits of the numbers it is adding. You are to imagine you are handling very large numbers (stored for example in 8MB text files). Of course, addition can be done very quickly compared to the value of the numbers; it runs in $\mathcal{O}(\log(N))$ time,...
The short and simple answer is: it doesn't. No modern mainstream CPU ISA works the way you think it does.For the CPU, it's just a bit pattern. It's up to you, the programmer, to keep track of what that bit pattern means.In general, ISAs do not distinguish between different data types, when it comes to storage. (Ignoring special-purpose registers such as ...
Gilles answer is a good one, except for the paragraph on the real numbers, which is completely false, except for the fact that the real numbers are indeed a different kettle of fish. Because this sort of misinformation seems to be quite widespread, I would like to record here a detailed rebuttal.It is not true that all inductive types are denumerable. For ...
Modular exponentiation is a well-known algorithm. It is routinely available in libraries and languages that can manipulate large integers, including Wolfram Alpha.When making computations modulo a large number, one does not first make the whole computation in $\mathbb{N}$ and then take the remainder of the result, because for something like an ...
Hint: Use Lucas's theorem.In general, any time a programming contest problem wants you to compute something mod $p$, check for opportunities to reduce everything mod $p$ before doing any further work.SpoilersDon't peek at any hint until you've spent a good amount of time thinking about the previous one!A more in-depth hint if the previous wasn't ...
Processors are clocked, so even if some instructions can clearly be done faster than others, they may well take the same number of cycles.You'll probably find that the circuitry required to transport data between registers and execution units is significantly more complicated than the adders.Note that the simple MOV (register to register) instruction ...
Addition is important enough to not have it wait for a carry bit to ripple through a 64-bit accumulator: the term for that is a carry-lookahead adder and they are basically part of 8-bit CPUs (and their ALUs) and upwards. Indeed, modern processors tend to need not much more execution time for a full multiplication either: carry-lookahead is actually a ...
I believe that NVidia GPUs they use a table lookup, followed by a quadratic interpolation. I think they are using an algorithm similar to the one described in: Oberman, Stuart F; Siu, Michael Y: "A High-Performance Area-Efficienct Mutlifunction Interpolator," _IEEE Int'l Symp Comp Arithmetic, (ARITH-17):272-279, 2005.The table lookup is indexed with the $...
If you treat your vectors as over the field $GF(2)$ rather than over the set $\{0,1\}$, then what you ask is to find a basis for the span of a set of vectors. This is a well-studied problem in linear algebra, which you probably know the solution for. (One option is Gaussian elimination.)
Multiplying by 10 is the same as multiplying by $(1010)_2$. To multiply a binary number $x$ by 10, we thus just have to add $x0$ and $x000$. For example, $6 \times 10 = 60$ is implemented by$$\begin{array}{ccccccc}&0&0&1&1&0&0 \\+&1&1&0&0&0&0 \\\hline&1&1&1&1&0&0\end{array}$$...
There are multiple ways to define a mathematical structure, depending on what properties you consider to be the definition. Between equivalent characterizations, which one you take to be the definition and which one you take to be an alternative characterization is not important.In constructive mathematics, it is preferable to pick a definition that makes ...
Since the factorial function grows so fast, your computer can only store $n!$ for relatively small $n$. For example, a double can store values up to $171!$. So if you want a really fast algorithm for computing $n!$, just use a table of size $171$.The question becomes more interesting if you're interested in $\log(n!)$ or in the $\Gamma$ function (or in $\...
This is just a refactoring (Python 3) of Andrej's code. In Andrej's code numbers are represented through a list of digits (scalars), while in the following code numbers are represented through a list of symbols taken from a custom string:def v2r(n, base): # value to representation"""Convert a positive number to its digit representation in a custom ...
To divide $b$ by $a$, you follow two steps: first you find $u$ such that $au \equiv 1 \pmod{p}$, and then you compute $bu \pmod{p}$. To find $u$, you run the extended GCD algorithm on $a$ and $p$. If it is indeed the case that $a \not\equiv 0 \pmod{p}$, then since $p$ is prime $(a,p) = 1$. So the GCD algorithm will come up with numbers $u,v$ such that $$ au +...
This can be proven, but you need some nontrivial tools to do it.Start with the set S = {0,3,5,6, ...} of non-negative integers having an even number of 1's in their base-2 expansion.It is well-known that this set is "2-automatic"; that is, there is a finite automaton accepting exactly the base-2 expansions of elements of S. Furthermore, it is well-... |
I don't know if you are looking for a purely
LaTeX approach. If that is what you want, then it seems Paul Gaborit's
ocgx package is the way to go. But you tagged your question as
acrotex which uses javascript to enable its mouseover features. The manuals of
acrotex also have mouseover features. In Mouseover events in beamer: hovering on \eqref and a comment containing the original equation popping up I used
fancytooltips which also uses
acrotex.
Looking at the
pdf's you linked to, it seems that some javascript magic was also implemented. My answer and Speravir's additional answer to Mouseover will greatly help in this case. There are also a lot of examples provided in the devloper's site: http://user.mendelu.cz/marik/fancy-preview/.
Here is a sample article. (I'm sorry for the inconsistency of topics in content and bibliography
:). I used different sources for it--the math is from http://en.wikipedia.org/wiki/Argument_principle while the bibliography is from
biblatex's
biblatex-examples.bib.
(I am lazy and almost proud of it).
Read through Mouseover for the explanation about compiling the code in different platforms.
% myfile1.tex
\documentclass[10pt]{article}
\usepackage{amsmath}
\usepackage[backend=bibtex]{biblatex}
\usepackage{hyperref}
\newtheorem{theorem}{Theorem}
\addbibresource{biblatex-examples.bib}
\begin{document}
This is an example of a citation \parencite{markey,westfahl:space,aksin}.
Here is a theorem.
\begin{theorem}[Argument Principle]
If $f(z)$ is a meromorphic function inside and on some closed contour $C$, and $f$ has no zeros or poles on $C$, then
\begin{equation}
\label{eq:1}
\oint \frac{f'(z)}{f(z)} dz = 2\pi i(N-P)
\end{equation}
\end{theorem}
$N$ and $P$ of Equation \eqref{eq:1} denote respectively the number of zeros and poles of $f(z)$ inside the contour $C$, with each zero and pole counted as many times as its multiplicity, respectively order, indicates.
\printbibliography.
\end{document}
I compiled this in terminal using
perl fancy-preview myfile1 --fancy_options="previewall,nosoap"
The
previewall option enables mouseover pop-ups even if the link being referenced to is on the same page. I used it here just for the sake of the example but I personally think that enabling mouseover for equations when one can already see the equations on the same page is nonsensical.
nosoap, the default, removes the soap-shaped figures (tooltipmarks) on the top of the links.
Here is the output.
To enable
tooltipmarks, compile with
perl fancy-preview myfile1 --fancy_options="previewall"
To use any of the four pre-defined
tooltipmarks of
fancytooltips, you may use
--fancy_options="previewall,tooltipmark=1 where you can replace
1 with
2,3, or
4.
Disadvantages You can only use
Acrobat or
Adobe Reader since other pdf readers like
evince have no javascript capabilities.
The files are bloated. The
tooltipmarks, according to the
fancytooltips manual can
disturb the text and I agree. The mouseover effects are sometimes annoying especially when you unintentionally mouse over the
tooltipmark.
As reminded by Paul Gaborit in comment, the developer of
AcroTex withdrew the bundle from
TeXLive and the newer versions are now commercial. Although an older version available from CTAN is included in
MikTeX. |
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 |
Problem 435
Let $\calF[0, 2\pi]$ be the vector space of all real valued functions defined on the interval $[0, 2\pi]$.
Define the map $f:\R^2 \to \calF[0, 2\pi]$ by \[\left(\, f\left(\, \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \,\right) \,\right)(x):=\alpha \cos x + \beta \sin x.\] We put \[V:=\im f=\{\alpha \cos x + \beta \sin x \in \calF[0, 2\pi] \mid \alpha, \beta \in \R\}.\] (a) Prove that the map $f$ is a linear transformation. (b) Prove that the set $\{\cos x, \sin x\}$ is a basis of the vector space $V$. (c) Prove that the kernel is trivial, that is, $\ker f=\{\mathbf{0}\}$. (This yields an isomorphism of $\R^2$ and $V$.) (d) Define a map $g:V \to V$ by \[g(\alpha \cos x + \beta \sin x):=\frac{d}{dx}(\alpha \cos x+ \beta \sin x)=\beta \cos x -\alpha \sin x.\] Prove that the map $g$ is a linear transformation. (e) Find the matrix representation of the linear transformation $g$ with respect to the basis $\{\cos x, \sin x\}$.
(Kyoto University, Linear Algebra exam problem)Add to solve later |
The standard Gibbs free energy of formation for copper(II) oxide, $\ce{CuO}$, is $\Delta_\mathrm fG (\pu{298.15 K},\pu{1 bar}) = \pu{-129.7 kJ mol-1}$.
How can I get $\Delta_\mathrm fG (\pu{0 K},\pu{0 bar})$, the limiting value of this?
I want to compare it with the approximate value obtained by density functional theory calculations:
$$\Delta_\mathrm fG (\pu{0 K},\pu{0 bar}) \approx E^\text{Total}_\ce{CuO} - E^\text{Total}_\ce{Cu} - \frac{1}{2}E^\text{Total}_\ce{O2}$$ |
Question:Let us consider the Poisson problem on the square with constant source $1$$$\begin{cases} - \Delta u &= 1, \qquad \text{ in } (0,1)^n \\\\ u &= 0, \qquad \text{ on } \partial (0,1)^n \end{cases}$$By symmetry the maximum is attained in $x_0:=(1/2,\dots,1/2)$. Does there exist a formula or precise estimate for$$ \max u((0,1)^n) = u(x_0) \quad ? $$How does this value scale in $n$? Can one recover some bound $o(1)$, for $n\to \infty$? Background:It is easy by comparison with $1/8-(x-x_0)^2/(2n)$ and the maximum principle to deduce the bound$$ u(1/2, \dots, 1/2 ) \leq 1/8 ,$$which is also sharp for $n=1$, where $u(x) = 1/2 \; x (1-x)$ is the solution. By the same method one finds for the solution of the Poisson problem on the unit ball the bound $1/(2n)$, which is the second part of the question. Strategy so far: By the product-ansatz one can deduce an explicit solution for $u$, given by$$ u(x) = \frac{4^n}{\pi^{n+2}}\sum_{\alpha\in (2 \mathbb{N}+1)^n} \frac{1}{\sum_{i=1}^n \alpha_i^2} \prod_{j=1}^n \frac{\sin(\pi \alpha_j x_j)}{\alpha_j} .$$Hereby, I denote with $2\mathbb{N}+1$ the odd integers starting with $1$, hence $\alpha\in (2\mathbb{N}+1)^n$ is a multiindex with odd entries. In principle, setting $x=(1/2,\dots,1/2)$ should give the solution.
How does one calculate the resulting alternating series or find estimates recovering at least the scaling in $n$?
Numerical results: For small $n$ from series representation:
n=1: 0.125n=2: 0.0737n=3: 0.0562n=4: 0.0473
Probabilistic interpretation of the solution: $u(x)$ is the mean hitting time of a Brownian particle starting in $x$, which may open estimates from the stochastic side. One more thought wrt. volume: It is not clear, if the comparison of unit cube and unit ball is the right comparison, since the volume of the unit ball goes to $0$ for $n\to \infty$. Maybe, one should consider the unit-volume-ball, for which I find the scaling of the maximum like $O(n^{-1/2})$. I can add details, if there is some indication in this direction. Bonus questions: Is there a more general method not directly based on explicit solutions, which would lead to estimates for general domains, for instance convex, star shaped or just bounded?
What is a reasonable quantity describing the scaling behaviour in $n$? For instance: Do all reasonable unit-volume domains scale like $O(n^{-1/2})$? |
I would like to preface this question by saying that I have asked a series of questions on this topic on Math Stack Exchange, but have almost never received any fruitful responses, with the exception of one (which I've linked to in this post).
The integral I would like to investigate is the following:
$$\displaystyle \int_{\mathbb{R}^2} \frac{J_{1}(\rho \|x\|)J_{1}(\rho \|b-x\|)}{\|x\| \|b-x\|} \ \mathrm{d}x,$$
where $J_{\nu}$ denotes the Bessel function of the first kind, $b \in \Gamma \backslash \{0\}$ for any rational lattice $\Gamma \subset \mathbb{R}^2$ of full rank, $\rho > 0$ is large and independent of $b$ and $x$, and $\|\cdot\|$ denotes the standard Euclidean norm. I'll try to explain from where this integral arises, and why I care about it.
This integral arises when considering a problem on the distribution of lattice points inside a ball in $\mathbb{R}^d$ with radius $\rho$, which can be thought of as a generalisation of some aspects of the Gauss circle problem. For the full details, I recommend this paper (particularly pages 10-11 and 15-16). The paper considers $\sigma_p$ and proves some asymptotic bounds for $p = 1$ and $p = 2$ which depend on the dimension $d$ and the radius $\rho$. I'm attempting to generalise that work to $p = 4$, and with some extra work, any $p \in \mathbb{N}.$ The derivation of this integral begins here, and uses the estimate at the bottom of this post to arrive at the aforementioned integral. With the help of this answer, I was able to prove that the integral
$$\displaystyle \int_{\mathbb{R^d}} \|x\|^{-d/2}\|b-x\|^{-d/2}J_{d/2}(\rho \|x\|)J_{d/2}(\rho \|b-x\|) \ \mathrm{d}x,$$
converges absolutely in all dimensions $d \geqslant 3,$ after using the asymptotic bound for the Bessel function, $|J_{\nu}(z)| \leqslant C|z|^{-1/2},$ for $z \rightarrow \infty.$ However, as the answerer explains, the proof he provides does not work for the case $d = 2$. The integral at the top of this post is therefore the remaining case left to verify. I would like to know if it converges, and how in particular it can be estimated.
I don't have much experience in dealing with integrals involving Bessel functions. I would appreciate any help anyone can offer on how to deal with this integral. |
Loss functions (Incorrect predictions penalty) Table of Contents 1 - About
Loss functions define how to penalize incorrect predictions. The optimization problems associated with various linear classifiers are defined as minimizing the loss on training points (sometime along with a regularization term).
They can also be used to evaluate the quality of models.
2 - Articles Related 3 - Type 3.1 - Regression Squared loss = <math>(y-\hat{y})^2</math> 3.2 - Classification 3.2.1 - 0-1
0-1 loss: Penalty is 0 for correct prediction, and 1 otherwise
As 0-1 loss is not convex, the standard approach is to transform the categorical features into numerical features: (See Statistics - Dummy (Coding|Variable) - One-hot-encoding (OHE)) and to use a regression loss.
3.2.2 - Log
Log loss is defined as: <MATH> \begin{align} \ell_{log}(p, y) = \begin{cases} -\log (p) & \text{if } y = 1\\\ -\log(1-p) & \text{if } y = 0 \end{cases} \end{align} </MATH> where
<math>p</math> is a probability between 0 and 1. A base probability for a binary event will be just the mean over the training targetThen it can be compared to the output of probabilistic model such as the logistic regression <math>y</math> is a label of either 0 or 1.
Log loss is a standard evaluation criterion when predicting rare-events such as click-through rate prediction
Python from math import log def computeLogLoss(p, y): """Calculates the value of log loss for a given probabilty and label. Args: p (float): A probabilty between 0 and 1. y (int): A label. Takes on the values 0 and 1. Returns: float: The log loss value. """ epsilon = 10e-12 if p==0: p = epsilon elif p==1: p = p - epsilon if y == 1: return -log(p) elif y == 0: return -log(1-p) |
Since there there several inequivalent ways to define the local compactness, I first give the defintion in Fred H. Croom's Principles of Topology.
Definition: A space $X$ is locally compact at a point $x$ if there is an open set $U$ such that $x\in U$ and $\bar U$ is compact. $X$ is said to be locally compact if it is locally compact at each of its points.
Now, let me write a question in Croom's book:
Question: Let $X$ be a Hausdorff space. Show that $X$ is locally compact if and only if each point of $X$ has a compact neighborhood; i.e., $X$ is locally compact if and only if each point $x\in X$ belongs to the interior of a compact set $K_x$.
I think I can prove this easily.
Proof: "$\Rightarrow$". Suppose $X$ is locally compact. By the definition, for each $x\in X$, there is an open set $U$ such that $x\in U$ and $\bar U$ is compact. So just let $K_x=\bar U$ and we have proved "$\Rightarrow$". Note that we did not use the fact that $X$ is Hausdorff.
"$\Leftarrow$". Suppose each point $x\in X$ belongs to the interior of a compact set $K_x$. Simply set $U=\operatorname{int} K_x$. So $x\in U$ and $\overline{U}$ is closed in $K_x$, which implies that $\bar U$ is compact. So $X$ is locally compact.
Again, we did not use $X$ being Hausdorff.
Therefore, it seems that $X$ needs not be Hausdorff. So there should be something wrong with my argument. Please find it. Thank you! |
You've identiifed the key problem. Certainly the primal
can be solved directly by, say, a quadratic programming solver. But typical QP solvers often don't scale well to large problem sizes. A projected gradient method can often scale to significantly larger problems---but only if the derivatives and projections are simple to compute. As I will show, the dual problem can be solved with inexpensive projected gradient iterations, while the primal cannot be.
First, let's simplify notation a bit: collect the vectors $y_ix_i$ into the
rows of a matrix $Z$, and the values $y_i$ into the elements of a vector $y$. Then we can write the problem as$$\begin{array}{ll} \text{minimize}_{w,b} & f(w,b) \triangleq \tfrac{1}{2}w^T w \\ \text{subject to} & Z w + y b \succeq \vec{1} \end{array}$$
A projected gradient algorithm will alternate between gradient steps and projections. The gradient is simply $\nabla f(w,b)=(w,0)$, so this is a pretty trivial operation. Let's denote by $(w_+,b_+)$ the result of a single gradient step. We must then project $(w_+,b_+)$ back onto the feasible set: find the nearest point $(w',b')$ that satisfies $Xw+y \succeq \vec{1}$. This means we must solve$$\begin{array}{ll}\text{minimize}_{w,b} & \|w-w_+\|_2^2+(b_+-b)^2 \\ \text{subject to} & Zw+by\succeq \vec{1} \\\end{array}$$
This is virtually the same problem as the original. In other words, each step of projected gradients for the primal problem is as expensive as the original problem itself.
Now examine the dual problem. To get a handle on this we need to simplify the dual function$$g(\lambda) = \min_{w,b} \tfrac{1}{2} w^T w - \lambda^T ( Z w + b y - \vec{1} )$$With a little calculus you can determine that the optimal value of $w=Z^T\lambda$. As for $b$, we observe that if $y^T\lambda\neq 0$, the right-hand expression can be driven to $-\infty$ by growing $b\cdot(y^T\lambda)\rightarrow +\infty$. So$$g(\lambda) = \begin{cases} - \vec{1}^T \lambda - \tfrac{1}{2} \lambda^T ZZ^T \lambda & y^T \lambda = 0 \\ -\infty & y^T \lambda \neq 0 \end{cases}$$So the effect of $b$ is to introduce an implicit constraint $y^T\lambda =0$, and the dual problem is equivalent to $$\begin{array}{ll} \text{maximize} & \bar{g}(\lambda) \triangleq \vec{1}^T \lambda - \tfrac{1}{2} \lambda^T ZZ^T \lambda \\ \text{subject to} & y^T \lambda = 0 \\ & \lambda \succeq 0 \end{array}$$We would not have been able to apply projected gradients to the original dual function, because it wasn't differentiable; but removing the implicit constraint $y^T\lambda=0$ changes that. Now the gradient is $\nabla\bar{g}(\lambda) = \vec{1} - ZZ^T\lambda$---a bit more complex than the primal gradient, but entirely manageable. (Note that we're maximizing now, so gradient steps are taken in the positive direction).
So what's left is to perform the projection. Given the point $\lambda_+$ that comes out of the gradient step, we need to solve$$\begin{array}{ll} \text{minimize} & \tfrac{1}{2} \|\lambda-\lambda_+\|_2^2 \\ \text{subject to} & y^T \lambda = 0 \\ & \lambda \geq 0 \end{array}$$This looks much easier than the primal projection, doesn't it? I haven't looked at the literature to see how people solve it now, but this is what I came up with: the value of $\lambda$ is$$\lambda_i = \max\{\lambda_{+,i}+sy,0\}, ~i=1,2,\dots, n$$where the parameter $s$ is chosen so that $y^T\lambda = 0$. I suspect there is a simple iterative method: start with $s=0$, compute $\lambda$ and $y^T\lambda$, adjust $s$, and repeat. A few $O(n)$ computations get the result. Don't trust me on this, I suspect the extant SVM literature has something solid here.
An important final step is to recover $(w,b)$ from the optimal solution $\lambda^*$. Our derivations for $g(\lambda)$ showed that $w=Z^T\lambda^*$. Once this has been recovered, $b$ is any value in the following interval:$$b \in \left[ 1 - \max_{i:y_i=1} w\cdot x_i, \max_{i:y_i=-1} w \cdot x_i - 1 \right] $$ |
Tagged: abelian group
Abelian Group Problems and Solutions.
The other popular topics in Group Theory are:
Problem 616
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$.
Add to solve later
(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 497
Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively. Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.
Also determine whether the statement is true if $G$ is a non-abelian group.Add to solve later
Problem 434
Let $R$ be a ring with $1$.
A nonzero $R$-module $M$ is called irreducible if $0$ and $M$ are the only submodules of $M$. (It is also called a simple module.) (a) Prove that a nonzero $R$-module $M$ is irreducible if and only if $M$ is a cyclic module with any nonzero element as its generator.
Add to solve later
(b) Determine all the irreducible $\Z$-modules. Problem 420
In this post, we study the
Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Add to solve later
Problem. Let $G$ be a finite abelian group of order $n$. If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic to the cyclic group $Z_n=\Zmod{n}$ of order $n$. Problem 343
Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.
Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of $G$. |
Problem:
A buffer was made by mixing $\pu{500 mL}$ of $\pu{1 M}$ acetic acid and $\pu{500 mL}$ of $\pu{0.5 M}$ calcium acetate.
What are the resulting concentrations of acetic acid (henceforth called $\ce{AcH}$), $\ce{Ca^2+}$ and acetate (henceforth called $\ce{Ac-}$, and what is the resulting $\mathrm{pH}$? $K_\mathrm{a}$(acetic acid) = $1.75\times10^{-5}.$
My attempted solution:
This problem gives me two equilibrium reactions to take into account.
\begin{align} \ce{AcH + H2O &<=> Ac- + H3O+}\tag{1}\\ \ce{Ac- + H2O &<=> AcH + OH- }\tag{2}\\ \end{align}
In (2), I assume that calcium acetate completely dissociates in water. I also assume that the concentration of $\ce{OH- }$ from the auto ionization of water is negligible.
First I calculate the concentrations of $\ce{AcH}$, $\ce{Ac- }$ and $\ce{H3O+}$ in (1), ignoring what I assume to be minuscule amounts of $\ce{H+}$ from the auto ionization of water. To do this, I use the equation for the acid dissociation constant:
$$K_\mathrm{a}=\frac{[\ce{Ac- }][\ce{H3O+}]}{[\ce{AcH}]}\tag{3}$$
Initial values for $\ce{AcH}$, $\ce{Ac- }$ and $\ce{H3O+}$ are set to 1, 0 and 0 M, respectively. Equilibrium concentrations will then be $1-x$, $x$ and $x$ M.
Plugging this into (3), I get $x = 4.17\times10^{-3}.$
I have now found (or can at least easily calculate) the concentrations of the compounds in (1) at equilibrium.
On to (2), I know that $K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}$, and I have the values for $K_\mathrm{a}$ and $K_\mathrm{w}$, which gives me the value $5.71\times10^{-10}$ for $K_\mathrm{b}.$
Initial concentrations in (2) for $\ce{Ac- }$, $\ce{AcH}$ and $\ce{OH- }$ are 1 (because 0.5 moles of calcium acetate gives $2\times0.5=1$ moles of acetate), 0 and 0 M, respectively. Equilibrium concentrations will then be $1-x$, $x$ and $x$. Plugging this into (3) (though using the equation for the base dissociation constant instead), I get $x = 2.39\times10^{-5}.$
I have now found all the values I need to calculate the final concentrations and pH when adding the two solutions to each other, and up till this point I am fairly sure that all my calculations are correct.
Questions:
How do I write the chemical equation for the mixing of the two solutions? I tried the following; $\ce{AcH + Ac- <=> Ac- + AcH},$ but this doesn’t seem to help at all.
From my younger days, I seem to remember a much simpler and quicker method of doing these calculations than what I've shown here; does such a method exist? Please note that if it does exist I’d like to learn about it, but I would also like answers based on the method I've used.
Let’s say you mix one weak acid with one weak base, and they aren't conjugate pairs. How would the method differ from the one in this problem? I am specifically wondering about whether both the dissociation constants must be provided to be able to solve, and also how to write the chemical equation, but other observations are more than welcome as well. |
Given the adjacency matrix $A_{ij}$ of a graph with $N$ vertices and $M$ links (or any binary symmetric matrix of size $N \times N$), is it possible to establish lower and upper boundaries of its eigenvalues? I mean, do $N$ and $M$ determine the lowest and the largest possible eigenvalues of the matrix?
In other words, let $L$ be the Lapacian of $A_{ij}$. We know that its eigenvalues obbey the following: $\lambda_1 = 0 \leq \lambda_2 \leq \lambda_3 \leq \ldots \leq \lambda_N$. The lowest eigenvalue is always $\lambda_1 = 0$ and hence, the eigenvalues of $L$ have a lower bound. Is there an upper bound for $\lambda_N$ that depends only on the size $N$ of the graph and/or on $M$?
Stated yet in another manner. Does anyone know a graph $G'$ whose $\lambda'_N$ is largest than the $\lambda_N$ of any other graph of the same size $N$ and/or number of links $M$?
Thank you! |
Command Summary
Calculates the inverse of the cumulative Student's t-distribution function with degrees of freedom ν.
Command Syntax
invT(
probability, ν)
Menu Location
Press:
2ND DISTR to access the distribution menu 4 to select invT(, or use arrows.
Calculator Compatibility
TI-84+/SE (OS 2.30 or greater)
Token Size
2 bytes
invT( is the inverse of the cumulative Student t distribution function: given a probability
p and a specified degrees of freedom ν, it will return the number x such that tcdf(-E99, x, ν) is equal to p Advanced
invT( is meant for use with so-called "one-tailed' tests; for two-tailed tests, the proper expression to use (corresponding to the inverse of tcdf(-
x, x, ν)) is invT(.5(1+ p), ν) Formulas
Unlike the tpdf( and tcdf( commands, the invT( command does not have a closed-form formula. However, it can be expressed in terms of the inverse incomplete beta function.
For one degree of freedom, invT( is expressible in terms of simpler functions:(1)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \operatorname{invT}(p,1)=\tan\left(\pi\left(p-\frac1{2}\right)\right) \end{align}
Related Commands |
Let $f: [a,b] \longrightarrow \mathbb{R}$ is a integrable function and $X \subset [a,b]$ dense in $[a,b]$. Suppose that $f(x) = 0$ for all $x \in X$. Prove that $\displaystyle \int\limits_{a}^{b}f(x)dx = 0$.
$\textbf{My idea:}$ Let $P = \lbrace a=t_{0}, t_{1}, ..., t_{n} = b \rbrace$ an arbitrary partition of $[a,b]$. Since $X$ in dense, in any partitions intervals $[t_{i-1}, t_{i}]$ of $[a,b]$, $\inf([t_{i-1}, t_{i}]) = 0$. Therefore, $s(|f|, P) = 0$, than $\displaystyle \int\limits_{a}^{b}f(x)dx = 0$, because $\sup s(|f|, P) =\inf S(|f|, P)$
$-$ $s(f, P)$ is the left riemann sum and $S(f, P)$ is the right riemann sum
Is the correct? |
Let $N$ be a natural number and $\mathbb{Z}_N := \mathbb{Z}/N\mathbb{Z}$. Let $\pi_{N'} : \mathbb{Z}_N \to \mathbb{Z}_{N'}$ be the projection, i.e. $\pi_{N'}(a + N\mathbb{Z}) := a + N'\mathbb{Z}$. A Dirichlet character (DC) mod $N$ is a group homomorphism $\chi : \mathbb{Z}_N^* \to \mathbb{C}^*$. For a divisor $N' | N$ we say that $\chi$ is defined mod $N'$ if for all $x, y \in \mathbb{Z}_N^*$ we have $\pi_{N'}(x) = \pi_{N'}(y) \Rightarrow \chi(x) = \chi(y)$. The conductor of $\chi$ is defined to be the smallest divisor $N'|N$ s.t. $\chi$ is defined modulo $N'$. Let me write $c(\chi)$ for its conductor. What i want to see now is the following:
$----$
If $N'|N$ is such that $\chi$ is defined modulo $N'$, then $c(\chi)|N'$.
$----$
I suspected the proof to go like this: Let us write $N' = vc(\chi) + r$ with $0 \leq r < c(\chi)$. We have to show $r=0$, so assume $r > 0$, then we claim that $\chi$ is defined modulo $r < c(\chi)$, a contradiction. For showing this, we simply compute for $a \in \mathbb{Z}_N^*$ and $k \in \mathbb{Z}$ such that $a + kr \in \mathbb{Z}_N^*$, that
$$\chi(a + kr) = \chi(a + k[N' - vc(\chi)]) = \chi( (a + kN') + (-kv)c(\chi)) = \chi(a + kN') = \chi(a)$$ where we have used that $\chi$ is defined modulo $N'$ and modulo $c(\chi)$. The problem with this proof is the following: It can happen that although $a$ and $a + kr$ are units in $\mathbb{Z}_N$, $a + kN'$ is not, so we cannot argue like this. I am still convinced that this assertion is true. Can somebody help?
Thanks in advance,
Fabian Werner |
if we just examine the translational dynamics (X and Y direction), the dynamics are coupled, So if I want to design a PID controller to control the position, the outcome will be unpredictable because of the coupling.I some research papers , I have seen people assume (psi =0 ) at the equilibrium point to remove the coupling in x and y dynamics. But I would like to design a controller along with the couple system? Is there any solution like decoupling the dynamics equations or something.
You can linearize and decouple the system about equilibrium under the following assumptions:
Small angle approximation: $\sin(x) \approx x$ and $\cos(x) \approx 1$
Angular acceleration due to actuators only: $\dot{p} = \frac{U_2}{I_{xx}}$, $\dot{r} = \frac{U_3}{I_{yy}}$, $\dot{p} = \frac{U_4}{I_{zz}}$
Quadrotors are coupled systems because you have 4 inputs for 6D motions.
Usually people are using cascade controllers to regulate both orientation and position and assume that desired orientation is achieved when considering the control of position.
see this famous RAM tutorial paper to get familiar with quadrotors: https://ieeexplore.ieee.org/document/6289431/?arnumber=6289431
Also, quadrotors are flat, in the control theory sens, for position and yaw, this means that the internal orientation dynamics are stable, if the flat output are stabilized. That is the reason why you can easily fly them controlling the position. |
The photon-correlation spectrometer Photocor Complex is based on the Photon Correlation Spectroscopy (PCS) technique and designed for measurements of sub-micron particle sizes, diffusion coefficients, viscosities, molecular weights of polymers in basic and applied studies.
The PCS method consists in determining the velocity distribution of particles movement by measuring dynamic fluctuations of intensity of scattered light. The disperse particles or macromolecules suspended in a liquid medium undergo Brownian motion which causes the fluctuations of the local concentration of the particles, resulting in local inhomogeneities of the refractive index. This in turn results in fluctuations of intensity of the scattered light. The linewidth of the light scattered spectrum Γ (defined as the half-width at half-maximum) is proportional to the diffusion coefficient of the particles D (Eq.1):$$ \Gamma = D_t q^2 $$
where$$ q = \frac{4\pi n}{\lambda} \sin \left(\frac{\theta}{2}\right) $$
n is the refractive index of the medium, λ the laser wavelength, and θ the scattering angle. With the assumption that the particles are spherical and non-interacting, the mean radius is obtained from the Stokes-Einstein equation$$ R_h = \frac{k_bT}{6\pi \eta D} $$
where k
B is the Boltzmann constant, T the temperature, and η the shear viscosity of the solvent. Information about the light-scattering spectrum can be obtained from the autocorrelation function G(τ) of the light-scattering intensity. For the simplest case of spherical monodisperse non-interacting particles in a dust-free fluid, the characteristic decay time of the correlation function is inversely proportional to the linewidth of the spectrum. Therefore, the diffusion coefficient and either particle size or viscosity can be found by fitting the measured correlation function to a single exponential function.
A characteristic autocorrelation function of the scattered light is shown below where the baseline b is proportional to the total intensity I, and it can be determined experimentally.
There exist two techniques of measuring the correlation function: heterodyning and homodyning. In heterodyne measurements, which are most suitable for small intensities, the scattered light is mixed coherently with a static light source at the incident wavelength, and the static field is added to the scattered fields at the photodetector. Eq.1 that connects the linewidth Γ and the diffusion coefficient D is given for a heterodyne spectrum. In homodyne measurements the photodetector receives scattered light only. Homodyning is most suitable for strong intensities (e.g. near the critical point of the fluid, or for colloid systems). For a homodyne spectrum the connection between Γ and D reads:$$ \Gamma = 2 D_t q^2 $$
The
Photocor Complex setup can operate in both homodyning and heterodyning regimes. Principle of operation and setup arrangement
The PCS method consists in analyzing the velocity distribution of particles movement by measuring dynamic fluctuations of light scattering intensity:
The disperse particles or macromolecules suspended in a liquid medium undergo Brownian motionwhich causes fluctuations of local concentration of the particles, resulting in local inhomogeneities of the refractive index. This in turn results in fluctuations of intensity of the scattered light. The diffusion coefficientof the particles is inversely proportional to the decay time of light scattering fluctuations. The decay time is obtained from the time-dependent correlation function of the scattered light. The particle sizeis calculated in accordance with Stokes-Einstein formula relating the particle size to the diffusion coefficient and viscosity. |
I have the following equation:
$$\frac{dx}{dt}+x=4\sin(t)$$
For solving, I find the homogenous part as: $$f(h)=C*e^{-t}$$
Then finding $f(a)$ and $df(a)$: $$f(a)=4A\sin(t)+4B\cos(t)$$ $$df(a)=4A\cos(t)-4B\sin(t)$$
Substituting in orginal equation:
$$4A\cos(t)-4B\sin(t)+4A\sin(t)+4B\cos(t)=4\sin(t)$$
I have to find numerical values of $A$ and $B$ but I absoloutly have no idea how can I solve this, I am also not sure if the steps I did are correct or not. Would somone please help with this equation?
The final answer should be substituted in: $$x=f(h)+f(a)$$ |
There's a bit of confusion here.
Whenever you work with a QFT you have to able to
define it first. It is not enough to just write down diverging integrals and assert that they correspond to transition amplitudes. This doesn't make sense.
So I assume that what you mean is: define a theory with an explicit momentum-scale cutoff $\Lambda$ such that all the integrals are finite. Consider $\Lambda$ just as physical as other constants like mass $m$ and the coupling constant $\lambda$.
There's a whole bunch of issues with this definition. Like, for example,
It isn't clear how to consistently impose momentum constraints on the loop integrals, as we can pass to different loop momenta integration variables for which different constraints have to apply. Note that you don't care about this subtlety when $\Lambda \gg p$. The theory with a momentum cut-off isn't Lorentz invariant. It simply isn't. You can, however, neglect this Lorentz violations when $\Lambda \gg p$.
The list can go on, but I think I've made my point already. But say that we somehow found relatively satisfactory answers to all of the questions above. What now?
1. Does this theory still require renormalization? If yes, why?
Yes, it does! Renormalization isn't about getting rid of infinities, and it isn't about getting rid of the unphysical $\Lambda$ (though it achieves both of these goals in the mean time). It is about making sense of the results that your theory gives.
Like, for example, you would want to give a particle interpretation to your theory, with an $S$-matrix corresponding to particle scattering. What do you require to give your theory a particle interpretation? One of the requirements is that the 2-point function has a pole when $p^2 = M^2$ with residue 1. This follows directly from normalization of states, and it allows you to talk about interacting particles of mass $M$ which your theory is supposed to describe.
If you calculate this 2-point function to some loop order you will find that both the location and the value of the pole are not $m$ and $1$ as you would've expect naively, but depend on $\Lambda$. But what's it mean? It means that your particles are of mass $M = M(m, \lambda, \Lambda)$ and are generated by a Fock-space operator associated to the physical observable $Z(m, \lambda, \Lambda) \cdot \phi$, not just the field operator $\phi$.
Again,
renormalization is about reinterpreting predictions in terms of interacting particles. 2. If yes, then instead of saying $\boldsymbol{\phi^6}$ to be a non-renormalizable theory, shouldn't we say that (i) it is renormalizable at low energy but (ii) non-renormalizable at high energies?
What happens is: the
higher-valency correlation functions become highly dependent on $\Lambda$ even in the $\Lambda \gg p$ regime. Physically, your theory becomes pathologically sensible to short-scale fluctuations.
With renormalizable theories we can say that $\Lambda$ is very large and corresponds to the boundary of the domain of applicability of our theory. But the exact details of this boundary aren't relevant for the long-range physics: we can just adopt the limiting value for the higher-valency correlations.
In case of $\phi^6$ in $4d$ though, this is not the case. Instead, we have the following
nonrenormalizable behaviour:
The long-range properties of your theory depend explicitly on the details of the cut-off procedure. It can be the value of $\Lambda$, the way you resolve the cutoff ambiguity in the loop momenta integration variables, masses of the Pauli-Villars regularizer fields, etc. The key fact is that - your results depend on something for which you can't really say how it works and if it is physical or not. That's what is bad about nonrenormalizability.
With nonrenormalizable theories you can tweak the theory to give whatever predictions you want by simply changing the cut-off mechanism a bit. Not a lot of predictive power there.
UPDATE: allright, I admit that it is not 100% true. I was trying to make a point in the context of HEP, but once you let go of your ambitions to describe arbitrary high-energy processes, you can actually do something useful with nonrenormalizable theories as well.
For instance, you can fix the perturbation theory order $k$
prior to renormalization, and then simply determine the values of counterterms from experiments. With renormalizable theories this could be done once, i.e. using a fixed finite number of counterterms independent of the perturbation theory order. But nonrenormalizable theories require more and more tweaking and adjustment with increasing order.
Of course one can claim that this is fine, since perturbation expansion is only an
assymptoitic series and thus even renormalizable theories can't be solved with an a-priori arbitrary precision using perturbation theory. And it is probably true.
There's another property of nonrenormalizable theories which has to do with Wilsonian renormalization group flow. Effective couplings used in perturbation theory blow up in the ultraviolet regime thus rendering the whole concept of perturbation theory meaningless. Thus we end up with phase transitions in the high-energy regime which we can't describe with perturbation theory.
It is also worth mentioning that such phase transitions aren't specific to nonrenormalizable theories. As a useful example, QED (Quantum Electrodynamics), though renormalizable, has an ultraviolet phase transition (the Landau Pole problem). Renormalizable theories without these phase transition are called asymptotically free.
And asymptotic freedom, together with renormalizability, is enough to ascertain that a HEP theory can be used to make sensible predictions up to whatever energy is associated with the boundary of its domain of validity. This is because the further into the UV you go, the less becomes the coupling, making the asymptotic expansion a better approximation for an even further order in perturbation theory (remember how the closer the coupling is to zero – the more orders of perturbation theory we can trust without worrying about asymptotic expansion blowing up?). |
Can someone please explain to me the intuition behind Positive recurrence. What does it mean and why is it different to normal recurrence?
If the probability of return or recurrence is $1$ then the process or state is
recurrent.
If the expected recurrence time is finite then this is called
positive-recurrent; if the expected recurrence time is infinite then this is called null-recurrent.
See the Wikipedia article on Markov chains for more details.
Added as an example:
In a simple symmetric 1D random walk, the probability of first return after $2n$ steps is $\dfrac{\frac{1}{2n-1}{2n \choose n}}{2^{2n}}$. Since $\sum_{n=1}^\infty \dfrac{\frac{1}{2n-1}{2n \choose n}}{2^{2n}} =1$, the probability of first return in finite time is $1$, so this is recurrent. But since $\sum_{n=1}^\infty 2n \dfrac{\frac{1}{2n-1}{2n \choose n}}{2^{2n}}$ is infinite, the expected time of the first return is infinite, so this is null-recurrent. |
What I am wondering is if mathematicians know whether (assuming consistency) the natural numbers are a definite object, without ambiguity. This seems intuitively obvious, but I don't know if its been shown. A formal question might be if two consistent axiomatic systems, each including all the axioms of the natural numbers and then some, can't contradict each other in statements about the natural numbers in the language of the natural numbers (not referring to objects guarunteed by the other axioms of the systems).
The overwhelming consensus is that the true natural numbers
is is a definite thing which objective properties -- not because this can be proved from anything more basic, but because mathematicians intuitively believe they have some kind of objective Platonic existence.
It is known, however, that not all properties of the Platonic naturals can be captured completely by any reasonable axiomatic system. Given any axiomatic system $T$ that proves no falsehoods about the integers, Gödel's incompleteness theorem shows how to construct an arithmetic sentence $\phi$ such that both $T\cup\{\phi\}$ and $T\cup\{\neg\phi\}$ are consistent theories.
A formal question might be if two consistent axiomatic systems, each including all the axioms of the natural numbers and then some, can't contradict each other in statements about the natural numbers in the language of the natural numbers (not referring to objects guarunteed by the other axioms of the systems).
If "all the axioms of the natural numbers" just means Peano Arithmetic, then it is completely possible to have two axiom systems that are each consistent, each contain all of Peano arithmetic, and are inconsistent with each other.
This is because there are many statements that are not provable nor disprovable in Peano arithmetic, even though the statements are in the language of Peano arithmetic. Therefore, if $A$ is such a statement, "Peano arithmetic plus $A$" and "Peano arithmetic plus the negation of $A$" are an example of the phenomenon I just mentioned.
One reason that many mathematicians think that the natural numbers are well defined (even if the powerset of the natural numbers is not) is that the natural numbers are all about finiteness, which we may think we have a good sense about. There are really three phenomena that are interrelated:
The natural numbers
The set of all finite strings on the alphabet $\{0,1\}$
The set of all formal proofs from an effective set of axioms
Any of these concepts can be interpreted, in a certain sense, in any of the others. For example, if we know what a finite proof is, we can interpret a natural number as the length of a proof. If we know what a natural number is, we can define proofs in terms of their Goedel numbers.
So, if the natural numbers are somehow vague, the concept of a finite sequence and the concept of a formal proof must be equally vague. Most mathematicians think we have some sort of perception of these objects, although a minority think the concepts are vague.
One of the most interesting recent developments in this respect is the potential in multiverse set theory (as developed by Joel David Hamkins and coworkers) there is
no "standard" model of the natural numbers, and that every model of set theory is not well founded relative to another model. I do not think anyone has yet written a philosophical argument about the consequences of that for foundations of mathematics.
HINT: Given two Peano systems $(N,0,S)$ and $(N',0',S')$, prove there exists a bijection $f:N\to N'$ such that $f(0)=0'$ and $f(S(x))=S'(f(x))$.
This answer is perhaps related to that of Carl Mummert. There is an informal but very inspiring article by Joel David Hamkins, about his caveats or doubts about if we are correct in thinking that we have an absolute concept of the finite. if Hamkins have doubts about it, I am pretty sure they should be taken seriously. All other number theorists that believe that the standard model of Peano arithmetic defines THE natural numbers beyond any doubts, should not be taken too seriously. You can read his inspiring thoughts at this link: http://jdh.hamkins.org/question-for-the-math-oracle/
There is very little if any ambiguity in this case. For all practical purposes (putting aside Gödel), Peano's axioms (2nd order) can be used as The Definition of the set of natural numbers:
$0\in N$
$S: N\to N$
$\forall x,y\in N:[S(x)=S(y)\implies x=y]$
$\forall x \in N: S(x)\ne 0$
$\forall P\subset N: [0\in P \land \forall x\in P: S(x)\in P \implies \forall x\in N: x\in P] $
It can be formally shown that if Peano's axioms hold on set $N$ with successor function $S$ and first element $0$ (as above), then the system $(N,S,0)$ is unique to within an isomorphism, i.e. all such systems are identical except for the names used. See: http://dcproof.com/EquivalentPeanoSystemsB.htm
EDIT: Here is a nice visualization based on the falling dominoes paradigm of the kind of "junk terms" (e.g. the disconnected side-loop shown) that is excluded by the induction axiom (5). |
From
Revuz and Yor - Continuous Martingales and Brownian Motion 1999
Chapter IV Proposition 1.13
it is proven, that for a continuous local martingale $M_t$ the intervals of constancy are equal with those of the predictable quadratic variation $<M>_t$ or optional quadratic variation $[M]_{t}$ (since they coincide due to the continuity of the local martingale).
I wonder if this stays true for $M_{t}$ being just càdlàg. I guess no.
So lets consider this setup: Given a square integrable Martingale $X_t=F_t-a\cdot K_{t}$ with predictable quadratic variation $b\cdot K_{t}$ where $a,b$ are constants and $K_{t}$ is continuous but $F_{t}$ only càdlàg. With the aim to conclude from $K_{t}$ being constant on some interval (predictable quadratic variation is continuous process of $K_{t}$ being constant) implies that $X_{t}$ is constant on that interval and thus $F_{t}$ on the interval. Where $K_{0}=0$, $K_{t}\rightarrow \infty$ a.s. and a non decreasing process. |
During October (2017) I will write a program per day for some well-known numerical methods in both Python and Julia. It is intended to be an exercise then don't expect the code to be good enough for real use. Also, I should mention that I have almost no experience with Julia, so it probably won't be idiomatic Julia but more Python-like Julia.
Finite Difference: Eigenproblems
Today we have a Finite difference method to compute the vibration modes of a beam. The equation of interest is
with
Following are the codes.
Python from __future__ import division, print_function import numpy as np from scipy.linalg import eigh as eigsh import matplotlib.pyplot as plt def beam_FDM_eigs(L, N): x = np.linspace(0, L, N) dx = x[1] - x[0] stiff = np.diag(6*np.ones(N - 2)) -\ np.diag(4*np.ones(N - 3), -1) - np.diag(4*np.ones(N - 3), 1) +\ np.diag(1*np.ones(N - 4), 2) + np.diag(1*np.ones(N - 4), -2) vals, vecs = eigsh(stiff/dx**4) return vals, vecs, x N = 1001 nvals = 20 nvecs = 4 vals, vecs, x = beam_FDM_eigs(1.0, N) #%% Plotting num = np.linspace(1, nvals, nvals) plt.rcParams["mathtext.fontset"] = "cm" plt.figure(figsize=(8, 3)) plt.subplot(1, 2, 1) plt.plot(num, np.sqrt(vals[0:nvals]), "o") plt.xlabel(r"$N$") plt.ylabel(r"$\omega\sqrt{\frac{\lambda}{EI}}$") plt.subplot(1, 2 ,2) for k in range(nvecs): vec = np.zeros(N) vec[1:-1] = vecs[:, k] plt.plot(x, vec, label=r'$n=%i$'%(k+1)) plt.xlabel(r"$x$") plt.ylabel(r"$w$") plt.legend(ncol=2, framealpha=0.8, loc=1) plt.tight_layout() plt.show() Julia using PyPlot function beam_FDM_eigs(L, N) x = linspace(0, L, N) dx = x[2] - x[1] stiff = diagm(6*ones(N - 2)) - diagm(4*ones(N - 3), -1) - diagm(4*ones(N - 3), 1) + diagm(1*ones(N - 4), 2) + diagm(1*ones(N - 4), -2) vals, vecs = eig(stiff/dx^4) return vals, vecs, x end N = 1001 nvals = 20 nvecs = 4 vals, vecs, x = beam_FDM_eigs(1.0, N) #%% Plotting num = 1:nvals # Missing line for setting the math font figure(figsize=(8, 3)) subplot(1, 2, 1) plot(num, sqrt.(vals[1:nvals]), "o") xlabel(L"$N$") ylabel(L"$\omega\sqrt{\frac{\lambda}{EI}}$") subplot(1, 2 ,2) for k in 1:nvecs vec = zeros(N) vec[2:end-1] = vecs[:, k] plot(x, vec, label="n=$(k)") end xlabel(L"$x$") ylabel(L"$w$") legend(ncol=2, framealpha=0.8, loc=1) tight_layout() show()
Both have (almost) the same result, as follows
Comparison Python/Julia
Regarding number of lines we have: 40 in Python and 39 in Julia. The comparisonin execution time is done with
%timeit magic command in IPython and
@benchmark in Julia.
For Python:
with result
For Julia:
with result
BenchmarkTools.Trial: memory estimate: 99.42 MiB allocs estimate: 55 -------------- minimum time: 665.152 ms (17.14% GC) median time: 775.441 ms (21.76% GC) mean time: 853.401 ms (16.86% GC) maximum time: 1.236 s (15.68% GC) -------------- samples: 6 evals/sample: 1
In this case, we can say that the Python code is roughly 4 times faster than Julia. |
I would like to show $\lim_{n \rightarrow \infty}\left(\frac{n - 1}{n}\right)^n = 1/e$.
I know the argument typically goes like this: Let $y = \left(\frac{n - 1}{n}\right)^n$. Then $\ln(y) = n\cdot{}\ln \left(\frac{n - 1}{n}\right)$. Taking the limit as $n \rightarrow \infty$, we have an indeterminant product of the form $\infty\cdot0$.
I think ideally I would like to use L'Hopital's Rule, so the issue is getting this into the correct form to apply it. I don't think the simplification $n\ln(n - 1) - n\ln(n)$ helps any.
But if we can establish that $\lim_{n\rightarrow\infty}\ln(y) = -1$, then using the identity $y = e^{\ln(y)}$, we'd arrive at the desired result.
Alternatively, could one use the definition of $e$? This might not help, but $e = \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n = \lim_{n \rightarrow \infty}(\frac{n + 1}{n})^n$, which looks similar to what we have, but not quite. |
berylium? really? okay then...toroidalet wrote:I Undertale hate it when people Emoji movie insert keywords so people will see their berylium page.
A forum where anything goes. Introduce yourselves to other members of the forums, discuss how your name evolves when written out in the Game of Life, or just tell us how you found it. This is the forum for "non-academic" content.
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
When xq is in the middle of a different object's apgcode. "That's no ship!"
Airy Clave White It Nay
When you post something and someone else posts something unrelated and it goes to the next page.
Also when people say that things that haven't happened to them trigger them.
Also when people say that things that haven't happened to them trigger them.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
Huh. I've never seen a c/posts spaceship before.drc wrote:"The speed is actually" posts
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
It could be solved with a simple PM rather than an entire post.Gamedziner wrote:What's wrong with them?drc wrote:"The speed is actually" posts
An exception is if it's contained within a significantly large post.
I hate it when people post rule tables for non-totalistic rules. (Yes, I know some people are on mobile, but they can just generate them themselves. [citation needed])
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
OK this is a very niche one that I hadn't remembered until a few hours ago.
You know in some arcades they give you this string of cardboard tickets you can redeem for stuff, usually meant for kids. The tickets fold beautifully perfectly packed if you order them one right, one left - zigzagging. When people fold them randomly in any direction giving a clearly low density packing with loads of strain, I just think
You know in some arcades they give you this string of cardboard tickets you can redeem for stuff, usually meant for kids.
The tickets fold beautifully perfectly packed if you order them one right, one left - zigzagging.
When people fold them randomly in any direction giving a clearly low density packing with loads of strain, I just think
omg why on Earth would you do that?!Surely they'd have realised by now? It's not that crazy to realise? Surely there is a clear preference for having them well packed; nobody would prefer an unwieldy mess?!
Also when I'm typing anything and I finish writing it and it just goes to the next line or just goes to the next page. Especially when the punctuation mark at the end brings the last word down one line. This also applies to writing in a notebook: I finish writing something but the very last thing goes to a new page.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: ... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature.
ON A DIFFERENT NOTE.
When i want to rotate a hexagonal file but golly refuses because for some reason it calculates hexagonal patterns on a square grid and that really bugs me because if you want to show that something has six sides you don't show it with four and it makes more sense to have the grid be changed to hexagonal but I understand Von Neumann because no shape exists (that I know of) that has 4 corners and no edges but COME ON WHY?! WHY DO YOU REPRESENT HEXAGONS WITH SQUARES?!
In all seriousness this bothers me and must be fixed or I will SINGLEHANDEDLY eat a universe.
EDIT: possibly this one.
EDIT 2:
IT HAS BEGUN.
HAS
BEGUN.
Last edited by 83bismuth38 on September 19th, 2017, 8:25 pm, edited 1 time in total.
Actually, I don't remember who I was referencing, but I don't think it was you, and if it was, it wasn't personal.83bismuth38 wrote:... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature.
x₁=ηx
V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$
http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: oh okay yeah of course sureA for awesome wrote:Actually, I don't remember who I was referencing, but I don't think it was you, and if it was, it wasn't personal.83bismuth38 wrote:... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature.
but really though, i wouldn't have cared.
When someone gives a presentation to a bunch of people and you
knowthat they're getting the facts wrong. Especially if this is during the Q&A section.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
When you watch a boring video in class but you understand it perfectly and then at the end your classmates dont get it so the teacher plays the borinh video again
Airy Clave White It Nay
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact:
when scientists decide to send a random guy into a black hole hovering directly above Earth for no reason at all.
hit; that random guy was me.
hit; that random guy was me.
83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact:
When I see a "one-step" organic reaction that occurs in an exercise book for senior high school and simply takes place under "certain circumstance" like the one marked "?" here but fail to figure out how it works even if I have prepared for our provincial chemistry olympiadEDIT: In fact it's not that hard.Just do a Darzens reaction then hydrolysis and decarboxylate.
Current status: outside the continent of cellular automata. Specifically, not on the plain of life.
An awesome gun firing cool spaceships:
An awesome gun firing cool spaceships:
Code: Select all
x = 3, y = 5, rule = B2kn3-ekq4i/S23ijkqr4eikry2bo$2o$o$obo$b2o!
When there's a rule with a decently common puffer but it can't interact with itself
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
-Terry Pratchett
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X
When that oscillator is just
When you're sooooooo close to a thing you consider amazing but miss... not sparky enough.
When you're sooooooo close to a thing you consider amazing but miss...
Airy Clave White It Nay
People posting tons of "new" discoveries that have been known for decades, showing that they've not observed standard netiquette by reading the forums a while before posting, nor done the most minimal research about whether things have been already known, despit repeated posts about where to find such resources (e.g. jslife, wiki, Life lexicon, etc.).
People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive"). Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades. Posts where the quoted text is substantially longer than added text. Especially "me too" posts. People whose signatures are longer than the actual text of their posts. People whose signatures include graphics or pattern files, especially ones that are just human-readable text. Improper grammar, spelling, and punctuation (although I've gotten used to that; long-term use of the internet has made me rather fluent in typo, both reading and writing). Imperfect English is not unreasonable from people for whom English is not a primary language, but from English speakers, it is a symptom of sloppiness that can also manifest in other areas.
People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive").
Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades.
Posts where the quoted text is substantially longer than added text. Especially "me too" posts.
People whose signatures are longer than the actual text of their posts.
People whose signatures include graphics or pattern files, especially ones that are just human-readable text.
Improper grammar, spelling, and punctuation (although I've gotten used to that; long-term use of the internet has made me rather fluent in typo, both reading and writing). Imperfect English is not unreasonable from people for whom English is not a primary language, but from English speakers, it is a symptom of sloppiness that can also manifest in other areas.
Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X That's G U S T A V O right theremniemiec wrote:People posting tons of "new" discoveries that have been known for decades, showing that they've not observed standard netiquette by reading the forums a while before posting, nor done the most minimal research about whether things have been already known, despit repeated posts about where to find such resources (e.g. jslife, wiki, Life lexicon, etc.). People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive"). Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades.
Also, when you walk into a wall slowly and carefully but you hit your teeth on the wall and it hurts so bad.
Airy Clave White It Nay |
Tagged: determinant of a matrix Problem 718
Let
\[ A= \begin{bmatrix} 8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2 \end{bmatrix} . \] Notice that $A$ contains every integer from $1$ to $9$ and that the sums of each row, column, and diagonal of $A$ are equal. Such a grid is sometimes called a magic square.
Compute the determinant of $A$.Add to solve later
Problem 686
In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not.
(a) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=0$.
Add to solve later
(b) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=-1$. Problem 582
A square matrix $A$ is called
nilpotent if some power of $A$ is the zero matrix. Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.
Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$.
Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample. Problem 571
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold). The time limit was 55 minutes.
This post is Part 2 and contains Problem 4, 5, and 6.
Check out Part 1 and Part 3 for the rest of the exam problems.
Problem 4. Let \[\mathbf{a}_1=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \mathbf{a}_2=\begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}, \mathbf{b}=\begin{bmatrix} 0 \\ a \\ 2 \end{bmatrix}.\]
Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.
Problem 5. Find the inverse matrix of \[A=\begin{bmatrix} 0 & 0 & 2 & 0 \\ 0 &1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}\] if it exists. If you think there is no inverse matrix of $A$, then give a reason. Problem 6. Consider the system of linear equations \begin{align*} 3x_1+2x_2&=1\\ 5x_1+3x_2&=2. \end{align*} (a) Find the coefficient matrix $A$ of the system. (b) Find the inverse matrix of the coefficient matrix $A$. (c) Using the inverse matrix of $A$, find the solution of the system.
(
Linear Algebra Midterm Exam 1, the Ohio State University) Read solution Problem 546
Let $A$ be an $n\times n$ matrix.
The $(i, j)$
cofactor $C_{ij}$ of $A$ is defined to be \[C_{ij}=(-1)^{ij}\det(M_{ij}),\] where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column.
Then consider the $n\times n$ matrix $C=(C_{ij})$, and define the $n\times n$ matrix $\Adj(A)=C^{\trans}$.
The matrix $\Adj(A)$ is called the adjoint matrix of $A$.
When $A$ is invertible, then its inverse can be obtained by the formula
For each of the following matrices, determine whether it is invertible, and if so, then find the invertible matrix using the above formula.
(a) $A=\begin{bmatrix} 1 & 5 & 2 \\ 0 &-1 &2 \\ 0 & 0 & 1 \end{bmatrix}$. (b) $B=\begin{bmatrix} 1 & 0 & 2 \\ 0 &1 &4 \\ 3 & 0 & 1 \end{bmatrix}$. Problem 509
Using the numbers appearing in
\[\pi=3.1415926535897932384626433832795028841971693993751058209749\dots\] we construct the matrix \[A=\begin{bmatrix} 3 & 14 &1592& 65358\\ 97932& 38462643& 38& 32\\ 7950& 2& 8841& 9716\\ 939937510& 5820& 974& 9 \end{bmatrix}.\]
Prove that the matrix $A$ is nonsingular.Add to solve later
Problem 505
Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix.
Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula: \[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]
Using the formula, calculate the inverse matrix of $\begin{bmatrix}
2 & 1\\ 1& 2 \end{bmatrix}$. Problem 486
Determine whether there exists a nonsingular matrix $A$ if
\[A^4=ABA^2+2A^3,\] where $B$ is the following matrix. \[B=\begin{bmatrix} -1 & 1 & -1 \\ 0 &-1 &0 \\ 2 & 1 & -4 \end{bmatrix}.\]
If such a nonsingular matrix $A$ exists, find the inverse matrix $A^{-1}$.
(
The Ohio State University, Linear Algebra Final Exam Problem) Read solution Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue Problem 419 (a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$.
Add to solve later
(b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. |
Difference between revisions of "State Feedback"
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{{chheader|Linear Systems|State Feedback|Output Feedback}}
{{chheader|Linear Systems|State Feedback|Output Feedback}}
−
This chapter describes how feedback can be used shape the local behavior of a system. The concept of reachability is introduced and used to investigate how to "design" the dynamics of a system through placement of its eigenvalues. In particular, it will be shown that under certain conditions it is possible to assign the system eigenvalues to arbitrary values by appropriate feedback of the system state.
+
This chapter describes how feedback can be used shape the local behavior of a system. The concept of reachability is introduced and used to investigate how to "design" the dynamics of a system through placement of its eigenvalues. In particular, it will be shown that under certain conditions it is possible to assign the system eigenvalues to arbitrary values by appropriate feedback of the system state.
{{chaptertable begin}}
{{chaptertable begin}}
{{chaptertable left}}
{{chaptertable left}}
== Textbook Contents ==
== Textbook Contents ==
−
{{am05pdf|
+
{{am05pdf|-statefbk||State Feedback}}
* 1. Reachability
* 1. Reachability
* 2. Stabilization by State Feedback
* 2. Stabilization by State Feedback
+
* 3. State Feedback Design Issues
* 3. State Feedback Design Issues
* 4. Integral Action
* 4. Integral Action
Line 20: Line 21:
== Supplemental Information ==
== Supplemental Information ==
* [[#Frequently Asked Questions|Frequently Asked Questions]]
* [[#Frequently Asked Questions|Frequently Asked Questions]]
+
* Wikipedia entries: [http://en.wikipedia.org/wiki/Controllability Controllability (reachability)], [http://en.wikipedia.org/wiki/State_space_%28controls%29#Feedback state feedback], [http://en.wikipedia.org/wiki/Optimal_control#Linear_quadratic_control LQR]
* Wikipedia entries: [http://en.wikipedia.org/wiki/Controllability Controllability (reachability)], [http://en.wikipedia.org/wiki/State_space_%28controls%29#Feedback state feedback], [http://en.wikipedia.org/wiki/Optimal_control#Linear_quadratic_control LQR]
* [[#Additional Information|Additional Information]]
* [[#Additional Information|Additional Information]]
Line 34: Line 36:
\end{aligned}
\end{aligned}
</amsmath></center>
</amsmath></center>
−
is said to be ''reachable'' if we can find an input <
+
is said to be ''reachable'' if we can find an input <>u(t)</> defined on the interval <>[0, T]</> that can steer the system from a given final point <>x(0) = x_0</> to a desired final point <>x(T) = x_f</>.
−
</p>
+
</p>
<li><p>The ''reachability matrix'' for a linear system is given by
<li><p>The ''reachability matrix'' for a linear system is given by
Line 41: Line 43:
W_r = \left[\begin{matrix} B & AB & \cdots & A^{n-1}B \end{matrix}\right].
W_r = \left[\begin{matrix} B & AB & \cdots & A^{n-1}B \end{matrix}\right].
</amsmath></center>
</amsmath></center>
−
A linear system is reachable if and only if the reachability matrix <
+
A linear system is reachable if and only if the reachability matrix <>W_r</> is invertible (assuming a single intput/single output system). Systems that are not reachable have states that are constrained to have a fixed relationship with each other.
−
</p>
+
</p>
<li><p>A linear system of the form
<li><p>A linear system of the form
Line 61: Line 63:
\det(sI-A) = s^n+a_1 s^{n-1} + \cdots + a_{n-1}s + a_n,
\det(sI-A) = s^n+a_1 s^{n-1} + \cdots + a_{n-1}s + a_n,
</amsmath></center>
</amsmath></center>
−
A reachable linear system can be transformed into reachable canonical form through the use of a coordinate transformation <
+
A reachable linear system can be transformed into reachable canonical form through the use of a coordinate transformation <>z = T x</>.
−
</p>
+
</p>
<li><p>A state feedback law has the form
<li><p>A state feedback law has the form
Line 68: Line 70:
u = -K x + k_r r
u = -K x + k_r r
</amsmath></center>
</amsmath></center>
−
where <
+
where <>r</> is the reference value for the output. The closed loop dynamics for the system are given by
<center><amsmath>
<center><amsmath>
\dot x = (A - B K) x + B k_r r.
\dot x = (A - B K) x + B k_r r.
</amsmath></center>
</amsmath></center>
−
The stability of the system is determined by the stability of the matrix <
+
The stability of the system is determined by the stability of the matrix <>A - BK</>. The equilibrium point and steady state output (assuming the systems is stable) are given by
<center><amsmath>
<center><amsmath>
x_e = -(A-BK)^{-1} B k_r r \qquad y_e = C x_e.
x_e = -(A-BK)^{-1} B k_r r \qquad y_e = C x_e.
</amsmath></center>
</amsmath></center>
−
Choosing <
+
Choosing <>k_r</> as
<center><amsmath>
<center><amsmath>
k_r = {-1}/\left(C (A-BK)^{-1} B\right).
k_r = {-1}/\left(C (A-BK)^{-1} B\right).
</amsmath></center>
</amsmath></center>
−
gives <
+
gives <>y_e = r</>.</p>
<li><p>If a system is reachable, then there exists a feedback law of the form
<li><p>If a system is reachable, then there exists a feedback law of the form
Line 87: Line 89:
</amsmath></center>
</amsmath></center>
the gives a closed loop system with an arbitrary characteristic polynomial. Hence the eigenvalues of a reachable linear system can be placed arbitrarily through the use of an appropriate feedback control law.
the gives a closed loop system with an arbitrary characteristic polynomial. Hence the eigenvalues of a reachable linear system can be placed arbitrarily through the use of an appropriate feedback control law.
−
</p>
+
</p>
−
<li><p>''Integral feedback'' can be used to provide zero steady state error instead of careful calibration of the gain <
+
<li><p>''Integral feedback'' can be used to provide zero steady state error instead of careful calibration of the gain <>K_r</>. An integral feedback controller has the form
<center><amsmath>
<center><amsmath>
u = - k_p (x - x_e) - k_i z + k_r r.
u = - k_p (x - x_e) - k_i z + k_r r.
</amsmath></center>
</amsmath></center>
where
where
−
<center><
+
<center><>
\dot z = y - r
\dot z = y - r
−
</
+
</></center>
−
is the integral error. The gains <
+
is the integral error. The gains <>k_p</>, <>k_i</> and <>k_r</> can be found by designing a stabilizing state feedback for the system dynamics augmented by the integrator dynamics.
−
</p>
+
</p>
<li><p>A ''linear quadratic regulator'' minimizes the cost function
<li><p>A ''linear quadratic regulator'' minimizes the cost function
Line 109: Line 111:
u = -Q_u^{-1} B^T P x.
u = -Q_u^{-1} B^T P x.
</amsmath></center>
</amsmath></center>
−
where <
+
where <>P \in R^{n \times n}</> is a positive definite, symmetric
matrix that satisfies the equation
matrix that satisfies the equation
<center><amsmath>
<center><amsmath>
P A + A^T P - P B Q_u^{-1} B^T P + Q_x = 0.
P A + A^T P - P B Q_u^{-1} B^T P + Q_x = 0.
</amsmath></center>
</amsmath></center>
−
This equation is called the''algebraic
+
This equation is called the ''algebraic
Riccati equation'' and can be solved numerically.
Riccati equation'' and can be solved numerically.
−
</p>
+
</p>
</ol>
</ol>
−
== Exercises ==
+ + + +
== Exercises ==
+
<ncl>State Feedback Exercises</ncl>
<ncl>State Feedback Exercises</ncl>
== Frequently Asked Questions ==
== Frequently Asked Questions ==
<ncl>State Feedback FAQ</ncl>
<ncl>State Feedback FAQ</ncl>
+ + + + + + + + + + + + + + + + + + +
== Additional Information ==
== Additional Information ==
+ + + Latest revision as of 04:06, 19 November 2012
Prev: Linear Systems Chapter 6 - State Feedback Next: Output Feedback
This chapter describes how feedback can be used to shape the local behavior of a system. The concept of reachability is introduced and used to investigate how to "design" the dynamics of a system through placement of its eigenvalues. In particular, it will be shown that under certain conditions it is possible to assign the system eigenvalues to arbitrary values by appropriate feedback of the system state.
Textbook Contents Lecture Materials Supplemental Information Chapter Summary
This chapter describes how state feedback can be used to design the (closed loop) dynamics of the system:
A linear system with dynamics
The
reachability matrixfor a linear system is given by
A linear system is reachable if and only if the reachability matrix is invertible (assuming a single intput/single output system). Systems that are not reachable have states that are constrained to have a fixed relationship with each other.
A linear system of the form
is said to be in
reachable canonical form. A system in this form is always reachable and has a characteristic polynomial given by
A state feedback law has the form
gives .
If a system is reachable, then there exists a feedback law of the form
the gives a closed loop system with an arbitrary characteristic polynomial. Hence the eigenvalues of a reachable linear system can be placed arbitrarily through the use of an appropriate feedback control law.
where
A
linear quadratic regulatorminimizes the cost function
The solution to the LQR problem is given by a linear control law of the form
This equation is called the
algebraicRiccati equation and can be solved numerically.
Additional Exercises
The following exercises cover some of the topics introduced in this chapter. Exercises marked with a * appear in the printed text.
Frequently Asked Questions
Errata
MATLAB code
The following MATLAB scripts are available for producing figures that appear in this chapter.
See the software page for more information on how to run these scripts.
Additional Information
More information on optimal control and the linear quadratic regulator can be found in the Optimization-Based Control supplement: |
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1. Homework Statement :A person aims a gun at a bird from a point at a horizontal distance of 100m. If the gun can impart a speed of 500m/s to the bullet , then above what height of the bird he should aim his gun to hit it ?2. Homework Equations :H_{max}=\frac{u^2\sin^2\theta}{2g}...
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Here is the numerical part with proper editing.Let A and C combine in the ratio \frac{a}{c1} and let B and C combine in the ratio \frac{b}{c2}=\frac{b*c1}{c1*c2}.(by mass). Then according to the law of reciprocal proportions, \frac{a*c2}{b*c1}=k \frac{a}{b} where k is whole number. If we...
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Background: Amount of substance is a fundamental physical quantity which has mole (mol) as it's SI unit. Therefore all expressions for amount of substance should have the unit mole on simplification.Therefore, if A = Amount of substance (normally in moles), m = Mass of the substance in a...
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Ocean dynamics define and describe the motion of water within the oceans. Ocean temperature and motion fields can be separated into three distinct layers: mixed (surface) layer, upper ocean (above the thermocline), and deep ocean.
Ocean dynamics has traditionally been investigated by sampling from instruments in situ.
[1]
The mixed layer is nearest to the surface and can vary in thickness from 10 to 500 meters. This layer has properties such as temperature, salinity and dissolved oxygen which are uniform with depth reflecting a history of active turbulence (the atmosphere has an analogous planetary boundary layer). Turbulence is high in the mixed layer. However, it becomes zero at the base of the mixed layer. Turbulence again increases below the base of the mixed layer due to shear instabilities. At extratropical latitudes this layer is deepest in late winter as a result of surface cooling and winter storms and quite shallow in summer. Its dynamics is governed by turbulent mixing as well as Ekman pumping, exchanges with the overlying atmosphere, and horizontal advection.
[2]
The upper ocean, characterized by warm temperatures and active motion, varies in depth from 100 m or less in the tropics and eastern oceans to in excess of 800 meters in the western subtropical oceans. This layer exchanges properties such as heat and freshwater with the atmosphere on timescales of a few years. Below the mixed layer the upper ocean is generally governed by the hydrostatic and geostrophic relationships.
[2] Exceptions include the deep tropics and coastal regions.
The deep ocean is both cold and dark with generally weak velocities (although limited areas of the deep ocean are known to have significant recirculations). The deep ocean is supplied with water from the upper ocean in only a few limited geographical regions: the subpolar North Atlantic and several sinking regions around the Antarctic. Because of the weak supply of water to the deep ocean the average residence time of water in the deep ocean is measured in hundreds of years. In this layer as well the hydrostatic and geostrophic relationships are generally valid and mixing is generally quite weak.
Contents Primitive equations 1 Mixed layer dynamics 2 Upper ocean dynamics 3 References 4 Primitive equations
Ocean dynamics are governed by Newton's equations of motion expressed as the Navier-Stokes equations for a fluid element located at (x,y,z) on the surface of our rotating planet and moving at velocity (u,v,w) relative to that surface:
the zonal momentum equation: \frac{Du}{Dt} = -\frac{1}{\rho} \frac{\partial p}{\partial x} + f v + \frac{1}{\rho} \frac{\partial \tau_x}{\partial z} the meridional momentum equation: \frac{Dv}{Dt} = -\frac{1}{\rho} \frac{\partial p}{\partial y} - f u + \frac{1}{\rho} \frac{\partial \tau_y}{\partial z} \frac{\partial p}{\partial z} = -\rho g \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0 \frac{\partial T}{\partial t} + u \frac{\partial T}{\partial x} + v \frac{\partial T}{\partial y} + w \frac{\partial T}{\partial z} = Q. [2] \frac{\partial S}{\partial t} + u \frac{\partial S}{\partial x} + v \frac{\partial S}{\partial y} + w \frac{\partial S}{\partial z} = (E-P)S(z=0). [2]
Here "u" is zonal velocity, "v" is meridional velocity, "w" is vertical velocity, "p" is pressure, "ρ" is density, "T" is temperature, "S" is salinity, "g" is acceleration due to gravity, "τ" is wind stress, and "f" is the Coriolis parameter. "Q" is the heat input to the ocean, while "P-E" is the freshwater input to the ocean.
Mixed layer dynamics
Mixed layer dynamics are quite complicated; however, in some regions some simplifications are possible. The wind-driven horizontal transport in the mixed layer is approximately described by Ekman Layer dynamics in which vertical diffusion of momentum balances the Coriolis effect and wind stress.
[3] This Ekman transport is superimposed on geostrophic flow associated with horizontal gradients of density. Upper ocean dynamics
Horizontal convergences and divergences within the mixed layer due, for example, to Ekman transport convergence imposes a requirement that ocean below the mixed layer must move fluid particles vertically. But one of the implications of the geostrophic relationship is that the magnitude of horizontal motion must greatly exceed the magnitude of vertical motion. Thus the weak vertical velocities associated with Ekman transport convergence (measured in meters per day) cause horizontal motion with speeds of 10 centimeters per second or more. The mathematical relationship between vertical and horizontal velocities can be derived by expressing the idea of conservation of angular momentum for a fluid on a rotating sphere. This relationship (with a couple of additional approximations) is known to oceanographers as the Sverdrup relation.
[3] Among its implications is the result that the horizontal convergence of Ekman transport observed to occur in the subtropical North Atlantic and Pacific forces southward flow throughout the interior of these two oceans. Western boundary currents (the Gulf Stream and Kuroshio) exist in order to return water to higher latitude. References
^ "Frontiers of Remote Sensing of the Oceans and Troposphere from Air and Space Platforms". Remote Sensing of Oceanography: Past, Present, and Future. NASA Technical Reports Server. Retrieved 22 September 2011. ^ a b c d DeCaria, Alex J., 2007: "Lesson 5 - Oceanic Boundary Layer." Personal Communication, Millersville University of Pennsylvania, Millersville, Pa. (Not a WP:RS) ^ a b Pickard, G.L. and W.J. Emery, 1990: Descriptive Physical Oceanography, Fifth Edition. Butterworth-Heinemann, 320 pp.
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This question already has an answer here:
I would like to read a very thorough and explained calculation process for a couple of integrals. For the life of me I just can't figure out the result on my own, and no resource on the web were able to help me.
First, a small question: Is a primitive of $\exp\left(\frac{-x^2}{2}\right)$, $-x\times\exp\left(\frac{-x^2}{2}\right)$ ? If so, what's the derivative ?
Now the real question. I would like to find the result of: $$\int \limits_{-\infty}^{+\infty}\exp\left(\dfrac{-x^2}{2}\right)\mathrm{d}x$$
And then, the result of: $$\int \limits_{-\infty}^{+\infty}x^2\times\exp\left(\dfrac{-x^2}{2}\right)\mathrm{d}x$$
Supposedly both are equal to $\sqrt{2\pi}$, but there's no way I can get there on my own. |
For any $z1, z2$ in $\mathbb{C} \setminus {0}$, $\log(z_1 z_2)=\log(z_1)+\log(z_2)$, but in general $\text{Log}(z_1 z_2)\ne \text{Log}(z_1)+\text{Log}(z_2)$.
Is $\log(z^2)=2\log(z)$?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
The answer to the question in the OP is that in general $\displaystyle \log(z^2)\ne 2\log(z)$.
This might seem paradoxical given the relationship expressed as
$$\log(z_1z_2)=\log(z_1)+\log(z_2) \tag1$$
But $(1)$ is interpreted as a set equivalence. It means that any value of $\log(z_1z_2)$ can be expressed as the sum of
value of $\log(z_1)$ and some
Note that $(1)$ is true since $\log(|z_1z_2|)=\log(|z_1|)+\log(|z_2|)$ and $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$.
EXAMPLE:
As an example, suppose that $z_1=z_2=-1$. Then, $z_1z_2=1$ and $\log(1)=i2n\pi$ for any integer $n$. For $n=0$, $\log(1) =0$. Then $(1)$ is certainly satisfied by $\log(z_1)=i\pi$ and $\log(z_2)=-i\pi$.
Note that although $z_1=z_2$ here, we needed to take two distinct values for $\log(z_1)$ and $\log(z_2)$. We were afforded that degree of freedom since we viewed $z_1$ and $z_2$ as independent.
In general,
$$\log(z^2)\ne 2\log(z) \tag 2$$
To see this, we note that for $z=re^{i\theta}$,
$$2\log(z)=2\log(r)+i2(\theta+2n\pi) \tag 3$$
while
$$\begin{align} \log(z^2)&=\log(r^2e^{i2\theta+i2n\pi})\\\\ &=2\log(r)+i2(\theta+n\pi)\tag 4 \end{align}$$
Comparing $(3)$ and $(4)$ we see that $\log(z^2)$ and $2\log(z)$ do not share the same set of values.
EXAMPLE:
As an example, suppose $z=i$. For the value $\log(i^2)=i3\pi$, there is no corresponding value of $2\log(z)$. Hence, $\log(i^2)\ne 2\log(i)$ in general.
If $p$ is a non-integer then $z^p$ is a complex multi-valued function and the principal value of $\ln z$ must lie in $- \pi < \Im \ln z< \pi$. From this it follows that \begin{align} \ln (z_1 z_2) &= \ln z_1 + \ln z_2 + 2 \pi i N_{+} \\ \ln \left( z_1 / z_2 \right) &= \ln z_1 - \ln z_2 + 2 \pi i N_{-}\\ \ln z^n &= n \ln z + 2 \pi i N_{n} \end{align} Where $N_{\pm} = 0, +1, -1$ and $$N_n = \frac{1}{2}+\left(\frac{n}{2 \pi}\right)\arg z$$ |
In a paper by Joos and Zeh, Z Phys B 59 (1985) 223, they say:This 'coming into being of classical properties' appears related to what Heisenberg may have meant by his famous remark [7]: 'Die "Bahn" entsteht erst dadurch, dass wir sie beobachten.'Google Translate says this means something ...
@EmilioPisanty Tough call. It's technical language, so you wouldn't expect every German speaker to be able to provide a correct interpretation—it calls for someone who know how German is used in talking about quantum mechanics.
Litmus are a London-based space rock band formed in 2000 by Martin (bass guitar/vocals), Simon (guitar/vocals) and Ben (drums), joined the following year by Andy Thompson (keyboards, 2001–2007) and Anton (synths). Matt Thompson joined on synth (2002–2004), while Marek replaced Ben in 2003. Oli Mayne (keyboards) joined in 2008, then left in 2010, along with Anton. As of November 2012 the line-up is Martin Litmus (bass/vocals), Simon Fiddler (guitar/vocals), Marek Bublik (drums) and James Hodkinson (keyboards/effects). They are influenced by mid-1970s Hawkwind and Black Sabbath, amongst others.They...
@JohnRennie Well, they repeatedly stressed their model is "trust work time" where there are no fixed hours you have to be there, but unless the rest of my team are night owls like I am I will have to adapt ;)
I think u can get a rough estimate, COVFEFE is 7 characters, probability of a 7-character length string being exactly that is $(1/26)^7\approx 1.2\times 10^{-10}$ so I guess you would have to type approx a billion characters to start getting a good chance that COVFEFE appears.
@ooolb Consider the hyperbolic space $H^n$ with the standard metric. Compute $$\inf\left\{\left(\int u^{2n/(n-2)}\right)^{-(n-2)/n}\left(4\frac{n-1}{n-2}\int|\nabla u|^2+\int Ru^2\right): u\in C^\infty_c\setminus\{0\}, u\ge0\right\}$$
@BalarkaSen sorry if you were in our discord you would know
@ooolb It's unlikely to be $-\infty$ since $H^n$ has bounded geometry so Sobolev embedding works as expected. Construct a metric that blows up near infinity (incomplete is probably necessary) so that the inf is in fact $-\infty$.
@Sid Eating glamorous and expensive food on a regular basis and not as a necessity would mean you're embracing consumer fetish and capitalism, yes. That doesn't inherently prevent you from being a communism, but it does have an ironic implication.
@Sid Eh. I think there's plenty of room between "I think capitalism is a detrimental regime and think we could be better" and "I hate capitalism and will never go near anything associated with it", yet the former is still conceivably communist.
Then we can end up with people arguing is favor "Communism" who distance themselves from, say the USSR and red China, and people who arguing in favor of "Capitalism" who distance themselves from, say the US and the Europe Union.
since I come from a rock n' roll background, the first thing is that I prefer a tonal continuity. I don't like beats as much as I like a riff or something atmospheric (that's mostly why I don't like a lot of rap)
I think I liked Madvillany because it had nonstandard rhyming styles and Madlib's composition
Why is the graviton spin 2, beyond hand-waiving, sense is, you do the gravitational waves thing of reducing $R_{00} = 0$ to $g^{\mu \nu} g_{\rho \sigma,\mu \nu} = 0$ for a weak gravitational field in harmonic coordinates, with solution $g_{\mu \nu} = \varepsilon_{\mu \nu} e^{ikx} + \varepsilon_{\mu \nu}^* e^{-ikx}$, then magic? |
In Asymptote, I'd like to draw two intersecting planes, one with a solid color and another with some pattern (e.g. checkerboard), so that it would look okay if printed black and white. However, the pattern module doesn't appear to work.
The code below is what I'm trying to get to work, with the statement that I wished drew that plane commented out (I understand of course that function doesn't exist, I just don't know which one to use).
settings.prc = false;defaultpen(fontsize(10pt));import three;import graph3;import patterns;size(11cm,0);real rot_param = 0.01;currentprojection=perspective((10000,-20000,100000),up=(-rot_param,1-rot_param,0));currentlight = (4,-1,4);render render = render(compression=Low, merge = true);limits((0,0,0), (3,3,3));real delta = 0.75;real dterm = (1-delta)/delta;draw(surface(O -- (dterm,0,0) -- (3,3-dterm,0) -- (3,3,0) -- cycle), orange+opacity(0.5), render);draw(surface(O -- (3,3,0) -- (0,3,0) -- cycle), orange+opacity(0.5), render);draw(surface(O -- (3,3,0) -- (3,3,3) -- (0,0,3) -- cycle), lightred+opacity(0.5), render);draw(surface((dterm,0,0) -- (3,3-dterm,0) -- (3,3-dterm,3) -- (dterm,0,3) -- cycle), lightblue+opacity(0.5), render);// Does not work!//filldraw(surface((dterm,0,0) -- (3,3-dterm,0) -- (3,3-dterm,3) -- (dterm,0,3) -- cycle), pattern("checker"));xaxis3("",Bounds,black+dashed,InTicks(3,1));yaxis3("",Bounds,black+dashed,InTicks(3,1));//zaxis3("",Bounds,black,InTicks(3,1));draw(O -- 3.5X,arrow=Arrow3);label("$\gamma(C)$",3.7X);draw(O -- 4Y,dashed, arrow=Arrow3);label("$\gamma(D)$",4.2Y);draw(O -- 4Z,arrow=Arrow3);label("$\phi(C,\theta_0)$",(-0.2,0,5));label("1",(-0.1,0,3));draw(box((0,0,3),(3,3,3)), black+linewidth(0.6pt));draw((0,3,0) -- (0,3,3), dashed);draw((3,3,0) -- (3,3,3));draw((3,0,0) -- (3,0,3));
This is how it looks (I'd like to make at least the blue plane checkerboard):
I would be very grateful if anyone knows how to do this. In case you can't tell I have some other problems too, so I'll just throw these out there if anyone happens to know:
I'm using perspective but I'd like to use the "oblique" perspective except from "below", so that the axis pointing "out of the screen" ($\phi(C,\theta_0)$ ) is actually pointing up and to the left a bit. I've tried to approximate it as you can see. Is there a way to do this? How do I get the last "3"s to show up as labels on the axis? I don't want the vertical axis to scale (I want it bigger than it appears, so I'm just relabeling, e.g. the 1). Is there a way to do this automatically? |
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Search for top squark pair production in final states with one isolated lepton, jets, and missing transverse momentum in √s = 8 TeV pp collisions with the ATLAS detector
(Springer, 2014-11)
The results of a search for top squark (stop) pair production in final states with one isolated lepton, jets, and missing transverse momentum are reported. The analysis is performed with proton-proton collision data at s√ ...
Search for supersymmetry in events with large missing transverse momentum, jets, and at least one tau lepton in 20 fb−1 of √s = 8 TeV proton-proton collision data with the ATLAS detector
(Springer, 2014-09-18)
A search for supersymmetry (SUSY) in events with large missing transverse momentum, jets, at least one hadronically decaying tau lepton and zero or one additional light leptons (electron/muon), has been performed using ...
Measurement of the top quark pair production charge asymmetry in proton-proton collisions at √s = 7 TeV using the ATLAS detector
(Springer, 2014-02)
This paper presents a measurement of the top quark pair ( tt¯ ) production charge asymmetry A C using 4.7 fb−1 of proton-proton collisions at a centre-of-mass energy s√ = 7 TeV collected by the ATLAS detector at the LHC. ...
Measurement of the low-mass Drell-Yan differential cross section at √s = 7 TeV using the ATLAS detector
(Springer, 2014-06)
The differential cross section for the process Z/γ ∗ → ℓℓ (ℓ = e, μ) as a function of dilepton invariant mass is measured in pp collisions at √s = 7 TeV at the LHC using the ATLAS detector. The measurement is performed in ...
Measurements of fiducial and differential cross sections for Higgs boson production in the diphoton decay channel at √s=8 TeV with ATLAS
(Springer, 2014-09-19)
Measurements of fiducial and differential cross sections are presented for Higgs boson production in proton-proton collisions at a centre-of-mass energy of s√=8 TeV. The analysis is performed in the H → γγ decay channel ...
Measurement of the inclusive jet cross-section in proton-proton collisions at \( \sqrt{s}=7 \) TeV using 4.5 fb−1 of data with the ATLAS detector
(Springer, 2015-02-24)
The inclusive jet cross-section is measured in proton-proton collisions at a centre-of-mass energy of 7 TeV using a data set corresponding to an integrated luminosity of 4.5 fb−1 collected with the ATLAS detector at the ...
ATLAS search for new phenomena in dijet mass and angular distributions using pp collisions at $\sqrt{s}$=7 TeV
(Springer, 2013-01)
Mass and angular distributions of dijets produced in LHC proton-proton collisions at a centre-of-mass energy $\sqrt{s}$=7 TeV have been studied with the ATLAS detector using the full 2011 data set with an integrated ...
Search for direct chargino production in anomaly-mediated supersymmetry breaking models based on a disappearing-track signature in pp collisions at $\sqrt{s}$=7 TeV with the ATLAS detector
(Springer, 2013-01)
A search for direct chargino production in anomaly-mediated supersymmetry breaking scenarios is performed in pp collisions at $\sqrt{s}$ = 7 TeV using 4.7 fb$^{-1}$ of data collected with the ATLAS experiment at the LHC. ...
Search for heavy lepton resonances decaying to a $Z$ boson and a lepton in $pp$ collisions at $\sqrt{s}=8$ TeV with the ATLAS detector
(Springer, 2015-09)
A search for heavy leptons decaying to a $Z$ boson and an electron or a muon is presented. The search is based on $pp$ collision data taken at $\sqrt{s}=8$ TeV by the ATLAS experiment at the CERN Large Hadron Collider, ...
Evidence for the Higgs-boson Yukawa coupling to tau leptons with the ATLAS detector
(Springer, 2015-04-21)
Results of a search for $H \to \tau \tau$ decays are presented, based on the full set of proton--proton collision data recorded by the ATLAS experiment at the LHC during 2011 and 2012. The data correspond to integrated ... |
Possible Duplicate: Spacing around \left and \right
I'm having some problems with writing down an equation in LaTeX. It's only a minor issue regarding appearance, but I'm kind of a perfectionist. And isn't that why we use LaTeX anyway?
Consider the following equation:
\begin{equation}f(x)=\exp[-\frac{x}{2}]\end{equation}
The brackets following the exponential indicator
\exp are closely placed next to "exp", clearly indicating that it's the variables for the exponential function. However, since the argument of the exponential function contains a fraction, it's appropriate to resize the brackets. So I let the brackets be resized automatically according to:
\begin{equation}f(x)=\exp\left[-\frac{x}{2}\right]\end{equation}
Indeed, the brackets are now automatically appropriately sized. However, for some reason, extra white space appears between "exp" and the first "[" bracket. It now looks like the function "exp" and the term in brackets are two independent components.
I haven't found a way to solve this issue. I'm quite surprised that a system like LaTeX, known for its excellent equation capabilities, exhibits this behavior. |
Cadabra
Computer algebra system for field theory problems
Python wrapper around the C++ core module, making the entire functionality of Cadabra accessible from Python. This consists of two parts. One is core/cadabra2_defaults.py, which is a pure-python file. It contains the display function to show Cadabra objects to the user in a UI-independent way (e.g. switching between a text representation on the command line and a LaTeX representation when running in a notebook). The other part are the various files in core/pythoncdb, which build the cadabra2 module for python, to be imported with 'from cadabra2 import *'.
file cadabra2_defaults.py
void pull_in (std::shared_ptr< cadabra::Ex > ex, cadabra::Kernel *) template<class Algo > Ex_ptr cadabra::apply_algo_base (Algo &algo, Ex_ptr ex, bool deep, bool repeat, unsigned int depth, bool pre_order=false) template<class Algo , typename... Args, typename... PyArgs> void cadabra::def_algo (pybind11::module &m, const char *name, bool deep, bool repeat, unsigned int depth, PyArgs... pyargs) template<class Algo , typename... Args, typename... PyArgs> void cadabra::def_algo_preorder (pybind11::module &m, const char *name, bool deep, bool repeat, unsigned int depth, PyArgs... pyargs) bool cadabra::Ex_compare (Ex_ptr one, Ex_ptr other) Ex_ptr cadabra::Ex_add (const Ex_ptr ex1, const ExNode ex2) Ex_ptr cadabra::Ex_mul (const Ex_ptr ex1, const Ex_ptr ex2) Ex_ptr cadabra::Ex_sub (const Ex_ptr ex1, const ExNode ex2) Ex_ptr cadabra::fetch_from_python (const std::string &nm) std::string cadabra::Ex_as_str (Ex_ptr ex) std::string cadabra::Ex_as_latex (Ex_ptr ex) Kernel * cadabra::create_scope () void cadabra::inject_defaults (Kernel *k) Kernel * cadabra::get_kernel_from_scope ()
Ex_ptr cadabra::apply_algo_base ( Algo & algo, Ex_ptr ex, bool deep, bool repeat, unsigned int depth, bool pre_order =
false
)
Generic internal entry point for the Python side to execute a C++ algorithm. This gets called by the various apply_algo functions below, which in turn get called by the def_algo functions.
Kernel * cadabra::create_scope ( )
Setup of kernels in current scope, callable from Python. When the decision was made to graft Cadabra onto Python, a choice had to be made about how Python variable scope would influence the visibility of Cadabra properties. It clearly makes sense to be able to declare properties which only hold inside a particular function. However Cadabra expressions and properties do not directly correspond to Python objects. Rather, declaring a property is more like a function call into the Cadabra module, which leaves its imprint on the state of the C++ part but does not change anything on the Python side, as you typically do not assign the created property to a Python symbol. Therefore, properties do not naturally inherit Python's scoping rules.\footnote{This is different from e.g.~SymPy, in which mathematical objects are always in one-to-one correspondence with a Python object.} A more fundamental problem is that properties can be attached to patterns, and those patterns can involve more than just the symbols which one passes into a function.
In order to not burden the user, properties are therefore by default global variables, stored in a single global Cadabra object \verb|__cdbkernel__| which is initialised at import of the Cadabra module. If you add new properties inside a function scope, these will go into this already existing \emph{global} property list by default. If you want to create a local scope for your computations, create a new \verb|__cdbkernel__| as in \begin{verbatim} def fun():
cdbkernel = cadabra.create_scope(); [your code here] \end{verbatim} Now computations will not see the global properties at all. If you want to import the global properties, use instead \begin{verbatim} def fun(): cdbkernel = cadabra.create_scope_from_global() [your code here] \end{verbatim} It is crucial that the \verb|__cdbkernel__| symbol is referenced from within Python and visible to the bytecompiler, because it is not possible to create new variables on the local stack at runtime. Internally, the second version above fetches, at runtime, the \verb|__cdbkernel__| from the globals stack, copies all properties in there into a new kernel, and returns the latter.
Both versions above do populate the newly created kernel with Cadabra's default properties. If you want a completely clean slate (for e.g.~testing purposes, or because you really do not want default rules for sums and products), use \begin{verbatim} def fun():
cdbkernel = cadabra.create_empty_scope() [your code here] \end{verbatim} Note that in all these cases, changes to properties remain local and do not leak into the global property list.
All Cadabra algorithms, when called from Python, will first look for a kernel on the locals stack (i.e.~what \verb|locals()| produces). If there is no kernel available locally, they will then revert to using the global kernel.
void cadabra::def_algo ( pybind11::module & m, const char * name, bool deep, bool repeat, unsigned int depth, PyArgs... pyargs )
Method to declare a Python function with variable number of arguments, and make that call a C++ algorithm as specified in the Algo template parameter. This will make the algorithm traverse post-order, that is to say, first on the innermost child (or leaf) of an expression tree, and then, if that fails, on parent nodes, and so on.
void cadabra::def_algo_preorder ( pybind11::module & m, const char * name, bool deep, bool repeat, unsigned int depth, PyArgs... pyargs )
Method to declare a Python function with variable number of arguments, and make that call a C++ algorithm as specified in the Algo template parameter. In contrast to def_algo, this one will apply the algorithm in pre-order traversal style, that is, it will first attempt to apply on a node itself before traversing further down the child nodes and attempting there.
Add two expressions, adding a top-level \sum node if required.
std::string cadabra::Ex_as_latex ( Ex_ptr )
The Python 'print' function always calls the 'str' member on objects to be printed. This one is required to produce output which looks readable but is also still valid input. In order to produce proper LaTeX output, this is therefore not the right function to use, because Cadabra only reads a restricted subset of LaTeX (for instance, we output spacing commands like '\,' but do not accept it on input). So we have a separate
latex() member on each object, which internally uses DisplayTeX to do the actual printing.
std::string cadabra::Ex_as_str ( Ex_ptr )
Generate the Python str() and repr() representation of the Ex object.
Multiply two expressions, adding a top-level \prod node if required.
Subtract two expressions, adding a top-level \sum node if required.
Ex_ptr cadabra::fetch_from_python ( const std::string & nm )
Fetch an Ex object from the Python side using its Python identifier.
Kernel * cadabra::get_kernel_from_scope ( )
Get a pointer to the currently visible kernel.
void cadabra::inject_defaults ( Kernel * )
Inject properties directly into the Kernel, even if the kernel is not yet on the Python stack (needed when we create a new local scope: in this case we create the kernel and pass it back to be turned into local
cdbkernel by Python, but we want to populate the kernel with defaults before we hand it back). |
Electrochemical Impedance Spectroscopy: Experiment, Model, and App
Electrochemical impedance spectroscopy is a versatile experimental technique that provides information about an electrochemical cell’s different physical and chemical phenomena. By modeling the physical processes involved, we can constructively interpret the experiment’s results and assess the magnitudes of the physical quantities controlling the cell. We can then turn this model into an app, making electrochemical modeling accessible to more researchers and engineers. Here, we will look at three different ways of analyzing EIS: experiment, model, and simulation app.
Electrochemical Impedance Spectroscopy: The Experiment
Electrochemical impedance spectroscopy (EIS) is a widely used experimental method in electrochemistry, with applications such as electrochemical sensing and the study of batteries and fuel cells. This technique works by first polarizing the cell at a fixed voltage and then applying a small additional voltage (or occasionally, a current) to perturb the system. The perturbing input oscillates harmonically in time to create an alternating current, as shown in the figure below.
An oscillating perturbation in cell voltage gives an oscillating current response.
For a certain amplitude and frequency of applied voltage, the electrochemical cell responds with a particular amplitude of alternating current at the same frequency. In real systems, the response may be complicated for components of other frequencies too — we’ll return to this point below.
EIS experiments typically vary the frequency of the applied perturbation across a range of mHz and kHz. The relative amplitude of the response and time shift (or phase shift) between the input and output signals change with the applied frequency.
These factors depend on the rates at which physical processes in the electrochemical cell respond to the oscillating stimulus. Different frequencies are able to separate different processes that have different timescales. At lower frequencies, there is time for diffusion or slow electrochemical reactions to proceed in response to the alternating polarization of the cell. At higher frequencies, the applied field changes direction faster than the chemistry responds, so the response is dominated by capacitance from the charge and discharge of the double layer.
The time-domain response is not the simplest or most succinct way to interpret these frequency-dependent amplitudes and phase shifts. Instead, we define a quantity called an
impedance. Like resistance in a static system, impedance is the ratio of voltage to current. However, it uses the real and imaginary parts of a complex number to represent the relation of both amplitude and phase to the input signal and output response. The mathematical tool that relates the impedance to the time-domain response is a Fourier transform, which represents the frequency components of the oscillating signal.
To explain the idea of impedance more fully for a simple case, consider the input voltage as a cosine wave oscillating at an angular frequency (
ω):
Then the response is also a cosine wave, but with a phase offset (
φ). Compared to the time shift in the image above, the phase offset is given as \phi = -\omega \,\delta t . The magnitude of the current and its phase offset depend on the physics and chemistry in the cell.
Now, let’s consider the resistance from Ohm’s law:
This quantity varies in time with the same frequency as the perturbing signal. It equals zero at times when the numerator also equals zero and becomes singular when the denominator equals zero. So unlike the resistance in a DC system, it’s not a very useful quantity!
Instead, from Euler’s theorem, let’s express the time-varying quantities as the real parts of complex exponentials, so that:
and
We denote the coefficients V_0 and I_0\,\exp(i\phi) as quantities \bar{V} and \bar{I}, respectively.
These are complex amplitudes that can be understood in terms of the Fourier transformation of the original time-domain sinusoidal signals. They express the distinct amplitudes and phase difference of the voltage and current. Because all of the quantities in the system are oscillating sinusoidally, we understand the physical effects by comparing these complex quantities, rather than the time-domain quantities. To describe the oscillating problem (often called
phasor theory), we define a complex analogue of resistance as:
This is the impedance of the system and, as the name suggests, it’s the quantity we measure in electrochemical impedance spectroscopy. It’s a complex quantity with a magnitude and phase, representing both resistive and capacitive effects. Resistance contributes the real part of the complex impedance, which is in-phase with the applied voltage, while capacitance contributes the imaginary part of the complex impedance, which is precisely out-of-phase with the applied voltage.
EIS specialists look at the impedance in the form of a spectrum, normally with a Nyquist plot. This plots the imaginary component of impedance against the real component, with one data point for every frequency at which the impedance has been measured. Below is an example from a simulation — we’ll discuss how it’s modeled in the next section.
Simulated Nyquist plot from an electrochemical impedance spectroscopy experiment. Points toward the top right are at lower frequencies (mHz), while those toward the bottom left are at higher frequencies (>100 Hz).
In the figure above, the semicircular region toward the left side shows the coupling between double-layer capacitance and electrode kinetic effects at frequencies faster than the physical process of diffusion. The diagonal “diffusive tail” on the right comes from diffusion effects observed at lower frequencies.
EIS experiments are useful because information about many different physical effects can be extracted from a single analysis. There is a quantitative relationship between properties like diffusion coefficients, kinetic rate constants, and dimensions of the features in Nyquist plots. Often, EIS experiments are interpreted using an “equivalent circuit” of resistors and capacitors that yields a similar frequency-dependent impedance to the one shown in the Nyquist plot above. This idea was discussed in my colleague Scott’s blog post on electrochemical resistances and capacitances.
When there is a linear relation between the voltage and current, only one frequency will appear in the Fourier transform. This simplifies the analysis significantly.
For the simple harmonic interpretation of the experiment in terms of impedance, we need the current response to oscillate at the same frequency as the voltage input. This means that the system must respond linearly. For an electrochemical cell, we can usually accomplish this by ensuring that the applied voltage is small compared to the quantity
RT/F — the ratio of the gas constant multiplied by the temperature to the Faraday constant. This is the characteristic “thermal voltage” in electrochemistry and is about 25 mV at normal temperatures. Smaller voltage changes usually induce a linear response, while larger voltage changes cause an appreciably nonlinear response.
Of course, with simulation to predict the time-domain current, we can always consider a nonlinear case and perform a Fourier transform numerically to study the effect on the impedance. In practice, the interpretation in terms of impedance illustrated above is best suited to the harmonic assumption. Impedance measurements are therefore often used in a complementary manner with transient techniques, such as amperometry or voltammetry, which are better suited for investigating nonlinear or hysteretic effects.
Let’s look at a simple example of the physical theory that underpins these ideas to see how the impedance spectrum relates to the real controlling physics.
Electrochemical Impedance Spectroscopy: The Model
To model an EIS experiment, we must describe the key underlying physical and chemical effects, which are the electrode kinetics, double-layer capacitance, and diffusion of the electrochemical reactants. In electroanalytical systems, a large quantity of artificially added supporting electrolytes keeps the electric field low so that solution resistance can be neglected. In this case, we can describe the mass transport of chemical species in the system using the diffusion equation (Fick’s laws) with suitable boundary conditions for the electrode kinetics and capacitance. In the COMSOL Multiphysics® software, we use the
Electroanalysis interface together with an Electrode Surface boundary feature to describe these equations.
For more details about how to set up this model, you can download the Electrochemical Impedance Spectroscopy tutorial example in the Application Library.
Model tree for the Electroanalysis interface in an EIS model.
Under
Transport Properties, we can specify the diffusion coefficients of the redox species under consideration. We at least need the reduced and oxidized species for a single redox couple, such as the common redox couple ferro/ferricyanide, to use as an analytical reference. The Concentration boundary condition defines the fixed bulk concentrations of these species. The Electrode Reaction and Double Layer Capacitance subnodes for the Electrode Surface boundary feature contribute Faradaic and non-Faradaic current, respectively. For the double-layer capacitance, we typically use an empirically measured equivalent capacitance and specify the electrode reaction according to a standard kinetic equation like the Butler-Volmer equation.
Note that we’re not referring to equivalent circuit properties at all here. In COMSOL Multiphysics, all of the inputs in the description of the electrochemical problem are physical or chemical quantities, while the output is a Nyquist plot. When analyzing the problem in reverse, we’re able to use an observed Nyquist plot from our experiments to make inferences about the real values of these physical and chemical inputs.
In the settings for the Electrode Surface feature, we represent the impedance experiment by applying a
Harmonic Perturbation to the cell voltage. Settings for the Electrode Surface boundary feature in an EIS model.
Here, the quantity
V_app is the applied voltage.
The harmonic perturbation is applied with respect to a resting steady voltage (or current) on the cell. In this case, we have set this to a reference value of zero volts. With more advanced models, we might consider using the results of another COMSOL Multiphysics model, one that’s significantly nonlinear for example, to find the resting conditions to which the perturbation is applied. If you’re interested in understanding the mathematics of the harmonic perturbation in greater detail, my colleague Walter discussed them in a previous blog post.
When studying lithium-ion batteries, for example, we can perform a time-dependent analysis of the cell’s discharge, studying its charge transport, diffusion and migration of the lithium electrolyte, and the electrode kinetics and diffusion of the intercalated lithium atoms. We can pause this simulation at various times to consider the impedance measured from a rapid perturbation. For further insight into the physics involved, you can read my colleague Tommy’s blog post on modeling electrochemical impedance in a lithium-ion battery.
Electrochemical Impedance Spectroscopy: The Simulation App
A frequent demand for electrochemical simulations is that they “fit” experimental data in order to determine unknown physical quantities or, more generally, to interpret the data at all. Even for experienced electroanalytical chemists, it can be difficult to intuitively “see” the physics and chemistry in the underlying graphs like the Nyquist plot. However, by simulating the plots under a range of conditions, the influence of different effects on the overall graph is revealed.
Simulation is helpful for analyzing EIS, but it can also be time consuming for the experts involved. As was the case with my old research group, these experts can spend more time writing programs and running models to fit data together with experimental researchers than on the science. Wouldn’t it be nice if all electrochemical researchers could load experimental data into a simple interface, simulate impedance spectra for a given physical model and inputs, and even perform automatic parameter fitting? The good news is that we can! With the Application Builder in COMSOL Multiphysics, we can create an easy-to-use EIS app based on an underlying model. As a model can contain any level of physical detail, the app provides direct access to the physical data and isn’t confined to simple equivalent circuits.
To highlight this, we have an EIS demo app based on the model available in the Application Library. The app user can set concentrations for electroactive species and tune the diffusion coefficients as well as the electrode kinetic rate constant and double-layer capacitance. After clicking the
Compute button, the app generates results that can be visualized through Nyquist and Bode plots. The EIS simulation app in action.
As well as enabling physical parameter estimation, this app is very helpful for teaching, since we can quickly change inputs and visualize the results that would occur in the experiment. A natural extension for the app is to import experimental data to the same Nyquist plot for direct comparison. We can also build up the underlying physical model to consider the influence of competing electrochemical reactions or follow-up homogeneous chemistry from the products of an electrochemical reaction.
Concluding Thoughts
Here, we’ve introduced electrochemical impedance spectroscopy and discussed some methods used to model it. We also saw how a simulation app built from a simple theoretical model can provide greater insight into the relationship between the theory of an electrochemical system and its behavior as observed in an experiment.
Further Reading Explore other topics related to electrochemical simulation on the COMSOL Blog Comentários (1) CATEGORIAS Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
L # 1
Show that
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Last edited by krassi_holmz (2006-03-09 02:44:53)
IPBLE: Increasing Performance By Lowering Expectations.
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It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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L # 2
If
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Let
log x = x' log y = y' log z = z'. Then:
x'+y'+z'=0.
Rewriting in terms of x' gives:
IPBLE: Increasing Performance By Lowering Expectations.
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Well done, krassi_holmz!
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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L # 3
If x²y³=a and log (x/y)=b, then what is the value of (logx)/(logy)?
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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loga=2logx+3logy
b=logx-logy loga+3b=5logx loga-2b=3logy+2logy=5logy logx/logy=(loga+3b)/(loga-2b). Last edited by krassi_holmz (2006-03-10 20:06:29)
IPBLE: Increasing Performance By Lowering Expectations.
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Very well done, krassi_holmz!
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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L # 4
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It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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You are not supposed to use a calculator or log tables for L # 4. Try again!
Last edited by JaneFairfax (2009-01-04 23:40:20)
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No, I didn't
I remember
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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You still used a calculator / log table in the past to get those figures (or someone else did and showed them to you). I say again:
no calculators or log tables to be used (directly or indirectly) at all!! Last edited by JaneFairfax (2009-01-06 00:30:04)
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log a = 2log x + 3log y
b = log x log y
log a + 3 b = 5log x
loga - 2b = 3logy + 2logy = 5logy
logx / logy = (loga+3b) / (loga-2b)
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Hi ganesh
for L # 1 since log(a)= 1 / log(b), log(a)=1 b a a we have 1/log(abc)+1/log(abc)+1/log(abc)= a b c log(a)+log(b)+log(c)= log(abc)=1 abc abc abc abc Best Regards Riad Zaidan
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Hi ganesh
for L # 2 I think that the following proof is easier: Assume Log(x)/(b-c)=Log(y)/(c-a)=Log(z)/(a-b)=t So Log(x)=t(b-c),Log(y)=t(c-a) , Log(z)=t(a-b) So Log(x)+Log(y)+Log(z)=tb-tc+tc-ta+ta-tb=0 So Log(xyz)=0 so xyz=1 Q.E.D Best Regards Riad Zaidan
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Gentleman,
Thanks for the proofs.
Regards.
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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log_2(16) = \log_2 \left ( \frac{64}{4} \right ) = \log_2(64) - \log_2(4) = 6 - 2 = 4, \,
log_2(\sqrt[3]4) = \frac {1}{3} \log_2 (4) = \frac {2}{3}. \,
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L # 4
I don't want a method that will rely on defining certain functions, taking derivatives,
noting concavity, etc.
Change of base:
Each side is positive, and multiplying by the positive denominator
keeps whatever direction of the alleged inequality the same direction:
On the right-hand side, the first factor is equal to a positive number less than 1,
while the second factor is equal to a positive number greater than 1. These facts are by inspection combined with the nature of exponents/logarithms.
Because of (log A)B = B(log A) = log(A^B), I may turn this into:
I need to show that
Then
Then 1 (on the left-hand side) will be greater than the value on the
right-hand side, and the truth of the original inequality will be established.
I want to show
Raise a base of 3 to each side:
Each side is positive, and I can square each side:
-----------------------------------------------------------------------------------
Then I want to show that when 2 is raised to a number equal to
(or less than) 1.5, then it is less than 3.
Each side is positive, and I can square each side:
Last edited by reconsideryouranswer (2011-05-27 20:05:01)
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Hi reconsideryouranswer,
This problem was posted by JaneFairfax. I think it would be appropriate she verify the solution.
It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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Hi all,
I saw this post today and saw the probs on log. Well, they are not bad, they are good. But you can also try these problems here by me (Credit: to a book):
http://www.mathisfunforum.com/viewtopic … 93#p399193
Practice makes a man perfect.
There is no substitute to hard work All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam
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JaneFairfax, here is a basic proof of L4:
For all real a > 1, y = a^x is a strictly increasing function.
log(base 2)3 versus log(base 3)5
2*log(base 2)3 versus 2*log(base 3)5
log(base 2)9 versus log(base 3)25
2^3 = 8 < 9
2^(> 3) = 9
3^3 = 27 < 25
3^(< 3) = 25
So, the left-hand side is greater than the right-hand side, because
Its logarithm is a larger number.
Offline |
The Annals of Statistics Ann. Statist. Volume 46, Number 5 (2018), 2125-2152. Estimation of a monotone density in $s$-sample biased sampling models Abstract
We study the nonparametric estimation of a decreasing density function $g_{0}$ in a general $s$-sample biased sampling model with weight (or bias) functions $w_{i}$ for $i=1,\ldots,s$. The determination of the monotone maximum likelihood estimator $\hat{g}_{n}$ and its asymptotic distribution, except for the case when $s=1$, has been long missing in the literature due to certain nonstandard structures of the likelihood function, such as nonseparability and a lack of strictly positive second order derivatives of the negative of the log-likelihood function. The existence, uniqueness, self-characterization, consistency of $\hat{g}_{n}$ and its asymptotic distribution at a fixed point are established in this article. To overcome the barriers caused by nonstandard likelihood structures, for instance, we show the tightness of $\hat{g}_{n}$ via a purely analytic argument instead of an intrinsic geometric one and propose an indirect approach to attain the $\sqrt{n}$-rate of convergence of the linear functional $\int w_{i}\hat{g}_{n}$.
Article information Source Ann. Statist., Volume 46, Number 5 (2018), 2125-2152. Dates Received: February 2016 Revised: May 2017 First available in Project Euclid: 17 August 2018 Permanent link to this document https://projecteuclid.org/euclid.aos/1534492831 Digital Object Identifier doi:10.1214/17-AOS1614 Mathematical Reviews number (MathSciNet) MR3845013 Zentralblatt MATH identifier 06964328 Keywords Density estimation empirical process theory Karush–Kuhn–Tucker conditions nonparametric estimation order statistics from multiple samples self-induced characterization shape-constrained problem $s$-sample biased sampling Citation
Chan, Kwun Chuen Gary; Ling, Hok Kan; Sit, Tony; Yam, Sheung Chi Phillip. Estimation of a monotone density in $s$-sample biased sampling models. Ann. Statist. 46 (2018), no. 5, 2125--2152. doi:10.1214/17-AOS1614. https://projecteuclid.org/euclid.aos/1534492831
Supplemental materials Supplement to “Estimation of a monotone density in $s$-sample biased sampling models”. In the supplementary paper, we provide the proofs for Propositions 3.1, 3.2, 4.1 and 5.13, Lemmas 5.1, 5.2, 5.4, 5.5, 5.9, 5.11, 5.12, 5.14, 5.15, 6.1, 6.2, 6.3, 6.4 and 6.5, Theorems 1.1 and 6.6. In addition, we also state and prove the fact that the function $\mathcal{\tilde{L}}_{n}$ defined in (3.3) is concave in $\boldsymbol{p}$ in Proposition 8.1, and hence establishes the unique existence of $\hat{g}_{n}$ in Proposition 8.2. |
As mentioned in other answers, hydrostatic pressure with no flow remains the same.Now, to estimate changes in pressure head with flow, you can use the
Darcy-Weisbach-equation:
\begin{equation} (1)\qquad\Delta z_E=\lambda\frac{L}{D}\frac{v^2}{2g} \end{equation}$L$ is the length of the pipe, $D$ its diameter, $v$ the flow velocity within the pipe, $g$ the gravitational acceleration and $\lambda$ the friction coefficient.Note, that $v$ is needed to solve this equation, thus this usually is an iterative process, where you calculate the velocity without any losses, then calculate the losses with
Darcy-Weisbach and so forth...
There is an empiric formula for $\lambda$, but its quite complicated and on an iterative basis as well. I think it will suffice to analyse the equations qualitatively. As you can see in equation (1), losses in pressure head $\Delta z_E$ due to friction are reduced, with an increasing pipe diameter.The same is generally true for the friction coefficient $\lambda$ as well.
In general, switching from multiple pipes to one with the same total cross-sectional area is beneficial, as there are fewer losses.
In your case, the initial total cross-sectional area is$$ A_0=3\cdot\left(\frac{5}{8}\mbox{in}\right)^2\frac{\pi}{4}=0.92 \mbox{sq in} $$
Afterwards it will be:
$$ A_1=(3\mbox{in})^2\frac{\pi}{4}=7.1\mbox{sq in} $$Which is more than seven times the initial cross-sectional area, resulting in a decreased velocity for the same flow, thus fewer losses.
Therefore it should be safe to say that there will be no drop in pressure compared to the system you had before. |
Here is what we know from virtual work:\begin{equation}\delta W=\sum_{i=1}^N{\vec F_i\cdot\delta\vec r_{i}}\end{equation}Where $N$ is the number of bodies in the system. I am considering a quadcopter, modeled as a rigid body so it is just one body and we have:$$\delta W=\vec F\cdot\delta\vec r$$My question concerns only rotational dynamics, so let's focus on just the torques. Let us rename $\vec F$ to $\vec M$ just to make it clear that we are talking about torques (moments):$$\delta W=\vec M \cdot \delta\vec r$$
Expressed in the quadcopter frame (call it frame $B$), the moments exerted by the propellers on the quadcopter are, generally:$$\vec M=\begin{bmatrix}\tau_x^B \\ \tau_y^B \\ \tau_z^B\end{bmatrix}$$Because we are talking about moments, the "position" vector $\vec r$ actually holds the rotations expressed in the same frame as $\vec M$. Because I express $\vec M$ in the $B$ frame, I need to express the rotations in the $B$ frame. I am using the Tait-Bryan Euler angle convention. Starting from the inertial frame $I$ then: Yawby $\psi$ around $z_I$, the result is the $\{x',y',z'\}$ frame Corresponding rotation matrix is $R_\psi$ Pitchby $\theta$ around $y'$, the result is the $\{x'',y'',z''\}$ frame Corresponding rotation matrix is $R_\theta$ Rollby $\phi$ around $x''$, the result is the $\{x_B,y_B,z_B\}$ frame Corresponding rotation matrix is $R_\phi$
You see that each Euler angle is in a different frame: $\psi$ is in the $I$ frame, $\theta$ is in the $(')$ frame and $\phi$ is in the $('')$ frame. From what I know, we transform between angular rates in the $B$ frame and the Euler angle time derivatives using:
$$ \omega=R_\phi\begin{bmatrix}\dot\phi\\0\\0\end{bmatrix}+ R_\phi R_\theta\begin{bmatrix}0\\\dot\theta\\0\end{bmatrix}+ R_\phi R_\theta R_\psi\begin{bmatrix}0\\0\\\dot\psi\end{bmatrix} $$
However, $\omega$ is not integrable (see sub-section 3.17 of http://people.mech.kuleuven.be/~bruyninc/tmp/HermanBruyninckx-robotics.pdf). Therefore angles in the body frame as such do not exist and I cannot simply write:
$$ \vec r=R_\phi\begin{bmatrix}\phi\\0\\0\end{bmatrix}+ R_\phi R_\theta\begin{bmatrix}0\\\theta\\0\end{bmatrix}+ R_\phi R_\theta R_\psi\begin{bmatrix}0\\0\\\psi\end{bmatrix} $$
Where $\vec r$ is our "position" vector in the virtual work equation containing three angles in the $B$ frame identifying the quad-copter's attitude. The end motivation is to apply the generalized force relation: $$ Q_j=\sum_{i=1}^N{\vec F_i \cdot \frac{\partial \vec r_i}{\partial q_j}} $$ Where $q_j$ are the generalized coordinates, for me these are:
$q_1=\phi$ $q_2=\theta$ $q_3=\psi$
I really appreciate your help for my question:
find the expression for the generalized forces $Q_j$ (well, generalized torque really in this case). |
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Change to browse by: References & Citations Bookmark(what is this?) Physics > Computational Physics Title: A comparison between simulation methods to compute interfacial tensions discussed at the example of the solid-liquid interface in hard spheres
(Submitted on 20 Sep 2019)
Abstract: We computed the interfacial tension of the solid-liquid interface in the hard-sphere model by means of three simulation methods, which are used frequently to determine interfacial tensions in materials science. We investigate the drawbacks (in theory and practice) of the capillary wave method and conclude that - while it is a useful method for liquid-liquid interfaces - a precise computation of a solid-liquid interfacial tension requires very large computational resources. The interfacial stiffness values we obtained by two versions of the capillary wave method for the (100)-interface are $\tilde{\gamma} = 0.458(9)k_{\mathrm{B}}\sigma^{-2}$ and $\tilde{\gamma} = 0.440(5)k_{\mathrm{B}} T\sigma^{-2}$, where $\sigma$ is the diameter of a sphere. The second method is thermodynamic integration with a specific integration scheme for liquids. We separated the bulk contribution to the total free energy from the interface contribution and obtained for the interfacial tension of the (100)-interface $\gamma=0.69(13)k_{\mathrm{B}} T\sigma^{-2}$, $\gamma=0.65(10)k_{\mathrm{B}} T\sigma^{-2}$ and $\gamma=0.67(5)k_{\mathrm{B}} T\sigma^{-2}$, where the differences are due to details of the numerical integration scheme. We also used the pressure-tensor method, but found it to be too inaccurate for the hard sphere system to present a value for $\gamma$ here. Submission historyFrom: Moritz Bültmann [view email] [v1]Fri, 20 Sep 2019 13:11:50 GMT (1415kb,D) |
Is it the same as the bits of the key (So a 2048 bit system will yield a 2048 bit signature)? At
most as the key? Or something else entirely?
If $d,N$ is the private key, signing a message $m$ is computed as $m^d\bmod N$. The $\bmod N$ makes it so that the signed message is between $0$ and $N$. So, it is no larger than $N$.
In most applications, however, there is usually some encoding or protocol fields that will make it larger.
PKCS#1, "the" RSA standard, describes how a signature should be encoded, and it is a sequence of bytes with big-endian unsigned encoding, always of the size of the modulus. This means that for a 2048-bit modulus, all signatures have length exactly 256 bytes, never more, never less.
PKCS#1 is the most widely used standard, but there are other standards in some areas which may decide otherwise. Mathematically, a RSA signature is an integer between $1$ and $N-1$, where $N$ is the modulus. In some protocols, there can be some wrapping around the signature, e.g. to indicate which algorithm was used, or to embed certificates. For instance, in CMS, a "signature" contains the RSA value itself, but also quite a lot of additional data, for a virtually unbounded total size (I have seen signatures of a size of several
megabytes, due to inclusion of huge CRL).
The RSA signature size is dependent on the key size, the RSA signature size is equal to the length of the modulus in bytes. This means that for a "n bit key", the resulting signature will be exactly n bits long. Although the computed signature value is not necessarily n bits, the result will be padded to match exactly n bits.
Now here is how this works: The RSA algorithm is based on modular exponentiation. For such a calculation the final result is the remainder of the "normal" result divided by the modulus. Modular arithmetic plays a large role in Number Theory. There the definition for congruence is (I'll use 'congruent' since I don't know how to get those three-line equal signs)
$m \equiv n \mod k$ if $k$ divides $m - n$
Simple example: Let $n = 2$ and $k = 7$, then
$2 \equiv 2 \mod 7$ ($7$ divides $2 - 2$)
$9 \equiv 2 \mod 7$ ($7$ divides $9 - 2$) $16 \equiv 2 \mod 7$ ($7$ divides $16 - 2$) ...
$7$ actually does divide $0$, the definition for division is
An integer $a$ divides an integer $b$ if there is an integer $n$ with the property that $b = n·a$.
For $a = 7$ and $b = 0$ choose $n = 0$. This implies that every integer divides $0$, but it also implies that congruence can be expanded to negative numbers (won't go into details here, it's not important for RSA).
So the gist is that the congruence principle expands our naive understanding of remainders, the modulus is the "number after mod", in our example it would be $7$. As there are an infinite amount of numbers that are congruent given a modulus, we speak of this as the congruence classes and usually pick one representative (the smallest congruent integer $\geq 0$) for our calculations, just as we intuitively do when talking about the "remainder" of a calculation.
In RSA, signing a message $m$ means exponentiation with the "private exponent" $d$, the result $r$ is the smallest integer with $0 \leq r < n$ so that
$$ m^d \equiv r \bmod n.$$
This implies two things:
The length of $r$ (in bits) is bounded by the length of $n$ (in bits). The length of $m$ (in bits) must $\leq$ length($n$) (in bits, too).
To make the signature exactly $n$ bits long, some form of padding is applied. Cf. PKCS#1 for valid options.
The second fact implies that messages larger than n would either have to be signed by breaking $m$ in several chunks $< n$, but this is not done in practice since it would be way too slow (modular exponentiation is computationally expensive), so we need another way to "compress" our messages to be smaller than $n$. For this purpose we use cryptographically secure hash functions such as SHA-1 that you mentioned. Applying SHA-1 to an arbitrary-length message m will produce a "hash" that is 20 bytes long, smaller than the typical size of a RSA modulus, common sizes are 1024 bits or 2048 bits, i.e. 128 or 256 bytes, so the signature calculation can be applied for any arbitrary message.
The cryptographic properties of such a hash function ensures (in theory - signature forgery is a huge topic in the research community) that it is not possible to forge a signature other than by brute force. |
My question is about the
challenge space size in Schnorr protocol. To be precise, I feel I've read all the Internet (twice) and I still don't understand why is it bad to allow challenge space to be large (say, why one shouldn't let $\text{Challenge} \in \mathbb{Z}_q$)
I'm interested only in situation of honest prover $P$ and
malicious verifier $\tilde{V}$.
To settle the notation, recall that (one round of) Schnorr protocol has the following form:
Prover generates random $r \in \mathbb{Z}_q$, calculates $\text{Commit}=g^r \pmod p$, then sends $\text{Commit}$ to verifier. Verifier generates $m$ random bits and forms the number $\text{Challenge}$ from them (thus $\text{Challenge}$ is the random number ranging from $0$ to $2^{m} - 1$), then sends $\text{Challenge}$ to prover. Prover calculates $\text{Response} = r+s \cdot \text{Challenge} \pmod q$, then sends $\text{Response}$ to verifier. Verifier checks that $g^{\text{Response}} = \text{Commit} \cdot y^{\text{Challenge}} \pmod p$
where $y = g^s \pmod p$ and $s$ is the prover's secret.
The
simulator for this protocol is as follows. The algorithm generates random $\text{Response} \in \mathbb{Z}_p$. The algorithm asks verifier for the number $\text{Challenge}$ and receives it. The algorithm sets $\text{Commit} = g^{\text{Response}} \cdot y^{-\text{Challenge}} \pmod p$ The algorithm appends $(\text{Commit}, \text{Challenge}, \text{Response})$ to the transcript.
The standard answer to my question is: if $\text{Challenge}$ is
pseudorandom -- namely, depends on $\text{Commit}$, say, is equal to $\text{hash} (\text{Commit} || M)$ -- this simulator protocol cannot work, since at step 2 it requires the knowledge of parameter generated at step 3. There are no other simulators known for this protocol, so in the case of malicious verifier there's just no simulator. And since any ZK-protocol has simulator, the conclusion follows that malicious verifier case is not ZK. Question 1. However, I still don't get how exactly can malicious verifier extract some information. Okay, there's no known simulator for malicious verifier case -- well, how does it help him? Question 2. What about the challenge space size? Mao, Wenbo in their "Modern Cryptography" state that, thus, because of this simulator argument, $\text{Challenge}$ space size should NOT be large (don't get the logic again) ($\text{Challenge} \in \mathbb{Z}_q$ is prohibited) and state that the best choice is $\text{Challenge} \in [0, \log_2 p)$ (or, equivalently, $m = \log_2 \log_2 p$). Why such an odd and strange value? |
An atom is prepared in the angular momentum state $$C\left(\begin{array}{c}1 \\ 2\end{array}\right)$$Here $C$ is a constant. This has benn written in the $S_z$ basis.
a)Find C
b)Work out $\langle S_y\rangle$ using matrices
c)Calculate the variance $\sigma_{S_{y}}^2$
I've calculated C to be $\frac{1}{\sqrt5}$ by normalization, and my $\langle S_y\rangle$ comes out to be zero while the variance is $\frac{\hbar^2}{4}$. Another part of the question asks for variances $\sigma_{S_{x}}^2$ (which I've calculated to be $\frac{9\hbar^2}{100}$) and $\sigma_{S_{z}}^2$(which I've calculated to be $\frac{4\hbar^2}{25}$). We are then asked whether the results are consistent with the uncertainty principle. Can anyone let me know how to show that the results are consistent?Any help would be appreciated. |
Let $f : [0,1] \to \left\{ 0, 1 \right\}$ be a function that has at each point a discontinuity of the second kind. Is $f$ measurable if we equip the domain with the Borel or even Lebesgue $\sigma$-algebra and the range with the discrete $\sigma$-algebra?
The indicator of the Cantor set is not a counterexample since it is not discontinuous everywhere. The indicator of the rationals is nowhere continuous and is measurable.
This can be helpful in order to search for counterexamples: If $A \subseteq [0,1]$ is a set such that both $A$ and $[0,1] \setminus A$ are dense in $[0,1]$ then the indicator of $A$ is nowhere continuous. So, the question can be weakened: Is every such set $A$ necessarily measurable?
Wikipedia says that the set of discontinuities is an $F_\sigma$ set, thus a countable union of closed sets and therefore Borel measurable. In our case it is the whole interval $[0,1]$, so this statement doesn't help. |
Takeuti (1987, 223) deduces a cut-elimination theorem for infinitary logic from the corresponding soundness-and-completeness theorems. However, is there a way to adapt the basic Gentzen-style argument?
The relevant cut rule says (this is a rephrasing from Takeuti, 215)
From proofs of $\Gamma\to \Delta, A_\alpha$ for all $\alpha<\lambda$ and a proof of $\{A_\alpha\}_{\alpha<\lambda},\Pi\vdash \Lambda$,
obtain a proof of $\Gamma,\Pi\to \Delta,\Lambda$.
The notion of rank used in the basic Gentzen argument assumes that an inference has only one upper-left sequent, whereas here we have maybe infinitely many. Perhaps one could redefine left-rank by a lexicographic ordering along the ranks of the upper-left sequents?
I was able to adapt the transformations needed for reduction in right-rank, but then got caught in a hairball on the upper left. The problem is that the transformations given in basic argument depend on the rule by which the upper-left sequent was obtained.
Has this been worked out someplace? A reference (or a hint!) would be much appreciated.
Thanks,
Max
PS. The relevant notion of infinitary logic is what Takeuti (213) calls 'a system of infinitary logic with homogeneous quantifiers'. Basically, this adjusts the logical rules to fit the infinitary connectives as one would expect, but also allows infinitely many simultaneous applications of any one logical rule. (This is also apparent in the generalized form of the cut rule above.) Please let me know if more detail would be helpful here.
PPS. If it makes any difference, for my purposes the result is needed only for propositional infinitary logic. |
Problem 435
Let $\calF[0, 2\pi]$ be the vector space of all real valued functions defined on the interval $[0, 2\pi]$.
Define the map $f:\R^2 \to \calF[0, 2\pi]$ by \[\left(\, f\left(\, \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \,\right) \,\right)(x):=\alpha \cos x + \beta \sin x.\] We put \[V:=\im f=\{\alpha \cos x + \beta \sin x \in \calF[0, 2\pi] \mid \alpha, \beta \in \R\}.\] (a) Prove that the map $f$ is a linear transformation. (b) Prove that the set $\{\cos x, \sin x\}$ is a basis of the vector space $V$. (c) Prove that the kernel is trivial, that is, $\ker f=\{\mathbf{0}\}$. (This yields an isomorphism of $\R^2$ and $V$.) (d) Define a map $g:V \to V$ by \[g(\alpha \cos x + \beta \sin x):=\frac{d}{dx}(\alpha \cos x+ \beta \sin x)=\beta \cos x -\alpha \sin x.\] Prove that the map $g$ is a linear transformation. (e) Find the matrix representation of the linear transformation $g$ with respect to the basis $\{\cos x, \sin x\}$.
(Kyoto University, Linear Algebra exam problem)Add to solve later |
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
Can we view the normal, nonrelativistic quantum mechanics as a classical fields?
Yes, you can view the wave function $\psi(x,t)$ as an ordinary complex-valued field in the spirit of, say, classical electrodynamics. This field describes the quantum mechanics of a single electron, but it is classical in the sense that it's an ordinary function $\psi(x,t) : \mathbb R^3\times\mathbb R \to \mathbb C$.
As usual, you can have a lot of fun with the Lagrangian. For instance, you can note that there it has a (global) $U(1)$ symmetry $\psi \mapsto e^{i}\psi$ and apply the Noether theorem. You will find a continuity equation for the quantity $\psi^*\psi$, which we commonly interpret as the probability density.
Of course, the Schrdinger equation is limited to non-relativistic physics, so people started to look for a relativistic equivalent. Dirac's eponymous equation was intended to be precisely that: an equation for a classical field that describes a quantum mechanical electron in a Lorentz-covariant way. Of course, there should be an equivalent of the probability density $\psi^*\psi$, which is always positive, but no matter how you spin it, this just didn't work out, even for the Dirac equation.
The solution to this problem is that electrons don't live in isolation, they are identical particles and linked together via the Pauli exclusion principle. Dirac could only make sense of his equation by considering a variable number of electrons. This is where the classical field $\psi$ has to be promoted to a
quantum field $\Psi$, a process known as second quantization. ("First quantization" refers to the fact that the classical field $\psi$ already describes a quantum mechanical particle.)
It turns out that second quantization is also necessary to explain certain corrections to the ordinary Schrdinger equation. In this light, the classical field $\psi$ is really an approximation as it does neglect the influence of a variable number of particles.
The process of considering a variable number of particles is actually quite neat. If you go from one to two particles, you would have to consider a classical field $\psi(x_1,x_2)$ that depends on two variables, the particle positions. Going to $N$ particles, you would have a field $\psi(x_1,\dots,x_N)$ depending on that many variables. You can get all particles at once by considering an
operator valued field $\Psi(x)$ instead, which creates a particle at position $x$. It turns out that you can just replace $\psi$ by $\Psi$ in the Lagrangian to get the right equations of motion for all particles at once.
Alas, I have to stop here, further details on second quantization and quantum field theory are beyond the scope of this answer.
Concerning literature, I found Sakurai's Advanced Quantum Mechanics to be a very clear if somewhat long-winded introduction to quantum field theory that starts where the Schrdinger equation left off.This post imported from StackExchange Physics at 2014-07-22 05:15 (UCT), posted by SE-user Greg Graviton |
Interface Issues¶ Background jobs¶
Yes, a Sage job can be run in the background on a UNIX system. The canonical thing to do is type
$ nohup sage < command_file > output_file &
The advantage of nohup is that Sage will continue running after you log out.
Currently Sage will appear as “sage-ipython” or “python” in the outputof the (unix)
top command, but in future versions of Sage it willappears as
sage.
Referencing Sage¶
To reference Sage, please add the following to your bibliography:
\bibitem[Sage]{sage}Stein, William, \emph{Sage: {O}pen {S}ource {M}athematical {S}oftware({V}ersion 2.10.2)}, The Sage~Group, 2008, {\tt http://www.sagemath.org}.
Here is the bibtex entry:
@manual{sage, Key = {Sage}, Author = {William Stein}, Organization = {The Sage~Group}, Title = {{Sage}: {O}pen {S}ource {M}athematical {S}oftware ({V}ersion 2.10.2)}, Note= {{\tt http://www.sagemath.org}}, Year = 2008}
If you happen to use the Sage interface to PARI, GAP or Singular, you should definitely reference them as well. Likewise, if you use code that is implemented using PARI, GAP, or Singular, reference the corresponding system (you can often tell from the documentation if PARI, GAP, or Singular is used in the implementation of a function).
For PARI, you may use
@manual{PARI2, organization = "{The PARI~Group}", title = "{PARI/GP, version {\tt 2.1.5}}", year = 2004, address = "Bordeaux", note = "available from \url{http://pari.math.u-bordeaux.fr/}" }
or
\bibitem{PARI2} PARI/GP, version {\tt 2.1.5}, Bordeaux, 2004,\url{http://pari.math.u-bordeaux.fr/}.
(replace the version number by the one you used).
For GAP, you may use
[GAP04] The GAP Group, GAP -- Groups, Algorithms, and Programming,Version 4.4; 2005. (http://www.gap-system.org)
or
@manual{GAP4, key = "GAP", organization = "The GAP~Group", title = "{GAP -- Groups, Algorithms, and Programming, Version 4.4}", year = 2005, note = "{\tt http://www.gap-system.org}", keywords = "groups; *; gap; manual"}
or
\bibitem[GAP]{GAP4} The GAP~Group, \emph{GAP -- Groups, Algorithms, and Programming, Version 4.4}; 2005, {\tt http://www.gap-system.org}.
For Singular, you may use
[GPS05] G.-M. Greuel, G. Pfister, and H. Sch\"onemann.{\sc Singular} 3.0. A Computer Algebra System for PolynomialComputations. Centre for Computer Algebra, University ofKaiserslautern (2005). {\tt http://www.singular.uni-kl.de}.
or
@TechReport{GPS05, author = {G.-M. Greuel and G. Pfister and H. Sch\"onemann}, title = {{\sc Singular} 3.0}, type = {{A Computer Algebra System for Polynomial Computations}}, institution = {Centre for Computer Algebra}, address = {University of Kaiserslautern}, year = {2005}, note = {{\tt http://www.singular.uni-kl.de}},}
or
\bibitem[GPS05]{GPS05}G.-M.~Greuel, G.~Pfister, and H.~Sch\"onemann.\newblock {{\sc Singular} 3.0}. A Computer Algebra System for Polynomial Computations.\newblock Centre for Computer Algebra, University of Kaiserslautern (2005).\newblock {\tt http://www.singular.uni-kl.de}.
Logging your Sage session¶
Yes you can log your sessions.
(a) Modify line 186 of the .ipythonrc file (or open .ipythonrc into an editor and search for “logfile”). This will only log your input lines, not the output.
(b) You can also write the output to a file, by running Sage in the background ( Background jobs ).
(c) Start in a KDE konsole (this only work in linux). Go to
Settings \(\rightarrow\)
History ... and selectunlimited. Start your session. When ready, go to
edit\(\rightarrow\)
save history as ....
Some interfaces (such as the interface to Singular or that to GAP)allow you to create a log file. For Singular, there is a logfileoption (in
singular.py). In GAP, use the command
LogTo.
LaTeX conversion¶
Yes, you can output some of your results into LaTeX.
sage: M = MatrixSpace(RealField(),3,3)sage: A = M([1,2,3, 4,5,6, 7,8,9])sage: print(latex(A))\left(\begin{array}{rrr} 1.00000000000000 & 2.00000000000000 & 3.00000000000000 \\ 4.00000000000000 & 5.00000000000000 & 6.00000000000000 \\ 7.00000000000000 & 8.00000000000000 & 9.00000000000000 \end{array}\right)
sage: view(A)
At this point a dvi preview should automatically be called to display in a separate window the LaTeX output produced.
LaTeX previewing for multivariate polynomials and rational functions is also available:
sage: x = PolynomialRing(QQ,3, 'x').gens()sage: f = x[0] + x[1] - 2*x[1]*x[2]sage: h = f /(x[1] + x[2])sage: print(latex(h))\frac{-2 x_{1} x_{2} + x_{0} + x_{1}}{x_{1} + x_{2}}
Sage and other computer algebra systems¶
If
foo is a Pari, GAP ( without ending semicolon), Singular,Maxima command, resp., enter
gp("foo") for Pari,
gap.eval("foo")}
singular.eval("foo"),
maxima("foo"), resp..These programs merely send the command string to the externalprogram, execute it, and read the result back into Sage. Therefore,these will not work if the external program is not installed and inyour PATH.
Command-line Sage help¶
If you know only part of the name of a Sage command and want toknow where it occurs in Sage, a new option for 0.10.11 has beenadded to make it easier to hunt it down. Just type
sage -grep <string> to find all occurrences of
<string> in theSage source code. For example,
$ sage -grep berlekamp_masseymatrix/all.py:from berlekamp_massey import berlekamp_masseymatrix/berlekamp_massey.py:def berlekamp_massey(a):matrix/matrix.py:import berlekamp_masseymatrix/matrix.py: g =berlekamp_massey.berlekamp_massey(cols[i].list())
Type
help(foo) or
foo?? for help and
foo.[tab] for searchingof Sage commands. Type
help() for Python commands.
For example
help(Matrix)
returns
Help on function Matrix in module sage.matrix.constructor:Matrix(R, nrows, ncols, entries = 0, sparse = False) Create a matrix. INPUT: R -- ring nrows -- int; number of rows ncols -- int; number of columns entries -- list; entries of the matrix sparse -- bool (default: False); whether or not to store matrices as sparse OUTPUT: a matrix EXAMPLES: sage: Matrix(RationalField(), 2, 2, [1,2,3,4]) [1 2] [3 4] sage: Matrix(FiniteField(5), 2, 3, range(6)) [0 1 2] [3 4 0] sage: Matrix(IntegerRing(), 10, 10, range(100)).parent() Full MatrixSpace of 10 by 10 dense matrices over Integer Ring sage: Matrix(IntegerRing(), 10, 10, range(100), sparse = True).parent() Full MatrixSpace of 10 by 10 sparse matrices over Integer Ring
in a new screen. Type q to return to the Sage screen.
Reading and importing files into Sage¶
A file imported into Sage must end in
.py, e.g.,
foo.py andcontain legal Python syntax. For a simple example see Permutation groupswith the Rubik’s cube group example above.
Another way to read a file in is to use the
load or
attachcommand. Create a file called
example.sage (located in the homedirectory of Sage) with the following content:
print("Hello World")print(2^3)
Read in and execute
example.sage file using the
load command.
sage: load("example.sage")Hello World8
You can also
attach a Sage file to a running session:
sage: attach("example.sage")Hello World8
Now if you change
example.sage and enter one blank line intoSage, then the contents of
example.sage will be automaticallyreloaded into Sage:
sage: !emacs example.sage& #change 2^3 to 2^4sage: #hit return*************************************************** Reloading 'example.sage'***************************************************Hello World16
Installation for the impatient¶
We shall explain the basic steps for installing the most recent version of Sage (which is the “source” version, not the “binary”).
Download
sage-*.tar(where
*denotes the version number) from the website and save into a directory, say
HOME. Type
tar zxvf sage-*.tarin
HOME.
cd
sage-*(we call this
SAGE_ROOT) and type
make. Now be patient because this process make take 2 hours or so.
Python language program code for Sage commands¶
Let’s say you want to know what the Python program is for the Sage command to compute the center of a permutation group. Use Sage’s help interface to find the file name:
sage: PermutationGroup.center?Type: instancemethodBase Class: <type 'instancemethod'>String Form: <unbound method PermutationGroup.center>Namespace: InteractiveFile: /home/wdj/sage/local/lib/python2.4/site-packages/sage/groups/permgroup.pyDefinition: PermutationGroup.center(self)
Now you know that the command is located in the
permgroup.py fileand you know the directory to look for that Python module. You canuse an editor to read the code itself.
“Special functions” in Sage¶
Sage has many special functions (see the reference manual at http://doc.sagemath.org/html/en/reference/functions/), and most of them can be manipulated symbolically. Where this is not implemented, it is possible that other symbolic packages have the functionality.
Via Maxima, some symbolic manipulation is allowed:
sage: maxima.eval("f:bessel_y (v, w)")'bessel_y(v,w)'sage: maxima.eval("diff(f,w)")'(bessel_y(v-1,w)-bessel_y(v+1,w))/2'sage: maxima.eval("diff (jacobi_sn (u, m), u)")'jacobi_cn(u,m)*jacobi_dn(u,m)'sage: jsn = lambda x: jacobi("sn",x,1)sage: P = plot(jsn,0,1, plot_points=20); Q = plot(lambda x:bessel_Y( 1, x), 1/2,1)sage: show(P)sage: show(Q)
In addition to
maxima,
pari and
octave also have specialfunctions (in fact, some of
pari’s special functions are wrappedin Sage).
Here’s an example using Sage’s interface (located insage/interfaces/octave.py) with
octave(http://www.octave.org/doc/index.html).
sage: octave("atanh(1.1)") ## optional - octave(1.52226,-1.5708)
Here’s an example using Sage’s interface to
pari’s specialfunctions.
sage: pari('2+I').besselk(3)0.0455907718407551 + 0.0289192946582081*Isage: pari('2').besselk(3)0.0615104584717420
What is Sage?¶
Sage is a framework for number theory, algebra, and geometry computation that is initially being designed for computing with elliptic curves and modular forms. The long-term goal is to make it much more generally useful for algebra, geometry, and number theory. It is open source and freely available under the terms of the GPL. The section titles in the reference manual gives a rough idea of the topics covered in Sage.
History of Sage¶
Sage was started by William Stein while at Harvard University in the Fall of 2004, with version 0.1 released in January of 2005. That version included Pari, but not GAP or Singular. Version 0.2 was released in March, version 0.3 in April, version 0.4 in July. During this time, support for Cremona’s database, multivariate polynomials and large finite fields was added. Also, more documentation was written. Version 0.5 beta was released in August, version 0.6 beta in September, and version 0.7 later that month. During this time, more support for vector spaces, rings, modular symbols, and windows users was added. As of 0.8, released in October 2005, Sage contained the full distribution of GAP, though some of the GAP databases have to be added separately, and Singular. Adding Singular was not easy, due to the difficulty of compiling Singular from source. Version 0.9 was released in November. This version went through 34 releases! As of version 0.9.34 (definitely by version 0.10.0), Maxima and clisp were included with Sage. Version 0.10.0 was released January 12, 2006. The release of Sage 1.0 was made early February, 2006. As of February 2008, the latest release is 2.10.2.
Many people have contributed significant code and other expertise, such as assistance in compiling on various OS’s. Generally code authors are acknowledged in the AUTHOR section of the Python docstring of their file and the credits section of the Sage website. |
Let $T: \R^n \to \R^m$ be a linear transformation.Suppose that the nullity of $T$ is zero.
If $\{\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k\}$ is a linearly independent subset of $\R^n$, then show that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.
Let $A$ be the matrix given by\[A=\begin{bmatrix}-2 & 0 & 1 \\-5 & 3 & a \\4 & -2 & -1\end{bmatrix}\]for some variable $a$. Find all values of $a$ which will guarantee that $A$ has eigenvalues $0$, $3$, and $-3$.
Let\[A=\begin{bmatrix}8 & 1 & 6 \\3 & 5 & 7 \\4 & 9 & 2\end{bmatrix}.\]Notice that $A$ contains every integer from $1$ to $9$ and that the sums of each row, column, and diagonal of $A$ are equal. Such a grid is sometimes called a magic square.
Define two functions $T:\R^{2}\to\R^{2}$ and $S:\R^{2}\to\R^{2}$ by\[T\left(\begin{bmatrix}x \\ y\end{bmatrix}\right)=\begin{bmatrix}2x+y \\ 0\end{bmatrix},\;S\left(\begin{bmatrix}x \\ y\end{bmatrix}\right)=\begin{bmatrix}x+y \\ xy\end{bmatrix}.\]Determine whether $T$, $S$, and the composite $S\circ T$ are linear transformations.
Let\[\mathbf{v}_{1}=\begin{bmatrix}1 \\ 1\end{bmatrix},\;\mathbf{v}_{2}=\begin{bmatrix}1 \\ -1\end{bmatrix}.\]Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?
Let $A$ be an $m \times n$ matrix.Suppose that the nullspace of $A$ is a plane in $\R^3$ and the range is spanned by a nonzero vector $\mathbf{v}$ in $\R^5$. Determine $m$ and $n$. Also, find the rank and nullity of $A$.
Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\]still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample.
For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]
For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by\[ (f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S . \]
(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?
(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.
(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.
(d) Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as\[ V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} . \]
(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?
(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$. |
Worst-case Hardness of NP-complete problems is not sufficient for cryptography. Even if NP-complete problems are hard in the worst-case ($P \ne NP$), they still could be efficiently solvable in the average-case. Cryptography assumes the existence of average-case intractable problems in NP. Also, proving the existence of hard-on-average problems in NP using ...
There have been.One such example is McEliece cryptosystem which is based on hardness of decoding a linear code.A second example is NTRUEncrypt which is based on the shortest vector problem which I believe is known to be NP-Hard.Another is Merkle-Hellman knapsack cryptosystem which has been broken.Note: I have no clue if the first two are broken/how ...
By the nondeterministic version of the time-hierarchy theorem, we have $\mathsf{NP} \subsetneq \mathsf{NEXP}$, where $\mathsf{NEXP}$ is the class of problems solvable in non-deterministic exponential-time. Thus it suffices to consider any problem which is $\mathsf{NP}$-hard and in $\mathsf{NEXP}$, but not in $\mathsf{NP}$. For instance, we may consider any $\...
As you said, there is no decision to make, so new complexity classes and new types of reductions are needed to arrive at a suitable definition of NP-hardness for optimization-problems.One way of doing this is to have two new classes NPO and PO that contain optimizations problems and they mimic of course the classes NP and P for decision problems. New ...
I can think of four major hurdles which are not entirely independent:NP-hardness only gives you information about complexity in the limit. For many NP-complete problems, algorithms exist that solve all instances of interest (in a certain scenario) reasonably fast. In other words, for any fixed problem size (e.g. a given "key"), the problem is not ...
${\rm D{\small OMINOSA}}$ is NP-hardPlaying the game is an optimization problem; finding a valid domino tiling such that it covers all the squares. The decision version of this problem is:Is there a perfect tiling covering a given a $(n+1) \times (n+2)$ grid with $n$ unique tiles?Obviously, the optimization problem, the problem of actually finding a ...
Usually what's shown is the NP-hardness of a "Gap" version of the problem. For example, suppose that you want to show that it's hard to approximate SET COVER to within a factor of 2.You define the following "promise" instance of SET COVER that we'll call 2-GAP-SET-COVER:Fix some number $\ell$. 2-GAP-SET-COVER consists of all instances of set cover where ...
Informally:In DNF, you can pick any clause to be true, to make the formula true. This means that a DNF that is equivalent to a certain CNF, is basically an enumeration of all the solutions to boolean sat on the CNF. Note, there can be an exponential number of solutions. Since solving boolean sat for CNF for a single solution is NP-complete, converting to ...
The problem isn't really well-posed. For any particular instance, there is a single solution, say $S$. Consequently, we can imagine an algorithm that has the answer $S$ hardcoded in: no matter what input you give it, all it does is just print $S$. This answer counts as a deterministic polynomial-time algorithm that solves this particular instance $I$....
There appears to be some considerable confusion in this community regarding this question. I'll give a detailed answer in the hope of clearing up the water and illuminating the relationship between computability and NP-hardness.First, I believe that being clear and explicit about the various definitions involved will resolve a lot of the confusion.A ...
Public-key cryptography as we know it today is built on one-way trapdoor permutations, and the trapdoor is essential.For a protocol to be publicly secure, you need a key available to anyone, and a way to encrypt a message using this key. Obviously, once encrypted, it should be hard to recover the original message knowing only its cipher and the public key :...
No, an $NP$-hard problem need not be computable. The definition is fairly complete: a problem $L$ is $NP$-hard if that problem having a poly-time solution implies every problem in $NP$ has a poly-time solution (that is, a reduction to $L$ exists for every problem in $NP$.).Uncomputable problems are then vacuously hard: suppose we could solve one in ...
It looks like you are trying to compute a hypergraph transversal of size $k$. That is, $\{T_1,\dots,T_m\}$ is your hypergraph, and $S$ is your transversal. A standard translation is to express the clauses as you have, and then translate the length restriction into a cardinality constraint.So use your existing encoding, i.e., $\bigwedge_{1 \le j \le m} \...
$(3,k)\text{-LSAT}$ is in P for all $k$. As you have indicated, locality is a big obstruction to NP-completeness.Here is a polynomial algorithm.Input: $\phi\in (3,k)\text{-LSAT}$, $\phi=c_1\wedge c_2\cdots \wedge c_m$, where $c_i$ is the $i$-th clause.Output: true if $\phi$ becomes 1 under some assignment of all variables.Procedure:Construct set $B_i$...
The differentiability requirement doesn't change the nature of the problem: requiring $\mathcal{C}^0$ (continuity) or $\mathcal{C}^{\infty}$ (infinite differentiability) gives the same lower bound for the length and the same order of points, and is equivalent to solving the traveling salesman problem.If you have a solution to the TSP, you have a $\mathcal{...
Your intuition about "relative hardness" is correct, the underlying mathematics is why III. is true. However your justification about I. is a little off (not wrong, but the reasoning is possibly not accurate).It might help to think about reductions in these terms (everything I'll talk about here will be polynomial time, so I will leave that out just so I ...
Actually, your version is correct and Wikipedia's is wrong! (Except that it has a tiny disclaimer at the bottom.)If $\mathrm{P}=\mathrm{NP}$, Wikipedia claims that every problem in $\mathrm{P}$ is $\mathrm{NP}$-complete. However, this is not true: in fact, every problem in $\mathrm{P}$ would be $\mathrm{NP}$-complete, except for the trivial languages $\...
You actually already have a reduction from special to general. By setting $s=0$, you are basically using the general algorithm to solve the special problem.For the other way round (i.e. a reduction from general to special):Suppose you are given a set $S = \{x_1, \dots, x_n\}$ and a number $K$ and you have to determine if there is some subset of $S$ which ...
The problem is NP-hard.We show this by reducing vertex cover:Given a graph $G=(V,E)$ and a threshold $k$, is there a subset $V' \subseteq V$ of cardinality at most $k$, so that each edge in $E$ is incident to at least one node in $V'$?We translate this into a regex crossword with $|E|+1$ columns and $|V|$ rows as follows:All columns, except for the ...
A problem $P$ is NP-complete if:$P$ is NP-hard and$P \in \textbf{NP}$.The authors give a proof of item number 1. Item number 2 is probably apparent (and should be clear to the paper's audience). For the proof of item number 1, you only need a (many-one) reduction from some NP-complete problem (e.g., SAT) to $P$; there is no need to construct a reduction ...
I'm assuming that you're thinking of the Euclidean traveling salesman problem, where $c$ and $v$ are vectors in $\mathbb{Z}^{2n}$, with two coordinates for each city. Let $minTSP(c)$ denote the length of the minimum traveling salesman tour for the cities with coordinates $c$. Then your problem asks whether$$minTSP(c + v) \ge min TSP(c) + x.$$But then ...
No. E.g. the Halting problem is a decision problem which is NP-hard but not in NP and therefore not NP-complete.In normal usage yes, because an NP-complete problem must be in NP and NP is a class of decision problems. But see Decision problems vs “real” problems that aren't yes-or-no.
You claim that every problem in NP can be reduced to its complement, and this is true for Turing reductions, but (probably) not for many-one reductions. A many-one reduction from $L_1$ to $L_2$ is a polytime function $f$ such that for all $x$, $x \in L_1$ iff $f(x) \in L_2$.If some problem $L$ in coNP were NP-hard, then for any language $M \in NP$ there ...
This is a sketch of a reduction from MONOTONE CUBIC PLANAR 1-3 SAT :Definition [1-3 SAT problem]:Input: A 3-CNF formula $\varphi = C_1 \land C_2 \land ... \land C_m$, in which every clause $C_j$ contains exactly three literals: $C_j = (\ell_{j,1} \lor \ell_{j,2} \lor \ell_{j,3})$.Question: Does there exist a satisfying assignment for $\varphi$ such that ...
Nope. NP-Hard means it is as hard, or harder, than the hardest NP-problems. Intuitively, being uncomputable will make it a lot more difficult than NP.Wikipedia:There are decision problems that are NP-hard but not NP-complete, for example the halting problem.Everyone knows that is not computable
For completeness, let us prove the following theorem:There exists an uncomputable language which is not NP-hard if and only if P$\neq$NP.If P=NP then any non-trivial language (one which differs from $\emptyset,\{0,1\}^*$) is NP-hard (exercise), and in particular any uncomputable language is NP-hard.Now suppose that P$\neq$NP. Let $T_i$ be some ...
See the definition of NP completeness. For a problem to be NP-complete, itneeds to be in NP andall NP problems need to be reducible to it in polynomial time.Condition 2 alone is what it means to be NP hard. Thus NP complete problems are the intersection of NP problems and NP hard problems.NP $\subseteq$ EXPTIME, thus NP problems can be solved in $2^...
Reduce from NP-complete SET-COVER instead.Let $S_1,\dots,S_m \subseteq\{1,\dots,n\}$ with $k\in\mathbb{N}$ an instance of set cover. Define an instance $(V,E,k')$ of DGD like this:$V= \{s_1,\dots,s_m,o_1,\dots,o_m,e_1,\dots,e_n,o\}$$E= \{(s_i,o_i) \mid i = 1,\dots,n \} \cup \{(s_i,e_j)\mid j \in S_i\} \cup\{(e_j,o)\mid j=1,\dots,n\}$$k' = m + k$It is ...
First we reduce the task of factorization to finding any factor of $n$ (or showing that it's prime). Once we know how to do this, we divide $n$ by such an factor and repeat the process until all factors are broken down to primes, and this takes $O(\log n)$ steps.Let $k=\lceil\log_2 n\rceil$ the number of bits in $n$.Next, we design a binary ...
Note: This is a continuation and revision of my other answer.Problems with the reductionRecall the decision problem:Is there a perfect tiling covering a given a $(n+1) \times (n+2)$ grid with $n$ unique tiles?So for an $(n+1) \times (n+2)$ grid, we can only use $n$ variables.But:Our reduction requires a lot of unique variables, much more ... |
I have a dataset with three variables, where all variables are quantitatives. Let call it $y$, $x_1$ and $x_2$. I'm fitting a regression model in a Bayesian perspective via MCMC with
rjags
I done a exploratory analysis and the scatterplot of $y\times x_2$ suggest that a quadratic term should be used. Then I fitted two models
(1) $y=\beta_0+\beta_1*x_1+\beta_2*x_2$
(2) $y=\beta_0+\beta_1*x1+\beta_2*x_2+\beta_3*x_1x_2+\beta_4*x_1^2+\beta_5*x_2^2$
In model 1 the effect size of each parameter is not small and the 95% credible interval not contains the value $0$.
In model 2 the effect size of parameters $\beta_3$ and $\beta_4$ are small and each of credible intervals for all parameters contains $0$.
The fact that a credible interval contains $0$ is enough to say that the parameter is not significant?
Then I adjusted the following model
(3)$y=\beta_0+\beta_1*x_1+\beta_2*x_2+\beta_3*x^2_2$
The effect size of each parameter is not small, but with exception of $\beta_1$ all credible intervals contains $0$.
Which is the right way to do variable selection in Bayesian statistics?
EDIT: I can use Lasso in any regression model, like Beta model? I'm using a model with variable dispersion where $$log(\sigma)=-\pmb{\delta}X$$where $\pmb{\delta}$ is a vector. I should use Laplace prior in $\pmb{\delta}$ too? EDIT2: I fitted two models, one with Gaussian priori for $\beta_j$, $\delta_j$ and one with Laplace(double-exponential).
The estimatives for the Gaussian model are
Mean SD Naive SE Time-series SEB[1] -1.17767 0.07112 0.0007497 0.0007498B[2] -0.15624 0.03916 0.0004128 0.0004249B[3] 0.15600 0.05500 0.0005797 0.0005889B[4] 0.07682 0.04720 0.0004975 0.0005209delta[1] -3.42286 0.32934 0.0034715 0.0034712delta[2] 0.06329 0.27480 0.0028966 0.0028969delta[3] 1.06856 0.34547 0.0036416 0.0036202delta[4] -0.32392 0.26944 0.0028401 0.0028138
The estimatives for the Lasso model are
Mean SD Naive SE Time-series SEB[1] -1.143644 0.07040 0.0007421 0.0007422B[2] -0.160541 0.05341 0.0005630 0.0005631B[3] 0.137026 0.05642 0.0005947 0.0005897B[4] 0.046538 0.04770 0.0005028 0.0005134delta[1] -3.569151 0.27840 0.0029346 0.0029575delta[2] -0.004544 0.15920 0.0016781 0.0016786delta[3] 0.411220 0.33422 0.0035230 0.0035629delta[4] -0.034870 0.16225 0.0017103 0.0017103lambda 7.269359 5.45714 0.0575233 0.0592808
The estimatives for $\delta_2$ and $\delta_4$ reduced a lot in Lasso model, it means that I should remove this variables from the model?
EDIT3: The model with double exponential prior (Lasso) gaves me bigger Deviance, BIC and DIC values than the model with Gaussian priors and I even get a smaller values after removing the dispersion coefficient $\delta_2$ in the Gaussian model. |
Sorry for overlooking the requirement that the functions are supported on $[0,1]$. If this condition is added, then the statement is true.
Straight from the definition, $f_n$ converges weakly to $f\in L^q$ iff for every $g\in L^{q^*}$,
$$\int f_ng\to \int fg,$$
where $1/q+1/q^*=1$. Since $p<q$, $p^*>q^*$, so $L^{p^*}([0,1])\subset L^{q^*}([0,1])$.Then the above convergence holds for every $g\in L^{p^*}$. This implies $f_n$ converges weakly to $f$ in $L^p$. As a sanity check, we see that $f$ is in $L^p$ because $L^q([0,1])\subset L^p([0,1])$.
More succinctly, this is because continuous linear maps between Banach spaces are also weakly continuous (See Theorem 1.1 of Chapter 6 of Conway,
A Course in Functional Analysis).
Just for reference, the result is false if the functions are supported on $\mathbb R$.
Let $r\in(p,q)$ and $f_n=n^{-1/r}$ on $[0,n]$ and $f_n=0$ elsewhere. Then
$$\|f_n\|_{q}=(n\cdot n^{-q/r})^{1/q}=n^{1/q-1/r}\to 0$$
as $n\to\infty$, so $f_n\to0$ in $L^q$. On the other hand, we replace $q$ by $p$ in the above inequality to get
$$\|f_n\|_{p}=n^{1/p-1/r}\to\infty$$
as $n\to\infty$. By the uniform boundedness principle (among other approaches), $f_n$ does not converge weakly in $L^p$. |
Why All These Stresses and Strains?
In structural mechanics you will come across a plethora of stress and strain definitions. It may be a
Second Piola-Kirchhoff Stress or a Logarithmic Strain. In this blog post we will investigate these quantities, discuss why there is a need for so many variations of stresses and strains, and illuminate the consequences for you as a finite element analyst. The defining tensor expressions and transformations can be found in many textbooks, as well as through some web links at the end of this blog post, so they will not be given in detail here.
The Tensile Test
When evaluating the mechanical data of a material, it is common to perform a uniaxial tension test. What is actually measured is a force versus displacement curve, but in order to make these results independent of specimen size, the results are usually presented as
stress versus strain. If the deformations are large enough, one question then is: do you compute the stress based on the original cross-sectional area of the specimen, or based on the current area? The answer is that both definitions are used, and are called Nominal stress and True stress, respectively.
A second, and not so obvious, question is how to measure the
relative elongation, i.e. the strain. The engineering strain is defined as the ratio between the elongation and the original length, \epsilon_{eng} = \frac{L-L_0}{L_0}. For larger stretches, however, it is more common to use either the stretch \lambda=\frac{L}{L_0} or the true strain (logarithmic strain) \epsilon_{true} = \log\frac{L}{L_0} = \log \lambda.
The true strain is more common in metal testing, since it is a quantity suitable for many plasticity models. For materials with a very large possible elongation, like rubber, the stretch is a more common parameter. Note that for the undeformed material, the stretch is \lambda=1.
In order to make use of the measured data in an analysis, you must make sure of the following two things:
How the stress and strain are defined in the test In what form your analysis software expects it for a specific material model
The transformation of the uniaxial data is not difficult, but it must not be forgotten.
Stress-strain curves for the same tensile test. Geometric Nonlinearity
Most structural mechanics problems can be analyzed under the assumption that the deformations are so small compared to the dimensions of the structure, that the equations of equilibrium can be formulated for the undeformed geometry. In this case, the distinctions between different stress and strain measures disappear. If displacements, rotations, or strains become large enough, then geometric nonlinearity must be taken into account. This is when we start to consider that area elements actually change, that there is a distinction between an original length and a deformed length, and that directions may change during the deformation. There are several mathematically equivalent ways of representing such finite deformations.
For the uniaxial test above, the different representations are rather straight-forward. In real life however, geometries are three-dimensional, have multiaxial stress states, and might rotate in space. Even if we just consider the same tensile test, keep the stress and strain fixed at a certain level, and then rotate the specimen, questions arise. What results can we expect? Are the values of the stress and strain components expected to change or not?
Stress Measures
The most fundamental and commonly used stress quantity is the
Cauchy stress, also known as the true stress. It is defined by studying the forces acting on an infinitesimal area element in the deformed body. Both the force components and the normal to the area have fixed directions in space. This means that if a stressed body is subjected to a pure rotation, the actual values of the stress components will change. What was originally a uniaxial stress state might be transformed into a full tensor with both normal and shear stress components. In many cases, this is neither what you want to use nor what you would expect.
Consider for example an orthotropic material with fibers having a certain orientation. It is much more plausible that you want to see the stress in the fiber direction, even if the component is rotated. The
Second Piola-Kirchhoff stress has this property. It is defined along the material directions. In the figure below, an originally straight cantilever beam has been subjected to bending by a pure moment at the tip. The xx-component of the Cauchy stress (top) and Second Piola-Kirchhoff stress (below) are shown. Since the stress is physically directed along the beam, the xx-component of the Cauchy stress (which is related to the global x-direction) decreases with the deflection. The Second Piola-Kirchhoff stress however, has the same through-thickness distribution all along the beam, even in the deformed configuration. Cauchy and Second Piola-Kirchhoff stress for an initially straight beam with constant bending moment.
Another stress measure that you may encounter is the
First Piola-Kirchhoff stress. It is a multiaxial generalization of the nominal (or engineering) stress. The stress is defined as the force in the current configuration acting on the original area. The First Piola-Kirchhoff is an unsymmetric tensor, and is for that reason less attractive to work with.
Sometimes you may also encounter the
Kirchhoff stress. The Kirchhoff stress is just the Cauchy stress scaled by the volume change. It has little physical significance, but can be convenient in some mathematical and numerical operations.
Unfortunately, even without a rotation, the actual values of all these stress representations are not the same. All of them scale differently with respect to local volume changes and stretches. This is illustrated in the graph below. The xx-component of several stress measures are plotted at the fixed end of the beam, where the beam axis coincides with the x-axis. In the center of the beam, where strains, and thereby volume changes are small, all values approach each other. So for a case with large rotation but small strains, the stress representations can be seen as pure rotations of the same stress tensor.
The distribution of axial stress at the fixed end of the beam.
If you want to compute the resulting force or a moment on a certain boundary, there are really only two possible choices: Either integrate the Cauchy stress over the deformed boundary, or integrate the First Piola-Kirchhoff stress over the same boundary in the undeformed configuration. In COMSOL Multiphysics this corresponds to selecting either “Spatial frame” or “Material frame” in the settings for the integration operator.
Strain Measures
When investigating the uniaxial tensile test above, three different representations of the strain were introduced. It is possible to generalize all of them to multiaxial cases, but for the true strain this is not trivial. It has to be done through a representation in the principal strain directions because that is the only way to take the logarithm of a tensor. The general tensor representation of the logarithmic strain is often called
Hencky strain.
There are also many other possible representations of the deformation. Any reasonable representation however, must be able to represent a rigid rotation of an unstrained body without producing any strain. The engineering strain fails here, thus it cannot be used for general geometrically nonlinear cases. One common choice for representing large strains is the
Green-Lagrange strain. It contains derivatives of the displacements with respect to the original configuration. The values therefore represent strains in material directions, similar to the behavior of the Second Piola-Kirchhoff stress. This allows a physical interpretation, but it must be realized that even for a uniaxial case, the Green-Lagrange strain is strongly nonlinear with respect to the displacement. If an object is stretched to twice its original length, the Green-Lagrange strain is 1.5 in the stretching direction. If the object is compressed to half its length, the strain would read -0.375.
An even more fundamental quantity is the
deformation gradient, \mathbf F, which contains the derivatives of the deformed coordinates with respect to the original coordinates, \mathbf F = \frac{\partial \mathbf x}{\partial \mathbf X}. The deformation gradient contains all information about the local deformation in the solid, and can be used to form many other strain quantities. As an example, the Green-Lagrange strain is \frac{1}{2} (\mathbf{F}^T \mathbf F-\mathbf I). A similar strain tensor, but based on derivatives with respect to coordinates in the deformed configuration, is the Almansi strain tensor, \frac{1}{2} ( \mathbf I-( \mathbf{F} \mathbf F^T)^{-1}). The Almansi strain tensor will then refer to directions fixed in space. Conjugate Quantities
A general way to express the continuum mechanics problem is by using a weak formulation. In mechanics this is known as the principle of virtual work, which states that the internal work done by an infinitesimal strain variation operating on the current stresses equals the external work done by a corresponding virtual displacement operating on the loads. The stress and strain measures must then be selected so that their product gives an accurate energy density. This energy density may be related either to the undeformed or deformed volume, depending on whether the internal virtual work is integrated over the original or the deformed geometry.
In the table below, some corresponding conjugate stress-strain pairs are summarized:
Strain Stress Symmetry Volume Orientation Engineering Strain (based on deformed geometry); True strain; Almansi strain Cauchy (True stress) Symmetric Deformed Spatial Engineering Strain (based on deformed geometry); True strain; Almansi strain Kirchhoff Symmetric Original Spatial Deformation gradient First Piola-Kirchhoff (Nominal Stress) Non-symmetric Original Mixed Green-Lagrange strain Second Piola-Kirchhoff (Material Stress) Symmetric Original Material
In the
Solid Mechanics interface in COMSOL Multiphysics, the principle of virtual work is always expressed in the undeformed geometry (the “Material frame”). Green-Lagrange strains and Second Piola-Kirchhoff stresses are then used. Such a formulation is sometimes called a “Total Lagrangian” formulation. A formulation that is instead based on quantities in the current configuration is called an “Updated Lagrangian” formulation. Additional Resources on Stresses and Strains Comentários (5) CATEGORIAS Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
Tagged: null space Problem 303
Let $A$ be a real $7\times 3$ matrix such that its null space is spanned by the vectors
\[\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \text{ and } \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}.\] Then find the rank of the matrix $A$.
(
Purdue University, Linear Algebra Final Exam Problem) Read solution Problem 270
Let
\[A=\begin{bmatrix} 4 & 1\\ 3& 2 \end{bmatrix}\] and consider the following subset $V$ of the 2-dimensional vector space $\R^2$. \[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\] (a) Prove that the subset $V$ is a subspace of $\R^2$.
Add to solve later
(b) Find a basis for $V$ and determine the dimension of $V$. Problem 260
Let \[A=\begin{bmatrix}
1 & 1 & 2 \\ 2 &2 &4 \\ 2 & 3 & 5 \end{bmatrix}.\] (a) Find a matrix $B$ in reduced row echelon form such that $B$ is row equivalent to the matrix $A$. (b) Find a basis for the null space of $A$. (c) Find a basis for the range of $A$ that consists of columns of $A$. For each columns, $A_j$ of $A$ that does not appear in the basis, express $A_j$ as a linear combination of the basis vectors.
Add to solve later
(d) Exhibit a basis for the row space of $A$. Problem 252
Let $W$ be the subset of $\R^3$ defined by
\[W=\left \{ \mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\in \R^3 \quad \middle| \quad 5x_1-2x_2+x_3=0 \right \}.\] Exhibit a $1\times 3$ matrix $A$ such that $W=\calN(A)$, the null space of $A$. Conclude that the subset $W$ is a subspace of $\R^3$. Problem 222
Suppose that $n\times n$ matrices $A$ and $B$ are similar.
Then show that the nullity of $A$ is equal to the nullity of $B$.
In other words, the dimension of the null space (kernel) $\calN(A)$ of $A$ is the same as the dimension of the null space $\calN(B)$ of $B$. Given Graphs of Characteristic Polynomial of Diagonalizable Matrices, Determine the Rank of Matrices Problem 217
Let $A, B, C$ are $2\times 2$ diagonalizable matrices.
The graphs of characteristic polynomials of $A, B, C$ are shown below. The red graph is for $A$, the blue one for $B$, and the green one for $C$.
From this information, determine the rank of the matrices $A, B,$ and $C$.
Read solution Add to solve later
Problem 211
In this post, we explain how to diagonalize a matrix if it is diagonalizable.
As an example, we solve the following problem.
Diagonalize the matrix
\[A=\begin{bmatrix} 4 & -3 & -3 \\ 3 &-2 &-3 \\ -1 & 1 & 2 \end{bmatrix}\] by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.
(Update 10/15/2017. A new example problem was added.)
Read solution Problem 200
Let
\[ A=\begin{bmatrix} 5 & 2 & -1 \\ 2 &2 &2 \\ -1 & 2 & 5 \end{bmatrix}.\]
Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix.
Your score of this problem is equal to that dimension times five.
(
The Ohio State University Linear Algebra Practice Problem) Read solution Problem 164
Let $T:\R^4 \to \R^3$ be a linear transformation defined by
\[ T\left (\, \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \,\right) = \begin{bmatrix} x_1+2x_2+3x_3-x_4 \\ 3x_1+5x_2+8x_3-2x_4 \\ x_1+x_2+2x_3 \end{bmatrix}.\] (a) Find a matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$. (b) Find a basis for the null space of $T$. (c) Find the rank of the linear transformation $T$.
(
The Ohio State University Linear Algebra Exam Problem) Read solution Problem 154
Define the map $T:\R^2 \to \R^3$ by $T \left ( \begin{bmatrix}
x_1 \\ x_2 \end{bmatrix}\right )=\begin{bmatrix} x_1-x_2 \\ x_1+x_2 \\ x_2 \end{bmatrix}$. (a) Show that $T$ is a linear transformation. (b) Find a matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ for each $\mathbf{x} \in \R^2$.
Add to solve later
(c) Describe the null space (kernel) and the range of $T$ and give the rank and the nullity of $T$. Problem 121
Let $A$ be an $m \times n$ real matrix. Then the
null space $\calN(A)$ of $A$ is defined by \[ \calN(A)=\{ \mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}_m\}.\] That is, the null space is the set of solutions to the homogeneous system $A\mathbf{x}=\mathbf{0}_m$.
Prove that the null space $\calN(A)$ is a subspace of the vector space $\R^n$.
(Note that the null space is also called the kernel of $A$.) Read solution Problem 38
Let $A$ be an $m \times n$ real matrix.
Then the of $A$ is defined as $\ker(A)=\{ x\in \R^n \mid Ax=0 \}$. kernel
The kernel is also called the
of $A$. null space
Suppose that $A$ is an $m \times n$ real matrix such that $\ker(A)=0$. Prove that $A^{\trans}A$ is invertible.
(
Stanford University Linear Algebra Exam) |
There is a fairly common example of a 2nd order formula that is only true in a universe with infinite domain:
$$\begin{align} \exists R \quad & \forall x && \lnot xRx \\ \land & \forall x \exists y && x R y \\ \land & \forall x \forall y && x R y \implies \lnot y R x \\ \land & \forall x \forall y \forall z && x R y \land y R z \implies x R z \\ \end{align}$$
And it is fairly easy to establish a formula that is only true in a finite universe:
$$\exists x \forall y~ x = y$$
which has only 1 element in the universe, or
$$\exists x_1 \exists x_2 \forall y ~x_1 = y \lor x_2 = y$$
which has no more than 2 elements in the universe. Is there a formula that only holds when the universe is finite without putting an upper finite limit on its size? In other words, it would be consistent with any formula of the form $\exists x_1 \dots \exists x_n \forall y ~ x_1 = y \lor \dots \lor x_n = y$ ? I would be interested even if the formula was a higher order (quantified over relations or functions) formula; however, I am mostly interested in formulas with no free variables and no unquantified constants. |
I would like to know what can we know about the following integral:
$$\intop_{0}^{\pi}f'(x)\cos x\ dx$$
whenever $f'(0)=f'(\pi)=0$.
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I would like to know what can we know about the following integral:
$$\intop_{0}^{\pi}f'(x)\cos x\ dx$$
whenever $f'(0)=f'(\pi)=0$.
Let us consider the vector space $V=\{f \in C1[0, \pi]: f'(0)=f'( \pi)=0\}$ and the linear functional $T:V \to \mathbb R$, defined by
$T(f)=\intop_{0}^{\pi}f'(x)\cos x\ dx$.
Since $T$ is not the zero-functional, $T(V)= \mathbb R$ |
I'm self-studying differential geometry using Frankel's ``The Geometry of Physics".
The first problem (1.1(1)) is about determining whether or not the locus $$x^2+y^2-z^2 = c $$ is a submanifold in $\mathbb{R}^3$, for $c\in\{1,0,-1\}$.
My solution was based on an example earlier in the text - that of the unit sphere. We can describe points in each hemisphere as $$ z = \pm F(x,y) =\pm \sqrt{x^2+y^2-c},$$ and we can see that the function is differentiable everywhere for $c<0$, which means (I gather from the aforementioned example).
My question is what happens for $c\geq0$. I would think that this is no longer a submanifold since around the origin, there is a ``hole", and the coordinates are not well defined. Is this correct? Is there a more rigorous approach to generally show whether a locus is a submanifold or not? |
Say a matrix A is positive semi-definite. Let B be a square matrix composed of replicas of A as sub-blocks, s.t. $$B=\begin{pmatrix} A & A \\ A & A \\ \end{pmatrix},$$ or $$\begin{pmatrix} A & A & A \\ A & A & A \\ A & A & A \\ \end{pmatrix},$$ etc. Would $B$ be semi-definite as well?
For any vector $x$, divide it into appropriately-sized subvectors $x_1,\ldots,x_n$ so that $$\begin{align} x^TBx &= \begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}^T\begin{bmatrix}A&\cdots& A\\\vdots&\ddots&\vdots\\A&\cdots&A\end{bmatrix}\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix} \\ &= x_1^TAx_1 + \cdots + x_1^TAx_n \\ &\phantom{=}+ \cdots \\ &\phantom{=}+ x_n^TAx_1 + \cdots + x_n^TAx_n \\ &= (x_1+\cdots+x_n)^TA(x_1+\cdots+x_n) \end{align}$$ which is nonnegative because $A$ is positive semidefinite.
Let $X$ and $Y$ be symmetric positive semidefinite matrices. Then $X\otimes Y$, where $\otimes$ denotes the Kronecker product, is symmetric positive semidefinite as well.
Let $$ X=E, \quad Y=A, $$ where $E$ is a square matrix of ones (for the matrices in question, $E$ is $2\times 2$ or $3\times 3$). Note that $E=ee^T$ with $e=[1,1,\ldots,1]^T$, so $E$ is obviously symmetric positive semidefinite. Now use the fact above.
Let $$B=\left( \begin{array}{cc} A & A \\ A & A \\ \end{array} \right)$$ Decomposit an Vector $z\in\mathbb{R}^{2n}$ to $$z=\left( \begin{array}{cc} x \\ y \\ \end{array}\right)\quad x,y\in \mathbb{R}^n$$ then $$Bz=\left( \begin{array}{cc} A & A \\ A & A \\ \end{array} \right) \left( \begin{array}{cc} x \\ y \\ \end{array} \right)=\left( \begin{array}{cc} Ax+Ay \\ Ax+Ay \\ \end{array} \right) $$ So $$\langle z,Bz\rangle=\left\langle\left( \begin{array}{cc} x \\ y \\ \end{array} \right),\left( \begin{array}{cc} Ax+Ay \\ Ax+Ay \\ \end{array} \right)\right\rangle=xAx+xAy+yAx+yAy=(x+y)A(x+y)\geq0$$
No, consider the 2-by-2 matrix with 1 in all entries.
Edit: I may have been too quick here, that matrix so defined actually is positive semi-definite, but it is not positive definite.
Edit2: I think you are correct: Let $A\geq 0$ and let $M$ be the '2-by-2' block matrix with $A$ in all 4 blocks. Then $$ \langle M(x,y),(x,y)\rangle = \langle Ax,x\rangle +\langle Ay,y\rangle +\langle Ax,y\rangle+\langle Ay,x\rangle=\langle Ax,x\rangle +\langle Ay,y\rangle +2Re \langle Ax,y\rangle. $$ Now, since $A\geq 0$, we can employ the Cauchy-Schwarz inequality to obtain
$$ |Re \langle Ax,y\rangle|\leq |\langle Ax,y\rangle|\leq \sqrt{\langle Ax,x\rangle\langle Ay,y\rangle}\leq \frac{1}{2}(\langle Ax,x\rangle + \langle Ay,y\rangle), $$
which implies that $M\geq 0$. I believe this will go through to higher orders as well.
Yes in special cases because the determinant of a block matrix is the product of the determinants of the blocks, if the blocks are placed in the the main diagonal. This is not true for general block matrices. Something like
[A 0 0]
[0 B 0]
[0 0 C]
is allowed. |
Jupyter Notebook here: https://git.io/fjRjL PDF Article here: http://dx.doi.org/10.13140/RG.2.2.17472.58886 For an explicit compressible algorithm the maximum timestep one can take is dictated by both the advective and acoustic speeds in the flow. This is given by the famous CFL condition \begin{equation} V \frac{\delta t}{\Delta x} \leq 1 \end{equation} where $V$ is the maximum speed in the …
If you’ve worked in computational fluid dynamics, then you’re probably aware of the Taylor-Green vortex – at least the two-dimensional case. A simple google search will land you on this wikipedia page. The classic solution there is presented in the form \begin{equation} u = \cos x \sin y F(t);\quad v = -\sin x \cos y …
LaTeX subequations produce a suprious space or indent immediately after. To get rid of this space, do one of two things Place your subequations label at the begining of the subequations environment, If you insist on placing the label at the end of the subequations environment, then place % sign after that.
Kronecker products can be used to efficiently and easily create 2D and 3D finite difference (and other) operators based on simple 1D operators for derivatives. Here’s a Jupyter notebook that shows you how to do this. You can find this notebook on nbviewer here.
This is from a short talk I gave at our group meeting last week.
In a recent article, I discussed how to sync your zotero library with your box account. However, as many have informed me, there seems to be a problem in setting up the “initial” folder. In fact, I just encountered this same exact problem today on a new computer that I am setting up. Today I …
For Faster video speeds use: ffmpeg -i input.mov -filter:v “setpts=0.5*PTS” output.mov For Slower video speeds use: ffmpeg -i input.mov -filter:v “setpts=2*PTS” output.mov Note the factor multiplying PTS. If that factor is less than 1, then you get a faster video. The opposite otherwise. Thanks to: Modify Video Speed with ffmpeg
May all the best that time and chance have to offer be yours – Jeff Bendock: friend, mentor, and role model. Today (Aug 10) my wife and I went through an extraordinary experience. But what we experienced was not unique. We’re not the first or last people to experience it by any means. There was …
Here’s a neat trick I learned today to display all matplotlib plots as vector format rather than raster in Jupyter notebooks:
You can view your (public) jupyter notebooks hosted by gitlab using nbviewer by changing the “blob” work in the hyperlink to raw.
Convert and display in place: jupyter-nbconvert –to slides mynotebook.ipynb –reveal-prefix=reveal.js –post serve Convert to html: jupyter-nbconvert –to slides mynotebook.ipynb –reveal-prefix=reveal.js as always, try: jupyter-nbconvert –help
This issue appears to have showed up on Sierra. Here’s a simple fix: Edit your bash profile (emacs ~/.bash_profile) Add export BROWSER=open Close your shell or source it (source ~/.bash_profile) Things should get back to normal now. Ref: https://github.com/conda/conda/issues/5408
make sure you don’t fall into the trap of copying pointers in Python: import numpy as np a = np.zeros(2) print(“a = “, a) b = np.zeros(2) print(“b = “, b) b = a # this assignment is simply a pointer copy – b points to the same data pointed to by a a[0]=33.33 # …
First install a utility called highlight. It is available through macports and homebrew. sudo port install highlight Now that highlight is installed, you can “apply” it to a file and pipe it to the clipboard: highlight -O rtf MyCode.cpp | pbcopy Now go to Kenote and simply paste from clipboard (or command + v). Highlight … |
In this post we take in our hands two simple tools. The first is the generating function of the harmonic numbers, you can see it in this Wikipedia, section 3, that holds for $|z|<1$. The second is the Cauchy product formula $$ \left( \sum_{n=1}^\infty a_n \right) \left( \sum_{n=1}^\infty b_n \right) = \sum_{n=1}^\infty a_k b_{n-k+1},$$ where the convergence is assumed, there are theorems that tell you when is convergent.
In our case our factors are convergents because are the same, and this is well defined: we consider the square of the generating function and after we take the integral $\int_0^{1/2}$ to get if there are no mistakes $$\sum_{n=1}^\infty\frac{1}{(n+2)2^{n+2}}\sum_{k=1}^{n}H_k H_{n-k+1}=\int_0^{1/2}\left(\sum_{n=1}^\infty H_n z^n\right)^2 dz=\int_0^{1/2}\left(\frac{\log(1-z)}{1-z}\right)^2 dz.$$ Notice that is required use the Cauchy product and swap the sign of the series and integral. The integral in RHS is computed in a closed form as $2(\log 2-1)^2$.
I don't know if is well known from the literature this sequence $$\sum_{k=1}^n H_k H_{n-k+1}.$$ The sequence starts as $1,3, \frac{71}{12}, \frac{29}{3}, \frac{638}{45}, \frac{349}{18}, \frac{14139}{560}, \frac{79913}{2520}\ldots$ And I would like to know what's about its asymptotic behaviour. One know that the plot of this arithmetic function is
smooth (with a well defined slope), see how Wolfram Alpha can show us the plot of partial sums if you type the code sum HarmonicNumber[k]HarmonicNumber[1000-k+1], from k=1 to 1000
in its online calculator.
Question.What's about the asymptotic behaviour of $$\sum_{k=1}^{n}H_k H_{n-k+1}$$ as $n\to\infty$? I am saying a big oh, or small oh statement or your answer as an asymptotic equivalence. You can provide me hints to get it with summation (I don't know if it is easy), or have you another idea? Thanks in advance.
Thus feel free to add hints, references if you need it, or a more detailed answer. |
I think that $H_1$ is the so called second quantized Hamiltonian, which lives in a multi-particle Hilbert space. While the other Hamiltonian, $H_2$, is a single-particle Hamiltonian, living in a single-particle Hilbert space.
If, as the question v4 says, the symbol $|m\rangle$ indicates a state of a particle localized to site $m$, then $H_2$ is indeed supposed to be viewed as a single particle Hamiltonian.In that case, the Hamiltonians are not necessarily the same.However, note that $H_1$ conserves the number of particles, so if you start with one particle, then practically speaking $H_1$ and $H_2$ are the same.
To what extent can one switch between the two Hamiltonians in the middle of a calculation? I believe that this often done every now and then, perhaps quite sloppily. What really happens as we switch between these formulations?
Often, that switch happens as follows:
Start with a multi-particle hamiltonian in second quantized language. Identify non-interacting modes in the system. Treat each non-interacting mode separately as its own single-particle system where the energy level of this fake single particle actually corresponds to the number of quanta (Fock state) in that mode.
That's the only case I can think of where we regularly switch from second quantization to single-particle quantum mechanics.
In chat, OP asked what is "given up" by going from second quantization to single-particle.in the case described by the bullet points, nothing is sacrificed.We treat each mode separately because they're non-interacting.However, suppose we just have a box which may or may not contain electrons.In full generality, we should use second quantization and allow for an arbitrary number of particles.However, if we know that there's only one (or more) electrons and we know that we're working with low enough energy that spontaneous electron creation will not happen, then we can use single particle (or two-particle or three-particle... based on how many electrons there are) quantum mechanics to describe the system.We do this all the time when describing e.g. the electrons in a helium atom.We know there are two electrons and we know there isn't going to be any particle creation/annihilation (unless we're at high enough energy).
Note that the energy to create a massive particle is $E = m c^2$ were $c$ is the speed of light.It's because this energy cost is large that we know the number of electrons in a helium atom won't change under normal conditions.On the other hand, if we're working with photons, which have zero mass, or we're working in a thermodynamic situation where we have a reservoir of particles, then the particle number isn't fixed and we can't restrict to a specific particle number.
If we interpret $|m\rangle$ to indicate a Fock state with one occupation on site $m$, and if we're working with Fermions, then the Hamiltonians are the same, but written in two different notations.The rest of this post elaborates on the previous sentence.
Representing linear transformations (operators) as sums of bras and kets
Given a vector space and a basis $e \equiv \{|e_1\rangle, |e_2\rangle,\ldots\}$, we can write
any linear transformation $T$ on that vector space as a sum$$T = \sum_{ij} T_{ji} |e_j \rangle \langle e_i | $$where $T_{ji}$ are literally the elements of the matrix representation of $T$ in the $e$ basis.We can show that trivially:$$ \langle e_j | T | e_i \rangle = \sum_{kl} T_{lk} \underbrace{\langle e_j | e_l \rangle}_{\delta_{jl}} \langle \underbrace{e_i | e_ k \rangle}_{\delta_{ik}} = T_{ji} \, .$$In other words, in the $e$ basis, the $j^\text{th}$ component of $T|e_i\rangle$ is $T_{ji}$.
An easy way to remember this is to use the fact that $\sum_j |e_j\rangle \langle e_j| = \text{identity}$, so$$T = \underbrace{\left( \sum_j |e_j \rangle \langle e_j| \right)}_\text{identity}T\underbrace{\left( \sum_i |e_i \rangle \langle e_i| \right)}_\text{identity}= \sum_{ij} \langle e_j | T | e_i \rangle |e_j\rangle\langle e_i| = \sum_{ij} T_{ji} |e_j \rangle \langle e_i| \, .$$
Fermion creation/annihilation
In second quantization language, we represent quantum states etc. with creation annihilation operators $c_m$ and $c_m^\dagger$, as noted in the original post.Those operators are of course just linear transformations and can be represented in the bra/ket way, just the same as the arbitrary transformation $T$ in the discussion above.
Physically, the operator $c_m$ removes one unit of excitation from the $m^\text{th}$ mode of the system, which in this case is a spatially localized mode, i.e. occupation of a lattice site.The operator $c_m^\dagger c_m$ removes one excitation from the $m^\text{th}$ mode and then puts it back.
Suppose we work in a basis consisting of two elements for each lattice site: $\{|0\rangle, |1\rangle\}$ corresponding to zero or one occupation on the site (these are Fermions so we can't have more than one).We can then write $c^\dagger c$ for a particular lattice site as\begin{align}c^\dagger c&= \langle 0 | c^\dagger c | 0 \rangle |0 \rangle \langle 0 | + \langle 0 | c^\dagger c | 1 \rangle |0 \rangle \langle 1 | + \langle 1 | c^\dagger c | 0 \rangle |1 \rangle \langle 0 | + \langle 1 | c^\dagger c | 1 \rangle |1 \rangle \langle 1 | \\&= \langle 1 | c^\dagger c | 1 \rangle |1 \rangle \langle 1 |\end{align}where in the second line we dropped terms that were zero.If a term in the Hamiltonian is $t (c^\dagger c)$, then we can write it as$t |1 \rangle \langle 1 |$.
All of that discussion was for a single mode.If we now go to all modes (sites on the lattice) and rename each $|1\rangle$ ket for that mode as $|m\rangle$, we can see that $H_1$ and $H_2$ are exactly the same thing. |
Hiroshima Mathematical Journal Hiroshima Math. J. Volume 46, Number 3 (2016), 333-341. Stable extendibility of some complex vector bundles over lens spaces and Schwarzenberger’s theorem Abstract
We obtain conditions for stable extendibility of some complex vector bundles over the $(2n + 1)$-dimensional standard lens space $L^n(p) \operatorname{mod} p$, where $p$ is a prime. Furthermore, we study stable extendibility of the bundle $\pi^*_n (\tau(\mathbf{C}P^n))$ induced by the natural projection $\pi_n : L^n(p)\to \mathbf{C}P^n$ from the complex tangent bundle $\tau(\mathbf{C}P^n)$ of the complex projective $n$-space $\mathbf{C}P^n$. As an application, we have a result on stable extendibility of $\tau(\mathbf{C}P^n)$ which gives another proof of Schwarzenberger’s theorem.
Article information Source Hiroshima Math. J., Volume 46, Number 3 (2016), 333-341. Dates Received: 2 February 2016 Revised: 30 August 2016 First available in Project Euclid: 25 February 2017 Permanent link to this document https://projecteuclid.org/euclid.hmj/1487991625 Digital Object Identifier doi:10.32917/hmj/1487991625 Mathematical Reviews number (MathSciNet) MR3614301 Zentralblatt MATH identifier 1367.55008 Subjects Primary: 55R50: Stable classes of vector space bundles, $K$-theory [See also 19Lxx] {For algebraic $K$-theory, see 18F25, 19-XX} Secondary: 55N15: $K$-theory [See also 19Lxx] {For algebraic $K$-theory, see 18F25, 19- XX} Citation
Hemmi, Yutaka; Kobayashi, Teiichi. Stable extendibility of some complex vector bundles over lens spaces and Schwarzenberger’s theorem. Hiroshima Math. J. 46 (2016), no. 3, 333--341. doi:10.32917/hmj/1487991625. https://projecteuclid.org/euclid.hmj/1487991625 |
Consider the simple symmetric random walk on $\mathbb{Z}$. That is, let $X_1, X_2, \dots$ be i.i.d. random variables with $$ P(X_i=1)=P(X_i=-1)=1/2, $$ and define $S_n=X_1+\dots+X_n$ with $S_0=0$. As is well known, the sum $S_n$ is (null) recurrent and satisfies the law of the iterated logarithm $$ P\left(\limsup_{n\to \infty} \frac{S_n}{\sqrt{2n \log \log n}}=1\right)=1, $$
Assume now that we restrict ourselves to the subset of realisations of $S_n$ where each realisation satisfies
$\frac{1}{n} S_n \to 0$.
For any integer $m$, there are infinitely many values of $n$ such that $S_n=m$ or $S_n=-m$.
Does the law of the iterated logarithm take a stronger form in this case, in the sense:
Q: If $S_n$ is satisfies the criteria above, does this imply $$ \limsup_{n\to \infty} \frac{|S_n|}{\sqrt{2n \log \log n}}\leq 1? $$
Or are there even in this case subsets of realisations for which this is not true? |
Recently, I came across a gem of a physics problem, which had the awesome combination of being easy to comprehend, yet being surprisingly difficult to solve.
The problem is thus:
The earth orbits around the sun because it has angular momentum. If we stopped the earth in orbit then let it fall straight towards the sun, then how long would it take to reach the sun in seconds?
Now, I love physics and maths, and I believed that it would be a simple matter to knock this one off. Hah... was I mistaken.
All right... first, I did succeed in finding the answer and it turned out to be worthy of all the trouble. But before I tell you what it is, why don't you join me in the journey that starts from the first principles and leads to this
beautiful result. If you are impatient, just scroll to the bottom for the answer... but if you can hold on to your horses (and don't mind some mathematics), you might enjoy yourself more.
Ok, lets start.
We want to find out the time taken by an object, \(m_e\) ( the earth ), to fall towards a much heavier object, \(M_s\) ( the sun ), purely under the force of gravity, given that \(m_e\) starts to fall at time \(t = 0\) from a stationary position a certain distance \((d = 150 \times 10^9 m)\) away. Let us assume that the earth and the sun lie on the \(x\) axis and at time \(t=0\), let the earth be located at at \(x=0\) and the sun at \(x=d\).
We know from Newton's Law of Gravitation that the force between the earth and the sun is given by
where
\(F\) : force in newtons (\(\text{kg}\text{ m}\text{ s}^{-2}\)) \(G\) : gravitational constant : \(6.67384 \times 10^{-11}\text{ kg}^{-1}\text{m}^3\text{s}^{-2}\) \(M_s\) : mass of the sun : \(1.989 \times 10^{30}\text{ kg}\) \(m_e\) : mass of the earth : \(5.972 \times 10^{24}\text{ kg}\) \(r\) : distance between earth and sun in meters (changes as earth falls). at \(t=0\), \(r = d = 1.496 \times 10^{11}\text{ m}\)
Also, let \(x\) be the position of the earth on the \(x\text{-axis}\), such that, at any given time, \((d-x)\) is the distance between the earth and the sun. Now, we try to solve this equation to find the time...
Now, we also know that
and we can write
The \(\frac{dv}{dt} = v \frac{dv}{dx}\) was one of the
key steps which enables us to separate the variables and solve the differential equation. Now,
Separating the variables (moving dx to the right side) and integrating, we get
Performing the integration, we get
where \(C\) is the constant of integration. We find \(C\) by using the known condition that \(v = 0\) at \(x = 0\),
Putting this value of \(C\) in the equation for \(v\),
If you look closely, you might notice that the above equation is the
law of conservation of gravitational potential and kinetic energy. Woohoo... we just derived a conservation law from first principles (Aside : this intermediate result probably keeps popping up in astronomical problems all the time, and was made into a law of it own. I could have used this law right in the beginning to find the velocity at any distance from the sun, but then, I would not have derived the conservation law by myself). Ok... back to the problem.
Moving \(dt\) to the right and variables involving \(x\) to the left and integrating, we get
This is where I hit another roadblock. The integral on the left side looked harmless enough, but was surprisingly difficult to solve (as we know well enough about integrals... if someone is getting too cocky about their mathematical abilities, give them a random integral, or a partial differential equation to solve). I finally took the help of Wolfram|Alpha to find the indefinite integral of the left side, and was pleasantly surprised that the function did exist (I differentiated it to make sure that it really was the right answer, and you should try it too. Just like in the spy movies... trust no one. In this case, not even yourself... so check your answer twice, and get it peer reviewed). We now have the following:
At both \(x=0\) and \(x=d\), the term inside the \(tan^{-1}()\) parentheses contains a division by zero. Usually, that is disastrous for a solution. However, in this case, its not too bad as \(tan^{-1}(\pm \infty)\) is well defined. And to know whether the division by zero leads to a \(+\infty\) or a \(-\infty\), we need to evaluate the LHS at \(x=0^+\) and \(x=d^-\):
The term \(\sqrt{x(d-x)}\) equals \(0^+\)at both \(x=0^+\) and \(x=d^-\)
Putting the LHS and RHS together, we get
Rearranging the variables around, we finally get
As often happens in interesting problems, we have journeyed from a simple equation (newton's law) to terrible looking intermediate results, divisions by zero, and challenging integrals... which all finally canceled out to give a simple answer.
Wait... there is a \(\pi\) in the answer!! How in the frigging cosmos did \(\pi\) make its way into the answer. Well, we can go back and look at where it exactly came in... through the \(tan^{-1}(+\infty)\) and \(tan^{-1}(-\infty)\), which basically means that it popped out from a couple of singularities. It might also have something to do with the periodic oscillation that the earth might have experienced were it not going to be evaporated on contact with the sun. This is as close to magic as we get in mathematics, and yet another beautiful example of unexpected places where \(\pi\) shows up.
Now that we have done all the hard work of finding the general solution, its time to put in the values to get the figure of...
or about 2.1 months.
All this is excellent but I also love programming. Its time to test this analytical solution with a numerical calculation.
I wrote a small python program to calculate the time by simulating the acceleration, velocity and position of the earth as it falls towards the sun. I have also plotted a graph denoting the distance of earth from the sun
vs time.
%matplotlib inline# calculate time for earth to fall into sun...from math import sqrtfrom time import time# physical constants/valuesG = 6.67384e-11 # gravitational constantMS = 1.98855e30 # mass of sunesd = 149597870700.0 # earth-sun distance in meters# program variablesn = 1000000myconst1 = (G*MS*n*n)/(esd*esd)myconst2 = esd/ndef getA(k): """ acceleration at distance (esd*k/n) from sun""" ak = myconst1/pow(n-k,2) return akdef getV(v, a, t): """ velocity at distance (esd*k/n) from sun""" vk = v + a*t return vkdef getVAT(k, vold, aold, told): """ time to get from position (esd*k/n) to (esd*(k+1)/n)""" vk = getV(vold, aold, told) ak = getA(k) tk = (sqrt(vk*vk + 2*ak*myconst2) - vk)/ak return vk, ak, tkstime = time()sts, v, a, t = 0, 0, 0, 0plotValues = []for x in range(n): v, a, t = getVAT(x, v, a, t) sts += t if x % 5000 == 0: plotValues.append((sts, n - x))plotValues.append((sts, n-x-1))print "Travel time = %.2f seconds\nDays = %5.3f"%(sts, sts/(60.0*60*24))print "Computation time = %5.4f seconds"%(time() - stime)
Travel time = 5578753.47 secondsDays = 64.569Computation time = 1.8074 seconds
Here's the plot of earth's approach towards the sun...
import matplotlib.pyplot as pltxvals = [sts for sts, x in plotValues]yvals = [149.597870*x for sts, x in plotValues]plt.xlabel("Travel time (seconds)")plt.ylabel("Distance between earth / sun (km)")plt.plot(xvals, yvals) |
Suppose non-trivial vacuum configuration of the Yang-Mills theory with the winding number $n$:
$$ \tag 1 A_{\mu}(x) = g_{(n)}(x)\partial_{\mu}g^{-1}_{(n)}(x) $$
The winding number is given by the surface integral of topological density over the 3-sphere:
$$ \tag 2 n = \frac{1}{24\pi^{2}}\int_{S^{3}} d\sigma_{\mu}\epsilon^{\mu\nu\alpha\beta}\text{tr}\big[(g_{(n)}\partial_{\nu}g_{(n)}^{-1})(g_{(n)}\partial_{\alpha}g_{(n)}^{-1} )(g_{(n)}\partial_{\beta}g_{(n)}^{-1} )\big] \ $$
In various literature sources (for example, in Rubakov's "Classical gauge fields. Bosons") people often rewrite the surface integral $(2)$ in terms of volume integral:
$$ \tag 3 n = -\frac{1}{16\pi^{2}}\int d^{4}x\text{tr}\big[F_{\mu\nu}\tilde{F}^{\mu\nu}\big], $$
where $F$ is the gauge field strength and $\tilde{F} = *F$ is its dual. They claim that $(3)$ and $(2)$ are equivalent. But in fact $(2)$ gives non-zero integer result for the pure gauge $(1)$, while $(3)$ vanishes! This can be seen by choosing the 4-dimensional
euclidean manifold to be the "cylinder", with the planes being defined by $\tau = \pm \infty$. Then in the gauge $A_{0} = 0$ we obtain from $(3)$ $$ \tag 4 n = n(\tau = \infty) - n(\tau = -\infty) $$
The precise reason is that we include $\epsilon^{\mu\nu\alpha\beta}\text{tr}\big[\partial_{\mu}F_{\nu\alpha}A_{\beta}\big]$ term in the action when converting the surface integral into the volume integral.
So why do people say that $(2)$ and $(3)$ are equivalent? |
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