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I'm trying to find the power function in the t test of two samples, (the variances are assumed to be equal ($\sigma_1=\sigma_2$ ), in the paper, on page 144 (5), I found that
"The power to detect a difference of $\delta=\mu_1-\mu_2$ with two-sided significance level $\alpha$ is given by:" $$1-\beta =T_v \left(t_{\alpha/2,v} ~ \left| \right. \frac{\delta}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}} \right)-T_v \left(-t_{\alpha/2,v} ~ |~\frac{\delta}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}} \right)$$
I tried to prove it, $1-\beta=P( |\bar{X_1}-\bar{X_2}|>cv ~|~\mu_1-\mu_2=\delta)$
$\beta=P(\frac{-cv-\delta}{s_p\sqrt{\sigma_1/n_1+\sigma_2/n_2}}<t<\frac{cv-\delta}{s_p\sqrt{\sigma_1/n_1+\sigma_2/n_2}})~~~~~cv =\mbox{critical value}$
what am I doing wrong? How can I prove it? |
Let $x_0$ be a fixed vector in a Hilbert space $H$ and suppose $\{x_1,...,x_n\}$ and $\{y_1,...,y_m\}$ are sets of linearly independent vectors in $H$. We seek the vector
$x^* = \operatorname{arg} \min\limits_x \|x-x_0\|$,
subject to
$x \in M=\text{span}(x_1,...,x_n)$ and $\langle x, y_i \rangle=c_i$ for $i=1,...,m$ where the $c_i$'s are constants.
ADDITIONAL INFO:
This question corresponds to exercise 22 in chapter 3 of Luenberger's book Optimization by Vector Space Methods. The exercise only asks for a set of linear equations which uniquely determine the solution (and not for the solution itself). The exercise should be solved by using the Hilbert projection theorem (and not other techniques, like Lagrange multipliers).
My attempt:
Let $N=\text{span}(y_1,...,y_m)$. Then, the linear variety defined by the equality constraints $\langle x, y_i \rangle=c_i$ is a translation of $N^\perp$.
Since $M$ and $N^\perp$ are closed, $M \cap N^\perp$ is also closed and, by the projection theorem, $x^*$ exists and is unique, assuming that $M \cap N^\perp$ is not empty. Furthermore, $x^*-x_0 \in (M \cap N^\perp)^\perp$.
$x^* \in M \iff x^*=\sum_{j=1}^n\alpha_j x_j$, for some scalars $\alpha_j$. Defining $X$ as the matrix with columns $x_j$ and $\alpha$ as the vector of the $\alpha_j$'s, we can write $x^*=X \alpha$.
Using this in the equality constraints yields $\sum_{j=1}^n \alpha_j \langle x_j, y_i \rangle=c_i$ for $i=1,...,m$. Defining the $n \times m$ matrix $G$ with elements $G_{j,i} =\langle x_j, y_i \rangle$ and $c$ as the vector of the $c_i$'s, we can write $G'\alpha = c$.
If $m \geq n$, the solution of $G'\alpha = c$ either does not exist or is unique, so assume $m < n$.
Now, I probably have to use the fact that $\langle x^*-x_0, z \rangle = 0$ for any $z \in M \cap N^\perp$. Using this, we have:
$z \in M \cap N^\perp \implies z \in M \iff z = X\beta$ for some $n$-dimensional vector $\beta$.
Also, $z \in M \cap N^\perp \implies z \in N^\perp \iff \langle z, y_i \rangle=0$ for $i=1,...,m$.
Replacing $z = X\beta$ in the equalities $\langle z, y_i \rangle = 0$ yields $G'\beta = 0$, where $G$ is naturally the same as above.
Finally, using $x^* = X\alpha$ and $z = X\beta$ in $\langle x^*-x_0, z \rangle = 0$, we obtain
$\langle X\alpha, X\beta \rangle = \langle x_0, X\beta \rangle$.
And this is where I am stuck right now. We have $2n$ unknowns (the vectors $\alpha$ and $\beta$) and $2m+1$ equations:
$G'\alpha = c$ ($m$ equations),
$G'\beta = 0$ ($m$ equations),
$\langle X\alpha, X\beta \rangle = \langle x_0, X\beta \rangle$ (one equation),
where the last equation is clearly non-linear. Therefore, this cannot be the solution... |
10 1
Hi,
I've had a question ever since my quantum classes that's pretty simple I guess, but still seems to elude me. So here it is:
One text I used for quantum (Liboff's "Introductory Quantum Mechanics") says that in classical mechanics, there is a "vector of the state" of a system, that contains all the information in a mechanical system. This vector he claims contains the positions and velocities of all particles of a system (this is all just chapter 1). Then he goes on to say that in quantum mechanics, the vector of the state of the system cannot contain all these quantities simultaneously--this is the nature of the Heisenberg uncertainty principle.
But wait a second--why do the positions and velocities of a system yield all the information in classical mechanics? Is Liboff talking about
[tex]\vec x(t)[/tex] and [tex]\frac{d \vec x(t)}{dt}[/tex]?
Because, basically, if you have determined the trajectory function
[tex] \vec x(t) [/tex] (a map [tex]\Re^3 \rightarrow \Re[/tex]),
which you
So then that's where my quantum mechanics questions come in. What would the Heisenberg uncertainty principle have to say about, for instance, the uncertainty in position and
If anyone has any insight, it would be greatly appreciated. Thanks a lot!
I've had a question ever since my quantum classes that's pretty simple I guess, but still seems to elude me. So here it is:
One text I used for quantum (Liboff's "Introductory Quantum Mechanics") says that in classical mechanics, there is a "vector of the state" of a system, that contains all the information in a mechanical system. This vector he claims contains the positions and velocities of all particles of a system (this is all just chapter 1). Then he goes on to say that in quantum mechanics, the vector of the state of the system cannot contain all these quantities simultaneously--this is the nature of the Heisenberg uncertainty principle.
But wait a second--why do the positions and velocities of a system yield all the information in classical mechanics? Is Liboff talking about
instantaneouspositions and velocities, or is he talking about the trajectories (world line functions)
[tex]\vec x(t)[/tex] and [tex]\frac{d \vec x(t)}{dt}[/tex]?
Because, basically, if you have determined the trajectory function
[tex] \vec x(t) [/tex] (a map [tex]\Re^3 \rightarrow \Re[/tex]),
which you
cando in mechanics, then you can also find the velocity [tex] \vec v(t) [/tex] at all points, the acceleration [tex] \vec a(t)[/tex]...hell, I could find the 20th derivative of position if I wanted to! So what's with saying the system is "totally specified" by giving positions and velocities? I would argue that all you need is the position function [tex] \vec x(t) [/tex].
So then that's where my quantum mechanics questions come in. What would the Heisenberg uncertainty principle have to say about, for instance, the uncertainty in position and
acceleration? What about jerk? What about the seventh derivative of position? If the seventh derivative of position were the "observable" I wanted to measure in a quantum system, how would I do it? What would the operator associated with this observable be? This is one of the main reasons quantum has always confused me. In classical mechanics, you really can't seperate position, velocity, acceleration, etc. They're all just derivatives of the same fundamental world line (trajectory) function [tex] x(t) [/tex].
If anyone has any insight, it would be greatly appreciated. Thanks a lot! |
One can interpret a homomorphic image of an algebraic structure as a "collapsed" or "low-resolution" version of it, since different elements of the original structure get blurred together into becoming the same pixel in the image. Thus if we have a chain of surjective homomorphisms, we are getting higher and higher resolution pictures of ...
something, "in the limit." That something is the inverse limit.
Consider real numbers for a moment. What are they? Most people have an intuitive idea of a continuum and maybe have been exposed to the phrase "number line," but only a fraction - people who study math - get to know how we actually construct real numbers. People can cite decimal expansions, for instance that of pi, $\pi=3.1415\dots$, but what does that mean? It means there is a series of approximations of indefinitely increasing precision, $3$, $3.1$, $3.14$, $3.141$, $3.1415$, $\cdots$. In fact one
formal construction of the real numbers is to identify a real number with a(n equivalence class) of Cauchy sequences. We can think of these Cauchy sequences (of rationals) as sequences of approximations of "ghosts" that are not really there in the set of rationals.
It is similar with inverse limits. Think about $\Bbb Z/p\Bbb Z\leftarrow\Bbb Z/p^2\Bbb Z\leftarrow\Bbb Z/p^3\Bbb Z\leftarrow\Bbb Z/p^4\Bbb Z\leftarrow\cdots$ with the canonical projection maps. Every positive integer can be written as $a_0+a_1p+\cdots+a_rp^r$, i.e. can be written in base $p$. What does the projection map $\Bbb Z/p^{r+1}\Bbb Z\to\Bbb Z/p^r\Bbb Z$ do to the digital representation of an integer residue? Simple: it simply deletes the $p^r$ digit. Thus, any "sequence of approximations" can be identified with an
infinite base $p$ expansion, $a_0+a_1p+a_2p^2+\cdots$ (the sequence of approximations tells us what every digit in the infinite expansion is, and vice-versa).
In general, given a chain $G_0\leftarrow G_1\leftarrow G_2\leftarrow\cdots$, and a sequence $(g_0,g_1,g_2,\cdots)$ which is appropriately "compatible" (i.e. $g_{i+1}\mapsto g_i$) we can think of it as a "convergent sequence" whose limit is an element of the inverse limit $\varprojlim G_i$. The operation(s) can be done pointwise. This construction works just as fine if we have a
nonlinear system of surjective homomorphisms.
Now let's think about Galois automorphisms. Given any tower $L/M/K$ in which all three extensions are Galois we know there is a surjection ${\rm Gal}(L/K)\to{\rm Gal}(M/K)$. One can consider the entire system of such projection maps for algebraic extensions of a given field $K$, and then consider $\varprojlim{\rm Gal}(L/K)$ (note that the groups are varying with the field $L$ in this limit). What does an element $\sigma$ of this inverse limit look like, tangibly? Well, for any Galois extension, there is a corresponding automorphism $\sigma|_L$ of the extension $L/K$, so ultimately $\sigma$, via all of these $\sigma|_L$s, "knows" where to send
every element which is algebraic over $K$, which means this element "knows" of an automorphism of $\overline{K}/K$. And vice-versa, since any element of ${\rm Gal}(\overline{K}/K)$ restricts to an automorphism of any Galois extension $L/K$.
If we view $\varprojlim{\rm Gal}(L/K)$ as the subgroup of $\prod_L{\rm Gal}(L/K)$ comprised of "compatible" elements (recall, that means our "convergent sequences"), then the obvious map ${\rm Gal}(\overline{K}/K)\to\varprojlim{\rm Gal}(L/K)$ (where the "$L$th" coordinate of where $\sigma$ is sent to is simply the restriction of $\sigma$ to $L$) is a group isomorphism. The corresponding inverse map is simply the process of patching together all of the compatible automorphisms of Galois extensions of $K$ into one giant automorphism of $\overline{K}/K$.
One can also endow these inverse limits with the profinite topology in which case we have an isomorphism of topological groups. Another way to define the inverse limit is via universal properties in category theory. Uniqueness of inverse limits (up to unique isomorphisms) follows from a simple abstract nonsense argument, and existence follows from using the explicit construction I discussed above. |
Answer:
Let the number of students in the class be \[x\]. Then rupees donated by each student = Rs \[x\]. \[\therefore \] Rupees denoted by \[x\] students \[=\,\text{Rs}\,x\times x\] \[=\text{Rs}\,{{x}^{2}}\] \[\therefore \] The students of class VIII of a school donated Rs 2401 for Prime Minister's National Relief Fund. \[\therefore \] \[{{x}^{2}}=2401\] \[\Rightarrow \] \[x=\sqrt{2401}\]. The prime factorisation of 2401 is
7 2401 7 343 7 49 7
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The ALICE Transition Radiation Detector: Construction, operation, and performance
(Elsevier, 2018-02)
The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ...
Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2018-02)
In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ...
First measurement of jet mass in Pb–Pb and p–Pb collisions at the LHC
(Elsevier, 2018-01)
This letter presents the first measurement of jet mass in Pb-Pb and p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV and 5.02 TeV, respectively. Both the jet energy and the jet mass are expected to be sensitive to jet ...
First measurement of $\Xi_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV
(Elsevier, 2018-06)
The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The ...
D-meson azimuthal anisotropy in mid-central Pb-Pb collisions at $\mathbf{\sqrt{s_{\rm NN}}=5.02}$ TeV
(American Physical Society, 2018-03)
The azimuthal anisotropy coefficient $v_2$ of prompt D$^0$, D$^+$, D$^{*+}$ and D$_s^+$ mesons was measured in mid-central (30-50% centrality class) Pb-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm ...
Search for collectivity with azimuthal J/$\psi$-hadron correlations in high multiplicity p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 and 8.16 TeV
(Elsevier, 2018-05)
We present a measurement of azimuthal correlations between inclusive J/$\psi$ and charged hadrons in p-Pb collisions recorded with the ALICE detector at the CERN LHC. The J/$\psi$ are reconstructed at forward (p-going, ...
Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(American Physical Society, 2018-02)
The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ...
$\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV
(Springer, 2018-03)
An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ...
J/$\psi$ production as a function of charged-particle pseudorapidity density in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2018-01)
We report measurements of the inclusive J/$\psi$ yield and average transverse momentum as a function of charged-particle pseudorapidity density ${\rm d}N_{\rm ch}/{\rm d}\eta$ in p-Pb collisions at $\sqrt{s_{\rm NN}}= 5.02$ ...
Energy dependence and fluctuations of anisotropic flow in Pb-Pb collisions at √sNN=5.02 and 2.76 TeV
(Springer Berlin Heidelberg, 2018-07-16)
Measurements of anisotropic flow coefficients with two- and multi-particle cumulants for inclusive charged particles in Pb-Pb collisions at 𝑠NN‾‾‾‾√=5.02 and 2.76 TeV are reported in the pseudorapidity range |η| < 0.8 ... |
I'm no graphics expert, but appreciate why square roots are useful. The Pythagorean theorem computes distance between points, and dividing by distance helps normalize vectors. (Normalizing is often just a fancy term for division.)
3D games like Quake divide by distance zillions (yes zillions) of times each second, so "minor" performance improvements help immensely. We don't want to take the square root and divide the regular way: exponentiation and division are really, really expensive for the CPU.
Given these conditions, here's the magic formula to get $1/\sqrt{x}$, as found in Quake (my comments inserted):
float InvSqrt(float x){ float xhalf = 0.5f * x; int i = *(int*)&x; // store floating-point bits in integer i = 0x5f3759df - (i >> 1); // initial guess for Newton's method x = *(float*)&i; // convert new bits into float x = x*(1.5f - xhalf*x*x); // One round of Newton's method return x; }
Yowza! Somehow, this code gets $1/\sqrt{x}$ using only
multiplication and bit-shift operations. There's no division or exponents involved -- how does it work? My Understanding: This incredible hack estimates the inverse root using Newton's method of approximation, and starts with a great initial guess.
To make the guess, it takes floating-point number in scientific notation, and negates & halves the exponent to get something close the the inverse square root. It then runs a round of Newton's approximation method to further refine the estimate and tada, we've got something near the inverse square root.
Newton's Method of Approximation
Newton's method can be used to find approximate roots of any function. You can keep iterating the method to get closer and closer to the root, but this function only uses 1 step! Here's a crash-course on Newton's method (it was new to me):
Let's say you have a function f(x) and you want to find its root (aka where f(x) = 0). Let's call your original guess "g". Newton's method gives you a way to get a new, better guess for the root:
You can keep repeating this process (plugging in your new guess into the formula) and get closer approximations for your root. Eventually you have a "new guess" that makes f(new guess) really, really close to zero -- it's a root! (Or close enough for government work, as they say).
In our case, we want the inverse square function. Let's say we have a number $i$ (that's all we start with, right?) and want to find the inverse square root: $1/\sqrt{i}$. If we make a guess "x" as the inverse root, the error between our original number and our guess "x" is:
This is because x is roughly $1/\sqrt{i}$. If we square x we get $1/i$, and if we take the inverse we should get something close to $i$. If we subtract these two values, we can find our error.
Clearly, we want to make our error as small as possible. That means finding the "x" that makes error(x) = 0, which is the same as finding the root of the error equation. If we plug error(x) into Newton's approximation formula:
and take the proper derivatives:
we can plug them in to get the formula for a better guess:
Which is exactly the equation you see in the code above, remembering that x is our new guess (g) and "xhalf" is half of the original value ($0.5 i$):
x = x*(1.5f - xhalf*x*x);
With this formula, we can start with a guess "g" and repeat the formula to get better guesses. Try this demo for using multiple iterations to find the inverse square:
In this demo, we start by guessing the square root is half the number: $\sqrt{n} \sim \frac{n}{2}$, which means $\frac{1}{\sqrt{n}} \sim \frac{2}{n}$. Running a few rounds of Newton's Method quickly converges on the real result. (Try n=2, 4, 10, etc.)
So my friends, the question becomes: "How can we make a good initial guess?"
Making a Good Guess
What's a good guess for the inverse square root? It's a bit of a trick question -- our best guess for the inverse square root is the inverse square root itself!
Ok hotshot, you ask, how do we actually get $1/\sqrt{x}$?
This is where the magic kicks in. Let's say you have a number in exponent form or scientific notation:
Now, if you want to find the regular square root, you'd just divide the exponent by 2:
And if you want the
inverse square root, divide the exponent by -2 to flip the sign:
So, how can we get the exponent of a number without other expensive operations?
Floats are stored in mantissa-exponent form
Well, we're in luck. Floating-point numbers are stored by computers in mantissa-exponent form, so it's possible to extract and divide the exponent!
But instead of explicitly doing division (expensive for the CPU), the code uses another clever hack: it shifts bits. Right-shifting by one position is the same as dividing by two (you can try this for any power of 2, but it will truncate the remainder). And if you want to get a negative number, instead of multiplying by -1 (multiplications are expensive), just subtract the number from "0" (subtractions are cheap).
So, the code converts the floating-point number into an integer. It then shifts the bits by one, which means the exponent bits are divided by 2 (when we eventually turn the bits back into a float). And lastly, to negate the exponent, we subtract from the magic number 0x5f3759df. This does a few things: it preserves the mantissa (the non-exponent part, aka 5 in: $5 \cdot 10^6$), handles odd-even exponents, shifting bits from the exponent
into the mantissa, and all sorts of funky stuff. The paper has more details and explanation, I didn't catch all of it the first time around. As always, feel free to comment if you have a better explanation of what's happening.
The result is that we get an initial guess that is really close to the real inverse square root! We can then do a single round of Newton's method to refine the guess. More rounds are possible (at an additional computational expense), but one round is all that's needed for the precision needed.
So, why the magic number?
The great hack is how integers and floating-point numbers are stored. Floating-point numbers like $5.4 \cdot 10^6$ store their exponent in a separate range of bits than "5.4". When you shift the entire number, you divide the exponent by 2, as well as dividing the number (5.4) by 2 as well. This is where the magic number comes in -- it does some cool corrections for this division, that I don't quite understand. However, there are several magic numbers that could be used -- this one happens to minimize the error in the mantissa.
The magic number also corrects for even/odd exponents; the paper mentions you can also find other magic numbers to use.
Resources
There's further discussion on reddit (user pb_zeppelin) and slashdot:
http://programming.reddit.com/info/t9zb/comments http://games.slashdot.org/article.pl?sid=06/12/01/184205 and my comment |
Many functions can be written in terms of a power series
\[ \sum _{k=0}^{\infty} a_k(x-x_0)^k\]
If we assume that a solution of a differential equation is written as a power series, then perhaps we can use a method reminiscent of undetermined coefficients. That is, we will try to solve for the numbers \(a_k\). Before we can carry out this process, let us review some results and concepts about power series.
7.1.1 Definition
As we said, a power series is an expression such as
\[ \sum_{k=0}^\infty a_k {(x-x_0)}^k =a_0 + a_1 (x-x_0) +a_2 {(x-x_0)}^2 +a_3 {(x-x_0)}^3 + \cdots ,\]
where \( a_0,a_1,a_2,\ldots,a_k,\ldots\) and \(x_0\) are constants. Let
\[ S_n(x) = \sum_{k=0}^n a_k {(x-x_0)}^k =a_0 + a_1 (x-x_0) + a_2 {(x-x_0)}^2 + a_3 {(x-x_0)}^3 + \cdots + a_n {(x-x_0)}^n ,\]
denote the so-called partial sum. If for some \(x\), the limit
\[ \lim_{n\to \infty} S_n(x) = \lim_{n\to\infty} \sum_{k=0}^n a_k {(x-x_0)}^k\]
exists, then we say that the series (7.1.2) converges at \(x\). Note that for \(x=x_0\), the series always converges to \(a_0\). When (7.1.2) converges at any other point \(x \neq x_0\), we say that (7.1.2) is a convergent power series. In this case we write
\[ \sum_{k=0}^\infty a_k {(x-x_0)}^k = \lim_{n\to\infty} \sum_{k=0}^n a_k {(x-x_0)}^k.\]
If the series does not converge for any point \(x \neq x_0\), we say that the series is divergent.
Example \(\PageIndex{1}\):
The series
\[ \sum_{k=0}^\infty \frac{1}{k!} x^k = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\]
is convergent for any \(x\). Recall that \(k! = 1\cdot 2\cdot 3 \cdots k\) is the factorial. By convention we define \(0!=1\). In fact, you may recall that this series converges to \(e^x\).
We say that (7.1.2) converges absolutely at \(x\) whenever the limit
\[ \lim_{n\to\infty} \sum_{k=0}^n \lvert a_k \rvert \, {\lvert x-x_0 \rvert}^k\]
exists. That is, the series \(\sum_{k=0}^\infty \lvert a_k \rvert \, {\lvert x-x_0 \rvert}^k\) is convergent.If (7.1.2) converges absolutely at \(x\), then it converges at \(x\). However, the opposite implication is not true.
Example \(\PageIndex{2}\):
The series
\[\sum_{k=1}^\infty \frac{1}{k} x^k\]
converges absolutely for all \(x\) in the interval \((-1,1)\).
It converges at \(x=-1\), as
\(\sum_{k=1}^\infty \frac{{(-1)}^k}{k}\) converges (conditionally)
by the alternating series test.
But the power series does not converge absolutely at \(x=-1\), because \(\sum_{k=1}^\infty \frac{1}{k}\) does not converge. The series diverges at \(x=1\).
7.1.2 Radius of convergence
If a power series converges absolutely at some \(x_1\), then for all \(x\) such that \(\lvert x - x_0 \rvert \leq \lvert x_1 - x_0 \vert\) (that is, \(x\) is closer than \(x_1\) to \(x_0\)) we have \(\left\lvert a_k {(x-x_0)}^k \right\rvert \leq \left\lvert a_k {(x_1-x_0)}^k \right\rvert\) for all \(k\). As the numbers \(\left\lvert a_k {(x_1-x_0)}^k \right\rvert\) sum to some finite limit, summing smaller positive numbers \(\left\lvert a_k {(x-x_0)}^k \right\rvert\) must also have a finite limit. Therefore, the series must converge absolutely at \(x\). We have the following result.
Theorem 7.1.1.
For a power series (7.1.2), there exists a number \(\rho\) (we allow \(\rho=\infty\)) called the
radius of convergence such that the series converges absolutely on the interval \((x_0-\rho,x_0+\rho)\) and diverges for \(x < x_0-\rho\) and \(x > x_0+\rho\). We write \(\rho=\infty\) if the series converges for all \(x\).
Figure 7.1: Convergence of a power series.
See Figure 7.1. In Example 7.1.1 the radius of convergence is \(\rho = \infty\) as the series converges everywhere. In Example7.1.2 the radius of convergence is \(\rho=1\). We note that \(\rho = 0\) is another way of saying that the series is divergent. A useful test for convergence of a series is the
ratio test. Suppose that
\[\sum_{k=0}^\infty c_k\]
is a series such that the limit
\[L = \lim_{n\to\infty} \left \lvert \frac{c_{k+1}}{c_k} \right \rvert \]
exists. Then the series converges absolutely if \(L < 1\) and diverges if \(L > 1\).
Let us apply this test to the series (7.1.2). That is we let \(c_k = a_k {(x - x_0)}^k\) in the test. Compute
\[ L = \lim_{n\to\infty} \left \lvert \frac{c_{k+1}}{c_k} \right \rvert = \lim_{n\to\infty} \left \lvert \frac{a_{k+1} {(x - x_0)}^{k+1}}{a_k {(x - x_0)}^k} \right \rvert = \lim_{n\to\infty} \left \lvert\frac{a_{k+1}}{a_k}\right \rvert \lvert x - x_0 \rvert .\]
Define \(A\) by
\[A =\lim_{n\to\infty} \left \lvert \frac{a_{k+1}}{a_k} \right \rvert . \]
Then if \(1 > L = A \lvert x - x_0 \rvert\) the series (7.1.2) converges absolutely. If \(A = 0\), then the series always converges. If \(A > 0\), then the series converges absolutely if \(\lvert x - x_0 \rvert < \frac{1}{A}\), and diverges if \(\lvert x - x_0 \rvert > \frac{1}{A}\). That is, the radius of convergence is \(\frac{1}{A}\). Let us summarize.
Theorem 7.1.2. Let
\[\sum_{k=0}^\infty a_k {(x-x_0)}^k\]
be a power series such that
\[A = \lim_{n\to\infty} \left \lvert \frac{a_{k+1}}{a_k} \right \rvert \]
exists. If \(A = 0\), then the radius of convergence of the series is \(\infty\). Otherwise the radius of convergence is \(\frac{1}{A}\).
Example \(\PageIndex{3}\):
Suppose we have the series
\[\sum_{k=0}^\infty 2^{-k} {(x-1)}^k .\]
First we compute,
\[A = \lim_{k\to\infty} \left \lvert \frac{a_{k+1}}{a_k} \right \rvert = \lim_{k\to\infty} \left \lvert \frac{2^{-k-1}}{2^{-k}} \right \rvert = 2^{-1} = \frac{1}{2}. \]
Therefore the radius of convergence is \(2\), and the series converges absolutely on the interval \((-1,3)\).
The ratio test does not always apply. That is the limit of \(\bigl \lvert \frac{a_{k+1}}{a_k} \bigr \rvert\) might not exist. There exist more sophisticated ways of finding the radius of convergence, but those would be beyond the scope of this chapter.
7.1.3 Analytic functions
Functions represented by power series are called analytic functions. Not every function is analytic, although the majority of the functions you have seen in calculus are. An analytic function \(f(x)\) is equal to its Taylor series near a point \(x_0\). That is, for \(x\) near \(x_0\) we have
\[f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!} {(x-x_0)}^k ,\]
where \(f^{(k)}(x_0)\) denotes the \(k^{\text{th}}\) derivative of \(f(x)\) at the point \(x_0\).
Figure 7.2: The sine function and its Taylor approximations around \(x_o=0\) of and degree.
For example, sine is an analytic function and its Taylor series around \(x_0 = 0\) is given by
\[ \sin(x) = \sum_{n=0}^\infty \frac{{(-1)}^n}{(2n+1)!} x^{2n+1} .\]
In Figure 7.2 we plot \(\sin(x)\) and the truncations of the series up to degree 5 and 9. You can see that the approximation is very good for \(x\) near 0, but gets worse for \(x\) further away from 0. This is what happens in general. To get a good approximation far away from \(x_0\) you need to take more and more terms of the Taylor series.
7.1.4 Manipulating power series
One of the main properties of power series that we will use is that we can differentiate them term by term. That is, suppose that \(\sum a_k {(x-x_0)}^k\) is a convergent power series. Then for \(x\) in the radius of convergence we have
\[\frac{d}{dx} \left[\sum_{k=0}^\infty a_k {(x-x_0)}^k\right]=\sum_{k=1}^\infty k a_k {(x-x_0)}^{k-1} .\]
Notice that the term corresponding to \(k=0\) disappeared as it was constant. The radius of convergence of the differentiated series is the same as that of the original.
Example \(\PageIndex{4}\):
Let us show that the exponential \(y=e^x\) solves \(y'=y\). First write
\[y = e^x = \sum_{k=0}^\infty \frac{1}{k!} x^k .\]
Now differentiate
\[y' = \sum_{k=1}^\infty k \frac{1}{k!} x^{k-1} =\sum_{k=1}^\infty \frac{1}{(k-1)!} x^{k-1} .\]
We reindex the series by simply replacing \(k\) with \(k+1\). The series does not change, what changes is simply how we write it. After reindexing the series starts
at \(k=0\) again.
\[\sum_{k=1}^\infty \frac{1}{(k-1)!} x^{k-1} =\sum_{k+1=1}^\infty \frac{1}{\bigl((k+1)-1\bigr)!} x^{(k+1)-1} =\sum_{k=0}^\infty \frac{1}{k!} x^k .\]
That was precisely the power series for \(e^x\) that we started with, so we showed that \(\frac{d}{dx} [ e^x ] = e^x\).
Convergent power series can be added and multiplied together, and multiplied by constants using the following rules. First, we can add series by adding term by term,
\[\left(\sum_{k=0}^\infty a_k {(x-x_0)}^k\right)+\left(\sum_{k=0}^\infty b_k {(x-x_0)}^k\right)=\sum_{k=0}^\infty (a_k+b_k) {(x-x_0)}^k .\]
We can multiply by constants,
\[\alpha \left(\sum_{k=0}^\infty a_k {(x-x_0)}^k\right)=\sum_{k=0}^\infty \alpha a_k {(x-x_0)}^k .\]
We can also multiply series together,
\[\left(\sum_{k=0}^\infty a_k {(x-x_0)}^k\right) \, \left(\sum_{k=0}^\infty b_k {(x-x_0)}^k\right)=\sum_{k=0}^\infty c_k {(x-x_0)}^k ,\]
where \(c_k = a_0b_k + a_1 b_{k-1} + \cdots + a_k b_0\). The radius of convergence of the sum or the product is at least the minimum of the radii of convergence of the two series involved.
7.1.5 Power series for rational functions
Polynomials are simply finite power series. That is, a polynomial is a power series where the \(a_k\) are zero for all \(k\) large enough. We can always expand a polynomial as a power series about any point \(x_0\) by writing the polynomial as a polynomial in \((x-x_0)\). For example, let us write \(2x^2-3x+4\) as a power series around \(x_0 = 1\):
\[2x^2-3x+4 = 3 + (x-1) + 2{(x-1)}^2 .\]
In other words \(a_0 = 3\), \(a_1 = 1\), \(a_2 = 2\), and all other \(a_k = 0\). To do this, we know that \(a_k = 0\) for all \(k \geq 3\) as the polynomial is of degree 2.
We write \(a_0 + a_1(x-1) + a_2{(x-1)}^2\), we expand, and we solve for \(a_0\), \(a_1\), and \(a_2\). We could have also differentiated at \(x=1\)and used the Taylor series formula (7.1.17).
Let us look at rational functions, that is, ratios of polynomials. An important fact is that a series for a function only defines the function on an interval even if the function is defined elsewhere. For example, for \(-1 < x < 1\) we have
\[\frac{1}{1-x} = \sum_{k=0}^\infty x^k = 1 + x + x^2 + \cdots\]
This series is called the
geometric series. The ratio test tells us that the radius of convergence is \(1\). The series diverges for \(x \leq -1\) and \(x \geq 1\), even though \(\frac{1}{1-x}\) is defined for all \(x \not= 1\).
We can use the geometric series together with rules for addition and multiplication of power series to expand rational functions around a point, as long as the denominator is not zero at \(x_0\). Note that as for polynomials, we could equivalently use the Taylor series expansion (7.1.17).
Example \(\PageIndex{5}\):
Expand \(\frac{x}{1+2x+x^2}\) as a power series around the origin (\(x_0 = 0\)) and find the radius of convergence. First, write \(1+2x+x^2 = {(1+x)}^2 = {\bigl(1-(-x)\bigr)}^2\). Now we compute
\[ \frac{x}{1+2x+x^2} =x {\left( \frac{1}{1-(-x)} \right)}^2 \\ =x { \left( \sum_{k=0}^{\infty} {(-1)}^k x^k \right)}^2 \\ =x \left(\sum_{k=0}^{\infty} c_k x^k \right) \\ = \sum_{k=0}^{\infty} c_k x^{k+1} , \]
where using the formula for the product of series we obtain, \(c_0 = 1\), \(c_1 = -1 -1 = -2\), \(c_2 = 1+1+1 = 3\), etc \(\ldots\).
Therefore
\[\frac{x}{1+2x+x^2}=\sum_{k=1}^\infty {(-1)}^{k+1} k x^k = x-2x^2+3x^3-4x^4+\cdots \]
The radius of convergence is at least 1. We use the ratio test
\[\lim_{k\to\infty} \left\lvert \frac{a_{k+1}}{a_k} \right\rvert = \lim_{k\to\infty} \left\lvert \frac{{(-1)}^{k+2} (k+1)}{{(-1)}^{k+1}k} \right\rvert= \lim_{k\to\infty} \frac{k+1}{k}=1.\]
So the radius of convergence is actually equal to 1.
When the rational function is more complicated, it is also possible to use method of partial fractions. For example, to find the Taylor series for \(\frac{x^3+x}{x^2-1}\), we write
\[ \frac{x^3+x}{x^2-1}=x + \frac{1}{1+x} - \frac{1}{1-x}=x + \sum_{k=0}^\infty {(-1)}^k x^k - \sum_{k=0}^\infty x^k= - x + \sum_{\substack{k=3 \\ k \text{ odd}}}^\infty (-2) x^k . \] |
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Now showing items 1-10 of 24
Production of Σ(1385)± and Ξ(1530)0 in proton–proton collisions at √s = 7 TeV
(Springer, 2015-01-10)
The production of the strange and double-strange baryon resonances ((1385)±, Ξ(1530)0) has been measured at mid-rapidity (|y|< 0.5) in proton–proton collisions at √s = 7 TeV with the ALICE detector at the LHC. Transverse ...
Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Inclusive photon production at forward rapidities in proton-proton collisions at $\sqrt{s}$ = 0.9, 2.76 and 7 TeV
(Springer Berlin Heidelberg, 2015-04-09)
The multiplicity and pseudorapidity distributions of inclusive photons have been measured at forward rapidities ($2.3 < \eta < 3.9$) in proton-proton collisions at three center-of-mass energies, $\sqrt{s}=0.9$, 2.76 and 7 ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Measurement of pion, kaon and proton production in proton–proton collisions at √s = 7 TeV
(Springer, 2015-05-27)
The measurement of primary π±, K±, p and p¯¯¯ production at mid-rapidity (|y|< 0.5) in proton–proton collisions at s√ = 7 TeV performed with a large ion collider experiment at the large hadron collider (LHC) is reported. ...
Two-pion femtoscopy in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(American Physical Society, 2015-03)
We report the results of the femtoscopic analysis of pairs of identical pions measured in p-Pb collisions at $\sqrt{s_{\mathrm{NN}}}=5.02$ TeV. Femtoscopic radii are determined as a function of event multiplicity and pair ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Charged jet cross sections and properties in proton-proton collisions at $\sqrt{s}=7$ TeV
(American Physical Society, 2015-06)
The differential charged jet cross sections, jet fragmentation distributions, and jet shapes are measured in minimum bias proton-proton collisions at centre-of-mass energy $\sqrt{s}=7$ TeV using the ALICE detector at the ...
Centrality dependence of high-$p_{\rm T}$ D meson suppression in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Springer, 2015-11)
The nuclear modification factor, $R_{\rm AA}$, of the prompt charmed mesons ${\rm D^0}$, ${\rm D^+}$ and ${\rm D^{*+}}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at a centre-of-mass ...
K*(892)$^0$ and $\Phi$(1020) production in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2015-02)
The yields of the K*(892)$^0$ and $\Phi$(1020) resonances are measured in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV through their hadronic decays using the ALICE detector. The measurements are performed in multiple ... |
Difference between revisions of "Model"
(→Mantle and large cardinals: -selflink)
(→Mantle and large cardinals: $$)
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If $\kappa$ is [[hyperhuge]], then $V$ has $<\kappa$ many [[ground]]s (so the mantle is a ground itself).<cite>Usuba2017:DDGandVeryLarge</cite>
If $\kappa$ is [[hyperhuge]], then $V$ has $<\kappa$ many [[ground]]s (so the mantle is a ground itself).<cite>Usuba2017:DDGandVeryLarge</cite>
−
If $κ$ is [[extendible]] then the $κ$-mantle of $V$ is its smallest ground (so of course the mantle is a ground of V).<cite>Usuba2018:ExtendibleCardinalsAndTheMantle</cite>
+
If $κ$ is [[extendible]] then the $κ$-mantle of $V$ is its smallest ground (so of course the mantle is a ground of V).<cite>Usuba2018:ExtendibleCardinalsAndTheMantle</cite>
On the other hand, it s consistent that there is a [[supercompact]] cardinal and class many grounds of $V$ (because of the indestructibility properties of supercompactness).<cite>Usuba2017:DDGandVeryLarge</cite>
On the other hand, it s consistent that there is a [[supercompact]] cardinal and class many grounds of $V$ (because of the indestructibility properties of supercompactness).<cite>Usuba2017:DDGandVeryLarge</cite>
Latest revision as of 12:22, 14 September 2019
A
model of a theory $T$ is a set $M$ together with relations (eg. two: $a$ and $b$) satisfying all axioms of the theory $T$. Symbolically $\langle M, a, b \rangle \models T$. According to the Gödel completeness theorem, in $\mathrm{PA}$ (Peano arithmetic) (so also in $\mathrm{ZFC}$) a theory has models iff it is consistent. According to Löwenheim–Skolem theorem, in $\mathrm{ZFC}$ if a countable first-order theory has an infinite model, it has infinite models of all cardinalities.
A
model of a set theory (eg. $\mathrm{ZFC}$) is a set $M$ such that the structure $\langle M,\hat\in \rangle$ satisfies all axioms of the set theory. If $\hat \in$ is base theory's $\in$, the model is called a transitive model. Gödel completeness theorem and Löwenheim–Skolem theorem do not apply to transitive models. (But Löwenheim–Skolem theorem together with Mostowski collapsing lemma show that if there is a transitive model of ZFC, then there is a countable such model.) See Transitive ZFC model. Contents Class-sized transitive models
One can also talk about class-sized transitive models. Inner model is a transitive class containing all ordinals. Forcing creates outer models, but it can also be used in relation with inner models.[1]
Among them are
canonical inner models like the core model the canonical model $L[\mu]$ of one measurable cardinal HOD and generic HOD (gHOD) mantle $\mathbb{M}$ (=generic mantle $g\mathbb{M}$) outer core the constructible universe $L$ Mantle $α$th inner mantle $\mathbb{M}^α$ is defined by $\mathbb{M}^0=V$, $\mathbb{M}^{α+1} = \mathbb{M}^{\mathbb{M}^α}$ (mantle of the previous inner mantle) and $\mathbb{M}^α = \bigcap_{β<α} \mathbb{M}^β$ for limit $α$. If there is uniform presentation of $\mathbb{M}^α$ for all ordinals $α$ as a single class, one can talk about $\mathbb{M}^\mathrm{Ord}$, $\mathbb{M}^{\mathrm{Ord}+1}$ etc. If an inner mantle is a ground, it is called the outer core.[1]
It is conjenctured (unproved) that every model of ZFC is the $\mathbb{M}^α$ of another model of ZFC for any desired $α ≤ \mathrm{Ord}$, in which the sequence of inner mantles does not stabilise before $α$. It is probable that in the some time there are models of ZFC, for which inner mantle is undefined (Analogously, a 1974 result of Harrington appearing in (Zadrożny, 1983, section 7), with related work in (McAloon, 1974), shows that it is relatively consistent with Gödel-Bernays set theory that $\mathrm{HOD}^n$ exists for each $n < ω$ but the intersection $\mathrm{HOD}^ω = \bigcap_n \mathrm{HOD}^n$ is not a class.).[1]
For a cardinal $κ$, we call a ground $W$ of $V$ a $κ$-ground if there is a poset $\mathbb{P} ∈ W$ of size $< κ$ and a $(W, \mathbb{P})$-generic $G$ such that $V = W[G]$. The
$κ$-mantle is the intersection of all $κ$-grounds.[3]
The $κ$-mantle is a definable, transitive, and extensional class. It is consistent that the $κ$-mantle is a model of ZFC (e.g. when there are no grounds), and if $κ$ is a strong limit, then the $κ$-mantle must be a model of ZF. However it is not known whether the $κ$-mantle is always a model of ZFC.[3]
Mantle and large cardinals $\kappa$-model
A
weak $κ$-model is a transitive set $M$ of size $\kappa$ with $\kappa \in M$ and satisfying the theory $\mathrm{ZFC}^-$ ($\mathrm{ZFC}$ without the axiom of power set, with collection, not replacement). It is a $κ$-model if additionaly $M^{<\kappa} \subseteq M$.[4, 5]
References Fuchs, Gunter and Hamkins, Joel David and Reitz, Jonas. Set-theoretic geology.Annals of Pure and Applied Logic 166(4):464 - 501, 2015. www arχiv DOI bibtex Usuba, Toshimichi. The downward directed grounds hypothesis and very large cardinals.Journal of Mathematical Logic 17(02):1750009, 2017. arχiv DOI bibtex Usuba, Toshimichi. Extendible cardinals and the mantle.Archive for Mathematical Logic 58(1-2):71-75, 2019. arχiv DOI bibtex Hamkins, Joel David and Johnstone, Thomas A. Strongly uplifting cardinals and the boldface resurrection axioms., 2014. arχiv bibtex Holy, Peter and Schlicht, Philipp. A hierarchy of Ramsey-like cardinals.Fundamenta Mathematicae 242:49-74, 2018. www arχiv DOI bibtex |
The Triangle Inequality for Inner Product Spaces
We will now look at a very important theorem known as the triangle inequality for inner product spaces. Suppose that $V$ i an inner product space. If we form a triangle with the vectors $u$, $v$, and $u + v$, then the shortest path from the initial point of $u$ to the terminal point of $v$ is the norm $\| u + v \|$ as summarized in the following diagram:
Thus for any vectors $u, v \in V$ we have that $\| u + v \| ≤ \| u \| + \| v \|$.
Theorem 1 (The Triangle Inequality for Inner Product Spaces): Let $V$ be an inner product space with $u, v \in V$. Then: a) $\| u + v \| ≤ \| u \| + \| v \|$. b) $\| u + v \| = \| u \| + \| v \|$ is and only if $u$ or $v$ is a nonnegative scalar multiple of the other. Proof of a):Let $V$ be an inner product space and let $u, v \in V$. Recall that if $z \in \mathbb{C}$. Then $z = a + bi$ and $\bar{z} = a - bi$ so $z + \overline{z} = 2a = 2 \mathrm{Re} (z)$. We will use this in the proof below: Now we will apply the Cauchy-Schwarz inequality to get that: We now take the square roots of both sides to get that $\| u + v \| ≤ \| u \| + \| v \|$. $\blacksquare$ Proof of b):$\Rightarrow$ Suppose that $\| u + v \| = \| u \| + \| v \|$. Then from the proof above we must have the following equality: We have that the use of the Cauchy-Schwarz inequality implies that then $u$ or $v$ must be a scalar multiple of the other, more precisely, a nonnegative scalar multiple. $\Leftarrow$ Suppose that one of $u$ or $v$ is a nonnegative scalar multiple of the other. Suppose $u = kv$ for $k \in \mathbb{F}$ ($\mathbb{R}$ or $\mathbb{C}$). Then: Since we have the equality above, we can see that from the proof of (a)that $\| u + v \| = \| u \| + \| v \|$. $\blacksquare$ |
Frequently we will want to estimate the empirical probability density function of real-world data and compare it to the theoretical density from one or more probability distributions. The following example shows the empirical and theoretical normal density for EUR/USD high-frequency tick data \(X\) (which has been transformed using log-returns and normalized via \(\frac{X_i-\mu_X}{\sigma_X}\)). The theoretical normal density is plotted over the range \(\left(\lfloor\mathrm{min}(X)\rfloor,\lceil\mathrm{max}(X)\rceil\right)\). The results are in the figure below. The discontinuities and asymmetry of the discrete tick data, as well as the sharp kurtosis and heavy tails (a corresponding interval of \(\approx \left[-8,+7\right]\) standard deviations away from the mean) are apparent from the plot.
We also show the theoretical and empirical density for the EUR/USD exchange rate log returns over different timescales. We can see from these plots that the distribution of the log returns seems to be asymptotically converging to normality. This is a typical empirical property of financial data.
The following R source generates empirical and theoretical density plots across different timescales. The data is loaded from files that are sampled at different intervals. I cant supply the data unfortunately, but you should get the idea.
[source lang=”R”]
# Function that reads Reuters CSV tick data and converts Reuters dates # Assumes format is Date,Tick readRTD <- function(filename) { tickData <- read.csv(file=filename, header=TRUE, col.names=c("Date","Tick")) tickData$Date <- as.POSIXct(strptime(tickData$Date, format="%d/%m/%Y %H:%M:%S")) tickData }
# Boilerplate function for Reuters FX tick data transformation and density plot
plot.reutersFXDensity <- function() { filenames <- c("data/eur_usd_tick_26_10_2007.csv", "data/eur_usd_1min_26_10_2007.csv", "data/eur_usd_5min_26_10_2007.csv", "data/eur_usd_hourly_26_10_2007.csv", "data/eur_usd_daily_26_10_2007.csv") labels <- c("Tick", "1 Minute", "5 Minutes", "Hourly", "Daily")
par(mfrow=c(length(filenames), 2),mar=c(0,0,2,0), cex.main=2)
tickData <- c() i <- 1 for (filename in filenames) { tickData[[i]] <- readRTD(filename) # Transform: `$Y = \nabla\log(X_i)$` logtick <- diff(log(tickData[[i]]$Tick)) # Normalize: `$\frac{(Y-\mu_Y)}{\sigma_Y}$` logtick <- (logtick-mean(logtick))/sd(logtick) # Theoretical density range: `$\left[\lfloor\mathrm{min}(Y)\rfloor,\lceil\mathrm{max}(Y)\rceil\right]$` x <- seq(floor(min(logtick)), ceiling(max(logtick)), .01) plot(density(logtick), xlab="", ylab="", axes=FALSE, main=labels[i]) lines(x,dnorm(x), lty=2) #legend("topleft", legend=c("Empirical","Theoretical"), lty=c(1,2)) plot(density(logtick), log="y", xlab="", ylab="", axes=FALSE, main="Log Scale") lines(x,dnorm(x), lty=2) i <- i + 1 } par(op) } [/source] |
The Method of Integrating Factors Examples 2
The Method of Integrating Factors Examples 2
Recall from The Method of Integrating Factors page that we can solve first order linear differential equations of the form $\frac{dy}{dt} + p(t) y = g(t)$ by multiplying both sides of the equation by the integrating factor $\mu (t) = e^{\int p(t) \: dt}$ so that:(1)
\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad \frac{d}{dt} \left ( \mu (t) y \right ) = \mu (t) g(t) \\ \quad \int \frac{d}{dt} \left ( \mu (t) y \right ) \: dt = \int \mu (t) g(t) \: dt \\ \quad \mu (t) y = \int \mu (t) g(t) \: dt \\ \quad y = \frac{\int \mu (t) g(t) \: dt}{\mu (t)} \end{align}
We will now look at some examples of applying this method of integrating factors.
More examples can be found on The Method of Integrating Factors Examples 1 page.
Example 1
Find all solutions to the differential equation $t\frac{dy}{dt} + 2y = t^2 - t + 1$.
We must first divide both sides of the equation by $t$ in order to get the differential equation above into the form $\frac{dy}{dt} + p(t) y = g(t)$. Upon doing so, we obtain:(2)
\begin{align} \quad \frac{dy}{dt} + \frac{2y}{t} = t - 1 + \frac{1}{t} \end{align}
Now we can clearly see that $p(t) = \frac{2}{t}$ and $g(t) = t - 1 + \frac{1}{t}$. Thus we obtain our integrating factor:(3)
\begin{align} \quad \mu (t) = e^{\int p(t) \: dt} = e^{\int \frac{2}{t} \: dt} = e^{2 \ln t} = e^{\ln (t^2)} = t^2 \end{align}
Thus for $C$ as a constant we get that:(4)
\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad t^2 \frac{dy}{dt} + 2ty = t^3 - t^2 + t \\ \quad \frac{d}{dt} \left ( t^2 y \right ) = t^3 - t^2 + t \\ \quad \int \frac{d}{dt} \left ( t^2 y \right ) \: dt = \int t^3 - t^2 + t \: dt \\ \quad t^2 y = \frac{t^4}{4} - \frac{t^3}{3} + \frac{t^2}{2} + C \\ \quad y = \frac{t^2}{4} - \frac{t}{3} + \frac{1}{2} + \frac{C}{t^2} \end{align}
Example 2
Find all solutions to the differential equation $(t^2 - 1) \frac{dy}{dt} + 2ty = t$.
We want to rewrite the differential equation above so that it is in the correct form to apply the method of integrating factors. If we divide both sides of the differential equation above by $(t^2 - 1)$ we get that:(5)
\begin{align} \quad \frac{dy}{dt} + \frac{2t}{t^2 - 1}y = \frac{t}{t^2 - 1} \end{align}
Therefore we have that $p(t) = \frac{2t}{t^2 - 1}$ and $g(t) = \frac{t}{t^2 - 1}$, and so our integrating factor is:(6)
\begin{align} \quad \mu (t) = e^{\int p(t) \: dt} = e^{\int \frac{2t}{t^2 - 1} \: dt} \end{align}
To solve the integral in the exponent of $\mu (t)$ we will use the substitution method for integration. Let $u = t^2 - 1$. Then $du = 2t \: dt$, and hence we have that $\int \frac{2t}{t^2 - 1} \: dt = \int \frac{1}{u} \: du = \ln (u) = \ln (t^2 - 1)$. Thus we have our integrating factor is:(7)
\begin{align} \quad \mu (t) = e^{\ln (t^2 - 1)} = t^2 - 1 \end{align}
Thus for $C$ as a constant we have:(8)
\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad (t^2 - 1) \frac{dy}{dt} + 2ty = t \\ \quad \frac{d}{dt} \left ( (t^2 - 1)y \right ) = t \\ \quad \int \frac{d}{dt} \left ( (t^2 - 1)y \right ) \: dt = \int t \: dt \\ \quad (t^2 - 1)y = \frac{t^2}{2} + C \\ \quad y = \frac{t^2}{2(t^2 - 1)} + \frac{C}{t^2 - 1} \end{align}
Example 3
Find all solutions to the differential equation $t \frac{dy}{dt} - 2y = t^4 \sin t$.
To get the differential equation above in the appropriate form, we will first divide both sides by $t$ to get that:(9)
\begin{align} \frac{dy}{dt} - \frac{2}{t}y = t^3 \sin t \end{align}
We can see that $p(t) = -\frac{2}{t}$ and $g(t) = t^3 \sin t$. Thus we have that our integrating factor is:(10)
\begin{align} \quad \mu (t) = e^{\int p(t) \: dt} = e^{\int -\frac{2}{t} \: dt} = e^{-2\ln t} = e^{\ln \left ( \frac{1}{t^2} \right )} = \frac{1}{t^2} \end{align}
Thus for $C$ as a constant we get that:(11)
\begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad \frac{1}{t^2} \frac{dy}{dt} - \frac{2}{t^3} y = t \sin t \\ \quad \frac{d}{dt} \left ( \frac{y}{t^2} \right ) = t \sin t \\ \quad \int \frac{d}{dt} \left ( \frac{y}{t^2} \right ) \: dt = \int t \sin t \: dt \\ \quad \frac{y}{t^2} = \int t \sin t \: dt \end{align}
To evaluate the integral on the righthand side, we will use integration by parts. Once again, recall that $\int u \: dv = uv - \int v \: du$. Let $u = t$ and let $dv = \sin t \: dt$. Then $du = 1$ and $v = -\cos t$ and thus we have that:(12)
\begin{align} \quad \int t \sin t \: dt = -t \cos t - \int -\cos t \: du = -t \cos t + \sin t + C = \sin t - t \cos t + C \end{align}
Therefore going back to our differential equation above, we have that:(13)
\begin{align} \quad \frac{y}{t^2} = \sin t - t \cos t + C\\ \quad y = t^2 \sin t - t^3 \cos t + Ct^2 \end{align} |
This post has been cross-posted on the Quansight LabsBlog.
As of November, 2018, I have been working at Quansight. Quansight is a new startup founded by the same people who started Anaconda, which aims to connect companies and open source communities, and offers consulting, training, support and mentoring services. I work under the heading of Quansight Labs. Quansight Labs is a public-benefit division of Quansight. It provides a home for a "PyData Core Team" which consists of developers, community managers, designers, and documentation writers who build open-source technology and grow open-source communities around all aspects of the AI and Data Science workflow.
My work at Quansight is split between doing open source consulting for various companies, and working on SymPy. SymPy, for those who do not know, is a symbolic mathematics library written in pure Python. I am the lead maintainer of SymPy.
In this post, I will detail some of the open source work that I have done recently, both as part of my open source consulting, and as part of my work on SymPy for Quansight Labs.
Bounds Checking in Numba
As part of work on a client project, I have been working on contributing codeto the numba project. Numba is a just-in-timecompiler for Python. It lets you write native Python code and with the use ofa simple
@jit decorator, the code will be automatically sped up using LLVM.This can result in code that is up to 1000x faster in some cases:
In [1]: import numbaIn [2]: import numpyIn [3]: def test(x): ...: A = 0 ...: for i in range(len(x)): ...: A += i*x[i] ...: return A ...:In [4]: @numba.njit ...: def test_jit(x): ...: A = 0 ...: for i in range(len(x)): ...: A += i*x[i] ...: return A ...:In [5]: x = numpy.arange(1000)In [6]: %timeit test(x)249 µs ± 5.77 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)In [7]: %timeit test_jit(x)336 ns ± 0.638 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)In [8]: 249/.336Out[8]: 741.0714285714286
Numba only works for a subset of Python code, and primarily targets code that uses NumPy arrays.
Numba, with the help of LLVM, achieves this level of performance through manyoptimizations. One thing that it does to improve performance is to remove allbounds checking from array indexing. This means that if an array index is outof bounds, instead of receiving an
IndexError, you will get garbage, orpossibly a segmentation fault.
>>> import numpy as np>>> from numba import njit>>> def outtabounds(x):... A = 0... for i in range(1000):... A += x[i]... return A>>> x = np.arange(100)>>> outtabounds(x) # pure Python/NumPy behaviorTraceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 4, in outtaboundsIndexError: index 100 is out of bounds for axis 0 with size 100>>> njit(outtabounds)(x) # the default numba behavior-8557904790533229732
In numba pull request #4432, I amworking on adding a flag to
@njit that will enable bounds checks for arrayindexing. This will remain disabled by default for performance purposes. Butyou will be able to enable it by passing
boundscheck=True to
@njit, or bysetting the
NUMBA_BOUNDSCHECK=1 environment variable. This will make iteasier to detect out of bounds issues like the one above. It will work like
>>> @njit(boundscheck=True)... def outtabounds(x):... A = 0... for i in range(1000):... A += x[i]... return A>>> x = np.arange(100)>>> outtabounds(x) # numba behavior in my pull request #4432Traceback (most recent call last): File "<stdin>", line 1, in <module>IndexError: index is out of bounds
The pull request is still in progress, and many things such as the quality of the error message reporting will need to be improved. This should make debugging issues easier for people who write numba code once it is merged.
removestar
removestar is a new tool I wrote toautomatically replace
import * in Python modules with explicit imports.
For those who don't know, Python's
import statement supports so-called"wildcard" or "star" imports, like
from sympy import *
This will import every public name from the
sympy module into the currentnamespace. This is often useful because it saves on typing every name that isused in the import line. This is especially useful when working interactively,where you just want to import every name and minimize typing.
However, doing
from module import * is generally frowned upon in Python. It isconsidered acceptable when working interactively at a
python prompt, or in
__init__.py files (removestar skips
__init__.py files by default).
Some reasons why
import * is bad:
It hides which names are actually imported. It is difficult both for human readers and static analyzers such aspyflakes to tell where a given name comes from when
import *is used. For example, pyflakes cannot detect unused names (for instance, from typos) in the presence of
import *.
If there are multiple
import *statements, it may not be clear which names come from which module. In some cases, both modules may have a given name, but only the second import will end up being used. This can break people's intuition that the order of imports in a Python file generally does not matter.
import *often imports more names than you would expect. Unless the module you import defines
__all__or carefully
dels unused names at the module level,
import *will import every public (doesn't start with an underscore) name defined in the module file. This can often include things like standard library imports or loop variables defined at the top-level of the file. For imports from modules (from
__init__.py),
from module import *will include every submodule defined in that module. Using
__all__in modules and
__init__.pyfiles is also good practice, as these things are also often confusing even for interactive use where
import *is acceptable.
In Python 3,
import *is syntactically not allowed inside of a function definition.
Here are some official Python references stating not to use
import * infiles:
In general, don’t use
from modulename import *. Doing so clutters the importer’s namespace, and makes it much harder for linters to detect undefined names.
PEP 8 (the official Python style guide):
Wildcard imports (
from <module> import *) should be avoided, as they make it unclear which names are present in the namespace, confusing both readers and many automated tools.
Unfortunately, if you come across a file in the wild that uses
import *, itcan be hard to fix it, because you need to find every name in the file that isimported from the
* and manually add an import for it. Removestar makes thiseasy by finding which names come from
* imports and replacing the importlines in the file automatically.
As an example, suppose you have a module
mymod like
mymod/ | __init__.py | a.py | b.py
with
# mymod/a.pyfrom .b import *def func(x): return x + y
and
# mymod/b.pyx = 1y = 2
Then
removestar works like:
$ removestar -i mymod/$ cat mymod/a.py# mymod/a.pyfrom .b import ydef func(x): return x + y
The
-i flag causes it to edit
a.py in-place. Without it, it would justprint a diff to the terminal.
For implicit star imports and explicit star imports from the same module,
removestar works statically, making use ofpyflakes. This means none of the code isactually executed. For external imports, it is not possible to work staticallyas external imports may include C extension modules, so in that case, itimports the names dynamically.
removestar can be installed with pip or conda:
pip install removestar
or if you use conda
conda install -c conda-forge removestar
sphinx-math-dollar
In SymPy, we make heavy use of LaTeX math in our documentation. For example, in our special functions documentation, most special functions are defined using a LaTeX formula, like
However, the source for this math in the docstring of the function uses RST syntax:
class besselj(BesselBase): """ Bessel function of the first kind. The Bessel `J` function of order `\nu` is defined to be the function satisfying Bessel's differential equation .. math :: z^2 \frac{\mathrm{d}^2 w}{\mathrm{d}z^2} + z \frac{\mathrm{d}w}{\mathrm{d}z} + (z^2 - \nu^2) w = 0, with Laurent expansion .. math :: J_\nu(z) = z^\nu \left(\frac{1}{\Gamma(\nu + 1) 2^\nu} + O(z^2) \right), if :math:`\nu` is not a negative integer. If :math:`\nu=-n \in \mathbb{Z}_{<0}` *is* a negative integer, then the definition is .. math :: J_{-n}(z) = (-1)^n J_n(z).
Furthermore, in SymPy's documentation we have configured it so that textbetween `single backticks` is rendered as math. This was originally done forconvenience, as the alternative way is to write
:math:`\nu` everytime you want to use inline math. But this has lead to many people beingconfused, as they are used to Markdown where `single backticks` produce
code.
A better way to write this would be if we could delimit math with dollarsigns, like
$\nu$. This is how things are done in LaTeX documents, as wellas in things like the Jupyter notebook.
With the new sphinx-math-dollarSphinx extension, this is now possible. Writing
$\nu$ produces $\nu$, andthe above docstring can now be written as
class besselj(BesselBase): """ Bessel function of the first kind. The Bessel $J$ function of order $\nu$ is defined to be the function satisfying Bessel's differential equation .. math :: z^2 \frac{\mathrm{d}^2 w}{\mathrm{d}z^2} + z \frac{\mathrm{d}w}{\mathrm{d}z} + (z^2 - \nu^2) w = 0, with Laurent expansion .. math :: J_\nu(z) = z^\nu \left(\frac{1}{\Gamma(\nu + 1) 2^\nu} + O(z^2) \right), if $\nu$ is not a negative integer. If $\nu=-n \in \mathbb{Z}_{<0}$ *is* a negative integer, then the definition is .. math :: J_{-n}(z) = (-1)^n J_n(z).
We also plan to add support for
$$double dollars$$ for display math so that
.. math :: is no longer needed either .
For end users, the documentation on docs.sympy.org will continue to render exactly the same, but for developers, it is much easier to read and write.
This extension can be easily used in any Sphinx project. Simply install it with pip or conda:
pip install sphinx-math-dollar
or
conda install -c conda-forge sphinx-math-dollar
Then enable it in your
conf.py:
extensions = ['sphinx_math_dollar', 'sphinx.ext.mathjax']
Google Season of Docs
The above work on sphinx-math-dollar is part of work I have been doing to improve the tooling around SymPy's documentation. This has been to assist our technical writer Lauren Glattly, who is working with SymPy for the next three months as part of the new Google Season of Docs program. Lauren's project is to improve the consistency of our docstrings in SymPy. She has already identified many key ways our docstring documentation can be improved, and is currently working on a style guide for writing docstrings. Some of the issues that Lauren has identified require improved tooling around the way the HTML documentation is built to fix. So some other SymPy developers and I have been working on improving this, so that she can focus on the technical writing aspects of our documentation.
Lauren has created a draft style guide for documentation at https://github.com/sympy/sympy/wiki/SymPy-Documentation-Style-Guide. Please take a moment to look at it and if you have any feedback on it, comment below or write to the SymPy mailing list. |
You most likely found this post for one of two reasons:
Either you haven’t heard of Z-Boxes and are interested in if they can somehow help you or you have to learn about Z-Boxes and you have absolutely no idea how to understand the mathematical definitions.
Either way, we’re going to investigate Z-Boxes – not using a box of formulas but using examples and Python code.
What are Z-Boxes?
If you have heard of Z-Boxes in a lecture or read about them in a book, you probably encountered\begin{array}{ll}(1)&\;\text{Let }s:=s_0\cdots s_{n-1}\in\Sigma^n\\(2)&\;\forall i\in\{1\cdots n-1\}:\;Z_i:=max\{j\in[0\cdots n]: s_i\cdots s_{i+j-1}=s_0\cdots s_{j-1}\}\\(3)&\;\text{Z-Box starting at position i}:=s_i\cdots s_{i+Z_i-1}\,\text{if }Z_i\neq0\end{array}
But let’s be honest. These definitions are mainly for people who already understand what this means. To most students this is not very useful in understanding an algorithm.
Let’s break it down anyway and see what we can salvage.
(1) This just says that s is a string of length n (this definition is used). Technically \Sigma is the
alphabet of the string – but for all practical intents and purposes this is defined by the programming context anyway, so we’ll just ignore \Sigma.
So let’s assume n=6. Then each of those would meet the definition:
s = "abc123" s = "fo bar" s = " " s = "abcabc"
In almost all defitions for algorithms dealing with strings, you’ll find a similar definition.
(2)\&(3) is slightly more complicated. I’ll explain it without talking about the definition at first.
Here’s the easy part: The
Z-Box starting at position i is just a substring of s, starting at position i. Z_i is its length.
Unfortunately, the hard part is that it’s not just
any substring starting at position i. But how can we check if a Z-Box is valid?
Let’s just have a look at a fragment of python code:
def is_valid_zbox(s, i, zi): # zi is the length of the zbox, compute the zbox zbox = s[i:i + zi] # Compute the prefix of the same length prefix = s[0:zi] return zbox == prefix is_valid_zbox("abcabc", 2, 2) # "ca" != "ab" => False is_valid_zbox("abcabc", 2, 3) # "cab" != "abc" => False is_valid_zbox("abcabc", 3, 2) # "ab" == "ab" => True is_valid_zbox("abcabc", 3, 3) # "abc" == "abc" => True
So a
Z-Box is valid if two strings are equal: The Z-Box itself and a prefix of s.
As we can see in the examples, there can be multiple valid
Z-boxes at any position i. Which one do we use? Due to the
max in (2), we always use the longest one.
A simple Z-Box algorithm
So here’s how to compute the
Z-boxes using what we know so far: def is_valid_zbox(s, i, zi): # zi is the length of the zbox, compute the zbox zbox = s[i:i + zi] # Compute the prefix of the same length prefix = s[0:zi] return zbox == prefix def zbox(s, i): # Maximum length of substring starting at pos. i # (s has only so many characters) maxlen = len(s) - i # Try out every zbox, starting at the longest # i.e. maxlen, maxlen-1, ..., 1 for zi in range(maxlen, 0, -1): if is_valid_zbox(s, i, zi): # Return the first valid zbox # As we're starting from the longest, # this is always the longest zbox zbox = s[i:i + zi] return zbox # Compute zboxes s = "abcabc" [zbox(s, i) for i in range(1, len(s))] # [None, None, 'abc', None, None]
Why are there so many
None values in the result? Because, as you can see in the string,
a, the first character in the string, only occurs again at position 3.
And why do we only compute
Z-boxes starting at [1\cdots n-1]? We could as well compute them for [0\cdots n-1]!?!? But if you take any substring of s of length j starting at 0, and compare it to the prefix of s of length j – they are the same, no matter how long they are! That’s because the prefix of s of length j is the s of length j starting at 0 ( pre means before, therefore we start at 0!)
Let’s have a look at another example:
# Compute zboxes s = "ananas" [zbox(s, i) for i in range(1, len(s))] # [None, 'ana', None, 'a', None]
And there are also examples where we can’t find
any Z-box! # Compute zboxes s = "foobar" [zbox(s, i) for i in range(1, len(s))] # [None, None, None, None, None]
But for some we can find a lot!
# Compute zboxes s = "abaabaabab" [zbox(s, i) for i in range(1, len(s))] # [None, 'a', 'abaaba', None, 'a', 'aba', None, 'ab', None] Additional reading:
Lecture-type material: |
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
That seems like what I need to do, but I don't know how to actually implement it... how wide of a time window is needed for the Y_{t+\tau}? And how on earth do I load all that data at once without it taking forever?
And is there a better or other way to see if shear strain does cause temperature increase, potentially delayed in time
Link to the question: Learning roadmap for picking up enough mathematical know-how in order to model "shape", "form" and "material properties"?Alternatively, where could I go in order to have such a question answered?
@tpg2114 For reducing data point for calculating time correlation, you can run two exactly the simulation in parallel separated by the time lag dt. Then there is no need to store all snapshot and spatial points.
@DavidZ I wasn't trying to justify it's existence here, just merely pointing out that because there were some numerics questions posted here, some people might think it okay to post more. I still think marking it as a duplicate is a good idea, then probably an historical lock on the others (maybe with a warning that questions like these belong on Comp Sci?)
The x axis is the index in the array -- so I have 200 time series
Each one is equally spaced, 1e-9 seconds apart
The black line is \frac{d T}{d t} and doesn't have an axis -- I don't care what the values are
The solid blue line is the abs(shear strain) and is valued on the right axis
The dashed blue line is the result from scipy.signal.correlate
And is valued on the left axis
So what I don't understand: 1) Why is the correlation value negative when they look pretty positively correlated to me? 2) Why is the result from the correlation function 400 time steps long? 3) How do I find the lead/lag between the signals? Wikipedia says the argmin or argmax of the result will tell me that, but I don't know how
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
Because I don't know how the result is indexed in time
Related:Why don't we just ban homework altogether?Banning homework: vote and documentationWe're having some more recent discussions on the homework tag. A month ago, there was a flurry of activity involving a tightening up of the policy. Unfortunately, I was really busy after th...
So, things we need to decide (but not necessarily today): (1) do we implement John Rennie's suggestion of having the mods not close homework questions for a month (2) do we reword the homework policy, and how (3) do we get rid of the tag
I think (1) would be a decent option if we had >5 3k+ voters online at any one time to do the small-time moderating. Between the HW being posted and (finally) being closed, there's usually some <1k poster who answers the question
It'd be better if we could do it quick enough that no answers get posted until the question is clarified to satisfy the current HW policy
For the SHO, our teacher told us to scale$$p\rightarrow \sqrt{m\omega\hbar} ~p$$$$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$And then define the following$$K_1=\frac 14 (p^2-q^2)$$$$K_2=\frac 14 (pq+qp)$$$$J_3=\frac{H}{2\hbar\omega}=\frac 14(p^2+q^2)$$The first part is to show that$$Q \...
Okay. I guess we'll have to see what people say but my guess is the unclear part is what constitutes homework itself. We've had discussions where some people equate it to the level of the question and not the content, or where "where is my mistake in the math" is okay if it's advanced topics but not for mechanics
Part of my motivation for wanting to write a revised homework policy is to make explicit that any question asking "Where did I go wrong?" or "Is this the right equation to use?" (without further clarification) or "Any feedback would be appreciated" is not okay
@jinawee oh, that I don't think will happen.
In any case that would be an indication that homework is a meta tag, i.e. a tag that we shouldn't have.
So anyway, I think suggestions for things that need to be clarified -- what is homework and what is "conceptual." Ie. is it conceptual to be stuck when deriving the distribution of microstates cause somebody doesn't know what Stirling's Approximation is
Some have argued that is on topic even though there's nothing really physical about it just because it's 'graduate level'
Others would argue it's not on topic because it's not conceptual
How can one prove that$$ \operatorname{Tr} \log \cal{A} =\int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr}e^{-s \mathcal{A}},$$for a sufficiently well-behaved operator $\cal{A}?$How (mathematically) rigorous is the expression?I'm looking at the $d=2$ Euclidean case, as discuss...
I've noticed that there is a remarkable difference between me in a selfie and me in the mirror. Left-right reversal might be part of it, but I wonder what is the r-e-a-l reason. Too bad the question got closed.
And what about selfies in the mirror? (I didn't try yet.)
@KyleKanos @jinawee @DavidZ @tpg2114 So my take is that we should probably do the "mods only 5th vote"-- I've already been doing that for a while, except for that occasional time when I just wipe the queue clean.
Additionally, what we can do instead is go through the closed questions and delete the homework ones as quickly as possible, as mods.
Or maybe that can be a second step.
If we can reduce visibility of HW, then the tag becomes less of a bone of contention
@jinawee I think if someone asks, "How do I do Jackson 11.26," it certainly should be marked as homework. But if someone asks, say, "How is source theory different from qft?" it certainly shouldn't be marked as Homework
@Dilaton because that's talking about the tag. And like I said, everyone has a different meaning for the tag, so we'll have to phase it out. There's no need for it if we are able to swiftly handle the main page closeable homework clutter.
@Dilaton also, have a look at the topvoted answers on both.
Afternoon folks. I tend to ask questions about perturbation methods and asymptotic expansions that arise in my work over on Math.SE, but most of those folks aren't too interested in these kinds of approximate questions. Would posts like this be on topic at Physics.SE? (my initial feeling is no because its really a math question, but I figured I'd ask anyway)
@DavidZ Ya I figured as much. Thanks for the typo catch. Do you know of any other place for questions like this? I spend a lot of time at math.SE and they're really mostly interested in either high-level pure math or recreational math (limits, series, integrals, etc). There doesn't seem to be a good place for the approximate and applied techniques I tend to rely on.
hm... I guess you could check at Computational Science. I wouldn't necessarily expect it to be on topic there either, since that's mostly numerical methods and stuff about scientific software, but it's worth looking into at least.
Or... to be honest, if you were to rephrase your question in a way that makes clear how it's about physics, it might actually be okay on this site. There's a fine line between math and theoretical physics sometimes.
MO is for research-level mathematics, not "how do I compute X"
user54412
@KevinDriscoll You could maybe reword to push that question in the direction of another site, but imo as worded it falls squarely in the domain of math.SE - it's just a shame they don't give that kind of question as much attention as, say, explaining why 7 is the only prime followed by a cube
@ChrisWhite As I understand it, KITP wants big names in the field who will promote crazy ideas with the intent of getting someone else to develop their idea into a reasonable solution (c.f., Hawking's recent paper) |
I am trying to solve the following exercise:
Prove that on a surface of constant curvature the geodesic circles have constant curvature.
"Constant curvature" in case of the surface I take to refer to the Gaussian curvature. Now, the geodesic curvature of a curve parameterized by arc length in orthogonal coordinates is given by
$$k_g(s) = \frac{1}{2 \sqrt{EG}} \left(G_u v'- E_v u' \right)+ \phi',$$
where $\cdot'$ denotes the derivative with respect to $s$, and $\phi$ is the angle the tangent of the curve makes with $x_u$.
Using geodesic polar coordinates (setting $u = \rho$ and $v = \theta$), a surface with constant Gaussian curvature $K$ satisfies
$$(\sqrt{G}_{\rho\rho}) + K \sqrt{G} = 0$$
Also, we get $E=1$, $F=0$, and a geodesic circle has the equation $\rho = \mathrm{const.}$ Therefore, the first equation above yields
$$ k_g(s) = \frac{G_\rho \theta'}{2\sqrt{G}} $$
It seems to prove that $k_g$ is constant, you would have to show that its derivative is 0. I tried that, but the derivative gets rather ugly and I don't see how to proceed. |
Suppose $X_1, \dots, X_n \overset{\text{iid}}{\sim}\dfrac{x}{\theta}\exp\left(-\dfrac{x^2}{2\theta}\right)\mathbf{1}_{(0, \infty)}(x)$, $\theta > 0$. At the end of the day, my goal is to calculate $$\max_{\theta > 16}L(\theta)$$ where $L$ is the likelihood function, i.e., $$L(\theta) = \dfrac{\prod_{i=1}^{n}x_i}{\theta^n}\exp\left(\dfrac{-1}{2\theta}\sum_{i=1}^{n}x_i^2\right)\text{.}$$ In order to do this, we have to find $\theta$ such that $L$ is maximized. Using the typical methods, I find that $$\hat{\theta}_n = \dfrac{\sum_{i=1}^{n}X_i^2}{2n}$$ is the maximum likelihood estimator of $\theta$, assuming $\theta > 0$.
However, when restricting $\theta$ to $\{\theta: \theta > 16\}$, the solution I have says that the maximum likelihood estimator is $$\hat{\theta}_n = \max\left(16, \dfrac{\sum_{i=1}^{n}X_i^2}{2n}\right)$$is the MLE.
Why is this? Intuitively, this doesn't make any sense to me because one would think that one wants $\hat\theta_n$ to be as small as possible, since increasing $\theta$ would decrease $L$, holding the $\{x_i\}$ constant. So I think $\max$ should be replaced with $\min$ in the equation for $\hat\theta_n$ above. However, yet at the same time, we are restricting $\theta > 16$.
Could someone please provide me with some insight for this? |
Tangential Continuous Displacement and Normal-Normal Continuous Stress Mixed Finite Elements for Linear Elasticity Dipl.-Ing. in Dr. in Astrid Pechstein Jan. 9, 2007, 3:30 p.m. T 1010
Abstract. We consider the mixed formulation of linear elasticity, which contains the displacement $u$ as well as the stress tensor $\sigma$ as unknowns,
$\begin{eqnarray} −div(\sigma) & = & f \\ A\sigma & = & \epsilon(u) = \frac 1 2 \left(\nabla u + \nabla u^T \right) \end{eqnarray}$
We derive a variational formulation of the problem, choosing $u \in H(curl)$. For the stress tensor $\sigma$ we do not only need $\sigma \in L^2$, but further
$div \sigma \in H(curl)^∗ = H^{−1}(div).$
We refer to this space as $H(div div)$. We see that the variational problem is well posed. We use the Finite Element Method to discretize the problem. For the displacement, we use tangential-continuous Nédélec finite elements, whereas for the stresses we propose a new family of finite elements. These elements are symmetric, tensor-valued, and normal-normal continuous. This formulation is suitable for nearly incompressible materials, where the Poisson ratio $ν$ approaches 1/2. Also, the elements do not suffer from shear locking when anisotropic elements are used.
We present shape functions of arbitrary order for the finite element spaces. To map the reference element onto the physical element, we use transformations that keep the tangential trace for the displacement and the normal-normal trace for the stresses. We see that the discrete system is stable without further stabilization.
In order to obtain a positive definite system matrix, we hybridize the stress space. This way we find a preconditioner that works for nearly incompressible materials.
The solution satisfies optimal order error estimates. We discuss the implementation of the new elements, and give numerical results. |
@user193319 I believe the natural extension to multigraphs is just ensuring that $\#(u,v) = \#(\sigma(u),\sigma(v))$ where $\# : V \times V \rightarrow \mathbb{N}$ counts the number of edges between $u$ and $v$ (which would be zero).
I have this exercise: Consider the ring $R$ of polynomials in $n$ variables with integer coefficients. Prove that the polynomial $f(x _1 ,x _2 ,...,x _n ) = x _1 x _2 ···x _n$ has $2^ {n+1} −2$ non-constant polynomials in R dividing it.
But, for $n=2$, I ca'nt find any other non-constant divisor of $f(x,y)=xy$ other than $x$, $y$, $xy$
I am presently working through example 1.21 in Hatcher's book on wedge sums of topological spaces. He makes a few claims which I am having trouble verifying. First, let me set-up some notation.Let $\{X_i\}_{i \in I}$ be a collection of topological spaces. Then $\amalg_{i \in I} X_i := \cup_{i ...
Each of the six faces of a die is marked with an integer, not necessarily positive. The die is rolled 1000 times. Show that there is a time interval such that the product of all rolls in this interval is a cube of an integer. (For example, it could happen that the product of all outcomes between 5th and 20th throws is a cube; obviously, the interval has to include at least one throw!)
On Monday, I ask for an update and get told they’re working on it. On Tuesday I get an updated itinerary!... which is exactly the same as the old one. I tell them as much and am told they’ll review my case
@Adam no. Quite frankly, I never read the title. The title should not contain additional information to the question.
Moreover, the title is vague and doesn't clearly ask a question.
And even more so, your insistence that your question is blameless with regards to the reports indicates more than ever that your question probably should be closed.
If all it takes is adding a simple "My question is that I want to find a counterexample to _______" to your question body and you refuse to do this, even after someone takes the time to give you that advice, then ya, I'd vote to close myself.
but if a title inherently states what the op is looking for I hardly see the fact that it has been explicitly restated as a reason for it to be closed, no it was because I orginally had a lot of errors in the expressions when I typed them out in latex, but I fixed them almost straight away
lol I registered for a forum on Australian politics and it just hasn't sent me a confirmation email at all how bizarre
I have a nother problem: If Train A leaves at noon from San Francisco and heads for Chicago going 40 mph. Two hours later Train B leaves the same station, also for Chicago, traveling 60mph. How long until Train B overtakes Train A?
@swagbutton8 as a check, suppose the answer were 6pm. Then train A will have travelled at 40 mph for 6 hours, giving 240 miles. Similarly, train B will have travelled at 60 mph for only four hours, giving 240 miles. So that checks out
By contrast, the answer key result of 4pm would mean that train A has gone for four hours (so 160 miles) and train B for 2 hours (so 120 miles). Hence A is still ahead of B at that point
So yeah, at first glance I’d say the answer key is wrong. The only way I could see it being correct is if they’re including the change of time zones, which I’d find pretty annoying
But 240 miles seems waaay to short to cross two time zones
So my inclination is to say the answer key is nonsense
You can actually show this using only that the derivative of a function is zero if and only if it is constant, the exponential function differentiates to almost itself, and some ingenuity. Suppose that the equation starts in the equivalent form$$ y'' - (r_1+r_2)y' + r_1r_2 y =0. \tag{1} $$(Obvi...
Hi there,
I'm currently going through a proof of why all general solutions to second ODE look the way they look. I have a question mark regarding the linked answer.
Where does the term e^{(r_1-r_2)x} come from?
It seems like it is taken out of the blue, but it yields the desired result. |
Difference between revisions of "Group cohomology of dihedral group:D8"
(→Over an abelian group)
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group = dihedral group:D8|
group = dihedral group:D8|
connective = of}}
connective = of}}
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==Homology groups for trivial group action==
==Homology groups for trivial group action==
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===Over the integers===
===Over the integers===
−
The homology groups
+
The homology groups the integers are as follows:
−
<math>
+
<math>(D_8;\mathbb{Z}) = \left \lbrace \begin{array}{rl} \mathbb{Z}, & = 0 \\\mathbb{Z}/2\mathbb{Z}, & \equiv 1 \pmod 4\\ \mathbb{Z}/\mathbb{Z}, & \equiv 3 \pmod 4 \\\\, \{ even }\\ \end{array}\right.</math>
− +
first few homology groups are :
−
{| class="
+
{| class="" border="1"
− +
<math></math> 012345678
|-
|-
−
| <math>
+
| <math>\mathbb{Z}</math> || <math>\mathbb{Z}</math> || <math>\mathbb{Z}/2\mathbb{Z}</math> || || <math>\mathbb{Z}/\mathbb{Z}</math> || <math>\mathbb{Z}/2\mathbb{Z}</math> || || <math>\mathbb{Z}/\mathbb{Z}</math> ||
|}
|}
===Over an abelian group===
===Over an abelian group===
−
The homology groups
+
The homology groups an abelian group <math>M</math> are as follows:
−
<math>
+
<math>(D_8;M) = \left \lbrace \begin{array}{rl} M, & = 0 \\M/2M, & \equiv 1 \pmod 4\\ \operatorname{Ann}()& \equiv 2 \pmod 4 \\M/, & \equiv 3 \pmod 4 \\\operatorname{Ann}(), & \equiv 0 \pmod 4\\ \end{array}\right.</math>
+ − + +
<math>
+ +
\operatorname{Ann}(M)</math> 2 <math>M</math> <math>\operatorname{Ann}(M)</math> <math>M</math>
+
==Cohomology groups for trivial group action==
==Cohomology groups for trivial group action==
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===Over the integers===
===Over the integers===
−
The cohomology groups
+
The cohomology groups the integers are as follows
+ + + +
:
−
<math>H^
+ + + +
<math>H^\mathbb{Z}\\\{}{}\mathbb{Z}\\\\mathbb{Z}/2\mathbb{Z} \\4\\mathbb{Z}/\mathbb{Z} \\\\4\\\\\{}\\\{}</math>
+
===Over an abelian group===
===Over an abelian group===
−
The cohomology groups
+
The cohomology groups an abelian group <math>M</math> are as follows:
−
<math>H^
+
<math>H^(D_8;M) = \left \lbrace \begin{array}{rl} M, & = 0 \\\operatorname{Ann}(M), & \equiv 1 \pmod 4\\ M/2M& \equiv 2 \pmod 4 \\\operatorname{Ann}(M), & \equiv 3 \pmod 4 \\M/, & \equiv 0 \pmod 4\\ \end{array}\right.</math>
− +
<math>
+ + + + +
\operatorname{Ann}(M)</math> 2<math>M</math> <math>\operatorname{Ann}(M)</math> <math>M</math>
+
==Cohomology ring with coefficients in integers==
==Cohomology ring with coefficients in integers==
Line 53: Line 80:
==Second cohomology groups and extensions==
==Second cohomology groups and extensions==
+ + + + + + + + + +
===Second cohomology groups for trivial group action===
===Second cohomology groups for trivial group action===
{| class="sortable" border="1"
{| class="sortable" border="1"
−
! Group acted upon !! Order !! Second part of GAP ID !! [[Second cohomology group for trivial group action]] !! Extensions !! Cohomology information
+
! Group acted upon !! Order !! Second part of GAP ID !! [[Second cohomology group for trivial group action]] !! Extensions !! Cohomology information
|-
|-
−
| [[cyclic group:Z2]] || 2 || 1 || [[elementary abelian group:E8]] || [[direct product of D8 and Z2]], [[SmallGroup(16,3)]], [[nontrivial semidirect product of Z4 and Z4]], [[dihedral group:D16]], [[semidihedral group:SD16]], [[generalized quaternion group:Q16]] || [[second cohomology group for trivial group action of D8 on Z2]]
+
| [[cyclic group:Z2]] || 2 || 1 || [[elementary abelian group:E8]] || [[direct product of D8 and Z2]], [[SmallGroup(16,3)]], [[nontrivial semidirect product of Z4 and Z4]], [[dihedral group:D16]], [[semidihedral group:SD16]], [[generalized quaternion group:Q16]] || [[second cohomology group for trivial group action of D8 on Z2]]
|-
|-
−
| [[cyclic group:Z4]] || 4 || 1 ||
+
| [[cyclic group:Z4]] || 4 || 1 || || || [[second cohomology group for trivial group action of D8 on Z4
+ +
]]
|}
|}
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Latest revision as of 00:27, 29 May 2013
Contents This article gives specific information, namely, group cohomology, about a particular group, namely: dihedral group:D8. View group cohomology of particular groups | View other specific information about dihedral group:D8 Family contexts
Family name Parameter value Information on group cohomology of family dihedral group of degree , order degree , order group cohomology of dihedral groups Homology groups for trivial group action FACTS TO CHECK AGAINST(homology group for trivial group action): First homology group: first homology group for trivial group action equals tensor product with abelianization Second homology group: formula for second homology group for trivial group action in terms of Schur multiplier and abelianization|Hopf's formula for Schur multiplier General: universal coefficients theorem for group homology|homology group for trivial group action commutes with direct product in second coordinate|Kunneth formula for group homology Over the integers
The homology groups over the integers are given as follows:
The first few homology groups are given below:
Over an abelian group
The homology groups over an abelian group are given as follows:
The first few homology groups with coefficients in an abelian group are given below:
Cohomology groups for trivial group action FACTS TO CHECK AGAINST(cohomology group for trivial group action): First cohomology group: first cohomology group for trivial group action is naturally isomorphic to group of homomorphisms Second cohomology group: formula for second cohomology group for trivial group action in terms of Schur multiplier and abelianization In general: dual universal coefficients theorem for group cohomology relating cohomology with arbitrary coefficientsto homology with coefficients in the integers. |Cohomology group for trivial group action commutes with direct product in second coordinate | Kunneth formula for group cohomology Over the integers
The cohomology groups over the integers are given as follows:
The first few cohomology groups are given below:
0 Over an abelian group
The cohomology groups over an abelian group are given as follows:
The first few cohomology groups with coefficients in an abelian group are:
Cohomology ring with coefficients in integers PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] Second cohomology groups and extensions Schur multiplier
This has implications for projective representation theory of dihedral group:D8.
Schur covering groups
The three possible Schur covering groups for dihedral group:D8 are: dihedral group:D16, semidihedral group:SD16, and generalized quaternion group:Q16. For more, see second cohomology group for trivial group action of D8 on Z2, where these correspond precisely to the stem extensions.
Second cohomology groups for trivial group action
Group acted upon Order Second part of GAP ID Second cohomology group for trivial group action (as an abstract group) Order of second cohomology group Extensions Number of extensions up to pseudo-congruence, i.e., number or orbits under automorphism group actions Cohomology information cyclic group:Z2 2 1 elementary abelian group:E8 8 direct product of D8 and Z2, SmallGroup(16,3), nontrivial semidirect product of Z4 and Z4, dihedral group:D16, semidihedral group:SD16, generalized quaternion group:Q16 6 second cohomology group for trivial group action of D8 on Z2 cyclic group:Z4 4 1 elementary abelian group:E8 8 direct product of D8 and Z4, nontrivial semidirect product of Z4 and Z8, SmallGroup(32,5), central product of D16 and Z4, SmallGroup(32,15), wreath product of Z4 and Z2 6 second cohomology group for trivial group action of D8 on Z4 Klein four-group 4 2 elementary abelian group:E64 64 [SHOW MORE] 11 second cohomology group for trivial group action of D8 on V4 Baer invariants
Subvariety of the variety of groups General name of Baer invariant Value of Baer invariant for this group abelian groups Schur multiplier cyclic group:Z2 groups of nilpotency class at most two 2-nilpotent multiplier groups of nilpotency class at most three 3-nilpotent multiplier any variety of groups containing all groups of nilpotency class at most three -- GAP implementation Computation of integral homology
The homology groups for trivial group action with coefficients in can be computed in GAP using the GroupHomology function in the
HAP package, which can be loaded by the command
LoadPackage("hap"); if it is installed but not loaded. The function outputs the orders of cyclic groups for which the homology or cohomology group is the direct product of these (more technically, it outputs the elementary divisors for the homology or cohomology group that we are trying to compute).
Here are computations of the first few homology groups:
Computation of first homology group gap> GroupHomology(DihedralGroup(8),1); [ 2, 2 ]
The way this is to be interpreted is that the first homology group (the abelianization) is the direct sum of cyclic groups of the orders listed, so in this case we get that is , which is the Klein four-group.
Computation of second homology group gap> GroupHomology(DihedralGroup(8),2); [ 2 ] Computation of first few homology groups
To compute the first eight homology groups, do:
gap> List([1,2,3,4,5,6,7,8],i->[i,GroupHomology(DihedralGroup(8),i)]); [ [ 1, [ 2, 2 ] ], [ 2, [ 2 ] ], [ 3, [ 2, 2, 4 ] ], [ 4, [ 2, 2 ] ], [ 5, [ 2, 2, 2, 2 ] ], [ 6, [ 2, 2, 2 ] ], [ 7, [ 2, 2, 2, 2, 4 ] ], [ 8, [ 2, 2, 2, 2 ] ] ] |
Nope. I like to think of ecological inference as creating confounds between contextual (and spatial) effects and individual level effects (in Sociological speak). So consider a set of equations at the individual level:
$$y_1 = \beta_1(x_1) + \beta_2(\bar{X})$$$$y_2 = \beta_1(x_2) + \beta_2(\bar{X})$$
Where $y_1$ and $x_1$ are individual level characteristics, and $\bar{X} = \frac{x_1 + x_2}{2}$ (so $\beta_2$ may be considered a contextual effect). If you aggregate (i.e. add the two individual level equations together) you get:
$$(y_1 + y_2) = \beta_1(x_1 + x_2) + 2 \cdot \beta_2(\bar{X})$$
Then dividing by two you have an equation for the group means:
$$\frac{y_1 + y_2}{2} = \beta_1(\frac{x_1 + x_2}{2}) + \frac{2 \cdot \beta_2(\bar{X})}{2}$$$$\bar{Y} = \beta_1(\bar{X}) + \beta_2(\bar{X})$$
Here you can see that you can't uniquely identify $\beta_1$ (the individual level effect) and $\beta_2$ (the contextual aggregate effect). All you can estimate is their combined effects, $(\beta_1 + \beta_2)$. So what exactly do fixed effects do? Lets start with our initial individual level equation and instead of an observed contextual effect, lets say we know a contextual effect exists but can't observe it, $\bar{Z}$.
$$y_1 = \beta_1(x_1) + \beta_2(\bar{Z})$$$$y_2 = \beta_1(x_2) + \beta_2(\bar{Z})$$
Using the same logic as before, we can estimate a group mean equation:
$$\bar{Y} = \beta_1(\bar{X}) + \beta_2(\bar{Z})$$
Now here the magic happens, we can subtract out the group mean equation from the individual level equation, and cancel out the unobserved contextual effect of $\beta_2(\bar{Z})$:
$$y_i - \bar{Y} = \beta_1(x_i - \bar{X}) + \beta_2(\bar{Z} - \bar{Z})$$
So to sum up; with the individual level equation fixed group effects can control for
all unobserved group level effects. They can not however undue confounds created from aggregation to begin with. |
How significant is a value compared to a list of values? In most cases statistical testing involves comparing a sample set to a population. In my case the sample is made by one value and we compare it to the population.
I am a dilettante in statistical hypothesis testing confronted with perhaps the most basic problem. It is not just one test but hundreds of them. I have a parameter space, and must do a significance test for every point. Both value and background list (population) are generated for each parameter combination. Then I am ordering this by p-value and find interesting parameter combinations. In fact, the finding of parameter combinations where this p-val is high (nonsignificance) is also important.
So let's take one single test: I have a computed value generated from a selected set and a background set of values computed by choosing a random training set. The computed value is 0.35 and the background set is (probably?) normally distributed with a mean of 0.25 and a very narrow std (e-7). I actually don't have knowledge on the distribution, because the samples are computed from something else, they are not random numbers samples from some distribution, so background is the correct word for it.
The null hypothesis would be that "the mean of the sample test equals my computed value, of 0.35". When should I consider this to be a Z-test or a T-test? I want the value to be significantly higher than the population mean, therefore it is a single-tailed test.
I am a bit confused as to what to consider as a sample: I either have a sample of one (the observation) and the background list as the population OR my sample is the background list and I am comparing that to the whole (unsampled) population which according to the null hypothesis should have the same mean. Once this is decided, the test goes to different directions I guess.
If it is a T-test, how do I compute its p-value? I would like to compute it myself rather than using an R/Python/Excel function (I already know how to do that) therefore I must establish the correct formula first.
To begin with, I suspect a T-test is a bit too general, since in my case the T-test would be linked to the sample size and would have the form: $$T=Z/s,$$ where $$Z=\frac{\bar{X}}{\frac{\sigma}{\sqrt{n}}}$$ and s is $$s=\hat{\sigma}/\sigma$$, the sample std versus the population std. So I have two cases: either my sample size is the size of the population, which I "guess" would mean I am dealing with a Z-test, or the population statistics (n and std) are unknown but the distribution can be in some way approximated and I am really dealing with a T-test. In any case my following questions are:
How do I compute a p-value? (i.e. not using an R/Python/Excel function or p-value table look-up but actually compute it based on a formula, because I want to know what I am doing) How do I decide a significance threshold based on my sample size? (a formula would be nice) |
S-shaped and broken s-shaped bifurcation curves for a multiparameter diffusive logistic problem with holling type-Ⅲ functional response
Department of Applied Mathematics, National University of Tainan, Tainan 700, Taiwan, ROC
${\left\{ {\begin{array}{*{20}{l}} {{u^{\prime \prime }}(x) + \lambda \left[ {ru(1 - \frac{u}{q}) - \frac{{{u^p}}}{{1 + {u^p}}}\% } \right] = 0{\text{,}} - {\text{1}} < x < 1{\text{,}}} \\ {u( - 1) = u(1) = 0{\text{, }}} \end{array}} \right.},$ uis the population density of the species, p> 1, q, rare two positive dimensionless parameters, and λ> 0 is a bifurcation parameter. For fixed p> 1, assume that q, rsatisfy one of the following conditions: (ⅰ) r≤ η 1, p * qand ( q, r) lies above the curve
$\begin{array}{l}{\Gamma _1} = \{ (q,r):q(a) = \frac{{a[2{a^p} - (p - 2)]}}{{{a^p} - (p - 1)}}{\rm{, }}\\\quad \quad \quad \quad \quad r(a) = \frac{{{a^{p - 1}}[2{a^p} - (p - 2)]}}{{{{({a^p} + 1)}^2}}}{\rm{, }}\sqrt[p]{{p - 1}}\% < a < C_p^*\} ;\end{array}$ r≤ η 2, p * qand ( q, r) lies on or below the curve Γ1, where η 1, p *and η 2, p *are two positive constants, and $C_{p}^{*}={{\left(\frac{{{p}^{2}}+3p-4+p\sqrt{%{{p}^{2}}+6p-7}}{4} \right)}^{1/p}}$. Then on the ( λ, || u|| ∞)-plane, we give a classification of threequalitatively different bifurcation curves: an S-shaped curve, a broken S-shaped curve, and a monotone increasing curve. Hence we are able to determine the exact multiplicity of positive solutions by the values of q, rand λ. Keywords:S-shaped bifurcation curve, broken S-shaped bifurcation curve, Holling type-Ⅲ functional response, diffusive logistic problem, time map. Mathematics Subject Classification:Primary: 34B15, 34B18. Citation:Tzung-shin Yeh. S-shaped and broken s-shaped bifurcation curves for a multiparameter diffusive logistic problem with holling type-Ⅲ functional response. Communications on Pure & Applied Analysis, 2017, 16 (2) : 645-670. doi: 10.3934/cpaa.2017032
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S-shaped and broken S-shaped bifurcation diagrams with hysteresis for a multiparameter spruce budworm population problem in one space dimension,
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J. Jiang and J. Shi,
Bistability dynamics in some structured ecological models, in , Chapman & Hall/CRC Press, Boca Raton, FL, 2009.
Google Scholar
Spatial Ecology [3] [4] [5] [6] [7] [8] [9]
J. D. Murray,
[10]
J. D. Murray,
[11] [12] [13] [14] [15] [16]
S.-H. Wang and T.-S. Yeh,
S-shaped and broken S-shaped bifurcation diagrams with hysteresis for a multiparameter spruce budworm population problem in one space dimension,
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Shao-Yuan Huang, Shin-Hwa Wang.
On S-shaped bifurcation curves for a
two-point boundary value problem arising in a theory of thermal explosion.
[2]
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S-shaped bifurcation curves for a combustion problem with general arrhenius reaction-rate laws.
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Stability results for discontinuous nonlinear elliptic and parabolic problems with a S-shaped bifurcation branch of stationary solutions.
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Inverse S-shaped probability weighting and its impact on investment.
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The effect of delay on a diffusive predator-prey system with Holling Type-II predator functional
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Positive steady state solutions of a diffusive Leslie-Gower predator-prey model with Holling type II functional response and cross-diffusion.
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Use the diagram to prove the double angle formula, where $t=\tan\theta$: $$\tan2\theta = {2t\over {1-t^2}},\quad \sin2\theta ={2t\over {1+t^2}},\quad \cos2\theta = {{1-t^2}\over {1+t^2}}$$
The point $P'=(p',q')$ is the image of the point $P=(p,q)$ afterreflection in the line $y=mx$. To find $(p',q')$ use the fact thatthe midpoint of $PP'$ is on the line $y=mx$ and the line segment$PP'$ is perpendicular to the line $y=mx$ and show that$$p'=p\cos2\theta + q\sin2\theta,\ q'=p\sin2\theta -q\cos2\theta\quad (1)$$ where $m=\tan\theta$. Hence establishanother proof that the matrix
$$T_2= \left( \begin{array}{cc} \cos 2\theta &\sin2\theta \\\sin2\theta &-\cos2\theta \end{array} \right) $$
gives a reflection in the line $y=x\tan\theta$. The point$P''=(p'',q'')$ is the image of the point $P'$ after reflection inthe line $y=x\tan\phi$. Apply the transformation $$T_2' = \left(\eqalign{\cos 2\phi &\sin2\phi \\ \sin2\phi &-\cos2\phi}\right)$$ to the point $P'=(p',q')$ to find thecoordinates of the point $P''$ in terms of $p, q, \theta$ and$\phi$. Hence show that the combination of two reflections indistinct intersecting lines is a rotation about the point ofintersection by twice the angle between the two mirror lines. Whatis the effect of the two reflections if the lines coincide (i.e.$\theta=\phi$)? |
The Existence/Uniqueness of Solutions to Second Order Linear Differential Equations
Recall that from The Existence/Uniqueness of Solutions to First Order Linear Differential Equations page that if $p$ and $g$ are continuous functions on an interval $I = (\alpha, \beta)$ and $t_0 \in I$, then the linear first order differential equations in the form $\frac{dy}{dt} + p(t) y = g(t)$ with the initial value condition $y(t_0) = y_0$ has a unique solution $y = \phi(t)$ for $t \in I$. We will now state an analogous theorem for linear second order differential equations.
Theorem 1: Let $p$, $q$, and $g$ be continuous functions on an open interval $I$ such that $t_0 \in I$. Then the second order linear differential equation $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = g(t)$ with the initial conditions $y(t_0) = y_0$ and $y'(t_0) = y'_0$ has a unique solution $y = \phi (t)$ throughout $I$.
We will now look at an example of applying Theorem 1.
Example 1 Determine the largest interval for which the second order linear differential equation $(t^2 - 4t) \frac{d^2y}{dt^2} + 3t \frac{dy}{dt} + 4y = 2$ with the initial conditions $y(3) = 0$ and $y'(3) = -1$ has a unique solution.
In order to apply Theorem 1, we will need to rewrite our differential equation as follows:(1)
Therefore $p(t) = \frac{3t}{t(t - 4)}$, $q(t) = \frac{4}{t(t-4)}$, and $g(t) = \frac{2}{t(t-4)}$. Collectively, it is not hard to see that all of these functions are continuous for all $t$ where their denominators are nonzero, that is for $t \neq 0$ and for $t \neq 4$. We note that our interval must contain our initial condition $y(3) = 0$, that is, the value $t = 3$ must be contained in our interval. Therefore a unique solution $y = \phi(t)$ to this second order linear differential equation exists on the interval $(0, 4)$.
Example 2 Determine the largest interval for which the second order linear differential equation $(t - 3) \frac{d^2y}{dt^2} + t \frac{dy}{dt} + \ln \mid t \mid y = 0$ with the initial conditions $y(1) = 0$ and $y'(1) = 1$ has a unique solution.
In order to apply Theorem 1, we will need to rewrite our differential equation as follows.(2)
Therefore $p(t) = \frac{t}{t -3}$, $q(t) = \frac{\ln \mid t \mid}{t - 3}$, and $g(t) = 0$. We have that $p(t)$ is continuous for $t \neq 3$, $q(t)$ is continuous for $t \ neq 3$ and $t \neq 0$, and $g(t)$ is continuous everywhere.
The possible intervals for the solution are therefore $(-\infty, 0)$, $(0, 3)$, or $(3, \infty)$.
With the initial condition $y(1) = 0$ we see that $1$ must be contained in our choice of interval, and so $(0, 3)$ is the largest interval for which a unique solution to our differential equation exists.
Remark 1: In both of the examples above, we needed the second order differential equation to be in the appropriate form $\frac{d^2y}{dt^2} + p(t) \frac{dy}{dt} + q(t)y = g(t)$ in order to apply Theorem 1. If we were to apply Theorem 1 without the second order differential equations from above in the correct form, then we would not obtain correct intervals for which a unique solution is guaranteed to be in. |
Tournaments and Rankings Tournaments
Definition: A Tournament is a directed graph $G$ so that for every pair $x, y \in V(G)$, then either $(x, y) \in E(D)$ (a directed edge from $x$ going to $y$) or $(y, x) \in E(D)$ (a directed edge from $y$ going to $x$).
By the definition above, a tournament on $n$-vertices can be thought of as an orientation of a complete graph $K_n$. For example, let's look at the following tournament on $4$ vertices:
The underlying graph is a complete graph on $4$ vertices, $K_4$, and there exists precisely one directed edge between each vertices, hence, the graph above is a tournament.
Number of Edges in Tournaments
Because a tournament is essentially just a directed complete graph, there will always be exactly $\frac{n!}{2(n - 2)!}$ arcs where n represents the number of vertices the tournament is on. For example, any tournament on $4$ vertices will have precisely $\frac{4!}{2 \cdot 2!} = 6$ edges.
Rankings
Definition: A Ranking for a tournament graph $G$ is an ordered list of vertices $v_1, v_2, ..., v_n$ of the vertex set so that each $i = 1, 2, ..., n-1$ then $(v_1, v_2) \in E(G)$.
Additionally, we can consider rankings to be directed Hamiltonian paths. Recall that a Hamiltonian path is a walk with no repeated edges or vertices that contains all vertices in the vertex set. Hence, a directed Hamiltonian path is the exact same thing with the only difference being directed edges dictating the Hamiltonian path.
In the example above, there exists a Hamiltonian path $acdb$ so the tournament in the earlier example has a ranking. |
Yes, but it may not be valid. The extrapolation will be valid for about 0.1 * PBL Height using the Log-Wind Profile
You will need:
PBL Height. A second Wind speed (within 0.1*PBL Height) Surface Sensible Heat Flux Surface Latent Heat Flux Potential Temperature
You can use the last three variables to calculate the Monin-Obukhov Length (MOL). Then use the MOL to calculate $\psi$:
If L>0 $\psi=\frac{z}{L}$
If L=0 $\psi=0$
If L<0 $\psi=2\ln(\frac{1+x}{2})+\ln(\frac{1+x^2}{2})-2\tan^{-1}(x)+\frac{\pi}{2}$
where $x=(1-15\frac{z}{L})^{\frac{1}{4}}$
Now that you have $\psi$, you can derive the friction velocity and surface roughness length using the log-wind profile and $\psi(\frac{z}{L})$. Once you have those two, you can extrapolate wind speed 600 meters up, provided the surface layer is that high. |
The output resistance \$r_a\$ is the parallel combination of \$R_E\$ and the resistance looking into the emitter of the transistor. As shown in the solution drawing, the resistance looking into the emitter of the transistor is \$1/g_m\$ plus some other resistance due to the source resistance \$R_G\$ and the biasing resistors \$R_B\$ and \$R_V\$. That "other resistance" caused by \$R_G\$, \$R_B\$, and \$R_V\$ is denoted by a convenience term that the solution calls \$r_a'\$. In equation form, this means that $$ r_a = R_E || \left(\frac{1}{g_m}+r_a'\right) $$
Now we need to calculate \$r_a'\$. In small signal analysis the DC sources are turned off so \$U_0 = U_G = 0\$, and this means that \$R_G\$, \$R_B\$, and \$R_V\$ end up in parallel with each other. The parallel combination of these three resistors is shown in the solution in the dotted box with the value \$R\$. \$r_a'\$ is the parallel combination of these resistors but divided by a factor of approximately \$\beta\$, the small signal current gain (since the current through the three resistors is \$i_b\$ but we were calculating the resistance looking into the emitter). Therefore $$ r_a'= \frac{R_G || R_B || R_V}{\beta}$$
You can substitute the equation for \$r_a'\$ into the equation for \$r_a\$ to get $$ r_a = R_E || \left(\frac{1}{g_m}+\frac{R_G || R_B || R_V}{\beta}\right) $$
The transistor model this solution uses is a little different than the commonly used hybrid-\$\pi\$ model which uses a resistance from base to emitter \$r_{\pi} = \beta/g_m\$. You might want to calculate the solution using the hybrid-\$\pi\$ model since this circuit's output resistance is a little easier to express in terms of \$r_{\pi}\$. |
Start with the unperturbed gravitational potential for a uniform sphere of mass M and radius R, interior and exterior:
$$ \phi^0_\mathrm{in} = {-3M \over 2R} + {M\over 2R^3} (x^2 + y^2 + z^2) $$$$ \phi^0_\mathrm{out} = {- M\over r} $$
Add a quadrupole perturbation, you get
$$ \phi_\mathrm{in} = \phi^0_\mathrm{in} + {\epsilon M\over R^3} D $$$$ \phi_\mathrm{out} = \phi^0_\mathrm{out} + {M\epsilon R^2\over r^5} D $$
$$ D = x^2 + y^2 - 2 z^2 $$
The scale factors of M and R are just to make $\epsilon$ dimensionless, the falloff of $D\over r^5$ is just so that the exterior solution solves Laplace's equation, and the matching of the solutions is to ensure that on
any ellipsoid near the sphere of radius R, the two solutions are equal to order $\epsilon$. The reason this works is because the $\phi^0$ solutions are matched both in value and in first derivative at x=R, so they stay matched in value to leading order even when perturbed away from a sphere. The order $\epsilon$ quadrupole terms are equal on the sphere, and therefore match to leading order.
The ellipsoid I will choose solves the equation:
$$ r^2 + \delta D = R^2 $$
The z-diameter is increased by a fraction $\delta$, while the x diameter decreased by $\delta/2$. So that the ratio of polar to equatorial radius is $3\delta/2$. To leading order
$$ r = R + {\delta D \over 2R}$$
We already matched the values of the inner and outer solutions, but we need to match the derivatives. taking the "d":
$$ d\phi_\mathrm{in} = {M\over R^3} (rdr) + {\epsilon M\over R^3} dD $$$$ d\phi_\mathrm{out} = {M\over r^3} (rdr) + {MR^2\epsilon \over r^5} dD - {5\epsilon R^2 M\over r^7} (rdr) $$
$$ rdr = x dx + y dy + z dz $$$$ dD = 2 x dx + 2ydy - 4z dz $$
To first order in $\epsilon$, only the first term of the second equation is modified by the fact that r is not constant on the ellipsoid. Specializing to the surface of the ellipsoid:
$$ d\phi_\mathrm{out}|_\mathrm{ellipsoid} = {M\over R^3} (rdr) + {3\delta \over 2 R^5}(rdr) + {\epsilon M \over R^3} dD - {5\epsilon M \over R^5} (rdr)$$
Equating the in and out derivatives, the parts proportional to $dD$ cancel (as they must--- the tangential derivatives are equal because the two functions are equal on the ellipsoid). The rest must cancel too, so
$$ {3\over 2} \delta = 5 \epsilon $$
So you find the relation between $\delta$ and $\epsilon$. The solution for $\phi_\mathrm{in}$ gives
$$ \phi_\mathrm{in} + {3M\over 2R} = {M\over 2R^3}( r^2 + {3\over 5} \delta D ) $$
Which means, looking at the equation in parentheses, that the equipotentials are 60% as squooshed as the ellipsoid.
Now there is a condition that this is balanced by rotation, meaning that the ellipsoid is an equipotential once you add the centrifugal potential:
$$ - {\omega^2\over 2} (x^2 + y^2) = -{\omega^2 \over 3} (x^2 + y^2 + z^2) -{\omega^2\over 6} (x^2 + y^2 - 2z^2) $$
To make the $\delta$ ellipsoid equipotential requires that $\omega^2\over 6$ equals the remaining ${2\over 5} {M\over 2R^2}$, so that, calling $M\over R^2$ (the acceleration of gravity) by the name "g", and $\omega^2 R$ by the name "C" (centrifugal)
$$\delta = {5\over 6} {C \over g} $$
The actual difference in equatorial and polar diameters is found by multiplying by 3/2 (see above):
$$ {3\over 2} \delta = {5\over 4} {C\over g} $$
instead of the naive estimate of ${C\over 2g}$. So the naive estimate is multiplied by two and a half for a uniform density rotating sphere.
Nonuniform interior: primitive model
The previous solution is both interior and exterior for a rotating uniform ellipsoid, and it is exact in r, it is only leading order in the deviation from spherical symmetry. So it immediately extends to give the shape of the Earth for a nonuniform interior mass distribution. The estimate with a uniform density is surprisingly good, and this is because there are competing effects largely cancelling out the correction for non-uniform density.
The two competing effects are:1. the interior distribution is more elliptical than the surface, because the interior solution feels all the surrounding elliptical Earth deforming it, with extra density deforming it more. 2. The ellipticity of the interior is suppressed by the $1/r^3$ falloff of the quadrupole solution of Laplace's equation, which is $1/r^2$ faster than the usual potential. So although the interior is somewhat more deformed, the falloff more than compensates, and the effect of the interior extra density is to make the Earth more spherical, although not by much.
These competing effects are what shift the correction factor from 2.5 to 2, which is actually quite small considering that the interior of the Earth is extremely nonuniform, with the center more than three times as dense as the outer parts.
The exact solution is a little complicated, so I will start with a dopey model. This assumes that the Earth is a uniform ellipsoid of mass M and ellipticity parameter $\delta$, plus a point source in the middle (or a sphere, it doesn't matter), accounting for the extra mass in the interior, of mass M'. The interior potential is given by superposition. With the centrifugal potential:
$$ \phi_{int} = - {M'\over r} - {2M\over 3R} + {M\over 2R^3}(r^2 - {3\over 5} \delta D) + {\omega^2\over 2} r^2 - {\omega^2\over 6} D $$
This has the schematic form of spherical plus quadrupole (including the centrifugal force inside F and G)
$$ \phi_{int} = F(r) + G(r) D $$
The condition that the $\delta$ ellipsoid is an equipotential is found by replacing $r$ with $R - {\delta D\over 2R}$ inside F(r), and setting the D-part to zero:
$$ {F'(R) \delta \over 2R} = G(r) $$
In this case, you get the equation below, which reduces to the previous case when $M'=0$:
$$ {M'\over M+M'}\delta + {M\over M+M'} (\delta - {3\over 5} \delta) = - {C\over 3 g } $$
where $C=\omega^2 R$ is the centrifugal force, and $ g= {M+M'\over R^2} $ is the gravitational force at the surface. I should point out that the spherical part of the centrifugal potential ${\omega^2\over 2} r^2$ always contributes a subleading term proportional to $\omega^2\delta$ to the equation and should be dropped. The result is
$$ {3\over 2} \delta = {1\over 2 (1 - {3\over 5} {M\over M+M'}) } {C\over g} $$
So that if you choose M' to be .2 M, you get the correct answer, so that the extra equatorial radius is twice the naive amount of ${C\over 2g}$.
This says that the potential at the surface of the Earth is only modified from the uniform ellipsoid estimate by adding a sphere with 20% of the total mass at the center. This is somewhat small, considering the nonuniform density in the interior contains about 25% of the mass of the Earth (the perturbing mass is twice the density at half the radius, so about 25% of the total). The slight difference is due to the ellipticity of the core.
Nonuniform mass density II: exact solution
The main thing neglected in the above is that the center is also nonspherical, and so adds to the nonspherical D part of the potential on the surface. This effect mostly counteracts the general tendency of extra mass at the center to make the surface more spherical, although imperfectly, so that there is a correction left over.
You can consider it as a superposition of uniform ellipsoids of mean radius s, with ellipticity parameter $\delta(s)$ for $0<s<R$ increasing as you go toward the center. Each is uniform on the interior, with mass density $|\rho'(s)|$ where $\rho(s)$ is the extra density of the Earth at distance s from the center, so that $\rho(R)=0$. These ellipsoids are superposed on top of a uniform density ellipsoid of density $\rho_0$ equal to the surface density of the Earth's crust:
I will consider $\rho(s)$ and $\rho_0$ known, so that I also know $|\rho'(s)|$, it's (negative) derivative with respect to s, which is the density of the ellipsoid you add at s, and I also know:
$$ M(r) = \int_0^r 4\pi \rho(s) s^2 ds $$
The quantity $M(s)$ is ${1\over 4\pi}$ times the additional mass in the interior, as compared to a uniform Earth at crust density. Note that $M(s)$ is not affected by the ellipsoidal shape to leading order, because all the nested ellipsoids are quadrupole perturbations, and so contain the same volume as spheres.
Each of these concentric ellipsoids is itself an equipotential surface for the centrifugal potential plus the potential from the interior and exterior ellipsoids. So once you know the form of the potential of all these superposed ellipsoids, which is of the form of spherical + quadrupole + centrifugal quadrupole (the centrifugal spherical part always gives a subleading correction, so I omit it):
$$ \phi_\mathrm{int}(r) = F(r) + G(r) D + {\omega^2 \over 6} D $$
You know that each of these nested ellipsoids is an equipotential
$$ F(s - {\delta(s) \over 2s}) D + G(s) D - {\omega^2\over 6} D $$
so that the equation demanding that this is an equipotential at any s is
$$ {\delta(s) F'(s) \over 2s} - G(s) + {\omega^2\over 6} = 0 $$
To find the form of F and G, you first express the interior/exterior solution for a uniform ellipsoid in terms of the density $\rho$ and the radius R:
$$ {\phi_\mathrm{int}\over 4\pi} = - {\rho R^2\over 2} + {\rho\over 6} r^2 + {\rho \delta\over 10} D $$
$$ {\phi_\mathrm{ext}\over 4\pi} = - {\rho R^3 \over 3 r} + {\rho\delta R^5\over 10 r^5} D $$
You can check the sign and numerical value of the coefficients using the 3/5 rule for the interior equipotential ellipsoids, the separate matching of the spherical and D perturbations at r=R, and dimensional analysis. I put a factor $4\pi$ on the bottom of $\phi$ so that the right hand side solves the constant free form of Laplace's equation.
Now you can superpose all the ellipsoids, by setting $\delta$ on each ellipsoid to be $\delta(s)$, setting $\rho$ on each ellipsoid to be $|\rho'(s)|$, and $R$ to be $s$. I am only going to give the interior solution at r (doing integration by parts on the spherical part, where you know the answer is going to turn out to be, and throwing away some additive constant C) is:
$$ {\phi_\mathrm{int}(r)\over 4\pi} - C = {\rho_0\over 6} r^2 + {\rho_0 \delta(R)\over 10} D - {M(r)\over 4\pi r} + {1\over 10r^5} \int_0^r |\rho'(s)| \delta(s) s^5 ds D + {1\over 10} \int_r^R |\rho'(s)|\delta(s) D $$
The first two terms are the interior solution for constant density $\rho_0$. The third term is the total spherical contribution, which is just as in the spherical symmetric case. The fourth term is the the superposed exterior potential from the ellipsoids inside r, and the last term is the superposed interior potential from the ellipsoids outside r.
From this you can read off the spherical and quadrupole parts:$$ F(r) = {\rho_0\over 6} r^2 + {M(r)\over r} $$$$ G(r) = {\rho_0\delta(R)\over 10} + {1\over 10r^5} \int_0^r |\rho'(s) |\delta(s) s^5 ds + {1\over 10} \int_r^R |\rho'(s)|\delta(s) $$
So that the integral equation for $\delta(s)$ asserts that the $\delta(r)$ shape is an equipotential at any depth.
$$ {F'(r)\delta(r)\over 2r} - G(r) + {\omega^2 \over 6} = 0 $$
This equation can be solved numerically for any mass profile in the interior, to find the $\delta(R)$. This is difficult to do by hand, but you can get qualitative insight.
Consider an ellipsoidal perturbation inside a uniform density ellipsoid. If you let this mass settle along an equipotential, it will settle to the same ellipsoidal shape as the surface, because the interior solution for the uniform ellipsoid is quadratic, and so has exact nested ellipsoids of the same shape as equipotentials. But this extra density will contribute less than it's share of elliptical potential to the surface, diminishing as the third power of the ratio of the radius of the Earth to the radius of the perturbation. But it will produce stronger ellipses inside, so that the interior is always more elliptical than the surface.
Oblate Core Model
The exact solution is too difficult for paper and pencil calculations, but looking [here]( http://www.google.com/imgres?hl=en&client=ubuntu&hs=dhf&sa=X&channel=fs&tbm=isch&prmd=imvns&tbnid=hjMCgNhAjHnRiM:&imgrefurl=http://www.springerimages.com/Images/Geosciences/1-10.1007_978-90-481-8702-7_100-1&docid=ijMBfCAOC1GhEM&imgurl=http://img.springerimages.com/Images/SpringerBooks/BSE%253D5898/BOK%253D978-90-481-8702-7/PRT%253D5/MediaObjects/WATER_978-90-481-8702-7_5_Part_Fig1-100_HTML.jpg&w=300&h=228&ei=ZccgUJCTK8iH6QHEuoHICQ&zoom=1&iact=hc&vpx=210&vpy=153&dur=4872&hovh=182&hovw=240&tx=134&ty=82&sig=108672344460589538944&page=1&tbnh=129&tbnw=170&start=0&ndsp=8&ved=1t:429,r:1,s:0,i:79&biw=729&bih=483 ), you see that it is sensible to model the Earth as two concentric spheres of radius $R$ and $R_1$ with total mass $M$ and $M_1$ and $\delta$ and $\delta_1$.
I will take
$$ R_1 = {R\over 2} $$
and
$$ M_1 = {M\over 4} $$
that is, the inner sphere is 3000 km across, with twice the density, which is roughly accurate. Superposing the potentials and finding the equation for the $\delta$s (the two point truncation of the integral equation), you find
$$ -\delta + {3\over 5} {M_0\over M_0 + M_1} \delta + {3\over 5} {M_1\over M_0 + M_1} \delta_1 ({R_1\over R})^2 = {C\over 3g} $$
$$ {M_0 \over M_0 + M_1} (-\delta_1 + {3\over 5} \delta) + {M_1 \over M_0 + M_1}( -\delta_1 + {3\over 5} \delta_1) = {C\over 3g} $$
Where
$$ g = {M_0+ M_1\over R^2}$$$$ C = \omega^2 R $$
are the gravitational force and the centrifugal force per unit mass, as usual. Using the parameters, and defining $\epsilon = {3\delta\over 2}$ and $\epsilon_1={3\delta_1\over 2}$, one finds:
$$ - 1.04 \epsilon + .06 \epsilon = {C\over g} $$$$ - 1.76 \epsilon_1 + .96 \epsilon = {C\over g} $$
(these are exact decimal fractions, there are denominators of 100 and 25). Subtracting the two equations gives:
$$ \epsilon_1 = {\epsilon\over .91} $$
(still exact fractions) Which gives the equation
$$ (-1.04 + {.06\over .91} ) \epsilon = {C\over g}$$
So that the factor in front is $.974$, instead of the naive 2. This gives an equatorial diameter of 44.3 km, as opposed to 42.73, which is close enough that the model essentially explains everything you wanted to know.
The value of $\epsilon_1$ is also interesting, it tells you that the Earth's core is 9% more eccentric than the outer ellipsoid of the Earth itself. Given that the accuracy of the model is at the 3% level, this should be very accurate. |
The Convergence of Newton's Method
Suppose that $f$ is a twice differentiable function on an interval containing the root of interest, $\alpha$ and suppose that $f'(\alpha) \neq 0$. Now consider the first order Taylor polynomial of $f$ about $x_n$ denoted $P_1(x) = f(x_n) + (x - x_n)f'(x_n)$. By Taylor's Theorem, there exists a $\xi_n$ between $\alpha$ and $x_n$ such that the remainder/error term is given by $E_n(x) = \frac{(\alpha - x_n)^2 }{2!} f''(\xi_n)$. We know that $f(x) = P_1(x) + E_n(x)$ and so:(1)
Plugging in $x = \alpha$ and noting that $f(\alpha) = 0$ since $\alpha$ is a root of $f$ and we get that:(2)
Note that Newton's iteration formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ and so $\frac{f(x_n)}{f'(x_n)} = x_n - x_{n+1}$. We will apply this substitution after dividing the equation above by $f'(x_n)$:(3)
So we have obtained a formula for our error, however, we do not necessarily know the value of $\xi_n$ so we want to replace this. Suppose that $x_n$ is near $\alpha$. Then $f'(x_n) \approx f'(\alpha)$. Furthermore, since $\xi_n$ is between $x_n$ and $\alpha$, then we must have that $\xi_n$ is near $\alpha$ and so $f''(c_n) \approx f''(\alpha)$. We then define $M$ as follows:(4)
Substituting $M$ into our error formula from above and then multiplying both sides of the equation by $M$ and we get that:(5)
Using induction, we have that:(6)
Now if $\lim_{n \to \infty} (\alpha - x_n) = 0$ then $\lim_{n \to \infty} x_n = \alpha$. To ensure convergence, we will need that $\mid M(\alpha - x_0) \mid < 1$, that is choose an initial approximation $x_0$ such that:(7)
We will now construct an upper bound for $M$. We start with $M = \frac{1}{2} \biggr \rvert \frac{f'(\alpha)}{f''(\alpha)} \biggr \rvert$. If we maximize $\frac{1}{2} \biggr \rvert \frac{f'(\alpha)}{f''(\alpha)} \biggr \rvert = M^*$ on the interval $[a, b]$ then:(8)
Thus if we can guarantee $M^* < 1$, then Newton's Method will converge with the initial approximation $x_0$. |
So let's start by creating a contingency table from the data. The rows will be the counts of females and males, the columns the locations.
$$\begin{bmatrix} 44 & 86 & 110 \\ 56 & 114 & 90 \end{bmatrix} $$
From thus table, we can compute the sum of the rows and the sum of the columns. The expected number cell frequencies are
$$\begin{bmatrix} 48 & 96 & 96 \\ 52 & 104 & 104 \end{bmatrix} $$
How did I get this? I'll just show one cell frequency, but this can be applied to all the cells. Let's take a look at the first column total, which is 100. We take the column total, multiply it by the row total (which is 240) and then divide by the total number of number of observations (which is 500). All in all
$$ 100 \times \dfrac{240}{500} = 48$$
And so that is the expected number in the first cell. Since we observed 100 individuals at site A, then we should expect 100-48 = 52 in the second cell in the first column. You can apply this logic to the remaining columns.
So now we have our expected cell frequencies. All we need to do now is compute the test statistic, which is
$$\sum \dfrac{(o_i - e_i)^2}{e^i} = \dfrac{(44-48)^2}{48} + \dfrac{(52-56)^2}{52} + \dots + \dfrac{(90-104)^2}{104} \approx 6.57 $$
We now need to compute the tail probability of this statistic from a chisquared distribution with $(R-1)\times(C-1) = 1 \times 2 = 2$ degrees of freedom. Here $R$ is the number of rows, and $C$ is the number of columns. You can look this up in a table.
The p value returned is 0.03743, so we can reject the null hypothesis at the $\alpha = 0.05$ level.
Here is some R code to verify we are correct in our calculations:
data = c(44,86,110, 56, 114, 90)
tbl = matrix(data, nrow = 2, byrow =T)
chisq.test(tbl, correct = F)
>> Pearson's Chi-squared test
data: tbl
X-squared = 6.5705, df = 2, p-value = 0.03743 |
Hints will display for most wrong answers; explanations for most right answers. You can attempt a question multiple times; it will only be scored correct if you get it right the first time.
I used the official objectives and sample test to construct these questions, but cannot promise that they accurately reflect what’s on the real test. Some of the sample questions were more convoluted than I could bear to write. See terms of use. See the MTEL Practice Test main page to view random questions on a variety of topics or to download paper practice tests.
MTEL General Curriculum Mathematics Practice
Question 1
The following story situations model \( 12\div 3\): I) Jack has 12 cookies, which he wants to share equally between himself and two friends. How many cookies does each person get? II) Trent has 12 cookies, which he wants to put into bags of 3 cookies each. How many bags can he make? III) Cicely has $12. Cookies cost $3 each. How many cookies can she buy? Which of these questions illustrate the same model of division, either partitive (partioning) or measurement (quotative)?
I and II I and III II and III
Hint:
Problem I is partitive (or partitioning or sharing) -- we put 12 objects into 3 groups. Problems II and III are quotative (or measurement) -- we put 12 objects in groups of 3.
All three problems model the same meaning of division
Question 2
Here is a student's work on several multiplication problems:
58 x 22
Hint:
This problem involves regrouping, which the student does not do correctly.
16 x 24
Hint:
This problem involves regrouping, which the student does not do correctly.
31 x 23
Hint:
There is no regrouping with this problem.
141 x 32
Hint:
This problem involves regrouping, which the student does not do correctly.
Question 3
Use the samples of a student's work below to answer the question that follows: This student divides fractions by first finding a common denominator, then dividing the numerators.\( \large \dfrac{2}{3} \div \dfrac{3}{4} \longrightarrow \dfrac{8}{12} \div \dfrac{9}{12} \longrightarrow 8 \div 9 = \dfrac {8}{9}\) \( \large \dfrac{2}{5} \div \dfrac{7}{20} \longrightarrow \dfrac{8}{20} \div \dfrac{7}{20} \longrightarrow 8 \div 7 = \dfrac {8}{7}\) \( \large \dfrac{7}{6} \div \dfrac{3}{4} \longrightarrow \dfrac{14}{12} \div \dfrac{9}{12} \longrightarrow 14 \div 9 = \dfrac {14}{9}\) Which of the following best describes the mathematical validity of the algorithm the student is using?
It is not valid. Common denominators are for adding and subtracting fractions, not for dividing them.
Hint:
Don't be so rigid! Usually there's more than one way to do something in math.
It got the right answer in these three cases, but it isn‘t valid for all rational numbers.
Hint:
Did you try some other examples? What makes you say it's not valid?
It is valid if the rational numbers in the division problem are in lowest terms and the divisor is not zero.
Hint:
Lowest terms doesn't affect this problem at all.
It is valid for all rational numbers, as long as the divisor is not zero.
Hint:
When we have common denominators, the problem is in the form a/b divided by c/b, and the answer is a/c, as the student's algorithm predicts.
Question 4
Use the samples of a student's work below to answer the question that follows:\( \large \dfrac{2}{3}\times \dfrac{3}{4}=\dfrac{4\times 2}{3\times 3}=\dfrac{8}{9}\) \( \large \dfrac{2}{5}\times \dfrac{7}{7}=\dfrac{7\times 2}{5\times 7}=\dfrac{2}{5}\) \( \large \dfrac{7}{6}\times \dfrac{3}{4}=\dfrac{4\times 7}{6\times 3}=\dfrac{28}{18}=\dfrac{14}{9}\) Which of the following best describes the mathematical validity of the algorithm the student is using?
It is not valid. It never produces the correct answer.
Hint:
In the middle example,the answer is correct.
It is not valid. It produces the correct answer in a few special cases, but it‘s still not a valid algorithm.
Hint:
Note that this algorithm gives a/b divided by c/d, not a/b x c/d, but some students confuse multiplication and cross-multiplication. If a=0 or if c/d =1, division and multiplication give the same answer.
It is valid if the rational numbers in the multiplication problem are in lowest terms.
Hint:
Lowest terms is irrelevant.
It is valid for all rational numbers.
Hint:
Can't be correct as the first and last examples have the wrong answers.
Question 5
Which of the following values of x satisfies the inequality \( \large \left| {{(x+2)}^{3}} \right|<3?\)
\( \large x=-3\)
Hint:
\( \left| {{(-3+2)}^{3}} \right|\)=\( \left | {(-1)}^3 \right | \)=\( \left | -1 \right |=1 \) .
\( \large x=0\)
Hint:
\( \left| {{(0+2)}^{3}} \right|\)=\( \left | {2}^3 \right | \)=\( \left | 8 \right | \) =\( 8\)
\( \large x=-4\)
Hint:
\( \left| {{(-4+2)}^{3}} \right|\)=\( \left | {(-2)}^3 \right | \)=\( \left | -8 \right | \) =\( 8\)
\( \large x=1\)
Hint:
\( \left| {{(1+2)}^{3}} \right|\)=\( \left | {3}^3 \right | \)=\( \left | 27 \right | \) = \(27\)
Question 6
Below is a pictorial representation of \(2\dfrac{1}{2}\div \dfrac{2}{3}\): Which of the following is the best description of how to find the quotient from the picture?
The quotient is \(3\dfrac{3}{4}\). There are 3 whole blocks each representing \(\dfrac{2}{3}\) and a partial block composed of 3 small rectangles. The 3 small rectangles represent \(\dfrac{3}{4}\) of \(\dfrac{2}{3}\). The quotient is \(3\dfrac{1}{2}\). There are 3 whole blocks each representing \(\dfrac{2}{3}\) and a partial block composed of 3 small rectangles. The 3 small rectangles represent \(\dfrac{3}{6}\) of a whole, or \(\dfrac{1}{2}\).
Hint:
We are counting how many 2/3's are in
2 1/2: the unit becomes 2/3, not 1.
The quotient is \(\dfrac{4}{15}\). There are four whole blocks separated into a total of 15 small rectangles.
Hint:
This explanation doesn't make much sense. Probably you are doing "invert and multiply," but inverting the wrong thing.
This picture cannot be used to find the quotient because it does not show how to separate \(2\dfrac{1}{2}\) into equal sized groups.
Hint:
Study the measurement/quotative model of division. It's often very useful with fractions.
Question 7
Which of the following inequalities describes all values of x with \(\large \dfrac{x}{2}\le \dfrac{x}{3}\)?
\( \large x < 0\)
Hint:
If x =0, then x/2 = x/3, so this answer can't be correct.
\( \large x \le 0\)
\( \large x > 0\)
Hint:
If x =0, then x/2 = x/3, so this answer can't be correct.
\( \large x \ge 0\)
Hint:
Try plugging in x = 6.
Question 8
The expression \( \large {{7}^{-4}}\cdot {{8}^{-6}}\) is equal to which of the following?
\( \large \dfrac{8}{{{\left( 56 \right)}^{4}}}\)
Hint:
The bases are whole numbers, and the exponents are negative. How can the numerator be 8?
\( \large \dfrac{64}{{{\left( 56 \right)}^{4}}}\)
Hint:
The bases are whole numbers, and the exponents are negative. How can the numerator be 64?
\( \large \dfrac{1}{8\cdot {{\left( 56 \right)}^{4}}}\)
Hint:
\(8^{-6}=8^{-4} \times 8^{-2}\)
\( \large \dfrac{1}{64\cdot {{\left( 56 \right)}^{4}}}\)
Question 9
The expression \( \large{{8}^{3}}\cdot {{2}^{-10}}\) is equal to which of the following?
\( \large 2\)
Hint:
Write \(8^3\) as a power of 2.
\( \large \dfrac{1}{2}\)
Hint:
\(8^3 \cdot {2}^{-10}={(2^3)}^3 \cdot {2}^{-10}\) =\(2^9 \cdot {2}^{-10} =2^{-1}\)
\( \large 16\)
Hint:
Write \(8^3\) as a power of 2.
\( \large \dfrac{1}{16}\)
Hint:
Write \(8^3\) as a power of 2.
Question 10
The student used a method that worked for this problem and can be generalized to any subtraction problem.
Hint:
Note that this algorithm is taught as the "standard" algorithm in much of Europe (it's where the term "borrowing" came from -- you borrow on top and "pay back" on the bottom).
The student used a method that worked for this problem and that will work for any subtraction problem that only requires one regrouping; it will not work if more regrouping is required.
Hint:
Try some more examples.
The student used a method that worked for this problem and will work for all three-digit subtraction problems, but will not work for larger problems.
Hint:
Try some more examples.
The student used a method that does not work. The student made two mistakes that cancelled each other out and was lucky to get the right answer for this problem.
Hint:
Remember, there are many ways to do subtraction; there is no one "right" algorithm.
Question 11
Which of the following is equivalent to \( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{2}{8}\times \dfrac{1}{2}?\)
\( \large \dfrac{7}{16}\)
Hint:
Multiplication comes before addition and subtraction in the order of operations.
\( \large \dfrac{1}{2}\)
Hint:
Addition and subtraction are of equal priority in the order of operations -- do them left to right.
\( \large \dfrac{3}{4}\)
Hint:
\( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{2}{8}\times \dfrac{1}{2}\)=\( \dfrac{3}{4}-\dfrac{1}{8}+\dfrac{1}{8}\)=\( \dfrac{3}{4}+-\dfrac{1}{8}+\dfrac{1}{8}\)=\( \dfrac{3}{4}\)
\( \large \dfrac{3}{16}\)
Hint:
Multiplication comes before addition and subtraction in the order of operations.
Question 12
On a map the distance from Boston to Detroit is 6 cm, and these two cities are 702 miles away from each other. Assuming the scale of the map is the same throughout, which answer below is closest to the distance between Boston and San Francisco on the map, given that they are 2,708 miles away from each other?
21 cm
Hint:
How many miles would correspond to 24 cm on the map? Try adjusting from there.
22 cm
Hint:
How many miles would correspond to 24 cm on the map? Try adjusting from there.
23 cm
Hint:
One way to solve this without a calculator is to note that 4 groups of 6 cm is 2808 miles, which is 100 miles too much. Then 100 miles would be about 1/7 th of 6 cm, or about 1 cm less than 24 cm.
24 cm
Hint:
4 groups of 6 cm is over 2800 miles on the map, which is too much.
Question 13
A solution requires 4 ml of saline for every 7 ml of medicine. How much saline would be required for 50 ml of medicine?
\( \large 28 \dfrac{4}{7}\) ml
Hint:
49 ml of medicine requires 28 ml of saline. The extra ml of saline requires 4 ml saline/ 7 ml medicine = 4/7 ml saline per 1 ml medicine.
\( \large 28 \dfrac{1}{4}\) ml
Hint:
49 ml of medicine requires 28 ml of saline. How much saline does the extra ml require?
\( \large 28 \dfrac{1}{7}\) ml
Hint:
49 ml of medicine requires 28 ml of saline. How much saline does the extra ml require?
\( \large 87.5\) ml
Hint:
49 ml of medicine requires 28 ml of saline. How much saline does the extra ml require?
Question 14
A
Hint:
\(\frac{34}{135} \approx \frac{1}{4}\) and \( \frac{53}{86} \approx \frac {2}{3}\). \(\frac {1}{4}\) of \(\frac {2}{3}\) is small and closest to A.
B
Hint:
Estimate with simpler fractions.
C
Hint:
Estimate with simpler fractions.
D
Hint:
Estimate with simpler fractions.
Question 15
Here is a mental math strategy for computing 26 x 16: Step 1: 100 x 16 = 1600 Step 2: 25 x 16 = 1600 ÷· 4 = 400 Step 3: 26 x 16 = 400 + 16 = 416 Which property best justifies Step 3 in this strategy?
Commutative Property.
Hint:
For addition, the commutative property is \(a+b=b+a\) and for multiplication it's \( a \times b = b \times a\).
Associative Property.
Hint:
For addition, the associative property is \((a+b)+c=a+(b+c)\) and for multiplication it's \((a \times b) \times c=a \times (b \times c)\)
Identity Property.
Hint:
0 is the additive identity, because \( a+0=a\) and 1 is the multiplicative identity because \(a \times 1=a\). The phrase "identity property" is not standard.
Distributive Property.
Hint:
\( (25+1) \times 16 = 25 \times 16 + 1 \times 16 \). This is an example of the distributive property of multiplication over addition.
If you found a mistake or have comments on a particular question, please contact me (please copy and paste at least part of the question into the form, as the numbers change depending on how quizzes are displayed). General comments can be left here. |
Bernoulli Bernoulli Volume 9, Number 6 (2003), 1003-1049. A quantization algorithm for solving multidimensional discrete-time optimal stopping problems Abstract
A new grid method for computing the Snell envelope of a function of an $\mathbb{R}^d$-valued simulatable Markov chain $(X_k)_{0\lambda \leq k\lambda \leq n}$ is proposed. (This is a typical nonlinear problem that cannot be solved by the standard Monte Carlo method.) Every $X_k$ is replaced by a `quantized approximation' $\widehat{X}_k$ taking its values in a grid $\Gamma_k$ of size $N_k$. The $n$ grids and their trans\-ition probability matrices form a discrete tree on which a pseudo-Snell envelope is devised by mimicking the regular dynamic programming formula. Using the quantization theory of random vectors, we show the existence of a set of optimal grids, given the total number $N$ of elementary $\mathbb{R}^d$-valued quantizers. A recursive stochastic gradient algorithm, based on simulations of $(X_k)_{0\lambda \leq k \lambda \leq n}$, yields these optimal grids and their transition probability matrices. Some a priori error estimates based on the $L^p$-quantization errors $\|X_k-\widehat X_k\|_{_p}$ are established. These results are applied to the computation of the Snell envelope of a diffusion approximated by its (Gaussian) Euler scheme. We apply these result to provide a discretization scheme for reflected backward stochastic differential equations. Finally, a numerical experiment is carried out on a two-dimensional American option pricing problem.
Article information Source Bernoulli, Volume 9, Number 6 (2003), 1003-1049. Dates First available in Project Euclid: 23 December 2003 Permanent link to this document https://projecteuclid.org/euclid.bj/1072215199 Digital Object Identifier doi:10.3150/bj/1072215199 Mathematical Reviews number (MathSciNet) MR2046816 Zentralblatt MATH identifier 1042.60021 Citation
Bally, Vlad; Pagès, Gilles. A quantization algorithm for solving multidimensional discrete-time optimal stopping problems. Bernoulli 9 (2003), no. 6, 1003--1049. doi:10.3150/bj/1072215199. https://projecteuclid.org/euclid.bj/1072215199 |
I'm working through Stancil and Prabhakar's 'Spin Waves', and am stuck with a vector identity which I am not sure how the authors have justified.
On page 34, we adopt the use of a scalar potential $\phi$, and a vector potential, $\vec{A}$. Then we use these to recast the electric and magnetic field in terms of the Coulomb gauge:
$$ \tag{1} B = \nabla \times \vec{A} $$ $$ \tag{2} E = - \frac{\partial \vec{A}}{\partial t} - \nabla \phi$$
Next we attempt to rewrite the Lorentz force in terms of the scalar and vector potentials. This looks like:
$$ \tag{3} m \frac{d \vec{v}}{dt} = q(-\frac{\partial \vec{A}}{dt} - \nabla \phi + \vec{v} \times (\nabla \times \vec{A})) $$
Here, we see a particular term, $\vec{v} \times (\nabla \times \vec{A})$.
If we use equation 3, and then use the following expression of $\frac{d A_x}{dt}$:
$$ \tag{4} \frac{d A_x}{dt} = \frac{\partial A_x}{\partial t} + (v_x \frac{\partial A_x}{\partial x} + v_y \frac{\partial A_x}{\partial y} + v_z \frac{\partial A_x}{\partial z}) $$
If we then take the x-component of $\vec{v} \times (\nabla \times \vec{A})$, we get:
$$ \tag{5} (\vec{v} \times (\nabla \times \vec{A}))_x = v_y (\frac{d A_y}{dx} - \frac{d A_x}{dy}) - v_z(\frac{d A_x}{dz} - \frac{d A_z}{dx}) $$
The next line is where my issue arises. The book states that then, this follows:
$$ \tag{6} (\vec{v} \times (\nabla \times \vec{A}))_x = \partial_x (\vec{v} \cdot \vec{A}) - \frac{d A_x}{dt} + \frac{\partial A_x}{\partial t} $$
However, working backwards from the RHS of equation 6's first term, I find that:
$$ \tag{7} \partial_x (\vec{v} \cdot \vec{A}) = (A_x \frac{\partial v_x}{\partial x} + A_y \frac{\partial v_y}{\partial x} + A_z \frac{\partial v_z}{\partial x}) + (v_x \frac{\partial A_x}{\partial x} + v_y \frac{\partial A_y}{\partial x} + v_z \frac{\partial A_z}{\partial x}) $$
And, rearranging equation 4:
$$ \tag{8} \frac{d A_x}{dt} - \frac{\partial A_x}{\partial t} = (v_x \frac{\partial A_x}{\partial x} + v_y \frac{\partial A_x}{\partial y} + v_z \frac{\partial A_x}{\partial z}) $$
Then, equation 6 becomes:
$$ \tag{9} \partial_x (\vec{v} \cdot \vec{A}) - \frac{d A_x}{dt} + \frac{\partial A_x}{\partial t} = (A_x \frac{\partial v_x}{\partial x} + A_y \frac{\partial v_y}{\partial x} + A_z \frac{\partial v_z}{\partial x})\\ + v_y \frac{\partial A_y}{\partial x} + v_z \frac{\partial A_z}{\partial x} - v_y \frac{\partial A_x}{\partial y} - v_z \frac{\partial A_x}{\partial z} $$
Which is the original statement made in equation 5, plus the bracketed terms $(A_x \frac{\partial v_x}{\partial x} + A_y \frac{\partial v_x}{\partial y} + A_z \frac{\partial v_x}{\partial y})$.
This means that, for equation 6 to be true, it would be necessary to state that
$$ \tag{10} A_x \frac{\partial v_x}{\partial x} + A_y \frac{\partial v_x}{\partial y} + A_z \frac{\partial v_x}{\partial z} = 0 $$
Is this justified? And if so, how? I'm not sure how I could prove this. The only way I can see this being possible, is by requiring $\vec{v}$ to be unchanging, but this is not an assumed detail in the textbook. |
Take a simple random walk $\gamma$ in the complex plane conditioned to start at point $a$ and end at point $b$. For this random walk, we can define the winding number $W_\gamma(a,b)$ around $b$ in the usual way for complex curves.
If instead we have a 2D Brownian motion $Z=X+iY$, then this definition becomes more complicated. For example, if we have a Brownian motion starting at the origin, we can talk about the winding number $\theta_t$ at time $t$ around the origin by solving the stochastic differential equation
$d\theta_t=\frac{X_tdY_t-Y_tdX_t}{|Z_t|^2}$
with initial condition $\theta_0=0$.
The issue is that for Brownian motion, we cannot condition on the path $Z$ to hit a particular point, because this has probability zero. Moreover, by considering annuli around $b$ and the fact that planar Brownian motion moves between concentric annuli with positive probability, it seems to me the situation becomes rather singular.
Question:Is there a sensible generalization for the winding number of a Brownian motion conditioned to hit a single point?
For example, can we look at the limit of the winding number around an annulus about point $b$ whose radius shrinks to zero? I would imagine we would require the Brownian motion to be conditioned to hit some region of positive area just outside the shrinking annulus. |
Fixed Points
So far we have looked at the Bisection Method and Newton's Method for approximating roots of functions. We are about to introduce another root finding method know as the Fixed Point Method, but before we do so, we will need to learn about special types of points on functions known as fixed points which we define below.
Definition: The value $\alpha$ is called a Fixed Point of the function $g$ if $\alpha$ if $\alpha = g(\alpha)$, that is $g$ maps $\alpha$ back to itself. The equation $x_{n+1} = g(x_n)$ is called the Fixed Point Iteration.
The following very simple lemma gives us a way to determine the fixed points of a function.
Lemma 1: A value $x$ is a fixed point of the function $g$ if and only if $y = g(x)$ intersects the diagonal line $y = x$. Proof:$\Rightarrow$ Suppose that $x$ is a fixed point of the function $g$. Then $x = g(x)$. But $g(x) = y$, so $x = y$, and so $(x, g(x))$ lies on the line $y = x$. $\Leftarrow$ Suppose that $y = g(x)$ intersects the diagonal line $y = x$. Then since $y = x$, we have that $x = g(x)$ so $x$ is a fixed point of $g$. $\blacksquare$
Thus far we have not even mentioned whether a fixed point to a function is guaranteed to exist. Theorem 1 below gives us a condition that guarantees the existence fixed points while Theorem 2 gives us a condition (though not necessary) that guarantees the uniqueness of a fixed point on an interval.
Theorem 1 (Condition for The Existence of Fixed Points): If $g$ is a continuous function on the interval $[a, b]$ and if for all $x \in [a, b]$ we have that $g(x) \in [a, b]$ then there exists at least one fixed point in $[a, b]$. Proof:Let $g$ be a continuous function on $[a, b]$ such that if $x \in [a, b]$ then $g(x) \in [a, b]$. Consider the following two cases. Case 1:If $a = g(a)$ and/or $b = g(b)$, the $g$ has a fixed point on at least one of endpoints of the interval $[a, b]$ and we are done. Case 2:Suppose that $a \neq g(a)$ and $b \neq g(b)$. Then $g$ does not have a fixed point on either of the endpoints of $[a, b]$. Therefore we have that $g(a) > a$ (since $a < g(x) < b$) and $g(b) < b$ (since $a < g(x) < b$. Define the function $h$ as: We note that $h$ is a continuous function on $[a, b]$ since $y = g(x)$ and $y = x$ are both continuous on $[a, b]$. Now $h(a) = g(a) - x > 0$ and $h(b) = g(b) - b < 0$. By the Intermediate Value Theorem, there exists a root $\alpha \in (a, b)$ such that $h(\alpha) = 0$, and so $h(\alpha) = 0$ implies that $g(\alpha) - \alpha = 0$, thus $\alpha = g(\alpha)$ and so $\alpha$ is a fixed point of $g$. $\blacksquare$
Theorem 2 (Condition for The Uniqueness of a Fixed Point): If $g$ is a continuous function on the interval $[a, b]$ and if for all $x \in [a, b]$ we have that $g(x) \in [a, b]$. Furthermore, if $g$ is differentiable of $(a, b)$ and there exists a number $k \in \mathbb{R}$ such that $0 < k < 1$ such that for all $x \in (a, b)$ we have that $\mid g'(x) \mid ≤ k$ , then the fixed point $\alpha \in [a, b]$ is unique. Proof:We are already guaranteed a fixed point $\alpha \in [a, b]$ from Theorem 1. Suppose that there exists a $k \in \mathbb{R}$ such that $0 < k < 1$ and such that $\mid g'(x) ≤ k < 1$. Suppose that $\alpha$ and $\beta$ are both fixed points in $[a,b]$ such that $\alpha \neq \beta$. By the Mean Value Theorem, there exists a real number $\xi$ between $\alpha$ and $\beta$ (and therefore $\xi \in (a, b)$) such that: Since $\xi \in (a, b)$ we have that $\mid g'(\xi) \mid ≤ 1$. Furthermore, since $\alpha$ and $\beta$ are fixed points, we have that $\mid \alpha - \beta \mid = \mid g(\alpha) - g(\beta)$ and so: But this is a contradiction which comes from assuming that $\alpha \neq \beta$. Thus, the fixed point $\alpha$ is unique. $\blacksquare$ Example 1 Show that the function $f(x) = x^2$ has two fixed points. Show that $\mid f'(x) \mid ≥ 1$ for some $x \in (-1, 2)$
The function $f(x) = x^2$ intersects the line $y = x$ when $x = 0$ and $x = 1$ since $f(0) = 0^2 = 0$ and $f(1) = 1^2 = 1$. Therefore $x = 0$ and $x = 1$ are the two fixed points of $g$.
Now $f'(x) = 2x$. This function is strictly increasing on $[-1, 2]$, and $f(-1) = -2$ and $f(2) = 4$ are the absolute maximum and absolute minimum (respectively) on $[-1, 2]$. Clearly there exists $x \in (-1, 2)$ such that $\mid f'(x) \mid ≥ 1$, take $x = -0.5$ for example. Then $\mid f'(-0.5) \mid = \mid 2(-0.5) \mid = 1 ≥ 1$. |
Conditioning shows up everywhere in probability theory and statistics. It is, perhaps, one of the most powerful tools from probability theory, since in science we're often interested in how two things are
related. As a simple example, linear regression is, in essence, a question of a conditional expectation.
Typically, when we see conditioning, we condition one random variable on different random variable. That is, we might have two random variables \(X\) and \(Y\), which really constitute a single object
1 \((X, Y)\) with a single distribution function \(F_{(X, Y)}(x, y)\). Technically, the distribution function \(F_{(X, Y)}\) tells us everything we need to know. We might then be interested in what values \(Y\) takes, given we see particular values of \(X\). That is, we're interested in the conditional distribution of a new object, \(Y | X = x.\) We can handle this by defining a conditional distribution function \(F_{Y | X}\) using the usual definitions from probability theory, \[ F_{Y | X} (y | x) = P(Y \leq y | X = x) = \frac{P(Y \leq y, X = x)}{P(X = x)}.\] You might be annoyed by the equality \(X = x\), since for a continuous random variable \(X\) the probability of this occurrence must be zero. But since we have the equality in both the numerator and the denominator, 'there is hope,' to quote two of my Greek professors. You can give this definition meaning by applying the Mean Value Theorem 2.
All of this to discuss something else. Conditioning a random variable on itself. This came up in two unrelated places: boxcar kernel smoothing and mean residual time. The basic idea is simple: we know something is true about a random variable. For example, it took on a value in some range. But that's
all we know. What can we now say about the distribution of the random variable?
To be more concrete, suppose we have a continuous random variable \(X\) whose range is the entire real line. We know that the random variable lies between zero and four, but
nothing else. What can we say about the distribution of the random variable now?
Intuitively
3, we could think of this as collecting a bunch of random draws \(X_{1}, \ldots, X_{n}\), looking at those that fall in the interval \((0, 4)\) (this is the conditioning 4 part), and then asking how those draws are distributed. Taking the limit to infinitely many draws, we recover what we're seeking.
We can write down the density function without much work. We're interested in the distribution of \(X | X \in (0, 4)\). Pretending like we just have two random variables, we write down \[f(x | X \in (0, 4)) = \frac{f(x, X \in (0, 4))}{P(X \in (0, 4))} = \frac{f(x) 1_{(0,4)}(x)}{\int_{(0, 4)} f(x) \, dx} = c f(x) 1_{(0,4)}(x)\] where \(1_{(0, 4)}(x)\) is the handy indicator function, taking the value 1 when \(x \in (0, 4)\) and 0 otherwise. Deconstructing this, it makes perfect sense. We're certainly restricted to the interval \((0, 4)\), which the indicator function takes care up, but we also have to integrate to 1, which the normalizing constant \(c = \frac{1}{\int_{(0, 4)} f(x) \, dx}\) takes care of. Basically, the density in the region, conditioned on the fact that we're in the region, is the same shape as the unconditional probability of being in the region, but scaled up.
To be super concrete, suppose \(X \sim N(0, 1)\). Then we can both estimate the conditional density (using histograms) and compute it analytically. In which case we get a figure like this:
Blue is the original density, red is the conditional density, and the histograms are, well, self-explanatory.
This probably seems face-palm obvious, but for some reason I felt the need to work it out.
This single object is called, unimaginatively, a random vector. I'm not sure why this vector idea isn't introduced earlier on in intermediate probability theory courses. One possible reason: showing that a random vector is a reasonable probabilistic object involves hairy measure theoretic arguments about showing that the Borel set of a product space is the product space of the Borel sets. If that doesn't make any sense, don't worry about it.↩
One of the nice things about the frequency interpretation of probability theory is that it does lend itself to intuition in these sorts of situations.↩
When I first learned probability theory, I frequently confused
conditioningand and-ing(for lack of a better term). That is, the difference between the probability that this andthat occur, versus the probability that this occurs giventhat occurs. For the first, you count up all of the times that both things occur. For the second, you first see how many times the second thing occurs, and of those count up how many times the first thing occurs. These are obvious when you write down their definitions. But the intuition takes some work.↩ |
Higher Order O.D.E.'s Complex Roots of The Characteristic Equation Examples 1
Consider the following $n^{\mathrm{th}}$ order linear homogenous differential equation:(1)
Recall from the Higher Order Homogenous Differential Equations - Complex Roots of The Characteristic Equation page that sometimes the characteristic equation $a_0r^n + a_1r^{n-1} + ... + a_{n-1}r + a_{n-1}r + a_n = 0$ does not solely have distinct real roots. Instead, there many be some roots of the characteristic equation that are complex.
We note that we can only have an even number of complex roots (since complex roots come in conjugate pairs). Recall that for each pair of complex roots we have that they can be written in the form $\lambda + \mu i$ where $\lambda, \mu \in \mathbb{R}$. If these complex roots are distinct, then we will have that $e^{\lambda} (\cos (\mu t) + \sin (\mu t))$ will be a part of our general solution.
We will now look at some examples of higher order linear homogenous differential equations of this form.
Example 1 Find the general solution to the differential equation $\frac{d^3y}{dt^3} + 2 \frac{d^2y}{dt^2} + x +2$.
The characteristic equation for this differential equation is $r^3 + 2r^2 + r + 2 = 0$. By trial and error, we can immediately see that $r_1 = -2$ is a solution to this differential equation, and by applying long division we have that:(2)
Note that the factor $r^2 + 1$ is irreducible as factors of real numbers. If we apply the quadratic formula, we'll see that this factor has two complex roots, namely $r_2 = i$ and $r_3 = -i$. For this pair of complex roots we have that $\lambda = 0$ and $\mu = 1$, and so the general solution to our differential equation is:(3)
Example 2 Find the general solution of the differential equation $\frac{d^3y}{dt^3} + 5 \frac{d^2y}{dt^2} + 17 \frac{dy}{dt} +13 = 0$.
The characteristic equation for this differential equation is $r^3 + 5r^2 + 17r + 13 = 0$. By trial and error we see that $r_1 = -1$ is a root to the characteristic equation, and upon factoring this root out by long division we get that:(4)
We will now use the quadratic formula on the term $r^2 +4r + 13$:(5)
Therefore we have that $r_2 = -2 + 3i$ and $r_3 = -2 - 3i$ are both roots to our characteristic equation. So $\lambda = -2$ and $\mu = 3$. Thus we have that the general solution to our differential equation is(6) |
We start this section by introducing an important number theoretic function. We proceed in defining some convenient symbols that will be used in connection with the growth and behavior of some functions that will be defined in later chapters.
The Function \([x]\)
The function \([x]\) represents the largest integer not exceeding \(x\). In other words, for real \(x\), \([x]\) is the unique integer such that
\[x-1<[x]\leq x<[x]+1.\]
We also define \(((x))\) to be the fractional part of \(x\). In other words \(((x))=x-[x]\).
Note
We now list some properties of \([x]\) that will be used in later or in more advanced courses in number theory.
\([x+n]=[x]+n\), if \(n\) is an integer. \([x]+[y]\leq [x+y]\). \([x]+[-x]\) is 0 if \(x\) is an integer and -1 otherwise. The number of integers \(m\) for which \(x<m\leq y\) is \([y]-[x]\). The number of multiples of \(m\) which do not exceed \(x\) is \([x/m]\).
Using the definition of \([x]\), it will be easy to see that the above properties are direct consequences of the definition.
We now define some symbols that will be used to estimate the growth of number theoretic functions. These symbols will be not be really appreciated in the context of this book but these are often used in many analytic proofs.
The "O" and "o" Symbols
Let \(f(x)\) be a positive function and let \(g(x)\) be any function. Then \(O(f(x))\) (pronounced "big-oh" of \(f(x)\))denotes the collection of functions \(g(x)\) that exhibit a growth that is limited to that of \(f(x)\) in some respect. The traditional notation for stating that \(g(x)\) belongs to this collection is: \[g(x)=O(f(x)).\] This means that for sufficiently large \(x\),
\[\frac{\mid g(x)\mid }{|f(x)|}<M,\]
here \(M\) is some positive number.
\(\sin (x)=O(x)\), and also \(\sin(x)=O(1)\).
Now, the relation \(g(x)=o(f(x))\), pronounced "small-oh" of \(f(x)\), is used to indicate that \(f(x)\) grows much faster than \(g(x)\). It formally says that
\[\lim_{x\rightarrow \infty}\frac{g(x)}{f(x)}=0.\]
More generally, \(g(x)=o(f(x))\) at a point \(b\) if
\[\lim_{x\rightarrow b}\frac{g(x)}{f(x)}=0.\]
\(\sin(x)=o(x)\) at \(\infty\), and \(x^k=o(e^x)\) also at \(\infty\) for every constant \(k\).
The notation that \(f(x)\) is asymptotically equal to \(g(x)\) is denoted by \(\sim\). Formally speaking, we say that \(f(x) \sim g(x)\) if
\[\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=1.\]
\([x] \sim x\).
The purpose of introducing these symbols is to make complicated mathematical expressions simpler. Some expressions can be represented as the principal part that you need plus a remainder term. The remainder term can be expressed using the above notations. So when you need to combine several expressions, the remainder parts involving these symbols can be easily combined. We will state now some properties of the above symbols without proof. These properties are easy to prove using the definitions of the symbols.
\(O(O(f(x)))=O(f(x))\),
\(o(o(f(x)))=o(f(x))\).
\(O(f(x))\pm O(f(x))=O(f(x))\),
\(o(f(x)\pm o(f(x))=o(f(x))\),
\(O(f(x))\pm O(g(x))=O(\max(f(x), g(x)))\),
There are some other properties that we did not mention here, properties that are rarely used in number theoretic proofs.
Exercises
Prove the five properties of the \([x]\)
Prove the five properties of the \(O\) and \(o\) notations in Example 24. |
I'm trying to understand Petersen's Theorem as a corollary to Tutte's Theorem.
Corollary 5.4. Every 3-regular graph without cut edges has a perfect matching.
Proof. Let $G$ be a 3-regular graph without cut edges, and let $S$ be a proper subset of $V.$ Denote by $G_1, G_2, ..., G_n$ the odd components of $G - S$ and let $m_i$ be the number of edges with one end in $G_i$ and one end in $S, 1 \leq i \leq n.$ Since $G$ is 3-regular \begin{align} \sum_{v \in G_i} d(v) &= 3 |(G_i)|, \tag{5.8} \end{align} for $1 \leq i \leq n$ and \begin{align} \sum_{v \in S} d(v) &= 3 |S|. \tag{5.9} \end{align}
By equation (5.8), $m_i = \sum_{v \in G_i} d(v) - 2 \epsilon(G_i)$ is odd, where $\epsilon(G_i)$ is the number of edges in $G_i$. Now $m(i) \neq 1$ since $G$ has no cut edges. Thus $m_i \geq 3$ for $1 \leq i \leq n$. It follows that
\begin{align} o(G - S) = n \leq 1/3 \sum_{i=1}^n m_i \leq 1/3 \sum_{v \in S} d(v) = |S| \end{align}
Therefore by Tutte's Theorem, G has a perfect matching.
Question. I can see that $m_i$ as given by the expression $\sum_{v \in G_i} d(v) - 2 \epsilon(G_i)$ is odd whenever G_i is an odd component. (I can also see that $m_i$ is even whenever $G_i$ is even. I can prove both facts. But I can't see how the equation $m_i = \sum_{v \in G_i} d(v) - 2 \epsilon(G_i)$ was deduced from equation 5.8.
I know that $\sum d(v) = 2 \epsilon(G_i)$. That's theorem 1.1 in the book. (The sum of the degrees of a graph equals 2 times its number of edges. A very intuitive theorem.) But I can't connect this fact and equation 5.8 to give the expression for $m_i$. Any direction is appreciated. Thank you. |
Joyal and Tierney's 1984 monograph,
An extension of the Galois theory of Grothendieck, is an example of a substantial piece of mathematics written using informal reasoning in internal logic. The main result is the following:
Theorem. Every open surjection of toposes is an effective descent morphism. In particular, every Grothendieck topos is equivalent to the category of equivariant sheaves on a localic groupoid.
If you look at Chapter I, you will find that reads just like any other mathematical text, save for the avoidance of classical logic and certain kinds of set-theoretic operations. The main difficulty with using internal logic is the interpretation of the conclusions – this requires much care! For example, the proposition scheme$$(\forall b : B . \exists a : A . f(a) = b) \to (\forall h : B^T . \exists g : A^T . h = g \circ f)$$where $B$ is fixed but $f : A \to B$ and $T$ are allowed to vary, says that "if $f$ is surjective, then for any $h : T \to B$, there exists $g : T \to B$ such that $h = g \circ f$", or in short, "$B$ is a projective object"... but in the canonical semantics, it is neither necessary nor sufficient that $B$ be projective for the statement to hold! That is because what the formula actually means is the following,
If $f : A \to B$ is an epimorphism, then $f^T : A^T \to B^T$ is also an epimorphism.
whereas $B$ being projective is the statement below:
If $f : A \to B$ is an epimorphism, then $\mathrm{Hom}(T, f) : \mathrm{Hom}(T, A) \to \mathrm{Hom}(T, B)$ is a split surjection of sets.
If the topos in question has a projective terminal object, then the first statement (internal projectivity) implies the second, and if the topos is well-pointed, then the second statement implies the first.
So much for projective objects. What about finitely generated modules? Again there are subtleties, but the most straightforward way to formulate it is to take a mixed approach. Let $R$ be an internal ring. Then $M$ is a finitely-generated $R$-module if there exist global elements $m_1, \ldots, m_n$ (i.e. morphisms $1 \to M$) such that$$\forall m : M . \exists r_1 : R . \cdots . \exists r_n : R . m = r_1 m_1 + \cdots + r_n m_n$$holds in the internal logic. This amounts to saying that the evident homomorphism $R^{\oplus n} \to M$ is an epimorphism, which is what we want. It is tempting to formulate the whole statement internally, but this cannot work: at best one will obtain an internal characterisation of modules that are
locally finitely generated.
Perhaps I should give a positive example. I'm afraid I can't think of anything
interesting, so I'll opt for something simple instead. It is well-known that a two-sided unit element of a magma is unique if it exists. This is also true for internal magmas in any topos, and the proof is exactly the same (so long as it is formulated directly). More explicitly:
Let $M$ be a magma. Suppose $u$ is a left unit element in $M$ and $v$ is a right unit element in $M$. Then, $u = u v = v$.
Formally, we are deducing that$$\forall u : M. \forall v : M. (\forall m : M. u m = m) \land (\forall m : M. m v = m) \to (u = v)$$which means that, for all $u : S \to M$ and $v : T \to M$, if $\mu \circ (u \times \mathrm{id}_M) = \pi_2$ and $\mu \circ (\mathrm{id}_M \times v) = \pi_1$, then $u \circ \pi_1 = v \circ \pi_2$ as morphisms $S \times T \to M$.
Now, suppose we have an internal magma $M$ for which$$\exists u : M . \forall m : M. (u m = m) \land (m u = m)$$holds in the internal logic, i.e. there exists a morphism $u : T \to M$ satisfying the relevant equations, such that the unique morphism $T \to 1$ is an epimorphism. (The latter is the true content of the quantifier $\exists$.) We wish to show that $M$ has a global unit element, i.e. a morphism $e : 1 \to M$ satisfying the obvious equations. Applying the above result in the case $u = v$, we deduce that $u$ must factor through the coequaliser of $\pi_1, \pi_2 : T \times T \to T$. But this coequaliser computes the coimage of the unique morphism $T \to 1$, and we assumed $T \to 1$ is an epimorphism, so $T \to 1$ is itself the coequaliser of $\pi_1$ and $\pi_2$. Thus $u$ factors through $1$ (in a unique way), yielding the required $e : 1 \to M$.
Of course, the above paragraph takes place in the external logic, but this is unavoidable: there is no way to formulate the existence of a
global element in the internal logic. I suppose the point is that, once you have built up a stock of these metatheorems that interpret statements in the internal logic, you can then prove various results using internal logic if so desired. |
Linear Dependence Lemma
We will now look at a very important lemma known as the linear dependence lemma.
Lemma (Linear Dependence Lemma): Let $\{ v_1, v_2, ..., v_m \}$ be a set of linearly dependent vectors in the vector space $V$ and $v_1 \neq 0$. Then there exists $j \in \{ 2, 3, ..., m \}$ such that: a) $v_j \in \mathrm{span} \{ v_1, ..., v_{j-1} \}$. b) $\mathrm{span} \{ v_1, ..., v_{j-1}, v_{j+1}, ..., v_m \} = \mathrm{span} \{ v_1, ..., v_m \}$.
The linear dependence lemma tells us that given a linear dependent set of vectors where the first vector is nonzero, then there exists a vector in the set $\{ v_1, v_2, ..., v_m \}$ such that $v_j$ can be written as a linear combination of $\{ v_1, v_2, ..., v_{j-1}$ (that is $v_j \in \mathrm{span} \{ v_1, v_2, ..., v_{j-1} \}$), and that the set of linear combinations of the set of vectors $\{ v_1, ..., v_{j-1}, v_{j+1}, ..., v_m \}$ is the same set as the combination of the set of vectors $\{ v_1, v_2, ..., v_m \}$.
Proof:Let $\{ v_1, v_2, ..., v_m \}$ be a set of vectors that are linearly dependent where $v_1 \neq 0$, and let $a_1, a_2, ..., a_m \in \mathbb{F}$. Since this set of vectors is not linearly independent, then: Where not all $a_i$ are zero (otherwise the set of vectors would be linearly independent). Now since $v_1 \neq 0$, it follows that not all $a_2, a_3, ..., a_m \in \mathbb{F}$ are equal to zero, and so there exists an $a_j \in \mathbb{F}$ such that $a_j \neq 0$ where $j$ is the largest index such that $a_j \neq 0$, in other words, $a_{j+1} = 0$, $a_{j+2} = 0$, etc…, and so so: Therefore $v_j$ can be written as a linear combination of the set of vectors $\{ v_1, v_2, ..., v_{j-1} \}$ so $v_j \in \mathrm{span} \{ v_1, v_2, ..., v_{j-1} \}$ which proves A. Now let $u \in \mathrm{span} \{ v_1, v_2, ..., v_m \}$, and so $u$ can be written as a linear combination of the vectors in this set, and so there exists $b_1, b_2, ..., b_m \in \mathbb{F}$ such that: Now we substitute that $\quad v_j = -\frac{a_1}{a_j}v_1 -\frac{a_2}{a_j}v_2 - ... - \frac{a_{j-1}}{a_j}v_{j-1}$ and so: Therefore $u$ can be written as a linear combination of the vectors $\{ v_1, ..., v_{j-1}, v_{j+1}, ..., v_m$ and so $u \in \mathrm{span} \{ v_1, ..., v_{j-1}, v_{j+1}, ..., v_m \}$. Therefore $\mathrm{span} \{ v_1, v_2, ..., v_m \} = \mathrm{span} \{ v_1, ..., v_{j-1}, v_{j+1}, ..., v_m \}$. $\blacksquare$ |
Notice:
If you happen to see a question you know the answer to, please do chime in and help your fellow community members. We encourage our fourm members to be more involved, jump in and help out your fellow researchers with their questions. GATK forum is a community forum and helping each other with using GATK tools and research is the cornerstone of our success as a genomics research community.We appreciate your help!
Test-drive the GATK tools and Best Practices pipelines on Terra Check out this blog post to learn how you can get started with GATK and try out the pipelines in preconfigured workspaces (with a user-friendly interface!) without having to install anything. Genotype Refinement workflow for germline short variants Contents Overview Summary of workflow steps Output annotations Example More information about priors Mathematical details 1. Overview
The core GATK Best Practices workflow has historically focused on variant discovery --that is, the existence of genomic variants in one or more samples in a cohorts-- and consistently delivers high quality results when applied appropriately. However, we know that the quality of the individual genotype calls coming out of the variant callers can vary widely based on the quality of the BAM data for each sample. The goal of the Genotype Refinement workflow is to use additional data to improve the accuracy of genotype calls and to filter genotype calls that are not reliable enough for downstream analysis. In this sense it serves as an optional extension of the variant calling workflow, intended for researchers whose work requires high-quality identification of individual genotypes.
While every study can benefit from increased data accuracy, this workflow is especially useful for analyses that are concerned with how many copies of each variant an individual has (e.g. in the case of loss of function) or with the transmission (or de novo origin) of a variant in a family.
If a “gold standard” dataset for SNPs is available, that can be used as a very powerful set of priors on the genotype likelihoods in your data. For analyses involving families, a pedigree file describing the relatedness of the trios in your study will provide another source of supplemental information. If neither of these applies to your data, the samples in the dataset itself can provide some degree of genotype refinement (see section 5 below for details).
After running the Genotype Refinement workflow, several new annotations will be added to the INFO and FORMAT fields of your variants (see below).
Note that GQ fields will be updated, and genotype calls may be modified. However, the Phred-scaled genotype likelihoods (PLs) which indicate the original genotype call (the genotype candidate with PL=0) will remain untouched. Any analysis that made use of the PLs will produce the same results as before. 2. Summary of workflow steps Input
Begin with recalibrated variants from VQSR at the end of the germline short variants pipeline. The filters applied by VQSR will be carried through the Genotype Refinement workflow.
Step 1: Derive posterior probabilities of genotypes Tool used: CalculateGenotypePosteriors
Using the Phred-scaled genotype likelihoods (PLs) for each sample, prior probabilities for a sample taking on a HomRef, Het, or HomVar genotype are applied to derive the posterior probabilities of the sample taking on each of those genotypes. A sample’s PLs were calculated by HaplotypeCaller using only the reads for that sample. By introducing additional data like the allele counts from the 1000 Genomes project and the PLs for other individuals in the sample’s pedigree trio, those estimates of genotype likelihood can be improved based on what is known about the variation of other individuals.
SNP calls from the 1000 Genomes project capture the vast majority of variation across most human populations and can provide very strong priors in many cases. At sites where most of the 1000 Genomes samples are homozygous variant with respect to the reference genome, the probability of a sample being analyzed of also being homozygous variant is very high.
For a sample for which both parent genotypes are available, the child’s genotype can be supported or invalidated by the parents’ genotypes based on Mendel’s laws of allele transmission. Even the confidence of the parents’ genotypes can be recalibrated, such as in cases where the genotypes output by HaplotypeCaller are apparent Mendelian violations.
Step 2: Filter low quality genotypes Tool used: VariantFiltration
After the posterior probabilities are calculated for each sample at each variant site, genotypes with GQ < 20 based on the posteriors are filtered out. GQ20 is widely accepted as a good threshold for genotype accuracy, indicating that there is a 99% chance that the genotype in question is correct. Tagging those low quality genotypes indicates to researchers that these genotypes may not be suitable for downstream analysis. However, as with the VQSR, a filter tag is applied, but the data is not removed from the VCF.
Step 3: Annotate possible de novo mutations Tool used: VariantAnnotator
Using the posterior genotype probabilities, possible de novo mutations are tagged. Low confidence de novos have child GQ >= 10 and AC < 4 or AF < 0.1%, whichever is more stringent for the number of samples in the dataset. High confidence de novo sites have all trio sample GQs >= 20 with the same AC/AF criterion.
Step 4: Functional annotation of possible biological effects Tool options: Funcotator (experimental)
Especially in the case of de novo mutation detection, analysis can benefit from the functional annotation of variants to restrict variants to exons and surrounding regulatory regions. Funcotator is a new tool that is currently still in development. If you would prefer to use a more mature tool, we recommend you look into SnpEff or Oncotator, but note that these are not GATK tools so we do not provide support for them.
3. Output annotations
The Genotype Refinement workflow adds several new info- and format-level annotations to each variant. GQ fields will be updated, and genotypes calculated to be highly likely to be incorrect will be changed. The Phred-scaled genotype likelihoods (PLs) carry through the pipeline without being changed. In this way, PLs can be used to derive the original genotypes in cases where sample genotypes were changed.
Population Priors
New INFO field annotation PG is a vector of the Phred-scaled prior probabilities of a sample at that site being HomRef, Het, and HomVar. These priors are based on the input samples themselves along with data from the supporting samples if the variant in question overlaps another in the supporting dataset.
Phred-Scaled Posterior Probability
New FORMAT field annotation PP is the Phred-scaled posterior probability of the sample taking on each genotype for the given variant context alleles. The PPs represent a better calibrated estimate of genotype probabilities than the PLs are recommended for use in further analyses instead of the PLs.
Genotype Quality
Current FORMAT field annotation GQ is updated based on the PPs. The calculation is the same as for GQ based on PLs.
Joint Trio Likelihood
New FORMAT field annotation JL is the Phred-scaled joint likelihood of the posterior genotypes for the trio being incorrect. This calculation is based on the PLs produced by HaplotypeCaller (before application of priors), but the genotypes used come from the posteriors. The goal of this annotation is to be used in combination with JP to evaluate the improvement in the overall confidence in the trio’s genotypes after applying CalculateGenotypePosteriors. The calculation of the joint likelihood is given as:
where the GLs are the genotype likelihoods in [0, 1] probability space.
Joint Trio Posterior
New FORMAT field annotation JP is the Phred-scaled posterior probability of the output posterior genotypes for the three samples being incorrect. The calculation of the joint posterior is given as:
where the GPs are the genotype posteriors in [0, 1] probability space.
Low Genotype Quality
New FORMAT field filter lowGQ indicates samples with posterior GQ less than 20. Filtered samples tagged with lowGQ are not recommended for use in downstream analyses.
High and Low Confidence De Novo
New INFO field annotation for sites at which at least one family has a possible de novo mutation. Following the annotation tag is a list of the children with de novo mutations. High and low confidence are output separately.
4. Example
Before:
1 1226231 rs13306638 G A 167563.16 PASS AC=2;AF=0.333;AN=6;… GT:AD:DP:GQ:PL 0/0:11,0:11:0:0,0,249 0/0:10,0:10:24:0,24,360 1/1:0,18:18:60:889,60,0
After:
1 1226231 rs13306638 G A 167563.16 PASS AC=3;AF=0.500;AN=6;…PG=0,8,22;… GT:AD:DP:GQ:JL:JP:PL:PP 0/1:11,0:11:49:2:24:0,0,249:49,0,287 0/0:10,0:10:32:2:24:0,24,360:0,32,439 1/1:0,18:18:43:2:24:889,60,0:867,43,0
The original call for the child (first sample) was HomRef with GQ0. However, given that, with high confidence, one parent is HomRef and one is HomVar, we expect the child to be heterozygous at this site. After family priors are applied, the child’s genotype is corrected and its GQ is increased from 0 to 49. Based on the allele frequency from 1000 Genomes for this site, the somewhat weaker population priors favor a HomRef call (PG=0,8,22). The combined effect of family and population priors still favors a Het call for the child.
The joint likelihood for this trio at this site is two, indicating that the genotype for one of the samples may have been changed. Specifically a low JL indicates that posterior genotype for at least one of the samples was not the most likely as predicted by the PLs. The joint posterior value for the trio is 24, which indicates that the GQ values based on the posteriors for all of the samples are at least 24. (See above for a more complete description of JL and JP.)
5. More information about priors
The Genotype Refinement Pipeline uses Bayes’s Rule to combine independent data with the genotype likelihoods derived from HaplotypeCaller, producing more accurate and confident genotype posterior probabilities. Different sites will have different combinations of priors applied based on the overlap of each site with external, supporting SNP calls and on the availability of genotype calls for the samples in each trio.
Input-derived Population Priors
If the input VCF contains at least 10 samples, then population priors will be calculated based on the discovered allele count for every called variant.
Supporting Population Priors
Priors derived from supporting SNP calls can only be applied at sites where the supporting calls overlap with called variants in the input VCF. The values of these priors vary based on the called reference and alternate allele counts in the supporting VCF. Higher allele counts (for ref or alt) yield stronger priors.
Family Priors
The strongest family priors occur at sites where the called trio genotype configuration is a Mendelian violation. In such a case, each Mendelian violation configuration is penalized by a de novo mutation probability (currently 10-6). Confidence also propagates through a trio. For example, two GQ60 HomRef parents can substantially boost a low GQ HomRef child and a GQ60 HomRef child and parent can improve the GQ of the second parent. Application of family priors requires the child to be called at the site in question. If one parent has a no-call genotype, priors can still be applied, but the potential for confidence improvement is not as great as in the 3-sample case.
Caveats
Right now family priors can only be applied to biallelic variants and population priors can only be applied to SNPs. Family priors only work for trios.
6. Mathematical details
Note that family priors are calculated and applied before population priors. The opposite ordering would result in overly strong population priors because they are applied to the child and parents and then compounded when the trio likelihoods are multiplied together.
Review of Bayes’s Rule
HaplotypeCaller outputs the likelihoods of observing the read data given that the genotype is actually HomRef, Het, and HomVar. To convert these quantities to the probability of the genotype given the read data, we can use Bayes’s Rule. Bayes’s Rule dictates that the probability of a parameter given observed data is equal to the likelihood of the observations given the parameter multiplied by the prior probability that the parameter takes on the value of interest, normalized by the prior times likelihood for all parameter values:
$$ P(\theta|Obs) = \frac{P(Obs|\theta)P(\theta)}{\sum_{\theta} P(Obs|\theta)P(\theta)} $$
In the best practices pipeline, we interpret the genotype likelihoods as probabilities by implicitly converting the genotype likelihoods to genotype probabilities using non-informative or flat priors, for which each genotype has the same prior probability. However, in the Genotype Refinement Pipeline we use independent data such as the genotypes of the other samples in the dataset, the genotypes in a “gold standard” dataset, or the genotypes of the other samples in a family to construct more informative priors and derive better posterior probability estimates.
Calculation of Population Priors
Given a set of samples in addition to the sample of interest (ideally non-related, but from the same ethnic population), we can derive the prior probability of the genotype of the sample of interest by modeling the sample’s alleles as two independent draws from a pool consisting of the set of all the supplemental samples’ alleles. (This follows rather naturally from the Hardy-Weinberg assumptions.) Specifically, this prior probability will take the form of a multinomial Dirichlet distribution parameterized by the allele counts of each allele in the supplemental population. In the biallelic case the priors can be calculated as follows:
$$ P(GT = HomRef) = \dbinom{2}{0} \ln \frac{\Gamma(nSamples)\Gamma(RefCount + 2)}{\Gamma(nSamples + 2)\Gamma(RefCount)} $$
$$ P(GT = Het) = \dbinom{2}{1} \ln \frac{\Gamma(nSamples)\Gamma(RefCount + 1)\Gamma(AltCount + 1)}{\Gamma(nSamples + 2)\Gamma(RefCount)\Gamma(AltCount)} $$
$$ P(GT = HomVar) = \dbinom{2}{2} \ln \frac{\Gamma(nSamples)\Gamma(AltCount + 2)}{\Gamma(nSamples + 2)\Gamma(AltCount)} $$
where Γ is the Gamma function, an extension of the factorial function.
The prior genotype probabilities based on this distribution scale intuitively with number of samples. For example, a set of 10 samples, 9 of which are HomRef yield a prior probability of another sample being HomRef with about 90% probability whereas a set of 50 samples, 49 of which are HomRef yield a 97% probability of another sample being HomRef.
Calculation of Family Priors
Given a genotype configuration for a given mother, father, and child trio, we set the prior probability of that genotype configuration as follows:
$$ P(G_M,G_F,G_C) = P(\vec{G}) \cases{ 1-10\mu-2\mu^2 & no MV \cr \mu & 1 MV \cr \mu^2 & 2 MVs} $$
where the 10 configurations with a single Mendelian violation are penalized by the de novo mutation probability μ and the two configurations with two Mendelian violations by μ^2. The remaining configurations are considered valid and are assigned the remaining probability to sum to one.
This prior is applied to the joint genotype combination of the three samples in the trio. To find the posterior for any single sample, we marginalize over the remaining two samples as shown in the example below to find the posterior probability of the child having a HomRef genotype:
This quantity P(Gc|D) is calculated for each genotype, then the resulting vector is Phred-scaled and output as the Phred-scaled posterior probabilities (PPs). |
In Figure 4.1.1 we see that a central angle of \(90^\circ \) cuts off an arc of length \(\tfrac{\pi}{2}\,r \), a central angle of \(180^\circ \) cuts off an arc of length \(\pi\,r \), and a central angle of \(360^\circ \) cuts off an arc of length \(2\pi\,r \), which is the same as the circumference of the circle. So associating the central angle with its intercepted arc, we could say, for example, that
\[\nonumber 360^\circ \quad\text{"equals''}\quad 2\pi\,r \quad\text{(or \(2\pi \) 'radiuses').} \] The radius \(r \) was arbitrary, but the \(2\pi \) in front of it stays the same. So instead of using the awkward "radiuses'' or "radii'', we use the term radians: \[\label{4.1} \boxed{360^\circ ~=~ 2\pi ~~\text{radians}} \] The above relation gives us any easy way to convert between degrees and radians:
\[\begin{alignat}{3}
\textbf{Degrees to radians:}&\quad x~~\text{degrees}\quad&=\quad \left( \frac{\pi}{180} \;\cdot\; x \right) ~~\text{radians}\label{eqn:deg2rad}\\ \textbf{Radians to degrees:}&\quad x~~\text{radians}\quad&=\quad \left( \frac{180}{\pi} \;\cdot\; x \right) ~~\text{degrees}\label{eqn:rad2deg} \end{alignat}\]
Equation \ref{eqn:deg2rad} follows by dividing both sides of Equation \ref{4.1} by \(360 \), so that \(1^\circ = \frac{2\pi}{360} = \frac{\pi}{180} \) radians, then multiplying both sides by \(x \). Equation \ref{eqn:rad2deg} is similarly derived by dividing both sides of Equation \ref{4.1} by \(2\pi \) then multiplying both sides by \(x \).
The statement \(\theta = 2\pi \) radians is usually abbreviated as \(\theta = 2\pi \) rad, or just \(\theta = 2\pi \) when it is clear that we are using radians. When an angle is given as some multiple of \(\pi \), you can assume that the units being used are radians.
Example 4.1
Convert \(18^\circ \) to radians.
Solution:
Using the conversion Equation \ref{eqn:deg2rad} for degrees to radians, we get
\[\nonumber 18^\circ ~=~ \frac{\pi}{180} \;\cdot\; 18 ~=~ \boxed{\frac{\pi}{10} ~~\text{rad}} ~.\]
Example 4.2
Convert \(\frac{\pi}{9} \) radians to degrees.
Solution:
Using the conversion Equation \ref{eqn:rad2deg} for radians to degrees, we get
\[\frac{\pi}{9} ~~\text{rad} ~=~ \frac{180}{\pi} \;\cdot\; \frac{\pi}{9} ~=~ \boxed{20^\circ} ~.\nonumber \]
Table 4.1 Commonly used angles in radians
Table 4.1 shows the conversion between degrees and radians for some common angles. Using the conversion Equation \ref{eqn:rad2deg} for radians to degrees, we see that
\[\nonumber
1 ~~\text{radian} ~~=~~ \frac{180}{\pi}~~\text{degrees} ~~\approx~~ 57.3^\circ ~. \]
Formally, a radian is defined as the central angle in a circle of radius \(r \) which intercepts an arc of length \(r \), as in Figure 4.1.2. This definition does not depend on the choice of \(r\) (imagine resizing Figure 4.1.2).
One reason why radians are used is that the scale is smaller than for degrees. One revolution in radians is \(2\pi \approx 6.283185307 \), which is much smaller than \(360 \), the number of degrees in one revolution. The smaller scale makes the graphs of trigonometric functions (which we will discuss in Chapter 5) have similar scales for the horizontal and vertical axes. Another reason is that often in physical applications the variables being used are in terms of arc length, which makes radians a natural choice.
The default mode in most scientific calculators is to use degrees for entering angles. On many calculators there is a button labeled \(\fbox{\( DRG\)}\) for switching between degree mode (D), radian mode (R), and
gradian mode (G). On some graphing calculators, such as the the TI-83, there is a \(\fbox{\(MODE\)}\) button for changing between degrees and radians. Make sure that your calculator is in the correct angle mode before entering angles, or your answers will likely be way off. For example,
\[\nonumber \begin{align*}
\sin\;4^\circ ~&=~ \phantom{-}0.0698 ~,\\ \nonumber \sin\;(4~\text{rad}) ~&=~ -0.7568 ~, \end{align*}\]
so the values are not only off in magnitude, but do not even have the same sign. Using your calculator's \(\fbox{\(\sin^{-1}\)}\), \(\fbox{\(\cos^{-1}\)}\), and \(\fbox{\(\tan^{-1}\)}\) buttons in radian mode will of course give you the angle as a decimal, not an expression in terms of \(\pi \).
You should also be aware that the math functions in many computer programming languages use radians, so you would have to write your own angle conversions. |
I wonder if someone knows any general rules of thumb regarding the number of bootstrap samples one should use, based on characteristics of the data (number of observations, etc.) and/or the variables included?
My experience is that statisticians won't take simulations or bootstraps seriously unless the number of iterations exceeds 1,000. MC error is a big issue that's a little under appreciated. For instance, this paper used
Niter=50 to demonstrate LASSO as a feature selection tool. My thesis would have taken a lot less time to run had 50 iterations been deemed acceptable! I recommend that you should always
inspect the histogram of the bootstrap samples. Their distribution should appear fairly regular. I don't think any plain numerical rule will suffice, and it would be overkill to perform, say, a double-bootstrap to assess MC error.
Suppose you were estimating the mean from a ratio of two independent standard normal random variables, some statistician might recommend bootstrapping it since the integral is difficult to compute. If you have basic probability theory under your belt, you would recognize that this ratio forms a Cauchy random variable with a non-existent mean. Any other leptokurtic distribution would require several additional bootstrap iterations compared to a more regular Gaussian density counterpart. In that case, 1000, 100000, or 10000000 bootstrap samples would be insufficient to estimate that which doesn't exist. The histogram of these bootstraps would continue to look irregular and wrong.
There are a few more wrinkles to that story. In particular, the bootstrap is only really justified when the moments of the data generating probability model exist. That's because you are using the empirical distribution function as a straw man for the actual probability model, and assuming they have the same mean, standard deviation, skewness, 99th percentile, etc.
In short, a bootstrap estimate of a statistic and its standard error is only justified when the histogram of the bootstrapped samples appears regular beyond reasonable doubt and when the bootstrap is justified.
edit:
If you are serious about having enough samples, what you should do is to run your bootstrap procedure with, what you hope are, enough samples a number of times and see how much the bootstrap estimates "jump around". If the repeated estimates does not differ much (where "much" depends on your specific situation) your are most likely fine. Of course you can estimate how much the repeated estimates jump around by calculating the sample SD or similar.
If you want a reference and a rule of thumb Wilcox(2010) writes "599 is recommended for general use." But this should be considered only a guideline or perhaps the minimum number of samples you should consider. If you want to be on the safe side there is no reason (if it is computationally feasible) why you should not generate an order of magnitude more samples.
On a personal note I tend to run 10,000 samples when estimating "for myself" and 100,000 samples when estimating something passed on to others (but this is quick as I work with small datasets).
Reference
Wilcox, R. R. (2010).
Fundamentals of modern statistical methods: Substantially improving power and accuracy. Springer.
There are a some situations where you can tell either beforehand or after a few iterations that huge numbers of bootstrap iterations won't help in the end.
You hopefully have an idea beforehand on the order of magnitude of precision that is required for meaningful interpretation of the results. If you don't maybe it is time to learn a bit more about the problem behind the data analysis. Anyways, after a few iterations you may be able to estimate how many more iterations are needed.
Obviously, if you have extremely few cases (say, the ethics committee allowed 5 rats) you don't need to think about tens of thousands of iterations. Maybe it would be better to look at all possible draws. And maybe it would be even better to stop and think how certain any kind of conclusion can (not) be based on 5 rats.
Think about the total uncertainty of the results. In my field, the part of uncertainty that you can measure and reduce by bootstrapping may only be a minor part of the total uncertainty (e.g. due to restrictions in the design of the experiments important sources of variation are often not covered by the experiment - say, we start by experiments on cell lines although the final goal will of course be patients). In this situation it doesn't make sense to run too many iterations -- it anyways won't help the final result and moreover it may indroduce a false sense of certainty.
A related (though not exactly the same) issue occurs during out-of-bootstrap or cross validation of models: you have two sources of uncertainty: the finite (and in my case usually very small number of independent cases) and the (in)stability of the bootstrapped models. Depending on your set up of the resampling validation, you may have only one of them contributing to the resampling estimate. In that case, you can use an estimate of the other source of variance to judge what certainty you should achieve with the resampling, and when it stops to help the final result.
Finally, while so far my thoughts were about how to do
feweriterations, here's a practical consideration in favor of doing more: In practice my work is not done after the bootstrap is run. The output of the bootstrap needs to be aggregated into summary statistics and/or figures. Results need to be interpreted the paper or report to be written. Much of these can already be done with preliminary results of a few iterations of the bootstrap (if the results are clear, they show already after few iterations, if they are borderline they'll stay borderline). So I often set up the bootstrapping in a way that allows me to pull preliminary results so I can go on working while the computer computes. That way it doesn't bother me much if the bootstrapping takes another few days. TLDR. 10,000 seems to be a good rule of thumb, e.g. p-values from this large or larger of bootstrap samples will be within 0.01 of the "true p-value" for the method about 95% of the time.
I only consider the percentile bootstrap approach below, which is the most commonly used method (to my knowledge) but also admittedly has weaknesses and shouldn't be used with small samples.
Reframing slightly. It can be useful to compute the uncertainty associated with results from the bootstrap to get a sense for the uncertainty resulting from the use of the bootstrap. Note that this does not address possible weaknesses in the bootstrap (e.g. see the link above), but it does help evaluate if there are "enough" bootstrap samples in a particular application. Generally, the error related to the bootstrap sample size
n goes to zero as
n goes to infinity, and the question asks, how big should
n be for the error associated with small bootstrap sample size to be small?
Bootstrap uncertainty in a p-value. The imprecision in an estimated p-value, say pv_est is the p-value estimated from the bootstrap, is about
2 x sqrt(pv_est * (1 - pv_est) / N), where
N is the number of bootstrap samples. This is valid if
pv_est * N and
(1 - pv_est) * N are both
>= 10. If one of these is smaller than 10, then it's less precise but very roughly in the same neighborhood as that estimate.
Bootstrap error in a confidence interval. If using a 95% confidence interval, then look at how variability of the quantiles of the bootstrap distribution near 2.5% and 97.5% by checking the percentiles at (for the 2.5th percentile)
2.5 +/- 2 * 100 * sqrt(0.025 * 0.975 / n). This formula communicates the uncertainty of the lower end of the 95% confidence interval based on the number of bootstrap samples taken. A similar exploration should be done at the top end. If this estimate is somewhat volatile, then be sure to take more bootstrap samples!
I start by responding to something raised in another answer: why such a strange number as "$599$" (number of bootstrap samples)?
This applies also to Monte Carlo tests (to which bootstrapping is equivalent when the underlying statistic is
pivotal), and comes from the following: if the test is to be exact, then, if $\alpha$ is the desired significance level, and $B$ is the number of samples, the following relation must hold:
$$\alpha \cdot (1+B) = \text{integer}$$
Now consider typical significance levels $\alpha_1 = 0.1$ and $\alpha_2 = 0.05$
We have
$$B_1 = \frac {\text{integer}}{0.1} - 1,\;\;\; B_2 = \frac {\text{integer}}{0.05} - 1$$
This "minus one" is what leads to proposed numbers like "$599$", in order to ensure an exact test.
I took the following information from Davidson, R., & MacKinnon, J. G. (2000). Bootstrap tests: How many bootstraps?. Econometric Reviews, 19(1), 55-68. (the working paper version is freely downloadable).
As regards rule of thumb, the authors examine the case of bootstrapping p-values and they suggest that for tests at the $0.05$ the minimum number of samples is about 400 (so $399$) while for a test at the $0.01$ level it is 1500 so ($1499$).
They also propose a pre-testing procedure to determine $B$ endogenously. After simulating their procedure they conclude:
"It is easy to understand why the pretesting procedure works well. When the null hypothesis is true, B can safely be small, because we are not concerned about power at all. Similarly, when the null is false and test power is extremely high, B does not need to be large, because power loss is not a serious issue. However, when the null is false and test power is moderately high, B needs to be large in order to avoid loss of power. The pretesting procedure tends to make B small when it can safely be small and large when it needs to be large."
At the end of the paper they also compare it to another procedure that has been proposed in order to determine $B$ and they find that theirs performs better.
Most bootstrapping applications I have seen reported around 2,000 to 100k iterations. In modern practice with adequate software, the salient issues with bootstrap are the statistical ones, more so than time and computing capacity. For novice users with Excel, one could perform only several hundreds before requiring the use of advanced Visual Basic programming. However, R is much simpler to use and makes generation of thousands of bootstrapped values easy and straightforward. |
Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be analytic and consider the ODE $$x'(t)=f(x(t)).$$ It is well-known that if $(t_{min},t_{max})$ is the maximal domain of a solution $x$ and $t_{max}<\infty$, then $$\lim_{t\to t_{max}}|x(t)|=\infty.$$ Let $t_0\in(t_{min},t_{max})$. What conditions on $f$ (appart from linearity) ensure that $$\int_{t_0}^{t_{max}}|x(t)|^2dt=\infty,\quad\text{when }t_{max}<\infty?$$
This is not an answer, but is a comment. (I can not give comment since I am under 50 reputation).
Linear vector fields are always complete vector field so they do not satisfy your condition.
But for higher order polynomial vector field, I guess that the solutions which are not a complete orbits, are not in $L ^2$. My motivation is that according to an interesting Paper of Chicone and Sotomayor, the solutions escape at infinity very fast(exponentially) since there is a hyperbolic singularity at equator.
On the other hand your question is very interesting for me since it implicitly suggests to consider some different function spaces to be acted by $D_f$, the derivational operator associated to the vector field $f$.
The motivations for study of this derivational operator is explained in the following posts:
Consider the case $n=1$ (and, say, $f>0$, wlog), then $t_{max}<\infty$ iff $\int\frac{dx}{f(x)}<\infty$. Then $\int x(t)^2\ dt=\int\frac{x^2}{f(x)}\ dx$.
Since you didn't mention that simple case, I hope it may help. |
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Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions
(Nature Publishing Group, 2017)
At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ...
K$^{*}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV
(American Physical Society, 2017-06)
The production of K$^{*}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ...
Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Springer, 2017-06)
The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ...
Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC
(Springer, 2017-01)
The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ...
Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC
(Springer, 2017-06)
We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ... |
I need to draw the derivation tree for $1-2-(3-4)*5*6$ from the grammar below. I want to know how many possible derivation trees are there from this grammar.
$$\begin{align}V_n&=\{expr,term,factor,number\}\\ V_t&= \{(,),-,*,0...9\}\\ P&=\left \{ \begin{aligned} expr&\to expr-expr\;\mid\;term\\ term&\to term*factor\;\mid\;factor\\ factor&\to number \;\mid\; (expr) \\ number&\to 0\mid1\mid2\mid3\mid4\mid5\mid6\mid7\mid8\mid9 \end{aligned} \right \}\\ S&=expr \end{align} $$
The possibilities that I can find are:
$$(1-2)-(((3-4)*5)*6)\\ 1-(2-( ((3-4)*5) *6))$$
Are there other possibilities? |
Solution Spaces to Systems of Linear Equations
We will now begin to look more in-depth into solution spaces for both homogenous and nonhomogenous linear systems of $m$ equations and $n$ unknowns.
Consider the following transformation $T : \mathbb{F}^{n} \to \mathbb{F}^{m}$ defined by:(1)
Homogenous Linear Systems
Suppose that the equation $T((x_1, x_2, ..., x_n)) = \underbrace{(0, 0, ..., 0)}_{\mathrm{m-times}}$. We can thus write this equation as a system of linear equations given as:(2)
Of course, this homogenous system of linear equations can be represented in matrix form as $\begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \vdots \\ 0\end{bmatrix}$
In earlier linear algebra pages, we noted that a homogenous system always has a solution, namely the trivial solution $(x_1, x_2, ..., x_n) = (0, 0, ..., 0)$, however, we were more interested in determining whether nontrivial solutions exist. We can easily answer this question in terms of linear transformations.
Theorem 1: A homogenous linear system of $m$ equations of $n$ unknowns has nontrivial solutions if $n > m$. Proof:Let $T \in \mathcal L (\mathbb{F}^{n}, \mathbb{F}^{m})$ be defined by $T((x_1, x_2, ..., x_n)) = \left ( \sum_{k=1}^{n} a_{1,k}x_k, \sum_{k=1}^{n} a_{2,k}x_k, ..., \sum_{k=1}^{n} a_{m,k}x_k \right )$. If $n > m$, then by Corollary 2 of The Dimension of The Null Space and Range page, we have that $T$ is not injective, and therefore $\mathrm{null} (T) \neq \{0\}$ and so there exists another vector $(x_1, x_2, ..., x_n) \in \mathbb{F}^{n}$ such that $T((x_1, x_2, ..., x_n)) = 0$ where not all $x_j$, $1 ≤ j ≤ n$ are zero, and so the system of linear equations has a nontrivial solution. $\blacksquare$ Nonhomogenous Linear Systems
Now consider the equation $T((x_1, x_2, ..., x_n)) = \underbrace{(b_1, b_2, ..., b_m)}_{\mathrm{m-times}}$ where not all $b_j$, $1 ≤ j ≤ n$ are zero. We can thus write this equation is a system of linear equations given as:(3)
Similarly, we can write this nonhomogenous system of linear equations in matrix form as $\begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix} = \begin{bmatrix} b_1\\ b_2\\ \vdots\\ b_m\end{bmatrix}$.
In such a case, we're not always guaranteed a solution as with homogenous linear systems. The question then becomes if a solution $(x_1, x_2, ..., x_n)$ always exists for any choice of a vector $(b_1, b_2, ..., b_m)$. The answer is not necessarily as summarized in the following theorem.
Theorem 2: There exists a vector $(b_1, b_2, ..., b_m)$ for which there is no solution $(x_1, x_2, ..., x_n)$ to the system of nonhomogenous linear equations of $m$ equations and $n$ unknowns if $n < m$. Proof:Let $T \in \mathcal L (\mathbb{F}^{n}, \mathbb{F}^{m})$ be defined by $T((x_1, x_2, ..., x_n)) = \left ( \sum_{k=1}^{n} a_{1,k}x_k, \sum_{k=1}^{n} a_{2,k}x_k, ..., \sum_{k=1}^{n} a_{m,k}x_k \right )$. If $n < m$, then by Corollary 3 The Dimension of The Null Space and Range page, we have that $T$ is not surjective, and so $\mathrm{range} (T) \neq \mathbb{F}^{m}$. Therefore, there exists a vector $(b_1, b_2, ..., b_n) \in \mathbb{F}^{m}$ such that no vector $(x_1, x_2, ..., x_n) \in \mathbb{F}^{n}$ maps to it, in other words, the nonhomogenous linear system $T((x_1, x_2, ..., x_n)) = (b_1, b_2, ..., b_m)$ has no solution. $\blacksquare$ |
Storage of Numbers in IEEE Single-Precision Floating Point Format
For 32 bit storage in computers, a real number $x$ can be stored in which is known as
IEEE Single-Precision Floating Point Format in the following format:
The sign of $x$, $\sigma$ will take up one bit. The significand can be represented in terms of $24$ digits, so the precision of the IEEE Single-Precision Floating Point Format is $24$ binary digits, though we will only specify $23$ binary digits, so the significand is of $23$ bits. We will restrict $e$ such that $-(126)_{10} ≤ e ≤ (127)_{10}$ which in binary corresponds to the inequality $-(11111110)_{2} ≤ e ≤ (11111111)_{2}$ which contains $8$ bits. Together, we can write a binary number $x$ in IEEE Single-Precision Floating Point Format in $32$ bits ($4$ bytes).
The number $x = \sigma \cdot (1.a_1a_2...a_{22}a_{23} ) \cdot 2^e$ can hence be stored in $32$ bits as $b_1b_2...b_{31}b_{32}$ where each of the bits are assigned as follows:
Bit $1$ The bit $b_1$ corresponds to the sign $\sigma$ of $x$ where $b_1 = \left\{\begin{matrix} 0 & \mathrm{if} \: \sigma = +1\\ 1 & \mathrm{if} \: \sigma = -1 \end{matrix}\right.$. Bits $2$-$9$ Most computers do not store the exponent $e$ of a floating point binary number directly. Instead, they define $E = e + 127$ which is a positive binary number (since $-126 ≤ e$). The eight bits $b_2b_3...b_8b_9$ correspond to this number $E$. Bits $10$-$32$ The $23$ succeeding digits $a_1a_2...a_{22}a_{23}$ of the significand of $x$, $1.a_1a_2...a_{22}a_{23}$ are stored here.
The following image represents how a binary number is stored in IEEE Single-Precision Floating Point Format.
Recall from the Floating Point Numbers page that if $x = 0$ then $x$ cannot be uniquely written in the form $\sigma \cdot \bar{x} \cdot 2^e$. We need to be able to store $0$ in a computer though, so we specially formalize $0$ to be stored where $b_1 = b_2 = ... = b_{31} = b_{32} = 0$, that is all $32$ bits are $0$.
Example 1 Suppose that $11011011011010000000000000000000$ represents an IEEE Single-Precision floating-point number. Determine what this number is in decimal.
We will first separate $1 10110110 11010000000000000000000$. Notice that $b_1 = 1$, $b_2b_3...b_8b_9 = 10110110$, and $b_{10}b_{11}...b_{31}b_{32} = 11010000000000000000000$. Therefore we have that $\sigma = 1$, $E = 10110110$, and $\bar{x} = 1.11010000000000000000000$.
Since $b_1 = 1$, this tells us that $\sigma = -1$, and so $x$ is a negative number.
Since $b_2b_3...b_9$ represent $E$, we have that $E = 10110110$, and $E = 182$, and so since $E = e + 127$, we have that $e = 55$.
Lastly, we have that $\bar{x} = 1.11010000000000000000000 = 1.1101 = \left ( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{16} \right ) = 1.8125$.
Putting this all together, we have that:(2)
Example 2 Suppose that $0001111111011100000000000000000$ represents an IEEE Single-Precision floating-point number. Determine what this number is in decimal.
Once again, we first separate $0011111111011100000000000000000$. Notice that $b_1 = 0$, $b_2b_3...b_9 = 01111111$, and $b_{10}b_{11}...b_{32} = 1011100000000000000000$.
Since $b_1 = 0$, this tells us that $\sigma = +1$, so $x$ is a positive number.
Now since $b_1b_2...b_9$ represents $E$, we have that $E = 00111111$. In decimal, we have that $E = 63$, and so since $E = e + 127$, we have that then $63 = e + 127$, so $e = -64$.
Lastly, we have that $\bar{x} = 1.b_{10}b_{11}...b_{32} = 1.1011100000000000000000$. Thus $\bar{x} = 1.71875$.
Therefore we have that:(3) |
The working principle of a BJT (Bipolar Junction Transistor), which makes it a useful thing, is that it
amplifies current. Throw a small current in, get a larger current out. The amplification factor is an important parameter of the transistor, and is called \$h_{FE}\$. A general purpose transistor may have an \$h_{FE}\$ of 100, for instance, sometimes higher. Power transistors have to do it with less, like 20 to 30. So if I inject a 1 mA current in the base of my general purpose NPN transistor I'll get 100 mA of collector current. That's amplification, right? Current amplification.
How about voltage amplification? Well, let's add a couple of resistors. Resistors are cheap, but if you want to make money you can try to sell them expensive by calling them
"voltage-to-current converters" :-).
We've added a base resistor, which will cause a base current of
\$ I_B = \dfrac{V_B - 0.7 V}{R_B} \$
And we know that the collector current \$I_C\$ is a factor \$h_{FE}\$ higher, so
\$ I_C = \dfrac{h_{FE} \cdot (V_B - 0.7 V)}{R_B} \$
Resistors are really great things, because next to
"voltage-to-current converters" you an also use them as "current-to-voltage converters"! (we can charge even more for them!) Due to Ohm's Law:
\$ V_{RL} = R_L \cdot I_C \$
and since \$V_C = V_{CC} - V_{RL}\$
we get
\$V_C = V_{CC} - R_L \cdot \dfrac{h_{FE} \cdot (V_B - 0.7 V)}{R_B}\$
or
\$V_C = - \dfrac{h_{FE} \cdot R_L}{R_B} \cdot V_B + \left(\dfrac{h_{FE} \cdot R_L}{R_B} \cdot 0.7 V + V_{CC}\right)\$
The term between the brackets is a constant which we're not interested in at the moment. The first term shows that \$V_C\$ is \$V_B\$ multiplied by some factor depending on three constants. Let's use concrete values: 100 for \$h_{FE}\$, 10 kΩ for \$R_B\$ and 1 kΩ for \$R_C\$. Then (again ignoring the constant factor)
\$V_C = - \dfrac{h_{FE} \cdot R_L}{R_B} \cdot V_B = - \dfrac{100 \cdot 1k\Omega}{10 k\Omega} \cdot V_B = - 10 \cdot V_B \$
So the output voltage is 10 times the input voltage plus a constant bias. Looks like we can use the transistor for
voltage amplification as well. |
So far the answers just (cleverly) elaborate on high school tricks and techniques. Therefore, I think it can be interesting to see, instead, how standard modern algorithms work in this special case. I will implement a small version of the Berlekamp-Zassenhaus algorithm. I will try to factor $F(x)=x^5+x+1$ over $\mathbb Z[x]$ as a product $f_1(x)f_2(x)$ of polynomials of degree 2 and 3; it will not be long (of course), nor difficult. I recall what is the plan:
Bound the coefficients of the factors $f_1,f_2$; Factor $F(x)=g_1(x)g_2(x)$ over modulo $p$ for some prime $p$; Lift (in essence, by Hensel's lemma) the factorization modulo higher $p^k$.
Bound the coefficients of $f_1(x)$: The leading coefficient of $F$ is $c=1$, the degree of $f_1$ is $\delta=2$, and the roots $\alpha$ of $F$ satisfy ${|\alpha|}^5\leq 1+|\alpha|$, so for sure, say, $|\alpha|<\rho=1.5$. Therefore the (absolute values of the) coefficients of $f_1(x)$ are all dominated by those of $(x+\rho)^2$. In particular they are all $<3$. This is called the binomial bound. The same estimate can be obtained with the Knuth-Cohen bound. See Abbott, John. "Bounds on Factors in Z [x]." Journal of Symbolic Computation 50 (2013): 532-563.
Find $g_i(x)\equiv f_i(x)$ modulo 2: Since $F(0)=1$ and $F(1)=3$ we have that $g_1(0)=g_1(1)=1\bmod 2$. Thus the only possibility is $g_1(x)=x^2+x+1$. By polynomial long division in $\mathbb F_2[x]$ we get $g_2(x)=\frac{F(x)}{g_1(x)}=x^3+x^2+1$. Well, if you are clever enough, and not a computer, you might finish the exercise here, by doing long division in $\mathbb Z[x]$.
Factor modulo 4 as $F=(g_1+2 h_1)(g_2 + 2 h_2)$: In other words, we need $g_1 h_2+g_2 h_1 = \frac {F(x)-g_1(x)g_2(x)}{2} = x^4+x^3+x^2 \bmod 2$. Of course the solution is $h_1=0$ and $h_2=x^2$.
Conclusion: We have $f_1(x)\equiv x^2+x+1\bmod 4$. Since the absolute value of the coefficients of $f_1(x)$ is at most $2$, we get $f_1(x)= x^2+x+1$. It works.
Supplement: actually I am deeply convinced that the most natural technique to factor $F(x)$ is the one provided by the OP (although all the other approaches, included the one I described above, are interesting). I'll try to justify this claim. Suppose you want to factor $7763073514021$ in prime numbers. The factor $7$ is easy to find (actually, this completes the factorization). Why? Because you decompose $7-7-63-0-7-35-14-0-21$ and you factor out termwise. "Piecewise" factorization is by far the most natural, and easy to spot, approach to factorization "by hand". Another example is $x^6-x^4-20x^3+14x^2+20 x -14$, where you may wish to take advantage of the pattern $(1,-1),(-20,20),(14,-14)$. Now, suppose you want to factor the number $636362236363$. It's very similar to the example before, only that you must be able to see the "negative": you are just subtracting a "14" from $636363636363$. Although the pattern of coefficients $[1,0,0,0,1,1]$ of $F(x)$ might be irregular at first glance, I find it very natural to see it as a $0-111-00$ subtracted from a $111-111$. |
Let $A \leftarrow C \rightarrow B$ be affinoid $K$-algebras, where $K$ is a non-archimedean field with non-trivial absolute value. Equipping $A$, $B$, $C$ with the supremum seminorms, there is a canonical seminorm $\nu_1$ on $A \otimes_C B$: $$\nu_1(f) = \inf \max_i |a_i|_{\sup} |b_i|_{\sup},$$ the infimum taken over all representations $f = \sum a_i \otimes b_i$. Then $\nu_1$ extends by continuity to a seminorm on the completed tensor product $A \widehat{\otimes}_C B$ (where the latter is endowed with any residue norm). On the other hand, $A \widehat{\otimes}_C B$ is known to be an affinoid algebra in its own right, so has its own supremum seminorm $\nu_2$.
Does $\nu_1 = \nu_2$?
I'm hopeful this is true, but doubt it (although I have no specific counter-example). It's clear that $\nu_2 \le \nu_1$, and the equality holds if and only if $\nu_1$ is power-multiplicative (i.e. $\nu_1(f^n) = \nu_1(f)^n$).
If it helps, we may assume $C$ is a Tate algebra and $\text{Sp} \, A$, $\text{Sp} \, B$ are affinoid subdomains of $\text{Sp} \, C$. |
Apparently the search term I was missing was "Brownian motion". With that, I found several leads. They contradict each other somewhat, but I can at least post a partial answer:
Geisler - Sound to Synapse: Physiology of the Mammalian Ear:
Estimates for the first of these sources, the pressure fluctuations due to the Brownian motion of air molecules impinging on the eardrum, are about 2 µPa (
−20 dB SPL), when the frequency bandwidth relevant for the detection of a 3 kHz tone is included (Harris. 1968). Calculations using this number suggest that the behavioral thresholds of humans for 3 kHz tones are not limited by this Brownian motion, but that those for the most sensitive of cats may approach it (Green. 1976)
Dallos - The Auditory Periphery Biophysics and Physiology:
By assuming a 1000-Hz bandwidth, Harris computed that the Brownian motion of air molecules generates a mean pressure fluctuation of 1.27×10
−5 dyne/cm 2 [ −24 dB SPL]. The usually accepted value of sound pressure corresponding to free-field listening threshold is 18 dB above the pressure level of thermal fluctuations. Thus one can immediately see that Brownian motion of air molecules is certainly not the limiting factor of our hearing sensitivity.
There's another available with more details:
Harris - Brownian motion and the threshold of hearing:
We can avoid the calculation of the Brownian noise at the eardrum by using the Brownian noise in a free field and comparing that with the minimum audible field (MAF) instead of the minimal audible pressure (MAP).
If we use frequency limits of 2500 Hz and 3500 Hz. we obtain a root mean square (rms) pressure fluctuation of 98 db below 1 dyne/cm
2 [ −24 dB SPL]. The MAF 2 is about 80 db below 1 dyne/cm 2 at 3000 Hz. This is 18 db above the estimate of Brownian noise. It seems clear from this calculation that Brownian noise in the air is not a limiting factor to the threshold of hearing.
2.5 kHz to 3.5 kHz is not the total bandwidth that would be picked up by a microphone, though.
Yost & Killian - Hearing Thresholds:
By making some assumptions about the acoustic energy present in the Brownian motion of air molecules, it can be shown that a sound presented at 0 dB SPL is only 20-30 dB more intense than that being produced by Brownian motion
So
−20 to −30 dB SPL.
Howard & Angus - Acoustics and Psychoacoustics:
At 4kHz, which is about the frequency of the sensitivity peak, the pressure amplitude variations caused by the Brownian motion of air molecules, at room temperature and over a critical bandwidth, correspond to a sound pressure level of about
−23 dB. Thus the human hearing system is close to the theoretical physical limits of sensitivity. In other words there would be little point in being much more sensitive to sound, as all we would hear would be a ”hiss” due to the thermal agitation of the air!
I would still like to know:
How this is derived What the spectrum is, and if it's different from the violet spectrum in water, why? What the 20 Hz-to-20 kHz and A-weighted values are
Update
I believe I've found an answer in these two papers, though both have errors that make it difficult to be sure:
Harris, G. G. Brownian motion in the cochlear partition. J Acoust. Soc. Am. 44: 176-186, 1968 L. J. Sivian and S. D. White, On minimum audible sound fields. Journal of the Acoustical Society of America, 1933, 4, 288-321
Harris's equation 1 is taken from Sivian-White, but seems erroneous. The original is dimensionally consistent, at least:
$$\overline P = \left [ \int^{f_2}_{f_1}{P_f}^2\cdot df \right ]^{1/2} = \left [ \frac{8 \pi \rho k T} {3c} ({f_2}^3-{f_1}^3)\right ]^{1/2}$$
where $\overline P$ is RMS pressure, $\rho$ is density of air, $k$ is Boltzmann's constant, $T$ is temperature, $c$ is speed of sound, and $f_1$ and $f_2$ are the bandwidth limits.
Sivian-White then calculate $5\times 10^{-5}$ bars for 1000–6000 Hz, which... also seems erroneous. That's equal to 5 Pa, or 108 dB SPL? If I calculate over the same range, I get 5.3×10
−11 bars = 5.3 µPa = −11.6 dB SPL, which seems more reasonable.
Now Harris says:
Also, a more accurate estimate of the Brownian noise would take into account the properties of a semi-rigid eardrum in an ear canal and not treat the tympanic membrane as an infinite reflecting wall. We can avoid this type of calculation by using the Brownian noise in a free field and comparing that with the minimum audible field (MAF) instead of the minimum audible pressure (MAP). The free-field Brownian motion is 3 dB less than that given by Eq. 1 owing to the fact that the waves traveling in opposite directions are not correlated in a free field, but are at a reflecting wall
When I use Sivian's equation and Harris' −3 dB with Harris' frequency band of 2500–3500 Hz, I get Harris' answer of 1.273×10
−5 dyne/cm 2 (= 1.27 μPa = −24 dB SPL) so it seems like I'm doing it right.
But they're interested only in whether the self-noise of air is close to the threshold of hearing in the most sensitive band. Calculating total SPL, in the same way but over 20 Hz to 20 kHz, I get 21.8 μPa, very close to 0 dB SPL. Coincidence?
This equation also lets us calculate the spectral density, which seems to be violet noise, increasing by 6 dB every octave, same as the underwater reference in the question: |
Question : How could I compute the (wave) kernel from the fact I have already found (wave) trace on unit circle?
The definitions are related to the page $25$ of the following pdf.
As the Spectrum$(S^1)=\{n^2 : n\ \in \mathbb{N}^*\}$, the trace (It this relevant for the question?) as distribution is simply $$w(t)=\sum_{k \geq 1} e^{it \sqrt{- \lambda_k}}=\int_{-\infty}^{\infty}W(t,x,x)dx=\frac{1}{e^t-1}.$$
From this fact, I would like to compute the kernel $$W(t,x,y)= \sum_{k \geq 1} e^{it \sqrt{- \lambda_k}} \mu_k(x) \mu_k(y),$$ where $\mu_k$ is the eigenfunctions of the eigenvalues $\lambda_k$, and I found $\mu_k (t) = a_k \cos kt + b_k \sin kt$, $k \in \mathbb{Z}$.
I think we have to use the Poisson Summation Formula, but it is unclear.
(The Poisson Summation Formula) Let $f(x)$ be any piecewise continuous function, defined for each $x$, $-\infty < x < \infty$, such that the sum $F(x)=\sum_{k=-\infty}^{\infty} f(x+2kL)$ converge (absolutely) to a continuous and piecewise $C^1$ function $F(x)$. Assume that this is uniform for $-L \leq x \leq L$. So that $F(x)$ is periodic of the period $2L$, and equals its Fourier series.
I don't understand how to do this, knowing that the exponential function isn't periodic. To help you better understand the concept, I think you can look at a case like the Heat Kernel on the following links #1 and #2.
Thanks!
P.S. Please, be aware that it is not for an homework, but rather for research work. |
The Transpose of a Matrix
Definition: If $A$ is an $m \times n$ matrix, then the Transpose of $A$ denoted $A^T$ is the $n \times m$ matrix resulting from interchanging both the rows and columns of $A$, that is $(A)_{ij} = (A^T)_{ji}$.
For example, suppose that we have the following $3 \times 4$ matrix $A = \begin{bmatrix}3 & 0 & 2 & 1\\ 2 & 3 & 1 & 7\\ -2 & 3 & 1 & 4 \end{bmatrix}$. If we interchange the rows and columns of $A$, then we will obtain the $4 \times 3$ matrix $A^T = \begin{bmatrix} 3 & 2 & -2\\ 0 & 3 & 3\\ 2 & 1 & 1\\ 1 & 7 & 4 \end{bmatrix}$.
One important point to note is that if $A$ is a square matrix, then the diagonal entries of $A^T$ will be the same as that of $A$. For example, if $A = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix}$, then $A^T = \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix}$. Notice that the entries in the main diagonal are the same between both matrices.
Example 1
(1)
Given the following matrix, determine $G^T$.
There are 2 rows and 5 columns to G, hence G is a $2 \times 5$ matrix, and $G^T$ will be a $5 \times 2$ matrix. Interchanging the rows and columns of G we obtain:(2)
Example 2 Prove that $G + G^T$ is only defined if $G$ is a square matrix. Proof:Suppose not, and let G be a matrix of size $m \times n$ where $m \neq n$. It thus follows that $G^T$ will be of size $n \times m$. Recall that matrix addition is only defined if all matrices in the sum are the same size. Hence $G_{m \times n} + G^T_{n \times m}$ is not defined since $m ≠ n$. $\blacksquare$ |
Feynman diagrams provide a very compact and intuitive way of representing interactions between particles. These diagrams can be included into LaTeX documents thanks to a few packages. One of the older packages is
feynmf which uses MetaPost in order to generate the diagrams. More recently, a new package called Ti
kZ-Feynman has been published which uses Ti kZ in order to generate Feynman diagrams.
Contents
Ti
kZ-Feynman is a LaTeX package allowing Feynman diagrams to be easily generated within LaTeX with minimal user instructions and without the need of external programs. It builds upon the Ti kZ package and its graph drawing algorithms in order to automate the placement of many vertices. Ti kZ-Feynman still allows fine-tuned placement of vertices so that even complex diagrams can be generated with ease.
Currently, Ti
kZ-Feynman is too new to have made it into ShareLaTeX's installation, but we are working to get it included soon. In the meantime, it is possible to include the package files manually in a ShareLaTeX project as shown in this template.
After installing the package, the Ti
kZ-Feynman package can be loaded with
\usepackage{tikz-feynman} in the preamble. It is recommend that you also specify the version of Ti
kZ-Feynman to use with the
compat package option:
\usepackage[compat=1.0.0]{tikz-feynman}. This ensures that any new versions of Ti
kZ-Feynman do not produce any undesirable changes without warning.
Feynman diagrams can be declared with the
\feynmandiagram command. It is analogous to the
\tikz command from Ti
kZ and requires a final semi-colon (
;) to finish the environment. For example, a simple
s-channel diagram is:
\feynmandiagram [horizontal=a to b] { i1 -- [fermion] a -- [fermion] i2, a -- [photon] b, f1 -- [fermion] b -- [fermion] f2, };
Let's go through this example line by line:
\feynmandiagram introduces the Feynman diagram and allows for optional arguments to be given in the brackets
[<options>]. In this instance,
horizontal=a to b orients the algorithm outputs such that the line through vertices
a and
b is horizontal.
i1,
a and
i2) and connecting them with edges
--. Just like the
\feynmandiagram command above, each edge also take optional arguments specified in brackets
[<options>]. In this instance, we want these edges to have arrows to indicate that they are fermion lines, so we add the
fermion style to them. As you will see later on, optional arguments can also be given to the vertices in exactly the same way.
a and
b with an edge styled as a photon. Since there is already a vertex labelled
a, the algorithm will connect it to a new vertex labeled
b.
f1 and
f2. It re-uses the previously labelled
b vertex.
;) is important.
The name given to each vertex in the graph does not matter. So in this example,
i1,
i2 denote the initial particles;
f1,
f2 denotes the final particles; and
a,
b are the end points of the propagator. The only important aspect is that what we called
a in line 2 is also
a in line 3 so that the underlying algorithm treats them as the same vertex.
The order in which vertices are declared does not matter as the default algorithm re-arranges everything. For example, one might prefer to draw the fermion lines all at once, as with the following example (note also that the way we named vertices is completely different):
\feynmandiagram [horizontal=f2 to f3] { f1 -- [fermion] f2 -- [fermion] f3 -- [fermion] f4, f2 -- [photon] p1, f3 -- [photon] p2, };
As a final remark, the calculation of where vertices should be placed is usually done through an algorithm written in Lua. As a result, LuaTeX is required in order to make use of these algorithms. If LuaTeX is not used, Ti
kZ-Feynman will default to a more rudimentary algorithm and will warn the user instead.
So far, the examples have only used the
photon and
fermion styles. The Ti
kZ-Feynman package comes with quite a few extra styles for edges and vertices which are all documented over in the package documentation. For example, it is possible to add momentum arrows with
momentum=<text>, and in the case of end vertices, the particle can be labelled with
particle=<text>. To demonstrate how they are used, we take the generic
s-channel diagram from earlier and make it a electron-positron pairs annihilating into muons:
\feynmandiagram [horizontal=a to b] { i1 [particle=\(e^{-}\)] -- [fermion] a -- [fermion] i2 [particle=\(e^{+}\)], a -- [photon, edge label=\(\gamma\), momentum'=\(k\)] b, f1 [particle=\(\mu^{+}\)] -- [fermion] b -- [fermion] f2 [particle=\(\mu^{-}\)], };
In addition to the style keys documented below, style keys from Ti
kZ can be used as well:
\feynmandiagram [horizontal=a to b] { i1 [particle=\(e^{-}\)] -- [fermion, very thick] a -- [fermion, opacity=0.2] i2 [particle=\(e^{+}\)], a -- [red, photon, edge label=\(\gamma\), momentum'={[arrow style=red]\(k\)}] b, f1 [particle=\(\mu^{+}\)] -- [fermion, opacity=0.2] b -- [fermion, very thick] f2 [particle=\(\mu^{-}\)], };
For a list of all the various styles that Ti
kZ provides, have a look at the Ti kZ manual; it is extremely thorough and provides many usage examples.
By default, the
\feynmandiagram and
\diagram commands use the
spring layout algorithm to place all the edges. The
spring layout algorithm attempts to `spread out' the diagram as much as possible which—for most simpler diagrams—gives a satisfactory result; however in some cases, this does not produce the best diagram and this section will look at alternatives. There are three main alternatives:
draw=none. The algorithm will treat these extra edges in the same way, but they are simply not drawn at the end;
The underlying algorithm treats all edges in exactly the same way when calculating where to place all the vertices, and the actual drawing of the diagram (after the placements have been calculated) is done separately. Consequently, it is possible to add edges to the algorithm, but prevent them from being drawn by adding
draw=none to the edge style.
This is particularly useful if you want to ensure that the initial or final states remain closer together than they would have otherwise as illustrated in the following example (note that
opacity=0.2 is used instead of
draw=none to illustrate where exactly the edge is located).
% No invisible to keep the two photons together \feynmandiagram [small, horizontal=a to t1] { a [particle=\(\pi^{0}\)] -- [scalar] t1 -- t2 -- t3 -- t1, t2 -- [photon] p1 [particle=\(\gamma\)], t3 -- [photon] p2 [particle=\(\gamma\)], };
% Invisible edge ensures photons are parallel \feynmandiagram [small, horizontal=a to t1] { a [particle=\(\pi^{0}\)] -- [scalar] t1 -- t2 -- t3 -- t1, t2 -- [photon] p1 [particle=\(\gamma\)], t3 -- [photon] p2 [particle=\(\gamma\)], p1 -- [opacity=0.2] p2, };
The graph drawing library from Ti
kZ has several different algorithms to position the vertices. By default,
\diagram and
\feynmandiagram use the
spring layout algorithm to place the vertices. The
spring layout attempts to spread everything out as much as possible which, in most cases, gives a nice diagram; however, there are certain cases where this does not work. A good example where the
spring layout doesn't work are decays where we have the decaying particle on the left and all the daughter particles on the right.
% Using the default spring layout \feynmandiagram [horizontal=a to b] { a [particle=\(\mu^{-}\)] -- [fermion] b -- [fermion] f1 [particle=\(\nu_{\mu}\)], b -- [boson, edge label=\(W^{-}\)] c, f2 [particle=\(\overline \nu_{e}\)] -- [fermion] c -- [fermion] f3 [particle=\(e^{-}\)], };
% Using the layered layout \feynmandiagram [layered layout, horizontal=a to b] { a [particle=\(\mu^{-}\)] -- [fermion] b -- [fermion] f1 [particle=\(\nu_{\mu}\)], b -- [boson, edge label'=\(W^{-}\)] c, c -- [anti fermion] f2 [particle=\(\overline \nu_{e}\)], c -- [fermion] f3 [particle=\(e^{-}\)], };
You may notice that in addition to adding the
layered layout style to
\feynmandiagram, we also changed the order in which we specify the vertices. This is because the
layered layout algorithm does pay attention to the order in which vertices are declared (unlike the default
spring layout); as a result,
c--f2, c--f3 has a different meaning to
f2--c--f3. In the former case,
f2 and
f3 are both on the layer below
c as desired; whilst the latter case places
f2 on the layer above
c (that, the same layer as where the W-boson originates).
In more complicated diagrams, it is quite likely that none of the algorithms work, no matter how many invisible edges are added. In such cases, the vertices have to be placed manually. Ti
kZ-Feynman allows for vertices to be manually placed by using the
\vertex command.
The
\vertex command is available only within the
feynman environment (which itself is only available inside a
tikzpicture). The
feynman environment loads all the relevant styles from Ti
kZ-Feynman and declares additional Ti kZ-Feynman-specific commands such as
\vertex and
\diagram. This is inspired from PGFPlots and its use of the
axis environment.
The
\vertex command is very much analogous to the
\node command from Ti
kZ, with the notable exception that the vertex contents are optional; that is, you need not have
{<text>} at the end. In the case where
{} is specified, the vertex automatically is given the
particle style, and otherwise it is a usual (zero-sized) vertex.
To specify where the vertices go, it is possible to give explicit coordinates though it is probably easiest to use the
positioning library from Ti
kZ which allows vertices to be placed relative to existing vertices. By using relative placements, it is possible to easily tweak one part of the graph and everything will adjust accordingly—the alternative being to manually adjust the coordinates of every affected vertex.
Finally, once all the vertices have been specified, the
\diagram* command is used to specify all the edges. This works in much the same way as
\diagram (and also
\feynmandiagram), except that it uses an very basic algorithm to place new nodes and allows existing (named) nodes to be included. In order to refer to an existing node, the node must be given in parentheses.
This whole process of specifying the nodes and then drawing the edges between them is shown below for the muon decay:
\begin{tikzpicture} \begin{feynman} \vertex (a) {\(\mu^{-}\)}; \vertex [right=of a] (b); \vertex [above right=of b] (f1) {\(\nu_{\mu}\)}; \vertex [below right=of b] (c); \vertex [above right=of c] (f2) {\(\overline \nu_{e}\)}; \vertex [below right=of c] (f3) {\(e^{-}\)}; \diagram* { (a) -- [fermion] (b) -- [fermion] (f1), (b) -- [boson, edge label'=\(W^{-}\)] (c), (c) -- [anti fermion] (f2), (c) -- [fermion] (f3), }; \end{feynman} \end{tikzpicture}
The
feynmf package lets you easily draw Feynman diagrams in your LaTeX documents. All you need to do is specify the vertices, the particles and the labels, and it will automatically layout and draw your diagram for you.
Let's start with a quick example:
\begin{fmffile*}{diagram} \begin{fmfgraph}(40,25) \fmfleft{i1,i2} \fmfright{o1,o2} \fmf{fermion}{i1,v1,o1} \fmf{fermion}{i2,v2,o2} \fmf{photon}{v1,v2} \end{fmfgraph} \end{fmffile*}
The
fmffile* environment must be put around all of your Feynman diagrams. You can use fmffile environment for multiple diagrams, so you can put one around your whole document and forget about it. The second argument to the fmffile environment tells LaTeX where to write the files that it uses to store the diagram. You can name this whatever you want, but you need to run metafont on your diagram between LaTeX runs in order for your diagram to show up (ShareLaTeX does this automatically):
The 'fmfgraph' environment starts a Feynman diagram, and the figures in brackets afterwards specify the width and height of the diagram.
The first thing you need to do is specify your external vertices, and where they should be positioned. You can name your vertices anything you like, and say where they should be positioned with the commands
\fmfleft, \fmfright, \fmftop, \fmfbottom. For example
% Creates two vertices on the left called i1 and i2 \fmfleft{i1,i2} % Creates two vertices on the right called o1 and o2 \fmfright{o1,o2}
You can connect up vertices with the
\fmf, which will create new vertices if you pass in names that haven't been created yet. For example
% Will create a fermion line between i1 and % the newly created v1, and between v1 and o1. \fmf{fermion}{i1,v1,o1} % Will create a photon line between v1 and the newly created v2 \fmf{photon}{v2,v2}
A vertex can be labelled using the
\fmflabel command, which takes two arguments: the label to apply to the vertex, and the name of the vertex to apply it to. For example, in the above diagram, if we add in the following labels, we get the updated diagram below:
Note that math mode can used inside the vertex labels, as we have done above.
We've seen the 'photon' and 'fermion' line styles above, but the
feynmf package support many more.
Appearance Name(s) gluon, curly dbl_curly dashes scalar, dashes_arrow dbl_dashes dbl_dashes_arrow dots ghost, dots_arrow dbl_dots dbl_dots_arrow phantom phantom_arrow vanilla, plain fermion, electron, quark, plain_arrow double, dbl_plain double_arrow, heavy, dbl_plain_arrow boson, photon, wiggly dbl_wiggly zigzag dbl_zigzag
For more information see: |
My question is that when we want to find the Lorentz force acted on a particle moving in an electric and magnetic field , the equation is invariant in any two inertial relativistic frames. Why is that so ? And only the electric and magnetic fields transform ?
The force f also transforms. In one frame you see f,E,v, and B. If you boost yourself wrt this frame, you will see f ',E',v',and B'. In the new frame the primed quantities satisfy the Lorentz force equation just like the unprimed quantities did in the original frame. The covariant form of the Lorentz force equation makes obvious how everything transforms:
$$ \frac{dp^{\alpha}}{d\tau}=qF^{\alpha \beta}U_{\beta} $$
The force on the charge q are the 3 spatial components of $\frac{dp^{\alpha}}{d\tau}$. The EM field tensor $F^{\alpha \beta}$ has the components of E and B in it. The 4-velocity $U^{\beta}=(\gamma,\gamma v_x, \gamma v_y, \gamma v_z)$. Please see "Relativistic form of the Lorentz force" down the page in https://en.wikipedia.org/wiki/Lorentz_force if you would like more details on all these covariant objects and how the vector form of the Lorentz force equation falls out. |
Urysohn metrization theorem says that every regular and second countable topological space is metrizable. My question, is the converse of this theorem ture ? If not, what are the counter examples?
Any reply kindly appreciated. Thanks.
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Urysohn metrization theorem says that every regular and second countable topological space is metrizable. My question, is the converse of this theorem ture ? If not, what are the counter examples?
Any reply kindly appreciated. Thanks.
The converse of this theorem does not hold.
As an example consider the set of all real numbers $ \mathbb{R} $ with the discrete topology $ \tau_{d} $. Clearly $ (\mathbb{R},\tau_{d}) $ is metrizable and the discrete metric $ \rho_{d} $ is the metrization of $ \tau_{d} $. Also $ \mathbb{R} $ is regular with respect to $ \tau_{d} $ since every closed subset of $ \mathbb{R} $ is open with respect to $ \tau_{d} $.
Notice that if a collection $ \mathcal{B} $ of subsets of $ \mathbb{R} $ is a base for $ (\mathbb{R},\tau_{d}) $ then $ \mathcal{B} $ contains $ \{\{x\}:x\in \mathbb{R}\} $. But $ \{\{x\}:x\in \mathbb{R}\} $ is uncountable. Hence $ \mathcal{B} $ is also uncountable. Therefore $ (\mathbb{R},\tau_{d}) $ has no countable base even though $ (\mathbb{R},\tau_{d}) $ is metrizable. Therefore the converse of the Urysohn metrization theorem does not hold.
A metrizable space is second countable if and only if it is separable.There are many examples of non-separable metrizable spaces,e.g.,in the previous answer, the discrete topology on any uncountable set |
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What is the rate of change of y=sinθ y = \sin \theta y=sinθ when θ=sin−1144180? \theta = \sin^{-1} \frac { 144}{180}? θ=sin−1180144?
What is the derivative of the function y=5cosx?y=5\cos x?y=5cosx?
What is the derivative of the function y=cotx?y=\cot x?y=cotx?
What is the derivative of the function y=cscx?y=\csc x?y=cscx?
What is the derivative of the function y=secx?y=\sec x?y=secx?
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EDIT: Thanks to hryghr I see that the starting assumptions were incorrect. The transfer function magnitude can't be found that simply.It is more than ten years since I considered my skills sharp on thistopic, and knives don't get sharper in the drawer! But I can't havethat I posted something formally incorrect, so here goes attempt #2:
I will derive the transfer function the dirty way .. using Kirchoff'sCurrent Law (KCL) (a very generic method). I call the output node \$V_{o}\$, and the middle node \$V_{x}\$. For the following equations i cut down on writing bywriting \$V_{o}\$ instead of the more accurate \$V_{o}(s)\$ :
I: KCL in \$V_{o}\$:
$$\frac{V_{o}-V_{x}}{R_{2}}+sC_{2}V_{o}=0$$
$$V_{x}=V_{o}(1+sR_{2}C_{2})$$II: KCL in \$V_{x}\$:
$$\frac{V_{x}-V_{i}}{R_{1}}+\frac{V_{x}-V_{o}}{R_{2}}+sC_{1}V_{x}=0$$
Rearranging terms:
$$R_{2}(V_{x}-V_{i})+R_{1}(V_{x}-V_{o})+sR_{1}R_{2}C_{1}V_{x}=0$$
Rearranging terms:
$$V_{x}(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{2}V_{i}-R_{1}V_{o}=0$$
Substituting \$V_{x}\$ with result of I:$$V_{o}(1+sR_{2}C_{2})(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{2}V_{i}-R_{1}V_{o}+sR_{1}R_{2}C_{1}V_{o}=0$$
Collecting terms for \$V_{o}\$
$$V_{o}((1+sR_{2}C_{2})(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{1})=R_{2}V_{i}$$
Rearranging:
$$\frac{V_{o}}{V_{i}}=\frac{R_{2}}{(1+sR_{2}C_{2})(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{1}}$$
Expanding terms:
$$\frac{V_{o}}{V_{i}}=\frac{R_{2}}{R_{1}+R_{2}+sR_{1}R_{2}C_{1}+sR_{1}R_{2}C_{2}+sR_{2}^{2}C_{2}+s^{2}R_{1}R_{2}^{2}C_{1}C_{2}-R_{1}}$$
\$R_{1}\$ cancels, then divide by \$R_{2}\$ top and bottom:
$$\frac{V_{o}}{V_{i}}=\frac{1}{1+sR_{1}C_{1}+sR_{1}C_{2}+sR_{2}C_{2}+s^{2}R_{1}R_{2}C_{1}C_{2}}$$
Prettified, the transfer function is:
$$H(s)=\frac{V_{o}(s)}{V_{i}(s)}=\frac{1}{s^{2}R_{1}R_{2}C_{1}C_{2}+s(R_{1}C_{1}+R_{1}C_{2}+R_{2}C_{2})+1}$$
This is probably a nice place to start converting to the standard form thathryghr mentions. It may be that the
corner frequency asked for relates to that form.I won't bother to much with that, but move on to find the -3dB point.
The magnitude of the transfer function can for instance be found bycalculating:
$$\left|H(\omega)\right|=\sqrt{H(s\rightarrow j\omega)H(s\rightarrow-j\omega)}$$
Setting \$A=R_{1}R_{2}C_{1}C_{2}\$ and \$B=(R_{1}C_{1}+R_{1}C_{2}+R_{2}C_{2})\$to simplify this calculation:
$$\left|H(\omega)\right|=\frac{1}{\sqrt{((j\omega)^{2}A+(j\omega)B+1)((-j\omega)^{2}A+(-j\omega)B+1)}}$$
$$\left|H(\omega)\right|=\frac{1}{\sqrt{(-\omega{}^{2}A+j\omega B+1)(-\omega{}^{2}A-j\omega B+1)}}$$
$$\left|H(\omega)\right|=\frac{1}{\sqrt{\omega{}^{4}A^{2}-\omega{}^{2}A(j\omega B-j\omega B+1+1)+\omega^{2}B^{2}+(j\omega B-j\omega B)+1}}$$
$$\left|H(\omega)\right|=\frac{1}{\sqrt{\omega{}^{4}A^{2}+\omega{}^{2}(B^{2}-2A)+1}}$$
Finding \$B^{2}-2A\$ gives you something like:
$$R_{1}^{2}(C_{1}+C_{2})^{2}+C_{2}^{2}(2R_{1}R_{2}+R_{2}^{2})$$
Then to find the -3dB point start at:
$$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{\omega{}^{4}A^{2}+\omega{}^{2}(B^{2}-2A)+1}}$$
$$2=\omega{}^{4}A^{2}+\omega{}^{2}(B^{2}-2A)+1$$
So far I have done it all by hand (hopefully no mistakes), but hereI call it a day, try mathematica, and get \$\omega\$ for the -3dB frequency as:
$$w\to\sqrt{\frac{1}{A}-\frac{B^{2}}{2A^{2}}+\frac{\sqrt{8A^{2}-4AB^{2}+B^{4}}}{2A^{2}}}$$ |
The ensuing development relies on the elementary inequality for the logarithm function
$$\log(x)\le x-1 \tag 1$$
for all $x>0$.
Let $f(t,h)$ be the function given by
$$f(t,h)=\frac{t^h-1}{h}\tag 2$$
for $h\ne 0$. Note that $f(t,h)>0$ for $t>1$ and $f(t,h)<0$ for $t<1$.
We seek to find a function $g(t)$ such that (i) $|f(t,h)|\le g(t)$ and (ii) $g(t)t^{x-1}e^{-t}$ is integrable for $t\in (0,\infty)$.
If successful, then the Dominated Convergence Theorem guarantees that the derivative of the Gamma function is given by
$$\begin{align}\Gamma'(x)&=\lim_{h\to 0}\int_0^\infty \left(\frac{t^h-1}{h}\right)t^{x-1}e^{-t}\,dt\\\\&=\int_0^\infty \lim_{h\to 0}\left(\frac{t^h-1}{h}\right)t^{x-1}e^{-t}\,dt\\\\&=\int_0^\infty \log(t)t^{x-1}e^{-t}\,dt \tag 3\end{align}$$
FINDING A BOUNDING FUNCTION
Proceeding to bound $f(t,h)$, we find immediately from $(1)$ to $(2)$ with $x=t^h$ that for $t<1$
$$|f(t,h)|\le |\log (t)| \tag 4$$
Next, note that for $h\ne 0$ and any fixed $t>0$, the partial derivative of $f(t,h)$ with respect to $h$ is given by
$$\frac{\partial f(t,h)}{\partial h}=\frac{\log(t^h)t^h-(t^h-1)}{h^2}\ge \frac{(t^h-1)^2}{h^2}\ge 0$$
Therefore $f(h)$ is an increasing function of $h$ and so for $h\le 1$, we have
$$\begin{align}f(t,h)&\le f(t,1)\\\\&=t-1\end{align}$$
So, for $t\ge 1$ we find that
$$|f(t,h)|\le t-1$$
Putting $(4)$ and $(5)$ together reveals
$$|f(t,h)|\le g(t)$$
where
$$g(t)=\begin{cases}|\log(t)|&,0<t\le 1\\\\t-1&,1<t\end{cases}$$
Inasmuch as $g(t)t^{x-1}e^{-t}$ is integrable and an upper bound for $\left|f(h) t^{x-1}e^{-t}\right|$, then application of the Dominated Convergence Theorem yields the result in $(3)$
$$\Gamma'(x)=\int_0^\infty \log(t)t^{x-1}e^{-t}\,dt$$
as was to be shown! |
Ab initio calculation of the $$np \to d ³$$ radiative capture process Abstract
In this study, lattice QCD calculations of two-nucleon systems are used to isolate the short-distance two-body electromagnetic contributions to the radiative capture process $$np \to d\gamma$$, and the photo-disintegration processes $$\gamma^{(\ast)} d \to np$$. In nuclear potential models, such contributions are described by phenomenological meson-exchange currents, while in the present work, they are determined directly from the quark and gluon interactions of QCD. Calculations of neutron-proton energy levels in multiple background magnetic fields are performed at two values of the quark masses, corresponding to pion masses of $$m_\pi \sim 450$$ and 806 MeV, and are combined with pionless nuclear effective field theory to determine these low-energy inelastic processes. Extrapolating to the physical pion mass, a cross section of $$\sigma^{lqcd}(np\to d\gamma)=332.4({\tiny \begin{array}{l}+5.4 \\ - 4.7\end{array}})\ mb$$ is obtained at an incident neutron speed of $$v=2,200\ m/s$$, consistent with the experimental value of $$\sigma^{expt}(np \to d\gamma) = 334.2(0.5)\ mb$$.
Authors: Univ. of Washington, Seattle, WA (United States) Massachusetts Inst. of Technology (MIT), Cambridge, MA (United States) Thomas Jefferson National Accelerator Facility (TJNAF), Newport News, VA (United States) Univ. of Barcelona, Barcelona (Spain) The City College of New York, New York, NY (United States); Brookhaven National Lab. (BNL), Upton, NY (United States) Publication Date: Research Org.: Oak Ridge National Laboratory, Oak Ridge Leadership Computing Facility (OLCF); Thomas Jefferson National Accelerator Facility (TJNAF), Newport News, VA (United States) Sponsoring Org.: USDOE Office of Science (SC) Contributing Org.: NPLQCD Collaboration OSTI Identifier: 1221536 Alternate Identifier(s): OSTI ID: 1224626 Report Number(s): JLAB-THY-15-2045; DOE/OR/23177-3346; arXiv:1505.02422 Journal ID: ISSN 0031-9007; PRLTAO Grant/Contract Number: OCI-1053575; PHY1206498; PHY12-05778; AC02-05CH11231; DOE DE-SC0013477; SC0010337-ER42045; FG02-04ER41302; FG02-00ER41132; AC05-00OR22725; SC0010495; AC05-06OR23177 Resource Type: Accepted Manuscript Journal Name: Physical Review Letters Additional Journal Information: Journal Volume: 115; Journal Issue: 13; Journal ID: ISSN 0031-9007 Publisher: American Physical Society (APS) Country of Publication: United States Language: English Subject: 72 PHYSICS OF ELEMENTARY PARTICLES AND FIELDS Citation Formats
Beane, Silas R., Chang, Emmanuel, Detmold, William, Orginos, Kostas, Parreño, Assumpta, Savage, Martin J., and Tiburzi, Brian C. Ab initio calculation of the $np \to d ³$ radiative capture process. United States: N. p., 2015. Web. doi:10.1103/PhysRevLett.115.132001.
Beane, Silas R., Chang, Emmanuel, Detmold, William, Orginos, Kostas, Parreño, Assumpta, Savage, Martin J., & Tiburzi, Brian C. Ab initio calculation of the $np \to d ³$ radiative capture process. United States. doi:10.1103/PhysRevLett.115.132001.
Beane, Silas R., Chang, Emmanuel, Detmold, William, Orginos, Kostas, Parreño, Assumpta, Savage, Martin J., and Tiburzi, Brian C. Thu . "Ab initio calculation of the $np \to d ³$ radiative capture process". United States. doi:10.1103/PhysRevLett.115.132001. https://www.osti.gov/servlets/purl/1221536.
@article{osti_1221536,
title = {Ab initio calculation of the $np \to d ³$ radiative capture process}, author = {Beane, Silas R. and Chang, Emmanuel and Detmold, William and Orginos, Kostas and Parreño, Assumpta and Savage, Martin J. and Tiburzi, Brian C.}, abstractNote = {In this study, lattice QCD calculations of two-nucleon systems are used to isolate the short-distance two-body electromagnetic contributions to the radiative capture process $np \to d\gamma$, and the photo-disintegration processes $\gamma^{(\ast)} d \to np$. In nuclear potential models, such contributions are described by phenomenological meson-exchange currents, while in the present work, they are determined directly from the quark and gluon interactions of QCD. Calculations of neutron-proton energy levels in multiple background magnetic fields are performed at two values of the quark masses, corresponding to pion masses of $m_\pi \sim 450$ and 806 MeV, and are combined with pionless nuclear effective field theory to determine these low-energy inelastic processes. Extrapolating to the physical pion mass, a cross section of $\sigma^{lqcd}(np\to d\gamma)=332.4({\tiny \begin{array}{l}+5.4 \\ - 4.7\end{array}})\ mb$ is obtained at an incident neutron speed of $v=2,200\ m/s$, consistent with the experimental value of $\sigma^{expt}(np \to d\gamma) = 334.2(0.5)\ mb$.}, doi = {10.1103/PhysRevLett.115.132001}, journal = {Physical Review Letters}, number = 13, volume = 115, place = {United States}, year = {2015}, month = {9} } Citation information provided by Web of Science
Web of Science |
Before answering the question more or less directly, I'd like to point out that this is a good question that provides an object lesson and opens a foray into the topics of
singular integral equations, analytic continuation and dispersion relations. Here are some references of these more advanced topics: Muskhelishvili, Singular Integral Equations; Courant & Hilbert, Methods of Mathematical Physics, Vol I, Ch 3; Dispersion Theory in High Energy Physics, Queen & Violini; Eden et.al., The Analytic S-matrix. There is also a condensed discussion of `invariant functions' in Schweber, An Intro to Relativistic QFT Ch13d.
The quick answer is that, for $m^2 \in\mathbb{R}$, there's no "shortcut." One must
choose a path around the singularities in the denominator. The appropriate choice is governed by the boundary conditions of the problem at hand. The $+i\epsilon$ "trick" (it's not a "trick") simply encodes the boundary conditions relevant for causal propagation of particles and antiparticles in field theory.
We briefly study the analytic form of $G(x-y;m)$ to demonstrate some of these features.
Note, first, that for real values of $p^2$, the singularity in the denominator of the integrand signals the presence of (a) branch point(s). In fact, [Huang,
Quantum Field Theory: From Operators to Path Integrals, p29] the Feynman propagator for the scalar field (your equation) may be explicitly evaluated:\begin{align}G(x-y;m) &= \lim_{\epsilon \to 0} \frac{1}{(2 \pi)^4} \int d^4p \, \frac{e^{-ip\cdot(x-y)}}{p^2 - m^2 + i\epsilon} \nonumber \\&= \left \{ \begin{matrix}-\frac{1}{4 \pi} \delta(s) + \frac{m}{8 \pi \sqrt{s}} H_1^{(1)}(m \sqrt{s}) & \textrm{ if }\, s \geq 0 \\ -\frac{i m}{ 4 \pi^2 \sqrt{-s}} K_1(m \sqrt{-s}) & \textrm{if }\, s < 0.\end{matrix} \right.\end{align}where $s=(x-y)^2$.
The first-order Hankel function of the first kind $H^{(1)}_1$ has a logarithmic branch point at $x=0$; so does the modified Bessel function of the second kind, $K_1$. (Look at the small $x$ behavior of these functions to see this.)
A branch point indicates that the Cauchy-Riemann conditions have broken down at $x=0$ (or $z=x+iy=0$). And the fact that these singularities are logarithmic is an indication that we have an endpoint singularity [eg. Eden et. al., Ch 2.1]. (To see this, consider $m=0$, then the integrand, $p^{-2}$, has a zero at the lower limit of integration in $dp^2$.)
Coming back to the question of boundary conditions, there is a good discussion in Sakurai,
Advanced Quantum Mechanics, Ch4.4 [NB: "East Coast" metric]. You can see that for large values of $s>0$ from the above expression that we have an outgoing wave from the asymptotic form of the Hankel function.
Connecting it back to the original references I cited above, the $+i\epsilon$ form is a version of the Plemelj formula [Muskhelishvili]. And the expression for the propagator is a type of Cauchy integral [Musk.; Eden et.al.]. And this notions lead quickly to the topics I mentioned above -- certainly a rich landscape for research.This post imported from StackExchange Physics at 2014-07-13 04:38 (UCT), posted by SE-user MarkWayne |
Let $Y^3$ be a handlebody with boundary $\Sigma$. By definition, there is some associated vector $v_{WRT}(Y^3)\in Z(\Sigma)$, the (finite dimensional) Hilbert space associated to $\Sigma$ by the Witten-Reshetikhin-Turaev TQFT. I'd like to understand what this vector is.
In short, $Z(\Sigma)$ is a space of sections of a line bundle over the $SU(2)$ character variety of $\Sigma$. I am hoping that the section $v_{WRT}(Y^3)$ achieves its maximum value (with respect to the canonical inner product on the line bundle) on the Lagrangian submanifold of the character variety consisting of those representations which extend to $Y^3$. [EDIT: there is a good reason to believe this holds, since then high powers of the section will concentrate on this Lagrangian, giving Volume Conjecture-like convergence to the classical Lagrangian intersection theory as the level of the TQFT goes to infinity]
In more detail, let's discuss an explicit description of $Z(\Sigma)$. There is a natural line bundle $\mathcal L$ over the character variety $X:=\operatorname{Hom}(\pi_1(\Sigma),SU(2))/\\!/SU(2)$. There is a natural symplectic form on $X$, and choosing a complex structure on $\Sigma$ equips $X$ with a complex structure which together with the symplectic form makes $X$ a Kahler manifold. Then $Z(\Sigma)$ is the Hilbert space of square integrable holomorphic sections of $\mathcal L$ ($\mathcal L$ carries a natural inner product, and the curvature form of the induced connection coincides with the natural symplectic form on $X$).
My question is then: how can one describe $v(Y^3)\in Z(\Sigma)$? Does the corresponding section achieve its maximum value on the Lagrangian subvariety of $X$ comprised of those characters of $\pi_1(\Sigma)$ extending to characters of $\pi_1(Y)$?
A comment: answering this question for an arbitrary $3$-manifold $Y^3$ seems unlikely to yield a clean answer, since it includes as a special case calculating the value of the WRT TQFT applied to $Y$ (and the description of this requires the introduction of a whole bunch of extra stuff, e.g. surgery diagrams for $Y^3$, etc.). This is why I am restricting to the case that $Y^3$ is a handlebody, in hopes that in this special case, there is a clean answer to this question.
This post imported from StackExchange MathOverflow at 2014-09-04 08:37 (UCT), posted by SE-user John Pardon |
For example, in this paper on page 21 the authors write the vev that breaks $SO(10)$ to $SU(4)\times SU(2) \times SU(2)$
$$ <54>= 1/5 \cdot diag( -2,-2,-2,-2,-2,-2,3,3,3,3) \omega_s$$
where $\omega_s$ denotes the scale.
What do the authors mean by this?
The Higgs field or Higgs fields that develop a vev are elements of the $54$ representation of $SO(10)$. Because $10 \otimes 10 = 1_s \oplus 54_s \oplus45_a$, we can write each element of $54$ as a $10 \times 10$ matrix.
This is similar how one determines the particle content of the $10$ representation of $SU(5)$, written as $5 \times 5$ matrix, by using $5 \otimes 5 =10 \oplus \ldots$.
The quantum numbers of the 10 inedpendent fields $\in$ 10 are given by
$$ QN(10_{ij})= QN(5_i) + QN(5_j) ,$$
where $10_{ij}$ denotes the $ij$ element of the $5 \times 5$ matrix.
This yields
Completely equivalently we can determine the quantum numbers of the Higgs fields in the $54$ representation of $S0(10)$ and write them in a $10 \times 10$ matrix.
How does $ <54>= 1/5 \cdot diag( -2,-2,-2,-2,-2,-2,3,3,3,3) \omega_s$ tell me which of these $54$ Higgs fields gets a vev?
For concreteness let's consider a simplified setup. A computation similar to $ QN(10_{ij})= QN(5_i) + QN(5_j) ,$, which yields for example the 10 of $SU(5)$ as quoted above, yields a matrix for the Higgs bosons. For example, in a 2×2 case: $\begin{pmatrix} A & B \\ C& D \end{pmatrix}$, where
$A,B,C,D $ denote four different Higgs fields. Now given a vev = diag(1,−1)w, does this mean the Higgs field A gets a vev w and the Higgs field D gets a vev −w, while C and D do not get a vev? |
On the variational representation of monotone operators
Dipartimento di Matematica, dell'Università degli Studi di Trento, via Sommarive 14,38050 Povo di Trento, Italy
$V$
$z'\in V'$
$\alpha: V\to {\mathcal P}(V')$
$\alpha(u) \ni z'$
$D_tu + \alpha(u) \ni z'$ representative function
$f_\alpha: V \!\times\! V'\to \mathbb{R}\cup \{+\infty\}$
$f_\alpha(v,v') \ge \langle v',v\rangle\quad\;\forall (v,v'), \qquad\quadf_\alpha(v,v') = \langle v',v\rangle\;\;\Leftrightarrow\;\;\; v'\in \alpha(v).$ Mathematics Subject Classification:47H05, 49J40, 58E. Citation:Augusto VisintiN. On the variational representation of monotone operators. Discrete & Continuous Dynamical Systems - S, 2017, 10 (4) : 909-918. doi: 10.3934/dcdss.2017046
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C. Baiocchi and A. Capelo,
[3] [4]
H. H. Bauschke and X. Wang,
The kernel average for two convex functions and its applications to the extension and representation of monotone operators,
[5]
H. Brezis,
[6]
H. Brezis and I. Ekeland, Un principe variationnel associé à certaines équations paraboliques. Ⅰ. Le cas indépendant du temps and Ⅱ. Le cas dépendant du temps,
[7]
F. Browder,
[8] [9] [10] [11]
I. Ekeland and R. Temam,
[12]
K. Fan, A minimax inequality and applications,
[13]
W. Fenchel,
[14]
S. Fitzpatrick,
Representing monotone operators by convex functions, Workshop/Miniconference on Functional Analysis and Optimization, Canberra, 1988,
[15]
N. Ghoussoub,
[16] [17] [18]
J. L. Lions,
[19] [20] [21] [22] [23] [24]
J. -P. Penot,
A representation of maximal monotone operators by closed convex functions and its impact on calculus rules,
[25] [26] [27]
T. Roche, R. Rossi and U. Stefanelli,
Stability results for doubly nonlinear differential inclusions by variational convergence,
[28]
R. T. Rockafellar,
Convex Analysis, Princeton University Press, Princeton, 1969.
Google Scholar
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A. Visintin, Structural compactness and stability of pseudo-monotone flows, forthcoming.Google Scholar
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Dalila Azzam-Laouir, Warda Belhoula, Charles Castaing, M. D. P. Monteiro Marques.
Multi-valued perturbation to evolution problems involving time dependent maximal monotone operators.
[2]
A. C. Eberhard, J-P. Crouzeix.
Existence of closed graph, maximal, cyclic pseudo-monotone relations and revealed preference theory.
[3]
JIAO CHEN, WEI DAI, GUOZHEN LU.
$L^p$ boundedness for maximal functions associated with multi-linear pseudo-differential operators.
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Xiao Ding, Deren Han.
A modification of the forward-backward splitting method for maximal monotone mappings.
[13]
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Separable Lyapunov functions for monotone systems: Constructions and limitations.
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Maximal dissipativity of a class of elliptic degenerate operators in
weighted $L^2$ spaces.
[15] [16] [17] [18] [19]
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Solving monotone inclusions involving parallel sums of linearly composed maximally monotone operators.
[20]
Anurag Jayswal, Ashish Kumar Prasad, Izhar Ahmad.
On minimax fractional programming problems involving generalized $(H_p,r)$-invex functions.
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Measuring K$^0_{\rm S}$K$^{\rm \pm}$ interactions using Pb-Pb collisions at ${\sqrt{s_{\rm NN}}=2.76}$ TeV
(Elsevier, 2017-11)
We present the first ever measurements of femtoscopic correlations between the K$^0_{\rm S}$ and K$^{\rm \pm}$ particles. The analysis was performed on the data from Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV measured ...
Measurement of D-meson production at mid-rapidity in pp collisions at ${\sqrt{s}=7}$ TeV
(Springer, 2017-08)
The production cross sections of the prompt charmed mesons ${\rm D^0}$, ${\rm D^+}$, ${\rm D^{*+}}$ and ${\rm D_s^+}$ were measured at mid-rapidity in proton-proton collisions at a centre-of-mass energy $\sqrt{s}=7$ TeV ...
Charged-particle multiplicity distributions over a wide pseudorapidity range in proton-proton collisions at √s = 0.9, 7, and 8 TeV
(Springer, 2017-12-09)
We present the charged-particle multiplicity distributions over a wide pseudorapidity range (−3.4<η<5.0 ) for pp collisions at √s=0.9,7 , and 8 TeV at the LHC. Results are based on information from the Silicon Pixel Detector ... |
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Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Multiplicity dependence of two-particle azimuthal correlations in pp collisions at the LHC
(Springer, 2013-09)
We present the measurements of particle pair yields per trigger particle obtained from di-hadron azimuthal correlations in pp collisions at $\sqrt{s}$=0.9, 2.76, and 7 TeV recorded with the ALICE detector. The yields are ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Coherent $\rho^0$ photoproduction in ultra-peripheral Pb-Pb collisions at $\mathbf{\sqrt{\textit{s}_{\rm NN}}} = 2.76$ TeV
(Springer, 2015-09)
We report the first measurement at the LHC of coherent photoproduction of $\rho^0$ mesons in ultra-peripheral Pb-Pb collisions. The invariant mass and transverse momentum distributions for $\rho^0$ production are studied ...
Inclusive, prompt and non-prompt J/ψ production at mid-rapidity in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2015-07-10)
The transverse momentum (p T) dependence of the nuclear modification factor R AA and the centrality dependence of the average transverse momentum 〈p T〉 for inclusive J/ψ have been measured with ALICE for Pb-Pb collisions ... |
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Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE
(Elsevier, 2017-11)
Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ... |
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J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ... |
Repeated Roots of The Characteristic Equation
Recall that if $a\frac{d^2y}{dt^2} + b \frac{dy}{dt} + c = 0$ is a second order linear homogenous differential equation, then the characteristic equation for this differential equation is $ar^2 + br + c = 0$. We saw that if the roots of this equation, call them $r_1$ and $r_2$ are both real and distinct, then for $C$ and $D$ as constants, $y = Ce^{r_1t} + De^{r_2t}$ gives us solutions to our differential equation. We also saw that if the roots of this equation are complex numbers with $r_1 = \lambda + \mu i$ and $r_2 = \lambda - \mu i$ then for $C$ and $D$ as constants, $y = Ce^{\lambda t} \cos (\mu t) + e^{\lambda t} \sin (\mu t)$ is the general solution to this differential equation.
We will now look at the case in which the roots of the characteristic polynomial are real and not distinct, that is $r_1 = r_2$. Suppose that $r_1 = r_2$. Then we must have that the descriminant of the characteristic equation $ar^2 + br + c = 0$ is zero, that is $b^2 - 4ac = 0$. Thus by applying the quadratic formula, we see that the solutions are:(1)
Thus we can see that $y_1 = e^{-bt/2a}$ is a solution to our differential equation. We need two different solutions to possibly form a fundamental set of solutions though. In order to find another solution, for some function $v(t)$ assume that $y = v(t)y_1(t) = v(t)e^{-bt/2a}$ is a solution to our differential equation. We note that the derivatives of this solution are:(2)
Substituting the values of $y$, $y'$, and $y''$ into our differential equation and we have that:(4)
Note that since $b^2 - 4ac = 0$ we have that $b^2 = 4ac$ and $\frac{b^2}{4a} = c$. Therefore $c - \frac{b^2}{4a} = 0$ and for some constants $C$ and $D$, the equation above becomes simply:(5)
Since $y = v(t) e^{-bt}{2a}$ we have that:(6)
Therefore we have obtained another solution to our differential equation, namely $y_2(t) = te^{-bt/2a}$. If we look at the Wronskian of $y_1(t) = e^{-bt/2a}$ and $y_2(t) = te^{-bt/2a}$ we see that:(7)
Note that the Wronskian $W(y_1, y_2)$ is never equal to zero, and so $y_1(t) = e^{-bt/2a}$ and $y_2(t) = te^{-bt/2a}$ form a fundamental set of solutions and so the general solution for the differential equation $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + cy = 0$ for some constants $C$ and $D$ is given by:(8) |
section 3.8 exercise
For each function, find a domain on which the function is one-to-one and non-decreasing, then find an inverse of the function on this domain.
1. \(f\left(x\right)=\left(x-4\right)^{2}\)
2. \(f\left(x\right)=\left(x+2\right)^{2}\)
3. \(f\left(x\right)=12-x^{2}\)
4. \(f\left(x\right)=9-x^{2}\)
5. \(f\left(x\right)=3x^{3} +1\)
6. \(f\left(x\right)=4-2x^{3}\)
Find the inverse of each function.
7. \(f\left(x\right)=9+\sqrt{4x-4}\)
8. \(f\left(x\right)=\sqrt{6x-8} +5\)
9. \(f\left(x\right)=9+2\sqrt[{3}]{x}\)
10. \(f\left(x\right)=3-\sqrt[{3}]{x}\)
11. \(f\left(x\right)=\frac{2}{x+8}\)
12. \(f\left(x\right)=\frac{3}{x-4}\)
13. \(f\left(x\right)=\frac{x+3}{x+7}\)
14. \(f\left(x\right)=\frac{x-2}{x+7}\)
15. \(f\left(x\right)=\frac{3x+4}{5-4x}\)
16. \(f\left(x\right)=\frac{5x+1}{2-5x}\)
Police use the formula \(v=\sqrt{20L}\) to estimate the speed of a car, \(v\), in miles per hour, based on the length, \(L\), in feet, of its skid marks when suddenly braking on a dry, asphalt road.
17. At the scene of an accident, a police officer measures a car’s skid marks to be 215 feet long. Approximately how fast was the car traveling?
18. At the scene of an accident, a police officer measures a car’s skid marks to be 135 feet long. Approximately how fast was the car traveling?
The formula \(v=\sqrt{2.7r}\) models the maximum safe speed, \(v\), in miles per hour, at which a car can travel on a curved road with radius of curvature \(r\), in feet.
19. A highway crew measures the radius of curvature at an exit ramp on a highway as 430 feet. What is the maximum safe speed?
20. A highway crew measures the radius of curvature at a tight corner on a highway as 900 feet. What is the maximum safe speed?
21. A drainage canal has a cross-section in the shape of a parabola. Suppose that the canal is 10 feet deep and 20 feet wide at the top. If the water depth in the ditch is 5 feet, how wide is the surface of the water in the ditch? [UW]
22. Brooke is located 5 miles out from the nearest point \(A\) along a straight shoreline in her sea kayak. Hunger strikes and she wants to make it to Kono’s for lunch; see picture. Brooke can paddle 2 mph and walk 4 mph. [UW]
a. If she paddles along a straight line course to the shore, find an expression that computes the total time to reach lunch in terms of the location where Brooke beaches her kayak.
b. Determine the total time to reach Kono’s if she paddles directly to the point \(A\). c. Determine the total time to reach Kono’s if she paddles directly to Kono’s. d. Do you think your answer to b or c is the minimum time required for Brooke to reach lunch? e. Determine the total time to reach Kono’s if she paddles directly to a point on the shore half way between point A and Kono’s. How does this time compare to the times in parts b or c? Do you need to modify your answer to part d?
23. Clovis is standing at the edge of a dropoff, which slopes 4 feet downward from him for every 1 horizontal foot. He launches a small model rocket from where he is standing. With the origin of the coordinate system located where he is standing, and the \(x\)-axis extending horizontally, the path of the rocket is described by the formula \(y=-2x^{2} +120x\). [UW]
a. Give a function \(h=f(x)\) relating the height \(h\) of the rocket above the sloping ground to its \(x\)-coordinate.
b. Find the maximum height of the rocket above the sloping ground. What is its \(x\)-coordinate when it is at its maximum height? c. Clovis measures the height \(h\) of the rocket above the sloping ground while it is going up. Give a function \(x=g\left(h\right)\) relating the \(x\)-coordinate of the rocket to \(h\). d. Does the function from (c) still work when the rocket is going down? Explain.
24. A trough has a semicircular cross section with a radius of 5 feet. Water starts flowing into the trough in such a way that the depth of the water is increasing at a rate of 2 inches per hour. [UW]
a. Give a function \(w=f\left(t\right)\) relating the width w of the surface of the water to the time \(t\), in hours. Make sure to specify the domain and compute the range too.
b. After how many hours will the surface of the water have width of 6 feet? c. Give a function \(t=f^{-1} \left(w\right)\) relating the time to the width of the surface of the water. Make sure to specify the domain and compute the range too. Answer
1. Domain \((4, \infty)\) Inverse \(f^{-1} (x) = \sqrt{x} + 4\)
3. Domain \((-\infty, 0)\) Inverse \(f^{-1} (x) = -\sqrt{12 - x}\)
5. Domain \((-\infty, 0)\) Inverse \(f^{-1} (x) = \sqrt[3]{\dfrac{x - 1}{3}}\)
7. \(f^{-1} (x) = \dfrac{(x - 9)^2}{4} + 1\)
9. \(f^{-1} (x) = (\dfrac{x - 9}{2})^3\)
11. \(f^{-1} (x) = \dfrac{2 - 8x}{x}\)
13. \(f^{-1} (x) = \dfrac{3 - 7x}{x - 1}\)
15. \(f^{-1} (x) = \dfrac{5x - 4}{3 + 4x}\)
17. 65.574 mph
19. 34. 073 mph
21. 14.142 feet |
February 13th, 2018, 02:49 PM
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Posterior Distribution from Beta Density with Exponential Prior
Let $X_1,...,X_n$ be iid random variables with a common density function given by:
$f(x|\theta)=\theta x^{\theta-1}$
for $x\in[0,1]$ and $\theta>0$.
Put a prior distribution on $\theta$ which is $EXP(2)$, where $2$ is the mean of the exponential distribution. Obtain the posterior density function of $\theta$. Do you recognize this distribution?
Clearly $f(x|\theta)$ is a beta distribution. So this implies that the posterior distribution is
$$f(\theta|x)\propto f(x|\theta)\pi(\theta)=\frac{\theta x^{\theta-1}e^{-\theta/2}}{2}$$
But I don't really recognize the distribution at all. Was my calculation incorrect? If not, what is the name of this distribution?
Tags beta, density, distribution, exponential, posterior, prior
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In the previous section, we derived two important properties of logarithms, which allowed us to solve some basic exponential and logarithmic equations.
properties of logs
Inverse Properties :
\(\log _{b} \left(b^{x} \right)=x\)
\(b^{\log _{b} x} =x\)
Exponential Property :
\(\log _{b} \left(A^{r} \right)=r\log _{b} \left(A\right)\)
Change of Base :
\(\log _{b} \left(A\right)=\frac{\log _{c} (A)}{\log _{c} (b)}\)
While these properties allow us to solve a large number of problems, they are not sufficient to solve all problems involving exponential and logarithmic equations.
properties of logs
Sum of Logs Property :
\(\log _{b} \left(A\right)+\log _{b} \left(C\right)=\log _{b} (AC)\)
Difference of Logs Property :
\(\log _{b} \left(A\right)-\log _{b} \left(C\right)=\log _{b} \left(\frac{A}{C} \right)\)
It’s just as important to know what properties logarithms do not satisfy as to memorize the valid properties listed above. In particular, the logarithm is not a linear function, which means that it does not distribute: log(\(A\) + \(B\)) \(\ne\) log(\(A\)) + log(\(B\)).
To help in this process we offer a proof to help solidify our new rules and show how they follow from properties you’ve already seen.
Let \(a=\log _{b} \left(A\right)\) and \(c=\log _{b} \left(C\right)\).
By definition of the logarithm, \(b^{a} =A\) and \(b^{c} =C\).
Using these expressions, \(AC=b^{a} b^{c}\)
Using exponent rules on the right, \(AC=b^{a+c}\)
Taking the log of both sides, and utilizing the inverse property of logs,
\(\log _{b} \left(AC\right)=\log _{b} \left(b^{a+c} \right)=a+c\)
Replacing \(a\) and \(c\) with their definition establishes the result
\(\log _{b} \left(AC\right)=\log _{b} A+\log _{b} C\)
The proof for the difference property is very similar.
With these properties, we can rewrite expressions involving multiple logs as a single log, or break an expression involving a single log into expressions involving multiple logs.
Example 1
Write \(\log _{3} \left(5\right)+\log _{3} \left(8\right)-\log _{3} \left(2\right)\) as a single logarithm.
Solution
Using the sum of logs property on the first two terms,
\(\log _{3} \left(5\right)+\log _{3} \left(8\right)=\log _{3} \left(5\cdot 8\right)=\log _{3} \left(40\right)\)
This reduces our original expression to \(\log _{3} \left(40\right)-\log _{3} \left(2\right)\)
Then using the difference of logs property,
\(\log _{3} \left(40\right)-\log _{3} \left(2\right)=\log _{3} \left(\frac{40}{2} \right)=\log _{3} \left(20\right)\)
Example 2
Evaluate \(2\log \left(5\right)+\log \left(4\right)\) without a calculator by first rewriting as a single logarithm.
Solution
On the first term, we can use the exponent property of logs to write
\(2\log \left(5\right)=\log \left(5^{2} \right)=\log \left(25\right)\)
With the expression reduced to a sum of two logs, \(\log \left(25\right)+\log \left(4\right)\), we can utilize the sum of logs property
\(\log \left(25\right)+\log \left(4\right)=\log (4\cdot 25)=\log (100)\)
Since 100 = 10\({}^{2}\), we can evaluate this log without a calculator:
\(\log (100)=\log \left(10^{2} \right)=2\)
Exercise
Without a calculator evaluate by first rewriting as a single logarithm:
\(\log _{2} \left(8\right)+\log _{2} \left(4\right)\)
Answer
\(\log _{2} \left(8\cdot 4\right)=\log _{2} \left(32\right)=\log _{2} \left(2^{5} \right)=5\)
Example 3
Rewrite \(\ln \left(\frac{x^{4} y}{7} \right)\) as a sum or difference of logs
Solution
First, noticing we have a quotient of two expressions, we can utilize the difference property of logs to write
\(\ln \left(\frac{x^{4} y}{7} \right)=\ln \left(x^{4} y\right)-\ln (7)\)
Then seeing the product in the first term, we use the sum property
\(\ln \left(x^{4} y\right)-\ln (7)=\ln \left(x^{4} \right)+\ln (y)-\ln (7)\)
Finally, we could use the exponent property on the first term
\(\ln \left(x^{4} \right)+\ln (y)-\ln (7)=4\ln (x)+\ln (y)-\ln (7)\)
Interestingly, solving exponential equations was not the reason logarithms were originally developed. Historically, up until the advent of calculators and computers, the power of logarithms was that these log properties reduced multiplication, division, roots, or powers to be evaluated using addition, subtraction, division and multiplication, respectively, which are much easier to compute without a calculator. Large books were published listing the logarithms of numbers, such as in the table to the right. To find the product of two numbers, the sum of log property was used. Suppose for example we didn’t know the value of 2 times 3. Using the sum property of logs:
\(\log (2\cdot 3)=\log (2)+\log (3)\)
Using the log table, \[\log (2\cdot 3)=\log (2)+\log (3)=0.3010300+0.4771213=0.7781513\]
We can then use the table again in reverse, looking for 0.7781513 as an output of the logarithm. From that we can determine:
\(\log (2\cdot 3)=0.7781513=\log (6).\)
By using addition and the table of logs, we were able to determine\(2\cdot 3=6\).
Likewise, to compute a cube root like \(\sqrt[{3}]{8}\)
\(\log (\sqrt[{3}]{8} )==\log \left(8^{1/3} \right)=\frac{1}{3} \log (8)=\frac{1}{3} (0.9030900)=0.3010300=\log (2)\)
So \(\sqrt[{3}]{8} =2\).
Although these calculations are simple and insignificant, they illustrate the same idea that was used for hundreds of years as an efficient way to calculate the product, quotient, roots, and powers of large and complicated numbers, either using tables of logarithms or mechanical tools called slide rules.
These properties still have other practical applications for interpreting changes in exponential and logarithmic relationships.
Example 4
Recall that in chemistry,\(pH=-\log \left(\left[H^{+} \right]\right)\). If the concentration of hydrogen ions in a liquid is doubled, what is the affect on pH?
Solution
Suppose \(C\) is the original concentration of hydrogen ions, and \(P\) is the original pH of the liquid, so \(P=-\log \left(C\right)\). If the concentration is doubled, the new concentration is \(2C\). Then the pH of the new liquid is
\(pH=-\log \left(2C\right)\)
Using the sum property of logs,
\(pH=-\log \left(2C\right)=-\left(\log (2)+\log (C)\right)=-\log (2)-\log (C)\)
Since \(P=-\log \left(C\right)\), the new pH is
\(pH=P-\log (2)=P-0.301\)
When the concentration of hydrogen ions is doubled, the pH decreases by 0.301.
Log properties in solving equations
The logarithm properties often arise when solving problems involving logarithms. First, we’ll look at a simpler log equation.
Example 5
Solve \(\log (2x-6)=3\).
Solution
To solve for \(x\), we need to get it out from inside the log function. There are two ways we can approach this.
Method 1: Rewrite as an exponential.
Recall that since the common log is base 10, \(\log (A)=B\) can be rewritten as the exponential \(10^{B} =A\). Likewise, \(\log (2x-6)=3\) can be rewritten in exponential form as
\(10^{3} =2x-6\)
Method 2: Exponentiate both sides.
If \(A=B\), then \(10^{A} =10^{B}\). Using this idea, since \(\log (2x-6)=3\), then \(10^{\log (2x-6)} =10^{3}\). Use the inverse property of logs to rewrite the left side and get \(2x-6=10^{3}\).
Using either method, we now need to solve \(2x-6=10^{3}\). Evaluate \(10^{3}\) to get
\(2x-6=1000\) Add 6 to both sides
\(2x=1006\) Divide both sides by 2 \(x=503\)
Occasionally the solving process will result in extraneous solutions – answers that are outside the domain of the original equation. In this case, our answer looks fine.
Example 6
Solve \(\log (50x+25)-\log (x)=2\).
Solution
In order to rewrite in exponential form, we need a single logarithmic expression on the left side of the equation. Using the difference property of logs, we can rewrite the left side:
\(\log \left(\frac{50x+25}{x} \right)=2\)
Rewriting in exponential form reduces this to an algebraic equation:
\(\frac{50x+25}{x} =10^{2} =100\) Multiply both sides by \(x\)
\(50x+25=100x\) Combine like terms \(25=50x\) Divide by 50 \(x=\frac{25}{50} =\frac{1}{2}\)
Checking this answer in the original equation, we can verify there are no domain issues, and this answer is correct.
Exercise
Solve \(\log (x^{2} -4)=1+\log (x+2)\).
Answer
\(\log (x^{2} -4)=1+\log (x+2)\) Move both logs to one side
\(\log \left(x^{2} -4\right)-\log \left(x+2\right)=1\) Use the difference property of logs
\(\log \left(\frac{x^{2} -4}{x+2} \right)=1\) Factor
\(\log \left(\frac{(x+2)(x-2)}{x+2} \right)=1\) Simplify
\(\log \left(x-2\right)=1\) Rewrite as an exponential
\(10^{1} =x-2\) Add 2 to both sides
\(x=12\)
Example 7
Solve \(\ln (x+2)+\ln (x+1)=\ln (4x+14)\).
Solution
\(\ln (x+2)+\ln (x+1)=\ln (4x+14)\) Use the sum of logs property on the right
\(\ln \left((x+2)(x+1)\right)=\ln (4x+14)\) Expand \(\ln \left(x^{2} +3x+2\right)=\ln (4x+14)\)
We have a log on both side of the equation this time. Rewriting in exponential form would be tricky, so instead we can exponentiate both sides.
\(e^{\ln \left(x^{2} +3x+2\right)} =e^{\ln (4x+13)}\) Use the inverse property of logs
\(x^{2} +3x+2=4x+14\) Move terms to one side \(x^{2} -x-12=0\) Factor \((x+4)(x-3)=0\) \(x = -4\) or \(x = 3\).
Checking our answers, notice that evaluating the original equation at \(x = -4\) would result in us evaluating \(\ln (-2)\), which is undefined. That answer is outside the domain of the original equation, so it is an extraneous solution and we discard it. There is one solution: \(x = 3\).
More complex exponential equations can often be solved in more than one way. In the following example, we will solve the same problem in two ways – one using logarithm properties, and the other using exponential properties.
Example 8a
In 2008, the population of Kenya was approximately 38.8 million, and was growing by 2.64% each year, while the population of Sudan was approximately 41.3 million and growing by 2.24% each year(World Bank, World Development Indicators, as reported on http://www.google.com/publicdata, retrieved August 24, 2010). If these trends continue, when will the population of Kenya match that of Sudan?
Solution
We start by writing an equation for each population in terms of \(t\), the number of years after 2008.
\(\begin{array}{l} {Kenya(t)=38.8(1+0.0264)^{t} } \\ {Sudan(t)=41.3(1+0.0224)^{t} } \end{array}\)
To find when the populations will be equal, we can set the equations equal
\(38.8(1.0264)^{t} =41.3(1.0224)^{t}\)
For our first approach, we take the log of both sides of the equation.
\(\log \left(38.8(1.0264)^{t} \right)=\log \left(41.3(1.0224)^{t} \right)\)
Utilizing the sum property of logs, we can rewrite each side,
\(\log (38.8)+\log \left(1.0264^{t} \right)=\log (41.3)+\log \left(1.0224^{t} \right)\)
Then utilizing the exponent property, we can pull the variables out of the exponent
\(\log (38.8)+t\log \left(1.0264\right)=\log (41.3)+t\log \left(1.0224\right)\)
Moving all the terms involving \(t\) to one side of the equation and the rest of the terms to the other side,
\(t\log \left(1.0264\right)-t\log \left(1.0224\right)=\log (41.3)-\log (38.8)\)
Factoring out the \(t\) on the left,
\(t\left(\log \left(1.0264\right)-\log \left(1.0224\right)\right)=\log (41.3)-\log (38.8)\)
Dividing to solve for \(t\)
\(t=\frac{\log (41.3)-\log (38.8)}{\log \left(1.0264\right)-\log \left(1.0224\right)} \approx 15.991\) years until the populations will be equal.
Example 8b
Solve the problem above by rewriting before taking the log.
Solution
Starting at the equation
\(38.8(1.0264)^{t} =41.3(1.0224)^{t}\)
Divide to move the exponential terms to one side of the equation and the constants to the other side
\(\frac{1.0264^{t} }{1.0224^{t} } =\frac{41.3}{38.8}\)
Using exponent rules to group on the left,
\(\left(\frac{1.0264}{1.0224} \right)^{t} =\frac{41.3}{38.8}\)
Taking the log of both sides
\(\log \left(\left(\frac{1.0264}{1.0224} \right)^{t} \right)=\log \left(\frac{41.3}{38.8} \right)\)
Utilizing the exponent property on the left,
\(t\log \left(\frac{1.0264}{1.0224} \right)=\log \left(\frac{41.3}{38.8} \right)\)
Dividing gives
\(t=\frac{\log \left(\frac{41.3}{38.8} \right)}{\log \left(\frac{1.0264}{1.0224} \right)} \approx 15.991\) years
While the answer does not immediately appear identical to that produced using the previous method, note that by using the difference property of logs, the answer could be rewritten:
\(t=\frac{\log \left(\frac{41.3}{38.8} \right)}{\log \left(\frac{1.0264}{1.0224} \right)} =\frac{\log (41.3)-\log (38.8)}{\log (1.0264)-\log (1.0224)}\)
While both methods work equally well, it often requires fewer steps to utilize algebra before taking logs, rather than relying solely on log properties.
Exercise
Tank A contains 10 liters of water, and 35% of the water evaporates each week. Tank B contains 30 liters of water, and 50% of the water evaporates each week. In how many weeks will the tanks contain the same amount of water?
Answer
Tank A: \(A(t)=10(1-0.35)^{t}\). Tank B: \(B(t)=30(1-0.50)^{t}\)
Solving A(t) = B(t),
\(10(0.65)^{t} =30(0.5)^{t}\) Using the method from Example 8b
\(\frac(0.65)^{t} }(0.5)^{t} } =\frac{30}{10}\) Regroup
\(\left(\frac{0.65}{0.5} \right)^{t} =3\) Simplify
\(\left(1.3\right)^{t} =3\) Take the log of both sides
\(\log \left(\left(1.3\right)^{t} \right)=\log \left(3\right)\) Use the exponent property of logs
\(t\log \left(1.3\right)=\log \left(3\right)\) Divide and evaluate
\(t=\frac{\log \left(3\right)}{\log \left(1.3\right)} \approx 4.1874\) weeks
Important Topics of this Section Inverse Exponential Change of base Sum of logs property Difference of logs property |
Determining Whether a Set is a Vector Space
We have looked at a variety of different vector spaces so far including:
The Vector Space of n-Component Vectors The Vector Space of m x n Matrices The Vector Space of Lines Through the Origin of R2 The Zero Vector Space The Vector Space of Polynomials of Arbitrary Degree The Vector Space of Polynomials of Degree ≤ n The Vector Space of Infinite Sequences The Vector Space of Real Valued Functions
We will now look at some contrived examples of sets under specified operations of addition and scalar multiplication and determine whether or not they are vector spaces.
Example 1 Let $V = \mathbb{R}^2$, and let $u, v \in \mathbb{V}$ such that $u = (u_1, u_2)$ and $v = (v_1, v_2)$. Define addition component-wise, that is $u + v = (u_1 + v_1, u_2 + v_2)$ and define scalar multiplication by $a \in \mathbb{F}$ to be $au = (au_1, 0)$. Determine whether or not this set under these operations is a vector space.
We first note that addition of vectors $u + v$ is defined as standard addition, however, multiplication is not defined standardly. Let $u$ be a vector whose second component is nonzero, that is $u_2 \neq 0$, and consider the existence of a multiplicative identity $1$ such that $1 u = u$. By the definition of scalar multiplication it follows that:(1)
Since $u_2 \neq 0$, it follows that $1u \neq u$ under the operation of scalar multiplication, and so axiom 6 (the existence of a multiplicative identity) does not hold, so $V$ is not a vector space under these prescribed operations.
Example 2 Let $V = \mathbb{R}^2$, and let $u, v, w \in \mathbb{V}$ such that $u = (u_1, u_2)$ and $v = (v_1, v_2)$. Define addition as $u + v = (u_1v_1, u_2v_2)$ and define scalar multiplication by $a \in \mathbb{F}$ to be $au = (au_1, au_2 + 1)$. Determine whether or not this set under these operations is a vector space.
In this example both addition and scalar multiplication are not standard. We will attempt to verify that all ten axioms hold, and will stop verifying if one axiom fails.
1.$u + v = (u_1v_1, u_2v_2) = (v_1u_1, v_2u_2) = v + u$. 2.$u + (v + w) = (u_1[v_1w_1], u_2[v_2w_2]) = (u_1v_1w_1, u_2v_2w_2) = ([u_1v_1]w_1,[u_2v_2]w_2) = (u + v) + w$. 3.The zero vector is $1 = (1, 1)$, that is $1 + u = (1u_1, 1u_2) = (u_1, u_2) = u$. 4.This is where $V$ under the specified operations fails. Notice that for any vector $u$, that additive inverse would have to be $-u = \left ( \frac{1}{u_1}, \frac{1}{u_2} \right)$. However notice that for any vector that contains at least one component that is zero, for example the vector $m = (2, 0)$, no such additive inverse exists since $-m = \left ( \frac{1}{2}, \frac{1}{0} \right)$ is undefined.
Therefore $V$ is not a vector space under the prescribed operations.
Example 3 Let $V = \mathbb{R}^3$, and let $u, v, w \in \mathbb{V}$ such that $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$. Define addition as $u + v = (u_1^2 + v_1^2, u_2^2 + v_2^2, u_3^2 + v_3^2)$ and define scalar multiplication by $a \in \mathbb{F}$ to be $au = (au_1, au_2 + 1)$. Determine whether or not this set under these operations is a vector space.
Once again, we will attempt to verify all ten axioms, and we will stop if at least one axiom fails.
1.$u + v = (u_1^2 + v_1^2, u_2^2 + v_2^2, u_3^2 + v_3^2) = (v_1^2 + u_1^2, v_2^2 + u_2^2, v_3^2 + u_3^2)$. 2.$u + (v + w) = (u_1^2 + [v_1^2 + w_1^2], u_2^2 + [v_2^2 + w_2^2], u_3^2 + [v_3^2 + w_3^2]) = ([u_1^2 + v_1^2] + w_1^2, [u_2^2 + v_2^2] + w_2^2, [u_3^2 + v_3^2] + w_3^2) = (u + v) + w$. 3.No such zero vector exists. Suppose that one did and call it $z$. Then $u + z = (u_1^2 + z_1^2, u_2^2 + z_2^2 + u_3^2 + z_3^2) = (u_1, u_2, u_3)$, or in other words, $u_i^2 + z_i^2 = u_i$ for $i = 1, 2, 3$ and so $z_i = \sqrt{u_i - u_i^2}$, and then $z = \left ( \sqrt{u_1 - u_1^2},\sqrt{u_2 - u_2^2},\sqrt{u_3 - u_3^2} \right)$. But this zero vector only holds for $u$. If $u \neq v$, then $v$ has a different zero vector. |
I assume that you are working over the complex numbers. Let $v_j = \frac{1}{\sqrt{n}}(1, \omega_j, \ldots, \omega_j^{n-1})^T$ for $1 \leq j \leq n$ where $\omega_j = e^{\frac{2\pi i j}{n}}$. The vectors $v_j$ form a basis of $\mathbb{C}^n$ and are eigenvectors of
all circulant matrices. If build a matrix $U$ whose columns are the vectors $v_j$, then $U$ will be invertible and $U^{-1} C U$ will be a diagonal matrix whenever $C$ is a circulant matrix.
Write $Av_j = \lambda_j v_j$ and $B v_j = \mu_j v_j$. Then
$$ \begin{pmatrix} U^{-1} & 0 \\ 0 & U^{-1} \end{pmatrix}\begin{pmatrix} 0 & A \\ B & 0 \end{pmatrix} \begin{pmatrix} U & 0 \\ 0 & U \end{pmatrix} = \begin{pmatrix} 0 & U^{-1} A U \\ U^{-1} B U & 0 \end{pmatrix} = \begin{pmatrix} 0 & \mathrm{diag}(\lambda_1, \ldots, \lambda_n) \\ \mathrm{diag}(\mu_1, \ldots, \mu_n) & 0 \end{pmatrix} = D $$
which shows that $M'$ is similar to $D$. For the matrix $D$ we can find the eigenvalues and eigenvectors explicitly. Note that we have
$$ De_i = \begin{cases} \mu_i e_{n+i} & 1 \leq i \leq n, \\ \lambda_{i-n} e_{i - n} & n + 1 \leq i \leq 2n. \end{cases} $$
and so each $\mathrm{span} \{ e_i, e_{n+i} \}$ (for $1 \leq i \leq n$) is a two-dimensional invariant subspace of $D$. Conjugating $D$ by a change of basis matrix, we see that $D$ is similar to the block diagonal matrix
$$ \begin{pmatrix} \begin{pmatrix} 0 & \lambda_1 \\ \mu_1 & 0 \end{pmatrix} & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \begin{pmatrix} 0 & \lambda_n \\ \mu_n & 0 \end{pmatrix} \end{pmatrix} $$
whose eigenvalues are $\pm \sqrt{\lambda_i \mu_i}$ and whose eigenvectors can be found explicitly (I leave that to you). Note that $M'$ is not necessarily diagonalizable which can be seen by taking $1 \times 1$ blocks and considering
$$ M' = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$
In fact, $M'$ will be diagonalizable if and only if $\lambda_i \mu_i \neq 0$ for all $1 \leq i \leq n$.
Regarding the generalization, I doubt you'll be able to say something nice. Even the case where the blocks are $1 \times 1$ is unpleasant and requires you to analyze the possible eigenvalues and eigenvectors of all $3 \times 3$ matrices having zeros on the diagonal. |
Let me come back to your question for more practical purposes. In my former theoretical approach (I keep all my previous notations), I gave a necessary and sufficient condition for an element $\alpha\in K^*$ to be a global $m$-th power, but in practice this criterion works well only to give a negative answer, i.e. to show that $\alpha$ is not an $m$-th power, because in that case, one needs only a finite number of trials and errows to find a prime $\mathcal L_v$ outside $S$ such that $\alpha$ is not a local $m$-th power in $K_v^{*}$. But a positive answer would require an infinity of checks, which is not very satisfying in practice.
A « finite » criterion for a « positive » answer when $m$ = an odd prime $p$ (because we want to avoid some « silly special cases », @franz lemmemermer dixit) can be derived from an « interesting » (Tate’s own words) local-global principle in chapter 7 of Cassels-Fröhlich’s book (p. 184, remark 9.3). A particular case is the following : let $E$ be a number field containing the group $\mu_p$ of $p$-th roots of unity ; pick an $\alpha \in E$ and let $S$ be a finite set of primes of $E$ containing (i) all archimedean primes (ii) all primes dividing $p$ and $\alpha$ (iii) all representatives of a system of generators for the ideal class group of $E$. Then any $S$-unit of $E$ which is a local $p$-th power at all primes
inside $S$ is a global $p$-th power. In our initial problem, $\alpha$ is in $K$ which does not necessarily contain $\mu_p$. Put $E = K(\mu_p)$ , $G = Gal(E/K)$, and try to relate $(K^*)^p$ and $(E^*)^p$. Taking $G$-cohomology of the exact sequence 1 --> $\mu_p$--> $E^*$--> $(E^*)^p$ -->1 , we get 1 -->$\mu_p^{G}$--> $K^*$ -->$ K^*\cap (E^*)^p$-->$H^1(G, \mu_p)$ ... If $K$ does not contain $\mu_p$, $G$ has order prime to $p$ and $H^1(G, \mu_p)$ is trivial. In any case we get $(K^*)^p \cong K^*\cap (E^*)^p$. Summarizing, Tate’s remark gives us a finite criterion for a positive answer (in the above sense) when $m = p$.
We may suspect that with the general local-global principle, we are going too far beyond the simple case of a quadratic field, for which the solution should be much less elaborate. In the case of $\mathbf Q$, the solution is immediate because of the factoriality of $\mathbf Z$, so the natural idea is to replace factoriality by the uniqueness of decomposition into prime ideals in a Dedekind domain. A necessary condition, as suggested by @Qiaochu Yuan, is obtained by taking norms down to $\mathbf Q$. But for a sufficient condition, it seems that we are definitely blocked by by the units of norm 1 in the totally real case. This is rather irritating. |
Problem.
Let $\{\lambda_n\}_{n\in\mathbb N}$ be a sequence of complex numbers . Let's call a family of exponential functions $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ $F$-independent (where $F$ is either $\mathbb C$ or $\mathbb R$) iff whenever the series with complex coefficients
$$f(s)=\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n s},\qquad s\in F,$$ converges to $f(s)\equiv 0$ uniformly on every compact subset of $F$, we have that $a_n=0$ for all $n\in\mathbb N$.
Question.Assume that a sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. Is it $\mathbb R$-independent? Background and motivation.
A particularly interesting case for applications is when $|\lambda_n|\sim n$. A.F. Leont'ev (whose work was mentioned in a previous MO question) proved that if $n=O(|\lambda_n|)$ then the corresponding family of exponentials is $\mathbb C$-independent (see also this note). It is relatively easy to construct a sequence of exponentials which is not $\mathbb C$-independent (see, e.g., here).
The question is related to the problem of uniqueness of solutions to the so called gravity equation $$f(x+h)-f(x-h)=2h f'(x),\qquad x\in \mathbb R,$$ where $h>0$ is fixed. The equation appears in the study of radially symmetric central forces (the long history of the gravity equation and some known results are presented in this article by S. Stein).
Titchmarsh proved that an arbitrary solution to the gravity equation has the form $$f(x)=Ax^2+Bx+c+\sum\limits_{n=1}^{\infty}a_n e^{\lambda_n x},\qquad x\in \mathbb R,$$ where $a_n\in\mathbb C$, $n\in \mathbb N$ and $\lambda_n$ are the solutions of the equation $\sinh hz=hz$. Thanks to the Leont'ev result, the sequence $\{\exp (\lambda_n s)\}_{n\in\mathbb N}$ is $\mathbb C$-independent. If the answer to the question above is positive, then every sufficiently smooth function satisfying the gravity equation with two different $h_1$ and $h_2$ is a quadratic polynomial. |
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Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider
(American Physical Society, 2016-02)
The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ...
Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Elsevier, 2016-02)
Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < |\eta| < 4.0$) and associated particles in the central range ($|\eta| < 1.0$) are measured with the ALICE detector in ...
Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Springer, 2016-08)
The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2016-03)
The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ...
Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2016-03)
Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ...
Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV
(Elsevier, 2016-07)
The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ...
$^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2016-03)
The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ...
Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV
(Elsevier, 2016-09)
The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ...
Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV
(Elsevier, 2016-12)
We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ...
Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV
(Springer, 2016-05)
Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ... |
Lephenixnoir af424d1baa update documentation after writing the wiki 3 mesi fa config 4 mesi fa include/TeX 3 mesi fa src 3 mesi fa .gitignore 3 mesi fa Makefile 4 mesi fa README.md 3 mesi fa TODO.md 3 mesi fa configure 3 mesi fa font5x7.bmp 4 mesi fa font8x9.bmp 4 mesi fa font10x12.bmp 4 mesi fa
This library is a customizable 2D math rendering tool for calculators. It can be used to render 2D formulae, either from an existing structure or TeX syntax.
\frac{x^7 \left[X,Y\right] + 3\left|\frac{A}{B}\right>} {\left\{\frac{a_k+b_k}{k!}\right\}^5}+ \int_a^b \frac{\left(b-t\right)^{n+1}}{n!} dt+ \left(\begin{matrix} \frac{1}{2} & 5 \\ -1 & a+b \end{matrix}\right)
List of currently supported elements:
\frac)
_ and
^)
\left and
\right)
\sum,
\prod and
\int)
\vec) and limits (
\lim)
\sqrt)
\begin{matrix} ... \end{matrix})
Features that are partially implemented (and what is left to finish them):
See the
TODO.md file for more features to come.
First specify the platform you want to use :
cli is for command-line tests, with no visualization (PC)
sdl2 is an SDL interface with visualization (PC)
fx9860g builds the library for fx-9860G targets (calculator)
fxcg50 builds the library for fx-CG 50 targets (calculator)
For calculator platforms, you can use
--toolchain to specify a different toolchain than the default
sh3eb and
sh4eb. The install directory of the library is guessed by asking the compiler, you can override it with
--prefix.
Example for an SDL setup:
% ./configure --platform=sdl2
Then you can make the program, and if it’s a calculator library, install it. You can later delete
Makefile.cfg to reset the configuration, or just reconfigure as needed.
% make% make install # fx9860g and fxcg50 only
Before using the library in a program, a configuration step is needed. The library does not have drawing functions and instead requires that you provide some, namely:
TeX_intf_pixel)
TeX_intf_line)
TeX_intf_size)
TeX_intf_text)
The three rendering functions are available in fxlib; for monospaced fonts the fourth can be implemented trivially. In gint, the four can be defined as wrappers for
dpixel(),
dline(),
dsize() and
dtext().
The type of formulae is
TeX_Env. To parse and compute the size of a formula, use the
TeX_parse() function, which returns a new formula object (or
NULL if a critical error occurs). The second parameter
display is set to non-zero to use display mode (similar to
\[ .. \] in LaTeX) or zero to use inline mode (similar to
$ .. $ in LaTeX).
char *code = "\\frac{x_7}{\\left\\{\\frac{\\frac{2}{3}}{27}\\right\\}^2}";struct TeX_Env *formula = TeX_parse(code, 1);
The size of the formula can be queried through
formula->width and
formula->height. To render, specify the location of the top-left corner and the drawing color (which will be passed to all primitives):
TeX_draw(formula, 0, 0, BLACK);
The same formula can be drawn several times. When it is no longer needed, free it with
TeX_free():
TeX_free(formula); |
I understand that in general if we're adding more planes of atoms (increasing thickness of sample) then the intensity would increase because we have more constructive interference. But isn't there a breaking point for this? Shouldn't there be a finite thickness past which the intensity decreases?
Each layer contributes an increase proportional to the intensity of the original beam and a decrease proportional to the intensity of the diffracted beam (from multiple diffractions and extinction). In the approximation that the original beam passes through the sample with minimal losses (and the diffracted beams escape with minimal losses), the signal increases linearly with thickness. If the sample is thick enough that the losses become significant, then you will need to use a more involved expression.
Yes, there is such a point. The precise formula varies as a function of the scattering geometry, but if we consider a special case:
normal incidence on a flat sample and small scattering/diffraction angle
it is quite simple: the scattered intensity is proportional with the sample thickness $d$ but it gets attenuated as $\exp(-\mu d)$ (the Beer-Lambert law with $\mu$ the absorbance).
The strongest signal is thus obtained at the maximum of $I_S \sim d \exp(-\mu d)$, which occurs at $d = 1/\mu$, i.e. a transmission of $1/\text{e}$. Beyond this value, the attenuation (an exponential effect) dominates the increase of scattering volume (a linear effect). |
I recently gave a tutorial at CMU about spectral learning for NLP. This tutorial was based on a tutorial I had given last year with Michael Collins, Dean Foster, Karl Stratos and Lyle Ungar at NAACL.
One of the algorithms I explained there was the spectral learning algorithm for HMMs by Hsu, Kakade and Zhang (2009). This algorithm estimates parameters of HMMs in the “unsupervised setting” — only from sequences of observations. (Just like the Baum-Welch algorithm — expectation-maximization for HMMs — does.)
I want to repeat this explanation here, and give some intuition about this algorithm, since it seems to confuse people quite a lot. At a first glance, it looks quite mysterious why the algorithm works, though its implementation is very simple. It is one of the earlier algorithms in this area of latent-variable learning using the method of moments and spectral methods, and promoted the creation of other algorithms for latent-variable learning.
So here are the main ideas behind it, with some intuition. In my explanation of the algorithm, I am going to forget about the “spectral” part. No singular value decomposition will be involved, or any type of spectral decomposition. Just plain algebraic and matrix multiplication tricks that require understanding what marginal probabilities are and how matrix multiplication and inversion work, and nothing more. Pedagogically, I think that’s the right thing to do, since introducing the SVD step complicates the understanding of the algorithm.
Consider a hidden Markov model. The parameters are represented in matrix form \( T \), \( O \) and \( \pi \). We assume \( m \) latent states, \( n \) observations. More specifically, \( T \) is an \( m \times m \) stochastic matrix where \( m \) is the number of latent states, such that \( T_{hh’} \) is the probability of transitioning to state \( h \) from state \( h’ \). \( O \) is an \( n \times m \) matrix such that \( O_{xh} \) is the probability of emitting symbol \( x \) — an observation — from latent state \( h \). \( \pi \) is an \( m \) length vector with \( \pi_h\) being the initial probability for state \( h \).
To completely get rid of the SVD step, and simplify things, we will have to make the assumption that \(m = n\). This means that the number of states equals the number of observations. Not a very useful HMM, perhaps, but it definitely makes the derivation more clear. The fact that \( m=n\) means that \( O \) is now a square matrix — and we will assume it is invertible. We will also assume that \(T \) is invertible, and that \( \pi \) is positive in all coordinates.
If we look at the joint distribution of \(p(X_1 = x_1,X_2 = x_2)\), the first two observations in the HMM, then it can written as:
\( p(X_1 = x_1, X_2 = x_2) = \sum_{h_1,h_2} p(X_1 = x_1, H_1 = h_1, X_2 = x_2, H_2 = h_2) = \sum_{h_1,h_2} \pi_{h_1} O_{x_1,h_1} T_{h_2,h_1} O_{x_2,h_2} \)
Nothing special here, just marginal probability summing out the first two latent states.
It is not hard to see that this can be rewritten in matrix form, i.e. if we define \( [P_{2,1}]_{x_2,x_1} = p(X_1 = x_1, X_2= x_2) \) then:
\( P_{2,1} = O T \mathrm{diag}(\pi)O^{\top} \)
where \( \mathrm{diag}(\pi) \) is just an \( m \times m \) diagonal matrix with \( \pi_h \) on the diagonal.
Just write down this matrix multiplication step-by-step explicitly, multiplying, say, from right to left, and you will be able to verify this identity for \( P_{2,1} \). Essentially, the matrix product, which involves dot-product between rows and vectors of two matrices, eliminates and sums out the latent states (and does other things, like multiplying in the starting probabilities).
Alright. So far, so good.
Now, what about the joint distribution of three observations?
\( p(X_1 = x_1, X_2 = x, X_3=x_3) = \sum_{h_1,h_2,h_3} p(X_1 = x_1, H_1 = h_1, X_2 = x_2, H_2 = h_2, X_3=x_3, H_3 = h_3) = \sum_{h_1,h_2,h_3} \pi_{h_1} O_{x_1,h_1} T_{h_2,h_1} O_{x_2,h_2} T_{h_3,h_2} O_{x_3,h_3} \)
Does this have a matrix form too? Yes, not surprisingly. If we fix \( x \), the second observation, and define \( [P_{3,x,1}]_{x_3,x_1} = p(X_1 = x_1, X_2 = x, X_3 = x_3) \), (i.e. \( P_{3,x,1} \) is an \( m \times m \) matrix defined for each observation symbol \( x \)), then
\( P_{3,x,1} = OT \mathrm{diag}(O_x) T \mathrm{diag}(\pi) O^{\top} \).
Here, \( \mathrm{diag}(O_x) \) is a diagonal matrix where the on the diagonal we have the \(x\)th row of \( O \).
Now define \( B_x = P_{3,x,1}P_{2,1}^{-1} \) (this is well-defined because \( P_{2,1} \) is invertible — all the conditions we had on the HMM parameters make sure that it is true), then:
\( B_x = OT \mathrm{diag}(O_x) T \mathrm{diag}(\pi) O^{\top} \times (O T\mathrm{diag}(\pi)O^{\top})^{-1} = OT\mathrm{diag}(O_x)O^{-1} \)
(just recall that \( (AB)^{-1} = B^{-1} A^{-1} \) whenever both sides are defined and \( A \) and \( B \) are square matrices.)
This part of getting \( B_x \) (and I will explain in a minute why we need it) is the hardest part in our derivation so far. We can also verify that \( p(X_1 = x_1) \) equals \( O\pi \). Let’s call \( b_1 \) a vector such that \([b_1]_x = p(X_1=x_1)\) — i.e. \( b_1 \) is exactly the vector \( P_1 \).
We can also rewrite \( P_1 \) the following way:
\( P_1^{\top} = 1^{\top} T \mathrm{diag}(\pi) O^{\top} = 1^{\top} O^{-1} \underbrace{O T \mathrm{diag}(\pi) O^{\top}}_{P_{2,1}} \)
where \( 1^{\top} \) is an \( 1 \times m \) vector with the value 1 in all coordinates. The first equality is the “surprising” one — we use \( T \) to calculate the distribution of \( p(X_1 = x_1) \) — but if you write down this matrix multiplication explicitly, you will discover that we will be summing over the elements of \( T \) in such a way that it does not play a role in the sum — that’s because each row of \( T \) sums to 1. (As Hsu et al. put it in their paper: this is an unusual but easily verified form to write \( P_1 \).)
The above leads to the identity \( P_1^{\top} = 1^{\top} O^{-1} P_{2,1} \).
Now, it can be easily verified from the above form of \( P_1 \) that for \( b_{\infty}^{\top} \) defined as \( (P^{\top}_{2,1})^{-1} P_1 \), an \(m\) length vector, then:
\( b_{\infty}^{\top} = 1^{\top} O^{-1} \).
So what do we have so far? We managed to define the following matrices and vectors based only on the joint distribution of the first three symbols in the HMM:
\( B_x = P_{3,x,1}P_{2,1}^{-1} = OT\mathrm{diag}(O_x)O^{-1}, \)
\( b_1 = P_1 = O\pi, \)
\( b_{\infty}^{\top} = (P^{\top}_{2,1})^{-1} P_1 = 1^{\top} O^{-1}. \)
The matrix \( B_x \in \mathbb{R}^{m \times m} \) and vectors \( b_{\infty} \in \mathbb{R}^m \) and \( b_1 \in \mathbb{R}^m \) will now play the role of our HMM parameters. How do we use them as our parameters?
Say we just observe a single symbol in our data, i.e. the length of the sequence is 1, and that symbol is \(x\). Let’s multiply \( b^{\top}_{\infty} B_x b_1 \).
According to the above equalities, it is true that this equals:
\( b^{\top}_{\infty} B_x b_1 = (1^{\top} O^{-1}) (O T \mathrm{diag}(O_x) O^{-1}) (O \pi) = 1^{\top} T \mathrm{diag}(O_x) \pi \).
Note that this quantity is a scalar. We are multiplying a matrix by a vector from left and right. Undo this matrix multiplication, and write it the way we like in terms of sums over the latent states, and what do we get? The above just equals:
\( b^{\top}_{\infty} B_x b_1 = \sum_{h_1,h_2} T_{h_2,h_1} O_{x,h_1} \pi_{h_1} = \sum_{h_1,h_2} p(H_1) p(X_1 = x | H_1 = h_1) p(H_2 = h_2 | H_1 = h_1) = p(X_1 = x_1) \).
So, this triplet-product gave us back the distribution over the first observation. That’s not very interesting, we could have done it just by using \( b_1 \) directly. But… let’s go on and compute:
\( b^{\top}_{\infty} B_{x_2} B_{x_1} b_1. \)
This can be easily verified to equal \( p(X_1 = x_1, X_2 = x_2) \).
The interesting part is that in the general case,
\( b^{\top}_{\infty} B_{x_n} B_{x_{n-1}}…B_{x_1} b_1 = p(X_1 = x_1, \ldots, X_n = x_n) \) –
we can now calculate the probability of any observation sequence in the HMM only by knowing the distribution over the first three observations! (To convince yourself about the general case above, just look at Lemma 1 in the Hsu et al. paper.)
In order to turn this into an estimation algorithm, we just need to estimate from data \( P_{2,1} \) and \( P_{3,x,1} \) for each observation symbol (all observed, just “count and normalize”), and voila, you can estimate the probability of any sequence of observations (one of the basic problems with HMMs according to this old classic paper, for example).
But… We made a heavy assumption. We assumed that \( n = m \) — we have as many observation symbols as latent states. What do we do if that’s not true? (i.e. if \( m < n \))? That’s where the “spectral” part kicks in. Basically, what we need to do is to reduce our \( O \) matrix into an \( m \times m \) matrix using some \( U \) matrix, while ensuring that \( U^{\top}O \) is invertible (just like we assumed \( O \) was invertible before). Note that \( U \) needs to be \( n \times m \).
It turns out that a \( U \) that will be optimal in some sense, and will also make all of the above algebraic tricks work is the left singular value matrix of \( P_{2,1} \). Understanding why this is the case requires some basic knowledge of linear algebra — read the paper to understand this! |
In my post Trigonometry Yoga, I discussed how defining sine and cosine as lengths of segments in a unit circle helps develop intuition for these functions.
I learned the circle definitions of sine and cosine in my junior year of high school, in the class that would now be called pre-calculus (it was called “Trig Senior Math”). Two years earlier, I’d learned the triangle definitions of sine, cosine, and tangent in geometry class. I don’t remember any of my teachers ever mentioning a circle definition of the tangent function.
The geometric definition of the tangent function, which predates the triangle definition, is the length of a segment tangent to the unit circle. The tangent really is a tangent! Just as for sine and cosine, this one-variable definition helps develop intuition. Here is the definition, followed by an applet to help you get a feel for it:
Let OA be a radius of the unit circle, let B = (1,0), and let \( \theta =\angle BOA\). Let C be the intersection of \(\overrightarrow{OA}\) and the line x=1, i.e. the tangent to the unit circle at B. Then \(\tan \theta\) is the y-coordinate of C, i.e. the signed length of segment BC.
Move the blue point below; the tangent is the length of the red segment. (If a label is getting in the way, right click and toggle “show label” from the menu).
The circle definition of the tangent function leads to geometric illustrations of many standard properties and identities. (If this were my class, I would stop here and tell you to explore on your own and with others).
Some things to notice:
\(\left| \tan \theta \right|\) gets big as \(\theta\) approaches \(\pm 90{}^\circ \).
\(\tan (\pm 90{}^\circ)\) is undefined, because at these angles, \(\overline{OA}\) is parallel to x=1, so the two lines don’t intersect, and point C doesn’t exist.
\(\tan 90{}^\circ\) tends toward \(+\infty\), \(\tan (-90{}^\circ)\) tends toward \(-\infty\).
\(\tan \theta\) is positive in the first and third quadrants, negative in the second and fourth quadrants.
\(\tan \theta\)=\(tan (\theta+180{}^\circ)\) — the angles \(\theta\) and \(\theta +180{}^\circ\) form the same line. Thus the period of the tangent function is \(180 {}^\circ = \pi\) radians.
\(\tan \theta\) = \(- \tan (-\theta)\). Moving from \(\theta\) to \(-\theta\) reflects \(OC\) about the x-axis.
\(\tan \theta\) is equal to the slope of OA (rise = \(\tan \theta\) , run =1), which is also equal to \(\dfrac{\sin\theta}{\cos\theta}\), as well as Opposite over Adjacent for angle \(\theta\) in right triangle CBO.
\(\tan (45{}^\circ)=1\). When \(\theta=45{}^\circ\), triangle CBO is a 45-45-90 triangle, and OB=1. Similarly, \(\tan (-45{}^\circ)=-1\), etc.
For small values of \(\theta\), \(\tan \theta\) is close to \(\sin \theta\), which is close to the arc length of AB, i.e. the measure of \(\theta\) in radians.
If we define \(\arctan \theta\) as the function whose input is the signed length of BC and whose output is the angle \(\theta\) corresponding to that tangent length, then the domain of that function is the reals, and it makes sense to define the range as \(-90 {}^\circ< \theta <90{}^\circ\) (in radians \(-\pi/2<\theta < \pi/2\) and arctan’s output is an arc length). This range includes all the angles we need and avoids the discontinuity at \(\theta= \pm 90{}^\circ =\pm \pi/2\) radians. For \(\left| \theta \right|\leq 45{}^\circ\), \(\left| \tan \theta \right|\leq 1\). Half of the input values of \(\tan \theta\) give outputs with absolute values less than or equal to 1, and the other half give values on the rest of the number line. This mapping also occurs with fractions and slopes, but there’s something very compelling about seeing the lengths change dynamically. Applets like the one above could also help students develop intuition about slopes. \(\tan (180{}^\circ-\theta) = -\tan \theta\). We reflect BC over the x-axis to form \(B{C}’\). Then \(\angle BO{C}’=\theta\) and \(\angle BOD =(180{}^\circ-\theta)\). \(B{C}’\) (the blue segment) is the tangent of \((180{}^\circ-\theta)\).
\(\tan (\theta \pm 90{}^\circ)\) = \(-1/\tan \theta\). The picture below illustrates the geometry of this identity when \(\theta\) is in the first quadrant.
The line formed at \(\theta + 90{}^\circ\) is perpendicular to OC and \(\triangle COB\sim \triangle ODB\). Thus \(\dfrac{BD}{OB}=\dfrac{OB}{BC}\), and with appropriate signs, \(\tan (\theta + 90{}^\circ)\) = \(-1/\tan \theta\). Since \(\tan \theta\)=\(\tan (\theta+180{}^\circ)\), \(\tan (\theta +90{}^\circ)=\tan(\theta-90{}^\circ)\).
The applet below shows the geometry in all quadrants, and it gives a dynamic sense of the relationship between \(\tan\theta\) and \(\tan(-\theta)\). Again, move the blue point:
Special Bonus: The Secant Function
The signed length of the segment OC is called the secant function, \(\sec\theta\).
Using similar triangles, we see that \(\sec \theta = \dfrac{1}{\cos \theta}\).
The Pythagorean Theorem applied to \(\triangle COB\) shows that \(\tan^2\theta+1=\sec^2 \theta\).
When the tangent function is big, so is the secant function, and when the tangent function is small, so is the secant function. Also \(\sec \theta\) is close to \(\pm 1\) when \(\theta\) is close to the x-axis and when \(\tan \theta\) is close to 0.
The graphs of the two functions look nice together: |
What are the advantages/disadvantages of using the arithmetic Sharpe Ratio vs the geometric Sharpe Ratio? Is one more correct? Or is one better in certain circumstances?
In addition to John's answer and just to make things clear:
The arithmetic mean is given by
$$\mu_a = \frac{1}{n} \sum_{i=1}^n x_i$$
The geometric mean is given by
$$\mu_g = \sqrt[n]{\prod_{i=1}^n (1+x_i)} -1$$
And we have that
$$\mu_g \leq \mu_a$$
So not only would the geometric sharp ratio be taking into account the "actual" return of the portfolio, but it is also a more conservative measure.
I'm not sure it makes sense to think of one as more correct than another. However, they do have significant differences. It may help to distinguish between ex-post evaluation of a strategy and ex-ante prediction of what the strategy's performance will be.
For simplicity, let's assume the log returns of the strategy are approximately i.i.d. univariate normal and the risk-free rate is a constant. If you were a mean-variance investor deciding between the risk-free rate and the strategy, you would estimate the mean and variance of the log returns, project it to the investor's horizon, and convert the normal to lognormal to obtain the arithmetic returns. So if you were calculating a Sharpe ratio that is consistent with the way it was originated in financial theory, i.e. the slope of the efficient frontier, would be this arithmetic ex ante expected Sharpe ratio.
However, the Sharpe ratio is also used in performance evaluation in different ways. I think one major reason that the geometric version is used is that the numerator will correspond with what the investor actually earned, the CAGR. This might be useful to some people, but personally I prefer to look at the CAGR by itself rather than in the Sharpe ratio. Further, the CAGR is the median of the lognormal distribution under some assumptions. I find it more intuitive to use the mean and keep ex-post consistent with ex-ante, which would bring me back to the arithmetic Sharpe.
Another reason to use the geometric version might be that it avoids the distributional issues with the lognormal distribution (since it has skewness/kurtosis). However, Opdyke (2007) provides the asympototic distribution of the Sharpe ratio under fairly general assumptions.
I think this is a no-brainer. Only log-returns make sense. The average return can only be computed by averaging the sum of individual log returns. Taking the average of standard (relative) returns does not give you an average of the individual returns. Consider a simple case where the value of an investment alternates between 100 and 50 an odd number of times. The standard return series would be: -0.5, 1, -0.5, …1 (-50% and +100%). The average of that sum gives us 0.25 (25%) – nonsense for an investment whose final value is the same as what we started with. The log returns, on the other hand give us alternating log returns of -0.6931, +0.6931, whose average is 0.
The difference between log returns and standard returns goes to zero as we shorten the period over which we evaluate the value of an investment: LN(P(n)/P(n-1)) is approximately equal to P(n)/P(n-1) – 1. Thus there would not be much difference between standard and log returns (and the computed Sharpe Ratio) if daily measurements were made. The scaling of that Sharpe Ratio from daily returns to annual returns is performed by the sqrt of the number of trade days (252), but that, of course assumes the return distribution is iid, which is not really the case.
Andrew W. Lo has a nice paper that considers the scaling of the Sharpe ratio when the return series is correlated ("The Statistics of Sharpe Ratios")
There are many variants proposed; some useful, some not so much. As an investor, the most important thing is to compare the exact same ratio, calculated in the exact same way, for each prospect. As the prospect/fund the most important thing is to be clear about the statistic you are reporting so your investors make well informed decisions. So let's start with some definitions specific to your question.
Sharpe ratio, sometimes called the "Modified Sharpe" ratio, is the arithmetic average of excess returns divided by the standard deviation of those returns, $r_t\in R$. $Sharpe Ratio = \frac{E[R_t - R_{free}]}{\sigma}$It is commonly calculated over annual periods, but monthly, or daily periods are common too. It is a class of signal-to-noise ratio indicating the expected reward per unit of risk. This is the most commonly cited variant and is the measure proposed by professor Sharpe in 1994. where $R_t\in[-1,\infty)$, which is a percentage, and $R_{free}$ is the risk-free return rate, typically taken as the current T-Bill rate. The Geometric Sharpe ratiois the geometric average of compounded excess returns divided by the standard deviation of those compounded returns. This is equivalent to the arithmetic average and standard deviation of $log(1+r_t)$, which is the more convienient calculation. The difference is mostly in the use of the average of the excess returns for the period or the compounded returns. Because the log returns are generally smaller, the "Geometric Sharpe ratios" generally suggests a higher number; about 12.5% higher for samples from a uniform distribution. Naturally, it is completely bogus to compare dissimilar ratios...so be sure you know which measure you are digesting.
The correct answer is "arithmetic mean, because Bill Sharpe says so". He invented the thing, and he's pretty clear on which one he was looking at.
If you use the geometric mean, which is lower the higher the volatility in the returns, and then you divide by standard deviation, you have essentially discounted your result TWICE for volatility.
The arithmetic average of +100% in Year 1 and -100% in Year 2 is 0%, but I we all know the result is not a 0% return. So arithmetic returns are absurd to use in any real life context. Maybe in another universe they can serve some purpose.
For any real world applications, the difference between the arithmetic and geometric Sharpe ratios is likely to 'fall under the noise floor',
i.e. be smaller, typically much smaller, than a standard error. This is even under the generous assumptions of stationarity and absence of omitted variables.
As a practical example, I have generated a sequence of returns where mean log returns are -12%, while the arithmetic mean is +1.1%.
This can happen with very heavy tailed distributions and, e.g. in evaluating hedge fund performance, the arithmetic return will mask this.
A glance at an equity curve would also highlight the issue so in reality no-one is likely to be fooled.
For unleveraged investing in stocks and indices it will make very little difference, but the arithmetic version is clearly wrong in some cases as it will give a positive sharpe for a series of returns resulting in a large loss of capital.
The only correct way is using log returns.
To keep everything consistent, take a arithmetic mean of log returns.
Then calculate it net of the risk free (how do you subtract properly using geometric returns?).
Then divide (how do you do this properly using geometric returns?)
by the standard deviation (how would you calculate this properly with geometric returns?).
A geometric mean is just the tip of the iceberg. Geometric returns implies a bunch of new calculations which are far from standard.
Log returns are the only way to go with time series analysis. |
Suppose $C$ is a 3-form, and $G$ is a 4-form defined by $G = dC$. Also, $M_{11}$ is an 11-dimensional manifold (without a boundary), $W_{6}$ is a 6-dimensional submanifold of $M_{11}$ and $D_{\epsilon}W_6 = -S_{\epsilon}W_{6}$ is the 4-sphere bundle over $W_6$.
Further, suppose $\rho$ is a 0-form and $e_{2}^{1}$ is a 2-form. Under a variation,
$$\delta C = -d(\rho e_{2}^{1})$$
I want to compute the variation $\delta S_{CS}$ in the Chern-Simons integral
$$S_{CS} = -\lim_{\epsilon\rightarrow 0}\int\limits_{M_{11}\backslash D_{\epsilon}W_6} C \wedge G \wedge G$$
Apparently, the
correct answer is
$$\delta S_{CS} = -\lim_{\epsilon \rightarrow}\int\limits_{S_{\epsilon}W_6}\rho e_{2}^{1}\wedge G \wedge G$$
But what I get is something else. Here is my detailed derivation.
$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[\int\limits_{M_{11}\backslash D_\epsilon W_6}\delta C \wedge G \wedge G + 2 \int\limits_{M_{11}\backslash D_\epsilon W_6}\delta G \wedge C \wedge G\right]$$
From integration by parts
$$\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G = \int\limits_{M_{11}\backslash D_\epsilon W_6}d(\delta C \wedge C \wedge G) = \int\limits_{M_{11}\backslash D_\epsilon W_6} \delta dC \wedge C \wedge G - \int\limits_{M_{11}\backslash D_\epsilon W_6} \delta C \wedge G \wedge G$$
So it should follow that
$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[2\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G + 3\int\limits_{M_{11}\backslash D_\epsilon W_6}\delta C \wedge G \wedge G\right]$$
As $M_{11}\backslash D_{\epsilon}W_6$, for finite $\epsilon$ cannot support an 11-form, the second integral vanishes inside the limit, and we are left with
$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[2\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G\right]$$
Substituting $\delta C = -d(\rho e_2^1)$ we get
$$\delta S_{CS} = +\lim_{\epsilon\rightarrow 0}\left[2\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)} d(\rho e_2^1) \wedge C \wedge G\right]$$
Now,
$$d(\rho e_2^1 \wedge (C\wedge G)) = d(\rho e_2^1) \wedge C\wedge G + \rho e_2^1 \wedge d(C\wedge G)$$
and $\partial(\partial(M_{11}\backslash D_\epsilon W_6)) \equiv 0$, so
$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[2\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\rho e_2^1 \wedge G \wedge G\right]$$
As a last step, using $\partial(M_{11}\backslash D_\epsilon W_6) = -S_{\epsilon} W_6$, one gets the
final expression
$$\delta S_{CS} = +\lim_{\epsilon\rightarrow 0}\left[2\int\limits_{S_\epsilon W_6}\rho e_2^1 \wedge G \wedge G\right]$$
This is off by a sign and a factor of 2.
What seems to be wrong in the derivation here?
Physics background: These algebraic manipulations are inspired by a calculation of the M5-brane anomaly in M-theory, perhaps discussed first in a paper by Freed, Minasian, Harvey and Moore (http://arxiv.org/abs/hep-th/9803205). It has been pointed out in some follow-up papers that there are an odd number of minus signs involved, and that the original paper may have overlooked one sign. (Of course the M5-brane anomaly is seen to cancel, but the cancellation is a bit involved and requires a careful understanding of minus signs and factors. Hence this question.)
This post imported from StackExchange Mathematics at 2015-08-09 22:38 (UTC), posted by SE-user leastaction |
One disadvantage of the fact that you have posted 5 identical answers (1, 2, 3, 4, 5) is that if other users have some comments about the website you created, they will post them in all these place. If you have some place online where you would like to receive feedback, you should probably also add link to that. — Martin Sleziak1 min ago
BTW your program looks very interesting, in particular the way to enter mathematics.
One thing that seem to be missing is documentation (at least I did not find it).
This means that it is not explained anywhere: 1) How a search query is entered. 2) What the search engine actually looks for.
For example upon entering $\frac xy$ will it find also $\frac{\alpha}{\beta}$? Or even $\alpha/\beta$? What about $\frac{x_1}{x_2}$?
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Is it possible to save a link to particular search query? For example in Google I am able to use link such as: google.com/search?q=approach0+xyz Feature like that would be useful for posting bug reports.
When I try to click on "raw query", I get curl -v https://approach0.xyz/search/search-relay.php?q='%24%5Cfrac%7Bx%7D%7By%7D%24' But pasting the link into the browser does not do what I expected it to.
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If I copy-paste search query into your search engine, it does not work. For example, if I copy $\frac xy$ and paste it, I do not get what would I expect. Which means I have to type every query. Possibility to paste would be useful for long formulas. Here is what I get after pasting this particular string:
I was not able to enter integrals with bounds, such as $\int_0^1$. This is what I get instead:
One thing which we should keep in mind is that duplicates might be useful. They improve the chance that another user will find the question, since with each duplicate another copy with somewhat different phrasing of the title is added. So if you spent reasonable time by searching and did not find...
In comments and other answers it was mentioned that there are some other search engines which could be better when searching for mathematical expressions. But I think that as nowadays several pages uses LaTex syntax (Wikipedia, this site, to mention just two important examples). Additionally, som...
@MartinSleziak Thank you so much for your comments and suggestions here. I have took a brief look at your feedback, I really love your feedback and will seriously look into those points and improve approach0. Give me just some minutes, I will answer/reply to your in feedback in our chat. — Wei Zhong1 min ago
I still think that it would be useful if you added to your post where do you want to receive feedback from math.SE users. (I suppose I was not the only person to try it.) Especially since you wrote: "I am hoping someone interested can join and form a community to push this project forward, "
BTW those animations with examples of searching look really cool.
@MartinSleziak Thanks to your advice, I have appended more information on my posted answers. Will reply to you shortly in chat. — Wei Zhong29 secs ago
We are open-source project hosted on GitHub: http://github.com/approach0Welcome to send any feedback on our GitHub issue page!
@MartinSleziak Currently it has only a documentation for developers (approach0.xyz/docs) hopefully this project will accelerate its releasing process when people get involved. But I will list this as a important TODO before publishing approach0.xyz . At that time I hope there will be a helpful guide page for new users.
@MartinSleziak Yes, $x+y$ will find $a+b$ too, IMHO this is the very basic requirement for a math-aware search engine. Actually, approach0 will look into expression structure and symbolic alpha-equivalence too. But for now, $x_1$ will not get $x$ because approach0 consider them not structurally identical, but you can use wildcard to match $x_1$ just by entering a question mark "?" or \qvar{x} in a math formula. As for your example, enter $\frac \qvar{x} \qvar{y} $ is enough to match it.
@MartinSleziak As for the query link, it needs more explanation, technologically the way you mentioned that Google is using, is a HTTP GET method, but for mathematics, GET request may be not appropriate since it has structure in a query, usually developer would alternatively use a HTTP POST request, with JSON encoded. This makes developing much more easier because JSON is a rich-structured and easy to seperate math keywords.
@MartinSleziak Right now there are two solutions for "query link" problem you addressed. First is to use browser back/forward button to navigate among query history.
@MartinSleziak Second is to use a computer command line 'curl' to get search results from particular query link (you can actually see that in browser, but it is in developer tools, such as the network inspection tab of Chrome). I agree it is helpful to add a GET query link for user to refer to a query, I will write this point in project TODO and improve this later. (just need some extra efforts though)
@MartinSleziak Yes, if you search \alpha, you will get all \alpha document ranked top, different symbols such as "a", "b" ranked after exact match.
@MartinSleziak Approach0 plans to add a "Symbol Pad" just like what www.symbolab.com and searchonmath.com are using. This will help user to input greek symbols even if they do not remember how to spell.
@MartinSleziak Yes, you can get, greek letters are tokenized to the same thing as normal alphabets.
@MartinSleziak As for integrals upper bounds, I think it is a problem on a JavaScript plugin approch0 is using, I also observe this issue, only thing you can do is to use arrow key to move cursor to the right most and hit a '^' so it goes to upper bound edit.
@MartinSleziak Yes, it has a threshold now, but this is easy to adjust from source code. Most importantly, I have ONLY 1000 pages indexed, which means only 30,000 posts on math stackexchange. This is a very small number, but will index more posts/pages when search engine efficiency and relevance is tuned.
@MartinSleziak As I mentioned, the indices is too small currently. You probably will get what you want when this project develops to the next stage, which is enlarge index and publish.
@MartinSleziak Thank you for all your suggestions, currently I just hope more developers get to know this project, indeed, this is my side project, development progress can be very slow due to my time constrain. But I believe its usefulness and will spend my spare time to develop until its publish.
So, we would not have polls like: "What is your favorite calculus textbook?" — GEdgar2 hours ago
@GEdgar I'd say this goes under "tools." But perhaps it could be made explicit. — quid1 hour ago
@quid I think that the type of question mentioned in GEdgar's comment is closer to book-recommendations which are valid questions on the main. (Although not formulated like that.) I also think that his comment was tongue-in-cheek. (Although it is a bit more difficult for me to detect sarcasm, as I am not a native speaker.) — Martin Sleziak57 mins ago
"What is your favorite calculus textbook?" is opinion based and/or too broad for main. If at all it is a "poll." On tex.se they have polls "favorite editor/distro/fonts etc" while actual questions on these are still on-topic on main. Beyond that it is not clear why a question which software one uses should be a valid poll while the question which book one uses is not. — quid7 mins ago
@quid I will reply here, since I do not want to digress in the comments too much from the topic of that question.
Certainly I agree that "What is your favorite calculus textbook?" would not be suitable for the main. Which is why I wrote in my comment: "Although not formulated like that".
Book recommendations are certainly accepted on the main site, if they are formulated in the proper way.
If there will be community poll and somebody suggests question from GEdgar's comment, I will be perfectly ok with it. But I thought that his comment is simply playful remark pointing out that there is plenty of "polls" of this type on the main (although ther should not be). I guess some examples can be found here or here.
Perhaps it is better to link search results directly on MSE here and here, since in the Google search results it is not immediately visible that many of those questions are closed.
Of course, I might be wrong - it is possible that GEdgar's comment was meant seriously.
I have seen for the first time on TeX.SE. The poll there was concentrated on TeXnical side of things. If you look at the questions there, they are asking about TeX distributions, packages, tools used for graphs and diagrams, etc.
Academia.SE has some questions which could be classified as "demographic" (including gender).
@quid From what I heard, it stands for Kašpar, Melichar and Baltazár, as the answer there says. In Slovakia you would see G+M+B, where G stand for Gašpar.
But that is only anecdotal.
And if I am to believe Slovak Wikipedia it should be Christus mansionem benedicat.
From the Wikipedia article: "Nad dvere kňaz píše C+M+B (Christus mansionem benedicat - Kristus nech žehná tento dom). Toto sa však často chybne vysvetľuje ako 20-G+M+B-16 podľa začiatočných písmen údajných mien troch kráľov."
My attempt to write English translation: The priest writes on the door C+M+B (Christus mansionem benedicat - Let the Christ bless this house). A mistaken explanation is often given that it is G+M+B, following the names of three wise men.
As you can see there, Christus mansionem benedicat is translated to Slovak as "Kristus nech žehná tento dom". In Czech it would be "Kristus ať žehná tomuto domu" (I believe). So K+M+B cannot come from initial letters of the translation.
It seems that they have also other interpretations in Poland.
"A tradition in Poland and German-speaking Catholic areas is the writing of the three kings' initials (C+M+B or C M B, or K+M+B in those areas where Caspar is spelled Kaspar) above the main door of Catholic homes in chalk. This is a new year's blessing for the occupants and the initials also are believed to also stand for "Christus mansionem benedicat" ("May/Let Christ Bless This House").
Depending on the city or town, this will be happen sometime between Christmas and the Epiphany, with most municipalities celebrating closer to the Epiphany."
BTW in the village where I come from the priest writes those letters on houses every year during Christmas. I do not remember seeing them on a church, as in Najib's question.
In Germany, the Czech Republic and Austria the Epiphany singing is performed at or close to Epiphany (January 6) and has developed into a nationwide custom, where the children of both sexes call on every door and are given sweets and money for charity projects of Caritas, Kindermissionswerk or Dreikönigsaktion[2] - mostly in aid of poorer children in other countries.[3]
A tradition in most of Central Europe involves writing a blessing above the main door of the home. For instance if the year is 2014, it would be "20 * C + M + B + 14". The initials refer to the Latin phrase "Christus mansionem benedicat" (= May Christ bless this house); folkloristically they are often interpreted as the names of the Three Wise Men (Caspar, Melchior, Balthasar).
In Catholic parts of Germany and in Austria, this is done by the Sternsinger (literally "Star singers"). After having sung their songs, recited a poem, and collected donations for children in poorer parts of the world, they will chalk the blessing on the top of the door frame or place a sticker with the blessing.
On Slovakia specifically it says there:
The biggest carol singing campaign in Slovakia is Dobrá Novina (English: "Good News"). It is also one of the biggest charity campaigns by young people in the country. Dobrá Novina is organized by the youth organization eRko. |
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let ai≻0a_i\succ0ai≻0 .Prove that∏i=1nai=1⇒∑i=1n1ai+1≥1\prod_{i=1}^{n}a_i=1\Rightarrow \sum_{i=1}^{n}\frac{1}{a_i+1}\ge1i=1∏nai=1⇒i=1∑nai+11≥1.
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This is a high pass T filter:
According to me, if input voltage is \$V_{\text{in}}\$, then voltage across the inductor \$L\$ should be $$V_x=\frac{j\omega L}{j\omega L + \frac{1}{2j\omega C}}V_{\text{in}}$$
Now, I think \$V_{\text{out}} = V_x\$ since we can only measure the EMF across the output terminals unless there's some load resistance places across those two output terminals.
In that case if we want the cut-off frequency, we'll need to set \$|V_{\text{out}}/V_{\text{in}}|=\frac{1}{\sqrt{2}}\$. I get \$\omega_{\text{cut-off}} = \sqrt{\frac{1}{2(\sqrt{2}+1)LC}}\$ from here. However, the actual cut-off (angular) frequency \$\omega_{\text{cut-off}}\$ (according to what we were told in class) should have been \$\frac{1}{2\sqrt{LC}}\$. I'm not sure why I'm not getting the same result.
Also, another question is: What is the use of the right hand side \$2C\$ capacitor of the T-section filter in determining the cut-off frequency? As long as no load is there, we should solely consider the voltage across the inductor \$L\$ to be the output voltage, isn't it? Or no? |
Exercise \(\PageIndex{1}\): Euler number
Consider the function \(f(x)=\left(1+\dfrac{1}{x}\right)^x\). Make a table showing \(f(x)\) for \(x=1,2,3, .....\) . Round your solutions to five decimal places. What can you say about the value of the function \(f(x)\) as \(x\) increases indefinitely?
Answer
\(\lim_{x \to \infty}\left(1+\dfrac{1}{x}\right)^x=2.7183=e.\) |
Evaluate the limit \[\lim_{h\rightarrow 0}\frac{2(-3+h)^{2}-18}{h}\] a) 12 b) 8 c) 14 d) 6
You can purchase MTH 102 TMA Solutions at the rate of #1000 only and all of your required Courses from us kindly send us a msg on whatsapp 08039407882
Evaluate the limit \[\lim_{h\rightarrow 0}\frac{2(-3+h)^{2}-18}{h}\] a) 12 b) 8 c) 14 d) 6
You can purchase MTH 102 TMA Solutions at the rate of #1000 only and all of your required Courses from us kindly send us a msg on whatsapp 08039407882 |
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
That seems like what I need to do, but I don't know how to actually implement it... how wide of a time window is needed for the Y_{t+\tau}? And how on earth do I load all that data at once without it taking forever?
And is there a better or other way to see if shear strain does cause temperature increase, potentially delayed in time
Link to the question: Learning roadmap for picking up enough mathematical know-how in order to model "shape", "form" and "material properties"?Alternatively, where could I go in order to have such a question answered?
@tpg2114 For reducing data point for calculating time correlation, you can run two exactly the simulation in parallel separated by the time lag dt. Then there is no need to store all snapshot and spatial points.
@DavidZ I wasn't trying to justify it's existence here, just merely pointing out that because there were some numerics questions posted here, some people might think it okay to post more. I still think marking it as a duplicate is a good idea, then probably an historical lock on the others (maybe with a warning that questions like these belong on Comp Sci?)
The x axis is the index in the array -- so I have 200 time series
Each one is equally spaced, 1e-9 seconds apart
The black line is \frac{d T}{d t} and doesn't have an axis -- I don't care what the values are
The solid blue line is the abs(shear strain) and is valued on the right axis
The dashed blue line is the result from scipy.signal.correlate
And is valued on the left axis
So what I don't understand: 1) Why is the correlation value negative when they look pretty positively correlated to me? 2) Why is the result from the correlation function 400 time steps long? 3) How do I find the lead/lag between the signals? Wikipedia says the argmin or argmax of the result will tell me that, but I don't know how
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
Because I don't know how the result is indexed in time
Related:Why don't we just ban homework altogether?Banning homework: vote and documentationWe're having some more recent discussions on the homework tag. A month ago, there was a flurry of activity involving a tightening up of the policy. Unfortunately, I was really busy after th...
So, things we need to decide (but not necessarily today): (1) do we implement John Rennie's suggestion of having the mods not close homework questions for a month (2) do we reword the homework policy, and how (3) do we get rid of the tag
I think (1) would be a decent option if we had >5 3k+ voters online at any one time to do the small-time moderating. Between the HW being posted and (finally) being closed, there's usually some <1k poster who answers the question
It'd be better if we could do it quick enough that no answers get posted until the question is clarified to satisfy the current HW policy
For the SHO, our teacher told us to scale$$p\rightarrow \sqrt{m\omega\hbar} ~p$$$$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$And then define the following$$K_1=\frac 14 (p^2-q^2)$$$$K_2=\frac 14 (pq+qp)$$$$J_3=\frac{H}{2\hbar\omega}=\frac 14(p^2+q^2)$$The first part is to show that$$Q \...
Okay. I guess we'll have to see what people say but my guess is the unclear part is what constitutes homework itself. We've had discussions where some people equate it to the level of the question and not the content, or where "where is my mistake in the math" is okay if it's advanced topics but not for mechanics
Part of my motivation for wanting to write a revised homework policy is to make explicit that any question asking "Where did I go wrong?" or "Is this the right equation to use?" (without further clarification) or "Any feedback would be appreciated" is not okay
@jinawee oh, that I don't think will happen.
In any case that would be an indication that homework is a meta tag, i.e. a tag that we shouldn't have.
So anyway, I think suggestions for things that need to be clarified -- what is homework and what is "conceptual." Ie. is it conceptual to be stuck when deriving the distribution of microstates cause somebody doesn't know what Stirling's Approximation is
Some have argued that is on topic even though there's nothing really physical about it just because it's 'graduate level'
Others would argue it's not on topic because it's not conceptual
How can one prove that$$ \operatorname{Tr} \log \cal{A} =\int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr}e^{-s \mathcal{A}},$$for a sufficiently well-behaved operator $\cal{A}?$How (mathematically) rigorous is the expression?I'm looking at the $d=2$ Euclidean case, as discuss...
I've noticed that there is a remarkable difference between me in a selfie and me in the mirror. Left-right reversal might be part of it, but I wonder what is the r-e-a-l reason. Too bad the question got closed.
And what about selfies in the mirror? (I didn't try yet.)
@KyleKanos @jinawee @DavidZ @tpg2114 So my take is that we should probably do the "mods only 5th vote"-- I've already been doing that for a while, except for that occasional time when I just wipe the queue clean.
Additionally, what we can do instead is go through the closed questions and delete the homework ones as quickly as possible, as mods.
Or maybe that can be a second step.
If we can reduce visibility of HW, then the tag becomes less of a bone of contention
@jinawee I think if someone asks, "How do I do Jackson 11.26," it certainly should be marked as homework. But if someone asks, say, "How is source theory different from qft?" it certainly shouldn't be marked as Homework
@Dilaton because that's talking about the tag. And like I said, everyone has a different meaning for the tag, so we'll have to phase it out. There's no need for it if we are able to swiftly handle the main page closeable homework clutter.
@Dilaton also, have a look at the topvoted answers on both.
Afternoon folks. I tend to ask questions about perturbation methods and asymptotic expansions that arise in my work over on Math.SE, but most of those folks aren't too interested in these kinds of approximate questions. Would posts like this be on topic at Physics.SE? (my initial feeling is no because its really a math question, but I figured I'd ask anyway)
@DavidZ Ya I figured as much. Thanks for the typo catch. Do you know of any other place for questions like this? I spend a lot of time at math.SE and they're really mostly interested in either high-level pure math or recreational math (limits, series, integrals, etc). There doesn't seem to be a good place for the approximate and applied techniques I tend to rely on.
hm... I guess you could check at Computational Science. I wouldn't necessarily expect it to be on topic there either, since that's mostly numerical methods and stuff about scientific software, but it's worth looking into at least.
Or... to be honest, if you were to rephrase your question in a way that makes clear how it's about physics, it might actually be okay on this site. There's a fine line between math and theoretical physics sometimes.
MO is for research-level mathematics, not "how do I compute X"
user54412
@KevinDriscoll You could maybe reword to push that question in the direction of another site, but imo as worded it falls squarely in the domain of math.SE - it's just a shame they don't give that kind of question as much attention as, say, explaining why 7 is the only prime followed by a cube
@ChrisWhite As I understand it, KITP wants big names in the field who will promote crazy ideas with the intent of getting someone else to develop their idea into a reasonable solution (c.f., Hawking's recent paper) |
A posteriori error estimates space-time in Isogeometric Analysis of parabolic problems Dr. Svetlana Matculevich Nov. 4, 2016, 10:15 a.m. S2 054
We are concern with guaranteed error control of space-time Isogeometric Analysis (IgA) numerical approximations of parabolic evolution equations in fixed and moving spatial computational domains. The approach is discussed within the paradigm of classical
linear parabolic initial-boundary value problem (I-BVP) as model problem: find $u: \overline{Q} \rightarrow ℝ^d$ such that
$$
\begin{alignat}{2} \partial_t u - \Delta_x u & = f \;\; {\rm in} \;\; Q, \qquad u = 0 \;\; {\rm on} \;\; \Sigma, \qquad u = u_0 \;\; {\rm on} \;\; \overline{\Sigma}_0, \label{eq:equation} \qquad (1) \end{alignat} $$
where $\overline{Q} := Q \cup \partial Q$, $Q := \Omega \times (0, T)$, denote the space-time cylinder with a bounded domain $\Omega \subset ℝ^d$, $d \in \{1, 2, 3\}$, having a Lipschitz boundary $\partial \Omega$, and $(0, T)$ is a given time interval, $0 < T < +\infty$. Here, the cylindrical surface is defined as $\partial Q := \Sigma \cup \overline{\Sigma}_{0} \cup \overline{\Sigma}_{T}$ with $\Sigma = \partial \Omega \times (0, T)$, $\Sigma_{0} = \Omega \times \{0\}$ and $\Sigma_{T} = \Omega \times \{T\}$.
Following the spirit of [1], we consider a
stable IgA space-time scheme for variation formulation of (1), which is obtained by testing it with auxilary function $v_h + \delta_h \, \partial_t v_h$, $\delta_h = \theta \ h$, $v_h \in V_{0h} \subset H^{1}_{0, \underline{0}}(Q),$ where $\theta$ is a positive constant and $h$ is the global mesh-size parameter (with mesh denoted by $\mathcal{K}_h$). The parameter $\delta_h$ in discrete bilinear form can be also localized such that $\delta_{K} = \theta \, h_K$, where $h_K := {\rm diam} (K)$, $K \in \mathcal{K}_h$ is the local mesh-size parameter. It is shown that for both cases the obtained discrete bilinear forms are $V_{0h}$-coercive on the IgA space with respect to corresponding discrete energy norms, which together with boundedness property, consistency and approximation results for the IgA spaces provides an a priori discretization error estimates.
Finally, we derive the functional a posteriori error estimates for the discussed schemes (see, e.g., [2]), which apart from the quantitatively efficient indicators provides the reliable and sharp error estimates. This type of error estimates can exploit the higher smoothness of NURBS basis functions to its advantage. Since the obtained approximations are generally $C^{p-1}$-continuous (provided that the inner knots have the multiplicity $1$), this automatically provides that its gradients are in $H(\Omega, {\rm div})$ space. Therefore, there is no need in projecting it from $\nabla u_h \in L^{2}(\Omega, ℝ^d)$ into $H(\Omega, {\rm div})$.
[1] U. Langer, S. Moore, and M. Neumüller,
Space-time isogeometric analysis of parabolic evolution equations, Comput. Methods Appl. Mech. Engrg., 2016, 306: 342–363. [2] S. I. Repin, Estimates of deviations from exact solutions of initial-boundary value problem for the heat equation, Rend. Mat. Acc. Lincei, 13(9):121–133, 2002. |
This question is based on a special case of the Coparaso Harris formula, as described in Counting curves on rational surfaces - R. Vakil.
Let $E$ be a non-singular planar conic.
Then every degree $d$ curve intersects it in $2d$ points (with mult.). Let us focus only on rational irreducible curves that have only simple tangencies with $E$ - these intersect $E$ at exactly $d$ points, each with multiplicity 2. If we fix $d-r$ of the intersection points-$\{p_i\}$ then it turns out that (using stable map terminology) the dimension of the moduli space of stable maps in $\overline{\mathcal{M}}_{0,d-r}(\mathbb{P}^2,dL)$ that map to curves satisfying our conditions (with the $k$-th marked point going to $p_k$) is $d+r-1$.
Fixing $d+r-1$ generic points on the plane for the curves to pass through, we are left with a finite set of points - $(d-r)!N^d(r)$ points to be exact.
$N^d(r)$ is the enumerative constant that counts the number of rational curves of degree $d$ which intersect $E$ only at simple tangencies, $d-r$ of them are fixed.
I'm looking for an asympotic upper bound for the sequence $N^d(r)$, $0\leq r\leq d$ as $d\to\infty$.
It turns out that there is a recursive formula which describes this sequence:
$N^1(0)=0$ and for all $d+r\geq 2$,
$N^d(r)=2(d-r+1)N^d(r-1)+\sum_{l=1}^{\min(r,d-r-2)}\{\sum\frac{1}{\sigma}\frac{(d+r-2)!}{(l+2)!}\prod_{i=1}^l (\frac{2r_iN^{d_i} (r_i)}{(d_i +r_i -1)!})\}$
Where,
$N^d(r)=0$ for $r<0$.
The inner sum runs over all sequences $d_1,\ldots,d_l$ and $r_1,\ldots,r_l$ where $d_i\geq r_i\geq 2,\,\forall i$ and $\sum _{i=1}^l r_i=r+l$, $\sum_{i=1}^l d_i=d-2$.
$\sigma$ equals to the order of the symmetry group of $<d_i| d_i=r_i>$
I'm pretty sure that $\log(N^d(r))<=2d\log d+O(d)$ or at most $\limsup\frac{\log N^d (r)}{2d\log d}\leq 1$, however, I don't know how to prove it.
Is this bound true, if not, what is the best bound that we can give? Any ideas? |
Accumulation Points of a Set in a Topological Space
Recall from the The Open Neighbourhoods of Points in a Topological Space page that if $(X, \tau)$ is a topological space and $x \in X$ then a open set $U \in \tau$ is called an open neighbourhood of $x$ if $x \in U$.
We will now define a very important type of point of a set in a topological spaces known as accumulation points.
Definition: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. A point $x \in X$ is called an Accumulation Point of $A$ if every open neighbourhood of $x$ contains a point in $A$ different from $x$. The Set of all Accumulation Points of $A$ (sometimes called the Derived Set of $A$) is denoted by $A'$. The terms " Limit Point" and "Cluster Point" are sometimes used to mean "accumulation point".
Proposition 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A \cup A'$ is closed. Proof:Let $x \in X \setminus (A \cup A')$. Then $x \not \in A$ and $x \not \in A'$. So $x$ is not in $A$ and $x$ is not an accumulation point of $A$. Since $x$ is not an accumulation point of $A$ there exists an open neighbourhood $U_x$ of $x$ such that $U_x$ does not contain a point in $A$ different from $x$. But also, $x \not \in A$, and so $U_x \cap A = \emptyset$. Since such a neighbourhood $U_x$ exists for every $x \in X \setminus (A \cup A')$ we see that $X \setminus (A \cup A')$ is open, and thus $A \cup A'$ is closed. $\blacksquare$
Let's now look at some examples.
Example 1
Consider the finite set $X = \{ a, b, c \}$ and the nested topology $\tau = \{ \emptyset, \{ a \}, \{ a , b \}, \{ a, b, c \} \}$. Let $A = X$ itself. The point $a \in X$ is NOT an accumulation point because the open neighbourhood $\{ a \} \in \tau$ of $a$ does not contain any points different from $a$.
However, the points $b, c \in X$ are accumulation points of $X$. The open neighbourhoods of $b$ are $\{ a, b \}$ and $\{a, b, c \}$, and in each of these open neighbourhoods there exists points different from $b$. Similarly, the only open neighbourhood of $c$ is $\{ a, b , c \}$ and there exists points different from $c$ in this open neighbourhood.
Example 2
Consider the set of real numbers $\mathbb{R}$ with the usual Euclidean topology. Let $A = (0, 1)$. Then every $x \in A$ is an accumulation point of $A$. Also, $0$ and $1$ (which are not in $A$) are accumulation points of $A$. In fact, $A' = [0, 1]$.
To partially verify this, observe that if $x \in [0, 1]$ then for any $\epsilon > 0$, $(x - \epsilon, x + \epsilon)$ is an open neighbourhood of $x$ that intersects $A = (0, 1)$ for which the intersection contains a member of $A$ different from $x$. This sufficiently shows that $[0, 1] \subseteq A'$ for if $U$ is any open neighbourhood of $x$ then $U$ must contain a set of the form $(x - \epsilon, x + \epsilon)$ for some $\epsilon > 0$.
Example 3
Consider the set of real numbers $\mathbb{R}$ where the topology $\tau$ is given by:(1)
Let $A = \mathbb{R}$. Consider the point $0 \in \mathbb{R}$. Then $(-1, 1)$, $(-2, 2)$, …, $(-n, n)$, …, $\mathbb{R}$ are all of the open neighbourhoods of $0$. Each of these open intervals contains points different from $0$, so $0 \in \mathbb{R}$ is an accumulation point in $\mathbb{R}$ with the topology $\tau$.
In fact, every element $x \in \mathbb{R}$ is an accumulation point. To prove this, fix $x \in \mathbb{R}$. Then $\mid x \mid > 0$ and by The Archimedean Property there exists a smallest natural number $n(x) \in \mathbb{N}$ (dependent on $x$) such that $\mid x \mid < n(x)$, so:(2)
By the density of the real numbers, there exists a $\xi \neq x$ such that $-n(x) < x < \xi < n(x)$. So $\xi \in (-n(x), n(x))$. All other larger open neighbourhoods of $x$ are of the form:(3)
Furthermore, the open neighbourhoods of $x$ are nested:(4)
Therefore, for each open neighbourhood of $x$ there exists a $\xi \in \mathbb{R}$, $\xi \neq x$ such that $\xi$ is contained in each of these open neighbourhoods. Therefore every $x \in \mathbb{R}$ is an accumulation point of $\mathbb{R}$ with this topology. |
Homotopically Equivalent Topological Spaces
Definition: Let $X$ and $Y$ be topological spaces. Then $X$ and $Y$ are said to be Homotopically Equivalent if there exists continuous functions $f : X \to Y$ and $g : Y \to X$ such that $g \circ f = \mathrm{id}_X$ and $f \circ g = \mathrm{id}_Y$.
For example, let $D^2$ denote the closed unit disk in $\mathbb{R}^2$. That is:(1)
And let $p$ be any point in $\mathbb{R}^2$.
Let $X = D^2$ and $Y = \{ p \}$. We claim that $X$ homotopically equivalent to $Y$, i.e., that the closed unit disk is homotopically equivalent to a single point in $\mathbb{R}^2$. Define $f : X \to Y$ for all $(x, y) \in X = D^2$ by:(2)
And define $g : Y \to X$ by:(3)
Then $f$ is continuous as it is a constant function, and $g$ is trivially continuous. Furthermore:(4)
And:(5)
So $g \circ f = (0, 0)$ and $f \circ g = \mathrm{id}_Y$.
We have
almost shown that $X$ is homotopically equivalent to $Y$. The only problem is that $g \circ f \neq \mathrm{id}_X$. We instead show that $g \circ f$ is homotopic to $\mathrm{id}_X$.
Let $H : X \times I \to Y$ be defined by:(6)
Then $H$ is clearly a continuous function, $H_0 = H(x, 0) = 1(x, y) = \mathrm{id}_X$, and $H_1 = H(x, 1) = 0(x, y) = (0, 0) = g \circ f$. So indeed, $g \circ f$ is homotopic to $\mathrm{id}_X$. So $D^2$ is homotopically equivalent to $\{ p \}$. |
Table of Contents
Isotopic and Non-Isotopic Embeddings on the Bounded Cone
Recall from the Isotopic Embeddings on Topological Spaces page that if $X$ and $Y$ are topological spaces and $f, g : X \to Y$ are embeddings then $f$ and $g$ are said to be isotopic if there exists a continuous function $H : X \times I \to Y$ such that:
1)$H_t : X \to Y$ is an embedding for each $t \in I$. 2)$H_0 = f$. 3)$H_1 = g$.
And such a function $H$ is called an isotopy. We now look at an example of isotopies of the circle embedded in a bounded cone. Let $X = S^1$ be the unit circle and let $Y$ be the bounded cone (sometimes referred to as the bounded double cone). First consider the following embeddings of $X$ into $Y$:
Observe that no isotopy can exist between these embeddings. If so, we can move the blue circle to the pink circle. But at some point we must cross the center point of the cone. But this cannot happen since then $H_t$ would not be an embedding for some $t \in [0, 1]$
Now consider the following embeddings of $X$ into $Y$:
For the same reason as above, there is no isotopy between the embedding of the red circle and the embedding of the yellow circle. However, there is an isotopy between the embedding of the red circle and the embedding of the orange circle. |
Polynomials Review
Polynomials Review Recall that a Polynomialif a function of the form $p(x) = a_0 + a_1x + a_2x^2 + .... + a_nx^n$ where $a_0, a_1, ..., a_n$ are coefficients from the field $\mathbb{F}$. If $a_n \neq 0$, then the Degreeof the polynomial $p$ is $\mathrm{deg} (p) = n$, that is, the largest exponent attached to the variable $x$. A Root(also known as a solution or zero) of the polynomial $p$ is the number $\lambda \in \mathbb{F}$ such that $p(\lambda) = 0$. Recall from the Properties of Polynomials page that if $p$ is a polynomial such that $\mathrm{deg} (p) = n ≥ 1$ then $\lambda \in \mathbb{F}$ is a root of $p$ if and only if we can factor $p$ with a polynomial $q$ with $\mathrm{deg} (q) = n - 1$ as a factor such that:
\begin{align} \quad p(x) = (x - \lambda) q(x) \end{align}
Furthermore, we saw that a polynomial $p$ with $\mathrm{deg} (p) = n$ has at most $n$ distinctroots. A quick way to see this is that if we have $m > n$ roots, then for each root $\lambda_i$ we have that $(x - \lambda_i)$ is a factor (for $i = 1, 2, ..., m > n$, and the product of these factors would be a polynomial $q(x) = (x - \lambda_1)(x - \lambda_2)...(x - \lambda_m)$ whose degree is $m$ which implies that $p \neq q$ - a contradiction. We also note that if $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ is a polynomial such that $p(x) = 0$ for each $x \in \mathbb{F}$ then $a_0 = a_1 = ... = a_n = 0$. We then looked at one of the most notable theorems in mathematics known as The Fundamental Theorem of Algebra which says that if $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ is a polynomial with $\mathrm{deg} (p) = n$ and the coefficients $a_0, a_1, ..., a_n \in \mathbb{C}$ (that is the coefficients of $p$ are complexnumbers) then $p$ has exactly $n$ roots. Furthermore, we saw that if $p$ is a polynomial with complexcoefficients then there exists a unique factorization of $p$ of the following form (for $\lambda_1$, $\lambda_2$, …, $\lambda_m$ as the roots of $p$ and $c \in \mathbb{C}$):
\begin{align} \quad c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m) \end{align}
On the Pairs of Complex Roots for Polynomials with Real Coefficients page we noted that if $p$ is a polynomial with realcoefficients and if $\lambda$ is a complex root of $p$ then $\bar{\lambda}$ is also a root of $p$. A prime example of this occurs in the polynomial $p(x) = 1 + x^2$. Clearly $p$ has real coefficients, and the roots of $p(x)$ are $\lambda_1 = i$ and $\lambda_2 = -i$. Note that $\lambda_1 = \bar{\lambda_2}$. As an important corollary, we saw that since complex roots come in pairs, then polynomials with odd degree must contain at least one real root since if all of the roots were complex, then there would be one complex root that was not paired with another contradicting what we just mentioned. Additionally, if $p$ is a nonconstant polynomial with realcoefficients then there exists a unique factorization of $p$ (where $\lambda_1, \lambda_2, ..., \lambda_m \in \mathbb{R}$ are roots of $p$, and $c, \alpha_1, \alpha_2, ..., \alpha_n, \beta_1, \beta_2, ..., \beta_n \in \mathbb{R}$):
\begin{align} \quad p(x) = c(x - \lambda_1)(x - \lambda_2)...(x - \lambda_m)(x^2 + \alpha_1x + \beta_1)(x^2 + \alpha_2x + \beta_2)...(x^2 + \alpha_nx + \beta_n) \end{align}
Each of the factors $(x^2 + \alpha_ix + \beta_i)$ for $i = 1, 2, ..., n$ are the irreducible quadratic factors for $p$ and $\alpha_i^2 < 4 \beta_i$ for each $i$. On the Root-Finding Techniques for Polynomials page we saw the Rational Root Theorem which says that if $r = \frac{p}{q}$ is a rational root of the polynomial $f(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ whose coefficients $a_0, a_1, ..., a_n \in \mathbb{Z}$ (having integer coefficients) then $p$ divides $a_0$ and $q$ divides $a_n$. We also mentioned Descarte's Rule of Signs which says that if $p$ is a polynomial with real coefficients written as $p(x) = a_nxn + a_{n-1}x^{n-1} + ... + a_1x + a_0$ then the number of positive real roots of $p$ is equal to the number of sign changes of $p(x)$ or the number of sign changes of $p(x)$ minus an even positive integer. Furthermore, the number of negative roots of $p(x)$ is equal to the number of sign changes in $p(-x)$ or the number of sign spaces of $p(-x)$ minus an even positive integer. |
When we are working with a new function, it is useful to know as much as we can about the function: its graph, where the function is zero, and any other special behaviors of the function. We will begin this exploration of linear functions with a look at graphs.
When graphing a linear function, there are three basic ways to graph it:
By plotting points (at least 2) and drawing a line through the points Using the initial value (output when x= 0) and rate of change (slope) Using transformations of the identity function \(f(x)=x\)
Example 1
Graph \(f(x)=5-\dfrac{2}{3} x\) by plotting points.
Solution
In general, we evaluate the function at two or more inputs to find at least two points on the graph. Usually it is best to pick input values that will “work nicely” in the equation. In this equation, multiples of 3 will work nicely due to the \(\dfrac{2}{3}\) in the equation, and of course using \(x = 0\) to get the vertical intercept. Evaluating \(f(x)\) at \(x = 0\), 3 and 6:
\(f(0) = 5 - \dfrac{2}{3} (0) = 5\)
\(f(3) = 5 - \dfrac{2}{3} (3) = 3\)
\(f(6) = 5 - \dfrac{2}{3} (6) = 1\)
These evaluations tell us that the points (0,5), (3,3), and (6,1) lie on the graph of the line. Plotting these points and drawing a line through them gives us the graph.
When using the initial value and rate of change to graph, we need to consider the graphical interpretation of these values. Remember the initial value of the function is the output when the input is zero, so in the equation \(f(x)=b+mx\), the graph includes the point \((0, b)\). On the graph, this is the vertical intercept – the point where the graph crosses the vertical axis.
For the rate of change, it is helpful to recall that we calculated this value as
\(m = \dfrac{change\ of\ output}{change\ of\ input}\)
From a graph of a line, this tells us that if we divide the vertical difference, or rise, of the function outputs by the horizontal difference, or run, of the inputs, we will obtain the rate of change, also called slope of the line.
\(m=\dfrac{change\ of \ output}{change\ of\ input} = \dfrac{rise}{run}\)
Notice that this ratio is the same regardless of which two points we use.
Definition: Graphical interpretation of a linear equation
Graphically, in the equation\(f(x)=b+mx\),
\(b\) is the
vertical intercept of the graph and tells us we can start our graph at (0, b)
\(m\) is the
slope of the line and tells us how far to rise & run to get to the next point
Once we have at least 2 points, we can extend the graph of the line to the left and right.
Example 2
Graph \(f(x)=5-\dfrac{2}{3} x\) using the vertical intercept and slope.
Solution
The vertical intercept of the function is (0, 5), giving us a point on the graph of the line.
The slope is \(-\dfrac{2}{3}\). This tells us that for every 3 units the graph “runs” in the horizontal, the vertical “rise” decreases by 2 units.
In graphing, we can use this by first plotting our vertical intercept on the graph, then using the slope to find a second point. From the initial value (0, 5) the slope tells us that if we move to the right 3, we will move down 2, moving us to the point (3, 3). We can continue this again to find a third point at (6, 1). Finally, extend the line to the left and right, containing these points.
Exercise
Consider that the slope \(-\dfrac{2}{3}\) could also be written as \(\dfrac{2}{-3}\). Using \(\dfrac{2}{-3}\), find another point on the graph that has a negative \(x\) value.
Answer
(-3,7) found by starting at the vertical intercept, going up 2 units and 3 in the negative horizontal direction. You could have also answered, (-6, 9) or (-9, 11) etc…
Another option for graphing is to use transformations of the identity function\(f(x)=x\).
In the equation\(f(x)=mx\), the m is acting as the vertical stretch of the identity function. When \(m\) is negative, there is also a vertical reflection of the graph. Looking at some examples:
In\(f(x)=mx+b\), the
b acts as the vertical shift, moving the graph up and down without affecting the slope of the line. Some examples:
Using Vertical Stretches or Compressions along with Vertical Shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way for you to graph this type of function, make sure you practice each method.
Example 3
Graph \(f(x)=-3+\dfrac{1}{2} x\) using transformations.
Solution
The equation is the graph of the identity function vertically compressed by ½ and vertically shifted down 3.
Notice how this nicely compares to the other method where the vertical intercept is found at (0, -3) and to get to another point we rise (go up vertically) by 1 unit and run (go horizontally) by 2 units to get to the next point (2, -2), and the next one (4, -1). In these three points (0, -3), (2, -2), and (4, -1), the output values change by +1, and the \(x\) values change by +2, corresponding with the slope \(m = 1/2\).
Example 4
Match each equation with one of the lines in the graph below
\(f(x) = 2x + 3\)
\(g(x) = 2x - 3\)
\(h(x) = -2x = 3\)
\(j(x) = \dfrac{1}{2} x + 3\)
Solution
Only one graph has a vertical intercept of - 3, so we can immediately match that graph with \(g(x)\).
For the three graphs with a vertical intercept at 3, only one has a negative slope, so we can match that line with \(h(x)\). Of the other two, the steeper line would have a larger slope, so we can match that graph with equation \(f(x)\), and the flatter line with the equation \(j(x)\).
In addition to understanding the basic behavior of a linear function (increasing or decreasing, recognizing the slope and vertical intercept), it is often helpful to know the horizontal intercept of the function – where it crosses the horizontal axis.
Definition: Finding Horizontal intercepts
The
horizontal intercept of the function is where the graph crosses the horizontal axis. If a function has a horizontal intercept, you can always find it by solving \(f(x) = 0\).
Example 5
Find the horizontal intercept of \(f(x)=-3+\dfrac{1}{2} x\)
Solution
Setting the function equal to zero to find what input will put us on the horizontal axis,
\(\begin{array} {rcl} {0} &= & {-3 + \dfrac{1}{2}x} \\ {3} &= & {\dfrac{1}{2} x} \\ {x} &= & {6} \end{array}\)
The graph crosses the horizontal axis at (6,0)
There are two special cases of lines: a horizontal line and a vertical line. In a horizontal line like the one graphed to the right, notice that between any two points, the change in the outputs is 0. In the slope equation, the numerator will be 0, resulting in a slope of 0. Using a slope of 0 in the \(f(x)=b+mx\), the equation simplifies to \(f(x)=b\).
Notice a horizontal line has a vertical intercept, but no horizontal intercept (unless it’s the line \(f(x) = 0\)).
In the case of a vertical line, notice that between any two points, the change in the inputs is zero. In the slope equation, the denominator will be zero, and you may recall that we cannot divide by the zero; the slope of a vertical line is undefined. You might also notice that a vertical line is not a function. To write the equation of vertical line, we simply write input=value, like \(x = b\).
Notice a vertical line has a horizontal intercept, but no vertical intercept (unless it’s the line \(x = 0\)).
Definition: horizontal and vertical lines
Horizontal lines have equations of the form \(f(x)=b\) Vertical lines have equations of the form \(x = a\)
Example 6
Write an equation for the horizontal line graphed above.
Solution
This line would have equation \(f(x)=2\)
Example 7
Write an equation for the vertical line graphed above.
Solution
This line would have equation \(x=2\)
Exercise
Describe the function \(f(x)=6-3x\) in terms of transformations of the identity function and find its horizontal intercept.
Answer
Vertically stretched by a factor of 3, Vertically flipped (flipped over the \(x\) axis),
Vertically shifted up by 6 units.
Horizontal intercept: \(6 - 3x = 0\) when \(x=23\).
Parallel and Perpendicular Lines
When two lines are graphed together, the lines will be
parallel if they are increasing at the same rate – if the rates of change are the same. In this case, the graphs will never cross (unless they’re the same line).
Definition: Parallel lines
Two lines are
parallel if the slopes are equal (or, if both lines are vertical).
In other words, given two linear equations \(f(x)=b+m_{1} x\) and \(g(x)=b+m_{2} x\), the lines will be parallel if \(m_{1} =m_{2}\).
Example 8
Find a line parallel to \(f(x)=6+3x\) that passes through the point (3, 0).
Solution
We know the line we’re looking for will have the same slope as the given line, \(m = 3\). Using this and the given point, we can solve for the new line’s vertical intercept:
\(g(x) = b + 3x\) then at (3, 0),
\(0 = b + 3(3)\)
\(b = -9\)
The line we’re looking for is \(g(x)=-9+3x\).
If two lines are not parallel, one other interesting possibility is that the lines are perpendicular, which means the lines form a right angle (90 degree angle – a square corner) where they meet. In this case, the slopes when multiplied together will equal -1. Solving for one slope leads us to the definition:
Definition: Perpendicular lines
Given two linear equations \(f(x)=b+m_{1} x\) and \(g(x)=b+m_{2} x\)
The lines will be
perpendicular if \(m_{1} m_{2} =-1\), and so \(m_{2} =\dfrac{-1}{m_{1} }\)
We often say the slope of a perpendicular line is the “negative reciprocal” of the other line’s slope.
Example 9
Find the slope of a line perpendicular to a line with:
a) a slope of 2. b) a slope of -4. c) a slope of \(\dfrac{2}{3}\).
Solution
If the original line had slope 2, the perpendicular line’s slope would be \(m_{2} =\dfrac{-1}{2}\)
If the original line had slope -4, the perpendicular line’s slope would be \(m_{2} =\dfrac{-1}{-4} =\dfrac{1}{4}\)
If the original line had slope \(\dfrac{2}{3}\), the perpendicular line’s slope would be \(m_2 = \dfrac{-1}{2/3} = \dfrac{-3}{2}\)
Example 10
Find the equation of a line perpendicular to \(f(x)=6+3x\) and passing through the point (3, 0).
Solution
The original line has slope m = 3. The perpendicular line will have slope \(m=\frac{-1}{3}\). Using this and the given point, we can find the equation for the line.
\(g(x)=b-\dfrac{1}{3} x\) then at (3, 0),
\(0 = b - \dfrac{1}{3} (3)\)
\(b = 1\)
The line we’re looking for is \(g(x)=1-\dfrac{1}{3} x\).
Exercise
Given the line \(h(t)=-4+2t\), find an equation for the line passing through (0, 0) that is:
parallel to h(t). perpendicular to h(t). Answer
Parallel: \(f(t)=2t\)
Perpendicular: \(g(t)=-\dfrac{1}{2} t\)
Example 12
A line passes through the points (-2, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).
Solution
From the two given points on the reference line, we can calculate the slope of that line:
\(m_1 = \dfrac{5-6}{4 - (-2) = \dfrac{-1}{6}\)
The perpendicular line will have slope
\(m_2 = \dfrac{-1}{-1/6} = 6\)
We can then solve for the vertical intercept that makes the line pass through the desired point:
\(g(x) = b + 6x\) then at (4, 5)
\(5 = b + 6(4)\)
\(b = -19\)
Giving the line \(g(x)=-19+6x\)
Intersections of Lines
The graphs of two lines will intersect if they are not parallel. They will intersect at the point that satisfies both equations. To find this point when the equations are given as functions, we can solve for an input value so that \(f(x)=g(x)\). In other words, we can set the formulas for the lines equal, and solve for the input that satisfies the equation.
Example 13
Find the intersection of the lines \(h(t)=3t-4\) and \(j(t)=5-t\).
Solution
Setting \(h(t)=j(t)\),
\(\begin{array} {rcl} {3t - 4} &= & {5 - t} \\ {4t} &= & {9} \\ {t} &= & {\dfrac{9}{4} \end{array}\)
This tells us the lines intersect when the input is \(\dfrac{9}{4}\).
We can then find the output value of the intersection point by evaluating either function at this input
\(j(\dfrac{9}{4} )=5-\dfrac{9}{4} =\dfrac{11}{4}\)
These lines intersect at the point \((\dfrac{9}{4} ,\dfrac{11}{4} )\). Looking at the graph, this result seems reasonable.
Two parallel lines can also intersect if they happen to be the same line. In that case, they intersect at every point on the lines.
Exercise
Look at the graph in example 13 above and answer the following for the function \(h(t)\):
a. Vertical intercept coordinates
b. Horizontal intercepts coordinates
c. Slope
d. Is \(j(t)\) parallel or perpendicular to \(h(t)\) (or neither)
e. Is h(t) an Increasing or Decreasing function (or neither)
f. Write a transformation description from the identity toolkit function \(f(x) = x\)
Answer
4. Given \(h(t)=3t-4\)
a. (0,-4)
b. \((\dfrac{4}{3} ,0)\)
c. Slope 3
d. Neither parallel nor perpendicular
e. Increasing function
f. Given the identity function, vertically stretch by 3 and shift down 4 units.
Finding the intersection allows us to answer other questions as well, such as discovering when one function is larger than another.
Example 14
Using the functions from the previous example, for what values of
t is \(h(t) > j(t)\). Solution
To answer this question, it is helpful first to know where the functions are equal, since that is the point where \(h(t)\) could switch from being greater to smaller than \(j(t)\) or vice-versa. From the previous example, we know the functions are equal at \(t=\dfrac{9}{4}\).
By examining the graph, we can see that \(h(t)\), the function with positive slope, is going to be larger than the other function to the right of the intersection. So \(h(t) > j(t)\) when \(t > \dfrac{9}{4}\)
Important Topics of this Section Methods for graphing linear functions Another name for slope = rise/run Horizontal intercepts (a,0) Horizontal lines Vertical lines Parallel lines Perpendicular lines Intersecting lines |
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