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# Klein-Gordon eqn: why dismiss messages at phase velocity

Hi All,

I've heard it said that the superluminal phase velocity of the KG eqn is not a problem for relativistic causality because signals travel at the packet/group velocity, which is the inverse of the phase velocity (c being 1). I'm a bit skeptical of this.

We can strip away all the quantum mystique and just consider the (1 dimensional) KG eqn to describe a guitar string with restoring springs along it, or (to put the same thing the other way around) a row of simple harmonic oscillators coupled together. Either way, we can bang on one end with a delta function and watch waves of all wavelengths spread along it at various superluminal speeds including infinity. To convince ourselves that signals can't travel at the phase velocity, we'll somehow have to prove that all those different components add up to zero at every event that's spacelike separated from the impulse. This would seem like a remarkable coincidence - there may be traveling nodes, but zero for the whole continuous spacetime region???

I'm not sure how to type formulae here, but I think I may have proved the opposite. We can consider an event at a spacelike interval from the impulse. Long waves travel faster under KG, so at that event, waves with k from 0 to K will be influencing psi at that event, but the shorter ones won't have arrived yet. We know K because we know that w^2 = k^2 + m^2. I'm not entirely sure what the amplitudes of the components of the fourier transform of the delta function are when the velocity isn't trivial, but let's just make it a constant. Whatever, you wind up with an integral over k of an exponential of i times something complicated but real. It doesn't really matter that it's complicated. The answer will always be some complex number of a known non-zero magnitude.

atyy

Avodyne
We can strip away all the quantum mystique and just consider the (1 dimensional) KG eqn to describe a guitar string with restoring springs along it, or (to put the same thing the other way around) a row of simple harmonic oscillators coupled together. Either way, we can bang on one end with a delta function and watch waves of all wavelengths spread along it at various superluminal speeds including infinity. To convince ourselves that signals can't travel at the phase velocity, we'll somehow have to prove that all those different components add up to zero at every event that's spacelike separated from the impulse.
Which is exactly what happens.

This has just been extensively discussed in this thread:

Also, if you consider the one-dimensional massless KG equation with an initial condition ##\varphi(x,0)=f(x)## and ##\dot\varphi(x,0)=0## (where the dot denotes the time derivative), then the exact solution is
$$\varphi(x,t)=\frac12\bigl[f(x-ct)+f(x+ct)\bigr].$$
So if ##\varphi(x,0)## is zero outside some interval, then ##\varphi(x,t)## remains zero everywhere that is spacelike separated from the interval at time ##t##.

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If it's massless I agree that everything travels at c and we have no problem. The massive field is the interesting case.

I read through that thread, but I still don't find an answer. I'm proposing that at t=x=0 I will flip a coin and bang on a KG field with either a positive or a negative delta function. It seems to me that a spacelike separated observer can tell the difference. I see nothing in that thread to say that those "phase-fronts" will all cancel out.

Avodyne
It's a theorem. Wave fronts do not travel faster than light, independent of the mass-squared (positive, negative, or zero).

Does this theorem have a name?

Demystifier
Gold Member
I read through that thread, but I still don't find an answer. I'm proposing that at t=x=0 I will flip a coin and bang on a KG field with either a positive or a negative delta function. It seems to me that a spacelike separated observer can tell the difference. I see nothing in that thread to say that those "phase-fronts" will all cancel out.
They will. One way to see it is to consider the appropriate Green function, which vanishes outside the light-cone. The appropriate Green function can be expressed as a sum of other "inappropriate" Green functions, so that the terms outside the light-cone cancel out. The full solution of the partial differential equation can be related to the initial condition via the appropriate Green function, so vanishing of the Green function implies also vanishing of the solution.

Of course, mathematically, one can also take a different Green function and in this way obtain a solution which propagates faster than light. However, for such a solution the initial condition is very constrained so the Cauchy problem with arbitrary initial condition is not well posed. In particular, you cannot have a delta-function as an initial condition. Such solutions are generally considered to be unphysical. Typically, the initial conditions related to such solutions cannot even be Fourier expanded.

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atyy
Demystifier
Gold Member
Does this theorem have a name?
Mathematicians usually refer to it as theorems regarding the domain of influence.

Can somebody refer me to some actual maths for this please? I did find the QFT stuff about the Feynmann propagator, etc, but that's a totally different model involving a bunch of coupled *quantum* simple harmonic oscillators, whereas the KG equation involves a bunch of coupled *classical* simple harmonic oscillators. I think the psi in the KG equation can be treated as real because even if it was complex, the real and imaginary terms would be independent systems in the absence of any first order diffs to mix them up. Somebody pointed me to the Kramers-Kronig relations but in that case the deal seems to be "If you want a causal fourier transform then you'll have to tolerate some friction in your system", but the KG equation doesn't have any friction so I guess I'll have to live without the causal fourier transform. But I still haven't found the maths to prove that those fast waves cancel out in a real, frictionless, 1D, KG system.

Demystifier
Gold Member
"If you want a causal fourier transform then you'll have to tolerate some friction in your system",
That's a nonsense. You can contruct the causal Green function by deforming the contour of integration over frequency, which effectively means that you add an infinitesimal imaginary energy to the theory, which mathematically looks similar to an infinitesimal friction. But this is merely a mathematical trick. You can also construct the causal Green function in a different way, without ever introducing the integration over frequency. Physicsts like the formalism with integration over frequency because that makes the formalism manifestly Lorentz invariant, but in the final expression there is no integration over frequency, only an integral over the spacial wave 3-vector k.

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atyy
Demystifier
Gold Member
But I still haven't found the maths to prove that those fast waves cancel out in a real, frictionless, 1D, KG system.
What you need is to find a proof that the causal Green function vanishes outside the light-cone. It is done in many textbooks. See e.g.
http://nptel.ac.in/courses/115106058/greenfunction.pdf
Eq. (9) and comments on it. For an exercise you can try to evaluate the two terms in (8) separately, and show that each of them treated separately does not vanish outside the lightcone. But outside the lightcone, as shown after (9), their sum is zero.

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atyy
Demystifier
Gold Member
I did find the QFT stuff about the Feynmann propagator, etc, but that's a totally different model involving a bunch of coupled *quantum* simple harmonic oscillators, whereas the KG equation involves a bunch of coupled *classical* simple harmonic oscillators.
That's not really totally different. You can construct the Feynman Green function for classical KG field, and it coincides with Feynman propagator for quantum KG field.

I've seen that kind of thing but it seems both overcomplicated and unsatisfactory. It hits a pole at the one and only point of interest and forces real quantities to be supposed complex for no other reason. So what's wrong with this approach...

$$\frac{\partial^2 \Psi }{\partial t^2} = \frac{\partial^2 \Psi }{\partial x^2} - m^2 \Psi$$
$$\Psi(k) = C(k).e^{i(\omega t-kx)}$$
where C is just an amplitude for that mode.
$$\omega^2 = k^2 + m^2$$
Phase velocity:
$$v(k) = \frac{\omega}{k} = \sqrt{1+\frac{m^2}{k^2}}$$

If I hit the origin with a very narrow gaussian impulse, I'll get a flat distribution of wavelengths spreading out in both directions, so ##C(k)## is constant. At an event t=1, x>1 where waves with ## k=K ## have only just arrived:

$$\Psi(k) = e^{i(\omega t - kx )} = e^{i(\omega - k.v(K) )}$$

$$\Psi(k) = e^{i(\sqrt{k^2 + m^2} - k.\sqrt{1 + m^2/K^2} )} = e^{ik(\sqrt{1 + m^2/k^2} - \sqrt{1 + m^2/K^2} )}$$

$$\Psi = \int_0^K dk.e^{ik(\sqrt{1 + m^2/k^2} - \sqrt{1 + m^2/K^2} )}$$

I don't know how to do that integral, but it doesn't look like zero to me. The ##k(\sqrt{1 + m^2/k^2} - \sqrt{1 + m^2/K^2} )## is 0 at ##k=0## and ##k=K## and positive in between. So the exponential heads off from 1 around the Argand plane and then turns around and goes back home. It'll take one hell of a coincidence to make that add up to zero.

BTW, the displacement of the string is the imaginary part of ##\Psi## because that narrow gaussian is an impulse.

Avodyne
That you get zero is indeed subtle from the point of view of these integrals.

As an example, here is an exact solution of the KG equation, valid for ##t>0##:
$$\varphi(x,t)=J_0\bigl(m\sqrt{t^2-x^2}\bigr)\theta(t-|x|)$$where ##J_0## is the ordinary Bessel function and ##\theta## is the unit step function. For ##t\ll 1/m##, this is a rectangular pulse (of unit height) that extends from ##x=-t## to ##x=+t##, with its leading edges moving out at the speed of light. Later it turns wavy behind the edges. This solution is the analog of the solution of the wave equation (KG with ##m=0##) of the form ##\theta(x+t)-\theta(x-t)##, which describes a unit height pulse that also expands outward, but with its edges staying sharp forever.

This solution illustrates that local disturbances have leading edges that propagate at the speed of light.

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Demystifier
Gold Member
I've seen that kind of thing but it seems both overcomplicated and unsatisfactory. It hits a pole at the one and only point of interest and forces real quantities to be supposed complex for no other reason. So what's wrong with this approach...

$$\frac{\partial^2 \Psi }{\partial t^2} = \frac{\partial^2 \Psi }{\partial x^2} - m^2 \Psi$$
$$\Psi(k) = C(k).e^{i(\omega t-kx)}$$
where C is just an amplitude for that mode.
$$\omega^2 = k^2 + m^2$$
Phase velocity:
$$v(k) = \frac{\omega}{k} = \sqrt{1+\frac{m^2}{k^2}}$$

If I hit the origin with a very narrow gaussian impulse, I'll get a flat distribution of wavelengths spreading out in both directions, so ##C(k)## is constant. At an event t=1, x>1 where waves with ## k=K ## have only just arrived:

$$\Psi(k) = e^{i(\omega t - kx )} = e^{i(\omega - k.v(K) )}$$

$$\Psi(k) = e^{i(\sqrt{k^2 + m^2} - k.\sqrt{1 + m^2/K^2} )} = e^{ik(\sqrt{1 + m^2/k^2} - \sqrt{1 + m^2/K^2} )}$$

$$\Psi = \int_0^K dk.e^{ik(\sqrt{1 + m^2/k^2} - \sqrt{1 + m^2/K^2} )}$$

I don't know how to do that integral, but it doesn't look like zero to me. The ##k(\sqrt{1 + m^2/k^2} - \sqrt{1 + m^2/K^2} )## is 0 at ##k=0## and ##k=K## and positive in between. So the exponential heads off from 1 around the Argand plane and then turns around and goes back home. It'll take one hell of a coincidence to make that add up to zero.

BTW, the displacement of the string is the imaginary part of ##\Psi## because that narrow gaussian is an impulse.
Nice try, but your argument contains a subtle and deep error. From your assumption that ##\Psi(t=0)## is a ##\delta##-function it does not follow that ##c(k)={\rm const}##. All what follows is that ##|c(k)|={\rm const}##, i.e.
$$c(k)={\rm const}\, e^{i\varphi(k)}$$
As you may guess, for most choices of the phase function ##\varphi(k)## there will be no cancellation. Yet, it is possible to choose ##\varphi(k)## such that the cancellation really happens.

What does such a special choice of the phase function corresponds to? You required that ##\Psi(t=0)## is a ##\delta##-function, but you said nothing about ##\Pi(t=0)##, where
$$\Pi=d\Psi/dt$$
is the canonical momentum. On the other hand, the KG equation is a second order equation, so you must specify both ##\Psi(t=0)## and ##\Pi(t=0)## to uniquelly determine the solution. In particular, if you want the initial condition to be fully localized, you must require that both ##\Psi(t=0)## and ##\Pi(t=0)## are ##\delta##-functions. But you can easily see that for most choices of the phase function this will not happen. So you must adjust the phase function very carefully to make both initial conditions proportional to a ##\delta##-function. And, the point is, precisely when you make such an adjustement, the wanted cancellation outside the litght-cone will take place.

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atyy
@Avodyne: You're telling me things, and I guess you're probably right, but I'm trying to understand where the bug in my logic is.

@Demystifier: I glossed over this, but I did think about it. I'm not requiring that the displacement of the string is a delta function. I'm setting that to zero everywhere. I'm setting the first diff of the displacement to a narrow gaussian. I think that sets my phases - no matter which wavelength, at the origin, the displacement is zero with maximum velocity in the positive direction. I'm reading the displacement off the imaginary part of ##\Psi##. Also, when they say that the FT of a narrow gaussian is a wide gaussian, is it not the case that the phase is determined?

Incidentally, I haven't really justified flattening |##c(k)##|, but I think that we can choose ##K## and ##m## such that the whole ##k(\sqrt{...}...)## expression never exceeds ##\Pi## in which case the contributions to the imaginary part by every mode are all positive (that's if we can nail the phase down as I argued above) and |##c(k)##| wouldn't matter.

Demystifier
Gold Member
I'm not requiring that the displacement of the string is a delta function. I'm setting that to zero everywhere.
You said "If I hit the origin with a very narrow gaussian impulse, I'll get a flat distribution of wavelengths spreading out in both directions". If "narrow" means "narrow in space" , then initial displacement (right after the hit) is a delta function.

Did you perhaps meant "narrow in time, but wide in space"?

It's narrow in space and time. I just whack a point on the string with a little hammer. I keep saying "narrow gaussian" instead of "delta function" because I feel like I know where I am and can just declare the FT to be that wide gaussian, but come to think of it:
$$\delta(x) = \frac{1}{2\pi} \int_{-\infty}^\infty dk.e^{ikx}$$
basically defines ##\delta## as that whose fourier transform is a constant so it's easier for me. ##x## is a vector including space and time here.

What I'm saying here is intuitively obvious. Suppose you drop a stone in a pond and the long ripples travel faster than the short ones. Suppose also that the water is initially pushed upwards by the stone displacing it. The first wavefronts to arrive are all pushing upwards, and it really doesn't matter exactly which wavelengths arrive when or how the amplitude depends on the wavelength. The maths above is saying merely that, so why expect it to have a flaw?

Demystifier
Gold Member
It's narrow in space and time.
Then what I said in #15 and #17 can be applied to your case. Your initial impulse picks a special phase function, which in turn is responsible for the cancellation.

Avodyne
If the initial conditions are ##\varphi(x,0)=\delta(x)## and ##\dot\varphi(x,0)=0##, then the solution is
$$\varphi(x,t)=\lim_{K\to\infty}\int_{-K}^{+K}{dk\over 2\pi}\cos(kx)\cos\bigl(\sqrt{k^2+m^2}t\bigr).$$ I haven't been able to find a way to evaluate this integral in closed form (which is quite frustrating). We know that it must vanish for ##|x|>t##, but how to prove this? (Other than the roundabout method of knowing that this is a solution to the KG eq, which must satisfy the domain-of-dependence theorem.)

atyy and Demystifier
Demystifier
Gold Member
If the initial conditions are ##\varphi(x,0)=\delta(x)## and ##\dot\varphi(x,0)=0##, then the solution is
$$\varphi(x,t)=\lim_{K\to\infty}\int_{-K}^{+K}{dk\over 2\pi}\cos(kx)\cos\bigl(\sqrt{k^2+m^2}t\bigr).$$ I haven't been able to find a way to evaluate this integral in closed form (which is quite frustrating). We know that it must vanish for ##|x|>t##, but how to prove this?
Here is the proof.

First take ##|x|>0##, ##t=0##. In this case the integral reduces to
$$\int_{-\infty}^{+\infty}{dk\over 2\pi}\cos(kx)=0 \;\;\; (Eq. 1)$$

Second consider general ##x## and ##t##. Using
$$\cos a \cos b = \frac{1}{2}[\cos(a-b)+\cos(a+b)]$$
it is easy to show that the subintegral function on the right-hand side is Lorentz invariant. (To see that, make also the change of integration variable ##k\rightarrow -k## when useful.)

Third, consider any ##x## and ##t## for which ##|x|>t##. For each such pair ##(x,t)## there is a Lorentz frame ##S'## for which this will mean that ##|x'|>0##, ##t'=0##. We have seen that the subintergral function is Lorentz invariant, so we can calculate it in ##S'##. Thus for each ##(x,t)##, ##|x|>t##, the integral can be reduced to
$$\int_{-\infty}^{+\infty}{dk\over 2\pi}\cos(k'x') \;\;\; (Eq. 2)$$

The measure ##dk## is not Lorentz invariant, so is not equal to ##dk'##. We have a Lorentz transformation
$$k=\gamma(k'-\beta\omega')$$
where ##\gamma## and ##\beta## are constants, while ##\omega'=\sqrt{k'^2+m^2}##. Hence
$$dk=\gamma(dk'-\beta d\omega')$$
so (Eq. 2) above splits into two integrals. One takes the same form as (Eq. 1), so it vanishes. The other is proportional to
$$\int_{-\infty}^{+\infty}{dk'\over 2\pi}{k'\over\sqrt{k'^2+m^2}}\cos(k'x')$$
which vanishes because the subintegral function is anti-symmetric. Therefore (Eq. 2) is zero, which finishes the proof. Q.E.D.

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atyy
Avodyne
Alas, this isn't right. When changing integration variables from ##k## to ##k'##, the transformation of the differential is (in general)
$$dk=\left|{\partial k\over\partial k'}\right|dk'$$ You left out the absolute-value sign. When you properly include it, your final integral does not vanish by symmetry.

EDIT: This is wrong! In fact, ##{\partial k/\partial k'}## is always positive, so the absolute-value sign can be safely omitted; see below.

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Demystifier
Gold Member
Alas, this isn't right. When changing integration variables from ##k## to ##k'##, the transformation of the differential is (in general)
$$dk=\left|{\partial k\over\partial k'}\right|dk'$$ You left out the absolute-value sign. When you properly include it, your final integral does not vanish by symmetry.
First, for a one-dimensional integral one does not need to take the absolute value because
$$dk={d k\over d k'} dk'$$
without the absolute value. However, one has to be careful because, in general, the limits of integration for ##k## do not need to be the same as those for ##k'##.

Second, in our case this is not really important, because even if one takes the absolute value, one gets the same final result. Taking the absolute value, the integration measure becomes proportional to
$$\left| 1-\beta \frac{k'}{\omega(k')}\right| dk'$$
But we have
$$|\beta|=\frac{v}{c}<1$$
$$\frac{|k'|}{\omega(k')}<1$$
Consequently
$$1-\beta \frac{k'}{\omega(k')}>0$$
so we may write
$$\left| 1-\beta \frac{k'}{\omega(k')}\right| dk' = \left( 1-\beta \frac{k'}{\omega(k')}\right) dk'$$
In this way, even if one takes the absolute value, it doesn't really change the result.

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atyy
Demystifier
Gold Member
First, for a one-dimensional integral one does not need to take the absolute value because
$$dk={d k\over d k'} dk'$$
without the absolute value. However, one has to be careful because, in general, the limits of integration for ##k## do not need to be the same as those for ##k'##.
For pedagogic purposes, let me explain these subtleties in more detail. As the simplest possible case, consider the integral
$$\int_{x=a}^{x=b}dx$$
We assume that ##b>a##, i.e. that the upper limit is larger than the lower limit, which is the standard form of the integral. Consider the change of variable ##x'=-x##. The Jacobian of this transformation is
$$J=\frac{\partial x}{\partial x'}=-1$$
The integral above is equal to
$$\int_{x=a}^{x=b}-dx' = \int_{x'=-a}^{x'=-b}-dx' = \int_{x'=-a}^{x'=-b}J dx'$$
Note that we have ##J## and not ##|J|##. However, we have ##-b<-a##, i.e. the upper limit of integration is not larger than the lower limit, so the integral is not written in the standard form. The standard form is
$$\int_{x'=-b}^{x'=-a}-J dx'=\int_{x'=-b}^{x'=-a}|J| dx'$$
where ##|J|##, rather than ##J##, appears.

This is nothing but a demonstration of the general rule, valid even for multi-dimensional integrals. If all limits of integration are taken in the standard form (the upper limits are larger than the lower limits), then one has to take ##|J|##. (Here ##J## is the determinant of the Jacobian matrix, so ##J## is a real function.) If the limits are not taken in the standard form, then one has to be careful about the ##\pm## sign in front of ##|J|##.

In practice, the standard form is used almost always for multi-dimensional integrals, but not so for one-dimensional integrals.

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atyy
Avodyne


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# Community probability

#### meph1st0pheles

A community organization consists of 20 families, of which 4 have one child, 8 have two children, 5 have three children, 2 have four children, and 1 has five children.
a. If one of these families is chosen at random, what is the probability it has 1 child? 5?
b. If one of the children is randomly chosen, what is the probability that child comes from a family having one child? 5?

Your help is always appreciated. Thanks

#### soroban

Math Team
Hello, meph1st0pheles!

First of all, probability is a fraction.

A community organization consists of 20 families
. . 4 families have one child,
. . 8 families have two children,
. . 5 families have three children,
. . 2 families have four children,
. . 1 family has five children.

(a) If one of these families is chosen at random,
. . . what is the probability it has 1 child?

Of the 20 families, 4 of them have one child.

$$\displaystyle P(\text{the family has 1 child}) \;=\;\frac{4}{20} \;=\;\frac{1}{5}$$

(b) If one of the children is randomly chosen,
. . . what is the probability that child comes from a family having one child?

There are 48 children, 4 of them come from families with one child.

$$\displaystyle P(\text{the child is from a family with 1 child}) \;=\;\frac{48}{48} \;=\;\frac{1}{12}$$

format: plain


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# Is the Hilbert-Smith conjecture still unsolved?

Conjecture Let $G$ be a locally compact topological group. If $G$ has a continuous faithful group action on an $n$-manifold, then $G$ is a Lie group.

Is this conjecture still unsolved? Is there any references to more details?

• I see three references to more details in the article you yourself have linked. – Chris Eagle Mar 17 '13 at 11:01
• These three references are a bit outdated. In february, 2011 Open Problem Garden announced this conjecture as an open problem. I'm looking for More recent attempts to solve this problem. – TXC Mar 17 '13 at 11:17
• Yes it's unsolved. It's known for a while that solving it (say in dimension $\le n$) reduces to proving that for every prime $p$ the group of $p$-adics $\mathbf{Z}_p$ has no continuous faithful action on any topological connected manifold of dimension $n$. I think it's also known to be enough to prove it for connected open subsets of $\mathbf{R}^n$. – YCor Apr 28 '19 at 12:01

According to J. PARDON paper (arXiv:1112.2324) it has been resolved in dimension $3$ but in dimension $\geq 4$ is open still.


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HW-8 new

HW-8 new - 1.0 predicted value Subject 14 0.8 0.6 0.4 0.2...

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# What is the electric field strength, the energy stored and the electric flux density of ?

## A capacitor consisting of two metal plates each of an area of 50cm^2 and spaced 0.2mm apart in air and also connected across a 120V supply

Apr 26, 2017

The electric field inside a parallel plate is uniform. The calculation follows from a more general relationship:

$E = \frac{\Delta V \left(x\right)}{\Delta x}$

Instead of defining terms, I'll just plug them in so you can see for yourself. So here:

$E = \frac{120}{0.2 \cdot {10}^{- 3}} = 0.6 \cdot {10}^{6} \setminus \text{V/m}$

One expression for the potential energy stored in the capacitor is:

$U = \frac{1}{2} C {V}^{2}$

We need the capacitance :

$C = \frac{{\varepsilon}_{r} {\varepsilon}_{o} A}{d}$, because the gap is air we use ${\varepsilon}_{r} = 1$

$\implies C = \frac{8.85 \cdot {10}^{- 12} \times 0.005}{0.2 \cdot {10}^{- 3}} = 2.2 \cdot {10}^{- 10} \setminus F$

$\implies U = \frac{1}{2} \times 2.2 \cdot {10}^{- 10} \times {120}^{2} \approx 1.6 \cdot {10}^{- 6} \setminus J$

In terms of electric flux density , $D$, one expression is:

$D = {\varepsilon}_{r} {\varepsilon}_{o} E = {\varepsilon}_{o} E = 8.85 \cdot {10}^{- 12} \times 0.6 \cdot {10}^{6} \setminus \text{ C/} {m}^{2}$


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# MinionPro and beamer: figure selection

MinionPro package options for figure selection don't appear to work under beamer. I'd like text figures in text mode (textosf) and lining figures in math mode (mathlf). In the example below, the numbers in math mode appear as text figures, not lining figures as expected.

I can hook figure selection into various commands/environments to get the desired effect, but this workaround (at least as I've implemented it) is messy. Is there a better solution?

\documentclass[serif,professionalfonts]{beamer}
%\documentclass{article}
\usepackage[mathlf,textosf]{MinionPro}
\DeclareMathAlphabet{\mathlf}{T1}{MinionPro-LF}{m}{n}

\begin{document}
\begin{frame}{Slide title}
Math mode: $012345$, $$012345$$,
$012345,$
\begin{equation*}
012345.
\end{equation*}
Text mode: 012345.
Select lining figures: \figureversion{lf}012345, $$\mathlf{012345}$$.
\end{frame}
\end{document}

-

You need to use the professionalfont option. You get lining figures in math mode if you replace

\documentclass[serif,professionalfonts]{beamer}

by

\documentclass[serif,professionalfont]{beamer}

(no 's'). Alternatively, you can use the professionalfonts font theme by issuing \usefonttheme{professionalfonts} in the preamble. By the way, the class option serif is obsolete; you should use \usefonttheme{serif} instead.

-
So either \documentclass[professionalfont]{beamer} and \usefonttheme{serif} or just \usefonttheme{serif,professionalfonts}. Clearly I am not reading the documentation closely enough. Thanks for this! – Audrey Jul 26 '11 at 14:25
Yes, exactly. Clearly, it is unfortunate that the class option has a different name than the font theme. – Michael Ummels Jul 26 '11 at 14:30


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# class not registered error in windows 8.1 Terryville, Connecticut

The posting of advertisements, profanity, or personal attacks is prohibited. And it'll work on most any other add-in that affects these 2 (again, as long as they are closed). - John Microsoft MVP - Windows Experience ( http://www.carrona.org/ ) **If you the ones you are deleting). Once reported, our moderators will be notified and the post will be reviewed.

If this user-level install was ever properly uninstalled (or went through self-destruct, see explanation of this above): it should have deleted this key. or inside your user profile - appdata\local\google\??. probably due to automatic purge by my house cleaning software. The following registry (.Reg) file was able to remove all references and things remain normal until the user logs out our the PC is rebooted..

Comment 71 by [email protected], May 31 2013 Processing As of 28.0.1500.29 beta-m I am no longer getting the "Class not registered" error when attempting to launch Chrome on Windows 8 (I Attached the procmon log 6. I am nervous so say its fixed, because it worked last night via start but not this morning, now its working again. In particular the "open\command\DelegateExecute" entry of the ProgId entry should point to the DelegateExecute's GUID (as described in the DelegateExecute COM registrations bullet point).

Here goes another Expense if I call them? From the reports I've seen here (and received offline by email), I think getting rid of the key in HKCU\Software\Classes\Wow6432Node\CLSID\{5C65F4B0-3651-4514-B207-D10CB699B14B} should fix it; the problem seems to be that users affected For anyone using Canary, the same experience can be replicated by navigating to the program folder, which is either under c:\program files (x86)\google\?? I tried with and without the Wow6432Node, and couldnt find it.

I cannot find this anywhere 3.a No to HKEY_LOCAL_MACHINE\Software\Classes\CLSID\{5C65F4B0-3651-4514-B207-D10CB699B14B} 3.a Yes to HKEY_LOCAL_MACHINE\Software\Classes$$Wow6432Node$$CLSID\{5C65F4B0-3651-4514-B207-D10CB699B14B} see attached 3.b No to HKEY_LOCAL_MACHINE\Software\Classes$$Wow6432Node$$\ChromeHTML\shell\command\open\DelegateExecute 3.b No to HKEY_LOCAL_MACHINE\Software\Classes$$Wow6432Node$$\ChromeHTML\shell\command\open\DelegateExecute Howevre I do have shell\open\command, but DelegateExecute isnt Just enter your email address. The issue is usually a DLL file which runs as a COM object (COM is an MS technology that provides interface for calling applications into an object from other applications that Instead, there seems to be a much easier solution. Â Bear in mind that I have only thus far tried this onÂ oneÂ machine, but it worked immediately, and it jives with other research

The message "class not registered" is too generic, since there are many COM objects that are installed within the system. Thanks, Gab Comment 31 by [email protected], Jan 29 2013 Processing Hello, I definitely NOT have C:\Users\\AppData\Local\Google\Chrome\Application\chrome.exe, I run Chrome from c:\Program Files (x86)\Google\Chrome\Application\chrome.exe. Thank you very much! Then launch chrome.exe by double-clicking the file.

In the list on the left find Microsoft Edge and click it. Comment 70 by [email protected], May 25 2013 Processing Since running v27.0.1453.94 m of Chrome, I'm no longer getting this issue. Cheers! Log out of your current account and switch to a new one.

perhaps this help to understand all the other default-problems? We will update Chrome soon to automatically delete this bad registry entry for users who might not be aware of this thread, but you can fix it yourself right now if The odd thing is that Chrome wouldn't open using the taskbar icon but would from it's location in Program Files. Delete (if present) the following registry keys: HKEY_CURRENT_USER\Software\Classes\Wow6432Node\CLSID\{5C65F4B0-3651-4514-B207-D10CB699B14B} HKLM\Software\Classes\Chrome HKLM\Software\Classes\ChromeHTML\open\command\DelegateExecute Reboot.

I click it or double click or run as admin, or delete those things in the reg. The start button is how I almost always invoke Chrome except when I'm having this problem. iTech 618,151 views 3:34 Google Chrome 0xc000005 Application Error Fix - Duration: 1:04. Ran from Command Prompt: ipconfig / release, and ipconfig / renew.

I'll close this Post here!Sugarbear246 Flag Permalink Reply This was helpful (0) Back to Windows Legacy OS forum 24 total posts Popular Forums icon Computer Help 51,912 discussions icon Computer Newbies by R. Thoughts. Proffitt Forum moderator / August 5, 2012 7:41 AM PDT In reply to: Class not Registered?

Working... Create new task window will appear. Solution - Restart explorer.exe If you have problems with Edge, Cortana or Start menu, itâ€™s advised that you try restarting explorer.exe. Generated Tue, 04 Oct 2016 23:19:45 GMT by s_hv987 (squid/3.5.20) HomeTermsPrivacyAbout Contact Us Menu How To Solution Reviews Fix: Google Chrome Class Not Registered Error in Windows 10/8.1/7 Jit Dutta Solution

We have tried: 1. I'm using Chrome Version 25.0.1364.172 m. Any reason to stick with this hard to support OS?Bob Flag Permalink Reply This was helpful (0) Collapse - Windows XP forum: Class not Registered? Flag Permalink Reply This was helpful (0) Collapse - Answer "I WAS NOT ABLE TO INSTALL AND RUN "SuperAntiSpyware" " by R.

There shouldn't be such a key in HKEY_CURRENT_USER if you don't have user-level Chrome, that's for sure. Just a clue that may mean nothing but... Comment 40 by [email protected], Feb 2 2013 Processing Status: Started Small rectification to points 1 and 2 above: you don't actually even need to have had Google Chrome Canary installed since Comment 93 by [email protected], Jan 11 2014 Processing I've been seeing this a lot on Windows 8.1 x64.

Deleting more from the registry could only make your problems worse and won't help.I tend to agree with Bob about 'start over', although I would start with a repair install from by sugarbear246 / August 1, 2012 10:24 AM PDT In reply to: Was there a recent virus or such? Gab Comment 37 by [email protected], Jan 30 2013 Processing Arg...! I do not have a user-level chrome install at C:\Users\big\AppData\Local\Google\Chrome\Application 3.

We have already diagnosed a few issues people were having, but it seems there might be other states a few users find themselves in which still makes this error appear. Some users have reported that they are unable to open anything on their desktop, and to fix this, itâ€™s advised to set Internet Explorer as default browser. I unzipped Chromium to the system-level C:\Program Files (x86)\Chromium\chrome-win32\ and created my shortcuts to chrome.exe in that directory 2) Do you have a user-level Chrome install (i.e. in the attachement you see the only-registry-keys wich are maked from portable-chrom. 2013-05-11_113350.jpg 29.2 KB View Download Comment 67 by Deleted [email protected], May 11 2013 Processing this two entries are from

Can you please open another bug at http://new.crbug.com to track this (I don't want to diverge this thread away from the "class not registered" error) -- cc me on the new That's all there was to it, no code to identify the exact problem, just "class not registered" in a non-descript dialog box. Yes, see attachment. 5) If you do get the issue again: (a) Can you send us a procmon log (using procmon.exe from http://technet.microsoft.com/en-us/sysinternals/bb842062) -- feel free to send the log only Triple-S Computers Blog - Louisville, KY computer repair specialist Tips from the trenches of advanced computer repair Search Main menu Skip to primary content Triple-S Computers Home Like me on Facebook!

Comment 8 by [email protected], Jan 24 2013 Processing The Start Menu you installed probably launches Chrome in the old Desktop way (i.e. Post navigation ← Previous Next → SOLUTION? "Class not registered" when trying to open Chrome in Windows 8.1/10 Posted on March 21, 2016 by Steve Schardein Recently I have been seeing Comment 99 by [email protected], Jan 29 2014 Processing @#98: Deleting HKCU\Software\Classes\Wow6432Node\CLSID\{5C65F4B0-3651-4514-B207-D10CB699B14B} should do it. But, nothing will work for you.

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## Monday, May 13, 2019

### Functional Coefficient Estimation: Part I (Nonparametric Estimation)

Recently, EViews 11 introduced several new nonparametric techniques. One of those features is the ability to estimate functional coefficient models. To help familiarize users with this important technique, we're launching a multi-part blog series on nonparametric estimation, with a particular focus on the theoretical and practical aspects of functional coefficient estimation. Before delving into the subject matter however, in this Part I of the series, we give a brief and gentle introduction to some of the most important principles underlying nonparametric estimation, and illustrate them using EViews programs.

1. Nonparametric Estimation
2. Global Methods
3. Local Methods
4. Conclusion
5. Files
6. References

### Nonparametric Estimation

Traditional least squares regression is parametric in nature. It confines relationships between the dependent variable $Y_{t}$ and independent variables (regressors) $X_{1,t}, X_{2,t}, \ldots$ to be, in expectation, linear in the parameter space. For instance, if the true data generating process (DGP) for $Y_{t}$ derives from $p$ regressors, the least squares regression model postulates that: $$Y_{t} = m(x_{1}, \ldots, x_{p}) \equiv E(Y_t | X_{1,t} = x_{1}, \ldots, X_{p,t} = x_{p}) = \beta_0 + \sum_{k=1}^{p}{\beta_k x_{k}}$$ Since this relationship holds only in expectation, a statistically equivalent form of this statement is: \begin{align} Y_t &= m\left(X_{1,t}, \ldots, X_{p,t}\right) + \epsilon_{t} \nonumber \\ &=\beta_0 + \sum_{k=1}^{p}{\beta_k X_{k,t}} + \epsilon_t \label{eq.1.1} \end{align} where the error term $\epsilon_{t}$ has mean zero, and parameter estimates are solutions to the minimization problem: $$\arg\!\min_{\hspace{-1em}\beta_{0}, \ldots, \beta_{p}} E\left(Y_{t} - \beta_0 + \sum_{k=1}^{p}{\beta_k X_{k,t}}\right)^{2}$$ Nevertheless, while this framework is typically sufficient for most applications, and is obviously very appealing and intuitive, when the true but unknown DGP is in fact non-linear, inference is rendered unreliable.

On the other hand, nonparametric modelling prefers to remain agnostic about functional forms. Relationships are, in expectation, simply functionals $m(\cdot)$, and if the true DGP for $Y_{t}$ is a function of $p$ regressors, then: $$Y_t = m\left(X_{1,t}, \ldots, X_{p,t}\right) + \epsilon_{t}$$ Here, estimators of $m(\cdot)$ can generally be cast as minimization problems of the form: \begin{align} \arg\!\min_{\hspace{-1em} m\in \mathcal{M}} E\left(Y_{t} - m\left(X_{1,t}, \ldots, X_{p,t}\right)\right)^{2} \label{eq.1.2} \end{align} where $\mathcal{M}$ is now a function space. In this regard, a nonparametric estimator can be thought of as a solution to a search problem over functions as opposed to parameters.

The problem in \eqref{eq.1.2}, however, is infeasible. It turns out the function space is effectively uncountable. In fact, even if arguing to the contrary, solutions would be unidentified since different functions in $\mathcal{M}$ can map to the same range. Accordingly, general practice is to reduce $\mathcal{M}$ to a lower dimensional countable space and optimize over it. This typically implies a reduction of the problem to a parametric framework so that the problem in \eqref{eq.1.2} is cast into: \begin{align} \arg\!\min_{\hspace{-1em} m\in \mathcal{M}} E\left(Y_{t} - h\left(X_{1,t}, \ldots, X_{p,t}; \mathbf{\Theta} \right)\right)^{2} \label{eq.1.3} \end{align} where $h(\cdot; \mathbf{\Theta}) \in \mathcal{H}$ is a function with associated parameters $\mathbf{\Theta} \in \mathbf{R}^{q}$ and $\mathcal{H}$ is a function space which is dense in $\mathcal{M}$; formally, $h^{\star} \in \mathcal{H} \rightarrow m^{\star} \in \mathcal{M}$ where $\rightarrow$ denotes asymptotic convergence. Recall that this means that any feasible estimate $h^{\star}$ must become arbitrarily close to the unfeasible estimate $m^{\star}$ as the space $\mathcal{H}$ grows to asymptotic equivalence with $\mathcal{M}$. In this regard, nonparametric estimators are typically classified into either global or local kinds.

### Global Methods

Global estimators, generally synonymous with the class of sieve estimators introduced by Grenander (1981), approximate arbitrary functions by simpler functions which are uniformly dense in the target space $\mathcal{M}$. A particularly important class of such estimators are linear sieves which are constructed as linear combinations of popular basis functions. The latter include Bernstein polynomials, Chebychev polynomials, Hermite polynomials, Fourier series, polynomial splines, B-splines, and wavelets. Formally, when the function $m(\cdot)$ is univariate, linear sieves assume the following general structure: \begin{align} \mathcal{H}_{J} = \left\{h \in \mathcal{M}: h(x; \mathbf{\Theta}) = \sum_{j=1}^{J}\theta_{k}f_{j}(x)\right\} \label{eq.1.4} \end{align} where $\theta_{j} \in \mathbf{\Theta}$, $f_{j}(\cdot)$ is one of the aforementioned basis functions, and $J \rightarrow \infty$.

For instance, if the sieve exploits the Stone-Weierstrass' Approximation Theorem which claims that any continuously differentiable function over a compact interval, can be uniformly approximated on that interval by a polynomial to any degree, then $f_{j}(x) = x^{j-1}$. In particular, if the unknown function of interest is $m(x)$, then choosing to approximate the latter with a polynomial of degree $J = J^{\star} < \infty$ (some integer), reduces the problem in \eqref{eq.1.4} to: $$\arg\!\min_{\hspace{-1em} m\in \mathcal{M}} E\left(Y_{t} - \theta_{0} + \sum_{j=1}^{J^{\star}}\theta_{j}X_{t}^{j} \right)^{2}$$ where $Y_{t}$ are the values we observe from the theoretical function $m(x)$, and $X_{t}$ is the regressor we're using to estimate it. Usual least squares now yields $\widehat{\theta}_{j}$ for $j=1,\ldots, J^{\star}$. Furthermore, $m(x)$ can be approximated as $$m(x) \approx \widehat{\theta}_{0} + \sum_{j=1}^{J^{\star}}\widehat{\theta}_{j}x^{j}$$ where $x$ is evaluated on some grid $[a,b]$, where it can have arbitrary length, or even on the original regressor values so that $x \equiv X_{t}$.

To demonstrates the procedure, define the true but unknown function $m(x)$ as: \begin{align} m(x) = \sin(x)\cos(\frac{1}{x}) + \log\left(x + \sqrt{x^2+1}\right) \quad x \in [-6,6]\label{eq.1.5} \end{align} Furthermore, generate observable data from $m(x)$ as $Y_{t} = m(x) + 0.5\epsilon_{t}$ and generate the regressor data as $X_{t} = x - 0.5 + \eta_{t}$ where $\epsilon_{t}$ and $\eta_{t}$ are mutually independent respectively standard normal and standard uniform random variables. Estimation is now summarized for polynomial degrees 1, 5, and 15, respectively.

Figure 1: Polynomial Sieve Estimation

Alternatively, if the sieve exploits Hermite polynomials, one can construct the Gaussian sieve which reduces the problem in \eqref{eq.1.4} to: $$\arg\!\min_{\hspace{-1em} m\in \mathcal{M}} E\left(Y_{t} - \theta_{0} + \sum_{j=1}^{J^{\star}}\theta_{j}\phi(X_{t})H_{j}(X_{t}) \right)^{2}$$ where $\phi(\cdot)$ is the standard normal density and $H_{j}(\cdot)$ are Hermite polynomials of degree $j$. The figure below demonstrates the procedure using sieve lengths 1, 3, and 10, respectively.

Figure 2: Gaussian Sieve Estimation

Clearly, both sieve estimators are very similar. So how does one select an optimal sieve? There really isn't a prescription for such optimization. Each sieve has its advantages and disadvantages, but the general rule of thumb is to choose a sieve that most closely resembles the function of interest $m(\cdot)$. For instance, if the function is polynomial, then using a polynomial sieve is probably best. Alternatively, if the function is expected to be smooth and concentrated around its mean, a Gaussian sieve will work well. On the other hand, the question of optimal sieve length lends itself to more concrete advice.

#### Optimal Sieve Length

Given the examples explored above, it is evident that sieve length plays a major role in fitting accuracy. For instance, estimation with a low sieve length resulted in severe underfitting, while a higher sieve length resulted in better fit. The question of course is whether an optimal length can be determined.

Li et. al. (1987) studied three well-known procedures, all of which are based on the mean squared forecast error of the estimated function over a search grid $\mathcal{J} \equiv \left\{J_{min},\ldots, J_{max}\right\}$, and all of which are asymptotically equivalent. In particular, let $J^{\star}$ the optimal sieve length and consider:

1. $C_{p}$ method due to Mallows (1973): $$J^{\star} = \min_{J \in \mathcal{J}} \frac{1}{T}\sum_{t=1}^{T}\left(Y_{t} - \widehat{m}(X_{t})\right)^{2} - 2\widehat{\sigma}^{2}\frac{J}{T}$$ where $\widehat{\sigma}^{2} = \frac{1}{n}\sum_{t=1}^{T}\left(Y_{t} - \widehat{m}(X_{t})\right)^{2}$
2. Generalized cross-validation method due to Craven and Wahba (1979): $$J^{\star} = \min_{J \in \mathcal{J}} \frac{1}{(1 - (J/2))^{2}T}\sum_{t=1}^{T}\left(Y_{t} - \widehat{m}(X_{t})\right)^{2}$$
3. Leave-one-out cross validation method due to Stone (1974): $$J^{\star} = \min_{J \in \mathcal{J}} \frac{1}{T}\sum_{t=1}^{T}\left(Y_{t} - \widehat{m}_{\setminus t^{\star}}(X_{t})\right)^{2}$$ where the subscript notation $\setminus t^{\star}$ indicates estimation after dropping observation $t^{\star}$.
Here we discuss the algorithm for the last of the three procedures. In particular, with the search grid $\mathcal{J}$ defined as before, iterate the following steps over $J \in \mathcal{J}$:
1. For each observation $t^{\star} \in \left\{1, \ldots, T \right\}$:
1. Solve the optimization problem in \eqref{eq.1.4} using data from the pair $(Y_{t}, X_{t})_{t \neq t^{\star}}$, and derive the estimated model as follows: $$\widehat{m}_{J,\setminus t^{\star}}(x) \equiv \widehat{\theta}_{_{J,\setminus t^{\star}}0} + \sum_{j=1}^{J}\widehat{\theta}_{_{J,\setminus t^{\star}}j}f_{j}(x)$$ where the subscript $J,\setminus t^{\star}$ indicates that parameters are estimated using sieve length $J$, after dropping observation $t^{\star}$.
2. Derive the forecast error for the dropped observation as follows: $$e_{_{J}t^{\star}} \equiv Y_{t^{\star}} - \widehat{m}_{J,\setminus t^{\star}}(X_{t^{\star}})$$
2. Derive the cross-validation mean squared error for sieve length $J$ as follows: $$MSE_{J} = \frac{1}{T}\sum_{t=1}^{T} e_{_{J}t}^{2}$$
3. Determine the optimal sieve length $J^{\star}$ as the minimum $MSE_{J}$ across $\mathcal{J}$. In other words $$J^{\star} = \min_{J\in\mathcal{J}} MSE_{J}$$
In words, the algorithm moves across the sieve search grid $\mathcal{J}$ and computes an out-of-sample forecast error for each observation. The optimal sieve length is that which minimizes the average mean squared error across the search grid. We demonstrate the selection criteria and accompanied estimation when using a grid search from 1 to 15.

Figure 3: Sieve Regression with Optimized Sieve Length Selection

Evidently, both the polynomial and Gaussian sieve models ought to use a sieve length of 15.

#### Critiques

While global nonparametric estimators are easy to work with, they exhibit several well recognized drawbacks. First, they leave little room for fine-tuning estimation. For instance, in the case of polynomial sieves, the polynomial degree is not continuous. In other words, if estimation underfits when sieve length is $J$, but overfits when sieve length is $J+1$, then there is no polynomial degree $J < J^{\star} < J+1$.

Second, global estimators are often subject to infeasibility since regressor values may not be sufficiently small. This is because increased sieve lengths can result in the values of the regressor covariance matrix to become extremely large. In turn, this can render the inverse of the covariance matrix nearly singular, and by extension, render estimation infeasible. In other words, at some point, increasing the polynomial degree further does not lead to estimate improvements.

Lastly, it is worth pointing out that global estimators fit curves by smoothing (averaging) over the entire domain. As such, they can have difficulties handling observations with strong influences such as outliers and regime switches. This is due to the fact that outlying observations will be averaged with the rest of the data, resulting in a curve that significantly under- or over- fits these observations. To illustrate this point, consider a modification of equation \eqref{eq.1.5} with outliers when $-1 < x \leq 1$ : \begin{align} m(x) = \begin{cases} \sin(x)\cos(\frac{1}{x}) + \log\left(x + \sqrt{x^2+1}\right) & \text{if } x\in [-6,1]\\ \sin(x)\cos(\frac{1}{x}) + \log\left(x + \sqrt{x^2+1}\right) + 4 & \text{if } x \in (-1,1]\\ \sin(x)\cos(\frac{1}{x}) + \log\left(x + \sqrt{x^2+1}\right) - 2 & \text{if } x \in (1,6] \end{cases}\label{eq.1.6} \end{align} We generate $Y_{t}$ and $X_{t}$ as before, and estimate this model using both polynomial and Gaussian sieves based on cross-validated sieve length selection.

Figure 4: Sieve Regression with Optimized Sieve Length Selection and Outliers

Clearly, both procedures have a difficult time handling jumps in the domain region $-1 < x \leq 1$. Nevertheless, it is evident that the Gaussian sieve does significantly better than polynomial regression. This is further corroborated by the leave-one-out cross-validation MSE values which indicate that the Gaussian sieve minimum MSE is roughly 6 times as small as the polynomial sieve minimum MSE.

It turns out that a number of these shortcomings can be mitigated by averaging locally instead of globally. In this regard, we turn to the idea of local estimation next.

### Local Methods

The general idea behind local nonparametric estimators is local averaging. The procedure partitions the functional variable $x$ into bins of a particular size, and estimates $m(x)$ as a linear interpolation of the average values of the dependent variable at the middle of each bin. We demonstrate the procedure when $m(x)$ is the function in \eqref{eq.1.5}.

In particular, define $Y_{t}$ as before, but let $X_{t} = x$. In other words, we consider deterministic regressors. We will relax the latter assumption later, but this is momentarily more instructive as it leads to contiguous partitions of the explanatory variable $X_{t}$. At last, define the bins as quantiles of $x$ and consider the procedure with bin partitions equal to 2, 5, 15, and 30, respectively.

Figure 5: Local Averaging with Quantiles

Clearly, when the number of bins is 2, the estimate is a straight line and severely underfits the objective function. Nevertheless, as the number of bins increases, so does the accuracy of the estimate. Indeed, local estimation here is shown to be significantly more accurate than global estimation used earlier on the same function $m(x)$. This is of course a consequence of local averaging which performs piecemeal smoothing on only those observations restricted to each bin. Naturally, high leverage observations and outliers are better accommodated as they are averaged only with those observations in the immediate vicinity which also fall in the same bin. In fact, we can demonstrate this using the function $m(x)$ in \eqref{eq.1.6}.

Figure 6: Local Averaging with Quantiles and Outliers

Evidently, increasing the number of bins leads to increasingly better adaptation to the presence of outlying observations.

It's worth pointing out here that unlike sieve estimation which can suffer from infeasibility with increased sieve length, in local estimation, there is in principle no limit to how finely we wish to define the bin width. Nevertheless, as is evident from the visuals, while increasing the number of bins will reduce bias, it will also introduce variance. In other words, smoothness is sacrificed at the expense of accuracy. This is of course the bias-variance tradeoff and is precisely the mechanism by which fine-tuning the estimator is possible.

#### Localized Kernel Regression

The idea of local averaging can be extended to accommodate various bin types and sizes. The most popular approaches leverage information of the points at which estimates of $m(x)$ are desired. For instance, if estimates of $m(x)$ are desired at a set of points $\left(x_{1}, \ldots, x_{J} \right)$, then the estimate $\widehat{m}(x_{j})$ can be the average of $Y_{t}$ for each point $X_{t}$ in some neighborhood of $x_{j}$ for $j=1,\ldots, J$. In other words, bins are defined as neighborhoods centered around the points $x_{j}$, with the size of the neighborhood determined by some distance metric. Then, to gain control over the bias-variance tradeoff, neighborhood size can be exploited with a penalization scheme. In particular, penalization introduces a weight function which disadvantages those $X_{t}$ that are too far from $x_{j}$ in any direction. In other words, those $X_{t}$ close to $x_{j}$ (in the neighborhood) are assigned larger weights, whereas those $X_{t}$ far from $x_{j}$ (outside the neighborhood) are weighed down.

Formally, when the function $m(\cdot)$ is univariate, local kernel estimators solve optimization problems of the form: \begin{align} \arg\!\min_{\hspace{-1em} \beta_{0}} E\left(Y_{t} - \beta_{0}\right)^{2}K_{h}\left(X_{t} - x_{j}\right) \quad \forall j \in \left\{1, \ldots, J\right\}\label{eq.1.7} \end{align} Here we use the traditional notation $K_{h}(X_{t} - x_{j}) \equiv K\left(\frac{|X_{t} - x_{j}|}{h}\right)$ where $K(\cdot)$ is a distributional weight function, otherwise known as a kernel, $|\cdot|$ denotes a distance metric (typically Euclidean), $h$ denotes the size of the local neighbourhood (bin), otherwise known as a bandwidth, and $\beta_{0} \equiv \beta_{0}(x_{j})$ due to its dependence on the evaluation point $x_{j}$.

To gain further insight, it is easiest to think of $K(\cdot)$ as a probability density function with support on $[-1,1]$. For instance, consider the famous Epanechnikov kernel: $$K(u) = \frac{3}{4}\left(1 - u^{2}\right) \quad \text{for} \quad |u| \leq 1$$ or the cosine kernel specified by: $$K(u) = \frac{\pi}{4}\cos(\frac{\pi}{2}u) \quad \text{for} \quad |u| \leq 1$$
 Figure 7A: Epanechnikov Kernel Figure 7B: Cosine Kernel
Now, if $|X_{t} - x| > h$, it is clear that $K(\cdot) = 0$. In other words, if the distance between $X_{t}$ and $x$ is larger than the bandwidth (neighborhood size), then $X_{t}$ lies outside the neighborhood and its importance will be weighed down to zero. Alternatively, if $|X_{t} - x| = 0$, then $X_{t} = x$ and $X_{t}$ will be assigned the highest weight, which in the case of the Epanechnikov and cosine kernels, is 0.75 and 0.8, respectively

To demonstrate the mechanics, consider a kernel estimator based on $k-$nearest neighbouring points, or the weighted $k-NN$ estimator. In particular, this estimator defines the neighbourhood as all points $X_{t}$, the distance of which to an evaluation point $x_{j}$, are no greater than the distance of the $k^{\text{th}}$ nearest point $X_{t}$ to the same evaluation point $x_{j}$. When used in the optimization problem \eqref{eq.1.7}, the resulting estimator is also sometimes referred to as LOWESS - LOcally Weighted Estimated Scatterplot Smoothing.

The algorithm used in the demonstration is relatively simple. First, define $k^{\star}$ as the number of neighbouring points to be considered and define a grid $\mathcal{X} \equiv \{x_{1}, \ldots, x_{J}\}$ of points at which an estimate of $m(\cdot)$ is desired. Next, define a kernel function $K(\cdot)$. Finally, for each $j \in \{1, \ldots, J\},$, execute the following:
1. For each $t \in \{1,\ldots, T\}$, compute $d_{t} = |X_{t} - x_{j}|$ -- the Euclidean distance between $X_{t}$ and $x_{j}$.
2. Order the $d_{t}$ in ascending order to form the ordered set $\{d_{(1)} \leq d_{(2)} \leq \ldots \leq d_{(T)}\}$.
3. Set the bandwidth as $h = d_{(k^{\star})}$.
4. For each $t \in \{1,\ldots, T\}$, compute a weight $w_{t} \equiv K_{h}(X_{t} - x_{j})$.
5. Solve the optimization problem: $$\arg\!\min_{\hspace{-1em} \beta_{0}} E\left(Y_{t} - \beta_{0}\right)^{2}w_{t}$$ to derive the parameter estimate: $$\widehat{m}_{\setminus t^{\star}}(x_{j}) \equiv \widehat{\beta}_{0}(x_{j}) = \frac{\sum_{t=1}^{T}w_{t}Y_{t}}{\sum_{t=1}^{T}w_{t}}$$
An estimate of $m(x)$ along the domain $\mathcal{X}$, is now the linear interpolation of the points $\{\widehat{\beta}_{0}(x_{1}), \ldots, \widehat{\beta}_{0}(x_{J})\}$.

For instance, suppose $m(x)$ is the curve defined in \eqref{eq.1.6}, the evaluation grid $\mathcal{X}$ consists of points in the interval $[-6,6]$, and $K(\cdot)$ is the Epanechnikov kernel. Furthermore, suppose $Y_{t} = m(x) + 0.5\epsilon_{t}$ and $X_{t} = x - 0.5 + \eta_{t}$. Notice that we're back to treating the regressor as a stochastic variable. Then, the $k-NN$ estimator of $m(\cdot)$ with 15, 40, 100, and 200 nearest neighbour points, respectively, is illustrated below.

Figure 8: k-NN Regression

Clearly, the estimator can be very adaptive to the nuances of outlying points but can suffer from both underfitting and overfitting. In this regard, observe that the number of neighbouring points is directly proportional to neighbourhood (bandwidth) size. In other words, as the number of neighbouring points increases, the bandwidth increases. This is evidenced by a very volatile estimator when the number of neighbouring points is 15, and a significantly smoother estimator when th number of neighbouring points is 200. Therefore, there must be some optimal middle ground between undersmoothing and oversmoothing. In general, notice that apart from the lower zero bound, the bandwidth is not bounded above. Thus, there is an extensive range of bandwidth possibilities. So how does one define what constitutes an optimal bandwidth?

#### Bandwidth Selection

While we will cover optimal bandwidth selection in greater detail in Part II of this series, it is not difficult to draw similarities between the role of bandwidth size in local estimation and sieve length in global methods. In fact similar methods for optimal bandwidth selection exist in the context of local kernel regression, and analogous to sieve methods, are also typically grid searches. In this regard, in order to avoid complicated theoretical discourse, consider momentarily the optimization problem in \eqref{eq.1.7}.

It is not difficult to demonstrate that the estimator $\widehat{\beta}_{0}(x)$ satisfies: \begin{align*} \widehat{\beta}_{0}(x) &= \frac{T^{-1}\sum_{t=1}^{T}K_{h}\left(X_{t} - x\right)Y_{t}}{T^{-1}\sum_{t=1}^{T}K_{h}\left(X_{t} - x\right)}\\ &=\frac{1}{T}\sum_{t=1}^{T}\left(\frac{K_{h}\left(X_{t} - x\right)}{T^{-1}\sum_{i=1}^{T}K_{h}\left(X_{i} - x\right)}\right)Y_{t} \end{align*} Accordingly, if $h\rightarrow 0$, then $\frac{K_{h}\left(X_{t} - x\right)}{T^{-1}\sum_{i=1}^{T}K_{h}\left(X_{i} - x\right)} \rightarrow T$ and is only defined on $x = X_{t}$. In other words, as the bandwidth approaches zero, $\widehat{\beta}_{0}(x) \equiv \widehat{\beta}_{0}(X_{t}) \rightarrow Y_{t}$, and the estimator is effectively an interpolation of the data. Naturally, this estimator has very small bias since it picks up every data point in $Y_{t}$, but also has very large variance for the same reason.

Alternatively, should $h \rightarrow \infty$, then $\frac{K_{h}\left(X_{t} - x\right)}{T^{-1}\sum_{i=1}^{T}K_{h}\left(X_{i} - x\right)} \rightarrow 1$ for all values of $x$, and $\widehat{\beta}_{0}(x) \rightarrow T^{-1}\sum_{t=1}^{T}Y_{t}$. That is, $\widehat{\beta}_{0}(x)$ is a constant function equal to the mean of $Y_{t}$, and therefore has zero variance, but suffers from very large modelling bias since it picks up only those points equal to the average.

Between these two extremes is an entire spectrum of models $\left\{\mathcal{M}_{h} : h \in \left(0, \infty\right) \right\}$ ranging from the most complex $\mathcal{M}_{0}$, to the least complex $\mathcal{M}_{\infty}$. In other words, the bandwidth parameter $h$ governs model complexity. Thus, the optimal bandwidth selection problem selects an $h^{\star}$ to generate a model $\mathcal{M}_{h^{\star}}$ best suited for the data under consideration. In other words, it reduces to the classical bias-variance tradeoff.

To demonstrate certain principles, we close this section by returning to the leave-one-out cross-validation procedure discussed earlier. As a matter of fact, the algorithm also applies to local kernel regression and we do so in the context of $k-NN$ regression, also discussed earlier.

In particular, define a search grid $\mathcal{K} \equiv \{k_{min}, \ldots, k_{max}\}$ of the number of neighbouring points, select a kernel function $K(\cdot)$, and iterate the following steps over $k \in \mathcal{K}$:
1. For each observation $t^{\star} \in \left\{1, \ldots, T \right\}$:
1. For each $t \neq t^{\star} \in \{1,\ldots, T\}$, compute $d_{t \neq t^{\star}} = |X_{t} - X_{t^{\star}}|$.
2. Order the $d_{t \neq t^{\star}}$ in ascending order to form the ordered set $\{d_{t \neq t^{\star} (1)} \leq d_{t \neq t^{\star} (2)} \leq \ldots d_{t \neq t^{\star} (T-1)}\}$.
3. Set the bandwidth as $h_{\setminus t^{\star}} = d_{t \neq t^{\star} (k)}$.
4. For each $t \neq t^{\star} \in \{1,\ldots, T\}$ , compute a weight $w_{_{\setminus t^{\star}}t} \equiv K_{h_{\setminus t^{\star}}}(X_{t} - X_{t^{\star}})$.
5. Solve the optimization problem: $$\arg\!\min_{\hspace{-1em} \beta_{0}} E\left(Y_{t} - \beta_{0}\right)^{2}w_{_{\setminus t^{\star}}t}$$ to derive the parameter estimate: $$\widehat{m}_{k,\setminus t^{\star}}(X_{t^{\star}}) \equiv \widehat{\beta}_{_{k,\setminus t^{\star}}0}(X_{t^{\star}}) = \frac{\sum_{t\neq t^{\star}}^{T}w_{_{\setminus t^{\star}}t}Y_{t}}{\sum_{t\neq t^{\star}}^{T}w_{_{\setminus t^{\star}}t}}$$ where we use the subscript $k,\setminus t^{\star}$ to denote explicit dependence on the number of neighbouring points $k$ and the dropped observation $t^{\star}$.
6. Derive the forecast error for the dropped observation as follows: $$e_{_{k}t^{\star}} \equiv Y_{t^{\star}} - \widehat{m}_{k,\setminus t^{\star}}(X_{t^{\star}})$$
2. Derive the cross-validation mean squared error when using $k$ nearest neighbouring points : $$MSE_{k} = \frac{1}{T}\sum_{t=1}^{T} e_{_{k}t}^{2}$$
3. Determine the optimal number of neighbouring points $k^{\star}$ as the minimum $MSE_{k}$ across $\mathcal{K}$. In other words $$k^{\star} = \min_{k\in\mathcal{K}} MSE_{k}$$
We close this section and blog entry with an illustration of the procedure. In particular, we again consider the function in \eqref{eq.1.6}, and use the cosine kernel to search for the optimal number of neighbouring points over the search grid $\mathcal{K} \equiv \{40, \ldots, 80\}$.

Figure 9: k-NN Regression with Optimized k

### Conclusion

Given the recent introduction of functional coefficient estimation in EViews 11, our aim in this multi-part blog series is to complement this feature release with a theoretical and practical overview. As first step in this regard, we've dedicated this Part I of the series to gently introducing readers to the principles of nonparametric estimation, and illustrated them using EViews programs. In particular, we've covered principles of sieve and kernel estimation, as well as optimal sieve length and bandwidth selection. In Part II, we'll extend the principles discussed here and cover the theory underlying functional coefficient estimation in greater detail.

### References

1. Peter Craven and Grace Wahba. Estimating the correct degree of smoothing by the method of generalized cross-validation. Numerische Mathematik, 31:377–403, 1979.
2. Ulf Grenander. Abstract inference. Technical report, 1981.
3. Ker-Chau Li and others. Asymptotic optimality for csub>p, csub>l , cross-validation and generalized cross-validation: Discrete index set. The Annals of Statistics, 15(3):958–975, 1987.
4. Colin L Mallows. Some comments on c p. Technometrics, 15(4):661–675, 1973.
5. Mervyn Stone. Cross-validation and multinomial prediction. Biometrika, 61(3):509–515, 1974.


text:
## Thursday, February 01, 2007

### Nobel Peace Prize downgraded

The Nobel Peace Prize has been downgraded from neutral to waste. The rating agency has decided after an updated analysis of the nominees.

Adolf Hitler was nominated in 1939. Other nominees included Joseph Stalin and Benito Mussolini. Yasser Arafat, Jimmy Carter, and Wangari Maathai - a Kenyan ecoterrorist - became winners.

Ladies and Gentlemen, it's here.

were nominated for the Nobel Peace Prize. How do the Norwegian lawmakers justify such an incredible absurdity?

• “A prerequisite for winning the Nobel Peace Prize is making a difference, and Al Gore has made a difference,” Conservative Member of Parliament Boerge Brende, a former minister of environment and then of trade, told The Associated Press.

That's nice but how does this explanation emphasizing the "difference" agree with the nomination of Sheila Watt-Cloutier, an Inuit I've never heard about who has 5 times less Google hits than your humble correspondent? What difference has she made? Is riding a dog sled enough?

Moreover, before the explanation above could be used to justify the nominations listed in this article except for Ms Watt-Cloutier who was treated separately, the Nobel Peace Prize would first have to be renamed to

• the Nobel Prize for Influential Mass Killers, Annihilators of Economies, Terrorists, Econuts, and Imbeciles.

Sounds pretty good. ;-)

Lindzen on CNN

Prof Richard Lindzen of MIT debated Dr Heidi Cullen and Bill Nye, a mechanical engineer and a TV clown, on
The clown tried to criticize Lindzen's science but King warned him: "Lindzen's from MIT, he knows what he's talking about". Nye backed down. Prof Lindzen has found out that the catastrophic global warming theorists were like little kids locking themselves in dark closets to see how much they can scare each other and themselves.


text:
## Random Matrix Models for the Hermitian Wilson-Dirac operator of QCD-like theories    [PDF]

Mario Kieburg, Jacobus J. M. Verbaarschot, Savvas Zafeiropoulos
We introduce Random Matrix Models for the Hermitian Wilson-Dirac operator of QCD-like theories. We show that they are equivalent to the $\epsilon$-limit of the chiral Lagrangian for Wilson chiral perturbation theory. Results are obtained for two-color QCD with quarks in the fundamental representation of the color group as well as any-color QCD with quarks in the adjoint representation. For $N_c=2$ we also have obtained the lattice spacing dependence of the quenched average spectral density for a fixed value of the index of the Dirac operator. Comparisons with direct numerical simulations of the random matrix ensemble are shown.
View original: http://arxiv.org/abs/1303.3242


text:
# Efficiently write in a file in C++

I've made a program which reads from the console and writes it in a .txt file. This programm should write in a file with \n.

#include <iostream>
#include <fstream>
#include <string>

using namespace std;

int main() {
ofstream fich("file.txt");

if (!fich)
cerr << "Error!\n";

char fst = 'l';
string c;
while (fst != '\0') {
getline(cin, c);
fst = c[0];
fich << c << endl;
}

fich.close();

return 0;
}

This program ends when the user inputs nothing and then a \n, because, doing that, the string c will only have a \0 end of string element.

I was wondering if there is any way of doing this more efficiently, maybe without needing a "void \n" line to finish the program.

• I guess main2() is a typo that should be main(), right? – uli May 21 at 6:37
• c[0] on an empty string is not guaranteed to be \0! Your test should be: while(getline(std::cin, c)) { Then you can end input by typing the EOF control code for your platform. <ctrl>-D on linux. This also makes it work nicely with files piped to the standard input. – Martin York May 26 at 22:40
• @uli It was a typo, i corrected it, aprecciate it – And9686 May 29 at 15:46
• @MartinYork Isn't the string empty when I create it? I thought that if it's empty then the \0 element should be in string[0]. – And9686 May 29 at 15:54
• You are thinking of a C-String. The C++ std::string class does not use a null terminator. Accessing at size() using operator[] is undefined behavior. Now there is a method that will give you a null terminated string that looks like a C-String call c_str() this guarantees a null terminated string. But why not just call empty() to test for an empty string. – Martin York May 29 at 16:50

# Error Handling

You check whether the output file was opened correctly, but all you do is output a message (you don't even say what's wrong!) and then you keep on ignoring it. What if file.txt is a directory or read-only? I guess you haven't tested these cases. By default, throw an exception if you can't continue at some point. In this case, throw std::runtime_error("failed to open output file").

# Loop Condition

You are reading until the line is empty. This is okay-ish, but why don't you just check whether c.empty() and then break from the loop? Another case is that the end of the input is reached. In that case, the streamstate is set to failed. I'm not sure what happens in getline() then (read the docs at cppreference.com), but if it just doesn't change the string passed by reference, you may end up in an endless loop.

As an alternative I would use getline(cin, c) as loop condition. If this succeeds, you have received a line which can then write to the output file. Otherwise, you either had an error or you reached the end of the input.

# Closing Filestreams

If you didn't call fich.close(), when would it be closed? However, consider the possibility that the file can't be written fully. You'll never notice this, neither with your code nor if you let the destructor do its work. So, what I'd rather do there is a call to fiche.flush() followed by a check of the streamstate.

# Efficiency

I wouldn't bother with that at the moment. Point is, you're still learning to walk, so don't try to run yet. Still, what you're doing is inefficient. Firstly, writing in size of lines is useless, because you're just moving bytes from one file to the other. Secondly, endl implies a newline and a flush. So this is not really a cheap operation! In addition, there is a lot of stuff going on in C++ IOStreams that there's a whole book about their internals. Lastly, copying files can sometimes be done more efficiently using OS features which will avoid looking at the data more than necessary.

• Yes a file will be closed correctly. The std::ofstream destructor calls close. In "most" cases not calling close is the correct action. Also manually calling flush is "usually" the wrong thing to do. The system implementation will auto flush when needed; manually calling flush is the cause of most speed issues in beginners C++ application. – Martin York May 26 at 22:41
• Speed issues in beginners' applications? Uselessly flushing is a source for bad performance, that much I agree with. However, if you don't flush your stream after you're done writing, you don't know whether the data was written. If there is buffered data that doesn't fit on the storage, you will only get corrupted data and you may not even notice until it's too late. – uli May 27 at 6:48
• "I'm not sure what happens in getline() then (read the docs at cppreference.com), but if it just doesn't change the string passed by reference, you may end up in an endless loop." The string is cleared first. So if there is an EOF or other problem, you will read an empty string. – JDługosz May 27 at 13:48

The \0 does not show up as a legal element of the string. In an empty string, doing c[0] is an error!

You don't need fst at all. You should use c.empty() to check for an empty string.

If you're not on a Unix-like system, be sure to open the file in Text mode, or you'll end up with stray \r characters in your string.

Your names, fst, fich, are confusing. Using c for a string is unusual and confusing to experienced programmers.

Don't write using namespace std;.

You can use a mid-decision loop to simplify your variables and logic. Something like:

for (;;) {
getline (cin, s);
if (s.empty()) break;
outfile << s << '\n';
}

As as already been pointed out, the whole idea of reading and writing a line at a time is not "efficient". It might be what you need, though, for an interactive program where the user is typing something. If you are implementing something like cat though it is less efficient than reading entire blocks using low-level primitives. The iostream implementation reads the file in a low-level way, and getline goes through that data (that's already been read into memory) just to locate the end-of-line characters; then copies that to another string. If you just wanted to copy everything, that's clearly extra overhead.


text:
# Homework Help: Small Poisson Question

1. Dec 5, 2006

### Hartogsohn26

Hello

I just found this forum, was recommended it by a person that I know and have a small probability question regarding Poisson distribution.

Let (T,U) be a to dimensional discrete stochastic vector with the probability function $$p_{T,U}$$ given by.

$$P(T=t, U = u) = \left{ \begin{array}{cccc} \frac{1}{3} \cdot e^{-\lambda} \frac{\lambda^u}{u!} & u \in \{-1,0,1\} & \mathrm{and} & t \in \{0, 1, \ldots\} \\0 & \mathrm{elsewhere.} \end{array}$$

where $$\lambda > 0$$

1)

describe the support for $$\mathrm{supp} (P_{T,U})$$

Solution (1) the support $$\mathrm{supp} (P_{T,U})$$ is the set of values for the real valued probability function P which produces non-negative values. therefore $$\mathrm{supp} (P_{T,U}) = \{0,1, \ldots \}$$

2)

show that the probability function $$p_T$$ and $$p_U$$ for T and U is.

$$P(T = t) = P_{T} = \left{ \begin{array}{cccc} \frac{1}{3} & t \in \{-1,0,1\} \\0 & \mathrm{elsewhere.} \end{array}$$

and

$$P(U = u) = P_{U} = \left{ \begin{array}{cccc} e^{-\lambda} \frac{\lambda^{u}}{u!} & u \in \{0,1,\ldots\} \\0 & \mathrm{elsewhere.} \end{array}$$

which inturn means $$U \sim po(\lambda)$$

Solution (2)

How do I show this ? If not as above. Or do I show that they have same variance??

(3) Assume that $$\lambda = 1$$ then $$P(T=U) = \frac{2}{3} e^{-\lambda}$$

Solution is $$P(T = U) = P(t \cup u)$$???

Best Regards
Hartogsohn


text:
# Volunteer radar gunslingers nail speeders

1. Oct 26, 2007

Just read this on yahoo. I dont know about you guys, but a 'warning letter' isnt going to stop me from going 50 in a 30.

2. Oct 27, 2007

### Ivan Seeking

Staff Emeritus
3. Oct 27, 2007

### BobG

I don't know how they handle traffic in your city, but most major thoroughfares have speed limits over 30. 30 is usually for residential neighborhoods (although that's been lowered to 25 in most neighborhoods in the Springs). If you're going 50 through residential neighborhoods then you deserve a whole lot worse than a warning letter.

4. Oct 27, 2007

### Moonbear

Staff Emeritus
Agreed. We have a few areas that are between 30-35 because it's a wider road NEAR a residential area, but not with houses directly lining it, or an industrial area with a lot of driveways entering, but if the speed limit is anywhere between 25-35, there's usually good reason for it...and it's that you can't stop in time for the little kid whose tricycle rolls out of the driveway, or chasing a ball without looking. If you're in a neighborhood and regularly driving 50 mph, your neighbors should want to do more than just send you a letter! :grumpy:

5. Oct 27, 2007

it was a JOKE. seriously though, whos going to be scared of a warning letter? whos to say that a neighbor that has something against you wont say that you were going 100 in the neighborhood (exaggerated). thats why it will never turn into an actual ticket, unless they catch it on film or something of the likes.

6. Oct 27, 2007

### slugcountry

what ***holes.... no ambitions and they hate life so they sit by the side of the road and try to make life miserable for others... losers.

7. Oct 27, 2007

### turbo

Actually, students at Unity College have been doing the same thing for some months in an effort to get people to stop speeding through the down-town. In rural towns, this isn't such a bad idea because we can't afford our own police forces and the county sheriff's deputies and the state police are spread VERY thin. I live on a back road. The speed limit is 45, and the weight restriction is 60,000 lbs gross weight. We are routinely dodging fully-loaded pulp trucks (easily exceeding 100,000 lbs) doing 55-60 mph. Such heavy trucks are murder on lightly-built back roads and they cost the taxpayers  in maintenance costs, and there is no way that a speeding 100,000+ lb truck is going to stop in time to avoid a collision with another vehicle or avoid a child chasing a frisbee or a ball.

Trucking is very important to our forestry industry, and the truckers need to stay efficient to make a decent living, but when they take high-speed short-cuts through these narrow back roads (no shoulders, not even painted center-lines) they will eventually kill someone and THEN there will be a crack-down. It will take a death to prompt action.

8. Oct 28, 2007

### slugcountry

Sucks to be you then but I will never never live down paying a 230 dollar ticket for going 55 in a 50 - I hate cops with a passion =X where I live there is virtually no crime and the local police spend ALL their time issuing minor traffic tickets and busting kids for smoking pot.

Like I said - losers.

9. Oct 28, 2007

### moose

Around where I used to live, there were tons of roads that were 30mph zones, and yet were very wide roads with fences on the sides and no incoming traffic or anything. Everyone goes around 45 down those roads...

10. Oct 28, 2007

### Hurkyl

Staff Emeritus
I'm sure not all people who break traffic law and use illegal substances are losers...

11. Oct 28, 2007

### turbo

Good one! I am pretty sure that people who would like to deter people from exceeding the speed limits past their homes by 15-20 mph are not losers, either. They have to be willing to give up their time and effort to make a point, and hope that their efforts will help protect pedestrians, playing kids, and other motorists. The State Police would not issue me a set of truck scales and give me the authority to stop heavy traffic on my back road, but I'll guarantee that if they did, I could stop 40,000+ lb overloaded rigs from shooting through here in a week or two. The scofflaws are stealing from us and endangering us.

12. Oct 28, 2007

### slugcountry

right, and I'm sure those same people take care to never ever exceed the speed limit as well, just like the cops who blaze by at 90+ mph on highways without even their emergency lights, and then turn around and issue speeding tickets - not just losers, but hypocrites I'll never have respect for people like that

13. Oct 28, 2007

### Kurdt

Staff Emeritus
I've never been fond of the strict adherence to speed limits. If one were to try and adhere to them all perfectly you'd do nothing but look at your speedo causing more crashes. In fact wasn't there a study saying far from lowering accidents speed cameras increased them because people were too busy looking at their dash boards?

14. Oct 29, 2007

Hmm. Now that i think about it, i understand what you mean because when im going through a 20 MPH school zone, i usually keep my eyes 50/50, on the road and the speedo. Good point.

And to the person talking about cops going 90 on the highway. today i saw a cop zoom by without emergency lights on. he was at least going 60 in a 45, hypocrites indeed. i always wondered what would happen if you were behind a cop car without its lights on going the same speed as he was. could they give you a speeding ticket knowing that you were going over the speed limit?

15. Oct 29, 2007

### slugcountry

Yes, they have a rear radar.

As far as going the same speed as you without their lights on, they do this on a routine basis and it is called pacing. As illegal as the practice may be on the surface, it is a very wide spread tactic and even has laws written that support it. An officer has to "pace" you (basically follow you within a certain distance) for a specific amount of time above the speed limit in order to issue a speeding ticket without the use of a radar gun.

Of course the courts literally always rule in favor of the officer in traffic citations. The only way to get out of a ticket is for the officer not to show up.

Honestly I hate hate hate cops =X. If they actually kept people safe it would be great, instead they just harass people.

16. Oct 29, 2007

### Hurkyl

Staff Emeritus
Literally? I'm sure there exists at least one counterexample.

I have absolutely no sympathy for someone complaining that they got penalized when they were caught breaking the law. I don't think you're condemnation of police is warranted; you're just bitter you got caught.

Last edited: Oct 29, 2007
17. Oct 29, 2007

### slugcountry

Hurkyl, just because a law exists doesn't mean it is just - don't assume that I've been caught for anything, you don't know a thing about me.

I simply make observations about my surroundings. I've been personally HARASSED by cops many many many times, though I've only ever received one ticket (for driving 55 in a 50 no less....)

I've been pulled over in the middle of the night because "my tires looked bald."

Yeah... right. I was pulled over because I was profiled as likely trouble (a younger man driving at night). Thats only one example among many - and I'm far from the only one who's fed up with this kind of behavior. Very very few people will stick up for someone who winds up on the wrong side of the law (prime example: you) - innocent or not. Just look at the number of innocent people on death row over the years. They do this kind of bull**** on a regular basis around here.

18. Oct 30, 2007

### Hurkyl

Staff Emeritus
I didn't assume. You said yourself that you have been ticketed for speeding. Also, you have been complaining about other people being caught.

In general, why would one attempt to defend an illegal act?

Last edited: Oct 30, 2007
19. Oct 30, 2007

### slugcountry

Ok. Again: Just because a law exists does NOT MAKE IT JUST.

Secondly: What I was trying to explain is that not everyone who winds up on the wrong side of the law is actually guilty.

20. Oct 30, 2007

### Hurkyl

Staff Emeritus
Unless you intend to argue that the notion of a speed limit is unjust, I don't see any relevance to this thread.

Unless I'm very much mistaken, being on the "wrong side of the law" means that you are guilty. (And a quick internet search confirms my understanding)

But again this is just a red herring, since you have been talking about people who were speeding (and were using illegal substances).

21. Oct 30, 2007

### Ivan Seeking

Staff Emeritus
When I was a teen I was hell on wheels - drove like a madman. Then, one day, on a road that I raced along every day, in my rear view mirror I saw a neighbor come running out of his house screaming at the top of his lungs as I passed by. Only when I saw the intense anger in his face did I realize what his perspective was like: What I saw as a nice little set of curves that was cool to hit hard. What he saw was an idiot who was threatening the life of his kids.

22. Oct 30, 2007

### slugcountry

Ok Hurkyl, I'll just assume that you've never exceeded a speed limit in your life. And you don't have to be guilty to be on the wrong side of the law, you just have to be on the wrong side of the law. Red herring is a nice buzz word, but as I was trying to explain to you police regularly harass people - they can lie that you were driving too fast, as an excuse to pull you over and investigate (as has happened to me more than once). Like I said I've been pulled over MANY times, but only ever received one ticket. This isn't because the cops were being nice or something.... its because they pulled me over when I wasn't doing anything wrong.

And as far as any "wrong doing" for that matter, going 60 in a 45 for example, on an abandoned road, is a victimless crime in my book so yes I would definitely say that in many cases a speed limit is a stupid and unjust notion.

23. Oct 30, 2007

### Staff: Mentor

No, we all speed sometimes in some places (including Hurkyl I'm sure), and we know it's against the law. But the point is that you try to be reasonable about it, and when you get pulled over, you man up and be honest and straightforward. If the police officer lets you off with a warning, that's nice. If s/he gives you a ticket, be straightforward and deal with it.

Doing 80mph in the #1 lane in a 65 zone on the freeway in commute traffic that's all mostly moving at the same pace is not unreasonable (but still plenty ticketable). Doing 50 in a 30 residential area is very marginal. And even driving 30 in a parking lot is stupid in most cases (quiz question -- why?).

The most unfortunate thing about the position you've taken in this thread is that you are trying to bash cops based on them enforcing the law. When they blow by you on the freeway and don't have their lights and sirens on, they still have a good reason for it and are within the law. Either "pacing" as already explained, or Code-2 hurry-up. They know what they are doing is dangerous, but it's part of the training, and part of the job.

I've been stopped probably a dozen times in my life, and gotten a total of 3 tickets, and every single stop was justified. Yes, I did get reamed once or twice verbally, but I stood there and took it, and finally was given a warning (because the police officer knew his case was marginal for a ticket).

Police officers do a lot more than just give out traffic tickets for violations of the vehicle code. You should see if your city offers ride-alongs with PD officers. I'd bet that after you've done a ride-along or two, your view of police officers will change some.

Last edited: Oct 30, 2007
24. Oct 30, 2007

### turbo

Either you are trolling, or you don't understand that speeding is not a victimless crime. People speed through my rural neighborhood every day, often exceeding the speed limit by 15-20 mph. They are not committing a victimless crime - they are indulging in criminal negligence that can (and eventually will) result in the death of innocent person(s).

25. Oct 30, 2007

### Moonbear

Staff Emeritus
Only until you find out that some other group of kids was parked along the side of that "abandoned" road to make out, or you end up rolling in a ditch because the "abandoned" road is not maintained well enough for you to do 60, or you smash into a farmer on his tractor because the "abandoned" road runs along his farm.

Speed limits are set for a reason, and that reason is safety. It includes safety margins for the curves on the road, safety margins for the amount of traffic on the road, safety margins for the location it is in and the number of driveways entering it. Just because you're young and reckless and have no regard for others on the road doesn't mean they are pointless.

And, if you do object to a law, think it's outdated, unjust, etc., then the way to fight it is NOT to just break the law. Take it to the legislature who writes the law. You may find out that your "unjust" law actually has a really good reason, even if it's not obvious to you.


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# $2^a+2^b=37k$ where $a(>1)$,$b(>1)$ and $k$ are integers is not possible.

I was trying do a problem I was stuck , the only remaining to prove was that $2^a+2^b=37k$ where $a(>1),b(>1)$ and $k$ are integers is not possible.

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$2^{18} + 1 = 37\cdot 7085$. Multiply by any power of $2$. –  Daniel Fischer Aug 12 '13 at 12:44
edit a and b must be greater than 1 –  maths lover Aug 12 '13 at 12:45
@DanielFischer thnx.. –  maths lover Aug 12 '13 at 12:46
Ah, well, maybe we can help you solve the problem in some other way? –  Daniel Fischer Aug 12 '13 at 12:46

Using $2^{18} + 1 = 37 \cdot 7085$ you can simply take $a = 20$ and $b = 2$ for example and you get $$2^{20}+2^2 = 37\cdot(7085\cdot 4)$$


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# Schnorr group

A Schnorr group, proposed by Claus P. Schnorr, is a large prime-order subgroup of ${\displaystyle \mathbb {Z} _{p}^{\times }}$, the multiplicative group of integers modulo ${\displaystyle p}$ for some prime ${\displaystyle p}$. To generate such a group, generate ${\displaystyle p}$, ${\displaystyle q}$, ${\displaystyle r}$ such that

${\displaystyle p=qr+1}$

with ${\displaystyle p}$, ${\displaystyle q}$ prime. Then choose any ${\displaystyle h}$ in the range ${\displaystyle 1 until you find one such that

${\displaystyle h^{r}\not \equiv 1\;({\text{mod}}\;p)}$.

This value

${\displaystyle g=h^{r}{\text{ mod }}p}$

is a generator of a subgroup of ${\displaystyle \mathbb {Z} _{p}^{\times }}$ of order ${\displaystyle q}$.

Schnorr groups are useful in discrete log based cryptosystems including Schnorr signatures and DSA. In such applications, typically ${\displaystyle p}$ is chosen to be large enough to resist index calculus and related methods of solving the discrete-log problem (perhaps 1024 to 3072 bits), while ${\displaystyle q}$ is large enough to resist the birthday attack on discrete log problems, which works in any group (perhaps 160 to 256 bits). Because the Schnorr group is of prime order, it has no non-trivial proper subgroups, thwarting confinement attacks due to small subgroups. Implementations of protocols that use Schnorr groups must verify where appropriate that integers supplied by other parties are in fact members of the Schnorr group; ${\displaystyle x}$ is a member of the group if ${\displaystyle 0 and ${\displaystyle x^{q}\equiv 1\;({\text{mod }}p)}$. Any member of the group except the element ${\displaystyle 1}$ is also a generator of the group.


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Hejhal's algorithm and computational methods for non-classical Maass wave forms

Hejhal's algorithm [1] was a little gadget invented in the 90's for calculating the Hecke eigenvalues and Fourier coefficients of Maass wave forms. Later, Booker, Strombergsson, and Venkatesh (BSV) [2] took Hejhal's paper and made it more viable to perform high-precision calculations.

The basic idea is to assume the Fourier expansion takes a certain form, and write down a finite linear system (where the finitude is possible due to assumptions on the final order of magnitude of desired precision). Now take a horocycle on the complex upper half plane, and use these points in the linear system, and you can almost solve it. In BSV's words, "to solve for r [the desired eigenvalue] the above linear system is repeatedly solved for two different Y-values [horocycles], successively adjusting r to make the two solution vectors as nearly equal as possible."

My question is whether such an algorithm or the germ of an idea for such an algorithm exists for non-classical Maass wave forms. What would take the analogue of horocycles, and would we have enough information to solve an analogous linear system? (Remember, Fourier expansions are a good deal more complicated in multiple dimensions)

[1] Dennis A. Hejhal. On eigenfunctions of the Laplacian for Hecke triangle groups. In Emerging applications of number theory (Minneapolis, MN, 1996), volume 109 of IMA Vol. Math. Appl., pages 291–315. Springer, New York, 1999.

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The second reference should be Booker, Strömbergsson and Venkatesh, not just Venkatesh – Johan Andersson Apr 14 '13 at 19:20
Thanks, I got blindsighted by the fact it was hosted on his website. – Robert K Apr 14 '13 at 20:07
What is an analogue of a Hecke eigenvalue, what is the Fourierexpansion, how to prove existence of cuspidal Maass forms....i do not expect something useful can be said in this generality. – Marc Palm Apr 14 '13 at 20:37
Are you asking for higher rank automorphic forms or non-congruence stuff. My previous comment applies to the later only. Please specify what you mean by non-classical Maass wave form. – Marc Palm Apr 15 '13 at 8:34
You're right that the term I used is shaky, but let's say anything non-holomorphic. For example, in the case of GSp4 we could refer to modular forms associated to generic representations gsp4.org/cat-browser.php?categoryID=199 Thanks for the response so far! – Robert K Apr 15 '13 at 18:32

I don't think it is practical to directly find higher rank Maass forms along the lines of what Hejhal did, because they are functions of several variables and their Fourier expansions involve multiple sums. Not to mention the need to implement the appropriate special functions that appear. Even if you have the Fourier expansion on the computer (which has been done for SL(3,Z) (by Boris Mezhericher) but not Sp(4,Z)), using the relations to make equations in the coefficients and/or the analog of integrating along horocycles is probably computationally prohibitive.

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Thank you! Do you know of a reference for the SL(3, Z) case? – Robert K Apr 18 '13 at 19:49
I think the paper on his home page is the same as the one he posted to the ArXiv: sites.google.com/site/mboris – David Farmer Apr 19 '13 at 10:05


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# Circumference and Area of a Circle

This applet illustrates formulas for the circumference by rolling a circle along a line, then measuring the length in units of the diameter, or the radius.[br][br]Check the "Show rolling circle" box, then roll the circle with the slider labeled "Roll circle".[br]Check "Show C measured with r, d."[br][br]This leads to a formula for the area of a circle, too. Cut it apart into sectors, and rearrange into a "parallelogram".[br][br]Check the last three boxes to see this.
Formulas:[br][br]Circumference = $\pi$ diameter (that is, $C=\pi d$)[br]Circumference = 2$\cdot\pi\cdot$radius (that is, $C=2\pi r$)[br][br]Area = Circumference $\cdot$ radius [br]Area = $\pi\cdot$ radius squared (that is, $A=\pi r^2$)


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Étude théorique et numérique du problème de la gestion de la diversité

Abstract : The diversity management problem is defined on a partially ordered set. The objective is to produce a subset of $k$ reference elements, $k$ being a given number, while minimizing the cost. Demands and unit costs of production are known. Each non produced element must be replaced by a reference which is higher in the partial order, and this implies an overcost. After a theoretical study on the complexity of this problem, we describe and justify the model we chose. This model is an integer linear program close to that commonly used for the $k$-median problem. To solve this program, we present a Lagrangean approach, and various criteria of variable fixing. We exploit also this approach to build feasible solutions of good quality. Finally, we design a Branch and Cut type algorithm to solve our problem to optimality. To do so, we first do a polyhedral study of the convex hull of the solutions. We present a particular type of cuts that eliminate fractional solutions, as well as a heuristic to generate them. We conclude by numerical tests carried out on real world instances.
Keywords :
Document type :
Theses
Domain :

Cited literature [45 references]

https://tel.archives-ouvertes.fr/tel-00004710
Contributor : Thèses Imag <>
Submitted on : Tuesday, February 17, 2004 - 10:59:18 AM
Last modification on : Friday, November 6, 2020 - 4:39:27 AM
Long-term archiving on: : Wednesday, September 12, 2012 - 1:25:32 PM

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• HAL Id : tel-00004710, version 1

Citation

Olivier Briant. Étude théorique et numérique du problème de la gestion de la diversité. Mathématiques [math]. Institut National Polytechnique de Grenoble - INPG, 2000. Français. ⟨tel-00004710⟩

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