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# Why The Big Bang? Part 1: Infinity

“Next contestant Sybil Fawlty from Torquay, special subject the bleeding obvious”

–   John Cleese, Fawlty Towers

In these posts I want to present a line of reasoning that is an explanation of why this universe exists, and in particular why it started with a Big Bang.

There are a few points to cover so it will require a bit of patience on your part. Along the way I will be attempting to answer three questions in particular:

What do we mean when we say infinity exists?

What do we mean when we say mathematics exists?

What do we mean when we say the universe exists?

A key part of my line of reasoning is the presumption that the universe is finite along the lines of digital physics. To try and persuade you of this I need to first talk about infinity as it is presented in mathematics and this requires me to talk about the axioms of mathematics, where infinity first makes its ugly appearance.

Although I am a mathematician myself, I find the study of axioms is particularly unnerving. It’s as if all my mathematical training has left me completely unprepared to grapple with axioms, as if I’ve ventured into a kind of Alice in Wonderland where things are not as they seem. What are the rules in the branch of mathematics that make the rules?

Axioms can be viewed as mathematical statements that are so simple that they can’t be proved. You’ve got to jump into the pool somewhere and axioms are it. They are the starting point that defines mathematics and you choose them because they are obvious.

The commonly accepted, fairly uncontroversial set of axioms that describe the natural numbers (0,1,2,3,…) are the Peano Axioms. There are only minor disagreements over these, for example people may argue about whether to include zero.

I will take the Wikipedia formulation of the Peano axioms, and what I want to focus on is where concepts of existence appear. Axiom 1 is very very simple indeed:

0 is a natural number.

This is saying that 0 is something that exists, and that it belongs to a collection of things named natural numbers. This innocuous statement is actually quite significant, we are beginning with a presumption of existence and that zero is the name we give to the thing we believe to exist, without any line of reasoning to support it.

The next few axioms aren’t quite so profound, for example axiom 4 says that if m,n and p are natural numbers that exist then:

If m=n and n=p then m=p

There is no presumption of m,n and p existing, it’s just saying that this holds if they happen to exist. After the first five axioms we still have only established the existence of zero. The following three axioms, in a subtle way, axiomatically presume the existence of an unlimited number of natural numbers. We now say that for every natural number (of which we initially know of only one – namely zero) that there exists an associated natural number which we’ll call S(n).

Let me cheat a bit here. S(n) is simply the next number. If n is 7 then S(7) is 8. Now forget what I just said, because you can’t do that when defining axioms. It’s considered bad form. Instead we’re going to define it in what appears to be the most perverse way a committee was able to come up with. Here it is

6. If n is a natural number                  then S(n) is too.

7. If n is a natural number,                 then S(n) isn’t zero.

8. If m and n are natural numbers     then if S(m)=S(n) then m=n.

I’m going to have a stab at explaining why these collectively axiomatize the existence of an indefinitely large number of natural numbers. Axiom 6 almost says that for every n another brand new natural number S(n) also exists, but not quite because S(n) might already exist by virtue it being the same as S(m) from a previous number m, hence axiom 8 plugs this loophole. Also, if S(n) ever equals zero then there would only be a finite cycle of natural numbers, so axiom 7 plugs this. It only remains now to arbitrarily symbolise every S(n): we denote S(0) by the symbol “1”, S(1) by the symbol “2” and so on. We now have a way of constructing an infinity of natural numbers.

So does the set of all natural numbers exist since we now know how to construct it? What does it even mean to say the set exists? Although we can embark on constructing it, to complete the construction would require an infinite number of steps, so we’d need an infinity in order to construct an infinite set. Not a lot of value added.

The Peano axioms do not, in fact, establish the existence of an infinitely large set. To axiomatize mathematics in a way that includes infinite sets you need to extend beyond the Peano axioms. The most common axiomatic foundations of more general mathematics are the Zermelo-Fraenkel axioms (with the axiom of choice) (ZFC). These are a good order of magnitude more difficult to grapple with. I only wish to refer to the seventh one, the axiom of infinity, an axiom that disturbs me. This axiom basically comes right out and states that the set of all natural numbers exists. Problem solved.

It turns out that much of mathematics beyond natural numbers doesn’t actually require the existence of infinite sets. This has been pioneered by Errett Bishop by the approach of constructivism. This approach asserts that it is necessary to construct a mathematical object to prove that it exists. In other words, infinite sets don’t exist. In practice, constructivist techniques are too gritty to be useful. Also they have their limitations. For example, if you want to take a solid sphere, cut it into 5 pieces, and reassemble these pieces (without changing their shapes) into two solid spheres of the same size then you do need to assume existence of infinite sets. Personally I’d sleep just a little bit easier at night knowing that no one really thinks that’s possible. But that’s just me.

So while the axioms that define finite numbers are obvious in that they formalise ‘ordinary’ mathematics, the axioms involving infinite sets are not obvious. It’s not even obvious what existence of infinite sets means let alone whether it’s true. There is arbitrariness about the axiom of infinity. In fact, you can equally axiomatize mathematics by using ZFC with the axiom of infinity reversed to say no infinite sets exist. Even if the full set of natural numbers doesn’t exist we can still say this: for any really big number N that you give me, I can find more than N natural numbers.

So what else doesn’t exist in maths? What about the number pi? No it doesn’t exist, or more specifically its full decimal expansion doesn’t exist. However, for any really big N, it is possible to calculate more than N decimal places. Also, the concept of pi certainly does exist and is used in mathematics without a problem. The algorithms for constructing its decimal expansion exist. But you will never encounter pi numerically, in mathematics or in the real world. Circles are approximations that take pi to many decimal places but that’s it.

$\sum_{i=1}^{\infty} \frac{1}{2^{i}}$

This exists and it equals one.  It does not need any concept of infinity either. Just interpret the above as “what is the number that we can get arbitrarily close to by adding ½ + ¼ + 1/8 + … a really long way?” It can be shown rigorously that such a number is 1. There is no infinity in the formula, only a summation sign on top of which there’s an 8 on its side that means something else. Taking the essence of this approach we can still do things like calculus.

To say that the axiom of infinity is wrong would be a mistake (unless it turns out to be inconsistent with the other axioms, something that in polite circles you assume not to be the case). Axioms aren’t right or wrong, they just lead to a different “mathematics”. The mathematics resulting from the ZFC is rich and interesting and I personally feel it is extremely important that it be studied. I just find it can be unhelpful sometimes.

In this post I’ve attempted to suggest that in mathematics presuming the existence of infinitely large sets is somewhat arbitrary, leads to results that don’t really make sense and is not really necessary. So I will answer the first of the three questions I set out at the start. My personal answer:

Infinity (i.e. a set of infinite size) does not exist. Unless you choose it to, and then you should feel responsible for the consequences.

I’ve laid out this philosophical groundwork in the hope that it might now make more sense to think that the universe is finite. In my next post I’ll address this further, and eventually I’ll get to why I think this all naturally leads to a Big Bang to kick off the universe.

I can also now explain the apparently self contradictory name of this blog Finitism Forever. I think that the universe is spatially finite but that time may well continue for more than N years, where N is as large a number as you’d like to throw at me.


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Soc., 1996. It is impossible to do this in such a way that all the distances are preserved. Outline 1. Sinh (-x) = -sinh x 2. hyperbolic geometry, quantum simulation: Abstract: We show how quantum many-body systems on hyperbolic lattices with nearest-neighbor hopping and local interactions can be mapped onto quantum field theories in continuous negatively curved space. A non-Euclidean geometry, also called Lobachevsky-Bolyai-Gauss geometry, having constant sectional curvature . Hyperbolic Trig Identities is like trigonometric identities yet may contrast to it in specific terms. Simply stated, this Euclidean postulate is: through a point not on a given line there is exactly one line parallel to the given line. Now no need to make paper notes to remember mathematics formulas(all maths formulas) just have this app put all the formulas on your favorite phones. Hyperbolic geometry for colour metrics Ivar Farup Faculty of Computer Science and Media Technology, Gjøvik University College, Norway ivar.farup@hig.no Abstract: It is well established from both colour difference and colour order perpectives that the colour space cannot be Euclidean. For example, they are related to the curve one traces out when chasing an object that is moving linearly. Visualization of Hyperbolic Geometry A more natural way to think about hyperbolic geometry is through a crochet model as shown in Figure 3 below. Klein constructed an analytic hyperbolic geometry in 1870 in which a point Hints help you try the next step on your own. . 2. Feb 8, 2015 @ 8:32am I also found that property, that the circumference of a circle grows exponentially with its diameter in hyperbolic geometry, very intriguing! extending straight lines that pass through and which do not Some of them are: The derivatives of hyperbolic functions are: Some relations of hyperbolic function to the trigonometric function are as follows: The hyperbolic function identities are similar to the trigonometric functions. For any infinite straight line and any point not on it, there are many other infinitely New York: Springer-Verlag, 1999. See more ideas about Hyperbolic geometry, Geometry, Escher art. Letp:(−1;1)! Dec 18, 2016 - Explore Pendarestan ☮ Math Art's board "Hyperbolic geometry", followed by 251 people on Pinterest. .117 ... geometry, for which I mostly followed O’Neil’s reference book on Semi-Riemannian geometry. The angles atA,B,Care denoted byα,β,γ, respectively. 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# Electric field's due to continuous charge distributions

1. Jan 7, 2009

### Spoony

I'm currently at uni, but have difficulty doing problems involving continuous charge distributions. Say theres a charge distribution dl, dA or dV on a length surface or volume respectively at a distance R away from a point i know i must integrate over total length area or volume (depending on if the charge is distributed in 1 2 or 3 dimensions) but i get stuck here.
I do not know how to set up the intergrand. could anyone elp me with some examples link or explanation?? thanks very much :)
spoony

2. Jan 7, 2009

### gabbagabbahey

What text(s) are you using in your E&M course? Different authors use different notation, so to avoid confusion it is probably best if you post a couple of example problems from whichever text you are using.

3. Jan 7, 2009

### Spoony

Well trouble is the course lecturer doesnt use online notes and i was a bit ill during that time, but ok i shall grab one from a past paper. (text book briefly covers this topic)

A charge q is located at a perpendicular distance B from an infinite earthed plate. For a point distance R from charge q, the induced charge per unit area is given by

$$\sigma$$ = -$$\frac{qa}{{2} {\pi} {r^{3}}}$$

Use this to verify total charge on plate is -q

This question is like a capacitor i spose then, but if it can be solve with taking consideration to the electric field at point P (distance B from plate) then itd help greatly

thanks, spoony

4. Jan 7, 2009

### gabbagabbahey

This question doesn't seem to be asking you what the electric field is. Rather it is asking you to compute the total charge on the plate.

That being said, the induced charge on the plate will create an electric field (so will the point charge!) and if you like, I will help you through trying to calculate that field. First, I recommend you try to find the total charge on the plate. Do you know how to do that?

5. Jan 7, 2009

### Spoony

intergrate the expression for charge density sigma = charge/area with respect to area over all area. But i dont know how to do the mathematics of this :S.

6. Jan 7, 2009

### gabbagabbahey

Okay, I would recommend your first step to be drawing a picture of this and choosing an appropriate coordinate system. My choice would be to use Cartesian coordinates and let the plate be the z=0 plane (i.e. the x-y plane) and place the point charge on the z-axis at $(0,0,a)$.

Now, the plate is infinite, so you need to integrate the charge density from x=-infinity to +infinity and likewise for y. In order to do this integration, you need to express $\sigma$ in terms of 'x' and 'y'. That means you need to express the distance $r$ from the point charge at $(0,0,a)$ to a general point on the plane $(x,y,0)$ in terms of x and y. Do you know how to do that? What is the general formula for the distance between to points?

7. Jan 7, 2009

### Spoony

Ok so the distance between a point on the plane x-y and z=a is sqrt((x-0)^2+(y-0)^2+(a)^2)) which is:
$$\sqrt{x^{2} + y^{2} + a^{2}}$$
so this is equal to r in the expression for charge density.
so i end up with:

area integral of $$\frac{-qa}{2 \pi (x^{2} + y^{2} + a^{2})^{\frac{3}{2}}$$
(for some reason its giving me a intergral of polar co-ords :S)

But if it still isnt displaying correctly it should be -qa/(2*pi*(x^2+y^2+a^2)^(3/2)) intergrated with respect to dx then dy with limits of infinity and minus infinity.
Question is now how to intergrate? normally i would express limits of x in terms of y then intergrate the first then the second... :S

8. Jan 7, 2009

### gabbagabbahey

x and y are independent for every point on the plane, so you can treat y as a constant and integrate with respect to x first, and then integrate with respect to y or vice versa.

9. Jan 7, 2009

### Spoony

now to intergrate 1/(x^2+y^2+a^2)^3/2 with respect to x. trouble is i cant see a way of doing it unless it was by inspection. tried substitution and parts doesnt work because its a composite function.

10. Jan 7, 2009

### gabbagabbahey

The easiest way is to use polar coordinates $\rho=\sqrt{x^2+y^2}$, $\phi=\tan^{-1}(y/x)$ and $da=\rho d\rho d\phi$ and integrate $\rho$ from 0 to infinity and $\phi$ from 0 to 2*Pi.

11. Jan 7, 2009

### Spoony

but the expression contains an extra a^2 term, so how does that compare to the expression for p?

12. Jan 7, 2009

### gabbagabbahey

$a^2$ is just a constant, you should be able to integrate $$\frac{\rho d\rho}{(\rho^2+a^2)^{(3/2)}}$$ easily.

13. Jan 7, 2009

### Spoony

i cant seem to intergrate that :\$ having an attack of the stupid today i really am, ok substitute u = p^2 then du = 2p so times the whole integrand by 1/2 then
= 1 / (u+a^2)^(3/2) du
still cant think how to :S

14. Jan 7, 2009

### gabbagabbahey

Try substituting $$u=(\rho^2+a^2)^{(-1/2)}$$ instead

15. Jan 7, 2009

### Spoony

DONE! thanks for all your help mate :) more questions to follow though :P

16. Jan 7, 2009

### gabbagabbahey

You're welcome Feel free to post more of these problems in the homework forum and I'll try to help with them.


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# Thread: Second derivative of infinite series gives 0^0 situation

1. ## Second derivative of infinite series gives 0^0 situation

What is the correct way to handle this problem?

Evaluate the second derivative wrt x of the summation of [(2x+1)^(n+1)]/(n!) from n=0..inf at x=-1/2

My solution was to take the derivative twice, giving:

summation of [4(n+1)(2x+1)^(n-1)]/[(n-1)!] from n=1..inf

then substituting x=-1/2 giving

summation of [4(n+1)*0^(n-1)]/[(n-1)!] from n=1..inf

Now the mistake I originally made here was to assume 0^(n-1) will cancel to zero making the solution zero, but the first term of the series creates a 0^0 situation which evidently should be treated as 1 -- resulting in the final solution being 8. I'm not very well versed on the proper way to handle 0^0 situations and was wondering if I should always treat 0^0 = 1 in problems like this?

I recognize that I could just break out the first two terms of the original series before taking the two derivatives to avoid getting to a 0^0 situation, but I wasn't sure if this was necessarily the best way to do it. Thanks.

2. Originally Posted by drumist
I recognize that I could just break out the first two terms of the original series before taking the two derivatives to avoid getting to a 0^0 situation, but I wasn't sure if this was necessarily the best way to do it. Thanks.
This is the wisest way to handle this kind of problem
I personally take a great care of the first terms of a series in order to be sure of what I am doing

3. Originally Posted by drumist
What is the correct way to handle this problem?

Evaluate the second derivative wrt x of the summation of [(2x+1)^(n+1)]/(n!) from n=0..inf at x=-1/2

My solution was to take the derivative twice, giving:

summation of [4(n+1)(2x+1)^(n-1)]/[(n-1)!] from n=1..inf

then substituting x=-1/2 giving

summation of [4(n+1)*0^(n-1)]/[(n-1)!] from n=1..inf

Now the mistake I originally made here was to assume 0^(n-1) will cancel to zero making the solution zero, but the first term of the series creates a 0^0 situation which evidently should be treated as 1 -- resulting in the final solution being 8. I'm not very well versed on the proper way to handle 0^0 situations and was wondering if I should always treat 0^0 = 1 in problems like this?

I recognize that I could just break out the first two terms of the original series before taking the two derivatives to avoid getting to a 0^0 situation, but I wasn't sure if this was necessarily the best way to do it. Thanks.
Originally Posted by running-gag
This is the wisest way to handle this kind of problem
I personally take a great care of the first terms of a series in order to be sure of what I am doing
Let's give a little stronger hint here. Your second derivative is

$y''(x) = \sum_{n = 0}^{\infty} \frac{4n(n + 1)}{n!}(2x + 1)^{n - 1}$

There is a problem with simplifying the coefficient the way you did. Compare what happens to these two "identical" expressions when n = 0:
$\frac{4n(n +1)}{n!}$ <--The unsimplified version

$\frac{4(n + 1)}{(n - 1)!}$ <--The simplification

The first expression is 0. The second expression is undefined because (-1)! is undefined. In order to correctly calculate the series at n = 0 you have to use the first expression. What does this do when you plug x = -1/2 into your series?

-Dan


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# What are the types of ab initio Molecular Dynamics?

I am new to the world of Molecular Dynamics, and am curious to know exactly what is considered to be an ab initio Molecular Dynamics (AIMD) method, and how do they work?

The seminal work by Roberto Car and Michele Parrinello, titled "Unified Approach for Molecular Dynamics and Density-Functional Theory" was published 35 years ago!

I have recently come across the following "types" of MD methods while reading some research papers:

If I am missing methods, which I am sure I am, please feel free to add them through an answer! Also, it would be appreciated if one method is explained per answer, and is summarized in 2-3 paragraphs.

• This might need to be split into two questions. You're mixing "AIMD" with "quantum MD". In one case we are going beyond Newton's laws and trying to include quantum forces for the MD. In the other case we're actually trying to solve the dynamics quantum mechanically, meaning solving the time-dependent Schroedinger equation. Classical MD on a quantum forcefield is different from quantum MD. I've commented our the "quantum MD" topics, they can be asked separately :) – Nike Dattani Jul 15 at 19:38
• Agree with @NikeDattani, Ehrenfest also introduces non-adiabatic effects so this further complicates things – Cody Aldaz Jul 15 at 19:43
• Ahh ok. This is exactly what confused me! I am new to MD as a whole so idk what is ab-initio and why it is (or not) quantum MD under these contexts? That was my confusion if Quantum MD was for solving QD. Thanks for the edit @NikeDattani. I am willing to ask the second question as well. – Etienne Palos Jul 15 at 19:47
• @NikeDattani I would want to save BOMD for someone else, because it should be fairly trivial to include. Would ab initio Ring Polymer/Path Integral MD be worth including or fit with this question? Technically, the ab initio part is basically the same as in BO/CPMD, but the nuclear part is done differently to approximate quantum effects. – Tyberius Jul 22 at 18:38
• @EtiennePalos can you please make a hyperlink to a paper, for last three bulleted items in the list? It may help people find the papers and explain them. – Nike Dattani Jul 24 at 17:23

## CPMD: Car-Parrinello Molecular Dynamics

An approximation of BOMD (Born-Oppenheimer MD) where fictitious dynamics is used on the electrons to keep them close to their ground state, so that we do not have to keep solving for their ground state at every single step. We start with Newton's 2nd law (as does classical MD), but instead of the force being calculated by a full-fledged ab initio calculation at every step, the force itself has an EOM (equation of motion) which below is given by Eq. \eqref{eq:fictitious}. For one nucleus with position $$\vec{r}$$ and several electrons with orbitals $$\{\psi_i\}$$ we get:

\begin{align} \tag{1} \vec{F} &= m\vec{\ddot{r}} \\ - \nabla \, E\left[\{ \psi_i \} , \vec{r} \right] &= m\vec{\ddot{r}}\tag{2}\\ \mu \ddot{\psi}_i(\vec{r},t) &= - \frac{\delta E}{\delta \psi_i^*(\vec{r},t)} + \sum_j \Lambda_{ij} \psi_j(\vec{r},t),\tag{3}\label{eq:fictitious} \end{align}

where $$\Lambda_{ij}$$ is a matrix of Lagrange multipliers to allow satisfaction of the constraint that the wavefunctions $$\psi_i$$ must be orthogonal; and $$E[\{\psi_i\},\vec{r}]$$ is an energy functional (usually a Kohn–Sham energy one). For several nucleii, just make a new subscript for $$\vec{r}$$ and change the functional to $$E[\{\psi_i\},\{\vec{r_I}\}]$$, then the equations are exactly the same.

Implemented in:

• CPMD: Literally named after the method! (open-source)
• CP2K: Might also be named after the method! (open-source)
• NWChem (open-source)
• Interestingly, CP2K doesn't (or at least no longer) implement traditional CP, only 2nd generation CP. – Tyberius Jul 31 at 22:34
• But does CP2K stand for "Car-Parinello 2000" ? – Nike Dattani Jul 31 at 22:43
• I just found where they explain the name: cp2k.org/faq:name – Tyberius Jul 31 at 22:44
• I guessed correctly! Okay so CP2K can be removed from my list, if you suggest. – Nike Dattani Jul 31 at 22:45

ab initio Ehrenfest Dynamics

"The Born Oppenheimer (BO) and extended Lagrangian (EL) trajectories are founded on the assumption that a single electronic potential surface governs the dynamics. .. A major limitation of adiabatic trajectories is that they are not applicable to reactions involving nonadiabatic electronic processes, i.e., multiple potential-energy surfaces." (Ex: Conical Intersections)

To account for electronic adiabaticity, we solve the full time-dependent Schrödinger equation for both nuclear and electronic degrees of freedom. In Ehrenfest methodology, the adiabatic potential energy surface

$$E_{eff} = \langle\Phi|\hat{H_{el}}|\Phi\rangle = \sum_i |a_i|^2 E_i^2$$ Thus, the atoms evolve on an effective potential representing an average over the adiabatic states weighted by their state populations $$|a_i|^2$$. The method is also therefore referred to as mean-field approach. As a comparison, for BOMD/Ehrenfest dynamics:

• BOMD

$$\hat{H}_{el} (\mathbf{r}; \mathbf{R}) \Phi_k(\mathbf{r}; \mathbf{R}) = E^{el}_k(\mathbf{R})\Phi_k(\mathbf{r}; \mathbf{R})$$

$$M_I\ddot{\mathbf{R}}_I =-\nabla_IE^{el}_k(\mathbf{R})=-\mathop{\nabla_I}_{\text{min }\Phi_k}\langle\Phi_k|\hat{H}_{el}|\Phi_k\rangle$$

The electronic wavefunction $$\Phi_k(\mathbf{r}; \mathbf{R})$$ is static (only implicit time-dependence) and the nuclear degrees of freedom are handled classically. The nuclear degrees of freedom are decoupled from electronic degrees of freedom, while for each MD step the electronic wavefunction has to be optimized for the ground state.

• Ehrenfest dynamics

$$i\hbar\frac{\partial \Phi(\mathbf{r};\mathbf{R},t)}{\partial t}= \hat{H}_{el} (\mathbf{r}; \mathbf{R}) \Phi(\mathbf{r};\mathbf{R},t)$$

$$M_I\ddot{\mathbf{R}}_I =-\nabla_I\langle\hat{H}_{el}(\mathbf{r};\mathbf{R})\rangle$$

Here we have an explicit time dependence of the electronic wavefunction. Electronic and nuclear time evolutions are propagated with a three-time-step integrator. The electronic wavefunction is evolved via TD-SCF approach.

New contributor
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• +1. A heroic effort for sure! Did you not consider answering this question: mattermodeling.stackexchange.com/q/1676/5 ? You could have got a 50-point bounty! – Nike Dattani Aug 4 at 21:36
• @NikeDattani I am not too knowledgeable in the Ehrenfest dynamics. I have studied and done some basic exercises, but I am not a professional or even ametuer user of Ehrenfest dynamics. I hope the contrasts between BOMD and Ehrenfest dynamics makes it a little bit easier to understand the basic needs and speciality of the theory. Plus I didn't see it earlier enough! – mykd Aug 4 at 21:42
• No problem. You can always see the active bounties available here: mattermodeling.stackexchange.com/?tab=bounties and you can also see all previous bounties here: chat.stackexchange.com/rooms/109905/bounty-land – Nike Dattani Aug 4 at 21:44

# 2nd Generation CPMD

Car-Parrinello MD avoids repeatedly solving the electronic problem by propagating the orbitals as if they were particles governed by Newton's equations. This is much more efficient than having to solve at each time step as is done in Born-Oppenheimer MD, though at the cost of decreasing the maximum timestep for the dynamics (too large a step will lose the ground state), slightly reduced accuracy (not exactly at the ground state for each time step) and introducing a spurious "mass parameter" to describe the electronic motion.

To address these problems, Thomas Kuhne et al. developed the "Car-Parrinello like approach to BOMD", also referred to as second-generation CPMD. The key differences of this approach are:

• Rather than propagating the orbitals (or rather the MO coefficients $$\mathbf{C}$$), 2nd-CPMD propagates the density $$\mathbf{P}$$ (or $$\mathbf{PS}$$ for nonorthogonal orbitals). The density seems to evolve more smoothly than the coefficients, making it easier to work with.
• A predictor-corrector method (in their paper, the Always Stable Predictor Corrector (ASPC) method, but in principle any such method) is used to propagate the density. This generates a prediction of the next coefficients $$\mathbf{C}^p(t_n)$$ based on the previous $$K$$ density matrices. A corrected set of coefficients $$\mathbf{C}(t_n)$$ is then formed as $$\mathbf{C}(t_n)=\omega \text{MIN}[\mathbf{C}^p(t_n)]+(1-\omega)\mathbf{C}^p(t_n)$$ $$\omega=\frac{K}{2K-1}$$ where $$\text{MIN}$$ is a minimization and $$K$$ is a parameter which determines the accuracy $$O(\Delta t^{2K-2})$$. This update procedure eliminates the need for the mass parameter.
• The nuclear dynamics become dissipative, possibly due to the nonsymplectic electron dynamics. This is corrected by performing a short validation run of dynamics to compute a damping coefficient $$\gamma$$ for the system.

There are some finer details to make this scheme work (specific parameterization of $$\mathbf{C}$$, how $$\gamma$$ is obtained) which are available in the original papers (both of which have arXiv preprint versions available).

Implementations:

• CP2K

References:

1. Thomas D. Kühne, Matthias Krack, Fawzi R. Mohamed, and Michele Parrinello Phys. Rev. Lett. 98, 066401 DOI: 10.1103/PhysRevLett.98.066401
2. Kühne, T.D. (2014), Second generation Car–Parrinello molecular dynamics. WIREs Comput Mol Sci, 4: 391-406. DOI: 10.1002/wcms.1176
• I forgot about the bounty and didn't get a reminder as I thought I would. Barely remembered, and manged to award it with a total of 11 minutes to spare! – Nike Dattani Jul 28 at 15:59


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## USF Tampa Graduate Theses and Dissertations

#### Title

Leonard Systems and their Friends

2016

Thesis

M.A.

#### Degree Name

Master of Arts (M.A.)

#### Degree Granting Department

Mathematics and Statistics

#### Major Professor

Brian Curtin, Ph.D.

#### Committee Member

Brendan Nagle, Ph.D.

#### Committee Member

Jean-Francois Biasse, Ph.D.

#### Keywords

Equitable sl_2 basis, Leonard Pairs, Linear Algebra

#### Abstract

Let $V$ be a finite-dimensional vector space over a field $\mathbb{K}$, and let

\text{End}$(V)$ be the set of all $\mathbb{K}$-linear transformations from $V$ to $V$.

A {\em Leonard system} on $V$ is a sequence

$(\A ;\B; \lbrace E_i\rbrace_{i=0}^d; \lbrace E^*_i\rbrace_{i=0}^d),$

where

$\A$ and $\B$ are multiplicity-free elements of \text{End}$(V)$;

$\lbrace E_i\rbrace_{i=0}^d$ and $\lbrace E^*_i\rbrace_{i=0}^d$

are orderings of the primitive idempotents of $\A$ and $\B$, respectively; and

for $0\leq i, j\leq d$, the expressions $E_i\B E_j$ and $E^*_i\A E^*_j$ are zero when $\vert i-j\vert > 1$ and

nonzero when $\vert i-j \vert = 1$.

%

Leonard systems arise in connection with orthogonal polynomials, representations of many nice algebras, and the study of some highly regular combinatorial objects. We shall use the construction of Leonard pairs of classical type from finite-dimensional modules of $\mathit{sl}_2$ and the construction of Leonard pairs of basic type from finite-dimensional modules of $U_q(\mathit{sl}_2)$.

Suppose $\Phi:=(\A ;\B; \lbrace E_i\rbrace_{i=0}^d; \lbrace E^*_i\rbrace_{i=0}^d)$ is a Leonard system.

For $0 \leq i \leq d$, let

$U_i = (E^*_0V+E^*_1V+\cdots + E^*_iV)\cap (E_iV+E_{i+1}V+\cdots + E_dV).$

Then $U_0$, $U_1$, \ldots, $U_d$ is the {\em split decomposition of $V$ for $\Phi$}.

%

The split decomposition of $V$ for $\Phi$ gives rise to canonical matrix representations of $\A$ and $\B$ in terms of useful parameters for the Leonard system. %These canonical matrix representations for $\A$, $\B$ are respectively lower bidiagonal and upper bidiagonal.

In this thesis, we consider when certain Leonard systems share a split decomposition.

We say that Leonard systems

$\Phi:=(\A ;\B; \lbrace E_i\rbrace_{i=0}^d; \lbrace E^*_i\rbrace_{i=0}^d)$

and

$\hat{\Phi}:=(\hat{\A} ;\hat{\B}; \lbrace \hat{E}_i\rbrace_{i=0}^d; \lbrace \hat{E^*}_i\rbrace_{i=0}^d)$

are {\em friends} when $\A = \hat{\A}$ and $\Phi$, $\hat{\Phi}$ have the same split decomposition.

%

We obtain Leonard systems which share a split decomposition by constructing them from closely related module structures for either $\mathit{sl}_2$ or $U_q(\mathit{sl}_2)$ on $V$. We then describe friends by a parametric classification.

In this manner we describe all pairs of friends of classical and basic types.

In particular, friendship is not entirely a property of isomorphism classes.

COinS


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# Each year, Johan typically does all his own landscaping and yard work. He spends $200 per year on... 1 answer below » Each year, Johan typically does all his own landscaping and yard work. He spends$200 per year on mulch for his flower beds, $225 per year on flowers and plants,$50 on fertilizer for his lawn, and $245 on gasoline and lawn mower maintenance. The lawn and garden store where he obtains his mulch and fertilizer charges other customers$500 for the service of spreading that much mulch in flower beds and $50 for the service of distributing fertilizer over a yard the size of Johan’s. Paying a professional yard care service to mow his lawn would require an expenditure of$1,200 per year, but in that case Johan would not have to buy gasoline or maintain his own lawn mower.

a. In a normal year, how much does Johan’s landscaping and yard work contribute to GDP?

b. Suppose that Johan has developed allergy problems this year and will have to reduce the amount of his yard work. He can wear a mask while running his lawn mower, so he will keep mowing his yard, but he will pay the lawn and garden center to spread mulch and distribute fertilizer. How much will all the work on Johan’s yard contribute to GDP this year?

c. As a follow-up to part (b), at the end of the year, Johan realizes that his allergies are growing worse and that he will have to arrange for all his landscaping and yard work to be done by someone else next year. How much will he contribute to GDP next year?

Q

Ashish

a)

GDP is the total desired market value of all the final output of goods and services produced within a given year and transacted in an economy.

In a traditional year John does his own trim, spreading mulch and fertilizer. So these activities cannot be included in GDP as there are no market transactions for these.

But for buying mulch, flowers and plants, fertilizer, gas and other maintenance can be included and only these transactions will...


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# Math Help - Dividing by zero

1. ## Dividing by zero

Why can a number divided by zero equal infinity? Apparently it has something to do with asymptotes.

Like, when change in x is zero and change in y is not zero, the result is infinity.

2. ## Re: Dividing by zero

Originally Posted by mathbeginner97
Why can a number divided by zero equal infinity? Apparently it has something to do with asymptotes.
Like, when change in x is zero and change in y is not zero, the result is infinity.
It is axiomatic, division is not allowed.

So a number divided by zero is not equal infinity. Actually it is not defined.

3. ## Re: Dividing by zero

What number system are you discussing? Are you talking about the nonnegative extended reals? The Riemann Sphere? Otherwise, the answer is, a number divided by zero does not equal infinity. Infinity is not a number in most number systems. It seems like you are discussing geometry. So, try this. Find the slope of the lines between the following points:

1. $(0,0)$ and $(1,1)$
2. $(0,0)$ and $\left(\dfrac{1}{2},1\right)$
3. $(0,0)$ and $\left(\dfrac{1}{10},1\right)$
4. $(0,0)$ and $\left(\dfrac{1}{100},1\right)$
5. $(0,0)$ and $\left(\dfrac{1}{10^{10}},1\right)$

Given a fraction where the numerator is greater than zero and the denominator is positive, but very close to zero, the number resulting from the division is larger than the numerator. The closer you get the denominator to zero, the larger the number becomes. I can keep making the denominator closer and closer to zero, and the closer I get, the larger the number becomes after division.

So, using the real numbers, division by zero does NOT equal infinity. It does not equal anything else, either. It is simply undefined.

4. ## Re: Dividing by zero

Originally Posted by SlipEternal
What number system are you discussing? Are you talking about the nonnegative extended reals? The Riemann Sphere? Otherwise, the answer is, a number divided by zero does not equal infinity. Infinity is not a number in most number systems. It seems like you are discussing geometry. So, try this. Find the slope of the lines between the following points:

1. $(0,0)$ and $(1,1)$
2. $(0,0)$ and $\left(\dfrac{1}{2},1\right)$
3. $(0,0)$ and $\left(\dfrac{1}{10},1\right)$
4. $(0,0)$ and $\left(\dfrac{1}{100},1\right)$
5. $(0,0)$ and $\left(\dfrac{1}{10^{10}},1\right)$

Given a fraction where the numerator is greater than zero and the denominator is positive, but very close to zero, the number resulting from the division is larger than the numerator. The closer you get the denominator to zero, the larger the number becomes. I can keep making the denominator closer and closer to zero, and the closer I get, the larger the number becomes after division.

So, using the real numbers, division by zero does NOT equal infinity. It does not equal anything else, either. It is simply undefined.
I understand your note as if you are saying the limit is infinity but nothing is equal to infinity, but you concluded corretly infinity is undefined. Your example of getting the denominator smaller and smalser the result of the division is a very large number that approches infinity. This is the intuitive mathematical argument that plunged philosophy into mathematics. at that level abstraction mathematics, as well as phyisics become the realm of philosophi. The notion of infinity is more a philosopy question than it is mathamatical. The reason we cannot devide by zero is simply axiomatic as Plato pointed out. The underlying reason for the axiom is because sero is nothing and deviding something by nothing is undefined. That axiom agrees with the notion of limit infinity, i.e. undefined. There are more phiplosphy books and thoughts about infinity in philosophy books than than there are discussions on infinity in math books.

5. ## Re: Dividing by zero

Originally Posted by votan
I understand your note as if you are saying the limit is infinity but nothing is equal to infinity, but you concluded corretly infinity is undefined. Your example of getting the denominator smaller and smalser the result of the division is a very large number that approches infinity. This is the intuitive mathematical argument that plunged philosophy into mathematics. at that level abstraction mathematics, as well as phyisics become the realm of philosophi. The notion of infinity is more a philosopy question than it is mathamatical. The reason we cannot devide by zero is simply axiomatic as Plato pointed out. The underlying reason for the axiom is because sero is nothing and deviding something by nothing is undefined. That axiom agrees with the notion of limit infinity, i.e. undefined. There are more phiplosphy books and thoughts about infinity in philosophy books than than there are discussions on infinity in math books.
I am not sure what you are saying. Infinity is equal to infinity. Especially in number systems that include infinity such as the extended reals or the Riemann Sphere...

6. ## Re: Dividing by zero

Originally Posted by SlipEternal
I am not sure what you are saying. Infinity is equal to infinity. Especially in number systems that include infinity such as the extended reals or the Riemann Sphere...
Anybody proved infinity is equal infinity? consider segment AB = 5 cm and a segment PQ = 10 cm. each contains an infinite real numbers. Are these infinities equal?

7. ## Re: Dividing by zero

Actually, not all infinities are considered equal. Georg Cantor developed a theory on this Georg Cantor - Wikipedia, the free encyclopedia

8. ## Re: Dividing by zero

Originally Posted by votan
Anybody proved infinity is equal infinity? consider segment AB = 5 cm and a segment PQ = 10 cm. each contains an infinite real numbers. Are these infinities equal?
As grillage mentioned, Georg Cantor proved that there are different infinite cardinalities. In the example you mention, line segments AB and PQ considered as sets of points have the same infinite cardinality. On the other hand, $\overline{PQ} \cap \mathbb{Q}$ has a smaller infinite cardinality than line segment AB.

If you are looking at the set of ordinal numbers, there are infinite distinct infinities.

If you are looking at the extended reals, there are two infinities: $\infty$ and $-\infty$. If you are looking at the projective reals $\left(\mathbb{R}\cup \{\infty\}\right)$ or the Riemann Sphere $\left(\mathbb{C} \cup \{\infty\}\right)$, there is only one infinity. In each of these contexts, however, the number system is not a field.


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# zeta and theta

Consider
$G = \prod_{m=2}^{\infty} \frac{1}{\zeta(m)}$
We can rewrite this:
$\prod_{m=2}^{\infty} \prod_{\mathrm{p prime}} \frac{1}{1-1/p^{m}}$
Consider at this point we can swap products:
$\prod_{\mathrm{p prime}} \prod_{m=2}^{\infty} \frac{1}{1-1/p^{m}}$
We can then rewrite the inner product:
$\prod_{\mathrm{p prime}} \left(\prod_{m=2}^{\infty} 1 - 1/p^{m}\right)^{-1}$

Written out fully, we get, after consulting mathworld:

The inner is almost a product of Jacobi theta functions:

$\prod_{m=2}^{\infty} \frac{1}{\zeta(m)} = \prod_{\mathrm{p prime}} \frac{p}{p-1} \left(\sqrt[24]{p}\sqrt[3]{\frac{1}{2}\theta_{1}'(0,1/\sqrt{p})}\right)$

It is interesting that this is almost, but not quite the product of the probabilities that a large random $p$-ary matrix is nonsingular, see A048651, and the logical question then becomes, page 27. How does enumerating finite nilpotent groups correspond to the product of the zeta function evaluated at $n\geq 2$: the question is, what is the more abstract viewpoint on the swapping of the products?

Moreover, it would be interesting to find a good Hecke correspondence angle on this, but I don’t know enough about Hecke correspondence to make the right statement about it.

Additional: I wonder if it is possible to employ the bewilderingly large range of identities for the theta function so that the side with the zeta on it turns into some kind of invariant.