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If he’s moving faster than light then that’s not unreasonable

! Moderator Note We're not doing this again. This speculation couldn't be supported before, and changing the diagrams doesn't change that fact. This is a waste of time and you've shown us this before, and your threads on it were closed, and you were asked not to do this again. Have a vacation, think about the rules you agreed to follow, and if you come back, we can discuss some science.

Good one. Here's the catch: (my emphasis on OP's source) Nearly a 100 years in the books. And still people, when in doubt between reality and causality, would rather sacrifice the latter. It is reality that is dead as a sharp concept, even though it is a very good approximate one. This goes to show how adhesive the concept of classical reality is: "It's either this or not this" is harder to give up than smoking.

! Moderator Note The account owners have been contacted to inform them some children have gained access the accounts

Yes I know for a fact that time dilation cannot account for the cosmological constant. When you do the math you will find that you get the wrong ratios of change. You will also discover as Migl mentioned Newtons Shell theorem that time dilation isn't involved when you have a homogeneous and isotropic mass distribution. Let's do a simple example if you use recessive velocity from Hubbles Law \[v_{r}=H_0d\] and apply that velocity term to the time dilation formula you will think it will work out. However once you get to the Hubble horizon it approach infinity. The Observable universe itself is far larger than the Hubble horizon. The recessive velocity formula will have a value for recessive velocity of 2.3 c. Thankfully recessive velocity is not a true velocity but an apparent velocity due to separation distance. As our time dilation formula would break down when velocity is greater than c. Leading to causality issues. Our light cone would be restricted to the Hubble horizon. The recessive velocity equals c at the hubble horizon. We observe beyond that. Does that help ?

Amen OP confused "theory of everything" with "everything I think is a theory".

Old house in Texas. Had a fly like this land on a bay window in the kitchen. Discovered quickly that day it was single pain glass when I smushed the fly with my now bleeding thumb. Taught myself that weekend how to glaze a new pane into a bay window. And squished the fly, so it was basically pure win.

ok well as far as the chart goes the details are correct the main reason why the SM model uses the SU(n) groups is that tis group is compact which becomes important for renormalization as well as Feymann path integrals. the SL(n) group is not compact which can lead to issues with renormalization even though both groups are closed groups the isomorphism for the Lorentz/Poincare group for example is \[S0(3.1)\simeq Sl(2,\mathbb{C})/\mathbb{Z}_2\] spinors are defined to transform under the action of the \( SL2(\mathbb{C}\) group. So yes that link is accurate where it gets used as opposed to other group types depends on the state being described. However as shown here \[S0(3.1)\simeq Sl(2,\mathbb{C}/\mathbb){Z}_2\] you can have isomorphisms with other groups the isomorphisms for SU(N) to the SL(2,c) group can be found here which is what reference 6 of the link you gave which corresponds to the chart you posted employs further details here https://diposit.ub.edu/dspace/bitstream/2445/121903/2/memoria.pdf I should also forewarn you though that the Schrodinger equation of QM is not Lorentz invariant although the Dirac equations are. The operators used in QM (position and momentum) are not employed in QFT (field and momentum). QM uses the Klein Gordon equations which are Lorentz invariant. This is the purpose of reference 6 ": Linear Canonical Transformations (LCTs) are known in signal processing and optics as the generalization of certain useful integral transforms. In quantum theory, they can be identified as the linear transformations which keep invariant the canonical commutation relations characterizing the coordinates and momenta operators. In this work, the possibility of considering LCTs to be the elements of a symmetry group for relativistic quantum physics is studied using the principle of covariance" reference 6 here https://arxiv.org/pdf/1804.10053

...and now a bible salesman! This short video is a good example of Trump's assumption that other people are as hateful and greedy as he is. A lot goes on in this short video. "The great John Paulson made plenty of money in Nevada. He doesn't live there but he makes a HELL of a lot of money. He makes money everywhere he goes actually. He's a money machine. You know what? Put him in treasury. You want to make a little money?" "....Then we go to South Carolina where we have done really well, where I have done well...." Notice how he contorts his mouth as he talks and how he belittles Tim Scott, but Tim just rolls with the punches. "Almost every one of them endorsed me. Two great senators. Which is hard. I mean did you ever think that she [Haley] actually appointed you, Tim? And think of it, appointed you, and you're the senator of her state, and she endorsed me. You must REALLY hate her." Laughter "No, it's a shame. It's a shame." Tim Scott steps forward to the mike. Trump is surprised to see Scott next to him and says "Uh...Oh..." as though he realized that he just insulted him by assumed Scott "REALLY hated" Niki Haley, because HE would hate in the same petty circumstances. But Scott meekly shows his subservience to Trump by saying "I just love YOU!" Trump replies, "That's why he's a great politician!" The Antichrist has the mouth of a lion – Rev. 13:2 – Trump contorts his mouth as he speaks, always showing he teeth, so you can read his lips from 100 yards, and often roars like a lion The Antichrist is arrogant, shall exalt himself above others – Daniel 8:25 - Check The Antichrist will love money as it is the LOVE of money, (not money itself), that is the “root of all kinds of evil.” – 1 Timothy 6:10 - Trump said “My whole life I’ve been greedy, greedy, greedy for money. I grabbed all the money I can get. I’m so greedy.” The Antichrist understands dark sentences or sinister schemes – Daniel 8:23 - “A king of fierce countenance, and understanding dark sentences, shall standup” - He is a master of intrigue and machination, which is scheming or crafty action, to accomplish an evil end. The Antichrist attains the kingdom by flattery – Daniel 11:21 - “And in His estate shall stand up a vile person, to whom they shall not give the honor of the kingdom: but He shall come in peacefully and obtain the kingdom by flatteries or fine promises (e.g. Mexico will pay for the wall) scheming with a small group of people (the electoral college). The Antichrist is connected to gold – Rev. 13:18 - King Solomon became obsessed with gold and pagan gods during his later years. Trump’s obsession with gold and “pleasant things” is comparable to King Solomon. Trump's hair is golden, his face is painted a darker shade of gold, his penthouse is of golden decor, and he changed the Oval Office curtains to gold. There you have it in less than 2 minutes, an illustration of Trump's greed, exalting himself, the mouth of a lion, flatterer, hatred, "man of gold" in his golden palace, all characteristics of the classical Antichrist.

LOL! It's interesting how much low value coinage gets thrown into lakes. I suppose people are seeing how many skips they can get a coin to do across the surface.

There are easier ways to catch fish. It takes 16 Teslas to levitate a frog, so you'd need at least that, and several megawatts to power the magnet, for the average fish. 😁

"single pain" appears to be a Freudian slip.😄

Developing on my previous post, I think that the metric \[diag(-1,1,1,1+H(z)z^2+H(-z)z^4)\] is not a valid solution to the EFE. *H() is Heaviside step function.

Have you been to the ophthalmologist about it? If not I suggest doing so without delay. New floaters can indicate posterior vitreous detachment (PVD). While common enough, these can result in retinal detachment in 10% of cases. I had a PVD last year and needed laser surgery to weld the flapping edge of my left retina back in place. This was regarded as very urgent by the doctors.

If the metric is twice differentiable everywhere, then its Einstein tensor is everywhere defined, and you can just take this Einstein tensor as the energy-momentum tensor of your equation. Then, this metric is a solution of this equation. P.S. Of course, the metric has to be locally Lorentz to start with.

This got me thinking - setting aside singularities for the moment, are there (in a purely mathematical sense) metrics that are not valid solutions to the EFE? IOW, could one write down a metric for which the equations cannot be worked backwards to obtain a corresponding energy-momentum tensor, given we are on a semi-Riemannian manifold? Again, I’m talking purely mathematically, never mind physical realisability. I suspect the answer is no.

You can but keep in mind that dissertation was back in the late 80's lol and publishing requirements will vary depending on the publisher. Swansont for example has a peer reviewed paper on arxiv. I studied it a few years back.

By fraudulent or shoddy science. It is important to understand that if I go to the trouble of gathering data in good faith then I am not going to ignore it. If others do the same experiment and it looks like they get really different data, then there will be intense peer review and scrutiny of our datasets and methods. A common fraud is someone getting data that doesn't support their favored hypothesis and so they massage the data to fit. Because very few scientists are the sole researchers in an area of inquiry, this kind of fraud has a way of being discovered. Someone reported that transcranial magnetic stimulus of the parietal lobe caused people to see God. Turned out to be bad interpretation of data and experimenter asking subjects leading questions and limiting the sample to religious people. Then others repeated the TCMS experiment with a better cross-section of population, more open and neutral questions as to what was felt, more longitudinal data (repeating experiment with a subject multiple times over a couple years) and so on. From those experiments it seemed the parietal lobe simply generated a sense of presence and then people filled in with details that, in their cultural context, made sense to them. Shoddy science and fraud, ignoring evidence, tends to reveal itself.

And yet there you are, crossing bridges, stepping aboard jetliners, flipping on lightswitches and computers and stoves, taking pharmaceuticals, reading weather forecasts before you select your day's wardrobe, etc as if they were going to work properly, based as they are on empirical science. You must not be that worried about bias. And when there is bias, like the over-dependence on computational models of mind, prominent scientists (like my cited one, Robert Epstein) make beaucoup de bruit about it and there are massive critical discussions all over the Web and in professional journals and pretty soon MIT Technology Review and Nature and Scientific American are devoting feature stories to how computers are a weak and sometimes misleading metaphor for how brains work. Indeed, it's hard to imagine any other human mode of knowledge that is so brutally mean towards bias. The glee that scientists bring to finding bias, in peer review, is breathtaking.

"Are all kinds of vinegar more or less the same?". Yes, but I checked once and discovered that my local supermarket was selling Balsamic vinegar at a higher price pre litre than the champagne. Fruit flies are typically attracted to alcohol which is released by decaying fruit, but also present in vinegar. It would be interesting to compare it with "non brewed condiment" in that regard.

Ah, so many 'wanna-be' Philosophers who think the universe resides in their mind, and its reasons and workings can be divined in their own heads. Good Philosophers temper their ideas with science and evidence based facts, before attempting to preach to us peons.

Well, to be fair, he did say and, true to his word, he didn't present any, and there is nothing to dispute or discuss on this discussion forum. Might as well close the thread, then.

Some have said that the universe appears to have been designed intelligently and this has been cited as an argument in favour of a divine creator. Arguments have gone something along the lines of "Take the complexity of the human eye, whether it came into being 6000 years ago or evolved over millions upon millions of years, it looks kind of like how we would design a camera." There is a reason this argument falls flat on it's face. The same reason an argument in favour of the chicken coming before the egg would fall flat on it's face. There is a temporal bias at play wherein you see how we have designed things first and you see those same patterns in nature around you second; there lies the mistake because those patterns came first and are the basis of how we design things intelligently. The universe is not modelled after intelligence, rather our intelligence is modelled after the universe. If you stumble upon a watch that may prove there is a watch maker but without time and space and the nature of those things being what they are, neither the watch or the watch maker would exist. Intelligence wouldn't exist. The universe was not designed intelligently, the universe designs intelligence. I am purely agnostic when it comes to the existence of the divine or some kind of creation. If something comes down from the sky with seemingly god like powers I'll be assuming technology that I don't understand before "This can only be a god." The Teleological argument has just never sat right with me due to this strange temporal bias at play within the minds of the religious and the spiritual. I'm open to hearing better arguments in favour of the existence of some kind of cosmic entity that actively cares about me as an individual but intelligent design just is not one of them. It would be like building a model of the golden gate bridge and then claiming the architect of golden gate bridge used your model. Unless you have a time machine it just doesn't make much sense.

OK. Then the Wikipedia article needs editing, because it's very confusing. It explains nothing of that. Not a thing.

Evidently your perception is mistaken. I assume that it is based on pop-science rather than actual science sources. In my direct experience, science is fun and exciting.

What then are the explicit links between your explanations and the classification that was given? and how about the part concerning what is called quantum phase space? I guess it's not just spacetime but also includes momentum space? I may somehow understand your answers but I still not understand the relations between all these things and the physics that is behind. Thank you

Understood and I'm glad you recognize that these symmetry relations are internal symmetries and not spacetime symmetries. To understand spin I would recommend taking time studying Cartan subalgebra. Here is the trick The synmmetry representations are expressed according to weights which correspond to eugenvalues. For example the quantum numbers of angular momentum all have their own weight under lie algebra however they also have their own weight diagram. (aka root diagram) for example the spin j of a particle is given by \[U(\vec{\theta})=e\frac{i}{\hbar}\vec{\theta}\cdot \hat{J}\] where \(\hat{J}\) is the three angular momentum operators whose representation will be given by \(2J+1\) dimensional and \(\vec{\theta}\) are the 3 parameters gives \[e^{\frac{i}{\hbar}\vec{\theta}\cdot J}|jm\rangle=\sum_{n}=-jC_nJn\rangle\] imposing \[U(\vec{\theta-1})U(\vec{\phi}(\theta_1\theta_2))\] results in subalgebra SU(2) \[[J_i,j_j]=i\hbar\epsilon_{i,j,k}J_k\] where raising and lowering operators are defined \[J_\pm=(J_1\pm iJ_2)/11/2\] there is the spin operations you mentioned. for SU(3) Now the Gell Mann matrices above has three basis states. \[|\Lambda\mu_1\rangle=\begin{pmatrix}1\\0\\0\end{pmatrix}\] \[|\Lambda\mu_2\rangle=\begin{pmatrix}0\\1\\0\end{pmatrix}\] \[|\Lambda\mu_2\rangle=\begin{pmatrix}0\\0\\1\end{pmatrix}\] where \({\mu_1,\mu_2,\mu_3}\) are called two component weight vectors given by eugenvalues \(H_1=\lambda_3/2\) and \(H_2=\lambda_8/2\) see Gell-Mann matrices above \[\Lambda \mu_1=(1/2,\sqrt{3}/6):|\Lambda \mu_2\rangle=(-1/2,\sqrt{3}/6):|\Lambda \mu_3\rangle=(0,-\sqrt{3}3/3)\] the above is your Dynkan spin representation of SU(3) SU(3) has an eight dimensional root diagram which is an adjoint representation not shown above https://en.wikipedia.org/wiki/Adjoint_representation#Structure_constants for the OP understandably this will likely be over your head but also for other readers Group theory is a theory of representations these representations gives us tools to find and organize symmetry relations and antisymmetric relations. These representations have their own algebras (lie Algebra, Clifford algebra, Cartan algebra, etc,etc). They often use internal symmetries which can be thought of as (mathematical symmetries) though these can be also be physical quantities or probability quantities. In particle physics the state is a typically a probability wavefunction same for QFT. Lie algebra involves raising and lowering of Operators an operator has a requirement of being a minimal 1 quanta of action. (Langrangian) Now we also have group symmetries homomorphism> a linear map between two lie algebras is homomorphic if it is non invertable.(useful for bosons aka symmetric) An isomophism is invertable (fermions aka antisymmetric). Now lie algebras have subalgebras. Dynkin diagrams help us organize all the simple and semi-simple representations. In a sense it forms an atlas of our mappings. So SU(3) has 8 generators The Gell Mann matrices above. Each matric has its own root and hence its own weight that has its own weight diagram (aka root diagram which is a map). These maps can be a sub group of a larger group and vice versa. Dynkin diagrams also provide these details.

It may be their personal belief. You can also say that some people who reside on fifth floors "do not even believe that the subjective even exists." This is as true.

The heuristics applied by an expert may appear as bias to a layperson. (I haven't seen the the word heuristics so far in this discussion)

I imagine the expensive ones are human-sourced and cheaper ones probably synthetic.

The Florida clemency board mentioned decides on restoration of voting rights of felons in that state. DeSantis is the chair. It has nothing to do with granting clemency for the conviction, just Florida voting rights. Not a jurisdiction issue.

One of the producers of The Apprentice has starting talking since the NDA he signed has lapsed. He said that it took hours of editing to make Trump seem sharp in that show. They would often resort to putting him into a recording booth to read lines from a script. They would then use those recordings in the shots when the camera was on a contestant and not Trump himself.

This doesn't make much sense to me; surely a clemency board only looks at state specific crimes? How can they jurisdictionally grant clemency to a different states conviction? Is it really that simple or is DeSantis wading into a hard legal battle in his home state?

Based on what we’re seeing, it depends more on who the felon is and whether they’re part of the governors in-group or out-group

Watch out for the ones with fuzzy butts!

Think yourself lucky, I got a Turkey who thinks I'm sexy enough to gobble at...

"Hell hath no fury like a woman scorned."

Today learned that female cicadas seem to confuse the sound of lawnmowers with the sound of male cicadas. I can walk through my yard and maybe have a couple of cicadas fly into me. When I cut the grass it is more on the order of 100 or so that land on me, one after the other.

Your welcome I wouldn't use the reference 6 it's not a method to learn particle physics from. While the LCTs and the SL(2C) group is part of particle physics the chart in reference 6 was generated using the SM Unitary groups. The groups the SM model primarily uses is U(1), SU(2) and SU(3). Focus on those groups first as well as the Poincare group. SO(3.1). The above is the groups for the SM model the LCTs while has uses that wiki page doesn't describe correctly where they are used in particle physics. Instead it's reference is a method in development It is the Unitary groups mentioned above that you will find in any particle physics textbook and not what is described by the LCT wiki link. Dynkin diagrams are advanced beyond the introductory level. While the Unitary groups will provide the details for another representation (Feymann path integrals)

for reference https://www.tigercheng.xyz/Dynkin_Diagrams.pdf see 4.2 for rank 2 roots. which I supplied the relations above. The roots can be thought of as 2 dimensional vectors in a plane some other helpful diagrams involving other groups as well. https://web.ma.utexas.edu/users/vandyke/notes/261_notes/lecture19.pdf this one includes the Coxeter diagrams (they act as symmetry reflections ) well explained in this link https://math.ucr.edu/home/baez/twf_dynkin.pdf hope that helps Edit I too find the 5 dimensional part used in reference 6 of the OPs link a bit fishy going to look into that particular article in greater detail

Lol I think you became too used to Unitary and orthogonal groups. Joigus Would it help to know SO(3.1) and SU(n) are both subgroups of SL(2,c)/Z_2 ? lets start with the following \[sl(2,\mathbb{C})=su(2)\oplus isu(2)\] generators denoted e,f,h [e,f=h] [h,e]=2e [h,f]=-2f the 2C is the linear combination of e,f,h \[\pi (h)=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\] \[\pi( e)=\begin{pmatrix}0&1\\0&0\end{pmatrix}\] \[\pi h=\begin{pmatrix}0&0\\-1&0\end{pmatrix}\] \[f_i,h_i,e_i\] i=1,2,3....r however the set of complex cannot all commute so you need commutations \[[h_ih_j]=0\] \[[h_i,e_j]=A_{ji}e_j\] \[h_i,f_i]=-A_{ji}f_j\] \[[e_i,f_j]=\delta_{ij}h_{ij}\] where \(A_{ij} \) is the Cartan matrix ( I won't go through the ladder operators as they are fairly lengthy) however it can be expressed as \[[h_i,e_i]=\langle\alpha_j\rangle=\frac{2}{\langle\alpha_j,\alpha_j}\langle\alpha_j,\alpha_i\rangle_j=A_{ji}e_j\] \[\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\] the above is for SL(2C) for sl(3,C) the Cartan matrix is an 8 dimensional algebra of rank 2 which means it has a 2 dimensional Cartan sub algebra given as follows \[\pi(t_1)= \begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}\] \[\pi(t_2)= \begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}\] \[\pi(t_3)= \begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\] \[\pi(t_4)= \begin{pmatrix}0&0&1\\0&-1&0\\0&0&0\end{pmatrix}\] \[\pi(t_5) =\begin{pmatrix}0&0&-i\\0&0&0\\i&0&0\end{pmatrix}\] \[\pi(t_6)= \begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}\] \[\pi(t_7)= \begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix}\] \[\pi(t_1)=\frac{1}{\sqrt{3}} \begin{pmatrix}0&1&0\\1&0&0\\0&0&-2\end{pmatrix}\] You may note the last is the Gell-Mann matrices if we take the commutator between \(\pi(t_1)\) and \(\pi(t_2)\) we get \([\pi(t_1),\pi(t_2)]=2i\pi(t_3)\) which is familiar in the su(2) algebra. Thus we can define the following \[x_1=\frac{1}{2}t_1\] \[x_2=\frac{1}{2}t_1\] \[x_3=\frac{1}{2}t_3\] \[y_4=\frac{1}{2}t_4\] \[y_5=\frac{1}{2}t_5\] \[z_6=\frac{1}{2}t_6\] \[z_7=\frac{1}{2}t_7\] \[z_8=\frac{1}{\sqrt{3}}t_8\] with change in basis \[e_1=x_1+ix_2\] \[e_2=y_4+iy_5\] \[e_3=z_6+iz_7\] \[f_1=x_1+ix_2\] \[f_2=y_4+iy_5\] \[f_3=z_6+iz_7\] Now I should inform everyone that the basis and coordinates I am describing apply to Dynken diagrams and what I am describing apply to the root diagrams... https://en.wikipedia.org/wiki/Dynkin_diagram the basis above in matrix form is \[\pi(e_1)=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}\] \[\pi(e_2)=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\] \[\pi(e_1)=\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}\] \[\pi(f_1)=\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix}\] \[\pi(f_2)=\begin{pmatrix}0&0&0\\0&0&0\\1&0&0\end{pmatrix}\] \[\pi(f_3)=\begin{pmatrix}0&0&0\\0&0&0\\0&1&0\end{pmatrix}\] \[\pi(x_3)=\frac{1}{2}\begin{pmatrix}1&1&0\\0&-1&0\\0&0&0\end{pmatrix}\] \[\pi(z_8)=\frac{1}{3}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}\] @joigus That should help better understand the special linear group of Real as well as complex. Now knowing the above applies to Dynken diagrams will also help better understand the validity of the OPs link as well as the methodology. @TheoM Hope this answers your question as well on the validity behind the LCT's and where they are applied in particle physics so yes the link overall you provided is valid

No. There’s no evidence that actually supports it, and depending on the version, there’s evidence that actively contradicts it. Of one is going to use “design” as a description, then there’s a lot of unintelligent design in nature.

Hrrm I can see these functions could have a broad range of applications. Thanks for sharing this, gives me something new to study myself. I always keep an eye out for useful mathematical methods that I could employ. I will have to look more into the Leal functions. You might this listing handy https://personal.math.ubc.ca/~cbm/aands/abramowitz_and_stegun.pdf There is a section listing transcendental functions. (The article simply has a good listing of various functions for the purpose of quick reference . It doesn't go into any particular details on any of them. I found it handy in the past you might as well. Edit doesn't seem to be a whole lot of information on those functions beyond the links you already posted.

[math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12 \phi^2 - 3 \phi^4}}{2 (1 - \phi^2)}[/math] From here, I took a shortcut and simply ignored the two [math]\phi^2[/math] outside the square root and the [math]3 \phi^4[/math] under the square root because I could see that these are not going to be a part of the final result. Thus: [math]1 + \alpha = \dfrac{4 - \sqrt{12 \phi^2}}{2}[/math] for [math]x \to 0[/math] (ignoring higher-order terms) [math]1 + \alpha = 2 - \sqrt{3} \phi[/math] However, one can verify that the shortcut leads to the same result as follows: [math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12 \phi^2 - 3 \phi^4}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12} \phi \sqrt{1 - \dfrac{1}{4} \phi^2}}{2 (1 - \phi^2)}[/math] The series expansion of [math]\dfrac{1}{1 + x}[/math] and [math]\sqrt{1 + x}[/math]: [math]\dfrac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots = 1 - x[/math] for [math]x \to 0[/math] (ignoring higher-order terms) [math]\sqrt{1 + x} = 1 + \dfrac{1}{2} x - \dfrac{1}{8} x^2 + \dfrac{1}{16} x^3 - \dfrac{5}{128} x^4 + \cdots = 1 + \dfrac{1}{2} x[/math] for [math]x \to 0[/math] (ignoring higher-order terms) Thus: [math]\dfrac{1}{1 - \phi^2} = 1 + \phi^2[/math] for [math]x \to 0[/math] (ignoring higher-order terms) [math]\sqrt{1 - \dfrac{1}{4} \phi^2} = 1 - \dfrac{1}{8} \phi^2[/math] for [math]x \to 0[/math] (ignoring higher-order terms) [math]1 + \alpha = \dfrac{1}{2} (4 - \phi^2) (1 + \phi^2) - \sqrt{3} \phi (1 - \dfrac{1}{8} \phi^2) (1 + \phi^2)[/math] [math]1 + \alpha = \dfrac{1}{2} (4 + 3 \phi^2 - \phi^4) - \sqrt{3} \phi (1 + \dfrac{7}{8} \phi^2 - \dfrac{1}{8} \phi^4)[/math] [math]1 + \alpha = 2 - \sqrt{3} \phi + \dfrac{3}{2} \phi^2 - \dfrac{7 \sqrt{3}}{8} \phi^3 - \dfrac{1}{2} \phi^4 + \dfrac{\sqrt{3}}{8} \phi^5[/math] [math]1 + \alpha = 2 - \sqrt{3} \phi[/math] for [math]x \to 0[/math] (ignoring higher-order terms)

Not stated but used in the solution is the series expansion of [math]\cos \phi[/math]: [math]\cos \phi = 1 - \dfrac{\phi^2}{2} + \dfrac{\phi^4}{24} - \dfrac{\phi^6}{720} + \cdots = \displaystyle \sum_{j=0}^{\infty} \dfrac{(-1)^j}{(2j)!} x^{2j}[/math] For [math]\phi \approx 0[/math], the terms after [math]\dfrac{\phi^2}{2}[/math] can be ignored, and the closer [math]\phi[/math] is to [math]0[/math], the better [math]1 - \dfrac{\phi^2}{2}[/math] is as an approximation of [math]\cos \phi[/math]. In the limit of [math]\phi \to 0[/math] ([math]n \to \infty[/math]), the approximation is effectively exact. But [math]\phi[/math] can't be exactly [math]0[/math] ([math]\cos 0 = 1[/math]) because then the ratio of the total area of the small circles to the area of the large circle in the limit of [math]n \to \infty[/math] becomes indeterminate. Thus: [math]\cos^2 \phi = (1 - \dfrac{\phi^2}{2})^2 = 1 - \phi^2 + \dfrac{\phi^4}{4} = 1 - \phi^2[/math] for [math]\phi \to 0[/math] (ignoring the [math]\dfrac{\phi^4}{4}[/math] term) [math]2(1 + \cos \phi) = 2 + 2(1 - \dfrac{\phi^2}{2}) = 4 - \phi^2[/math] for [math]\phi \to 0[/math] Also, solving the quadratic equation involved carefully ignoring the higher order terms, leaving only the term that is linear in [math]\phi[/math].

I don't know where you got those numbers from, but have you interpreted [math]\alpha (R-r)[/math] as [math]\alpha[/math] as a function of [math]R-r[/math] instead of [math]\alpha[/math] multiplied by [math]R-r[/math]? [math]\alpha[/math] is the ratio of the radius of the second outermost circle to the radius of the outermost circle. It is also the ratio of the distance between the centre of the large circle and the center of the second outermost circle to the distance between the centre of the large circle and the center of the outermost circle. Thus, [math]\alpha r[/math] is the radius of the second outermost circles, and [math]\alpha (R-r)[/math] is the distance between the centre of the large circle and the center of the second outermost circle. I chose to define the ratio [math]\alpha[/math] rather than the radius and distance themselves. Much of the solution was about finding the value of [math]\alpha[/math], and even then, only for very large values of [math]n[/math]. It should be noted that the same value [math]\alpha[/math] describes both ratios because the second and subsequent outermost layers of circles are scaled down versions of the outermost layer of circles.

That's ok, I was curious. How did you solve the second problem?

The following is a solution to the second problem: Let [math]R[/math] be the radius of the large circle. Let [math]r[/math] be the radius of the small circles in the outermost layer. Let [math]\alpha r[/math] be the radius of the small circles in the second outermost layer. Let [math]n[/math] be the number of small circles around the circumference of the large circle. Let [math]P[/math] be the proportion of the area of the large circle covered by small circles for [math]n[/math] small circles around the circumference of the large circle. Partitioning the large circle into [math]n[/math] sectors, the area of each sector is: [math]= \dfrac{\pi R^2}{n}[/math] The total area of small circles covering a sector: [math]= \pi r^2 (1 + \alpha^2 + \alpha^4 + \alpha^6 + \cdots) = \dfrac{\pi r^2}{1 - \alpha^2}[/math] Then the proportion of the area of a sector covered by small circles: [math]P = \dfrac{n r^2}{(1 - \alpha^2) R^2}[/math] [math]\dfrac{r}{R - r} = \sin\dfrac{\pi}{n}[/math] [math]r = (R - r) \sin\dfrac{\pi}{n}[/math] [math]r (1 + \sin\dfrac{\pi}{n}) = R \sin\dfrac{\pi}{n}[/math] [math]\dfrac{r}{R} = \dfrac{\sin\dfrac{\pi}{n}}{1 + \sin\dfrac{\pi}{n}}[/math] [math]P = \dfrac{n\>\sin^2\dfrac{\pi}{n}}{(1 - \alpha^2) (1 + \sin\dfrac{\pi}{n})^2}[/math] Let [math](1 - \alpha^2) = \beta \dfrac{\pi}{n}[/math] [math]\displaystyle \lim_{n\to\infty} P = \dfrac{n\>\left( \dfrac{\pi}{n} \right)^2}{\beta\>\dfrac{\pi}{n}} = \dfrac{\pi}{\beta}[/math] Applying the law of cosines: Note that [math](1 + \alpha) r[/math] is the distance between the centres of the outermost and nearest second outermost circles, and that [math](R - r)[/math] and [math]\alpha (R - r)[/math] are the distances from the centre of the large circle to the centres of the outermost circle and second outermost circle, respectively. [math](1 + \alpha)^2 r^2 = (1 + \alpha^2) (R - r)^2 - 2 \alpha (R - r)^2 \cos\dfrac{\pi}{n}[/math] [math](1 + \alpha)^2 \dfrac{r^2}{(R - r)^2} = 1 + \alpha^2 - 2 \alpha \cos\dfrac{\pi}{n}[/math] [math]\dfrac{r}{R - r} = \sin\dfrac{\pi}{n}[/math] [math](1 + \alpha)^2 \sin^2\dfrac{\pi}{n} = 1 + \alpha^2 - 2 \alpha \cos\dfrac{\pi}{n}[/math] [math](1 + \alpha)^2 \sin^2\dfrac{\pi}{n} = (1 + \alpha)^2 - 2 \alpha (1 + \cos\dfrac{\pi}{n})[/math] [math](1 + \alpha)^2 \cos^2\dfrac{\pi}{n} - 2 \alpha (1 + \cos\dfrac{\pi}{n}) = 0[/math] [math](1 + \alpha)^2 \cos^2\dfrac{\pi}{n} - 2 (1 + \alpha) (1 + \cos\dfrac{\pi}{n}) + 2 (1 + \cos\dfrac{\pi}{n}) = 0[/math] Let [math]\phi = \dfrac{\pi}{n}[/math] [math](1 + \alpha)^2 \cos^2\phi - 2 (1 + \alpha) (1 + \cos\phi) + 2 (1 + \cos\phi) = 0[/math] Also let [math]\phi \approx 0[/math] [math]\cos\phi \approx 1 - \dfrac{1}{2} \phi^2[/math] [math](1 + \alpha)^2 (1 - \phi^2) - (1 + \alpha) (4 - \phi^2) + (4 - \phi^2) = 0[/math] This is a quadratic equation to be solved for [math](1 + \alpha)[/math]. However, only the lesser of the two solutions is of interest. [math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{(4 - \phi^2)^2 - 4 (1 - \phi^2)(4 - \phi^2)}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{(16 - 8 \phi^2 + \phi^4) - (16 - 20 \phi^2 + 4 \phi^4)}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12 \phi^2 - 3 \phi^4}}{2 (1 - \phi^2)}[/math] [math]1 + \alpha = 2 - \sqrt{3} \phi[/math] [math]\alpha = 1 - \sqrt{3} \phi[/math] [math]\alpha^2 = 1 - 2 \sqrt{3} \phi + 3 \phi^2[/math] [math]1 - \alpha^2 = \beta \dfrac{\pi}{n} = 2 \sqrt{3} \phi = 2 \sqrt{3} \dfrac{\pi}{n}[/math] [math]\beta = 2 \sqrt{3}[/math] Therefore: [math]\displaystyle \lim_{n\to\infty} P = \dfrac{\pi}{2 \sqrt{3}}[/math]