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My brushes have never been that wide. I acknowledge my conservative leanings when it comes to several areas, as well as my liberal leanings in others. Wrt guns in the US, I'm a gun owner who would happily participate in a national buyback program with the aim of removing all the guns. I'd also support Japanese style background checks done by the police for anyone who claims to need a gun. I think that's extremely liberal (for the US). I'd pay for all the extra police work by moving their funding around and instituting passive radar checks and automatic ticketing for running red lights, which actually enforce the law on vehicles and don't care what color/gender/other privilege flavor you are. I think that's extremely conservative, but I'm not interested in being exempted from punishment just because I'm normally a lawful driver.

Books don't matter. What matters is how you spend your free time. Do you write your own apps for Android/iOS/Web etc., or do you drink beer with friends in a pub/club? Install Android Studio, make Android app, install iOS XCode, make app, install Xamarin, make app, and so on, so on.. (You can drink, if you're able to think clearly... )

From the second Friedmann equation or acceleration equation, [math] \frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) = - (\frac{{4\pi G}}{3})(\rho + 3P)[/math] Since ρ is the mass density in the acceleration equation, the pressure term P has a dimension of energy density (c=1). Current acceleration equation inevitably require negative mass density. Without negative (gravitational) mass density, it is impossible to create acceleration expansion. The mainstream produces acceleration expansion by setting the pressure of the cosmological constant or vacuum energy to P=−ρ . Currently, dark energy is described as a being that exerts a negative pressure while having a positive energy density. However, there is a serious problem when considering a physical existence with these characteristics. ρ+3P=ρ+3(−ρ)=−2ρ 1) Sign of the negative pressure Since the universe is treated as a fluid with uniform density, the pressure between regions is zero. In general relativity, pressure is equivalent energy density of the kinetic energy. In the acceleration equation, 3P has the idea of an equivalent energy density corresponding to the kinetic energy of the particle. So, assuming that the pressure P term has a negative energy density is same assuming that it has negative kinetic energy. In order to have negative kinetic energy, it must have negative inertial mass or imaginary velocity. But, because they assumed a positive inertial mass, it is a logical contradiction. [math]K = \frac{1}{2}m{v^2} < 0[/math] m<0 or v=Vi : negative mass or imaginary speed. Negative mass contradicts the assumption of positive energy density, and energy density with imaginary speed is far from physical reality. 2) Size of the negative pressure In the ideal gas state equation, we obtain, [math]P = \frac{1}{3}(\frac{{{v^2}}}{{{c^2}}})\rho = \omega \rho[/math] In the case of matter, v << c, So [math]P = \frac{1}{3}{(\frac{v}{c})^2}\rho \simeq 0[/math] In the case of radiation, v=c, So [math]P = \frac{1}{3}{(\frac{v}{c})^2}\rho = \frac{1}{3}\rho [/math] However, in the case of dark energy [math]P = \frac{1}{3}{(\frac{v}{c})^2}\rho = - \rho [/math] a) Only considering the size (absolute value) [math]P = \frac{1}{3}{(\frac{v}{c})^2}\rho = | - \rho |[/math] [math]v = \sqrt 3 c[/math] We need energy density with super-luminous speed.(FTL) b) Considering even the negative sign [math]P = \frac{1}{3}{(\frac{v}{c})^2}\rho = - \rho[/math] [math]v = (\sqrt 3 c)i[/math] We need energy density with imaginary super-luminous speed. We can avoid this problem by assuming that cosmological constant or vacuum energy does not apply to the results (ideal gas state eq.) obtained through physics. However, ideal gas state equation applies to "massless ~ infinite mass'' particles, and the velocity ranges from "0 ~ c'' are all included. If we seriously consider the physical presence of P=-p=-3(p/3), we will find that there is a serious problem. 3) Incorrect application of dU = - PdV In the equation dU = - PdV, researchers claim that negative pressure exists if the energy density remains constant as the volume increases. Is this claim correct? Going to the root of the problem, the analysis of this formula seemed simple (dU=pdV), so I had the illusion that the logic of negative pressure was certain. But~ a) This explanation is an inverted explanation Since pressure is a property of an object, pressure exists first, and because of this pressure, changes in internal energy according to volume change appear. That is, since pressure is positive, if dV>0, then dU<0. Since the pressure is positive, if dV<0, then dU>0. By the way, we use the logic "if dV>0, dU>0, then P<0''. How are you sure that this logic is correct? b) ρ+3P=ρ+3(−ρ)=−2ρ Mass density ρ and pressure P are properties of the object to be analyzed. Both mass density ρ and pressure P are sources of gravity. It means that even if the region maintains a constant size without expanding or contracting, gravitational force is applied as much as ρ+3P . In other words, it suggests that the object (or energy density) has a gravity with a negative mass density of −2ρ . This is different from a vacuum with a positive energy density ρ , which we think of. c) dU = - PdV is the expression obtained when the law of conservation of energy is established dU=dQ-dW, if dQ=0, dU=-dW=-PdV However, in the case of vacuum energy and the cosmological constant, energy conservation does not hold. As the universe expands, the total energy in the system increases. Therefore, we cannot guarantee that dU = - PdV holds. dU = - PdV is an equation that holds when the energy of the system is conserved. However, researchers are applying this equation to vacuum energy or cosmological constant where the energy of the system is not conserved. "In the equation dU = - PdV, even if the universe expands, if it has a uniform density, it should have a negative pressure...'' What's wrong here? I am not sure if this equation holds even for negative pressure. However, although this equation holds even in the case of negative pressure, its interpretation is as follows. This equation holds true when substances in radius r_1 expand from r_1 to r_2 (r_2 > r_1), and have the same uniform density in r_1 and r_2. In other words, it is argued that a negative pressure is required to create a uniform density effect only with the material present in radius r_1. But, vacuum energy is a form in which energy is newly generated by an increased volume. It is also energy that can be assumed to have an initial speed of 0 ~ c. Considering the positive energy density, pressure term seems reasonable to assume 0 ~ (1/3)p. If the cosmological constant term is shifted to the right side of the field equation, it becomes a negative equivalent mass. It is not an object that has a positive energy density (positive inertial mass) and acts on a negative gravitational mass, but just a negative mass. It is not necessary for an entity to exist that exerts a negative pressure while having a positive energy density. Only a negative mass component may exist, or a positive energy component may exist and a negative energy component may exist. And, the negative energy component may be the energy of the gravitational field or the gravitational potential energy. Q1. Is it possible to have negative kinetic energy while having positive mass density? Is it possible for a physical entity to have an imaginary velocity while having a positive energy density? Q2. [math]P = - \rho = - 3(\frac{1}{3}\rho ) [/math] Can such an entity (an entity with a pressure(kinetic energy component) three times greater than light) physically exist? Q3. Is dark energy energy with FTL (Faster Than Light) or an imaginary FTL? Q4. Does the dU=-PdV equation hold for a system in which energy is not conserved? Q5. Even if the universe expands, If an entity with uniform mass (energy) density exists, it is said that it will have a negative pressure. Is this claim correct? We can considered the vacuum energy density with P=0. However, in order to have a uniform energy density even when space expands, does it have to suddenly have negative pressure? Shouldn't it be newly created, with P=0, and filling the larger universe? Can we not think of a vacuum energy density with P=0? Can't there be a vacuum energy density with P=0 ~ (1/3)p? Even if vacuum energy is present, it is not certain whether it creates a negative pressure. ρ+3P=ρ+3(−ρ)=−2ρ There seems to be a serious problem with this argument.

Why invent a word everyone understands and yet has no definition? In fact, it is defined in ever bill and charter of rights. It means that nobody can legally be deprived of opportunity, political franchise or freedom of speech and action based on their race, creed, colour or gender. While it's certainly open to debate in its ramifications, the concept is solidly embedded in the democratic ideology. Oddly enough, nobody brings this up when it's a question of two white men being unequal: nobody seems to think the dumb blond middle-aged one should have different rights from the smart red-haired old one. Or that a short man ought to be paid more than a tall man. Not in the voting booth. No struggle: just make an unequivocal X against a name.

This is the kind of insanity perpetrated by your Government GOP passes bill aiming to root out 'suspected' transgender female athletes with genital inspection - Ohio Capital Journal "House Republican lawmakers in Ohio passed a bill at 11:15 p.m. Wednesday night that would ban transgender girls and women from participating in high school and college athletics. It also comes with a “verification process” of checking the genitals of those “accused” of being trans." Genital inspections ???? Your daughter ran a little too fast at the high school track meet ? Well, drop your pants young lady, so we can see what you have between your legs ! Why do you guys keep electing such morons ?

Glad you posted this. Deadwood, which I live thirty miles from, had the "coat check" policy the article mentions. Municipalities didn't worry about parsing the Constitution, they just did what was necessary to keep the peace. And it worked most of the time. Deadwood, these days, is a boring tourist trap. There is little I would say is interesting, except the neutrino lab a few miles south in Lead (down in the former Homestake gold mine).

This Smithsonian gun control history article seems to paint a different picture to the current Right-wing narrative of it becoming more restrictive with time. It seems it's actually the reverse.

I think that this a very US-conservative view, though. While it is exported increasingly (Canada gets a fair bit of it) and others have high similarity in other aspects of it (who the deserving ones are, for example), gun attitudes are special in the US.

It also makes the baddies job a lot easier when attacking the goodies. And if neither has guns, easier still.

Sorry Koti. I do hope you choose to stay. Your opinion has always been respected, if not always agreed with, by me. I would hope you respect mine, even if you disagree with it.

The USB slots are likely to be 2 or 3 unless it's an old pc. On my laptop, the usb 3 sockets have a blue coloured insert, don't know if that's standard. edit : Another option is a USB memory stick. Less cabling and works fine with the USB supply voltage. Might be a bit slower, but ideal for occasional use. https://i.ebayimg.com/images/g/QvAAAOSwuAtfSSYK/s-l1600.jpg

I computed the volume inside both the sphere [math] x^2 + y^2 + z^2 =1[/math] and cone [math] z= \sqrt{x^2 + y^2}[/math] as follows: [math]\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\frac{\pi}{4}}\displaystyle\int_0^1 \rho^2 \sin{\phi}d\rho d\phi d\theta= 0.61343412301 =\frac{(2-\sqrt{2})\pi}{3}[/math]. This answer is correct.

The approach above should give you the max(f())=ln(n) quite fast. To get min(f), assume that ln(xj) is the smallest of all ln(xi), take the condition x1+x2+... xn = 1, multiply it by -ln(xj): -(x1+x2+... xn)*ln(xj)=-ln(xj). Because all ln(xi)>=ln(xj), -(x1*ln(x1)+...+xn*ln(xn))>=-ln(xj). The left side is -ln(x1^x1*x2^x2*... xn^xn), the right side -ln(xj)>=0. Thus -ln(x1^x1*x2^x2*... xn^xn)>=0. And we know how to get it =0. Thus 0 is the minimum.

I'd start with simplifying the expression and then finding an extremum using Lagrange multiplier method.

A cool little example of common confusion with partial derivatives, from Penrose's "The Road to Reality" (he attributes the words in the title to Nick Woodhouse.) Let's consider a function of two coordinates, f(x,y), and a coordinate change X = x, Y = y + x Because the X coordinate didn't change and is the same as the x coordinate, one could expect that the corresponding partial derivatives are the same, fX=fx. And, because the Y coordinate is different from the y, these partial derivatives, fY and fy, could be expected to differ. In fact, this is just opposite: fX=fx-fy fY=fy The confusion is caused by the notation: fX does not mean a derivative along X, but rather a derivative with a constant Y; and fY is not a derivative along Y, but a derivative with a constant X.

Evidence for which claims? Not like it matters because your mind is immune to the fact that Johny Depp was abused by a woman but since you ask please be specific.

Facepalm.