No, because it would create a contradiction. If a "perfect, endless repeat of pi" were eventually found (say, starting at the nth digit), then you can construct a rational number (a fraction with an integer numerator and denominator) that precisely matches it. However, pi is provably irrational, meaning no such pair of integers exists. That produces a contradiction, so the initial assumption that a "perfect, endless repeat of pi" exists cannot be true.
Yes and that contradiction is already present in my premise which is the point. Pi, if an infinite stream of digits and with the prime characteristic it is normal/random, will, at some point include itself, by chance. Unless, not random...
This applies to every normal, "irrational" number, the name with which I massively agree, because the only way they can be not purely random suggests they are compressible further and so they have to be purely random, and thus... can't be.
It is a completely irrational concept, thinking rationally.
> Pi, if an infinite stream of digits and with the prime characteristic it is normal/random, will, at some point include itself, by chance.
What you are essentially saying is that pi = 3.14....pi...........
If that was the case, wouldn't it mean that the digits of pi are not countably infinite but instead is a continuum. So you wouldn't be able to put the digits of pi in one to one correspondence with natural numbers. But obviously we can so shouldn't our default be to assume our premise was wrong?
> It is a completely irrational concept, thinking rationally.
The belief that a normal number must eventually contain itself arises from extremely flawed thinking about probability. Like djkorchi mentioned above, if we knew pi = 3.14....pi..., that would mean pi = 3.14... + 10^n pi for some n, meaning (1 - 10^n) pi = 3.14... and pi = (3.14...) / (1 - 10^n), aka a rational number.
> The belief that a normal number must eventually contain itself arises from extremely flawed thinking about probability.
Yes. There is an issue with the premise as it leads to a contradiction.
> Like djkorchi mentioned above, if we knew pi = 3.14....pi..., that would mean pi = 3.14... + 10^n pi for some n, meaning (1 - 10^n) pi = 3.14... and pi = (3.14...) / (1 - 10^n), aka a rational number.
Yes. If pi = 3.14...pi ( pi repeats at the end ), then it is rational as the ending pi itself would contain an ending pi and it would repeat forever ( hence a rational number ). I thought the guy was talking about pi contain pi somewhere within itself.
pi = 3.14...pi... ( where the second ... represents an infinite series of numbers ). Then we would never reach the second set of ... and the digits of pi would not be enumerable.
So if pi cannot be contained within ( anywhere in the middle of pi ) and pi cannot be contained at the end, then pi must not contain pi.
> If that was the case, wouldn't it mean that the digits of pi are not countably infinite but instead is a continuum.
No; combining two countably infinite sets doesn't increase the cardinality of the result (because two is finite). Combining one finite set with one countably infinite set won't give you an uncountable result either. The digits would still be countably infinite.
Looking at this from another direction, it is literally true that, when x = 1/7, x = 0.142....x.... , but it is obviously not true that the decimal expansion of 1/7 contains uncountably many digits.
> Pi, if an infinite stream of digits and with the prime characteristic it is normal/random, will, at some point include itself, by chance.
A normal number would mean that every finite sequence of digits is contained within the number. It does not follow that the number contains every infinite sequence of digits.
In general, something that holds for all finite x does not necessarily hold for infinite x as well.
Exactly - and when you remove the assumptions, what's left?
Pi is assumed to be infinite, random, and normal. The point here is not these assumptions may be wrong. Underneath them may sit a greater point; that irrationality is defined in a contradictory way - which may be correct, or not, or, both.
Given proof Pi is infinite lay on irrationality, it is rather an important issue. Pi may not be infinite, and a great place to observe that may be Planck.
> A normal number would mean that every finite sequence of digits is contained within the number.
Is that true? I don't see how that could be true. The sequence 0-9 repeated infinitely is, by definition, a normal number (in that the distribution of digits is uniform)
...and yet nowhere in that sequence does "321" appear ...or "654" ...or "99"
There are an infinite number of combinations of digits that do not appear in that normal number I've just described. So, I don't think your statement is true.
> I don't see how that could be true. The sequence 0-9 repeated infinitely is, by definition, a normal number (in that the distribution of digits is uniform)
Well, your first problem is that you don't know the definition of a normal number. Your second problem is that this statement is clearly false.
Here's Wolfram Alpha:
> A normal number is an irrational number for which any finite pattern of numbers occurs with the expected limiting frequency in the expansion in a given base (or all bases). For example, for a normal decimal number, each digit 0-9 would be expected to occur 1/10 of the time, each pair of digits 00-99 would be expected to occur 1/100 of the time, etc. A number that is normal in base-b is often called b-normal.
Your "counterexample" is not a normal number in any sense, most obviously because it isn't irrational, but only slightly less obviously because, as you note yourself, the sequences "321", "654", and "99" do not ever appear.
> Your "counterexample" is not a normal number in any sense, most obviously because it isn't irrational, but only slightly less obviously because, as you note yourself, the sequences "321", "654", and "99" do not ever appear.
lol. Your counterargument is a tautology because it contains "the sequences "321", "654", and "99" do not ever appear."
It's like if you claim, "A has the property B" then I say, "based on this definition, I don't think A has property B"
Then you say, "if it doesn't have property B, then it's not A"
...okay, but my point is, the definition that I had (from wikipedia) doesn't imply B. So for you to say, "if it doesn't have B, then it's not A" is just circular.
Now, you can point out that the definition I got from wikipedia is different from the one you got from wolfram. That's fine. That's also true. And you can argue that the definition you used does indeed imply B.
But what you cannot do is use B as part of the definition, when that's the thing I'm asking you to demonstrate.
You: all christians are pro-life
Me: I don't see how that's true. Here's the definition of christianity. I don't see how it necessarily implies being against abortion.
You: your """"counterexample"""" (sarcastic quotes to show how smart I am) is obviously wrong because, as you note yourself, that person is pro-choice, therefore, not a christian.
^^^^^ do you see how this exchange inappropriately uses the thing you're being asked to prove, which is that christians are pro-life, as a component of the argument?
Again, it's totally cool if you fine a different definition of christian that explicitly requires they be pro-life. But given that I didn't use that definition, that doesn't make it the slam dunk you imagine.
> But given that I didn't use that definition, that doesn't make it the slam dunk you imagine.
You might have a better argument if there were more than one relevant definition of a normal number. As you should have read in the other responses to your comment, the definition given on wikipedia does not differ from the one given on Wolfram Alpha.
> And you can argue that the definition you used does indeed imply B.
Given that the implication of "B" is stated directly within the definition ("For example, ..."), this seemed unnecessary.
> but my point is, the definition that I had (from wikipedia) doesn't imply B. So for you to say, "if it doesn't have B, then it's not A" is just circular.
Look at it this way:
1. You provided a completely spurious definition, which you obviously did not get from wikipedia.
2. You provided a number satisfying your spurious definition, which - not being normal - didn't have the properties of a normal number.
3. I responded that you weren't using the definition of a normal number.
4. And I also responded that it's easy to see that the number you provided is not normal, because it doesn't have the properties that a normal number must have.
Try to identify the circular part of the argument.
And, consider whether it's cause for concern that you believe you got a definition of "normal number" from wikipedia when that definition of "normal number" is not available on wikipedia.
It depends on your definition of "normal number". You seem to be using what wikipedia[1] calls "simply normal", which is that every digit appears with equal probability.
What people usually call "normal number" is much stronger: a number is normal if, when you write it in any base b, every n-digit sequence appears with the same probability 1/b^n.
IIRC the property ‘each single digit has the same density’ is the definition for a ‘simply normal number’ (in a given base), while ‘each finite string of a particular length has the same density as all other strings of that length’ is the definition for a ‘normal number’ (in a given base). And then ‘normal in all bases’ is sometimes called ‘absolutely normal’, or just ‘normal’ without reference to a base.
Is there actually a balance needed between pressure and collapse? Radiation pressure presumably doesn’t do anything to the constituent dm particles. Similarly, wouldn’t the particles in the star be on various elliptical trajectories and not collapse?
The DM particles would indeed be on random/chaotic orbits, unlike visible matter which can shed kinetic energy via EM interaction (collisions). But DM density near a mass concentration would still be higher than far away from anything massive.
Normal stars are in hydrostatic equilibrium, a density where the inward force exerted by gravity and the outward force exerted by the pressure of the hot plasma are balanced. In a dark star the situation would be similar, except the heat would be generated by DM annihilation rather than fusion (the heat from annihilation would keep the star too "puffy" to reach the core pressure and temperature required for fusion.
> Similarly, wouldn’t the particles in the star be on various elliptical trajectories and not collapse?
That's a common misunderstanding. Orbits around many bodies do not work that way, and the particles exchanging momentum so they collide or escape the cloud is normal.
You're forgetting about friction. A diffuse cloud is eventually going to become tighter and tighter as gravity draws it together and the individual orbiting particles within that cloud lose momentum from hitting each other.
Well, they say the DS would also have a decent amount of baryonic matter in it - some diffuse hydrogen and helium. If the pressure from the annihilations pushes the hydrogen and helium outwards, that could create some outward gravity to pull on the dark matter, right? Wait, except the outward gravitational pull inside of a hollow shell is zero because it all cancels.
Would there even be friction though? It sounds like the only interactions these hypothetical particles have is gravity and annihilation.
Yes, no friction for the DM particles. WIMP DM cannot collapse by shedding kinetic energy as heat like visible matter can. Thus the dark matter halos around galaxies. But gravity would still cause there to be a higher average density of DM inside and near the dark star (or indeed a modern-day galaxy) than far away from any massive objects.
> If the pressure from the annihilations pushes the hydrogen and helium outwards, that could create some outward gravity to pull on the dark matter, right?
But it's not on net being pushed outwards. It's in equilibrium. The outward push is on average exactly canceled out by the baryonic matter falling inwards due to friction. If it weren't, then everything would get either denser/sparser until equilibrium were regained. The point is that the forces are working in such a way as to maintain a stable equilibrium, like in normal stars.
The precession of mercury has a few contributions. The largest is the tug of planets (500 arc seconds/century), the next largest is from general relativity (50 arc seconds /century). There are other, smaller contributions stemming from the sun being oblate. Knowledge about the anomaly in the precession was partly what motivated Einstein, and was one of the early predictions of the theory.
Wow! It's amazing that I can run python in my browser using an IDE that feels like my daily driver. Even the little keyboard shortcuts, ctrl+} to indent a line of code worked as expected. I became so immersed I accidentally used alt+f4 to close a terminal window, and instead closed my browser!
Thanks! That's great to hear it was able to trick you a bit. In fullscreen I use keyboard lock to bind the Windows/Meta key, I'll add ALT+F4 also so at least people in fullscreen can continue the illusion a little longer.
Geosynchronous orbit is at 37,000km (well above ISS which iirc is at 400km), so by most definitions that is in space. Often that threshold is put as low as ~100km, when the atmosphere becomes too thin to support winged flight.
The idea of assembling large space telescopes is very exciting, although there are obviously many engineering challenges to solve. What that doesn't yet address is large, ground based radio telescopes. Consider the square kilometer array, which will have one million square meters of observing surface. How many launches would it take to get something comparable into orbit? Oddly, the bandwidth requirements (about one terabyte/second) for the SKA might be the hardest thing to meet in space.
Professional astronomers observe during the daytime too, using radio or microwave. Mega-constellations impinge on their ability to do science, especially for techniques like VLBI, where there's no obvious way to remove satellites from the data.
That applies to all satellites not simply large arrays. Large constellations all on the same frequencies aren’t even close to 1:1 as bad as normal satellites.
Are you sure this is true? You can frequently see satellites in the dead of night. Further, they show up quite plainly in non-visible wavelengths like radio or infrared even during the day.
The parent post said that only observations at dusk and dawn are affected. I'm pointing out that, even at astronomical midnight, large fractions of the sky aren't in earth's shadow. For other wavelengths the shadow is irrelevant, since satellites contaminate data even without the sun's direct illumination.
Well sure, at high latitudes and during certain seasons that may be true for higher altitude LEO satellites (Starlink is fairly low), but if we’re talking corner cases, sometimes the surface of the Earth isn’t in Earth’s shadow at astronomical midnight (land of the midnight sun).
Corner cases don’t really count as “frequent,” however.
