< Earlier Kibitzing · PAGE 303 OF 303 ·
|Oct-18-17|| ||SwitchingQuylthulg: <al wazir> WolframAlpha's calculator, https://apfloat.appspot.com/ (with lots of extra 0s after the decimal points to encourage precision) and http://keisan.casio.com/calculator. The first two gave the same result, while the last only displayed 50 digits, all identical with pi.|
|Oct-18-17|| ||al wazir: <beatgiant: <there must be a place where this occurs>... is actually not <definitely known> but is an open conjecture, unless you've got a major new result up your sleeve.> No, it's not a new result.|
The first 0 occurs only in the 40th place. If no zeros ever occurred, the sequence would fail the tests for "randomness."
The same thing is true if 00, 000, 0000, 00000, or any other finite sequence you care to choose never occurs. It's just a little less blatant.
|Oct-18-17|| ||beatgiant: <al wazir>
We can only ever have finite subsets of the decimal expansion for pi to test for randomness. That is good evidence, but not proof.
It's widely believed, but not as far as I know proven, that <any finite sequence you care to choose> does eventually occur.
|Oct-18-17|| ||al wazir: <beatgiant: We can only ever have finite subsets of the decimal expansion for pi to test for randomness. That is good evidence, but not proof.> You're right.|
Pi has been calculated out to one trillion digits or so. But that means we can't know anything about the frequency of occurrence of sequences two trillion digits long.
However, I'm sure that the frequency with which short sequences like 00000 occur has been checked -- though not beyond the trillionth digit.
|Oct-19-17|| ||beatgiant: Updated leaderboard with the contributions of our world-beater <SwitchingQuylthulg>:|
3 + (the next 1755524129972 digits of pi)
Method: Look for a long string of 0's in the digits of pi.
This method will scale to arbitrarily high efficiency if the conjecture about the normality of pi is true. That would kill the whole contest or motivate a new definition of efficiency.
(7.85 + 738 / (8!*8874.6+51.23114723657983449515923235473)) ^ (5/9)
Method: fine-tune a previous efficient solution.
Method: fine-tune a previous efficient solution.
And one up-and-comer:
((n!*(e/n)^n)^2)/2n = pi
Method: Stirling's formula. The initial test with n=29 and using 6 digits of e was not very efficient, but it might scale with very large values of n, which we can generate with a single digit via repeated factorials. But I lack the resources to test the idea.
|Oct-19-17|| ||Count Wedgemore: <al wazir: The first 0 occurs only in the 40th place>|
Actually, the 32nd..
<beatgiant: It's widely believed, but not as far as I know proven, that <any finite sequence you care to choose> does eventually occur.>
While it is fascinating to ponder the distinct possibilty that pi may contain the answer to every question one could possibly want to ask, it is also not of any practical use, is it?
Ater all, a string that contains everything contains nothing.
|Oct-19-17|| ||john barleycorn: < Count Wedgemore: ...
Ater all, a string that contains everything contains nothing.>
Amen to that. However, the irrationality of pi may be contagious to the mind of its researchers.
Anyway, I never gave up upon a search for next week's lotto numbers under the known digits of pi. :-)
|Oct-19-17|| ||al wazir: <Count Wedgemore: <al wazir: The first 0 occurs only in the 40th place<<>>> Actually, the 32nd.>|
I slipped and wrote that in octal there. Sorry about that.
|Oct-19-17|| ||beatgiant: Here's a new spinoff stumper based on <al wazir>'s stumper. As we saw, the <al wazir>-efficiency of representations of pi is increasing without bound but apparently very slowly. |
The new stumper is:
Define AW(b) as the best attainable al wazir efficiency for a given budget of b digits. The rules are same as before - we can use +, -, x, /, ^, !, ., (, ), and we have an nth-root oracle. I show the first few values below.
The new stumper is, what's the smallest number b such that AW(b) > 2?
AW(1) = 0, solution 3
AW(2) = 0, solutions 3.1, 3 + .1
AW(3) = 0, solutions 3.14, 3 + .14, 22/7, (4!-2)/7
AW(4) = 1, solution 2 + 4!^(1/4!)
AW(5) = 2, solutions 7.85^(5/9), 355/(4! - 7)
|Oct-20-17|| ||WannaBe: Okay, any predictions on when SkyNet will come online? 2030? 2040?|
|Oct-20-17|| ||diceman: <WannaBe:
Okay, any predictions on when SkyNet will come online? 2030? 2040?>
August 13 @ 8:38AM EST.
(all Fibonacci numbers)
|Oct-25-17|| ||WannaBe: Garbage in, garbage out... Guess programmers still have not yet learned.|
|Oct-25-17|| ||Marmot PFL: Bet cat owners and atheists scored even worse but I am not about to give them my CC number just to find out.|
|Nov-10-17|| ||Sneaky: I know it's been brought up before here, but I've got a story to go with it now. The stumper is this:|
"You are in an automobile, windows rolled up, with a helium balloon floating in the backseat. When you accelerate rapidly, which direction will the balloon point: forward, backward, or stay the same?"
The answer is... (spoiler ahead)
The answer is that it moves *forward* — for exactly the same reason it floats, it's lighter than air. When you accelerate, there is a force acting on everything in the cabin moving it backwards, but since the air around the balloon has more mass than the balloon, it displaces it, making the balloon point forward.
At first I didn't believe it, after all it's rather unintuitive, but after some pondering I realized that yes, that must be the correct answer.
Now the story: yesterday I took my sweetheart to her birthday dinner and happened to have a party balloon in the car. I told her "I know this sounds really weird, but I just have to try something. Hang on, we're going to zip up this road really fast."
So I zoomed through an empty backstreet at 50-60mph. Sure enough, the ballon pointed toward the hood of the car. It's not just theory, it's empirical fact.
|Nov-10-17|| ||al wazir: <Sneaky: Sure enough, the ballon pointed toward the hood of the car. It's not just theory, it's empirical fact.>|
All right, since you're into empirical observations, here's another one for you to explain.
You've seen birds soaring. They keep their wings spread out without flapping them. Soaring hawks, vultures, etc., typically execute huge circles or near-circles, so half the time they are going *against* the wind.
How do they do it?
|Nov-13-17|| ||al wazir: Leonardo gave an explanation. So did Rayleigh. (Both were researching the possibility of human flight.)|
|Nov-15-17|| ||WannaBe: I heard that Blake Sheldon have just been named Sexiest Man Alive.|
How come no one ever gets named Sexiest Man (or Woman) Dead?
Enquiring minds want to know.
|Nov-15-17|| ||al wazir: <WannaBe:How come no one ever gets named Sexiest Man (or Woman) Dead? Enquiring minds want to know.> You're in the wrong forum. Try https://www.redhotpawn.com/forum/ge....|
|Nov-17-17|| ||Sneaky: <al wazir> That is really interesting. I never would have figured it out... but upon hearing the explanation it makes sense.|
The question of “how do birds fly?” is a very hard question in itself, IMO.
|Nov-20-17|| ||al wazir: <Sneaky: That is really interesting. I never would have figured it out... but upon hearing the explanation it makes sense.> I think that explanation is oversimplified. I thought of trying to "improve" it, but I'm afraid everyone would think I had just overcomplicated it.|
But the gist of it is that for a bird to keep flying or soaring, the forces on it (lift, gravity, drag, propulsion) don't have to balance exactly. They only have to balance *on the average*.
|Dec-05-17|| ||Sneaky: I've seen some stout beers, especially Guinness, exhibit a well-known and strange phenomenon. When you pop the cap and closely look at the bubbles in your glass, you'll notice quite a few errant bubbles which are not going up, but clearly going from the top to the bottom—sinking!|
The explanation I've heard explain it is ... well, I'll let y'all ponder it first.
|Dec-06-17|| ||WannaBe: AI creates AI Child:
I think Terminator had it wrong, the name is not SkyNet, but GoogleNet.
|Dec-06-17|| ||john barleycorn: <al wazir: ...I think that explanation is oversimplified. I thought of trying to "improve" it, but I'm afraid everyone would think I had just overcomplicated it. ...>|
<“Any intelligent fool can make things bigger, more complex, and more violent. It takes a touch of genius — and a lot of courage to move in the opposite direction.”
― Ernst F. Schumacher>
Let us skip the "intelligent" and we have <al wazir>
|Dec-06-17|| ||WannaBe: Tensor Processing Unit (TPU)
|Dec-06-17|| ||al wazir: <john barleycorn>: I take that as an invitation. Here it is, just for you.|
The energy of a bird aloft is the sum of two kinds of energy, kinetic and potential. Kinetic energy is proportional to the square of the velocity and potential energy is proportional to the distance above a reference point, say, sea level. They are interchangeable. When a bird dives (stoops), its potential energy decreases but its kinetic energy increases by the same amount. If it it coasts upward its kinetic energy decreases but its potential energy increases by that amount. It can increase its energy by flapping its wings (propulsion), but air resistance (drag) reduces it.
In order to stay aloft, the bird has to maintain its energy. Since drag is always present, there must be some way to replenish the energy that is lost. If the bird is soaring without flapping its wings, this energy comes from fluctuations in the local wind.
A few hundred feet above the ground there is always some wind. Moreover, there are always fluctuations: updrafts, downdrafts, shear winds, etc. They result from the flow of the prevailing wind over irregular ground terrain and variations in surface temperature and from local variations in air pressure, which can be caused, e.g., by clouds covering the sun. Observers on the ground see the motion of a bird relative to the fixed surface of the Earth, but what matters most to the soaring bird is its motion relative to the air.
The key to replenishing the energy lost through drag is for the bird to maneuver so that part of the time when it is coasting *with* the wind the air mass moves *faster* than the average wind speed, which accelerates it and therefore *adds to its energy*, while at least part of the time when it is coasting into a head wind, that wind is *slower* than average.
Soaring is possible only in fairly calm weather. In very high winds a soaring bird cannot both remain aloft and maintain its station by executing circles; soaring speeds relative to the ground are always comparable to local airspeed. To maintain station it has to be flying slower (relative to the ground) when it has a tail wind and faster when it has a head wind. This can always be done by combining changes in altitude with changes in direction, since by changing altitude (thereby interchanging potential and kinetic energy) it can adjust its speed without losing energy.
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