Without a formal system like this, you can end up with paradoxes by defining numbers like "the smallest positive real number without a definition".
The example is a bit unfortunate. Unlike the integer case, the subset of positive undefined reals doesn't need to have a minimum. It does have an infimum, but the infimum doesn't need to be a member of the set, and could be a defined number.
Good catch, you're perfectly right here. And in fact if we take our formal system to be algebraic numbers, the smallest non-algebraic positive real number doesn't exist. Hmm. I'm pretty certain one *can* make a paradox here, but it's proving harder to achieve than I thought. I think you could use the axiom of choice and say "an arbitrary undefined number" but that's somehow unsatisfying, and also I feel I'm getting above my pay grade (I'm a computer scientist really, not a mathematician).
Well, the formal system itself must be definable with a finite amount of information (or else it's not definable in practice; we cannot read an infinite definition in order to know all of its axioms. A formal system with an infinite definition would not be very useful.)
I can immediately define an uncountable set of formal systems: choose a real number in [0,1), call it r. Then formal system r consists of the countable set of definitions which define a number r+i, for integer i.
For any one of those formal systems, almost all real numbers are undefinable within the system. But all real numbers are definable in exactly one of those formal systems.
Depending on what you allow as a "formal system", there's always one where the number is definable.
Well, the formal system itself must be definable with a finite amount of information (or else it's not definable in practice; we cannot read an infinite definition in order to know all of its axioms. A formal system with an infinite definition would not be very useful.)
Sound reasonable. But to define a formal system with a finite string, you need - guess what? - another formal system to describe it in. You can happily construct an infinite formal turtle tower, but in the end, you have to start *somewhere*, and the very first formal system you use cannot be restricted in the way you like.
And depending on the initial formal system (and the amount of recursion you did), you get a different set of definable numbers.
Warp wrote:
Thus couldn't we define our set as "all the real numbers definable by a finite statement in any possible (finite) formal system"?
no, see above. Since "finite formal system" is ill-defined, so is the set.
Warp wrote:
There have to be some numbers that are impossible to define using a finite amount of information.
"finite amount of information" is another one of those intuitive things that translate poorly to mathematical definitions.
If you're a constructivist/finitist/intuitionist/whatever (and there are many examples of excellent mathematicians who subscribe to these views), you'd simply think that uncountable sets are bullshit and people should try to redefine analysis in a way that doesn't create a jungle of undefinable entities.
Makes me wonder.. did anyone even come close to that goal? To avoid undefinable entities, one would not only need to restrict analysis to contain only a countable number of numbers, one would also need to restrict analysis to allow only a countable number of statements. That seems.. very limited; possibly useless.
Language theory, for example, works on strictly countable objects: sets of words over a finite alphabet. Still, most properties of languages are undefinable and/or undecidable simply because the amount of properties of languages is larger than the (countable) amount of automatons we can construct to decide them. I cannot imagine a theory of analysis that doesn't inevitably run into the same problem.
I agree 100% with everything you said. In fact, asking me if someone has come close to developing a constructive version of analysis is the same as asking Milton Friedman if communism can provide large economic growth :)
We'll inevitably lose some didactic if I keep talking about this matter, but I'll do so anyway. Restricting ourselves to only countable sets is useless because every formal system will contain some limitation. The possibility of a set containing elements that cannot be defined in any way is simply the manifestation that a formal system will contain sentences that are unprovable, and that is a logical fact. It does not go away if you simply ban real numbers. The only difference is that, unlike in arithmetic, this weirdness is not restricted to esoteric Godel sentences or stuff no one cares about, but it manifests itself in the set where we define functions, and that apparently makes some people psychologically uncomfortable.
Some theorems in analysis can be derived countably by some clever redefinitions, although most of the time there's more than one possible redefinition and they don't work for all cases. However, there are many "harmless" theorems, like the intermediate value theorem, that are simply wrong in constructivist math, because it's inconsistent to talk about the sets at which they are defined. To fix this problem you'd have to constrain analysis to weaker versions of them.
The ugliest case, imo, is the fundamental theorem of algebra. Every proof we know of it depends on real analysis. It's not difficult to see why. It's impossible to find all roots of polynomials of degrees larger than 4 using a finite expression of square roots. So, do we use an iterative procedure? At some point, you have to prove that the series converges, and so we need analysis. What this means is that in constructivist mathematics, even statements such "a polynomial of degree n has n complex roots" is simply a conjecture, in order to prove an analogue without uncountable sets someone would have to use an amazingly complicated construction or to modify it in a way that it has little similarity to the original formulation.
And the list of problems goes on and on, many useful proof techniques are downright wrong in a constructivist approach, which gives a lot more work for the mathematician, and a much higher probability of errors. Even if someone managed to overcome these formidable technicalities and finished the program, at the end we would just have exactly what we already have with standard analysis, so research-wise, there's zero benefit.
Still, some people try to formulate these theories, and I admit that some of the work is insightful, and someone could always use their techniques to prove novel stuff. Although, to be honest, I never understood why the search for alternative models of analysis is so popular. For me, it's like insisting that Newton had to formulate his theory of gravity in a geocentric frame. People have been debating these issues since the invention of calculus, it's likely that the opposition to standard analysis will never go away.
I'm not sure I disagree with anything p4wn3r says, but I'd like to say something in defence of constructivism: constructive systems are particularly noteworthy in the Curry-Howard correspondence. Many interesting type systems are no more powerful than intuitionist logic, and only constructive results are of interest to these type systems.
That said, there's an interesting result which shows that call/cc's type corresponds to Peirce's Law, from which classical logic follows. It then follows that if your system doesn't have a call/cc-like primitive baked in, you can't have a well-typed implementation of it from the simply typed lambda calculus.
Here's a "fun" integral I saw in a quantum mechanics book. Prove that
Here delta(x) is the Dirac delta function. The formalism for this kind of integrals is pretty hard, so understand "prove" as "give a sensible explanation of that formula".
EDIT: Ah, and please don't use Fourier transforms, as that would be almost assuming what you want to prove :P
On Penn and Teller's: Fool us, there was a neat card trick which predicts the location of 2 chosen cards in a deck. But, as shown by explanation videos, the solution is that the positions are worked out in advance. Then, through manipulation of each section; without the need to know where the 2 cards originally were, they can be forced into the arranged locations and thus the trick always works.
However, that vid only explains how to replicate it, not the logic behind why it works. How could the joker positions force cards of unknown locations into a known one? Prove it however you can.
By marking the cut. Or, in more proof-y terms:
Let a number, x or y simply indicate an amount of cards, a J is a joker and A and B are the spectator's cards. The deck is set up like this:
9, J, 18, J, 25
By letting the spectators cut to 1/3 and 2/3, the deck becomes three piles.
1: 9, J, x, A
2: (17 - x), J, y, B
3: (24 - y)
These piles are restored in order 2 - 1 - 3:
2: (17 - x), J, y, B
1: 9, J, x, A
3: (24 - y)
Resulting in the following deck:
(17 - x), J, y, B, 9, J, x, A, (24 - y)
The deck is cut at the jokers (which are removed), resulting in three new piles:
1: (17 - x)
2: y, B, 9
3: x, A, (24 - y)
These are restored in order 1 - 3 - 2:
1: (17 - x)
3: x, A, (24 - y)
2: y, B, 9
Resulting in:
(17 - x), x, A, (24 - y), y, B, 9
or:
17, A, 24, B, 9
As such, the spectator's cards are in the 18th and 43th position, respectively. This works for other numbers than 9, 18 and 25, as long as the original setup can (almost) guarantee the jokers will be in the first two cut piles.
Mersenne primes are prime numbers of the form 2n-1. There are quite many of them (ok, only 48 have been found so far, but there doesn't seem to be a good reason to believe there aren't infinitely many, although that's still an unproven conjecture).
So what about prime numbers of the form 2n+1?
If you test the primality of such numbers, you'll find out that they are prime for values of n of 1, 2, 4, 8 and 16. There seems to be a clear pattern here. It seems that whenever n is a power of 2, 2n+1 is a prime.
But then you try with n=32, and it's composite. Bummer. How about n=64? Composite. 128? 256? 512? All composite. In fact, no other n has ever been found (be it a power of 2 or not).
For some reason n follows a nice pattern up to 16, and then it just stops. I find it curious. Is it just coincidence?
(Btw, primes of the form 2n+1 where n is a power of 2 are called Fermat primes. I think it has been proven that it can only be prime if n is a power of 2, but it has not been proven that it goes only up to 16. In other words, it's yet unknown if there are more, or even infinitely many.)
Yes, 2n+1 can only be prime if n is a power of two (or if n=0).
From the information on this page, 2^(2^m)+1 is composite when 5≤m≤32. The smallest open case is m=33.
In this video there's a Texas hold'em showdown where one of the players has four aces and the other has a royal flush. What is the probability of this happening?
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Warp wrote:
In this video there's a Texas hold'em showdown where one of the players has four aces and the other has a royal flush. What is the probability of this happening?
Impossible? Doesn't a Royal Flush use the Ace as well?
When TAS does Quake 1, SDA will declare war.
The Prince doth arrive he doth please.
In this video there's a Texas hold'em showdown where one of the players has four aces and the other has a royal flush. What is the probability of this happening?
Impossible? Doesn't a Royal Flush use the Ace as well?
In Texas hold'em players have a shared card pool. And both players can use the same card simultaneously from that.
All syllogisms have three parts, therefore this is not a syllogism.
In this video there's a Texas hold'em showdown where one of the players has four aces and the other has a royal flush. What is the probability of this happening?
Impossible? Doesn't a Royal Flush use the Ace as well?
The ace of the royal flush is on the table (the "community cards") rather than his hand. (The same ace is part of the four aces hand.)
This will be assuming the game consists of cards just getting blindly dealt to the players, then they show them. Without being too familiar with the rules, whether or not they could swap cards, exchange etc, which just adds too much complexity, here's a rough model.
Easy answer: Given those 5 shared cards first, then deal 2 cards each, we have only one set of specific cards completing the 4A/RF hands. Thus P(4A n RF)=47C2 * 45C2, multiplied by 2 if you also count Person B beating Person A as a desired outcome.
Difficult answer: If you also want to factor the probability of having community cards which allow for such a setup in the first place, then that complicates things. One way to work this out is by using the probability of having the cards needed for this setup being the 9 cards in play, and then the probability of them being in locations between players/middle which allows for a climax.
P(Climax)=P(Cards being dealt)*P(Dealt to right players|given these cards)
Clearly there are 52C9 cards which could be in play at any time, and those could be split between the player/table/player in 9C2*7C5*2C2=9!/2!5!2!=756 combinations, only a few of which allow for the climax. We need to work out how many ways this is.
For the significant cards required for each hand, the royal flush requires 5, the 4xAce requires 4, but they are not independent. It requires an intersection of 1 card, thus 5+4-1=8 unique cards needed for this setup, leaving 1 random dummy card (9s in video). Also remember though that there is only 1 unique set of 4 significant cards which produce the 4xAces, yet the significant cards forming a royal flush have 4 different variations; one for each suit. Thankfully though, each of these still allows for 4A combo, as they would not require additional cards in play.
Total combinations for cards being in play=52C9
Amount of 4A combos=1
Amount of RF combos=4
Amount of Dummy Cards=44
P(Cards being dealt)=(1*4*44)/52C9
This is the chance that the interesting cards are dealt onto the table in some fashion, but we still need to work out the probability of them actually being distributed within the correct slots. For example the assignment AA...TJQKA...A9 would not give the climax we saw, despite having the correct cards in play.
A Royal Flush uses 5 cards, the player holding at most 2 of these means we need at least 3 in the middle. 4 Aces uses 4 cards, the player holding at most 2 means we need at least 2 in the middle. One card needed in both sets, IE must be in middle, thus 3+2-1=4 significant cards in the middle minimum. The 5th card could be part of the player's pattern, or the dummy. So given 9 cards in play, we know 1 dummy card might be in the middle, or in player A or B's hand. Therefore to work out the total probability, we can consider just one of the three locations, and consider the probabilities of creating a good setup with or without the dummy card in that particular pile. (The 'without' option would be the sum of the other 2, as a small shortcut). Here we will consider the effect of having the potential dummy in Player A's hand, thus having only 1 RF card with it.
P(Dealt to right players)=P(Setup with Dummy in A)+P(Setup without Dummy in A)
1 Dummy card to be in first hand
5 potential RF cards to be paired with it =5C1=5
4 remaining RF cards plus 1 random ace in middle =3C1=3
Other 2 aces in final hand =2C2=1
P(Setup with Dummy in A)=5/9C2 * 3/7C5 * 1/1 =~ 0.02
Next, we consider climactic hands with the dummy not in Player A's hand, which could be broken down into specific locations within Middle/B, but quicker to consider it as a single calculation. This is because having 2xRFs in Hand A means there are no Aces, and thus they all must lie somewhere within Mid/B. Thus the dummy's exact location in either Mid or B makes no difference to player B's ability to make 4A using the Mid/B selection; and so won't affect if it's a climax or not.
We have potentially 9C2 combos for first hand, 7C5 for middle, 2C2=1 for second hand.
2 RF cards in hand =5C2 ways
Amount of ways to pair 3 remaining RF cards with 2 others=4C2 ways
Therefore we have
P(1st person getting 2xRF cards)=5C2/9C2
P(3xRF cards in middle, plus 2 random)=4C2/7C5
P(2 others going to 2nd player)=1
P(Setup without Dummy in A)=5C2/9C2 * 4C2/7C5 * 1 =~ 0.08
So finally this should give us
P(Climax)=P(Cards being dealt)*P(Dealt to right players|given these cards)
=176/52C9 * ( P(Setup with Dummy in A) + P(Setup without Dummy in A) )
=176/52C9 * ( ( 5/9C2 * 3/7C5 ) + ( 5C2/9C2 * 4C2/7C5 ) )
=~4.7x10^-9
That finally gives us the probability that Player A makes a Royal Flush, and B gets 4 Aces using the provided community cards to help. Again, multiply this by 2 if you want to consider the other dude winning instead.
(PS, **** me this seems overly complicated, there must be an easier way)
My method of calculation for the 4A/RF situation is as follows. Same assumptions as in Flip's opening paragraph.
We wish to find all possible permutations of 4A/RF cards/dummy in the form
xx yyyyy zz
where xx is Player A's hand, yyyyy is the board (the middle), and zz is Player B's hand, and Player A cards + board form a royal flush, and Player B cards + board form a 4-aces hand. The order of cards is important for this calculation.
Assume for the moment that spades is the RF suit. Then the ace of spades must be on the board (since it is shared by both 4A and RF. Remove this card for the moment, so we count
xx yyyy zz
where the 8 cards are the RF cards except the ace (KS, QS, JS, TS), the other aces (AC, AD, AH), and the dummy card (DU). KS, QS, JS, TS must occur on x or y positions, and AC, AD, AH on y or z positions.
Suppose we are given the order from left to right of {KS, QS, JS, TS} as well as of {AC, AD, AH}. There are two cases. If DU is in A's hand, then one RF card is in A's hand, and the other three are on the board, and the aces fill out the rest. Then there are 2C1*4C3=8 arrangements. On the other hand, if DU is not in A's hand, then two RF cards are in A's hand, the other two are on the board, and the aces and DU fill out the rest. Then there are 4C2*4C3=24 arrangements.
So, given the order as above, there are 32 arrangements. Multiply this by:
- all possible permutations of {KS, QS, JS, TS} (4!)
- the position of the ace of spades when put back (5)
- permutation of all suits (this permutes all aces as well as the RF suit) (4!)
- the assignment of the dummy card (44)
This results in a total of 4055040.
Divided by all permutations of 9 cards, P(52,9)=1335062881152000, this gives 3.037x10-9, or 1 in 329 million. Multiply by 2 if it doesn't matter who has the RF and who has the 4A.
Note that Texas hold'em poker is generally played at tables with up to 10 players. For a full table, the probability that this situation occurs in each hand (each deal) goes up by a factor of 90 (choose one player as A, another as B, and the rest are irrelevant).
Flip wrote:
1 Dummy card to be in first hand
5 potential RF cards to be paired with it =5C1=5
...
2 RF cards in hand =5C2 ways
The ace of the RF cannot be placed in a hand; it must be in the middle. That leaves 4C1=4 for the first case and 4C2=6 for the second. Doing so eventually results in the same value as I calculated above.
The numberphile youtube channel posted recently some interesting videos about the golden ratio and fibonacci sequences.
The videos prove that given any sequence of numbers where each number is the sum of the two previous numbers, if you take the ratio between two consecutive numbers in the sequence, it will approach the golden ratio the further you go in the sequence. And it doesn't matter what the initial two numbers are. They can be anything, even real numbers.
That got me thinking, is it also true if the two initial numbers are complex numbers?
That got me thinking, is it also true if the two initial numbers are complex numbers?
Although I lack the skills to actually prove it, I'm sure the answer is yes. You can consider the series as two separate series: one for the real part, one for the imaginary part. Both will converge to the golden ratio, as such the real part will converge to the golden ratio too, while the imaginary part converges to zero.
You could look at the proof shown at numberphile and convert each
and you would still get
1+(1/Δ) = Δ
where Δ turns out to be the golden ratio.
Another way of looking at this is using Scepheo's approach.
Scepheo wrote:
You can consider the series as two separate series: one for the real part, one for the imaginary part. Both will converge to the golden ratio
So for n -> ∞ that ratio
or for the sake of simplicity
has the property that a/c = b/d = golden ratio (since both parts can be seen as separate series).
Multiplying that equation with c and d we get ad = bc.
By following this tutorial on how to divide complex numbers I was able to form this equation:
(wolfram alpha)
Take a look at the imaginary part. We already said ad = bc which means bc - ad = 0 so it will indeed converge to 0.
Now to the real part. Multiplying a/c = b/d with c² will give us ac = bc²/d. Now we replace that ac with the one in the real part, giving us:
Warning: Might glitch to creditsI will finish this ACE soon as possible
(or will I?)
By the way, one of the videos also pointed out that round(φn) will always be the same as the nth Lucas number (well, (n+2)th or something). Is there a simple proof of this?
(φ is the golden ratio, and Lucas numbers are the sequence 2,1,3,4,7,11...)
The Lucas numbers are defined as: L0=2, L1=1, Ln=Ln-1+Ln-2 for n≥2.
We claim that Ln=φn+(1-φ)n. Proof is as follows:
The generating function L(x) satisfies
(1-x-x2)L(x) = L0+(L1-L0)x = 2-x,
and so L(x)= (2-x) / (1-x-x2).
Note that φ is a root of x2-x-1, so φ2-φ-1=0. We can write this as φ(φ-1)=1.
Then 1-x-x2 is factored as (1-φx)(1+(φ-1)x) and so we write
L(x)= (2-x) / (1-x-x2) = 1/(1-φx) + 1/(1+(φ-1)x).
Therefore, Ln=[xn]L(x)=φn+(1-φ)n.
(The above can also be proved using induction instead of generating functions.)
Since -0.7<1-φ<0, then |(1-φ)n|<0.5 for n≥2, and so Ln=round(φn) for n≥2.
A little about string art. I'm thinking few days already about it.
Here is picture
[URL=http://imgur.com/bZTZIgX][/URL]
It's made in 2D, but if you may imagine that there some surface.
Question is: What That Surface?! (in 3D)
How it's made: just draw line between points (0,i) (1-i,0) for i = 0 to 1 by 1/n
:)
I can't say "what it is in 3D" (more on that later), but I can say what the 2D envelope (the curve that is the upper bound) is.
The envelope is one section of a parabola. Note that it is related to the first example on the Wikipedia page for envelope, and it even explains how to find it. It is also a quadratic Bézier curve with control points (1,0), (0,0), (0,1).
As for "what it looks like in 3D", it looks similar to the standard hyperbolic paraboloid diagram (also the standard saddle point diagram). Conveniently, the envelope itself is a section of a parabola, just like the cross section of the hyperbolic paraboloid z=y2-x2 in the plane x=0.
Yep, I know what envelope there, and yes, it's parabola.
I already told that I'm thinking few days already, so I know something :)
If you assume that all straight lines in 2D is stright in 3D then it class of ruled surfaces (hyperbolic paraboloid there too)
And then, you can setup many of such surfaces...
But, when you see image in 2D you imagine exactly some certain surface.
And then, different question - how to understand what you see?
:)
Here delta(x) is the Dirac delta function. The formalism for this kind of integrals is pretty hard, so understand "prove" as "give a sensible explanation of that formula".
EDIT: Ah, and please don't use Fourier transforms, as that would be almost assuming what you want to prove :P
and you would still get
1+(1/Δ) = Δ
where Δ turns out to be the golden ratio.
Another way of looking at this is using Scepheo's approach.
or for the sake of simplicity
has the property that a/c = b/d = golden ratio (since both parts can be seen as separate series).
Multiplying that equation with c and d we get ad = bc.
By following this tutorial on how to divide complex numbers I was able to form this equation:
(wolfram alpha)
Take a look at the imaginary part. We already said ad = bc which means bc - ad = 0 so it will indeed converge to 0.
Now to the real part. Multiplying a/c = b/d with c² will give us ac = bc²/d. Now we replace that ac with the one in the real part, giving us:
[/URL]
It's made in 2D, but if you may imagine that there some surface.
Question is: What That Surface?! (in 3D)
How it's made: just draw line between points (0,i) (1-i,0) for i = 0 to 1 by 1/n
:)
