Let's discuss Mathematics

[meromorph]

Ok.

The motivation behind this question were symbolic calculation programs like Mathematica. I was wondering, how they handle with numbers, and thought they could think numbers as "multidimensional": having "rational dimension" and "irrational dimensions".

I mean, suppose you sum numbers like 4+2sqrt(5) and 3 +4\pi. The machine would sum rationals separately and irrationals separately, so that the number would be (7,2,4), where first digit is "rational dimension", the second one "sqrt(5) dimension" and the third one "\pi dimension". Otherwise the machine would have to operate with rounded numbers, and I suppose that won't do in symbolic maths program.

This was just something I though upon when walking to school at morning.

Mathematica stores symbolic expressions as nested functions. For example, 1 + 2 Pi would be stored as:

Plus[1, Times[2, Pi]]

Expressions with rational powers like 4 + 2 sqrt(5) expand to:

Plus[1, Times[2, Power[5, Rational[1,2] ] ] ]

Nested functions can be visualized as trees, f[a,b] becomes a parent node f with two child nodes a and b.

The same thing happens with rational numbers that don't easily simplify. It will keep them as symbolic values and do arithmetic with them, but it can't perform boolean tests on them or treat them as numeric values.

To my knowledge there is no other way to handle this, and it may even be impossible to improve on it at all.

Interesting. In mathematica, constants like Pi are kept as symbols but have an attribute marking that they can be evaluated to a number, and the tests 4 > Pi and 4.0 > Pi both return true. Mathematica uses dynamic typing, so if an expression combines symbols and floating-point like 2 + 2.0 then the result is output as floating-point 4.0 instead of returning a type error.

 

[Blue Emu]

... once stable they should be tidally locked eg always face each other with the same side. At that point tidal effects should no longer significantly effect the orbits.

Not entirely true, since a body can only tidally lock on one other object at a time.

Consider the Earth-Moon-Sun system. Early in its history, the Moon became tidally locked on the Earth, with the excess angular momentum transferred from the Moon's own rotation to its revolution... thus forcing the Moon out into a wider orbit. Over the next few billion years, the tides that the Moon exerts on the Earth will slow the Earth's own rotation towards a tidal lock, again transferring angular momentum to the Moon and forcing it into a wider orbit.

But the Sun also exerts tides on both the Earth and the Moon, roughly 1/3rd as strong as the Moon's tidal couple. If the Earth-Moon-Sun system lasts long enough... if the Moon isn't lost to space and if the Sun's expansion into a Red Giant doesn't destroy the Earth or Moon... then the Solar tides will eventually become dominant over the decreasing Earth-Moon tides; and the Moon will start spiralling inward again, and eventually impact the Earth. The angular momentum lost in that process will be transferred to the Earth, widening its orbit around the Sun.

 

[Atticus]

Thanks, meromorph! Is your name from meromorphic functions, btw?

And thanks to you too, nc-1701.

 

[meromorph]

Thanks, meromorph! Is your name from meromorphic functions, btw?

Haha, yes it is. Looking at my join date it was around the time that I was trying to learn algebraic geometry from this book:

It was fun to learn the algebraic definition of meromorphic functions: the set of holomorphic functions on a manifold forms a ring, and the set of meromorphic functions is the fraction field of that ring. It made me wish they used more geometric examples when teaching abstract algebra!

 

[PlutonianEmpire]

I found this: http://www.universetoday.com/93494/is-venus-rotation-slowing-down/

6.5 minutes in 16 years. Which equals to about 0.40625 minutes, or 24.375 seconds, per year. With that in mind, and if the number is a constant, how long ago would Venus have had a rotation period of 24 hours? Its current rotation period is 5832.443616 hours.

 

[Petek]

I found the following in a book and thought it was interesting. Let N denote the natural numbers {1,2,3,...}. It's well-known that NxN is countably infinite. The usual proof of this fact is to consider the lattice points (m,n), draw a series of finite diagonal lines through the points, and observe that all lattice points are eventually enumerated by this process. Here's another method to show this fact:

Define a function F: NxN --> N by

F(m,n) = .5(m + n - 2)(m + n -1) + n

Then F is one-to-one onto and is thus a bijection between NxN and N. Since N is countably infinite, so is NxN.

I'll post a proof in a couple of days if no one else supplies one.

 

[IdiotsOpposite]

Isn't NxN equal to Q?

 

[Petek]

Isn't NxN equal to Q?

They are not equal as sets because (1,2) and (2,4) are distinct elements of NxN, but both correspond to the same element (1/2) of Q. They have the same cardinality, but that fact requires a proof too.

 

[dutchfire]

I found the following in a book and thought it was interesting. Let N denote the natural numbers {1,2,3,...}. It's well-known that NxN is countably infinite. The usual proof of this fact is to consider the lattice points (m,n), draw a series of finite diagonal lines through the points, and observe that all lattice points are eventually enumerated by this process. Here's another method to show this fact:

Define a function F: NxN --> N by

F(m,n) = .5(m + n - 2)(m + n -1) + n

Then F is one-to-one onto and is thus a bijection between NxN and N. Since N is countably infinite, so is NxN.

I'll post a proof in a couple of days if no one else supplies one.

If you calculate F(m,n) on a grid, you'll see it equals:

1 3 6 10 15 21 28 36 45 55
2 5 9 14 20 27 35 44 54 65
4 8 13 19 26 34 43 53 64 76
7 12 18 25 33 42 52 63 75 88
11 17 24 32 41 51 62 74 87 101
16 23 31 40 50 61 73 86 100 115
22 30 39 49 60 72 85 99 114 130
29 38 48 59 71 84 98 113 129 146
37 47 58 70 83 97 112 128 145 163
46 57 69 82 96 111 127 144 162 181

In other words, this function does exactly what is done in the usual proof, counting along diagonals.

Spoiler matlab code to do this :

N=10 ;
c = [] ;
for i=1:N
for j=1:N
c(i,j) = 0.5 * (i+j-2)*(i+j -1) + j ;
end
end
c

 

[Petek]

Nice presentation! Your observation is spot on.

 

[meromorph]

Here is a pattern to list the rationals between 0 and 1:

0, 1, 1/2, 1/3, 2/3, 1/4, 2/4, 3/4,...

I find it simpler than the usual method based on the diagonals of the square lattice, but I guess it appears less frequently because it only lists the rationals between 0 and 1, rather than all of Q.

I think of countability in terms of lists. I mean, one can never "count" the rationals, but we can list them!

In contrast, we cannot list the real numbers. Cauchy's diagonalization argument is often phrased as a proof by contradiction, but it is really a proof by contrapositive. First I will explain the difference:

Theorem: P implies Q

Proof by contradiction: Assume P and not Q, derive a contradiction.

Proof by contrapositive: Assume not Q and derive not P.

Now to apply this to Cauchy's proof. Assume S is a list of real numbers. Then we can construct a real number x not in S by making the nth digit of x always differs from the nth digit of the nth element of S. Since x is a real number not in S, S cannot contain all real numbers.

The point is, we didn't have to assume S was all real numbers and derive a contradiction. We just assumed S was a list, and derived that S was not equal to R.

Can anyone give an example of a genuine proof by contradiction that cannot be reformulated this way, as a proof by contrapositive? (hint: there are many).

 

[Atticus]

This requires more thinking, at least from me, since on the first look it looks impossible:
-(p & -q) is equivalent to
p => q is equivalent to
-q => -p.

So you'd probably have to drop the lists altogether. I remember encountering some proof(s) of this in topology.

 

[meromorph]

This requires more thinking, at least from me, since on the first look it looks impossible:
-(p & -q) is equivalent to
p => q is equivalent to
-q => -p.

Yes those propositions are equivalent, but they don't capture the notion "q is provable from p." In logic this is written p ⊢ q ("p yields q"), and a contradiction is written ⊥.

If q is a tautology, then we can write ⊢ q. This is equivalent to the statement that the negation of q is a contradiction -q ⊢ ⊥.

So in proving p => q by contradiction we show:

(p & -q) ⊢ ⊥

which is equivalent to ⊢ -(p & -q), so p => q is a tautology.

In the case of proof by contrapositive, we show -Q ⊢ -P, which by soundness is equivalent to ⊢ P => Q.

So you'd probably have to drop the lists altogether. I remember encountering some proof(s) of this in topology.

I'm not sure if there is a proof for the uncountability of the reals that essentially requires contradiction, the example I have in mind of a popular proof that does make full use of P & -Q is in another domain.

 

[PlutonianEmpire]

Ok. Let's say we have a hypothetical individual, who has a curse placed on him (bear with me, please. ), where his life is on "rewind" at an accelerated pace, culminating in his death on the infamous doomsday date at the end of this year. His body ages normally, but the rewind is at an accelerated pace, with a start date of today (April 2nd, 2012), and the rewind is at 2.61 years per real life month.

With this in mind, two life events took place in this loser's history about 116 days apart in RL. With this number compressed into the accelerated rewind, how many real life days will pass between the repeats of said two life events?

And how did you get this number?

 

[Atticus]

First you calculate the multiplicator:

2.61 years/month= 952 days/30 days= 31.8 (about)

So one real day corresponds to 31.8 days in his life. Then 116 days is 31.8 * 116 ... ans so on.

BTW, doesn't "rewind" mean that his life is going backwards?

 

[PlutonianEmpire]

First you calculate the multiplicator:

2.61 years/month= 952 days/30 days= 31.8 (about)

So one real day corresponds to 31.8 days in his life. Then 116 days is 31.8 * 116 ... ans so on.

BTW, doesn't "rewind" mean that his life is going backwards?

Yes. But his body ages normally. Meaning, that the rewind is enough so that December 21st, 2012 is the equivalant of him being born, indicating he will die by whatever the voodoo witch wanted him to die by.

I figured out that "31.8 * 116" should actually be "116 / 31.8", which would be about 3.6478 days.

Anyways, thanks.

 

[PlutonianEmpire]

Google has failed me. Wolfram Alpha has failed me. EVERYTHING has failed me.

I've been looking for the past bleeping hour for an easy calculation for the roche limits of celestial objects, to no avail, and at this point, I am way beyond pissed off.

So I've come here as a last resort. My request is simple.

Can someone translate the formula for celestial roche limits into a single line of code-ish text, with info on which numbers go where? (Preferrably I'd be able to copypasta into Win 7's Calculator.) The images on Wikipedia (for example) containing the formulas are too confusing.

Example (the formula for calculating the semi-major axis of a given planet):

semi-major axis = (perihelion + aphelion) /2

 

[LulThyme]

What exactly is confusing about the Wikipedia formulas?

To be really honest, I do think everybody's time would be spent more productively if you tried to learn how to parse such formulas. Especially in the long run, you wouldn't have to ask again for every new formula (teach a man to fish and all that..) .

As far as I can tell, the formulas on Wikipedia only have the four basic operations plus exponentiation, so you only need to understand a bit of notation and order of operations, it should not be too hard and you can come and ask here for help I suppose.

 

[Mise]

I looked on wiki for "Roche limit" and here is the first formula in that page:

Which is d = R*(2*p_M/p_m)^(1/3) if you're typing it into a computer. (I'm using p's instead of rho's.)

 

[Chairman Meow]

Mathematica stores symbolic expressions as nested functions. For example, 1 + 2 Pi would be stored as:

Plus[1, Times[2, Pi]]

Expressions with rational powers like 4 + 2 sqrt(5) expand to:

Plus[1, Times[2, Power[5, Rational[1,2] ] ] ]

Nested functions can be visualized as trees, f[a,b] becomes a parent node f with two child nodes a and b.



Interesting. In mathematica, constants like Pi are kept as symbols but have an attribute marking that they can be evaluated to a number, and the tests 4 > Pi and 4.0 > Pi both return true. Mathematica uses dynamic typing, so if an expression combines symbols and floating-point like 2 + 2.0 then the result is output as floating-point 4.0 instead of returning a type error.

One thing you have to watch out for in Mathematica if you use the Units and PhysicalConstants packages is that Mathematica treats pi as a unit rather than part of the number. This means if you try to strip the units off a physical constant by taking the first part (e.g. VacuumPermittivity[[1]]), then you can end up with a number that's off by a factor of pi. I've been bitten by that bug in the past...