Let's discuss Mathematics

[Souron]

Thanks Petek.

Makes sense. Though that wouldn't work for proving that sqrt (5) is irrational, for instance. Easy to change though, you simply change the (-k) part to be (-4k), and then you get n<k iff sqrt (5) < 5.

That doesn't work, because n is not a positive integer, which is needed for there to be a contradiction.

It does, however, work if that value is 2:
sqrt(5) ~= 2.24, so 2 < sqrt(5) < 3
n = k*sqrt(5)-2k
2 < sqrt(5)
#n>0
sqrt(5) < 3
#n<k
#n*sqrt(5) is an integer. (this part is the same)

Both these proofs require a proof of a method for computing which two integers the square root lies between, which is not hard to find.

EDIT: or generically:
n = k*sqrt(x)-floor(sqrt(x))k

Then you don't need the extra step above. Instead x must be shown not to be a perfect square.

 

[Spoonwood]

Say that we have x^(x^(x^...)) where "^" represents exponentiation and the "..." part indicates that we have a countable infinity of x raised to the x power. Solve x^(x^(x^...))=2 for x.

 

[IdiotsOpposite]

Say that we have x^(x^(x^...)) where "^" represents exponentiation and the "..." part indicates that we have a countable infinity of x raised to the x power. Solve x^(x^(x^...))=2 for x.

But, but...

EDIT: More reasonably, I believe that the left side of the equation will be infinity for any positive number greater than 1, 1 for x=1, and 0 for any positive number less than 1. But that's just what I believe.

 

[ParadigmShifter]

From the first few terms it looks like the answer is going to be sqrt(2).

 

[IdiotsOpposite]

EDIT: More reasonably, I believe that the left side of the equation will be infinity for any positive number greater than 1, 1 for x=1, and 0 for any positive number less than 1. But that's just what I believe.

From the first few terms it looks like the answer is going to be sqrt(2).

I stand corrected.

 

[sanabas]

Looks like x = n^(1/n) will have the whole thing = n

 

[ParadigmShifter]

Is that a contraction mapping? That would show it has a fixed point. Off to work in a mo.

http://en.wikipedia.org/wiki/Contraction_mapping

 

[Spoonwood]

Looks like x = n^(1/n) will have the whole thing = n

I think you mean that x^(x^(...))=n will have n^(1/n), n a natural number, and that will work. Such a solution, of course, isn't unique though. I find the reasoning interesting. In x^(x^(...)) x^(...) has just as many terms as does x^(x^(...)). So, since x^(x^(...))=n, x^(...)=n also. Thus, x^(x^(...))=x^n=n. So, there exists cases where we have an irrational number raised to an irrational number raised to... which equals a rational number.

 

[Petek]

Let's see if we can come up with a rigorous proof that if x^(x^(x^...) = 2, then x = sqrt(2). For convenience, write sqrt(2) = s.

First, let's write down a formal definition of x^(x^(x^...) as a sequence: Define a₁(s) = s, a₂(s) = s^a₁(s), ,,, , a_n(s) = s^a_n-1(s) and so on. (From now on I'll usually omit the (s) when writing a_n.) Then we can define

x^(x^(x^...) = lim (n--> infinity) a_n.

(From now on I'll omit the (n-->infinity); it should be understood that all limits go to infinity.)

I claim that {a_n} is a monotone increasing sequence that's bounded from above. It's a standard theorem that such sequences converge, so we can conclude that {a_n} converges.

Monotone increasing: By induction on n. a₁ = s and a₂ = s^a₁ = s² = 2, so a₁ < a₂. Now assume that a_n-1 < a_n. Then a_n = s^a_n-1 < s^a_n = a_n+1. This concludes the proof that {a_n} is monotone increasing.

Bounded from above: It's easy to show that a_n < 2 for all n by an argument similar to the above, so I'll omit the proof. Thus, {a_n} is bounded from above and so the sequence converges.

Since the sequence converges, we can use standard limit theorems to compute the limit as follows:

Let L = lim a_n. Then

L = lim s^a_n-1
= s^{lim a_n-1}
= s^{lim a_n}
= s^L

That is, L = sqrt(2)^L. This equation has only the two solutions L = 2, 4 (left as an exercise). Since {a_n} clearly can't converge to 4 (all the terms are less than 2), we conclude that it converges to 2, and so

s^(s^(s^...) = 2.

It's much more difficult to determine the range of values for which x^(x^(x^...) converges. I think I know the answer, but not the proof.

 

[ThinkTank]

Monotone increasing: By induction on n. a₁ = s and a₂ = s^a₁ = s² = 2, so a₁ < a₂.

Here you have a₂ = 2 which is wrong. But this is easily repaired.
As a₁ = s > 1, we have s^s > s¹ = s, so a₂ > a₁.

 

[Petek]

Here you have a₂ = 2 which is wrong. But this is easily repaired.
As a₁ = s > 1, we have s^s > s¹ = s, so a₂ > a₁.

Thanks for the correction.

 

[Atticus]

Bounded from above: It's easy to show that a_n < 2 for all n by an argument similar to the above, so I'll omit the proof.

Sorry, but I don't quickly see why it's easy. Usually I've found it hard to prove by induction that something is below given real number.

 

[Petek]

Sorry, but I don't quickly see why it's easy. Usually I've found it hard to prove by induction that something is below given real number.

Good thing I kept my notes! As above, s = sqrt(2), a₁ = s and a_n = s^a_n-1. We have that a₁ = s < 2, so the result is true for n = 1. Now assume inductively that a_n-1 < 2. Then a_n = s^a_n-1 < s² = 2. OK?

 

[Atticus]

Ok, it's fine. Should have spent a few seconds more with it.

 

[Spoonwood]

I'll remark that I found this problem in a book called The Art of the Infinite by Robert and Ellen Kaplan. I didn't find the following though.

Alright, let's suppose we've got things correct here (the following makes me think not, though Petek's reasoning seems good). Now let us consider an infinite product say equal to 2, like x*(x*(...))=2, where x only belongs to the positive reals (the negative case comes out worse since you can't tell if an infinite product of negative numbers is positive or negative). Were we to use one-to-one pairing like the above, we would have x*2=2. So, x=1, which of course is pure rubbish. Also, if we have x+(x+(...))=2, by one-to-one pairing we would have x+2=2. So, x=0, which again is a lode of nonsense. These cases illustrate can get generalized also, since x*(x*(...))=y by one-to-one pairing yields x=1, and x+(x+(...))=y yields x=0, which only work when y=1 for the infinite product and y=0 for the infinite summation. So, how does such an infinite exponentation converge, while a seemingly similar infinite product and seemingly similar summation both diverge?

Thoughts? Opinions? Any mistakes I've made?

 

[sanabas]

I'll remark that I found this problem in a book called The Art of the Infinite by Robert and Ellen Kaplan. I didn't find the following though.

Alright, let's suppose we've got things correct here (the following makes me think not, though Petek's reasoning seems good). Now let us consider an infinite product say equal to 2, like x*(x*(...))=2, where x only belongs to the positive reals (the negative case comes out worse since you can't tell if an infinite product of negative numbers is positive or negative). Were we to use one-to-one pairing like the above, we would have x*2=2. So, x=1, which of course is pure rubbish. Also, if we have x+(x+(...))=2, by one-to-one pairing we would have x+2=2. So, x=0, which again is a lode of nonsense. These cases illustrate can get generalized also, since x*(x*(...))=y by one-to-one pairing yields x=1, and x+(x+(...))=y yields x=0, which only work when y=1 for the infinite product and y=0 for the infinite summation. So, how does such an infinite exponentation converge, while a seemingly similar infinite product and seemingly similar summation both diverge?

Thoughts? Opinions? Any mistakes I've made?

Because the infinite product simplifies to be x^inf, and the summation can be simplified to inf*x, which are obviously both divergent for most positive numbers. The exponent doesn't simplify that way. Though it does if you make it (((((((x^x)^x)^x)^x)... which simplifies to x^(x^inf), which will only not be divergent for x <= 1

The only way for a*x = x is for a to be the multiplicative identity, i.e. 1, or for x to be 0, which is what the product converges to for a < 1. The only way for a+x=x is for a to be the additive identity, i.e. 0. But x^y = y has an infinite number of solutions with y>0. x = [e^(ln y)/y] Plug in a positive value for y, and you get a positive value for x that will give you a series that converges to y. So the product & sum only have one solution that doesn't diverge, but using exponents has infinite solutions that don't diverge.

 

[Petek]

In general. you can't reason that if (x^(x^(x^ ...) = a, then x^a = a and so x = a^1/a. You need to show, as I did above, that the sequence converges. For example, suppose that

(x^(x^(x^ ...) = 4

then x^4 = 4, which implies that x^2 = 2, or x = sqrt(2), contradicting the above result. This section of a Wikipedia article discusses the convergence of (x^(x^(x^...). It converges for e^-e < x < e^1/e.

Another example:

Suppose that x = 1 + 2 + 4 + 8 + .... Then x = 1 + 2( 1 + 2 + 4 + 8 + ...) = 1 + 2x, and solving for x gives x = -1, which is obviously invalid. Again, the series 1 + 2 + 4 + 8 + ... doesn't converge.

 

[sanabas]

In general. you can't reason that if (x^(x^(x^ ...) = a, then x^a = a and so x = a^1/a. You need to show, as I did above, that the sequence converges. For example, suppose that

(x^(x^(x^ ...) = 4

then x^4 = 4, which implies that x^2 = 2, or x = sqrt(2),

Fair point. Didn't think it through enough. Though I think it's still the case that if it converges to a, then x = [e^(ln a)/a]

 

[Spoonwood]

Thanks guys!

 

[azzaman333]

Is it just my university, or are linear algebra and vector calculus normally taught as the one subject? Because it seems like way too much to learn for a single subject.