Buster's math problem!

[ColinDavies]

Nish - Native CPian wrote: Solve for n :- n^n^n^n.........(infinite) = 2 n = 1 + ( 1 / infinity ) Regardz Colin J Davies

Sonork ID 100.9197:Colin

More about me :-)

[Nish Nishant]

****Colin Davies wrote: n = 1 + ( 1 / infinity ) Reasonable :-) Nish

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Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.

[ColinDavies]

Nish - Native CPian wrote: Reasonable It appalls me :-( Regardz Colin J Davies

Sonork ID 100.9197:Colin

More about me :-)

[David Wengier]

How do you know? He is dividing both sides by "n^n^n^n^n.." to infinity. Surely that would take an infinite amount of time to compute :) What I love about this stuff, is how the infinite amount of numbers between 0 and 1, is larger than the infinte amount of numbers between 0 and infinity. -- David Wengier Sonork ID: 100.14177 - Ch00k

[Alexandru Savescu]

Do you really mean that solution or are you just taking the piss? Best regards, Alexandru Savescu

[Richard Deeming]

It is correct, as long as you take the equation to read:

n ^ (n ^ (...

and not:

((n ^ n) ^ n)...

Proof:

∀ x > 1:
∀ y > 1, x < x^y
∀ 0 < y < 1, 1 < x^y < x
Limy → 1 x^y = x

Let x=2, p = 2y, n = √2 [= 2^½].

1 < n < 2
x^y = 2^(p/2) = n^p
∀ 0 < p < 2, 1 < n^p < 2
∀ p > 1, n < n^p
∴ ∀ 1 < p < 2, n < n^p < 2
Limp → 2 n^p = 2

Let p0 = n, pi = n ^ pi-1:

1 < p0 < 2 ⇒ p0 < [n^p0 = p1] < 2
∴ 1 < p0 < ... < pi-1 < pi < ... < 2
Limi → ∞ pi = 2

Elementary! :wtf: Richard

[ColinDavies]

I agree if But It would have been documented differently, Also I admit I automatically assumed that, and considered it far too simple and logical. :-( Regardz Colin J Davies

Sonork ID 100.9197:Colin

More about me :-)

[Lost User]

Nish - Native CPian wrote: n^n^n^n.........(infinite) = 2 equals to 2 or tends to 2?

[Alexandru Savescu]

Well it is not elementary enough :wtf: All your statements from above are correct except for the last one Richard_D wrote: 1 < p0 < 2 ⇒ p0 < [n^p0 = p1] < 2∴ 1 < p0 < ... < pi-1 < pi < ... < 2Limi → ∞ pi = 2 I agree for every i, 1 < pi < 2, but this does not mean that the limit is 2, It may be very well 1.8. You must also prove that for every x>0 there exists and k so that 2-x < pk < 2. ;) Another classic example is the definition of e: limit of (1 + 1/n)^n. All you can tell that for every n: 2 < (1 + 1/n)^n < 3 but the limit is not 3 is 2.718182.... Yet I tend to agree (my intuition) that the limit of our sequence p0, p1,...pn is 2. I'll look up into it. Best regards, Alexandru Savescu

[Nish Nishant]

Anonymous wrote: equals to 2 or tends to 2? Tends to 2 of course!!! Nish

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Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.

[Richard Deeming]

OK, I haven't done real maths for four years! This may be total crap, but I'm thinking along these lines: [Trying to extend this to n^n^... = X ⇒ n = X ^ (1/X) ∀ X ∈ N]

p0 = X^(1/X), pi = p0 ^ pi-1

We already have:

∀ i, 1 < pi-1 < pi < X

The proof requires:

∀ 0 < y, ∃ k : X - y < pk < X

The case where y ≥ X - 1 is elementary:

∀ y ≥ X - 1, X - y ≤ 1 < pi < X

So we are left with showing:

∀ 0 < y < X - 1, ∃ k : X - y < pk < X

Thr section (0,1) seems to be easy. Since:

∀ x,y > 0, b > 1 : x < y ⇔ Logb(x) < Logb(y)

I can take the Log base p0 [call this LP] of the inequality ( :confused: ) to get:

LP(X) = ln(X) / ln(X ^ (1/X)) = ln(X) / ((1/X) ln (X)) = X
LP pk = pk-1
LP(X - y) = X × [ln(X - y) / ln(X)]

Let α = ln(X) - ln(X - y)
= ln(X) - ln(X) × ln(-y)
= ln(X) × [1 - 1/ln(y)]

∴ LP(X - y) = X × [1 - α / ln(X)]
= X × [1 - ln(X) × (1 - 1/ln(y)) / ln(X))
= X × [1 - (1 - 1/ln(y))]
= X / ln(y)

So the inequality is:

∀ 0 < y < 1, ∃ k : X / ln(y) < pk-1 < X
∀ 0 < y < 1, ln(y) < 0
∴ X / ln(y) < 0 < pk < X

So any value of k satisfies the inequality for (0,1). This leaves the section [1,X-1). Since the lower bound is included, we simply need to show:

∃ k : X - 1 < pk < X

In the case of X=2, any value of k satisfies this. Now, I just need proof for X > 2! I hope this make at least some sense, and doesn't read like the insane ramblings of a deranged lunatic! :) Cheers, Richard