Buster's math problem!
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Solve for n :-
n^n^n^n.........(infinite) = 2
Solution :- Let x = n^n^n^n^n....(infinite) = 2 Now n^x = n^(n^n^n^n....) = n^2 But also n^x = n^(n^n^n^n...) = n^n^n^n^n... [we take out the brackets] Thus n^2 = n^n^n^n^n... Since n^n^n^n^n... = 2 We now have :- n^2 = 2 n = √2 (Square Root of 2 ) Nish
Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.
Sorry Nish your Answer doesn't compute with me :-( Regardz Colin J Davies
Sonork ID 100.9197:Colin
More about me :-)
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Solve for n :- n^n^n^n.........(infinite) = 2 Regards Nish p.s. You have only 300 seconds. After that the solution will be posted if no one has solved this
Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.
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 :-)
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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 :-)
****Colin Davies wrote: n = 1 + ( 1 / infinity ) Reasonable :-) Nish
Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.
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****Colin Davies wrote: n = 1 + ( 1 / infinity ) Reasonable :-) Nish
Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.
Nish - Native CPian wrote: Reasonable It appalls me :-( Regardz Colin J Davies
Sonork ID 100.9197:Colin
More about me :-)
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Sorry Nish your Answer doesn't compute with me :-( Regardz Colin J Davies
Sonork ID 100.9197:Colin
More about me :-)
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
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Solve for n :-
n^n^n^n.........(infinite) = 2
Solution :- Let x = n^n^n^n^n....(infinite) = 2 Now n^x = n^(n^n^n^n....) = n^2 But also n^x = n^(n^n^n^n...) = n^n^n^n^n... [we take out the brackets] Thus n^2 = n^n^n^n^n... Since n^n^n^n^n... = 2 We now have :- n^2 = 2 n = √2 (Square Root of 2 ) Nish
Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.
Do you really mean that solution or are you just taking the piss? Best regards, Alexandru Savescu
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Do you really mean that solution or are you just taking the piss? Best regards, Alexandru Savescu
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 = xLet 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 = 2Let p0 = n, pi = n ^ pi-1:
1 < p0 < 2 ⇒ p0 < [n^p0 = p1] < 2
∴ 1 < p0 < ... < pi-1 < pi < ... < 2
Limi → ∞ pi = 2Elementary! :wtf: Richard
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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 = xLet 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 = 2Let p0 = n, pi = n ^ pi-1:
1 < p0 < 2 ⇒ p0 < [n^p0 = p1] < 2
∴ 1 < p0 < ... < pi-1 < pi < ... < 2
Limi → ∞ pi = 2Elementary! :wtf: Richard
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 :-)
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Solve for n :- n^n^n^n.........(infinite) = 2 Regards Nish p.s. You have only 300 seconds. After that the solution will be posted if no one has solved this
Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.
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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 = xLet 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 = 2Let p0 = n, pi = n ^ pi-1:
1 < p0 < 2 ⇒ p0 < [n^p0 = p1] < 2
∴ 1 < p0 < ... < pi-1 < pi < ... < 2
Limi → ∞ pi = 2Elementary! :wtf: Richard
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
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Anonymous wrote: equals to 2 or tends to 2? Tends to 2 of course!!! Nish
Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.
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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
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 < XSo 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
