My brother came up with a fun problem:1) Solve x^x^x^x^... = 22) Solve x^x^x^x^... = 4Resolve the paradox.

1) x = 2^(1/2)2) x = 4^(1/4)What is the paradox?

QuoteOriginally posted by: ymous1) x = 2^(1/2)2) x = 4^(1/4)What is the paradox?4^1/4 = 2^1/2 If i call y=x^x^x..., then ln(y)=y*ln(x), hence x=exp(ln(y)/y) < e for y >1, as ln(y)<y for those. Thus there could be no solution for the second case

Does this limit even exist for x > 1?That was a rhetorical question The limit of x^x^x^... is 1 for |x| <= 1 and does not exist for x > 1. Thus there is no paradox because the original problem is ill posed.

QuoteOriginally posted by: UltravioletDoes this limit even exist for x > 1?That was a rhetorical question The limit of x^x^x^... is 1 for |x| <= 1 and does not exist for x > 1. Thus there is no paradox because the original problem is ill posed.And the proof of this is...? Alternatively, a simple sketch of proof that 2^(1/2) solves x^x^x... =2 could look like

QuoteOriginally posted by: zerdnaQuoteOriginally posted by: UltravioletDoes this limit even exist for x > 1?That was a rhetorical question The limit of x^x^x^... is 1 for |x| <= 1 and does not exist for x > 1. Thus there is no paradox because the original problem is ill posed.And the proof is...?in the pudding? :-)x^x^x^... = exp(x^n * ln x)lim(n->inf) exp(x^n * ln x) = a and lim(n->inf) (x^n * ln x) = ln a. Since x^n diverges for x > 1, ln a is +infinite and hence so is a.

QuoteOriginally posted by: UltravioletQuoteOriginally posted by: zerdnaQuoteOriginally posted by: UltravioletDoes this limit even exist for x > 1?That was a rhetorical question The limit of x^x^x^... is 1 for |x| <= 1 and does not exist for x > 1. Thus there is no paradox because the original problem is ill posed.And the proof is...?in the pudding? :-)x^x^x^... = exp(x^n * ln x)lim(n->inf) exp(x^n * ln x) = a and lim(n->inf) (x^n * ln x) = ln a. Since x^n diverges for x > 1, ln a is +infinite and hence so is a.I am not sure what you mean. Normally, what's called x^n is x*x*x...(n times). x^n is certainly diverging, but that's a completely different function than x^x^x...(n times), isn't it?why is x^x^x^... = exp(x^n * ln x) or even lim(n->inf) exp(x^n * ln x)?I agree that x^x^x^... = exp(ln(x) * x^x^x ... ) or lim(n->inf) exp(ln(x) * x^x^x...^x (n times)) , but that's a different story

QuoteOriginally posted by: zerdnaQuoteOriginally posted by: ymous1) x = 2^(1/2)2) x = 4^(1/4)What is the paradox?4^1/4 = 2^1/2 If i call y=x^x^x..., then ln(y)=y*ln(x), hence x=exp(ln(y)/y) < e for y >1, as ln(y)<y for those. Thus there could be no solution for the second caseI don't get this, we already knew x=2^1/2<e.

Quote I don't get this, we already knew x=2^1/2<e. Indeed, i meant to prove that x^x.. = a has only solutions at a < e. What i wrote first doesn't prove it, but what i wrote later below seems to be the sketch of the proof which points why x=a^(1/a) converges after repeated exponentiation x^x^x.. to a if a <e, and why it doesn't when a > e edit: crossed with your post.

QuoteOriginally posted by: zerdnaIndeed, what i meant to say is x = exp(ln(y)/y) < e^ (1/e) since Max[ ln(y)/y] =1/e at y=e.That still only places a bound on x not y, so y is not restricted from being 4 in this equation. But it it is bounded by 2.