Infinity problem
Posted: Thu May 21, 2015 8:56 pm
Suppose we have x to the x to the x to the x (ad infinitum)
= 2. Then what does x equal?
PhilX
= 2. Then what does x equal?
PhilX
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Interesting. Do you have an answer to the problem I posed? There is a method that will easily let you solve it.Dalek Prime wrote:http://www.had2know.com/academics/power ... ation.html
I will let it run for another week before I give the method with the answer to let others have a chance to solve this problem.Dalek Prime wrote:No, but it's a set, isn't it? I'm terrible at math. Hence I, paradoxically, program.
Honestly, I don't know.
I'll post the method here. The square root of 2 is equal to 1.414, right? Now multiply numerator and denominator of 1 divided by sq rt 2 to get sq rt 2 divided by 2. Substitute 1.414 for the sq rt 2 to get 1.414/2 = .707 which is your answer. Simple? (but a bit wordy)Dalek Prime wrote:I think I'm ill-equiped to to give you value, and would not wish to take up your time. But thank you.
I wasn't even able to divide 1 by 1.414 in my head, though the answer made me smack it lol!
But it isn't?Philosophy Explorer wrote:I'll post the method here. The square root of 2 is equal to 1.414, right? ...
To the third decimal point, it is.Arising_uk wrote:But it isn't?Philosophy Explorer wrote:I'll post the method here. The square root of 2 is equal to 1.414, right? ...
Which hat did you pick that from?Arising_uk wrote:Since when does 1.999396 = 2?
I multiplied 1.414 by itself? I thought a square root was the number obtained by multiplying a number by itself? Can't seem to get 2 from your number?Philosophy Explorer wrote:Which hat did you pick that from? PhilX
Round it off.Arising_uk wrote:I multiplied 1.414 by itself? I thought a square root was the number obtained by multiplying a number by itself? Can't seem to get 2 from your number?Philosophy Explorer wrote:Which hat did you pick that from? PhilX