 My Math Forum Solving Square Super-Root with Elementary Functions
 User Name Remember Me? Password

 Number Theory Number Theory Math Forum

 November 29th, 2017, 03:21 PM #1 Newbie   Joined: Nov 2017 From: Boulder, Colorado Posts: 1 Thanks: 0 Solving Square Super-Root with Elementary Functions First, a little background. I joined to ask this question, so apologies if it is in the wrong sub-forum. I am a senior in high school in Calc 1, but I have recently been intrigued with hyper operations. With a little experimenting in the calculator and some logic, I came to the conclusion that I can find the square super-root of any real number on the domain (~.6923, ∞) using a chain (not sure if right word) of elementary functions. Specifically, if we define x^x = y, then: a = y^(1/y) b = y^(1/a) c = y^(1/b) d = y^(1/c) ... and so on. Eventually, the value of y^(1/n) converges to the correct square super-root of y. If x<1, it asymptotes to the square super-root from above. If x>1, it oscillates above and below the correct value and eventually converges. However, computing the square super-root of numbers larger than 15 requires carrying a higher number of digits than I currently can in a computer program. Here is an example. I only wrote down 3 digits each time for brevity, but I carried 10 digits in the calculator. if x^x = 3: a = 3^(1/3) = 1.44 b = 3^(1/1.44) = 2.14 c = 3^(1/2.14) = 1.67 d = 3^(1/1.67) = 1.93 e = 3^(1/1.93) = 1.77 f = 3^(1/1.77) = 1.86 g = 3^(1/1.86) = 1.80 h = 3^(1/1.80) = 1.84 i = 3^(1/1.84) = 1.82 j = 3^(1/1.82) = 1.83 k = 3^(1/1.83) = 1.82 1.82^1.82 ≈ 3 So after 11 iterations, we have 3 correct digits of the 2nd super-root of 3. After infinitely many iterations, we have infinitely many correct digits. Thoughts? Has this been recognized before, and if so, where can I read about it? As far as I have read, square super-roots should not be able to be calculated with elementary functions. December 1st, 2017, 02:54 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Such a function can not be calculated using a finite number of elementary functions. But any function can be calculated as a limit of an infinite sequence of elementary functions. June 23rd, 2018, 06:47 PM   #3
Member

Joined: Mar 2017
From: venezuela

Posts: 36
Thanks: 3

Quote:
 Originally Posted by eliboardman First, a little background. I joined to ask this question, so apologies if it is in the wrong sub-forum. I am a senior in high school in Calc 1, but I have recently been intrigued with hyper operations. With a little experimenting in the calculator and some logic, I came to the conclusion that I can find the square super-root of any real number on the domain (~.6923, ∞) using a chain (not sure if right word) of elementary functions. Specifically, if we define x^x = y, then: a = y^(1/y) b = y^(1/a) c = y^(1/b) d = y^(1/c) ... and so on. Eventually, the value of y^(1/n) converges to the correct square super-root of y. If x<1, it asymptotes to the square super-root from above. If x>1, it oscillates above and below the correct value and eventually converges. However, computing the square super-root of numbers larger than 15 requires carrying a higher number of digits than I currently can in a computer program. Here is an example. I only wrote down 3 digits each time for brevity, but I carried 10 digits in the calculator. if x^x = 3: a = 3^(1/3) = 1.44 b = 3^(1/1.44) = 2.14 c = 3^(1/2.14) = 1.67 d = 3^(1/1.67) = 1.93 e = 3^(1/1.93) = 1.77 f = 3^(1/1.77) = 1.86 g = 3^(1/1.86) = 1.80 h = 3^(1/1.80) = 1.84 i = 3^(1/1.84) = 1.82 j = 3^(1/1.82) = 1.83 k = 3^(1/1.83) = 1.82 1.82^1.82 ≈ 3 So after 11 iterations, we have 3 correct digits of the 2nd super-root of 3. After infinitely many iterations, we have infinitely many correct digits. Thoughts? Has this been recognized before, and if so, where can I read about it? As far as I have read, square super-roots should not be able to be calculated with elementary functions.

There are other simpler ways to manage all this stuff on roots and super-roots:
Take a look at this thread:

The Mediant and new high-order root approximating methods.

and this:
https://domingogomezmorin.wordpress.com/

Regards,
Domingo Gomez Morin Tags elementary, functions, solving, square, superroot Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post pirateprogrammer Complex Analysis 3 September 11th, 2016 03:46 PM mathbalarka Real Analysis 2 August 30th, 2012 04:37 AM success123 Algebra 8 December 13th, 2011 10:09 PM jared_4391 Algebra 3 August 8th, 2007 09:06 AM success123 Calculus 2 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      