Approximation method for super square root
#1
I came up with this method for calculating x where x^x=y. It's similar to a method for calculating square roots. The procedure is as follows:

1. Pick the number y that you would like to find the super root of.

2. Find the self-power that is closest to and less than y. We call this t^t. For example, if y=4000 then t=5 and t^t=3125.

3. Calculate the difference between y and t^t. So for y=4000, t=5, we get 875.

4. Divide that difference by the interval between the self-power below y and the one above it. So for y=4000, this would be (6^6-5^5)=46656-3125=43531.

5. Add this to t.

So with our example the result is 5+875/43531 which is about 5.0201. The actual value is 5.094, so the approximation is 1.5% off the actual value.

The approximation generally gets more accurate as y and t get higher. Within the regime of a specific t, the most inaccurate value will occur when y is about 27.4% through the interval (example, for t=5, the approximation is most inaccurate at y=11927+3125=15092). I haven't figured out why this is.

The accuracy at this point decreases with increasing t. The integer for which this yields the least accurate value is 2, at about 85%. This method is much more accurate than a first order Taylor approximation of the actual solution.

I haven't yet been able to apply this to higher order tetrations.
Reply
#2
To compute the inverse function of a strictly increasing function a method that always works is bisection.
perhaps you start with an integer number as you described.
Then you know the real value such that must lie in the interval , set .
Next you divide the interval into two halfes by , and you know that must either be in the left half or in the right half ; in the first case must and in the second case . You choose the new interval accordingly.
And do again bisection on it.
By repetition of bisection you can compute the to arbitrary precision (in the above argumentation I assumed that the solution is never on the boundary of the interval, in which case one can abort the bisection, having found the solution).

For a more concise description see wikipedia.
There are also other root-finding algortithms, like Newton method, etc.
Reply
#3
(03/23/2010, 10:54 AM)bo198214 Wrote: To compute the inverse function of a strictly increasing function a method that always works is bisection.
perhaps you start with an integer number as you described.
Then you know the real value such that must lie in the interval , set .
Next you divide the interval into two halfes by , and you know that must either be in the left half or in the right half ; in the first case must and in the second case . You choose the new interval accordingly.
And do again bisection on it.
By repetition of bisection you can compute the to arbitrary precision (in the above argumentation I assumed that the solution is never on the boundary of the interval, in which case one can abort the bisection, having found the solution).

For a more concise description see wikipedia.
There are also other root-finding algortithms, like Newton method, etc.

Yeah it's not the most accurate method, but it's done in one step without repetition or recursion, so it would be good more mental math enthusiasts.
Reply


Possibly Related Threads…
Thread Author Replies Views Last Post
  tommy's "linear" summability method tommy1729 2 88 01/24/2023, 11:26 AM
Last Post: jacob
  another infinite composition gaussian method clone tommy1729 2 46 01/24/2023, 12:53 AM
Last Post: tommy1729
  Semi-group iso , tommy's limit fix method and alternative limit for 2sinh method tommy1729 1 155 12/30/2022, 11:27 PM
Last Post: tommy1729
  [MSE] short review/implem. of Andy's method and a next step Gottfried 4 470 11/03/2022, 11:51 AM
Last Post: Gottfried
  [NT] primitive root conjecture tommy1729 0 264 09/02/2022, 12:32 PM
Last Post: tommy1729
  Is this the beta method? bo198214 3 539 08/18/2022, 04:18 AM
Last Post: JmsNxn
  Describing the beta method using fractional linear transformations JmsNxn 5 697 08/07/2022, 12:15 PM
Last Post: JmsNxn
Question The Etas and Euler Numbers of the 2Sinh Method Catullus 2 488 07/18/2022, 10:01 AM
Last Post: Catullus
Question Functional Square Root Catullus 24 2,422 07/01/2022, 09:17 PM
Last Post: tommy1729
  Tommy's Gaussian method. tommy1729 34 11,528 06/28/2022, 02:23 PM
Last Post: tommy1729



Users browsing this thread: 1 Guest(s)