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 Approximation method for super square root Ztolk Junior Fellow Posts: 21 Threads: 4 Joined: Mar 2010 03/23/2010, 02:33 PM (03/23/2010, 10:54 AM)bo198214 Wrote: To compute the inverse function of a strictly increasing function $f$ a method that always works is bisection. perhaps you start with an integer number $t_0$ as you described. Then you know the real value $t$ such that $f(t)=y$ must lie in the interval $(t_0,t_0+1)$, set $u_0=t+1$. Next you divide the interval $(t_0,u_0)$ into two halfes by $w_0=\frac{t_0+u_0}{2}$, and you know that $t$ must either be in the left half $(t_0,w_0)$ or in the right half $(w_0,u_0)$; in the first case must $f(t_0) and in the second case $f(w_0). You choose the new interval $(t_1,u_1)$ accordingly. And do again bisection on it. By repetition of bisection you can compute the $t$ 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. « Next Oldest | Next Newest »

 Messages In This Thread Approximation method for super square root - by Ztolk - 03/23/2010, 01:09 AM RE: Approximation method for super square root - by bo198214 - 03/23/2010, 10:54 AM RE: Approximation method for super square root - by Ztolk - 03/23/2010, 02:33 PM

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