HP50G Bisection methode

07242015, 12:01 PM
(This post was last modified: 07312015 03:51 PM by tigger.)
Post: #1




HP50G Bisection methode
I would like to program the Bisection Methode.
Is there anybody who can help me with it. I am an absolute beginner. I have the function y=x^55. I assume the result between 1 and 2: We first define the function F(X). Now save 2 in a variable called B and 1 in a variable called A. We have to find the midpoint between A and B, our new approximation, C. Can anbody help me to go on. I would like to program other easy stuff. 

07242015, 04:35 PM
(This post was last modified: 07242015 04:43 PM by Gilles.)
Post: #2




RE: HP50G Bisection methode
(07242015 12:01 PM)tigger Wrote: I would like to program the Bisection Methode.Hi It seems there is a problem with your example as there is no root for this function between 1 and 2 here is an example (not optimised), in RPL, aprox mode with a modified function (X^55) : Code: 'F(X)=X^55' DEFINE 

07292015, 11:31 AM
Post: #3




RE: HP50G Bisection methode
(07242015 12:01 PM)tigger Wrote: I would like to program the Bisection Methode. The root for y=x^5 (or for any y=x^n) is zero, not in the interval [1, 2]. You can start with an interval of, say, [1, 1]. Namir 

07292015, 11:37 AM
(This post was last modified: 07292015 11:39 AM by Namir.)
Post: #4




RE: HP50G Bisection methode
(07242015 04:35 PM)Gilles Wrote:(07242015 12:01 PM)tigger Wrote: I would like to program the Bisection Methode.Hi Your code assumes F(A)<0 and that's risky, because if F(A)>0 your implementation will not work. The traditional Bisection algorithm uses the test F(A)*F(C)>0 instead to determine if we set A=C (when that condition is true) and B=C otherwise. I assume the RPL code should be: Code: 'F(X)=X^55' DEFINE Namir 

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