Formula of Bolzano

Hi there
Today @ math we learned a new formula, it's like typing alwasy the same thing over again, to get closer to a point.
Now i thougt to put it in a program, but the program won't work. What i got is this:

(A+B)/2 => M
Y1(M) => C
If C>0
Then
M => B
Else
M => A
Disp "A=",A
Disp "B=",B
Disp "M=",M
Disp "F(m)=",C

The problem is in the if-then-else part.. When F(m) outcomes negative, it should store M in A, when it turns op postive it should store it in B..

The problem is, it won't Disp anything? :s

What i figured out now, it SOMETIMES gives the Displays.. sometimes it just says Done without giving nothing.. What's causing this bug?

For the ones who wanna figure it out, stop here, because i got it already, sorry for making the forum dirty for it
I had to add an end in it, because in the first version, it only displays when it's negative.. Programming is just logics i guess

(A+B)/2 => M
Y1(M) => C
If C>0
Then
M => B
Else
M => A
End
Disp "A=",A
Disp "B=",B
Disp "M=",M
Disp "F(m)=",C

hmm, would this be a program that you ran over and over until the value converged? if so, why not create a While() loop? Then you wouldn't have to rerun the program over and over again.

and then what if f(x)= -x^2 + 4 with A=0 and B=5,
(0+5)/2=2.5
f(2.5)= -2.25

according to your code (if C>0, then replace B else replace A) you'd replace the wrong value...correct? causing a squeeze between 2.5 and 5 when the zero is between 0 and 2.5

i think you'd need more to the program first, in order to decide with way to squeeze.

edit: ah, the teacher wants to see your work, thus the steps shown...got it. still the error in the replacement.

check this out!