Bisection method-numerical analysis

[I like Serena]

So,if they are not equal,why do we use the termination criteria |a-b|<TOL?I don't get it... :(

We don't.
See your previous post:

Nice.. And..something else..I found implementations of the bisection method and there the termination criteria is

 while(fabs((b-a)/2)>TOL)

.

Why is it like that??

See?

 

[evinda]

So,is it wrong when I write fabs(b-a)<TOL?

 

[I like Serena]

So,is it wrong when I write fabs(b-a)<TOL?

Depends on the rest of your algorithm.
If you return (a+b)/2, then your result will still be within TOL.
So then it is right!

 

[evinda]

Could you give me an example for this condition?
If for example the maximum number of iterations is 15,TOL is 0.001
and the initial interval is [0,2]

what is equal to the first

$\big|x_k-x_{k-1}\big|$ we have to use?