Bisection method-numerical analysis

[evinda]

Could you explain it further to me?
Because I found this:
When the interval is [$$a_{k-1}$$,$$b_{k-1}$$]

$$x_{k-1}$$=$$\frac{a_{k-1}+b_{k-1}}{2}$$

when the interval is [$$a_{k}$$,$$b_{k}$$]

$$x_{k}=\frac{a_{k-1}+\frac{a_{k-1}+b_{k-1}}{2}}{2}=\frac{3a_{k-1}+b_{k-1}}{4}$$

So,|$$x_{k}-x_{k-1}$$|=|$$\frac{3a_{k-1}+b_{k-1}}{4}-(\frac{a_{k-1}+b_{k-1}}{2})$$|=|$$\frac{a_{k-1}-b_{k-1}}{4}$$|

But |$$a_{k}-b_{k}$$|=|$$\frac{a_{k-1}-b_{k-1}}{2}$$|

so they are not equal..what have I done wrong?

 

[I like Serena]

Could you explain it further to me?
Because I found this:
When the interval is [$$a_{k-1}$$,$$b_{k-1}$$]

$$x_{k-1}$$=$$\frac{a_{k-1}+b_{k-1}}{2}$$

when the interval is [$$a_{k}$$,$$b_{k}$$]

$$x_{k}=\frac{a_{k-1}+\frac{a_{k-1}+b_{k-1}}{2}}{2}=\frac{3a_{k-1}+b_{k-1}}{4}$$

So,|$$x_{k}-x_{k-1}$$|=|$$\frac{3a_{k-1}+b_{k-1}}{4}-(\frac{a_{k-1}+b_{k-1}}{2})$$|=|$$\frac{a_{k-1}-b_{k-1}}{4}$$|

But |$$a_{k}-b_{k}$$|=|$$\frac{a_{k-1}-b_{k-1}}{2}$$|

so they are not equal..what have I done wrong?

I see nothing wrong.
That looks entirely correct! ;)

What you have matches with what I wrote:

$$| x_k - x_{k-1} | = \frac{b_{k-1} - a_{k-1}}{4} = \frac{b_{k} - a_{k}}{2} = b_{k+1} - a_{k+1}$$

 

[evinda]

I see nothing wrong.
That looks entirely correct! ;)

What you have matches with what I wrote:

|$$x_{k}-x_{k-1}$$|=|$$\frac{3a_{k-1}+b_{k-1}}{4}-(\frac{a_{k-1}+b_{k-1}}{2})$$|=|$$\frac{a_{k-1}-b_{k-1}}{4}$$|

|$$a_{k}-b_{k}$$|=|$$\frac{a_{k-1}-b_{k-1}}{2}$$|

How can they be equal?I don't understand :(

 

[I like Serena]

|$$x_{k}-x_{k-1}$$|=|$$\frac{3a_{k-1}+b_{k-1}}{4}-(\frac{a_{k-1}+b_{k-1}}{2})$$|=|$$\frac{a_{k-1}-b_{k-1}}{4}$$|

|$$a_{k}-b_{k}$$|=|$$\frac{a_{k-1}-b_{k-1}}{2}$$|

How can they be equal?I don't understand :(

They are not equal.
|$$x_{k}-x_{k-1}$$| is half of |$$a_{k}-b_{k}$$|.

If you want equality, pick |$$a_{k+1}-b_{k+1}$$|.

 

[evinda]

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

 

[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?