1. Bisection method problem ????

Can somebody clarify me how to estimate number of iterations required for a particular accuracy in bisection method ?
i.e
think we want to know a real solution of x+sin(x)=pi/2 between pi/2 n 2pi
How many iterations do we need to get x with an accuracy of 3 decimal places using bisection method ? (without doing any iterations)

please tell me if you dont get the idea of my Qn .

Thank You

2. Originally Posted by K A D C Dilshan
Can somebody clarify me how to estimate number of iterations required for a particular accuracy in bisection method ?
i.e
think we want to know a real solution of x+sin(x)=pi/2 between pi/2 n 2pi
How many iterations do we need to get x with an accuracy of 3 decimal places using bisection method ? (without doing any iterations)

please tell me if you dont get the idea of my Qn .

Thank You
The length of the interval containing the root after $\displaystyle n$ itterations is $\displaystyle L/2^n$ where $\displaystyle L$ is the length of the initial interval.

After n itterations, if we take the mid point as our estimate of the root, the error $\displaystyle \le L/2^{n+1}$, so you need the smallest $\displaystyle n$ such that

$\displaystyle L/2^{n+1} \le 0.0005$

CB

Originally Posted by CaptainBlack
The length of the interval containing the root after $\displaystyle n$ itterations is $\displaystyle L/2^n$ where $\displaystyle L$ is the length of the initial interval.

After n itterations, if we take the mid point as our estimate of the root, the error $\displaystyle \le L/2^{n+1}$, so you need the smallest $\displaystyle n$ such that

$\displaystyle L/2^{n+1} \le 0.0005$

CB
so is it the the maximum absolute error ?
how can i develop an inequality to calculate for an accuracy upto 3 decimal places ?????

4. Originally Posted by CaptainBlack
The length of the interval containing the root after $\displaystyle n$ itterations is $\displaystyle L/2^n$ where $\displaystyle L$ is the length of the initial interval.

After n itterations, if we take the mid point as our estimate of the root, the error $\displaystyle \le L/2^{n+1}$, so you need the smallest $\displaystyle n$ such that

$\displaystyle L/2^{n+1} \le 0.0005$

CB

sry thanx