[SOLVED] Halving the interval (bisection)?

• Jan 30th 2009, 01:14 PM
noobonastick
[SOLVED] Halving the interval (bisection)?
can someone help me with halving the interval, i know the theory, method, etc but i dont know how to arrive at the final answer when it asks to solve to two decimal places.
e.g Question:
Use the method of bisection to solve the following equation to 2 decimal places given that:
x^3 - 3x - 20 = 0 has a root between 3 and 3.1

thanks
• Jan 30th 2009, 11:15 PM
earboth
Quote:

Originally Posted by noobonastick
can someone help me with halving the interval, i know the theory, method, etc but i dont know how to arrive at the final answer when it asks to solve to two decimal places.
e.g Question:
Use the method of bisection to solve the following equation to 2 decimal places given that:
x^3 - 3x - 20 = 0 has a root between 3 and 3.1

thanks

Use a table: \$\displaystyle f(x)=x^3-3x-20\$

\$\displaystyle 1. f(3)=-2~~~~f(3.1)=0.491~~~~\longrightarrow~3<x<3.1\$

\$\displaystyle 2. f(3.05)=-0.777...~~~~f(3.1)=0.491~~~~\longrightarrow~3.05<x <3.1\$

\$\displaystyle 3. f(3.075)=-0.1489... ~~~~f(3.1)=0.491~~~~ \longrightarrow~ 3.075<x<3.1\$

Because \$\displaystyle f(3.0875)=\bold{+0.165...}\$ you now have to examine:

\$\displaystyle f(3.08125)=\bold{+0.0099...}\$ It is still positive. Therefore you have to examine now:

\$\displaystyle 4. f(3.078125)=-0.695...~~~~f(3.08125)=0.0099~~~~ \longrightarrow~3.078125<x<3.08125\$

Thus the result to 2 dp is 3.08

By the way: A linear interpolation yields the same result in one step.
• Jan 31st 2009, 01:26 AM
noobonastick
Thanks, so basically I have to keep halving until the upper and lower x values round off to the same two decimal places?
• Jan 31st 2009, 10:06 AM
earboth
Quote:

Originally Posted by noobonastick
Thanks, so basically I have to keep halving until the upper and lower x values round off to the same two decimal places?

Correct! (Clapping)