It's a method used to apptoximate roots of an equation by continually "narrowing" the interval on which a "root" occurs......here's an example.....
Let us suppose that we wanted to solve this equation....
x^2 - 5 = 0
And let us suppose that we know that one root lies between 2 and 3......divide this interval in half = 2.5 and substitue this value into the equation. So we have (2.25)^2 - 5 = 1.25.
Then, since this value is positive and the value "2" would produce a "negative," then we know that the "root occurs between 2 and 2.25. So, divide this interval in half = 2.125, and substitue this value into the equation . So we have (2.125)^2 - 5 = a negative vlaue. So, we know that the root must lie between 2.125 and 3.
And dividing this interval in half we have...2.5625....and substituting this value into the equation we have...(2.5625)^2 - 5 = a positive......so now we know that the root lies btween 2.125 and 2.5625. So, we divide this interval in half, etc.
Notice that width of the interval has decreased from 1 to .4375 in just three steps......by continuing this iterative puocess just a few more times, we would get close to the "real" root of about 2.236...
This method is based on the Intermediate Value Theorem.....which says that if we have a continuous interval [a,b] and f(a) and f(b) have opposite signs, then there must be some "c" on [a,b] such that f(c) = 0.......
lol !melody that just made me laugh , the way you said it " what is the rest of the question" lol!
There you go Rosala, I am here to amuse you. I am glad i am doing a good job. LOL
yes you are!some phrases in spite of being so simple, cheer the mind and heart!
and it also depends on the situation!
It's a method used to apptoximate roots of an equation by continually "narrowing" the interval on which a "root" occurs......here's an example.....
Let us suppose that we wanted to solve this equation....
x^2 - 5 = 0
And let us suppose that we know that one root lies between 2 and 3......divide this interval in half = 2.5 and substitue this value into the equation. So we have (2.25)^2 - 5 = 1.25.
Then, since this value is positive and the value "2" would produce a "negative," then we know that the "root occurs between 2 and 2.25. So, divide this interval in half = 2.125, and substitue this value into the equation . So we have (2.125)^2 - 5 = a negative vlaue. So, we know that the root must lie between 2.125 and 3.
And dividing this interval in half we have...2.5625....and substituting this value into the equation we have...(2.5625)^2 - 5 = a positive......so now we know that the root lies btween 2.125 and 2.5625. So, we divide this interval in half, etc.
Notice that width of the interval has decreased from 1 to .4375 in just three steps......by continuing this iterative puocess just a few more times, we would get close to the "real" root of about 2.236...
This method is based on the Intermediate Value Theorem.....which says that if we have a continuous interval [a,b] and f(a) and f(b) have opposite signs, then there must be some "c" on [a,b] such that f(c) = 0.......
ok so you wanted the bisection method to approximate roots. That is what CPhill thinks anyway. Umm
Why are there so many unfinished questions these days?
Are people's fingers to sore to type a few more key strokes?
Apparently, rosala doesn't care WHAT the question is.....she just doesn't want to be bothered with LONG/BORING answers...LOL!!!