# Interval Bisection: Solve for Root in [1, 2]

• aurao2003
For the record, I was getting my information from the original poster, who wrote that they obtained an answer of 1.3 after 3 intervals. If you look at the attempt at a solution, you can see that the original poster was doing 3 bisections (and then they did another one after that, which is why I wrote "technically, you did 4 bisections").

## Homework Statement

Hi
Please, I need clarification on this queston

Show that the equation
0 = x/2 -1/x
,x>0, has a root in the interval [1, 2].

b Obtain the root, using interval bisection two times. Give your answer to two significant figures.

## Homework Equations

Change in sign of values between (a+b)/2

## The Attempt at a Solution

I am actually wondering how many times do I find (a+b)/2. The question obviously states twice. But the answer was based on 4 interval bisections! I obtained 1.3 and they obtained 1.4. Can anyone please comment?

If you bisect the interval once you get two halves. If you bisect one of these halves again, you get two quarters. Is that what you're asking?

Show what you did to get 1.3...

For the a part
I said let f(x) = x/2 - 1/x
f(1) = 0.5-1= -0.5
f(2) = 1-0.5= 0.5
There is a change of sign betwen f(1) and f(2). Therefore a root exists in the interval
[1,2]
I now used these values to obtain f(a+b/2)
So, for example (a+b)/2 = 1.5 (When considering Interval 1,2). I noticed where the signs changed and took a new interval. This led to my result of 1.3.

If you bisect the interval [1, 2], you get two intervals: [1, 1.5] and [1.5, 2]. Since the function changes sign in the first interval, there's a root in [1, 1.5].

If you bisect the interval [1, 1.5], you get two more intervals: [1, 1.25] and [1.25, 1.5]. In one of these two intervals, the function changes sign, so an estimate for the root is the midpoint of that interval.

You bisect two intervals, then take the midpoint of the next interval.

Mark44 said:
If you bisect the interval [1, 2], you get two intervals: [1, 1.5] and [1.5, 2]. Since the function changes sign in the first interval, there's a root in [1, 1.5].

If you bisect the interval [1, 1.5], you get two more intervals: [1, 1.25] and [1.25, 1.5]. In one of these two intervals, the function changes sign, so an estimate for the root is the midpoint of that interval.

I am not doubting the answer in the book. But I will like to know how many bisection intervals you applied. I obtained 1.3 after 3 intervals. The root is closer to 1.4 as the book states. My diiference of opinion is regarding how many intervals were used.

Count how many times I wrote "bisect the interval" in post #4. The root estimate is the middle of the interval [1.25, 1.5]. It might seem like we're bisecting the interval for a third time to do this, but we're not. Notice that the number in the middle of [1.25, 1.5] is not 1.3.

The first interval is [1, 2] f(1)= 1/2- 1= -1/2< 0. f(2)= 2/2- 1/2= 1/2> 0.

The midpoint of that interval is 3/2= 1.5. f(3/2)= 3/4- 2/3> 0 so the new interval is [1, 3/2].

The midpoint of that interval is 5/4= 1.25. f(5/4)= 5/8- 8/5= (25- 64)/8< 0 so our new interval is [5/4, 3/2]

The midpoint of that interval is 11/8 =1.375. f(11/8)= 11/16- 8/11= (121- 128)/176< 0 so our new interval [5/4, 11/8].

The midpoint of that interval is 21/16= 1.3125. f(21/16)= 21/32- 16/21= (441- 512)/672< 0 so our new interval is [5/4. 21/16].

The midpoint of that interval is 41/32= 1.28125.

Since the last two both round to 1.3, that is the correct answer to two significant figures.
x/2 - 1/x

It seems to come down to how you count. In my way of thinking, the first bisection gives you [1, 3/2] and [3/2, 2], with the root being in the first subinterval.

When you bisect the [1, 3/2] interval, you get [1, 5/4] and [5/4, 3/2]. Since f(5/4) and f(3/2) are opposite in sign, but the same is not true for f(1) and f(5/4), the root is in the interval [5/4, 3/2]. To get an estimate of the root take the midpoint of that interval, which is 11/8, or 1.375. Rounded to two significant figures, this is 1.4. I am distinguishing between taking the midpoint of an interval for the root estimate and doing another bisection. Since the 1.4 result agrees with the answer in the book, I believe this is what the authors had in mind. 1.4 is also closer to the analytic solution, which is $\sqrt{2} \approx 1.414$.

BTW, Halls, f(5/4) = 5/8 - 4/5, not 5/8 - 8/5, as you had.

## 1. What is interval bisection?

Interval bisection is a numerical method used to find a root (or solution) of a mathematical equation within a given interval. It involves dividing the interval into smaller subintervals and narrowing down the range until the root is found.

## 2. How does interval bisection work?

Interval bisection works by evaluating the function at the endpoints of the interval and determining which subinterval contains the root. The subinterval that contains the root is then divided in half and the process is repeated until the root is found within a desired level of accuracy.

## 3. What is the benefit of using interval bisection?

Interval bisection is a simple and reliable method for finding roots of equations. It is also easy to implement and does not require any prior knowledge of the function.

## 4. Is interval bisection always guaranteed to find a root?

Yes, interval bisection is guaranteed to find a root if the function is continuous and changes sign within the given interval. However, it may not always converge to the root within a specified number of iterations.

## 5. Can interval bisection find multiple roots within an interval?

No, interval bisection is only able to find one root within a given interval. If there are multiple roots, the method may converge to any one of them depending on the initial choice of subintervals.