I don't know this method of solving, but I think that since you got -1/2 for x = 1 and 1/2 for x = 2, you would get a more accurate answer taking 1.5, and maybe they used that initial value to get 1.4, or something of the like.
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.
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?
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.
I think you have misunderstood my post. All the solutions I looked at calculated up to 5 approximations for the interval bisection. This was despite specifying using the bisection 2 or 3 times. My value was slightly different from the textbooks because I kept to the letter. Do I need to find the closest approximation or stick to the question? For example, the question I posed had up to 5 approximations before they obtained 1.4. Whereas, I obtained 1.3 because the question specified only 2 approximations.
hmmm. yeah I see your point. Except I get 1.4 after 3 bisections. First bisection = 1.5, second bisection = 1.25 and third = 1.375 rounded to 1.4.
The only thing I can think of is that they don't count 1.5 as a bisection but a starting point. If in other similar questions you find that it is off by 1 bisection then I say they don't count the first one. If you are correct in other questions then maybe a mistake?
also I'd say 0 = x/2 -1/x
,x>0, has a solution in the interval [1, 2].
and the equation has a root in the interval...
I get a different result. Just one result that rounds to 1.4 is not enough.
f(1)= 1/2- 1/1 - -1/2, f(2)= 2/2- 1/2= 1/2 so there is a root between 1 and 2.
The midpoint of [1, 2] is 3/2= 1.5. f(3/2)= 3/4- 2/3= 9/12- 6/12= 3/12= 1/4> 0 so there is a root between 1 and 3/2.
The midpoint of [1, 3/2] is 5/4= 1.25. f(5/4)= 5/8- 4/5= (20- 32)/40< 0 so there is a root between 5/4 and 3/2.
The midpoint of [5/4, 3/2] is 11/8= 1.375. f(11/8)= 11/16- 8/11= (121- 128)/176< 0 so there exist a root between 5/4 and 11/8.
The midpoint of [5/4, 11/8] is 21/16= 1.3125. f(21/16)= 21/32- 16/21= (441- 512)/672< 0 so there exist a root between 5/4 and 21/16.
The midpoint of [5/4, 21/16] is 41/32= 1.28125.
Those last two answers both round to 1.3 so the solution, to two 2 significant figures, is 1.3.
yes Halls. That's what the OP also gets. But the problem with the question is that it states to use only 2 bisections and it gives the answer of 1.4.
What you've done is shown that the one that rounds to 1.4 must be the one that the book claims is the 2nd bisection