00:01
Hello students, today we will discuss about this question.
00:04
In this question, we are given that in approximating the roots of the equation that are given, 2x cube minus 2x minus 5 is equals to 0 using the second method at what and the iteration, the value of x that is equals to 1 .6 .005 -9 -04.
00:31
We need to use x 0 that is equals to 1.
00:35
So here we need to find after how much iteration the value of x will be this.
00:43
So here first of all we are given f of x that is equals to 2 x cube minus 2 x minus 5.
00:52
So f of 1 is equal to 2 minus 2 minus 5 which is less than 0.
00:59
Then f of 2 that is equal to 16 minus 4 minus 5 that is equals to 7 which is greater than 0 now roots lies between 1 1 and 2 so therefore x 0 is equal to 1 and x 1 that is equals to 2 now by the second method we can we know that x n plus 1 that is equal to x n minus 1 f of x 0 minus x n f of x 0 minus x n f f of x n minus x n f f f of x n minus 1 divided by f of x 0 minus f of x and minus 1.
01:36
So therefore x2 that is equal to x0 f of x1 minus x1 f of x0 divided by f of x1 minus f of x0 so that is equals to 1 .416666667 so from the above table so after here we can see that now we will use, we can write a table that is here it is n, here it is n minus 1, here it is x of n and here x of n minus n plus 1.
02:18
So therefore here n is equal to 1, 2, 3, 4, 5, 6, 7, 8 and 9, n minus 1, that is 1, 2, x of n minus 1, 2, x of n minus one, that will be 1 ,2, 1 .41, 666666, 7, 1 .25.
02:46
So here, sorry, here it will be 1 .253585, then 1 .60762935, then 1 .6 -25, then 1 .6 -20 -356 -24, 1 .6 -203 -256 -24, 1 .6 -204, 1 .6 -0 .0 .0 .5.
03:07
59733, 1 .6 -0 -05985, 1 .6 -0 -0 -59 -855...