00:01
In program 70, we want to test newton's method conversions for this function, and we have already the exit solution of this function, where the zeros are 0, 1 and 2.
00:21
We will test newton's method using many initial values.
00:27
Newton's method depends on iterations based on these initial values.
00:31
The next root of the iteration x n plus 1 equals the previous root x n minus f of x n divided by f dash of x n.
00:44
We have f of x n which is the main function in the problem.
00:49
We can evaluate f dash of x because we will use it.
00:53
F dash of x the differentiation of x cubed which is 3x squared minus 2 multiplied by 3 equals 6 x plus 2 we can start by the initial guess of part a x node equals 0 .54 for clear calculations we make our calculations in a table the number of iteration the value of x the value of x the value of f dash of x we can evaluate f of x by substituting in the main function here can evaluate f dash of x of any number by substituting here the initial value was given in the problem 0 .54 we can get f of x node as we said from here by substituting x equals 0 .54 we get the value of the function 0 .36 26 the value of the differential we can get it from here by substituting x equals 0 .54 equals minus 0 .3652.
02:20
This means we can get x1 by using this equation here x1 equals x node minus f of x node divided by f dash of x node we have these three values in the table by substitution we get x1 equals 1 .5333 we can can see that x1 is not equal to x node then we must continue iterations f of x1 equals minus 0 .382 the value of f dash of x1 equals minus 0 .14755 by using newton's method equation we can get x2 which equals minus 1 .053.
03:20
This sound doesn't sound good because we have we have increased here then we have decreased in the same direction but we will continue iterations until we see what happens the value of f f f of x2 equals minus 6 .6 004 the value of f dash of x2 equals 11 .645 you using nettance method equation, x3 equals minus 0 .4862.
04:06
It sounds good to now.
04:10
We continue iterations, we get f of x4 minus 1 .79665...