Sir, in Example 3, another g(x) could've been obtained using the ln x term of the f(x) as follows: g(x) -> 2 - ln x = 0 g(x) -> ln x = 2 g(x) -> x = e^2 g'(x) = 0 < 1, so x (subscript n) would converge However, when we start the iterations using x (subscript n) = g (x subscript (n-1)), we obtain the same constant e^2 in every iteration, which does not lead us to the desired root.
@SkanCityAcademy_SirJohn9 күн бұрын
I see, I guess you omitted something, From x + In(x) - 2 = 0 Using In(x), x = e^(2-x) Derivative w.r.t.x will be -e^(2-x) putting in xo considering the absolute value, you will get 1.6487 which is not less than 1 at xo hence it will not converge
@Lemon_205028 күн бұрын
Oh yes, sir. Sorry, I missed that 2. My bad! I understood it now. Thanks a lot for clarifying it.
@SkanCityAcademy_SirJohn8 күн бұрын
Most welcome
@TurkishDoorsEnterpriseGH6 ай бұрын
Great. Please do a video on the bracketing method for us. Thank you ❤
@SkanCityAcademy_SirJohn6 ай бұрын
Not yet please
@solomonmule34142 ай бұрын
so what is the criteria for choosing those numbers to put in the equation to get the interval[a,b]. and can they be different from person to person🙏
@SkanCityAcademy_SirJohn2 ай бұрын
the rule is that f(a) < 0, and f(b) > 0, yes yours can be different from others, but the closer the interval, the closer and more accurate the iterated value
@CillaDarko9 ай бұрын
Very helpful 🙏🏽👏
@SkanCityAcademy_SirJohn9 ай бұрын
Thank you Priscy😍
@iamtheroze78934 ай бұрын
Excuse me, but 2cos(1.9) is greater than 1 in example 1.
@SkanCityAcademy_SirJohn3 ай бұрын
It's because your calculator was in the degree mode. Kindly put your calculator in the radian mode, and you will have an absolute value of 0.6465.... which is less than 1
@darindana10452 ай бұрын
@@SkanCityAcademy_SirJohnallways must in radian or not?
@SkanCityAcademy_SirJohn2 ай бұрын
@darindana1045 yes please, in radians
@darindana10452 ай бұрын
@@SkanCityAcademy_SirJohn my teacher said in degree🥲idk why
@SkanCityAcademy_SirJohn2 ай бұрын
It's supposed to be radians, maybe the question he gave in class requires that it's be solved in degrees.